Category: ML Aggarwal Solutions

Well-structured ML Aggarwal Class 12 Solutions Chapter 6 Indeterminate Forms Ex 6.4 facilitate a deeper understanding of mathematical principles.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Ex 6.4

Typical Problems:

Question 1.
(i) Evaluate the following limits:
(i) \(\underset{x \rightarrow \infty}{\mathbf{L t}}\) x tan^-1 (\(\frac{2}{x}\))
(ii) \(\underset{x \rightarrow \infty}{\mathbf{L t}}\) \(\frac{\sin ^{-1} \frac{1}{x}}{\tan \frac{1}{x}}\)
Solution:
(i) Put \(\frac{1}{x}\) = t
as x → ∞ ⇒ t → 0
∴ \(\underset{x \rightarrow \infty}{\mathbf{L t}}\) x tan^-1 (\(\frac{2}{x}\)) = \(\underset{t \rightarrow 0}{\mathrm{Lt}}\) \(\frac{\tan ^{-1}(2 t)}{t}\)
(\(\frac{0}{0}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{t \rightarrow 0} \frac{2}{1+4 t^2}\)
= \(\frac{2}{1+4 \times 0}\) = 2.

(ii) Put \(\frac{1}{x}\) = t
⇒ x = \(\frac{1}{t}\)
as x → ∞ ⇒ t → 0
∴ \(\ {Lt}_{x \rightarrow \infty} \frac{\sin ^{-1} \frac{1}{x}}{\tan \frac{1}{x}}=\underset{t \rightarrow 0}{\mathrm{Lt}} \frac{\sin ^{-1} t}{\tan t}\)
(\(\frac{0}{0}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{t \rightarrow 0} \frac{\frac{1}{\sqrt{1-t^2}}}{\sec ^2 t}\)
= \(\ {Lt}_{t \rightarrow 0} \frac{1}{\sec ^2 t \sqrt{1-t^2}}\)
= \(\frac{1}{(1)^2 \times \sqrt{1-0^2}}\) = 1.

Question 2.
If \(\frac{\sin 2 x+k \sin x}{x^3}\) is finite, find k and the limit.
Solution:
\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{\sin 2 x+k \sin x}{x^3}\)
(\(\frac{0}{0}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow 0} \frac{2 \cos 2 x+k \cos x}{3 x^2}\) …………(1)
Now deno 3x² → 0 as x → 0 and limit is given to be finite.
So it is necessary that Numerator of eqn. (1)
i.e. 2 cos 2x + k cos x vanishes as x → 0
and this happens when 2 + k = 0
⇒ k = – 2
∴ From (1) ; we have
given limit = \(\ {Lt}_{x \rightarrow 0} \frac{2 \cos 2 x-2 \cos x}{3 x^2}\) ((\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{-4 \sin 2 x+2 \sin x}{6 x}\) ((\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{-8 \cos 2 x+2 \cos x}{6}\)
= \(\frac{-8 \times 1+2 \times 1}{6}\) = – 1.

Question 3.
Find the values of a and b \(\ {Lt}_{x \rightarrow 0} \frac{x(1-a \cos x)+b \sin x}{x^3}\) exists and equals \(\frac{1}{3}\).
Solution:
\(\ {Lt}_{x \rightarrow 0} \frac{x(1-a \cos x)+b \sin x}{x^3}\)
(\(\frac{0}{0}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow 0} \frac{(1-a \cos x)+a x \sin x+b \cos x}{3 x^2}\) ………….(1)
Now the denominator of eqn. (1)
i.e. 3x² → 0 as x → 0.
∴ the given limit exists finitely, it is necessary that Numerator of eqn. (1) is also goes to 0 as x → 0 and this happen when
1 – a × 1 + a × 0 + b = 0
⇒ 1 – a + b = 0
⇒ a – b = 1 ………..(2)
Let eqn. (2) happens, Then given limit
= \(\ {Lt}_{x \rightarrow 0} \frac{(1-a \cos x)+a x \sin x+b \cos x}{3 x^2}\left(\frac{0}{0}\right)\)
= \(\ {Lt}_{x \rightarrow 0} \frac{a \sin x+a \sin x+a x \cos x-b \sin x}{6 x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{2 a \sin x+a x \cos x-b \sin x}{6 x}\left(\frac{0}{0}\right)\)
= \(\ {Lt}_{x \rightarrow 0} \frac{2 a \cos x+a \cos x-a x \sin x-b \cos x}{6}\)
= \(\frac{2 a+a-b}{6}\)
Also given limit = \(\frac{1}{3}\)
∴ \(\frac{2 a+a-b}{6}\) = \(\frac{1}{3}\)
⇒ 3a – b = 2 ………..(3)
On solving eqn. (2) and eqn. (3) ; we have
a = \(\frac{1}{2}\)
and b = – \(\frac{1}{2}\).

Question 4.
Find the values of a, b and c so that \(\ {Lt}_{x \rightarrow 0} \frac{a e^x-b \cos x+c e^{-x}}{x \sin x}\) = 2.
Solution:
\(\ {Lt}_{x \rightarrow 0} \frac{a e^x-b \cos x+c e^{-x}}{x \sin x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{a e^x-b \cos x+c e^{-x}}{x^2}\left(\frac{x}{\sin x}\right)\)
= \(\ {Lt}_{x \rightarrow 0} \frac{a e^x-b \cos x+c e^{-x}}{x^2}\) . 1 ……….(1)
[∵ \(\underset{\theta \rightarrow 0}{\ {Lt}} \frac{\theta}{\sin \theta}\) = 1]
Here denominator of eqn. (1) = x² → 0 as x → 0.
Now the limit exists finitely if Numerator of eqn. (1)
i.e. ae^x – b cos x + c e^-x as x → 0
and thos happen if a – b + c = 0 ……….(2)
suppose eqn.(2) is holds. Then given limit
= \(\ {Lt}_{x \rightarrow 0} \frac{a e^x-b \cos x+c e^{-x}}{x^2}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{a e^x+b \sin x-c e^{-x}}{2 x}\) ………..(3)
Now denominator of eqn. (3) i.e. 2x → 0 as x → 0
∴ in order that the limit exists finitely it is necessary that Numerator of eqn. (3) vanishes as x → 0.
This happens if a × 1 + b × 0 – c = 0
⇒ a – c = 0 ………..(4)
Now eqn. (4) holds.
Then given limit = \(\ {Lt}_{x \rightarrow 0} \frac{a e^x+b \sin x-c e^{-x}}{2 x}\left(\frac{0}{0}\right)\)
= \(\ {Lt}_{x \rightarrow 0} \frac{a e^x+b \cos x+c e^{-x}}{2}\)
= \(\frac{a+b+c}{2}\)
Also given limit = 2
∴ a + b + c = 4 ……..(5)
from (4) ;
a = c
∴ from (1) ;
2c – b = 0 …………..(6)
and from (5) ;
b + 2c = 4 …………..(7)
On solving eqn. (6) and (7) ; we have
c = 1; a = 1; b = 2.

Question 5.
Evaluate the following limits :
(i) \(\underset{x \rightarrow \infty}{\ {Lt}} \frac{\log _a x}{x^k}\), k > 0
(ii) \(\ {Lt}_{x \rightarrow \infty} 2^x \sin \frac{a}{2^x}\), a ≠ 0
Solution:
(i) \(\underset{x \rightarrow \infty}{\ {Lt}} \frac{\log _a x}{x^k}\), k > 0
(\(\frac{\infty}{\infty}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow \infty} \frac{\frac{1}{x} \frac{1}{\log a}}{k x^{k-1}}\)
= \(\ {Lt}_{x \rightarrow \infty} \frac{1}{(\log a) k x^k}\), k > 0 = 0

(ii) put \(\frac{a}{2^x}\) = t
as x → ∞
⇒ 2^x → ∞
⇒ \(\frac{a}{2^x}\) → 0
⇒ t → 0 [a ≠ 0]
∴ \(\underset{x \rightarrow \infty}{\mathbf{L t}}\) 2^x sin \(\frac{a}{2^x}\)
= \(\underset{t \rightarrow 0}{\mathrm{Lt}}\) \(\frac{a}{t}\) sin t
= a \(\underset{x \rightarrow \infty}{\mathbf{L t}}\) \(\frac{sin t}{t}\)
= a × 1 = a.

Question 6.
Evaluate the following limits:
(i) \(\ {Lt}_{x \rightarrow 0}\left(\frac{\tan x}{x}\right)^{1 / x}\)
(ii) \(\ {Lt}_{x \rightarrow a}\left(2-\frac{a}{x}\right)^{\tan \frac{\pi x}{2 a}}\).
Solution:
(i) Let F(x) = \(\ {Lt}_{x \rightarrow 0}\left(\frac{\tan x}{x}\right)^{1 / x}\)
⇒ log F(x) = \(\frac{1}{x}\) log \(\left(\frac{\tan x}{x}\right)\)

(ii) Let F(x) = \(\left(2-\frac{a}{x}\right)^{\tan \frac{\pi x}{2 a}}\)

Question 7.
Evaluate the following limits:
(i) \(\ {Lt}_{x \rightarrow 0}\left(\frac{\tan x}{x}\right)^{1 / x^2}\)
(ii) \(\ {Lt}_{x \rightarrow 0}(1+2 x)^{\frac{x+5}{x}}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0}\left(\frac{\tan x}{x}\right)^{1 / x^2}\)

∴ log (\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) F(x)) = \(\frac{1}{3}\)
⇒ F(x) = e^1/3

(ii) Let F(x) = \(\ {Lt}_{x \rightarrow 0}(1+2 x)^{\frac{x+5}{x}}\)
⇒ log F(x) = \(\left(\frac{x+5}{x}\right)\) log (1 + 2x)
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) log F(x) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\left(\frac{x+5}{x}\right)\) log (1 + 2x)
= \(\ {Lt}_{x \rightarrow 0} \frac{(x+5) \log (1+2 x)}{x}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{\frac{(x+5) 2}{1+2 x}+\log (1+2 x)}{1}\)
= \(\frac{(0+5) 2}{1+0}\) + log (1 + 0) = 0
∴ log \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) F(x) = 10
⇒ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\ {Lt}_{x \rightarrow 0}(1+2 x)^{\frac{x+5}{x}}\) = e¹⁰

Question 8.
Evaluate : \(\ {Lt}_{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{\sin x}{x-\sin x}}\)
Solution:
\(\ {Lt}_{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{\sin x}{x-\sin x}}\) = \(\ {Lt}_{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{\frac{\sin x}{x}}{1-\frac{\sin x}{x}}}\)

Students can track their progress and improvement through regular use of Class 12 ISC Maths Solutions Chapter 6 Indeterminate Forms Ex 6.3.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Ex 6.3

Evaluate the following (1 to 6) limits:

Question 1.
(i) \(\underset{x \rightarrow \alpha}{\mathbf{L t}}\) (x – α)^{x – α}
(ii) \(\underset{x \rightarrow 1^{-}}{\mathbf{L t}}\) \(\left(1-x^2\right)^{\frac{1}{\log (1-x)}}\)
S0lution:
(i) Let F(x) = (x – α)^{x – α}
⇒ log F(x) = (x – α) log (x – α)
∴ \(\underset{x \rightarrow \alpha}{\mathbf{L t}}\) log (F(x)) = \(\underset{x \rightarrow \alpha}{\mathbf{L t}}\) (x – α) log (x – α)
= \(\ {Lt}_{x \rightarrow \alpha} \frac{\log (x-\alpha)}{\frac{1}{x-\alpha}}\)
= \(\ {Lt}_{x \rightarrow \alpha} \frac{\frac{1}{x-\alpha}}{-\frac{1}{(x-\alpha)^2}}\)
= \(\underset{x \rightarrow \alpha}{\mathbf{L t}}\) – (x – α) = 0
⇒ log (\(\underset{x \rightarrow \alpha}{\mathbf{L t}}\) F(x)) = 0
⇒ \(\underset{x \rightarrow \alpha}{\mathbf{L t}}\) (x – α)^{x – α} = e⁰ = 1

(ii) Let F(x) = \(\underset{x \rightarrow 1^{-}}{\mathbf{L t}}\) \(\left(1-x^2\right)^{\frac{1}{\log (1-x)}}\)
⇒ log F(x) = \(\frac{\log \left(1-x^2\right)}{\log (1-x)}\)
∴ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) log (F(x)) = \(\ {Lt}_{x \rightarrow 1^{-}} \frac{\log \left(1-x^2\right)}{\log (1-x)}\) (\(\frac{\infty}{\infty}\) form)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{\frac{1}{1-x^2}(-2 x)}{\frac{1}{1-x}(-1)}\)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{2 x(1-x)}{1-x^2}\)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{2 x}{1+x}\)
∴ log (\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) F(x)) = 1
⇒ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) F(x) = e
⇒ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) \(\left(1-x^2\right)^{\frac{1}{\log (1-x)}}\) = e.

Question 2.
(i) \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) (tan x)^{sin 2x}
(ii) \(\underset{x \rightarrow \frac{\pi^{-}}{2}}{\text { Lt }}\) (tan x)^{sin 2x}
Solution:
(i) Let F(x) = sin 2x log tan x
⇒ log F(x) = sin 2x log tan x
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) log (F(x)) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) sin 2x log tan x (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\log \tan x}{\ {cosec} 2 x}\) (\(\frac{\infty}{\infty}\) form)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\frac{1}{\tan x} \sec ^2 x}{-\cot 2 x \ {cosec} 2 x \cdot 2}\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{2}{\frac{\sin 2 x}{-2 \cot 2 x \{cosec} {2 x}}\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\ {cosec} 2 x}{-\cot 2 x \ {cosec} 2 x}\)
= \(\ {Lt}_{x \rightarrow 0}\) (- tan 2x) = 0
⇒ log (\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) F(x)) = 0
⇒ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) F(x) = e⁰ = 1
⇒ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) (tan x)^{sin 2x} = 1

(ii) Let F(x) = (tan x)^{sin 2x}
⇒ log F(x) = sin 2x log tan x
∴ \(\underset{x \rightarrow \frac{\pi^{-}}{2}}{\text { Lt }}\) F(x) = \(\underset{x \rightarrow \frac{\pi^{-}}{2}}{\text { Lt }}\) sin 2x log tan x (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow \frac{\pi^{-}}{2}} \frac{\log \tan x}{\ {cosec} 2 x}\) (\(\frac{\infty}{\infty}\) form)
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}^{-}} \frac{\frac{\sec ^2 x}{\tan x}}{2 \cot 2 x \ {cosec} 2 x}\)
= \(\ {Lt}_{x \rightarrow \frac{\pi^{-}}{-}} \frac{2 \ {cosec} 2 x}{-2 \cot 2 x \ {cosec} 2 x}\)
= – \(\underset{x \rightarrow \frac{\pi^{-}}{2}}{\text { Lt }}\) tan 2x = 0
⇒ log (\(\underset{x \rightarrow \frac{\pi^{-}}{2}}{\text { Lt }}\) F(x)) = 0
⇒ \(\underset{x \rightarrow \frac{\pi^{-}}{2}}{\text { Lt }}\) F(x) = e⁰ = 1
⇒ \(\underset{x \rightarrow \frac{\pi^{-}}{2}}{\text { Lt }}\) (tan x)^{sin 2x} = 1.

Question 3.
(i) \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) \((\cot x)^{\frac{1}{\log x}}\)
(ii) \(\underset{x \rightarrow \infty}{\mathbf{L t}}\) (1 + x)^1/x
Solution:
Let F(x) = \((\cot x)^{\frac{1}{\log x}}\)
⇒ log f(x) = \(\frac{\log \cot x}{\log x}\)
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) log F(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) \(\frac{\log \cot x}{\log x}\) (\(\frac{\infty}{\infty}\) form)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\frac{1}{\cot x}\left(-\ {cosec}^2 x\right)}{\frac{1}{x}}\) [using L ‘Hopital’s rule]
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{-x \tan x}{\sin ^2 x}\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{-x}{\cos x \sin x}\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{x}{\sin x} \cdot \underset{x \rightarrow 0^{+}}{\ {Lt}} \frac{1}{\cos x}\)
∴ log (\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) F(x)) = – 1. 1 = – 1
⇒ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) F(x) = e^-1
= \(\frac{1}{e}\)
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) \((\cot x)^{\frac{1}{\log x}}\) = \(\frac{1}{e}\).

(ii) Let F(x) = (1 + x)^1/x
⇒ log F(x) = \(\frac{1}{x}\) log (1 + x)
\(\underset{x \rightarrow \infty}{\mathbf{L t}}\) log F(x) = \(\ {Lt}_{x \rightarrow \infty} \frac{\log (1+x)}{x}\) (\(\frac{\infty}{\infty}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow \infty} \frac{\frac{1}{1+x}}{1}=\frac{1}{\infty}\) = 0
∴ log (\(\underset{x \rightarrow \infty}{\mathbf{L t}}\) F(x)) = 0
⇒ \(\underset{x \rightarrow \infty}{\mathbf{L t}}\) F(x) = e⁰ = 1
\(\underset{x \rightarrow \infty}{\mathbf{L t}}\) (1 + x)^1/x = 1.

Question 4.
(i) \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) \(\left(\frac{1}{x}\right)^{\tan x}\)
(ii) \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) (cot x)^{sin 2x}
Solution:
(i) Let f(x) = \(\left(\frac{1}{x}\right)^{\tan x}\)
⇒ log F(x) = tan x log (\(\frac{1}{x}\))
= – tan x log x
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) log F(x) = – \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) tan x log x
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) – \(\frac{\log x}{\cot x} \quad\left(\frac{\infty}{\infty}\right)\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{-\frac{1}{x}}{-\ {cosec}^2 x}\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\sin ^2 x}{x}\left(\frac{0}{0}\right)\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{2 \sin x \cos x}{1}\)
= 2 × 0 × 1 = 0
∴ log (\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) log F(x)) = 0
⇒ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) F(x) = e⁰ = 1
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) \(\left(\frac{1}{x}\right)^{\tan x}\) = 1

(ii) Let F(x) = (cot x)^{sin 2x}
⇒ log F(x) = sin 2x log cot x
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) log F(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) sin 2x log cot x (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\log \cot x}{\ {cosec} 2 x}\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\frac{1}{\cot x}\left(-\ {cosec}^2 x\right)}{-2 \ {cosec} 2 x \cot 2 x}\)
= \(\underset{x \rightarrow 0^{+}}{\ {Lt}} \frac{2 \ {cosec} 2 x}{2 \ {cosec} 2 x \cot 2 x}\)
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) tan 2x = 0
∴ log (\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) F(x)) = 0
⇒ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) F(x) = e⁰ = 1
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) (cot x)^{sin 2x} = 1.

Question 5.
(i) \(\underset{x \rightarrow \frac{\pi}{2}^{-}}{\text {Lt }}\) (sec x)^{cot x}
(ii) \(\begin{gathered}
\text { Lt } \\
x \rightarrow \frac{\pi}{4}
\end{gathered}\) (tan x)^{tan 2x}
Solution:
(i) Let F(x) = (sec x)^{cot x}
⇒ log F(x) = \(\underset{x \rightarrow \frac{\pi}{2}^{-}}{\text {Lt }}\) cot x log sec x
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}^{-}} \frac{\log (\sec x)}{\tan x}\) (\(\frac{\infty}{\infty}\) form)
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}^{-}} \frac{\frac{1}{\sec x} \sec x \tan x}{\sec ^2 x}\)
= \(\underset{x \rightarrow \frac{\pi}{2}^{-}}{\text {Lt }}\) cos x sin x
= 0 × 1 = 0
⇒ log (\(\underset{x \rightarrow \frac{\pi}{2}^{-}}{\text {Lt }}\) F(x)) = 0
⇒ \(\underset{x \rightarrow \frac{\pi}{2}^{-}}{\text {Lt }}\) F(x) = e⁰ = 1
⇒ \(\underset{x \rightarrow \frac{\pi}{2}^{-}}{\text {Lt }}\) (sec x)^{cot x} = 1

(ii) Let F(x) = (tan x)^{tan 2x}
⇒ log F(x) = tan 2x log tan x
\(\mathrm{Lt}_{x \rightarrow \frac{\pi}{4}}\) log F(x) = \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{4}}\) tan 2x log tan x (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow \frac{\pi}{4}} \frac{\log \tan x}{\cot 2 x}\left(\frac{0}{0} \text { form }\right)\)
using L’Hopital’s rule, we have
= \(\ {Lt}_{x \rightarrow \frac{\pi}{4}} \frac{\frac{1}{\tan x} \sec ^2 x}{-2 \ {cosec}^2 2 x}\)
= \(\frac{\frac{1}{1} \cdot(\sqrt{2})^2}{-2(1)^2}\) = – 1
⇒ log (\(\mathrm{Lt}_{x \rightarrow \frac{\pi}{4}}\) F(x)) = – 1
⇒ \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{4}}\) F(x) = e^-1 = \(\frac{1}{e}\)
∴ \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{4}}\) (tan x)^{tan 2x} = \(\frac{1}{e}\).

Question 6.
(i) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (1 + x)^1/x
(ii) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (e^x + 4x)^1/x
Solution:
(i) Let F(x) = (1 + x)^1/x
⇒ log F(x) = \(\frac{\log (1+x)}{x}\)
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) log F(x) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{\log (1+x)}{x}\)
(\(\frac{0}{0}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{i \rightarrow 0} \frac{1}{\frac{1+x}{1}}\)
= \(\frac{1}{1+0}\) = 1
⇒ log (\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) F(x)) = 1
⇒ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) F(x) = e⁰ = 1
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (1 + x)^1/x = 1.

(ii) Let F(x) = (e^x + 4x)^1/x
⇒ log F(x) = \(\frac{1}{x}\) log (e^x + 4x)
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) log F(x) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{\log \left(e^x+4 x\right)}{x}\) (\(\frac{0}{0}\) form)
= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{\frac{1}{e^x+4 x}\left(e^x+4\right)}{1}\)
= \(\frac{e^0+4}{e^0+4 \times 0}\) = 5
⇒ log (\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) F(x)) = 5
⇒ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) F(x) = e⁵
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (e^x + 4x) = e⁵.

Parents can use ISC Maths Class 12 Solutions Chapter 6 Indeterminate Forms Ex 6.2 to provide additional support to their children.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Ex 6.2

Evaluate the following (1 to 14) limits:

Question 1.
(i) \(\underset{x \rightarrow 0^{+}}{\ {Lt}} \frac{\log \sin x}{\cot x}\)
(ii) \(\ {Lt}_{x \rightarrow \frac{\pi}{2}^{-}} \frac{\tan x}{\log \cos x}\)
Solution:
(i) \(\underset{x \rightarrow 0^{+}}{\ {Lt}} \frac{\log \sin x}{\cot x}\) (\(\frac{\infty}{\infty}\) form)
using L’Hopital’s rule
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\frac{\cos x}{\sin x}}{-\ {cosec}^2 x}\)
= – \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) cos x sin x
= – 1 × 0 = 0

(ii) \(\ {Lt}_{x \rightarrow \frac{\pi}{2}^{-}} \frac{\tan x}{\log \cos x}\)
(\(\frac{\infty}{\infty}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow \frac{\pi^{-}}{2}} \frac{\sec ^2 x}{-\tan x}\)
= \(\ {Lt}_{x \rightarrow \frac{\pi^{-}}{2}}-\frac{1}{\cos ^2 x} \times \frac{\cos x}{\sin x}\)
= \(\ {Lt}_{x \rightarrow \frac{\pi^{-}}{2}} \frac{-1}{\cos x \sin x}\) → – ∞

Question 2.
(i) \(\ {Lt}_{x \rightarrow \frac{\pi}{2}^{+}} \frac{\log \left(x-\frac{\pi}{2}\right)}{\tan x}\)
(ii) \(\ {Lt}_{x \rightarrow 1^{-}} \frac{\log (1-x)}{\cot \pi x}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow \frac{\pi}{2}^{+}} \frac{\log \left(x-\frac{\pi}{2}\right)}{\tan x}\)
(\(\frac{\infty}{\infty}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow \frac{\pi^{+}}{2}} \frac{1}{\left(x-\frac{\pi}{2}\right) \sec ^2 x}\)
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}^{+}} \frac{\cos ^2 x}{x-\frac{\pi}{2}}\left(\frac{0}{0}\right)\)
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}^{+}} \frac{-2 \cos x \sin x}{1}\)
= – 2 × 0 × 1 = 0

(ii) \(\ {Lt}_{x \rightarrow 1^{-}} \frac{\log (1-x)}{\cot \pi x}\) (\(\frac{\infty}{\infty}\) form)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{-1}{(1-x) \pi\left(-\ {cosec}^2 \pi x\right)}\)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{\sin ^2 \pi x}{1-x}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{2 \pi \sin \pi x \cos \pi x}{-1}\)

Question 3.
(i) \(\underset{x \rightarrow 0^{+}}{\ {Lt}} \frac{\log \cot x}{e^{\ {cosec}^2 x}}\)
(ii) \(\underset{x \rightarrow 0^{+}}{\mathbf{L t}}\) log_{tan x} tan 2x.
Solution:
(i) \(\underset{x \rightarrow 0^{+}}{\ {Lt}} \frac{\log \cot x}{e^{\ {cosec}^2 x}}\)
(\(\frac{\infty}{\infty}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\frac{1}{\cot x}\left(-\ {cosec}^2 x\right)}{-e^{\ {cosec}^2 x} 2 \ {cosec}^2 x \cot x}\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{1}{2 \cot ^2 x e^{\ {cosec}^2 x}}\)
= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) \(\frac{1}{2}\) tan² x e^{– cosec2 x}
= \(\frac{1}{2}\) × 0 × 0 = 0
[∵ e^{– cosec2 x} → e^{– ∞} = 0, ae, x → 0⁺]

(ii) \(\underset{x \rightarrow 0^{+}}{\mathbf{L t}}\) log_{tan x} tan 2x

Question 4.
(i) \(\underset{x \rightarrow \infty}{\ {Lt}} \frac{\log x}{x}\)
(ii) \(\underset{x \rightarrow \infty}{\ {Lt}} \frac{5 x+2 \log x}{x+3 \log x}\)
Solution:
(i) \(\underset{x \rightarrow \infty}{\ {Lt}} \frac{\log x}{x}\)
(\(\frac{\infty}{\infty}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow \infty} \frac{\frac{1}{x}}{1}\)
= \(\ {Lt}_{x \rightarrow \infty} \frac{1}{x}\) = 0.

(ii) \(\underset{x \rightarrow \infty}{\ {Lt}} \frac{5 x+2 \log x}{x+3 \log x}\) (\(\frac{\infty}{\infty}\) form)
= \(\ {Lt}_{x \rightarrow \infty} \frac{5+\frac{2}{x}}{1+\frac{3}{x}}\)
= \(\frac{5+0}{1+0}\)
= 5

Question 5.
(i) \(\ {Lt}_{x \rightarrow \infty} \frac{x^4+x^2}{e^x+1}\)
(ii) \(\ {Lt}_{x \rightarrow 0} \frac{\cot x}{\cot 2 x}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow \infty} \frac{x^4+x^2}{e^x+1}\)
(\(\frac{\infty}{\infty}\) form, using L’Hopital’s rule)

(ii) \(\ {Lt}_{x \rightarrow 0} \frac{\cot x}{\cot 2 x}\)
(\(\frac{\infty}{\infty}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow 0} \frac{\tan 2 x}{\tan x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{2 \sec ^2 2 x}{\sec ^2 x}\)
= \(\frac{2 \times 1^2}{1^2}\) = 2.

Question 6.
(i) \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) cosec πx log x
(ii) \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) (1 – x) tan \(\frac{\pi}{2}\) x
Solution:
(i) \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) cosec πx log x (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow 1} \frac{\log x}{\sin \pi x}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 1} \frac{\frac{1}{x}}{\pi \cos \pi x}\)
= \(\frac{1}{1 \times \pi \times \cos \pi}=\frac{1}{-\pi}\)

(ii) \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) (1 – x) tan \(\frac{\pi}{2}\) x (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow 1} \frac{1-x}{\cot \frac{\pi x}{2}}\) [\(\frac{0}{0}\) form]
= \(\ {Lt}_{x \rightarrow 1} \frac{-1}{-\frac{\pi}{2} \ {cosec}^2 \frac{\pi x}{2}}\)
= \(\ {Lt}_{x \rightarrow 1} \frac{2}{\pi} \sin ^2 \frac{\pi x}{2}\)
= \(\frac{2}{\pi} \sin ^2 \frac{\pi}{2}\)
= \(\frac{2}{\pi}(1)^2\)
= \(\frac{2}{\pi}\).

Question 7.
(i) \(\ {Lt}_{x \rightarrow 1}\left(\frac{1}{x^2-1}-\frac{2}{x^4-1}\right)\)
(ii) \(\ {Lt}_{x \rightarrow 1}\left(\frac{1}{\log x}-\frac{1}{x-1}\right)\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 1}\left(\frac{1}{x^2-1}-\frac{2}{x^4-1}\right)\) (∞ – ∞ form)
= \(\ {Lt}_{x \rightarrow 1}\left[\frac{x^2+1-2}{x^4-1}\right]\)
= \(\ {Lt}_{x \rightarrow 1} \frac{x^2-1}{x^4-1}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 1} \frac{2 x}{4 x^3}\)
= \(\frac{2}{4}=\frac{1}{2}\)

(ii) \(\ {Lt}_{x \rightarrow 1}\left(\frac{1}{\log x}-\frac{1}{x-1}\right)\)

Question 8.
(i) \(\ {Lt}_{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{x^2} \log (1+x)\right)\)
(ii) \(\ {Lt}_{x \rightarrow 2}\left(\frac{1}{\log (x-1)}-\frac{1}{x-2}\right)\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{x^2} \log (1+x)\right)\) (∞ – ∞ form)
= \(\ {Lt}_{x \rightarrow 0}\left[\frac{x-\log (1+x)}{x^2}\right]\) (\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule)
= \(\ {Lt}_{x \rightarrow 0} \frac{1-\frac{1}{1+x}}{2 x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{1+x-1}{2 x(1+x)}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{1}{(1+x)^2}\)
= \(\frac{1}{2(1+0)}=\frac{1}{2}\)

(ii) \(\ {Lt}_{x \rightarrow 2}\left(\frac{1}{\log (x-1)}-\frac{1}{x-2}\right)\) (∞ – ∞ form)

Question 9.
(i) \(\ {Lt}_{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{e^x-1}\right)\)
(ii) \(\ {Lt}_{x \rightarrow 0}\left(\frac{1}{2 x}-\frac{1}{x\left(e^{\pi x}+1\right)}\right)\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{e^x-1}\right)\) (∞ – ∞ form)

 

(ii) \(\ {Lt}_{x \rightarrow 0}\left(\frac{1}{2 x}-\frac{1}{x\left(e^{\pi x}+1\right)}\right)\) (∞ – ∞ form)
= \(\ {Lt}_{x \rightarrow 0}\left[\frac{e^{\pi x}+1-2}{2 x\left(e^{\pi x}+1\right)}\right]\)
= \(\ {Lt}_{x \rightarrow 0} \frac{e^{\pi x}-1}{2 x\left(e^{\pi x}+1\right)}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{e^{\pi x}-1}{2 x} \cdot \frac{1}{1+1}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{e^{\pi x}-1}{4 x}\) (\(\frac{0}{0}\) form)
using L’Hopital’s rule, we have
= \(\ {Lt}_{x \rightarrow 0} \frac{e^{\pi x} \cdot \pi}{4}=\frac{\pi}{4}\)

Question 10.
(i) \(\underset{x \rightarrow 0}{\ {Lt}}\left(\cot x-\frac{1}{x}\right)\)
(ii) \(\underset{x \rightarrow 0}{\ {Lt}}\left(\cot ^2 x-\frac{1}{x^2}\right)\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0}\left(\cot x-\frac{1}{x}\right)\) (∞ – ∞ form)

 

(ii) \(\underset{x \rightarrow 0}{\ {Lt}}\left(\cot ^2 x-\frac{1}{x^2}\right)\)

 

Question 11.
(i) \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{2}}\) (sec x – tan x)
(ii) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (cosec x – cot x)
Solution:
(i) \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{2}}\) (sec x – tan x) (∞ – ∞ form)
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}}\left[\frac{1-\sin x}{\cos x}\right]\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}} \frac{-\cos x}{-\sin x}\)
= \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{2}}\) cot x = 0.

(ii) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) (cosec x – cot x)
= \(\ {Lt}_{x \rightarrow 0} \frac{1-\cos x}{\sin x}\) (\(\frac{0}{0}\) form)
using ‘L’ Hopitals’s rule
= \(\ {Lt}_{x \rightarrow 0} \frac{\sin x}{\cos x}\)
= \(\frac{0}{1}\) = 0.

Question 12.
(i) \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x log x
(ii) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) sin x log x².
Solution:
(i) \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x log x (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\log x}{\frac{1}{x}}\)
(\(\frac{\infty}{\infty}\) form, using L’Hopital’s rule)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{\frac{1}{x}}{-\frac{1}{x^2}}\)
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) (- x) = 0.

(ii) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) sin x log x²
= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) 2 sin x log x (0 . ∞ form)
[∵ log a^b = b log a]

 

Question 13.
(i) \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x log (tan x)
(ii) \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{2}}\) (1 – sin x) tan x
Solution:
(i) \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x log (tan x) (0 . ∞ form)

(ii) \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{2}}\) (1 – sin x) tan x (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\cot x}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow \frac{\pi}{2}} \frac{-\cos x}{-\ {cosec}^2 x}\)
= \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{2}}\) cos x sin² x
= 0 × 1² = 0.

Question 14.
(i) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) x tan (\(\frac{\pi}{2}\) – x)
(ii) \(\underset{x \rightarrow c}{\mathrm{Lt}}\) (c – x) tan \(\frac{\pi x}{2 c}\)
Solution:
(i) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) x tan (\(\frac{\pi}{2}\) – x)
= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) x cot x (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow 0} \frac{x}{\tan x}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 0} \frac{1}{\sec ^2 x}\)
= \(\frac{1}{1^2}\) = 1.

(ii) \(\underset{x \rightarrow c}{\mathrm{Lt}}\) (c – x) tan \(\frac{\pi x}{2 c}\) (0 . ∞ form)
= \(\ {Lt}_{x \rightarrow c} \frac{c-x}{\cot \frac{\pi x}{2 c}}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow c} \frac{-1}{-\ {cosec}^2 \frac{\pi x}{2 c} \cdot \frac{\pi}{2 c}}\)
= \(\frac{2 c}{\pi} \ {Lt}_{x \rightarrow c} \sin ^2 \frac{\pi x}{2 c}\)
= \(\frac{2 c}{\pi} \times 1^2\)
= \(\frac{2 c}{\pi}\)

Question 15.
Show that \(\ {Lt}_{x \rightarrow \infty} \frac{e^x-e^{-x}}{e^x+e^{-x}}\) = 1 and \(\ {Lt}_{x \rightarrow-\infty} \frac{e^x-e^{-x}}{e^x+e^{-x}}\) = – 1.
Solution:
\(\ {Lt}_{x \rightarrow \infty} \frac{e^x-e^{-x}}{e^x+e^{-x}}\) (\(\frac{\infty}{\infty}\) form)
= \(\ {Lt}_{x \rightarrow \infty} \frac{e^x\left(1-e^{-2 x}\right)}{e^x\left(1+e^{-2 x}\right)}\)
= \(\ {Lt}_{x \rightarrow \infty} \frac{1-e^{-2 x}}{1+e^{-2 x}}\)
= \(\frac{1-0}{1+0}\) = 1
[as x → ∞, e^-2x → e^{– ∞} = \(\frac{1}{e^{\infty}}\) = 0]
and \(\ {Lt}_{x \rightarrow-\infty} \frac{e^x-e^{-x}}{e^x+e^{-x}}\)
= \(\ {Lt}_{x \rightarrow-\infty} \frac{e^{2 x}-1}{e^{2 x}+1}\)
= \(\frac{0-1}{0+1}\)
= 1.

The availability of ISC Mathematics Class 12 Solutions Chapter 6 Indeterminate Forms Ex 6.1 encourages students to tackle difficult exercises.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 6 Indeterminate Forms Ex 6.1

Evaluate the following (1 to 13) limits:

Question 1.
(i) \(\ {Lt}_{x \rightarrow 3} \frac{x^4-81}{x-3}\)
(ii) \(\ {Lt}_{x \rightarrow 0} \frac{(1+x)^n-1}{x}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 3} \frac{x^4-81}{x-3}\)
= \(\ {Lt}_{x \rightarrow 3} \frac{4 x^3-0}{1}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= 4 × 3³
= 108

(ii) \(\ {Lt}_{x \rightarrow 0} \frac{(1+x)^n-1}{x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{n(1+x)^{n-1}}{n}\)
= n (1 + 0)^n-1
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= n × 1^n-1 = n.

Question 2.
(i) \(\underset{x \rightarrow 0}{\ {Lt}} \frac{\sin a x}{\sin b x}\)
(ii) \(\ {Lt}_{x \rightarrow 2} \frac{e^x-e^2}{x-2}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{\sin a x}{\sin b x}=\ {Lt}_{x \rightarrow 0} \frac{a \cos a x}{b \cos b x}\)
= \(\frac{a \times 1}{b \times 1}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\frac{a}{b}\)

(ii) \(\ {Lt}_{x \rightarrow 2} \frac{e^x-e^2}{x-2}\) = \(\ {Lt}_{x \rightarrow 2} \frac{e^x-0}{1-0}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= e²

Question 3.
(i) \(\ {Lt}_{x \rightarrow 0} \frac{x e^x}{1-e^x}\)
(ii) \(\ {Lt}_{x \rightarrow 0} \frac{e^x-1}{\tan 2 x}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{x e^x}{1-e^x}=\ {Lt}_{x \rightarrow 0} \frac{x e^x+e^x}{-e^x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{e^x(x+1)}{-e^x}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) – (x + 1)
= – (0 + 1)
= – 1.

(ii) \(\ {Lt}_{x \rightarrow 0} \frac{e^x-1}{\tan 2 x}=\ {Lt}_{x \rightarrow 0} \frac{e^x}{2 \sec ^2 2 x}\)
= \(\frac{e^0}{2 \sec ^2 0}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\frac{1}{2 \times 1}=\frac{1}{2}\)

Question 4.
(i) \(\ {Lt}_{x \rightarrow 0} \frac{e^x-(1+x)}{x^2}\)
(ii) \(\ {Lt}_{x \rightarrow 1} \frac{x^2-x \log x+\log x-1}{x-1}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{e^x-(1+x)}{x^2}\) = \(\ {Lt}_{x \rightarrow 0} \frac{e^x-1}{2 x}\) (\(\frac{0}{0}\) form)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 0} \frac{e^x}{2}\)
= \(\frac{e^0}{2}=\frac{1}{2}\)

(ii) \(\ {Lt}_{x \rightarrow 1} \frac{x^2-x \log x+\log x-1}{x-1}\) = \(\ {Lt}_{x \rightarrow 1} \frac{2 x-\left(x \times \frac{1}{x}+\log x \cdot 1\right)+\frac{1}{x}}{1}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\underset{x \rightarrow 1}{\mathrm{Lt}}\) 2x – (1 + log x) + \(\frac{1}{x}\)
= 2 × 1 – (1 + log 1) + 1
= 2 – 1 – 0 + 1
= 2

Question 5.
(i) \(\ {Lt}_{x \rightarrow 0} \frac{\cos x-1}{\cos 2 x-1}\)
(ii) \(\ {Lt}_{x \rightarrow 0} \frac{8^x-2^x}{4 x}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{\cos x-1}{\cos 2 x-1}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 0} \frac{-\sin x}{-2 \sin 2 x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{\sin x}{2 \sin 2 x}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{\cos x}{4 \cos 2 x}\)
= \(\frac{1}{4 \times 1}=\frac{1}{4}\)

(ii) \(\ {Lt}_{x \rightarrow 0} \frac{8^x-2^x}{4 x}\)
Solution:
\(\ {Lt}_{x \rightarrow 0} \frac{8^x-2^x}{4 x}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 0} \frac{8^x \log 8-2^x \log 2}{4}\)
= \(\frac{8^0 \log 8-2^0 \log 2}{4}\)
[∵ log a^b = b log a]
= \(\frac{\log 8-\log 2}{4}\)
= \(\frac{1}{4}\) log 4
= \(\frac{1}{4}\) × 2 log 2
= \(\frac{1}{2}\) log 2

Question 6.
(i) \(\ {Lt}_{x \rightarrow 0} \frac{x-\tan x}{x-\sin x}\)
(ii) \(\ {Lt}_{x \rightarrow 0} \frac{2 \tan ^{-1} x-x}{2 x-\sin ^{-1} x}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{x-\tan x}{x-\sin x}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 0} \frac{1-\sec ^2 x}{1-\cos x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{-2 \sec ^2 x \tan x}{\sin x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{-2 \sec ^2 x \sin x}{\sin x \cos x}\)
= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) – 2 sec³ x
= – 2 × 1 = – 2

(ii) \(\ {Lt}_{x \rightarrow 0} \frac{2 \tan ^{-1} x-x}{2 x-\sin ^{-1} x}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 0} \frac{\frac{2}{1+x^2}-1}{2-\frac{1}{\sqrt{1-x^2}}}\)
= \(\frac{2-1}{2-1}\)
= 1.

Question 7.
(i) \(\underset{x \rightarrow 0}{\ {Lt}} \frac{\log \sec 2 x}{\log \sec x}\)
(ii) \(\ {Lt}_{x \rightarrow 0} \frac{\cos 2 x-\cos x}{\sin ^2 x}\)
Solution:
(i) \(\underset{x \rightarrow 0}{\ {Lt}} \frac{\log \sec 2 x}{\log \sec x}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 0} \frac{\frac{1}{\sec 2 x} \sec 2 x \tan 2 x \cdot 2}{\frac{1}{\sec x} \sec x \tan x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{2 \tan 2 x}{\tan x}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{4 \sec ^2 2 x}{\sec ^2 x}\)
= \(\frac{4 \times 1}{1}\) = 4.

(ii) \(\ {Lt}_{x \rightarrow 0} \frac{\cos 2 x-\cos x}{\sin ^2 x}\)

Question 8.
(i) \(\ {Lt}_{x \rightarrow 0} \frac{x-\sin x}{x^3}\)
(ii) \(\ {Lt}_{x \rightarrow 0} \frac{e^x-e^{-x}-2 \log (1+x)}{x \sin x}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{x-\sin x}{x^3}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 0} \frac{1-\cos x}{3 x^2}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{\sin x}{6 x}\)
= \(\frac{1}{6} \ {Lt}_{x \rightarrow 0} \frac{\sin x}{x}\)
= \(\frac{1}{6} \times 1=\frac{1}{6}\)

(ii) \(\ {Lt}_{x \rightarrow 0} \frac{e^x-e^{-x}-2 \log (1+x)}{x \sin x}\)

Question 9.
(i) \(\ {Lt}_{x \rightarrow 0} \frac{e^x+e^{-x}+2 \cos x-4}{x^3}\)
(ii) \(\ {Lt}_{x \rightarrow 0} \frac{x^2+2 \cos x-2}{x \sin ^3 x}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{e^x+e^{-x}+2 \cos x-4}{x^3}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 0} \frac{e^x-e^{-x}-2 \sin x}{3 x^2}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{e^x+e^{-x}-2 \cos x}{6 x}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{e^x-o^{-x}+2 \sin x}{6}\)
= \(\frac{1-1+2 \times 0}{6}\)
= 0

(ii) \(\ {Lt}_{x \rightarrow 0} \frac{x^2+2 \cos x-2}{x \sin ^3 x}\)

Question 10.
(i) \(\ {Lt}_{x \rightarrow 0} \frac{\log (1-x)}{\tan \frac{\pi}{2} x}\)
(ii) \(\ {Lt}_{x \rightarrow 0} \frac{\left(\tan ^{-1} x\right)^2}{\log \left(1+x^2\right)}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{\log (1-x)}{\tan \frac{\pi}{2} x}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 0} \frac{\frac{-1}{1-x}}{\frac{\pi}{2} \sec ^2 \frac{\pi}{2} x}\)
= \(\frac{\frac{-1}{1-0}}{\frac{\pi}{2} \times 1^2}\)
= – \(\frac{2}{\pi}\)

(ii) \(\ {Lt}_{x \rightarrow 0} \frac{\left(\tan ^{-1} x\right)^2}{\log \left(1+x^2\right)}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 0} \frac{2 \tan ^{-1} x \cdot \frac{1}{1+x^2}}{\frac{2 x}{1+x^2}}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{\tan ^{-1} x}{x}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{\frac{1}{1+x^2}}{1}\)
= \(\frac{1}{1+0^2}\) = 1.

Question 11.
(i) \(\ {Lt}_{x \rightarrow 0} \frac{x-\sin x \cos x}{x^3}\) (ISC 2013)
(ii) \(\ {Lt}_{x \rightarrow 0} \frac{e^x-e^{\sin x}}{x-\sin x}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{x-\sin x \cos x}{x^3}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{x-\frac{1}{2} \sin 2 x}{x^3}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 0} \frac{1-\frac{1}{2} \cos 2 x \times 2}{3 x^2}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{2 \sin 2 x}{6 x}\)
= \(\frac{2}{3} \ {Lt}_{x \rightarrow 0} \frac{\sin 2 x}{2 x}\)
= \(\frac{2}{3} \times 1=\frac{2}{3}\)
[∵ \(\ {Lt}_{\theta \rightarrow 0} \frac{\sin \theta}{\theta}\) = 1]

(ii) \(\ {Lt}_{x \rightarrow 0} \frac{e^x-e^{\sin x}}{x-\sin x}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]

Question 12.
(i) \(\ {Lt}_{x \rightarrow 0} \frac{\log \left(1-x^2\right)}{\log \cos x}\)
(ii) \(\ {Lt}_{x \rightarrow 0} \frac{e^x \sin x-x-x^2}{x^3}\)
Solution:
(i) \(\ {Lt}_{x \rightarrow 0} \frac{\log \left(1-x^2\right)}{\log \cos x}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 0} \frac{\frac{-2 x}{1-x^2}}{-\frac{\sin x}{\cos x}}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{2 x}{\left(1-x^2\right) \tan x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{2}{1-x^2} \cdot \underset{x \rightarrow 0}{\ {Lt}} \frac{x}{\tan x}\)
= \(\frac{2}{1-0}\) . 1 = 2
[∵ \(\ {Lt}_{\theta \rightarrow 0} \frac{\theta}{\tan \theta}\) = 1]

(ii) \(\ {Lt}_{x \rightarrow 0} \frac{e^x \sin x-x-x^2}{x^3}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 0} \frac{e^x \cos x+\sin x e^x-1-2 x}{3 x^2}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{e^x(-\sin x)+\cos x e^x+e^x \cos x+\sin x e^x-2}{6 x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{2 \cos x e^x-2}{6 x}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{2 \cos x e^x-2 \sin x e^x}{6}\)
= \(\frac{2 \times 1 \times 1-2 \times 0 \times 1}{6}=\frac{1}{3}\).

Question 13.
(i) \(\ {Lt}_{x \rightarrow 0^{+}} \frac{3^x-2^x}{\sqrt{x}}\)
(ii) \(\ {Lt}_{x \rightarrow 0^{+}} \frac{(1+x)^n-n x-1}{x^2}\), n > 1
Solution:
(i) \(\ {Lt}_{x \rightarrow 0^{+}} \frac{3^x-2^x}{\sqrt{x}}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{3^x \log 3-2^x \log 2}{\frac{1}{2 \sqrt{x}}}\)
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) 2√x [3^x log 3 – 2^x log 2]
= 2 × 0 [1 × log 3 – 1 × log 2] = 0

(ii) \(\ {Lt}_{x \rightarrow 0} \frac{(1+x)^n-n x-1}{x^2}\) ; n > 1
[Using L Hopital’s rule and (\(\frac{0}{0}\) form)]
= \(\ {Lt}_{x \rightarrow 0} \frac{n(1+x)^{n-1}-n}{2 x}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{n(n-1)(1+x)^{n-2}-0}{2}\)
= \(\frac{n(n-1)}{2}\)

Question 14.
What is the fallacy in the following use of L’Hop[ital’s rule ?
\(\ {Lt}_{x \rightarrow 2} \frac{x^3-x^2-x-2}{x^3-3 x^2+3 x-2}\) = \(\ {Lt}_{x \rightarrow 2} \frac{3 x^2-2 x-1}{3 x^2-6 x+3}\) = \(\ {Lt}_{x \rightarrow 2} \frac{6 x-2}{6 x-6}=\ {Lt}_{x \rightarrow 2} \frac{6}{6}\) = 1
Solution:
\(\ {Lt}_{x \rightarrow 2} \frac{x^3-x^2-x-2}{x^3-3 x^2+3 x-2}\)
[\(\frac{0}{0}\) form, using ‘L’ Hopitals’s rule]
= \(\ {Lt}_{x \rightarrow 2} \frac{3 x^2-2 x-1}{3 x^2-6 x+3}\)
which is not (\(\frac{0}{0}\)) form so we can’t apply L’Hopital’s rule in this step.

Question 15.
If \(\underset{x \rightarrow 0}{\ {Lt}} \frac{\sin 2 x+k \sin x}{x^3}\) is finite, find k and the limit.
Solution:
\(\ {Lt}_{x \rightarrow 0} \frac{\sin 2 x+k \sin x}{x^3}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{2 \cos 2 x+k \cos x}{3 x^2}\) ……….(1)
(using L’Hopital’s rule)
Since denominator of eqn. (1)
i.e. 3x² → 0 as x → 0 so in order that given limit exist finitely.
Numerator of eqn. (1)
i.e. 2 cos 2x + k cos x → 0
as x → 0 and this happen.
When 2 × 1 + k × 1 = 0
⇒ k = – 2
= \(\ {Lt}_{x \rightarrow 0} \frac{2 \cos 2 x-2 \cos x}{3 x^2}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{-4 \sin 2 x+2 \sin x}{6 x}\) (\(\frac{0}{0}\))
= \(\ {Lt}_{x \rightarrow 0} \frac{-8 \cos 2 x+2 \cos x}{6}\)
= \(\frac{-8 \times 1+2 \times 1}{6}\)
= \(\frac{-6}{6}\) = – 1.

Question 16.
Find the values of a and b \(\ {Lt}_{x \rightarrow 0} \frac{x(1-a \cos x)+b \sin x}{x^3}\) exists and equals \(\frac{1}{3}\).
Solution:
\(\ {Lt}_{x \rightarrow 0} \frac{x(1-a \cos x)+b \sin x}{x^3}\) (\(\frac{0}{0}\) form)
= \(\ {Lt}_{x \rightarrow 0} \frac{(1-a \cos x)+a x \sin x+b \cos x}{3 x^2}\) …………….(1)
Since denominator of eqn. (1) i.e. 3x² → 0 as x → 0
so in order that given limit ecxists finitely, the numerator of eqn. (1)
i.e. (1 – a cos x) + ax sin x + b cos x → 0
as x → 0 and this happen
When 1 – a + b =0
Now eqn. (2) is satisfied.
When given limit
= \(\ {Lt}_{x \rightarrow 0} \frac{(1-a \cos x)+a x \sin x+b \cos x}{3 x^2}\left(\frac{0}{0} \text { form }\right)\)
= \(\ {Lt}_{x \rightarrow 0} \frac{a \sin x+a \sin x+a x \cos x-b \sin x}{6 x}\left(\frac{0}{0} \text { form }\right)\)
= \(\ {Lt}_{x \rightarrow 0} \frac{a \cos x+a \cos x+a \cos x-a x \sin x-b \cos x}{6}\)
= \(\frac{a+a+a-b}{6}\)
= \(\frac{3 a-b}{6}=\frac{1}{3}\) (given)
⇒ 3a – b = 2
On solving (2) and (3) ; we have
1 + 2a = 2
⇒ a = \(\frac{1}{2}\)
and b = – \(\frac{1}{2}\)

Regular engagement with Understanding ISC Mathematics Class 12 Solutions Chapter 5 Continuity and Differentiability MCQs can boost students’ confidence in the subject.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability MCQs

Choose the correct answer from the given four options in questions (1 to 55) :

Question 1.
The number of points of discontinuity of the rational function f(x) = \(\frac{x^2-3 x+2}{4 x-x^3}\) is
(a) 1
(b) 2
(c) 3
(d) none
Solution:
(c) 3

Given f(x) = \(\frac{x^2-3 x+2}{4 x-x^3}\)
Clearly f(x) is discontinuous
when 4x – x³ = 0
⇒ x (4 – x²) = 0
⇒ x = 0, ± 2
Thus required no. of points of discontinuity of f(x) be three.

Question 2.
The number of points of discontinuity of the function f(x) = |x – 1| + |x – 2| + sin x, x ∈ [0, 4], is
(a) 1
(b) 2
(c) 3
(d) 0
Solution:
(d) 0

Given f(x) = |x – 1| + |x – 2| + sin x
When x < 1
∴ |x – 1| = – (x – 1) ;
When 1 ≤ x < 2
∴ |x – 1| = x – 1 ;
When x ≥ 2
∴ |x – 1| = x – 1 ;
|x – 2| = x – 2
∴ f(x) = \(\left\{\begin{array}{ccc}
-(x-1)-(x-2)+\sin x & ; & x<1 \\
x-1-(x-2)+\sin x & ; & 1 \leq x<2 \\
x-1+x-2+\sin x & ; & x \geq 2
\end{array}\right.\)
= \(\left\{\begin{array}{clc}
-2 x+3+\sin x & ; & x<1 \\
1+\sin x & ; & 1 \leq x<2 \\
2 x-3+\sin x & ; & x \geq 2
\end{array}\right.\)

at x = 1
\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) – 2x + 3 + sin x
= 1 + sin 1
\(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) 1 + sin x
= 1 + sin 1
∴ f(x) is continous at x = 1.

at x = 2:
\(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) 1 + sin x
= 1 + sin 2
\(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) (2x – 3) + sin x
= 1 + sin 2
∴ f(x) is continous at x = 2.

Since every polynomial function and trigonometric function is continuous in its domain.
Thus given function is everywhere continuous.

Question 3.
The functionf(x) = cot x is discontinuous on the set
(a) {x = nπ, n ∈ N}
(b) {x = 2nπ, n ∈ Z}
(c) {x = \(\frac{n \pi}{2}\), n ∈ Z}
(d) {x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z}
Solution:
(d) {x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z}

Given f(x) = cot x
= \(\frac{\cos x}{\sin x}\)
f(x) is not defined when sin x = 0
⇒ x = nπ ∀ n ∈ Z
Thus, f(x) is discontinuous on set {x = nπ, n ∈ N}.

Question 4.
The domain of continuity of the function f(x) = tan x is
(a) R – {nπ, n ∈ Z}
(b) R – {2nπ, n ∈ Z}
(c) R – {(2n + 1)\(\frac{\pi}{2}\), n ∈ Z}
(d) R – {\(\frac{n \pi}{2}\), n ∈ Z}
Solution:
(c) R – {(2n + 1)\(\frac{\pi}{2}\), n ∈ Z}

Given f(x) = tan x
= \(\frac{\sin x}{\cos x}\)
Now f(x) is not defined when cos x = 0
⇒ x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z
So f(x) is discontinuous at x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z
Hence domain of continuity of function f(x) be R – {(2n + 1) \(\frac{\pi}{2}\), n ∈ Z}.

Question 5.
The functionf(x) = [x], where [x] denotes the greatest integer function, is continuous at
(a) – 2
(b) 1
(c) 4
(d) 1.5
Solution:
We know that,
\(\underset{x \rightarrow a^{-}}{\mathrm{Lt}}\) [x] = a – 1
\(\underset{x \rightarrow a^{+}}{\mathrm{Lt}}\) [x] = a, where a ∈ I
Thus, f(x) is discontinuous for all integral points
and \(\underset{x \rightarrow a}{\mathrm{Lt}}\) = [a]
Thus f(x) is continuous for all non-integral points.
Clearly f(x) is continuous at x = 1.5 = \(\frac{3}{2}\) ∉ I.

Question 6.
The function f(x) = \(\left\{\begin{array}{cc}
x-1, & x<2 \\
2 x-3, & x \geq 2
\end{array}\right.\) is continous function
(a) at x = 2 only
(b) for all real values of x
(c) for all real values of x except 2
(d) for all integral values of X only
Solution:
When x < 2 ; f (x) = x – 1 which is a polynomial function and hence continuous everywhere. When x > 2 ;
f (x) = 2x – 3, which is a polynomial function and hence continuous everywhere.
at x = 2
\(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) x – 1
= 2 – 1 = 1
\(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) 2x – 3
= 4 – 3 = 1
and f(2) = 1
∴ f(x) is continuous at x = 2
Thusf(x) is continuous for all real values of x.

Question 7.
If f(x) = \(\left\{\begin{array}{cc}
5 x-4, & 0 4 x^2+3 a x, & 1<x<2
\end{array}\right.\) is continuous for all x ∈ (0, 2), then the value of a is
(a) 1
(b) 0
(c) – 1
(d) \(\frac{1}{3}\)
Solution:
(c) – 1

Since function f(x) is continuous for all x ∈ (0, 2)
∴ f(x) is continuous at x = 1
\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) 5x – 4
= 5 – 4 = 1
and \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) (4x² + 3ax)
= 4 + 3a
⇒ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x)
⇒ 1 = 4 + 3a
⇒ – 3 = 3a
⇒ a = – 1.

Question 8.
If f(x) = \(\left\{\begin{array}{cc}
\frac{\sqrt{4+x}-2}{x}, & x \neq 0 \\
k, & x=0
\end{array}\right.\) is continous at x = 0, then the value of k is
(a) 1
(b) 4
(c) \(\frac{1}{4}\)
(d) 0
Solution:
(c) \(\frac{1}{4}\)

\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = \(\ {Lt}_{x \rightarrow 0} \frac{\sqrt{4+x}-2}{x} \times \frac{\sqrt{4+x}+2}{\sqrt{4+x}+2}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{4+x-4}{x[\sqrt{4+x}+2]}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{1}{\sqrt{4+x}+2}\)
= \(\frac{1}{2+2}=\frac{1}{4}\)
and f(0) = k
Since f(x) is continuous at x = 0
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\)

Question 9.
If f(x) = \(\left\{\begin{array}{cc}
\frac{\sqrt{x^2+5}-3}{x+2}, & x \neq-2 \\
k, & x=-2
\end{array}\right.\) is continuous at x = – 2, then the value of k is
(a) – \(\frac{2}{3}\)
(b) 0
(c) \(\frac{2}{3}\)
(d) none of these
Solution:
(a) – \(\frac{2}{3}\)

Question 10.
If f(x) = \(\left\{\begin{array}{cc}
\frac{\sin \pi x}{5 x}, & x \neq 0 \\
k, & x=0
\end{array}\right.\) is continuous at x = 0, then k is equal to
(a) \(\frac{\pi}{5}\)
(b) \(\frac{5}{\pi}\)
(c) 1
(d) 0
Solution:
(b) \(\frac{5}{\pi}\)

\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = \(\ {Lt}_{x \rightarrow 0} \frac{\sin \pi x}{5 x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{\sin \pi x}{\pi x} \times \frac{\pi}{5}\)
= \(\frac{\pi}{5} \times 1=\frac{\pi}{5}\)
[∵ \(\underset{\theta \rightarrow 0}{\ {Lt}} \frac{\sin \theta}{\theta}\) = 1]
Since f(x) is continuous at x = 0
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = f(0)
⇒ \(\frac{\pi}{5}\) = k.

Question 11.
The value of function f at x = 0, so that the function f(x) = \(\frac{2^x-2^{-x}}{x}\), x ≠ 0 is continuous at x = 0, is
(a) 0
(b) log 2
(c) log 4
(d) 2⁴
Solution:
(c) log 4

\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = \(\ {Lt}_{x \rightarrow 0} \frac{2^x-2^{-x}}{x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{2^x-1-\left(2^{-x}-1\right)}{x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{2^x-1}{x}+\underset{x \rightarrow 0}{\ {Lt}} \frac{2^{-x}-1}{-x}\)
= log 2 + log 2
= 2 lo0g 2
= log 2²
= log 4
Since f(x) is continuous at x = 0
∴ f(0) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = log 4.

Question 12.
If f(x) = \(\frac{2 x+\sin ^{-1} x}{2 x-\tan ^{-1} x}\) is continuous for all x in (- 1, 1), then the value of f(0) is
(a) 2
(b) 3
(c) – 3
(d) \(\frac{3}{2}\)
Solution:
(b) 3

\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = \(\ {Lt}_{x \rightarrow 0} \frac{2 x+\sin ^{-1} x}{2 x-\tan ^{-1} x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{2+\frac{\sin ^{-1} x}{x}}{2-\frac{\tan ^{-1} x}{x}}\)
= \(\frac{2+1}{2-1}\) = 3
[∵ \(\underset{\theta \rightarrow 0}{\mathrm{Lt}} \frac{\sin ^{-1} \theta}{\theta}=\underset{\theta \rightarrow 0}{\mathrm{Lt}} \frac{\tan ^{-1} \theta}{\theta}\) = 1]
Since f(x) is continuous ∀ x ∈ (- 1, 1)
∴ f(x) is continuous at x = 0 ∈ (- 1, 1).
Thus, f(0) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = 3.

Question 13.
If f(x) = \(\left\{\begin{array}{cc}
\frac{1-\tan x}{4 x-\pi}, & x \neq \frac{\pi}{4} \\
k, & x=\frac{\pi}{4}
\end{array}\right.\) is continuous at x = \(\frac{\pi}{4}\), then the value of k is
(a) 1
(b) – 1
(c) \(\frac{1}{2}\)
(d) – \(\frac{1}{2}\)
Solution:
(d) – \(\frac{1}{2}\)

Given f(x) is continuous at x = \(\frac{\pi}{4}\)
∴ \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{4}}\) f(x) = f(latex]\frac{\pi}{4}[/latex])
⇒ – \(\frac{1}{2}\) = k.

Question 14.
If f(x) = \(\left\{\begin{array}{cc}
\tan \left(\frac{\pi}{4}-x\right) & x \neq \frac{\pi}{4} \\
k, & x=\frac{\pi}{4}
\end{array}\right.\) is continuous at x = \(\frac{\pi}{4}\), then the value of k is
(a) 1
(b) 2
(c) \(\frac{1}{2}\)
(d) none of these
Solution:
(c) \(\frac{1}{2}\)

Given f(x) is continuous at x = \(\frac{\pi}{4}\)
∴ \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{4}}\) f(x) = f(\(\frac{\pi}{4}\))
⇒ \(\frac{1}{2}\) = k.

Question 15.
If f(x) = \(\left\{\begin{array}{cc}
\frac{1-\cos p x}{x \sin x}, & x \neq 0 \\
\frac{1}{2}, & x=0
\end{array}\right.\) is continuous at x = 0, then p is equal to
(a) 2
(b) – 2
(c) 1, – 1
(d) none of these
Solution:
(c) 1, – 1

∴ \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) f(x) = f(0)
⇒ \(\frac{p^2}{2}=\frac{1}{2}\)
⇒ p = 1.

Question 16.
If f(x) = \(\left\{\begin{array}{cc}
\frac{\sqrt{1-\cos 2 x}}{\sqrt{2} x}, & x \neq 0 \\
k, & x=0
\end{array}\right.\), then which value of k will make function f continuous at x = 0?
(a) 1
(b) – 1
(c) 0
(d) none of these
Solution:
(d) none of these

When x → 0^–
⇒ sin x < 0 as x lies in IVth quadrant
⇒ |sin x| = – sin x
= \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = – 1
⇒ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) ≠ \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x)
Thus f(x) is discontinuous function.

Question 17.
If f(x) = \(\left\{\begin{array}{cc}
x^2 \sin \frac{1}{x}, & x \neq 0 \\
k, & x=0
\end{array}\right.\) is continuous at x = 0, then the value of k
(a) 1
(b) – 1
(c) 0
(d) none of these
Solution:
(c) 0

Let g(x) = x²
and h(x) = sin \(\frac{1}{x}\)
Here \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) g(x) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) x² = 0
and |h(x)| = |sin \(\frac{1}{x}\)| ≤ 1 ∀ x ∈ R – {0}
Thus h(x) is bounded in the ngd of o.
⇒ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) g(x) h(x) = 0
⇒ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) x² sin \(\frac{1}{x}\) = 0
⇒ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = 0
and f(0) = k
Given f(x) is continuous at x = 0
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = f(0)
⇒ 0 = k.

Question 18.
If f(x) = \(\left\{\begin{array}{rr}
m x+1, & x \geq \frac{\pi}{2} \\
\sin x+n, & x<\frac{\pi}{2}
\end{array}\right.\) is continuous at x = \(\frac{\pi}{2}\), then
(a) m = 1, n = 0
(b) m = \(\frac{n \pi}{2}\) + 1
(c) n = \(\frac{m \pi}{2}\)
(d) m = n = \(\frac{\pi}{2}\)
Solution:
\(\begin{gathered}
\mathrm{Lt} \\
x \rightarrow \frac{\pi^{+}}{2}
\end{gathered}\) f(x) = \(\begin{gathered}
\mathrm{Lt} \\
x \rightarrow \frac{\pi^{+}}{2}
\end{gathered}\) mx + 1
= \(\frac{m \pi}{2}\) + 1
\(\begin{gathered}
\mathrm{Lt} \\
x \rightarrow \frac{\pi^{-}}{2}
\end{gathered}\) f(x) = \(\begin{gathered}
\mathrm{Lt} \\
x \rightarrow \frac{\pi^{-}}{2}
\end{gathered}\) sin x + n
= sin \(\frac{\pi}{2}\) + n
= 1 + n
Since f(x) is continuous at x = \(\frac{\pi}{2}\)
∴ \(\begin{gathered}
\mathrm{Lt} \\
x \rightarrow \frac{\pi^{-}}{2}
\end{gathered}\) f(x) = \(\begin{gathered}
\mathrm{Lt} \\
x \rightarrow \frac{\pi^{-}}{2}
\end{gathered}\) f(x)
⇒ \(\frac{m \pi}{2}\) + 1 = 1 + n
⇒ n = \(\frac{m \pi}{2}\).

Question 19.
The functionf(x) = | x | at x = 0 is
(a) continuous but not differentiable
(b) differentiable but not continuous
(c) continuous and differentiable
(d) discontinuous and not differentiable
Solution:
(a) continuous but not differentiable

\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) |x|
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x = 0
and \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) |x|
= \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) – x = 0
and f(0) = |0| = 0
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = f(0)
∴ f(x) is continuous at x = 0
Lf'(0) = \(\ {Lt}_{x \rightarrow 0^{-}} \frac{|x|-0}{x-0}\)
= \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\)
= – 1
Rf'(0) = \(\underset{x \rightarrow 0^{+}}{\ {Lt}} \frac{|x|-0}{x-0}\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{x}{x}\)
= 1
∴ Lf'(0) ≠ Rf'(0)
⇒ f(x) is not differentiable at x = 0.

Question 20.
The function f(x) = |x| at x = 0 is
(a) continuous but not differentiable
(b) differentiable but not continuous
(c) continuous and differentiable
(d) neither continuous nor differentiable
Solution:
(c) continuous and differentiable

f(x) = x |x|
= \(\left\{\begin{aligned}
-x^2 & ; \quad x<0 \\
x^2 & ; \quad x \geq 0
\end{aligned}\right.\)

Continuity at x = 0:
\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x² = 0
\(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) – x² = 0
and f(0) = 0
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = f(0)
⇒ f(x) is continuous at x = 0.

Differentiability at x = 0:
Lf'(0) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) \(\frac{-x^2-0}{x}\)
= \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) x = 0
and Rf'(0) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) \(\frac{x^2-0}{x}\)
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x = 0
∴ Lf'(0) = Rf'(0)
⇒ f(x) is continuous at x = 0.

Question 21.
The derivative of the fiunction f(x) = x |x| at x = 0 is
(a) 1
(b) 0
(c) 2
(d) – 2
Solution:
(c) 2

f(x) = x |x|
= \(\left\{\begin{aligned}
-x^2 & ; \quad x<0 \\
x^2 & ; \quad x \geq 0
\end{aligned}\right.\)

Continuity at x = 0:
\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x² = 0
\(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) – x² = 0
and f(0) = 0
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = f(0)
⇒ f(x) is continuous at x = 0.

Differentiability at x = 0:
Lf'(0) = \(\ {Lt}_{x \rightarrow 0^{-}} \frac{-x^2-0}{x}\)
= \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) x = 0
and Rf'(0) = \(\ {Lt}_{x \rightarrow 0^{+}} \frac{x^2-0}{x}\)
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x = 0
∴ Lf'(0) = Rf'(0)
⇒ f(x) is continuous at x = 0.

Question 22.
If f(x) = \(\left\{\begin{array}{cc}
x, & x \leq 0 \leq 1 \\
2 x-1, & x>1
\end{array}\right.\), then
(a) f is discontinuous at x = 1
(b) f is differentiable at x = 1
(c) f is continuous but not differentiable at x = 1
(d) none of these
Solution:
(c) f is continuous but not differentiable at x = 1

\(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) 2x – 1
= 2 × 1 – 1 = 1
\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) x = 1
and f(1) = 1
∴ \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = f(1)
⇒ f(x) is continuous at x = 1.

Differentiability at x = 1:
Lf'(1) = \(\ {Lt}_{x \rightarrow 1^{-}} \frac{x-1}{x-1}\) = 1
Rf'(1) = \(\ {Lt}_{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{+}} \frac{2 x-1-1}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{-1}} \frac{2(x-1)}{x-1}\)
= 2
∴ Lf'(1) ≠ Rf'(1)
⇒ f(x) is not differentiable at x = 1.

Question 23.
If f(x) = \(\left\{\begin{array}{cc}
a x^2+1, & x>1 \\
x+a, & x \leq 1
\end{array}\right.\) is derivable at x = 1, then the value of a is
(a) 0
(b) 1
(c) \(\frac{1}{2}\)
(d) 2
Solution:
(c) \(\frac{1}{2}\)

at x = 1
Lf'(1) = \(\ {Lt}_{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{x+a-1-a}{x-1}\) = 1
and Rf'(1) = \(\ {Lt}_{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{+}} \frac{a x^2+1-1-a}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{+}} \frac{a\left(x^2-1\right)}{x-1}\)
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) a(x + 1) = 2a
Since f(x) is derivable at x = 1
∴ Lf'(1) = Rf'(1)
⇒ 1 = 2a
⇒ a = \(\frac{1}{2}\)

Question 24.
The function f(x) = |x| + |x – 1|
(a) differentiable at x = 0 but not at x = 1
(b) differentiable at x = 1 but not at x = 0
(c) neither differentiable at x = 0 nor at x = 1
(d) differentiable at x = 0 as well as at x = 1
Solution:
(c) neither differentiable at x = 0 nor at x = 1

Given f(x) = |x| + |x – 1|
= \(\left\{\begin{array}{ccc}
-x-(x-1) & ; & x<0 \\
x-(x-1) & ; & 0 \leq x<1 \\
x+x-1 & ; & x \geq 1
\end{array}\right.\)
⇒ f(x) = \(\left\{\begin{array}{ccc}
-2 x+1 & ; & x<0 \\
1 & ; & 0 \leq x<1 \\
2 x-1 & ; & x \geq 1
\end{array}\right.\)

at x = 0:
Lf'(0) = \(\ {Lt}_{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}\)
= \(\ {Lt}_{x \rightarrow 0^{-}} \frac{-2 x+1-1}{x}\)
= – 2
Rf'(0) = \(\ {Lt}_{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{1-1}{x}\) = 0
∴ Lf'(0) ≠ Rf'(0)
∴ f is not differentiable at x = 0.

at x = 1:
Lf'(1) = \(\ e{Lt}_{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{1-1}{x-1}\) = 0
and Rf'(1) = \(\ {Lt}_{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{+}} \frac{2 x-1-1}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{+}} \frac{2(x-1)}{x-1}\) = 2
∴ Lf'(1) ≠ Rf'(1)
∴ f is not differentiable at x = 1.

Question 25.
The derivative of f(x) = 3 |2 + x| at x = – 3 is
(a) 3
(b) – 3
(c) 0
(d) d0es not exist
Solution:
(b) – 3

Given f(x) = 3 |2 + x|
∴ f'(x) = \(\frac{3(2+x)}{|(2+x)|}\) × 1
⇒ f'(- 3) = \(\frac{3(2-3)}{|2-3|}\) = – 3.

Question 26.
The derivative of f(x) = |x – 1| + |x – 3| at x = 2 is
(a) 2
(b) – 2
(c) 0
(d) does not exist
Solution:
(c) 0

Given f(x) = |x – 1| + |x – 3|
∴ f'(x) = \(\frac{x-1}{|x-1|}+\frac{x-3}{|x-3|}\)
⇒ f'(2) = \(\frac{2-1}{|2-1|}+\frac{2-3}{|2-3|}\)
= \(\frac{1}{|1|}-\frac{1}{|-1|}\)
= 1 – 1 = 0.

Question 27.
The function f(x) = e^|x| is
(a) continuous everywhere but not differentiable at x = 0
(b) continuous and differentiable everywhere
(e) not continuous at x = 0
(d) none of these
Solution:
(a) continuous everywhere but not differentiable at x = 0

\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) e^x
= e⁰ = 1
and \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) e^{– x}
= e⁰ = 1
and f(0) = e^|0|
= e = 1
Thus, \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = f(0)
∴ f(x) is continuous at x = 0.
Since exponential function is continuous for all x < 0 and x > 0
Thus f(x) is continuous everywhere.
Further when x < 0 ; f (x) = e^{– x} is differentiable function.
when x > 0 ; f(x) = e^x is differentiable function.

at x = 0:
Lf'(0) = \(\ {Lt}_{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}\)
= \(\ {Lt}_{x \rightarrow 0^{-}} \frac{e^{-x}-1}{x}\)
= – \(\ {Lt}_{x \rightarrow 0^{-}} \frac{e^{-x}-1}{-x}\)
= – 1
and Rf'(0) = \(\ {Lt}_{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{e^x-1}{x}\)
= 1
∴ Lf'(0) ≠ Rf'(0)
Thus f(x) is not differentiable at x = 1.

Question 28.
The function f(x) = |sin x|
(a) f is differentiable everywhere
(b) f is continuous everywhere but not differentiable at x = nπ, n ∈ Z
(c) f is continuous everywhere but not differentiable atx=(2n+1) \(\frac{\pi}{2}\), n ∈ Z
(d) none of these
Solution:
(b) f is continuous everywhere but not differentiable at x = nπ, n ∈ Z

f(x) = |sin x|
= \(\left\{\begin{aligned}
-\sin x ; & x<n \pi \\
\sin x ; & x \geq n \pi
\end{aligned}\right.\) ;
x = even
x = nπ (n even)

∴ L.H.D. ≠ R.H.D.
Thus f(x) is not differentiable at x = nπ (n even)

∴ L.H.D. = R.H.D.
Thus f(x) is not differentiable at x = nπ (n = odd)
Hence f(x) is not differentiable at x = nπ, n ∈ Z.

Question 29.
The set of numbers where the function f given by f (x) = | 2x – 1 | cos x is differentiable is
(a) R
(b) R – {\(\frac{1}{2}\)}
(c) (0, ∞)
(d) none of these
Solution:
Refer to MCQ-28 (RD Sharma 10 + 2) Chapter 9.

Question 30.
If y = log (sec e^x2), then \(\frac{d y}{d x}\) =
(a) x² e^x2 tan e^x2
(b) e^x2 tan e^x2
(c) 2x e^x2 tan e^x2
(d) none of these
Solution:
(c) 2x e^x2 tan e^x2

Given y = log (sec e^x2)
∴ \(\frac{d y}{d x}\) = \(\frac{1}{\sec e^{x^2}} \frac{d}{d x}\) sec e^x2
= \(\frac{1}{\sec e^{x^2}}\) sec e^x2 . tan e^x2 . e^x2 . 2x
= e^x2 . 2x tan e^x2.

Question 31.
If y = log \(\left(\frac{1-x^2}{1+x^2}\right)\), |x| < 1, then \(\frac{d y}{d x}\) =
(a) \(\frac{4 x^3}{1-x^4}\)
(b) \(\frac{-4 x}{1-x^4}\)
(c) \(\frac{1}{4-x^4}\)
(d) \(\frac{-4 x^3}{1-x^4}\)
Solution:
(b) \(\frac{-4 x}{1-x^4}\)

Given y = log \(\left(\frac{1-x^2}{1+x^2}\right)\)
= log (1 – x²) – log (1 + x²)
∴ \(\frac{d y}{d x}\) = \(\frac{-2 x}{1-x^2}-\frac{2 x}{1+x^2}\)
= \(\frac{-2 x\left(1+x^2+1-x^2\right)}{1-x^4}\)
= \(\frac{-4 x}{1-x^4}\).

Question 32.
If f(x) = log x, then the derivative of f (log x) w.r.t. x is
(a) \(\frac{\log x}{x}\)
(b) \(\frac{x}{\log x}\)
(c) x log x
(d) \(\frac{1}{x \log x}\)
Solution:
(d) \(\frac{1}{x \log x}\)

Given f(x) = log x
⇒ f(log x) = log (log x)
⇒ \(\frac{d}{d x}\) f (log x) = \(\frac{1}{\log x}\) \(\frac{d}{d x}\) log x
= \(\frac{1}{x \log x}\).

Question 33.
For the curve √x + √y = 4, \(\frac{d y}{d x}\) at (\(\left(\frac{1}{4}, \frac{1}{4}\right)\)) is
(a) – 1
(b) 1
(c) 2
(d) \(\frac{1}{2}\)
Solution:
(a) – 1

Given √x + √y = 4
Diff. both sides w.r.t. x ; we have
\(\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}\) = 0
\(\frac{d y}{d x}=-\sqrt{\frac{y}{x}}\)
at (\(\left(\frac{1}{4}, \frac{1}{4}\right)\)) ;
\(\frac{d y}{d x}\) = – \(\sqrt{\frac{\frac{1}{4}}{\frac{1}{4}}}\) = – 1

Question 34.
If y = \(\sqrt{\sin x+y}\), then \(\frac{d y}{d x}\) =
(a) \(\frac{\cos x}{2 y-1}\)
(b) \(\frac{\cos x}{1-2 y}\)
(c) \(\frac{\sin x}{2 y-1}\)
(d) \(\frac{\sin x}{1-2 y}\)
Solution:
(a) \(\frac{\cos x}{2 y-1}\)

Given y = \(\sqrt{\sin x+y}\)
⇒ y² = sin x + y ;
diff. both sides w.r.t. x ;
2y \(\frac{d y}{d x}\) = cos x + \(\frac{d y}{d x}\)
⇒ (2y – 1) \(\frac{d y}{d x}\) = cos x
⇒ \(\frac{d y}{d x}\) = \(\frac{\cos x}{2 y-1}\).

Question 35.
The derivative of sec (tan^-1 x) w.r.t. x is
(a) \(\frac{x}{1+x^2}\)
(b) \([\frac{1}{\sqrt{1+x^2}}/latex]
(c) [latex]\frac{x}{\sqrt{1+x^2}}\)
(d) x \(\sqrt{1+x^2}\)
Solution:
(c) \(\frac{x}{\sqrt{1+x^2}}\)

Let y = sec (tan^-1 x)
put tan^-1 x = θ
⇒ x = tan θ

∴ y = sec θ
= \(\sqrt{x^2+1}\)
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{2}\) (x² + 1)^{\(-\frac{1}{2}\)} \(\frac{d}{d x}\) (x² + 1)
= \(\frac{2 x}{2 \sqrt{x^2+1}}\)
= \(\frac{x}{\sqrt{x^2+1}}\)

Question 36.
If f(x) = x tan^-1 x, then f'(1) is equal to
(a) \(\frac{\pi}{4}+\frac{1}{2}\)
(b) \(\frac{\pi}{4}-\frac{1}{2}\)
(c) \(\frac{1}{2}-\frac{\pi}{4}\)
(d) none of these
Solution:
(a) \(\frac{\pi}{4}+\frac{1}{2}\)

Given f(x) = x tan^-1 x
f'(x) = tan^-1 x . 1 + \(\frac{x}{1+x^2}\)
f'(1) = tan^-1 1 + \(\frac{1}{1+1^2}\)
= \(\frac{\pi}{4}+\frac{1}{2}\).

Question 37.
The derivative of tan^-1 \(\left(\frac{3 x-4 x^3}{1-3 x^2}\right)\) w.r.t. x is
(a) \(\frac{1}{1+x^2}\)
(b) \(\frac{3}{1+9 x^2}\)
(c) \(\frac{3}{1+x^2}\)
(d) 3 sec² 3x
Solution:

Let y = tan^-1 \(\left(\frac{3 x-4 x^3}{1-3 x^2}\right)\)
put x = tan θ
⇒ θ = tan^-1 x
⇒ y = tan^-1 \(\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right)\)
= tan^-1 (tan 3θ)
Thus y = 3θ
= 3 tan^-1 x
⇒ \(\frac{d y}{d x}\) = \(\frac{3}{1+x^2}\).

Question 38.
The derivative of tan^-1 x w.r.t. cot^-1 x is
(a) \(\frac{\pi}{2}\)
(b) – 1
(c) 1
(d) none of these
Solution:
(b) – 1

Let y = tan^-1 x
and z = cot^-1 x
∴ \(\frac{d y}{d x}=\frac{1}{1+x^2}\)
and \(\frac{d z}{d x}=\frac{-1}{1+x^2}\)
we want to diff. y w.r.t. z i.e. to find \(\frac{d y}{d z}\)
∴ \(\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}\)
= \(– \frac{\frac{1}{1+x^2}}{\frac{1}{1+x^2}}\) = – 1

Question 39.
The derivative of cos^-1 (2x² – 1) w.r.t. cos^-1 x is
(a) 2
(b) \(\frac{2}{x}\)
(c) 1 – x²
(d) \(\frac{-1}{2 \sqrt{1-x^2}}\)
Solution:
(a) 2

Let y = cos^-1 (2x² – 1) ………..(1)
and z = cos^-1 x ………..(2)
we want to diff. y w.r.t. z i.e. to find \(\frac{d y}{d z}\)
putting x = cos θ
∴ θ = cos^-1 x in eqn. (1) ;
y = cos^-1 (2 cos² θ – 1)
= cos^-1 (cos 2θ)
= 2θ
= 2 cos^-1 x
∴ y = 2z
⇒ \(\frac{d y}{d z}\) = 2.

Question 40.
The derivative of sin^-1 \(\left(\frac{2 x}{1+x^2}\right)\) w.r.t. tan^-1 \(\left(\frac{2 x}{1-x^2}\right)\) is
(a) \(\frac{1}{2}\)
(b) 2
(c) \(\frac{1-x^2}{1+x^2}\)
(d) 1
Solution:
(d) 1

Let y = sin^-1 \(\left(\frac{2 x}{1+x^2}\right)\)
= 2 tan^-1 x
and z = tan^-1 \(\left(\frac{2 x}{1-x^2}\right)\)
= 2 tan^-1 x
∴ y = z
⇒ \(\frac{d y}{d z}\) = 1.

Question 41.
The derivative of tan^-1 \(\left(\frac{x}{\sqrt{1-x^2}}\right)\) w.r.t. x is
(a) \(\frac{1}{\sqrt{1-x^2}}\)
(b) \(\frac{1}{\sqrt{1-x^2}}\)
(c) \(– \frac{1}{\sqrt{1-x^2}}\)
(d) none of these
Solution:
(b) \(\frac{1}{\sqrt{1-x^2}}\)

Let y = tan^-1 \(\left(\frac{x}{\sqrt{1-x^2}}\right)\)
put x = sin θ
∴ θ = sin^-1 x
∴ y = tan^-1 \(\left(\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}\right)\)
= tan^-1 (tan θ)
= θ
= sin^-1 x
⇒ \(\frac{d y}{d x}=\frac{1}{\sqrt{1-x^2}}\)

Question 42.
The derivative of sin^-1 \(\left(\frac{x}{\sqrt{1+x^2}}\right)\) w.r.t. x is
(a) \(\frac{1}{\sqrt{1-x^2}}\)
(b) \(\frac{1}{\left(1+x^2\right)^{\frac{3}{2}}}\)
(c) \(\frac{1}{1+x^2}\)
(d) none of these
Solution:
(c) \(\frac{1}{1+x^2}\)

Let y = sin^-1 \(\left(\frac{x}{\sqrt{1+x^2}}\right)\)
put x = tan θ
∴ θ = tan^-1 x
⇒ y = sin^-1 \(\left(\frac{\tan \theta}{\sqrt{1+\tan ^2 \theta}}\right)\)
= sin^-1 \(\left(\frac{\tan \theta}{\sqrt{1+\tan ^2 \theta}}\right)\)
= sin^-1 \(\left(\frac{\tan \theta}{\sec \theta}\right)\)
⇒ y = θ = tan^-1 x
Thus, \(\frac{d y}{d x}=\frac{1}{1+x^2}\).

Question 43.
If y = cos^-1 \(\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\) + cosec^-1 \(\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\) then \(\frac{d y}{d x}\) is equal to
(a) \(\frac{\pi}{2}\)
(b) 0
(c) 1
(d) none of these
Solution:
(b) 0

Given y = cos^-1 \(\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\) + cosec^-1 \(\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\)
⇒ y = cos^-1 \( \left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\) + sin^-1 \( \left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
[∵ cosec^-1 x = sin^-1 \(\frac{1}{x}\)]
⇒ y = \(\frac{\pi}{2}\)
[∵ cos^-1 x + sin^-1 x = \(\frac{\pi}{2}\) ∀ x ∈ [- 1, 1]]
∴ \(\frac{d y}{d x}\) = 0.

Question 44.
If y = a (1 + cos t) and x = a (t – sin t) then \(\frac{d y}{d x}\) is equal to
(a) tan \(\frac{t}{2}\)
(b) – tan \(\frac{t}{2}\)
(c) – cot \(\frac{t}{2}\)
(d) none of these
Solution:
(c) – cot \(\frac{t}{2}\)

Given y = a (1 + cos t)
∴ \(\frac{d y}{d t}\) = – a sin t
and x = a (t – sin t)
∴ \(\frac{d x}{d t}\) = a (1 – cos t)
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{-a \sin t}{a(1-\cos t)}\)
= \(\frac{-2 \sin \frac{t}{2} \cos \frac{t}{2}}{2 \sin ^2 \frac{t}{2}}\)
= – cot \(\frac{t}{2}\).

Question 45.
If x = a cos³ t and y = a sin³ t, then \(\frac{d y}{d x}\) is equal to
(a) cos t
(b) sec t
(c) cosec t
(d) – tan t
Solution:
(d) – tan t

Given x = a cos³ t
⇒ \(\frac{d x}{d t}\) = 3a cos² t (- sin t)
and y = a cos³ t
⇒ \(\frac{d y}{d t}\) = 3a sin² t (cos t)
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{3 a \sin ^2 t \cos t}{-3 a \cos ^2 t \sin t}\)
= – tan t.

Question 46.
If x = t² and y = t³, then \(\frac{d^2 y}{d x^2}\) is equal to
(a) \(\frac{3}{2}\)
(b) \(\frac{3}{2}\) t
(c) \(\frac{3}{2 t}\)
(d) \(\frac{3}{4 t}\)
Solution:
(d) \(\frac{3}{4 t}\)

Given x = t²
⇒ \(\frac{d x}{d t}\) = 2t
and y = t³
⇒ \(\frac{d y}{d t}\) = 3t²
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{3 t^2}{2 t}=\frac{3}{2} t\)
⇒ \(\frac{d^2 y}{d x^2}=\frac{d}{d t}\left(\frac{3}{2} t\right) \frac{d t}{d x}\)
= \(\frac{3}{2} \times \frac{1}{2 t}=\frac{3}{4 t}\).

Question 47.
The derivative of log x with respect to \(\frac{1}{x}\) is
(a) – \(\frac{1}{x^3}\)
(b) – \(\frac{1}{x}\)
(c) – x
(d) \(\frac{1}{x^3}\)
Solution:
(c) – x

Let y = log x
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{x}\)
and z = \(\frac{1}{x}\)
⇒ \(\frac{d z}{d x}\) = \(-\frac{1}{x^2}\)
we want to diff. y w.r.t. z i.e. to find \(\frac{d y}{d z}\)
∴ \(\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}\)
= \(\frac{\frac{1}{x}}{-\frac{1}{x^2}}\)
= – x.

Question 48.
The function f : R → R given by f(x) = – |x – 1| is
(a) continuous as well as differentiable at x = 1
(b) not continuous but differentiable at x = 1
(c) continuous but not differentiable at x = 1
(d) neither continuous nor differentiable atx= 1.
Solution:
(c) continuous but not differentiable at x = 1

Given f(x) = – |x – 1|
= \(\left\{\begin{array}{ccc}
(x-1) & ; & x<1 \\
-(x-1) & ; & x \geq 1
\end{array}\right.\)

Continuity at x = 1:
\(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) – (x – 1)
= – (1 – 1) = 0
\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) (x – 1)
= (1 – 1) = 0
and f(1) = – (1 – 1) = 0
∴ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = f(1)
Thus f(x) is continuous at x = 1.

Differentiability at x = 1:
Lf'(1) = \(\ {Lt}_{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{x-1-0}{x-1}\)
= 1
Rf'(0) = \(\ {Lt}_{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{+}} \frac{-(x-1)-0}{x-1}\)
= – 1
∴ Lf'(1) ≠ Rf'(1)
Thus, f(x) is not differentiable at x = 1.

Question 49.
If y = f(x²) and f'(x) = e^√x, then \(\frac{d y}{d x}\) is equal to
(a) 2x e^2√x
(b) 2x e^x
(c) 4x e^√x
(d) 4x e^x
Solution:
(b) 2x e^x

Given y = f(x²)
∴ \(\frac{d y}{d x}\) = f'(x²) . 2x
= \(e^{\sqrt{x^2}}\) . 2x
= 2x e^x
[∵ f'(x) = e^√x
⇒ f'(x²) = \(e^{\sqrt{x^2}}\).

Question 50.
If y = log_e \(\left(\frac{x^2}{e^2}\right)\), then \(\frac{d^2 y}{d x^2}\) is equal to
(a) \(-\frac{1}{x}\)
(b) \(-\frac{1}{x^2}\)
(c) \(\frac{2}{x^2}\)
(d) \(-\frac{2}{x^2}\)
Solution:
(d) \(-\frac{2}{x^2}\)

Given y = log_e \(\left(\frac{x^2}{e^2}\right)\)
= log x² – log e²
= 2 log x – log e²
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}=\frac{2}{x}\)
∴ \(\frac{d^2 y}{d x^2}=-\frac{2}{x^2}\).

Question 51.
In Rolle’s theorem, the value of c for the function f (x) = x³ – 3x in the interval [0, √3] is
(a) 1
(b) – 1
(c) \(\frac{1}{3}\)
(d) \(\frac{2}{3}\)
Solution:
(a) 1

Given f(x) = x³ – 3x
∴ f’(x) = 3x² – 3
which exists for all x ∈ R
Since f (x) is polynomial in x so it is continuous and differentiable everywhere.
f(x) is continuous in [0, π] and f(x) is diff. in (0, √3)
Also, f(0) = 0 ;
f(√3) = (√3)³ – 3√3
= 3√3 – 3√3 = 0
∴ f(0) = f(5)
Thus, all the three conditions of Rolle’s theorem are satisfied so ∃ atleast one real number c ∈ (0, √3) s.t f’ (c) = 0
⇒ 3c² – 3 = 0
⇒ c = ± 1
∴ c = 1 ∈ (0, √3).

Question 52.
The value of c is Rolle’s theorem for the function f(x) = e^x sin x, x ∈[0, π], is
(a) \(\frac{\pi}{6}\)
(b) \(\frac{\pi}{4}\)
(c) \(\frac{\pi}{2}\)
(d) \(\frac{3 \pi}{4}\)
Solution:
(d) \(\frac{3 \pi}{4}\)

Given f(x) = e^x sin x
Since exponential function is continuous and differentiable everywhere.
f’(x) = e^x cos x + sin x e^x
= e^x (cos x + sin x)
Further the product of continuous and differentiable is continuous and differentiable.
∴ f(x) is continuous in [0, π] and differentiable in (0, π).
Now f(0) = e⁰ sin 0 = 0
f(π) = e^π sin π = 0
∴ f(0) = f(π)
Thus all the three conditions of Rolle’s theorem are satisfied so ∃ atleast one real number c ∈ (0, π).
s.t. f'(c) = 0
⇒ e^c (cos c + sin c)= 0
⇒ cos c + sin c = 0
[∵ e^c > 0]
⇒ tan c = – 1
= – tan \(\frac{\pi}{4}\)
= tan (- \(\frac{\pi}{4}\))
⇒ c = nπ – \(\frac{\pi}{4}\), n ∈ Z
⇒ c = \(\frac{3 \pi}{4}\), – \(\frac{\pi}{4}\), ………….. but c ∈ (0, π)
∴ c = \(\frac{3 \pi}{4}\).

Question 53.
Rolle’s theorem is applicable in the interval [- 1, 1] for the function
(a) f (x) = x
(b) f(x) = x²
(c) f(x) = x³ + 2
(d) f(x) = |x|
Solution:
(b) f(x) = x²

Now f(x) = x², which is polynomial in x and hence continuous and differentiable everywhere
∴ f (x) is continuous in [- 1, 1] and differentiable in (- 1, 1).
Further f(1) = 1² = 1 ;
f(- 1) = (- 1)² = 1
∴ f(1) = f(- 1)
So all three conditions of Rolle’s theorem
so ∃ atleastonerealnumber c ∈ (- 1, 1)
s.t f’(c) = 0
⇒ 2c = 0
⇒ c = 0 ∈ (- 1, 1)
Rolle’s theorem is applicable.

Question 54.
The value of c in Lagrange’s Mean Value= theorem for the function f(x) = x (x – 2) in the interval [1, 2] is
(a) \(\frac{3}{2}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{3}{4}\)
(d) \(\frac{5}{4}\)
Solution:
(a) \(\frac{3}{2}\)

Given f(x) = x (x – 2)
which is a polynomial in x and hence continuous and differentiable everywhere.
Now f’ (x) = 2x – 2 which exists ∀ x ∈ R
Thus f (x) is continuous in [1, 2] and differentiable in (1,2).
So all the conditions of Lagrange’s mean value theorem are satisfied so ∃atleast one real no c ∈ (1, 2)
s.t. \(\frac{f(2)-f(1)}{2-1}\) = f’9c) ………….(1)
⇒ Now f(2) = 0;
f(1) = 1 (1 – 2) = – 1
∴ from (1) ;
\(\frac{0-(-1)}{1}\) = 2x – 2
⇒ 2c – 2 = 1
⇒ 2c = 3
⇒ c ∈ (1, 2).

Question 55.
The value of c in Lagrange’s Mean Value theorem for the function f(x) = x + \(\frac{1}{x}\) in the interval [1, 3] is
(a) 1
(b) 2
(c) √3
(d) – √3
Solution:
(c) √3

Given f(x) = x + \(\frac{1}{x}\)
∴ f'(x) = 1 – \(\frac{1}{x^2}\)
which exists for all x ∈ R – {0}
Thus f(x) is continuous in [1, 3] and differentiable in (1, 3).
So all the conditions of Lagrange’s mean value theorem arte satisfied.
So ∃ atleast one real number c ∈ (1, 3).
s.t. \(\frac{f(3)-f(1)}{3-1}\) = f'(3) …………(1)
Now f(1) = 1 + \(\frac{1}{1}\) = 2 ;
f(3) = 3 + \(\frac{1}{3}\) = \(\frac{10}{3}\)
∴ from (1) ;
⇒ \(\frac{\frac{10}{3}-2}{2}=1-\frac{1}{c^2}\)
⇒ \(\frac{4}{6}=1-\frac{1}{c^2}\)
⇒ \(\frac{1}{c^2}=1-\frac{2}{3}=\frac{1}{3}\)
⇒ c² = 3
⇒ c = ± √3
Here c = √3 ∈ (1, 3).

Peer review of ML Aggarwal Maths for Class 12 Solutions Chapter 5 Continuity and Differentiability Chapter Test can encourage collaborative learning.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Chapter Test

Question 1.
Is the function defined by f(x) = x² – sin x + 5 continuous at x = π? (NCERT)
Solution:
Given, f(x) = x² – sin x + 5
\(\underset{x \rightarrow \pi^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{h \rightarrow 0^{+}}{\ {Lt}}\) f(π + h)
= \(\underset{h \rightarrow 0^{+}}{\ {Lt}}\) [(π + h)² – sin (π + h) + 5]
= \(\underset{h \rightarrow 0^{+}}{\ {Lt}}\) [(π + h)² + sin h + 5]
= π² + 5
and \(\underset{h \rightarrow 0^{-}}{\ {Lt}}\) f(x) = \(\underset{h \rightarrow 0^{+}}{\ {Lt}}\) f(π – h)
= \(\underset{h \rightarrow 0^{+}}{\ {Lt}}\) [(π – h)² – sin (π – h) + 5)]
= \(\underset{h \rightarrow 0^{+}}{\ {Lt}}\) [(π – h)² – sin h + 5]
= π² + 5
= f(π)
∴ \(\mathrm{Lt}_{x \rightarrow {\pi}}\) f(x) = f(π)
Thus f(x) is continuous at x = π.

Question 2.
(i) Find the value of k so that f(x) = \(\begin{cases}\frac{\sin k x}{x}, & x<0 \\ 8-3 x & x \geq 0\end{cases}\) may be continuous at x = 0.
(ii) If the function f defined by f(x) = \(\left\{\begin{array}{cc}
\frac{1-\cos 2 x}{2 x^2}, & x \neq 0 \\
k & , x=0
\end{array}\right.\) is continuous at x = 0, find the value of k.
(iii) If the function f is defined by f(x) = \(\left\{\begin{array}{cc}
\frac{1-\tan x}{4 x-\pi}, & 0<x<\frac{\pi}{2}, x \neq \frac{\pi}{4} \\
k, & x=\frac{\pi}{4}
\end{array}\right.\) is continuous at x = \(\frac{\pi}{4}\), find the value of k.
Solution:
(i) L.H.Limit = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) \(\frac{\sin k x}{x}\)
= k . \(\frac{\sin k x}{k x}\)
= k × 1 = k
[∵ \(\underset{\theta \rightarrow 0}{\ {Lt}} \frac{\sin \theta}{\theta}\) = 1]
R.H.Limit = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) 8 – 3x
= 8 – 3 × 0 = 8
and f(0) = 8 – 3 × 0 = 8
Now f(x) is continuous at x = 0
∴ \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = f(0)

(ii) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = \(\ {Lt}_{x \rightarrow 0} \frac{1-\cos 2 x}{2 x^2}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{2 \sin ^2 x}{2 x^2}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{\sin ^2 x}{x^2}\)
= \(\left[\ {Lt}_{x \rightarrow 0} \frac{\sin x}{x}\right]^2\)
= 1² = 1
also f(0) = k
since f(x) is continuous at x = 0
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = f(0)
⇒ 1 = k

(iii) \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{4}}\) f(x) = \(\ {Lt}_{x \rightarrow \frac{\pi}{4}} \frac{1-\tan x}{4 x-\pi}\)
put x = \(\frac{\pi}{4}\) + h
so that as x → \(\frac{\pi}{4}\)
⇒ h → 0

Also, f(\(\frac{\pi}{4}\)) = k
Since f(x) be continuous at x = \(\frac{\pi}{4}\)
∴ \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{4}}\) f(x) = f(\(\frac{\pi}{4}\))
⇒ – \(\frac{1}{2}\) = k
Hence k = – \(\frac{1}{2}\).

Question 3.
Find the points of discontinuity (if any) of the following functions:
(i) f(x) = \(\begin{cases}\frac{x}{|x|}, & \text { if } \quad x<0 \\ -1, & \text { if } x \geq 0\end{cases}\) (NCERT)
(ii) f(x) = |x| – |x + 1| (NCERT)
Solution:
(i) Let c ∈ R be any arbitrary element
Since D_f = R

Case – I:
When c < 0,
f(c) = \(\frac{c}{|c|}\)
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) \(\frac{x}{|x|}\)
= \(\frac{c}{|c|}\)
= f(c)
∴ f is continuous for all c < 0. Case – II: when c > 0 ,
f(c) = – 1
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) – 1
= – 1
= f(c)
∴ f(c) is continuous for all c > 0.

Case – III:
at c = 0
\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) – 1 = – 1
and \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) \(\frac{x}{|x|}\)
= \(\ {Lt}_{x \rightarrow 0^{-}} \frac{x}{-x}\) = – 1
[∵ if x < 0 then |x| = – x]
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = f(0) = 1
Thus f is continuous at x = 0.
Thus, by combining all three cases, f is continuous at everywhere on R and has no points of discontinuity.

(ii) Given f(x) = |x| – |x – 1|
= \(\left\{\begin{array}{ccc}
-x+x-1 & ; & x<0 \\
x+x-1 & ; & 0 \leq x<1 \\
x-(x-1) & ; & x \geq 1
\end{array}\right.\)
= \(\left\{\begin{array}{ccc}
-1 & ; & x<0 \\
2 x-1 & ; & 0 \leq x<1 \\
1 & ; & x \geq 1
\end{array}\right.\)
Hence D_f = R.
So we examine f for continuity at all x ∈ R.
Let c ∈ R be any arbitrary real number.
Thus threee cases arises.

Case – I:
When c < 0 then f(c) = – 1
∴ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) – 1
= – 1 = f(c)
∴ f is continuous at all c < 0.

Case – II:
When 0 < c < 1
then f(c) = 2c – 1
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) 2x – 1
= 2c – 1 = f(c)
Thus, f is continuous for all c, where 0 < c < 1. Case – III: When c > 1 then f(c) = 1
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) 1 = 1 = f(c)
∴ f is continuous for all c > 1.

When x = 0:
\(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) – 1 = – 1
\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) 2x – 1
= 2 × 0 – 1 = – 1
and f(0) = 0 – 1 = – 1
∴ \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = f(0)
Thus f(x) is continuous at x = 0.

When x = 1:
\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) 2x – 1
= 2 – 1 = 1
\(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) 1 – 1 ;
f(1) = 1
∴ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = f(1)

Thus,f is continuous at x = 1
Hence on combining all cases, f is continuous everywhere in R and has no point of discontinuity.

Question 3 (old).
Discuss the continuity of the function f defined by
f(x) = \(\left\{\begin{array}{l}
3, \text { if } \quad 0 \leq x \leq 1 \\
4, \text { if } 1<x<3 \\
5, \text { if } 3 \leq x \leq 10
\end{array}\right.\). (NCERT)
Solution:
Since constant function is continuous everywhere
when 0 ≤ x < 1 ;
f (x) = 3, which is a constant function
and hence continuous in 0 ≤ x < 1
When 1 < x < 3 ;
f(x) = 4, which is a constant function
and hence continuous for all x ∈ (1, 3)
When 3 f (x) = 5, which is a constant function
and hence continuous is 3 < x ≤ 10.

at x = 1:
\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) 3 = 3
and \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) 4 = 4
∴ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) ≠ \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x)
Thus f is discontinuous at x = 1.

at x = 3:
\(\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}\) 4 = 4
and \(\underset{x \rightarrow 3^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 3^{+}}{\mathrm{Lt}}\) 5 = 5
∴ \(\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 3^{+}}{\mathrm{Lt}}\) f(x)
Thus f is discontinuous at x = 3.
Hence f is continuous in [0, 10] except at x = 1 and x = 3.

Question 4.
Examine the function f(x) = |x – 1| + |x – 3| for continuity and differentiability at x I and x = 3.
Solution:
Given f(x) = |x – 1| + |x – 3|
= \(\left\{\begin{array}{ccc}
-(x-1)-(x-3) & ; & x<1 \\
(x-1)-(x-3) & ; & 1 \leq x<3 \\
(x-1)+x-3 & ; & x \geq 3
\end{array}\right.\)
= \(\left\{\begin{array}{ccc}
-2 x+4 & ; & x<1 \\
2 & ; & 1 \leq x<3 \\
2 x-4 & ; & x \geq 3
\end{array}\right.\)

Continuity at x = 1:
\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) (- 2x + 4)
= – 2 + 4 = 2
\(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) 2 = 2
and f(1) = 2
∴ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = f(1)
∴ f is continuous at x = 1.

Differentiability at x = 1:

∴ Lf'(1) ≠ Rf'(1)
Thus f(x) is not differentiable at x = 1.

Continuity at x = 3:
\(\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}\) 2 = 2
and \(\underset{x \rightarrow 3^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 3^{+}}{\mathrm{Lt}}\) 2x – 4
= 6 – 4 = 2
Also f(3) = 2 × 3 – 4
= 6 – 4 = 2
Thus, \(\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 3^{+}}{\mathrm{Lt}}\) f(x) = f(3)
∴ f is continuous at x = 3.

Differentiability at x = 3:
Lf'(3) = \(\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}\) \(\frac{f(x)-f(3)}{x-3}\)
= \(\ {Lt}_{x \rightarrow 3^{-}} \frac{2-2}{x-3}\)
= 0
Rf'(3) = \(\ {Lt}_{x \rightarrow 3^{+}} \frac{f(x)-f(3)}{x-3}\)
= \(\ {Lt}_{x \rightarrow 3^{+}} \frac{2 x-4-2}{x-3}\)
= \(\frac{2(x-3)}{x-3}\)
∴ Lf'(3) ≠ Rf'(3)
Thus f(x) is not differentiable at x = 3.

Question 5.
Find the derivative of the following functions w.r.t. x:
(i) \(\frac{\sqrt{a}+\sqrt{x}}{\sqrt{a}-\sqrt{x}}\)
(ii) \(\frac{1-\tan x}{1+\tan x}\)
(iii) sin (\(\sqrt{\sin \sqrt{x}}\)).
Solution:
(i) Let y = \(\frac{\sqrt{a}+\sqrt{x}}{\sqrt{a}-\sqrt{x}}\)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{(\sqrt{a}-\sqrt{x})\left(\frac{1}{2 \sqrt{x}}\right)-(\sqrt{a}+\sqrt{x})\left(-\frac{1}{2 \sqrt{x}}\right)}{(\sqrt{a}-\sqrt{x})^2}\)
= \(\frac{\sqrt{a}-\sqrt{x}+\sqrt{a}+\sqrt{x}}{2 \sqrt{x}(\sqrt{a}-\sqrt{x})^2}\)
= \(\frac{\sqrt{a}}{\sqrt{x}(\sqrt{a}-\sqrt{x})^2}\)

(ii) Let y = \(\frac{1-\tan x}{1+\tan x}\)
= tan (\(\frac{\pi}{4}\) – x)
\(\frac{d y}{d x}=\sec ^2\left(\frac{\pi}{4}-x\right) \frac{d}{d x}\left(\frac{\pi}{4}-x\right)\)
= \(-\frac{1}{\cos ^2\left(\frac{\pi}{4}-x\right)}\)
= \(\frac{-2}{1+\cos \left(\frac{\pi}{2}-2 x\right)}\)
= \(\frac{-2}{1+\sin 2 x}\)

(iii) Let y = sin (\(\sqrt{\sin \sqrt{x}}\))
Diff. both sides w.r.t. x, we have

Question 6.
Differentiate the following functions w.r.t. x:
(i) \(\frac{\tan 2 x}{1-\cot 2 x}\)
(ii) sin \(\left(\frac{1+x^2}{1-x^2}\right)\)
(iii) \(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\)
Solution:
(i) Let y = \(\frac{\tan 2 x}{1-\cot 2 x}\) ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{(1-\cot 2 x) \sec ^2 2 x \cdot 2-2 \tan 2 x \ {cosec}^2 2 x}{(1-\cot 2 x)^2}\)
= \(\frac{2\left[\sec ^2 2 x-\cot 2 x \sec ^2 2 x-\tan 2 x \ {cosec}^2 2 x\right]}{(1-\cot 2 x)^2}\)
= \(\frac{2\left[\sec ^2 2 x-2 \ {cosec} 4 x-2 \ {cosec} 4 x\right]}{(1-\cot 2 x)^2}\)
= \(\frac{2\left[\sec ^2 2 x-4 \ {cosec} 4 x\right]}{(1-\cot 2 x)^2}\)

(ii) Let y = sin \(\left(\frac{1+x^2}{1-x^2}\right)\)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\cos \left(\frac{1+x^2}{1-x^2}\right)\left[\frac{\left(1-x^2\right) 2 x-\left(1+x^2\right)(-2 x)}{\left(1-x^2\right)^2}\right]\)
= \(\cos \left(\frac{1+x^2}{1-x^2}\right)\left[\frac{2 x\left(1-x^2+1+x^2\right)}{\left(1-x^2\right)^2}\right]\)
= \(\frac{4 x}{\left(1-x^2\right)^2} \cos \left(\frac{1+x^2}{1-x^2}\right)\)

(iii) Let y = \(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\)

Question 7.
(i) y = log \(\left(\frac{a+b \tan \frac{x}{2}}{a-b \tan \frac{x}{2}}\right)\), prove that \(\frac{d y}{d x}=\frac{a b}{a^2 \cos ^2 \frac{x}{2}-b^2 \sin ^2 \frac{x}{2}}\).
(ii) If 2y = x \(\sqrt{x^2-a^2}\) – a² log (x + \(\sqrt{x^2-a^2}\)), prove that \(\frac{d y}{d x}=\sqrt{x^2-a^2}\).
(iii) If x = 2a sin^-1 \(\sqrt{\frac{y}{2 a}}-\sqrt{2 a y-y^2}\), prove that \(\frac{d y}{d x}=\sqrt{\frac{2 a-y}{y}}\).
Solution:
(i) Given y = log \(\left(\frac{a+b \tan \frac{x}{2}}{a-b \tan \frac{x}{2}}\right)\)
= log (a + b tan \(\frac{x}{2}\)) – log (a – b tan \(\frac{x}{2}\))
Diff. both sides w.r.t. x ; we have

(ii) Given 2y = x \(\sqrt{x^2-a^2}\) – a² log (x + \(\sqrt{x^2-a^2}\))
Diff. both sides w.r.t. x ; we have

(iii) Given x = 2a sin^-1 \(\sqrt{\frac{y}{2 a}}-\sqrt{2 a y-y^2}\)
Diff. both sides w.r.t. y ; we have

Question 8.
Differentiate the following functions w.r.t. x:
(i) sin^-1 \(\left(\frac{a+b \cos x}{b+a \cos x}\right)\)
(ii) cot^-1 \(\left(\frac{\sqrt{1+x^2}-1}{x}\right)\)
(iii) tan^-1 \(\sqrt{\frac{a-x}{a+x}}\)
Solution:
(i) Let y = sin^-1 \(\left(\frac{a+b \cos x}{b+a \cos x}\right)\)
Diff. both sides w.r.t. x ; we have

(ii) Let y = cot^-1 \(\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) ……….(1)
put x = tan θ
⇒ θ = tan^-1 x
∴ from (1) ;
y = cot^-1 \(\left(\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right)\)
= cot^-1 \(\left(\frac{\sec \theta-1}{\tan \theta}\right)\)
⇒ y = cot^-1 \(\left(\frac{1-\cos \theta}{\sin \theta}\right)\)
= cot^-1 \(\left(\frac{2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)\)
⇒ y = cot^-1 \(\left(\tan \frac{\theta}{2}\right)\)
= cot^-1 (cot \(\left(\frac{\pi}{2}-\frac{\theta}{2}\right)\))
⇒ y = \(\frac{\pi}{2}-\frac{\theta}{2}\)
= \(\frac{\pi}{2}-\frac{1}{2} \tan ^{-1} x\)

(iii) Let y = tan^-1 \(\sqrt{\frac{a-x}{a+x}}\) ………..(1)
put x = a cos θ
θ = cos^-1 \(\frac{x}{a}\)
∴ from eqn. (1) ; we have
y = tan^-1 \(\sqrt{\frac{a-a \cos \theta}{a+a \cos \theta}}\)
= tan^-1 \(\sqrt{\frac{2 \sin ^2 \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}}}\)
y = tan^-1 (tan \(\frac{\theta}{2}\))
= \(\frac{\theta}{2}\)
= \(\frac{1}{2}\) cos^-1 \(\frac{x}{a}\)
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}=-\frac{1}{2} \frac{1}{\sqrt{1-\frac{x^2}{a^2}}} \frac{1}{a}\)
= – \(\frac{1}{2 \sqrt{a^2-x^2}}\).

Question 9.
If y = 2 tan^-1 \(\sqrt{\frac{x-a}{b-x}}\), a < x < b, prove that \(\left(\frac{d y}{d x}\right)^2+\frac{1}{(x-a)(x-b)}\) = 0.
Solution:
Given y = 2 tan^-1 \(\sqrt{\frac{x-a}{b-x}}\)
put x = a cos² θ + b sin² θ ………..(1)
∴ x – a = a cos² θ + b sin² θ – a
= – a sin² θ + b sin² θ
⇒ x – a = (b – a) sin² θ ………..(2)
and b – x = b – a cos² θ – b sin² θ
= b cos² θ – a cos² θ
= (b – a) cos² θ ………..(3)
∴ y = 2 tan^-1 \(\sqrt{\frac{(b-a) \sin ^2 \theta}{(b-a) \cos ^2 \theta}}\)
= 2 tan^-1 (tan θ)
= 2θ
⇒ \(\frac{d y}{d x}=2 \frac{d \theta}{d x}\)
= \(\frac{2}{(b-a) \sin 2 \theta}\)
= \(\frac{1}{(b-a) \sin \theta \cos \theta}\) ………..(4)
[from (1) ;
\(\frac{d x}{d \theta}\) = 2a cos θ (- sin θ) + 2b sin θ cos θ = (b – a) sin 2θ]
On squaring both sides of eqn. (4) ; we have
\(\left(\frac{d y}{d x}\right)^2=\frac{1}{(b-a)^2 \sin ^2 \theta \cos ^2 \theta}\)
= \(\frac{1}{(b-a)^2 \frac{(x-a)}{b-a}\left(\frac{b-x}{b-a}\right)}\) [using (2) and (3)]
⇒ \(\left(\frac{d y}{d x}\right)^2=\frac{1}{(x-a)(b-x)}\)
⇒ \(\left(\frac{d y}{d x}\right)^2+\frac{1}{(x-a)(x-b)}\) = 0.

Question 10.
Differentiate the following functions w.r.t. x:
(i) tan^-1 \(\left(\frac{\sqrt{x}(3-x)}{1-3 x}\right)\)
(ii) tan^-1 \(\left(\frac{3 a^2 x-x^3}{a\left(a^2-3 x^2\right)}\right)\)
Solution:
(i) Let y = tan^-1 \(\left(\frac{\sqrt{x}(3-x)}{1-3 x}\right)\)
put √x = tan θ
⇒ θ = tan^-1 √x
∴ y = tan^-1 \(\left(\frac{\tan \theta\left(3-\tan ^2 \theta\right)}{1-3 \tan ^2 \theta}\right)\)
= tan^-1 (tan 3θ)
= 3θ
= 3 tan^-1 √x
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}=\frac{3}{1+(\sqrt{x})^2} \frac{d}{d x} \sqrt{x}\)
= \(\left(\frac{3}{1+x}\right)\left(\frac{1}{2 \sqrt{x}}\right)\)
= \(\frac{3}{2 \sqrt{x}(1+x)}\)

(ii) Let y = tan^-1 \(\left(\frac{3 a^2 x-x^3}{a\left(a^2-3 x^2\right)}\right)\)
put x = a tan θ
θ = tan^-1 \(\frac{x}{a}\)
∴ y = tan^-1 \(\left(\frac{3 a^3 \tan \theta-a^3 \tan ^3 \theta}{a^3-3 a^3 \tan ^2 \theta}\right)\)
= tan^-1 \(\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^3 \theta}\right)\)
= tan^-1 (tan 3θ)
= 3θ
= 3 tan^-1 \(\frac{x}{a}\)
Diff. both sides w.r.t. x ;
\(\frac{d y}{d x}=\frac{3}{1+\frac{x^2}{a^2}}\left(\frac{1}{a}\right)\)
= \(\frac{3 a}{a^2+x^2}\)

Question 11.
If tan y = \(\frac{2 t}{1-t^2}\) and sin x = \(\frac{2 t}{1+t^2}\), find \(\frac{d y}{d x}\). (NCERT Exampler)
Solution:
Given sin x = \(\frac{2 t}{1+t^2}\)
⇒ x = sin^-1 \(\frac{2 t}{1+t^2}\) ………….(1)
and tan y = \(\frac{2 t}{1-t^2}\)
⇒ y = tan^-1 \(\frac{2 t}{1-t^2}\) ………….(2)
putting t = tan θ
and t = tan Φ in eqn. (1) and (2) ; we get
x = sin^-1 \(\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\)
and y = tan^-1 \(\left(\frac{2 \tan \phi}{1-\tan ^2 \phi}\right)\)
⇒ x = sin^-1 (sin 2θ)
and y = tan^-1 (tan 2Φ)
⇒ x = 2θ
and y = 2Φ
⇒ x = 2 tan^-1 t
⇒ \(\frac{d x}{d t}\) = \(\frac{2}{1+t^2}\)
and y = 2 tan^-1 t
⇒ \(\frac{d y}{d t}=\frac{2}{1+t^2}\)
Thus, \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\) = 1.

Question 12.
If y = sin^-1 \(\left(\frac{1}{\sqrt{1+x^2}}\right)\) + tan^-1 \(\left(\frac{\sqrt{1+x^2}-1}{x}\right)\), find \(\frac{d y}{d x}\).
Solution:
Given y = sin^-1 \(\left(\frac{1}{\sqrt{1+x^2}}\right)\) + tan^-1 \(\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) ………..(1)
put x = tan θ
⇒ θ = tan^-1 x
∴ from (1) ;

Question 12 (old).
If y = a^{t + \(\frac{1}{t}\)} and x = (t + \(\frac{1}{t}\))^a, a > 0, find \(\frac{d y}{d x}\). (NCERT)
Solution:
Given y = a^{t + \(\frac{1}{t}\)} …………(1)
and x = (t + \(\frac{1}{t}\))^a …………(2)
Differentiate eqn. (1) and (2) w.r.t. ‘t’ ; we have
\(\frac{d y}{d t}\) = a^{t + \(\frac{1}{t}\)} log a . [1 – \(\frac{1}{t^2}\)]
and \(\frac{d x}{d t}\) = a \(\left(t+\frac{1}{t}\right)^{a-1}\left(1-\frac{1}{t^2}\right)\)
∴ \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}\)
= \(\frac{a^{\left(t+\frac{1}{t}\right)} \log a\left(1-\frac{1}{t^2}\right)}{a\left(t+\frac{1}{t}\right)^{a-1}\left(1-\frac{1}{t^2}\right)}\)
= \(\frac{a^{\left(t+\frac{1}{t}\right)} \log a}{a\left(t+\frac{1}{t}\right)^{a-1}}\).

Question 13.
If y \(\sqrt{1-x^2}\) + x \(\sqrt{1-y^2}\) = c, prove that \(\frac{d y}{d x}=-\sqrt{\frac{1-y^2}{1-x^2}}\).
Solution:
Given y \(\sqrt{1-x^2}\) + x \(\sqrt{1-y^2}\) = c ……….(1)
put x = sin θ i.e. θ = sin^-1 x
and y = sin Φ i.e. Φ = sin^-1 y
∴ from (1) ;
we have
sin Φ cos θ + sin θ cos Φ = c
⇒ sin (θ + Φ) = c
⇒ θ + Φ = sin^-1 c
⇒ sin^-1 x + sin^-1 y = sin^-1 c
Diff. both sides w.r.t. x ; we have
\(\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-y^2}} \frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=-\sqrt{\frac{1-y^2}{1-x^2}}\).

Question 14.
Differentiate the following functions w.r.t. x :
(i) tan (x^x)
(ii) (5x)^{3 cos 2x} (NCERT)
(iii) (sin x)^x + sin^-1 √x
(iv) (x + \(\frac{1}{x}\))^x + x^{(1 + \(\frac{1}{x}\))}.
Solution:
(i) Let y = tan (x^x)
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}\) = sec² (x^x) \(\frac{d}{d x}\) (x^x)
= sec² (x^x) \(\frac{d}{d x}\) e^{x log x}
= sec² (x^x) e^{x log x} [x × \(\frac{1}{x}\) + log x – 1]
= x^x sec² (x^x) (1 + log x)

(ii) Let y = (5x)^{3 cos 2x}
Taking logarithm on both sides ; we have
log y = 3 cos 2x log 5x
Diff. both sides w.r.t. x ; we have
\(\frac{1}{y} \frac{d y}{d x}\) = 3 [cos 2x \(\frac{1}{5 x}\) . 5 + log 5x . (- 2 sin 2x)]
⇒ \(\frac{d y}{d x}\) = (5x)^{3 cos 2x} [\(\frac{3 \cos 2 x}{x}\) – 6 sin 2x log 5x]

(iii) Let y = (x + \(\frac{1}{x}\))^x + x^{(1 + \(\frac{1}{x}\))}
⇒ y = u + v …………(1)
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) …………(2)
where u = (sin x)^x
⇒ log u = x log sin x
diff. both sides w.r.t. x ; we have
\(\frac{1}{u} \frac{d u}{d x}\) = log sin x + \(\frac{x}{sin x}\) cos x
\(\frac{d u}{d x}\) = (sin x)^x [log sin x + x cot x] ……….(3)
and v = sin^-1 √x
⇒ \(\frac{d v}{d x}=\frac{1}{\sqrt{1-(\sqrt{x})^2}} \times \frac{1}{2 \sqrt{x}}\)
\(\frac{d v}{d x}=\frac{1}{2 \sqrt{x} \sqrt{1-x}}=\frac{1}{2 \sqrt{x-x^2}}\) ……….(4)
putting eqn. (3) and eqn. (4) in eqn. (2) ;
∴ \(\frac{d y}{d x}\) = (sin x)^x [log sin x + x cot x] + \(\frac{1}{2 \sqrt{x-x^2}}\)

(iv) Let y = (x + \(\frac{1}{x}\))^x + x^{(1 + \(\frac{1}{x}\))}
⇒ y = u + v ……….(1)
where u = (x + \(\frac{1}{x}\))^x
and v = x^{(1 + \(\frac{1}{x}\))}
Diff. eqn. (1) both sides w.r.t. x ; we have
\(\frac{d y}{d u}=\frac{d u}{d x}+\frac{d v}{d x}\) …………(2)
Since u = (x + \(\frac{1}{x}\))^x ;
Taking logarithm on both sides ; we have
log u = x log (x + \(\frac{1}{x}\)) ;
diff. both sides w.r.t. x,

Question 14 (old).
(iii) (sin x – cos x)^{sin x – cos x}, \(\frac{\pi}{4}\) < x < \(\frac{3 \pi}{4}\) (NCERT).
Solution:
Let y = (sin x – cos x)^{sin x – cos x}, \(\frac{\pi}{4}\) < x < \(\frac{3 \pi}{4}\)
Taking logarithm on both sides ; we have
log y = (sin x – cos x) log (sin x – cos x)
On differentiating both sides ; we have
\(\frac{1}{y} \frac{d y}{d x}\) = \(\frac{(\sin x-\cos x)}{\sin x-\cos x}\) (cos x + sin x) + (cos x + sin x) log (sin x – cos x)
\(\frac{1}{y} \frac{d y}{d x}\) = (cos x + sin x) [1 + log (sin x – cos x)]
⇒ \(\frac{d y}{d x}\) = (sin x – cos x)^{sin x – cos x} (cos x + sin x) [1 + log (sin x – cos x)]

Question 15.
(i) If x^y – y^x = a^b, find \(\frac{d y}{d x}\).
(ii) If y = \((\cos x)^{(\cos x)^{(\cos x) \ldots \infty}}\), prove that \(\frac{d y}{d x}=-\frac{y^2 \tan x}{1-y \log (\cos x)}\). (NCERT Exampler)
Solution:
(i) Given x^y – y^x = a^b
⇒ u – v = a^b …………(1)
where u = x^y ………..(2)
and v = y^x …………..(3)
Diff. eqn. (1) both sides w.r.t. x ;
\(\frac{d u}{d x}-\frac{d v}{d x}\) = 0 ………..(4)
Taking logarithm on both sides of eqn. (2) ;
log u = log x^y = y log x
Diff. both sides w.r.t. x ;
∴ \(\frac{1}{u} \frac{d u}{d x}=\frac{y}{x}+\log x \cdot \frac{d y}{d x}\)
⇒ \(\frac{d u}{d x}=x^y\left[\frac{y}{x}+\log x \cdot \frac{d y}{d x}\right]\) ………..(5)
Taking logarithm on both sides of eqn. (3) ;
log v = log y^x = x log y
Diff. both sides w.r.t. x ; we have
\(\frac{1}{v} \frac{d v}{d x}=\frac{x}{y} \frac{d y}{d x}\) + log y
⇒ \(\frac{d v}{d x}\) = y^x [\(\frac{x}{y} \frac{d y}{d x}\) + log y] ……….(6)
putting eqn. (5) and eqn. (6) in eqn. (4) ; we have
\(x^y\left[\frac{y}{x}+\log x \frac{d y}{d x}\right]-y^x\left[\frac{x}{y} \frac{d y}{d x}+\log y\right]\) = 0
⇒ (x^y log x – xy^{x – 1}) \(\frac{d y}{d x}\) = y^x log y – yx^{y – 1}
⇒ \(\frac{d y}{d x}=\frac{y^x \log y-y x^{y-1}}{x^y \log x-x y^{x-1}}\).

(ii) Given y = (cos x)^y ;
Taking logarithm on both sides ; we have
log y = y log cos x ;
diff. both sides w.r.t. x, we get

Question 16.
If \(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}\) = 6, prove that \(\frac{d y}{d x}=\frac{y}{x}\).
Solution:
Given \(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}\) = 6
⇒ x + y = 6 \(\sqrt{xy}\)
Taking logarithm on both sides ; we have
log (x + y) = log 6 + \(\frac{1}{2}\) (log x + log y)
Diff. both sides w.r.t. x ; we have
\(\frac{1}{x+y}\left[1+\frac{d y}{d x}\right]=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y} \frac{d y}{d x}\right)\)
⇒ \(\frac{d y}{d x}\left[\frac{1}{x+y}-\frac{1}{2 y}\right]=\frac{1}{2 x}-\frac{1}{x+y}\)
⇒ \(\frac{d y}{d x}\left(\frac{y-x}{2 y(x+y)}\right)=\frac{y-x}{2 x(x+y)}\)
⇒ \(\frac{d y}{d x}=\frac{y}{x}\).

Question 17.
Find the second derivative of the following functions:
(i) tan² (3x – 2)
(ii) \(\frac{x^2+2 x-1}{x^2-3 x+2}\)
Solution:
(i) Let y = tan² (3x – 2) ;
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}\) = 2 tan (3x – 2) sec² (3x – 2) . 3
Again diff. both sides w.r.t. x : we have
\(\frac{d^2 y}{d x^2}\) = 6 [3 sec² (3x – 2) sec² (3x – 2) + 2 tan² (3x – 2) sec² (3x – 2) × 3]
= 18 sec⁴ (3x – 2) + 36 tan² (3x – 2) sec² (3x – 2)

(ii) Let y = \(\frac{x^2+2 x-1}{x^2-3 x+2}\)
= \(\frac{x^2+2 x-1}{(x-1)(x-2)}\)
= 1 – \(\frac{2}{x-1}+\frac{7}{x-2}\) ………..(1)
[by using partial fractions]
Let \(\frac{x^2+2 x-1}{x^2-3 x+2}=1+\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x-2}\) ………….(2)
Multiply both sides of eqn. (2) ; we have
x² + 2x – 1 = x² – 3x + 2 + A (3x – 2) + B (x – 1)
putting x = 1 in eqn. (3) ; we have
2 = – A
⇒ A = – 2
putting x = 2 in eqn. (3) ; we have
7 = B
Diff. both sides of eqn. (1) w.r.t. x, we have
\(\frac{d y}{d x}=\frac{2}{(x-1)^2}-\frac{7}{(x-2)^2}\)
Again differentiating w.r.t. x, we have
\(\frac{d^2 y}{d x^2}=\frac{-4}{(x-1)^3}+\frac{14}{(x-2)^3}\).

Question 18.
If y = 3x sin 3x + cos 3x, prove that x \(\frac{d^2 y}{d x^2}\) + 9xy = 2 \(\frac{d y}{d x}\).
Solution:
Given y = 3x sin 3x + cos 3x ………..(1)
Diff. eqn. (1) w.r.t. x both sides ; we have
\(\frac{d y}{d x}\) = 3 [3x cos 3x + sin 3x] – 3 sin 3x
= 9 x cos 3x ……….(2)
again differentiating eqn. (2) w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = 9 [- 3x sin 3x + cos 3x]
⇒ x \(\frac{d^2 y}{d x^2}\) = – 27 x² sin 3x + 9x cos 3x
⇒ x \(\frac{d^2 y}{d x^2}\) = – 9x [3x sin 3x + cos 3x] + 18x cos 3x
⇒ x \(\frac{d^2 y}{d x^2}\) = – 9 xy + 2 \(\frac{d y}{d x}\)
[using eqn. (1) and eqn. (2)]
⇒ x \(\frac{d^2 y}{d x^2}\) + 9xy = 2 \(\frac{d y}{d x}\)

Question 19.
If y = cos (3 cos^-1 x), then prove that \(\frac{d^2 y}{d x^2}\) = 24x.
Solution:
Given y = cos (3 cos^-1 x)
put cos^-1 x = θ
⇒ x = cos θ
⇒ y = cos 3θ
= 4 cos³ θ – 3 cos θ
= 4x³ – 3x
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}\) = 12x² – 3 ;
again differentiating w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = 24x.

Question 20.
If sin (x + y) = ky, prove that y₂ + y (1 + y₁)³ = 0.
Solution:
Given sin (x + y) = ky …………(1)
T.P. y₂ + y (1 + y₁)³ = 0
Diff. eqn. (1) both sides w.r.t. x, we have
cos (x + y) (1 + y₁) = ky₁ …………(2)
Differentiating eqn. (2) w.r.t. x ; we have
cos (x + y) y₂ + (1 + y₁) {- sin (x + y) (1 + y₁)} = ky₂
⇒ \(\) y₂ – (1 + y₁)² ky = ky₂
[using eqn. (1) and (2)]
⇒ y₁y₂ – (1 + y₁)² y = y₂ (1 + y₁)
⇒ y₂ (y₁ – 1 – y₁) – (1 + y₁)³ y = 0
⇒ y₂ + y (1 + y₁)³ = 0.

Question 21.
If xy = sin x, prove that \(\frac{d^2 y}{d x^2}\) + \(\frac{2}{x} \frac{d y}{d x}\) + y = 0.
Solution:
Given xy = sin x ……….(1)
Diff. eqn. (1) both sides w.r.t. x, we have
x \(\frac{d y}{d x}\) + y . 1 = cos x ………(2)
Differentiating eqn. (2) w.r.t. x ; we have
x \(\frac{d^2 y}{d x^2}\) + \(\frac{d y}{d x}\) . 1 + \(\frac{d y}{d x}\) = – sin x
⇒ x \(\frac{d^2 y}{d x^2}\) + 2 \(\frac{d y}{d x}\) = – xy
On dividing both sides throughout by x, we have
\(\frac{d^2 y}{d x^2}\) + \(\frac{2}{x} \frac{d y}{d x}\) + y = 0.

Question 22.
If y = \(\frac{2}{\sqrt{a^2-b^2}}\) tan^-1 \(\left(\sqrt{\frac{a \cdot b}{a+b}} \tan \frac{x}{2}\right)\), prove that \(\frac{d^2 y}{d x^2}=\frac{b \sin x}{(a+b \cos x)^2}\).
Solution:
Given y = \(\frac{2}{\sqrt{a^2-b^2}}\) tan^-1 \(\left(\sqrt{\frac{a \cdot b}{a+b}} \tan \frac{x}{2}\right)\) ;
Diff. both sides w.r.t. x ; we have

Question 23.
If x = (a + bt) e^{– nt}, prove that \(\frac{d^2 x}{d t^2}\) + 2n \(\frac{d x}{d t}\) + n²x = 0.
Solution:
Given x = (a + bt) e^{– nt} ……….(1)
Diff. eqn. (1) both sides w.r.t. t ; we have
\(\frac{d x}{d t}\) = (a + bt) e^{– nt} (- n) + e^{– nt} b
⇒ \(\frac{d x}{d t}\) = – nx + be^{– nt} ………..(2)
again differentiating eqn. (2) w.r.t. t, we have
\(\frac{d^2 x}{d t^2}\) = – n \(\frac{d x}{d t}\) + be^{– nt} (- n)
⇒ \(\frac{d^2 x}{d t^2}\) = – n \(\frac{d x}{d t}\) – n [\(\frac{d x}{d t}\) + nx] [using eqn. (2)]
⇒ \(\frac{d^2 x}{d t^2}\) + 2n \(\frac{d x}{d t}\) + n²x = 0

Question 24.
If y = \(\frac{3 a t}{1+t}\) and x = \(\frac{2 a t^2}{1+t}\), find \(\frac{d^2 y}{d x^2}\).
Solution:
Given y = \(\frac{3 a t}{1+t}\) ………….(1)
and x = \(\frac{2 a t^2}{1+t}\) …………(2)
Differentiating eqn. (1) and eqn. (2) w.r.t. t, we have

Question 25.
If x = sin t and y = sin 2t, prove that
(i) (1 – x²) (\(\frac{d y}{d x}\))² = 4 (1 – y²)
(ii) (1 – x²) \(\frac{d^2 y}{d x^2}\) – x \(\frac{d y}{d x}\) + 4y = 0.
Solution:
Given x = sin t ……….(1)
and y = sin 2t ………..(2)
diff. eqns. (1) and (2) ; w.r.t. t ; we have
\(\frac{d x}{d t}\) = cos t ;
\(\frac{d y}{d t}\) = 2 cos 2t
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{2 \cos 2 t}{\cos t}\) ………..(3)
⇒ \(\left(\frac{d y}{d x}\right)^2=4 \frac{\cos ^2 2 t}{\cos ^2 t}\)
= \(\frac{4\left(1-\sin ^2 2 t\right)}{1-\sin ^2 t}\)
= \(\frac{4\left(1-y^2\right)}{1-x^2}\)
[using (1) and (2)]
⇒ (1 – x²) (\(\frac{d y}{d x}\))² = 4 (1 – y²) ………..(4)

(ii) Diff. eqn. (3) both sides w.r.t. x ;
\(\frac{d^2 y}{d x^2}=\frac{2[\cos t(-2 \sin 2 t)-\cos 2 t(-\sin t)]}{\cos ^2 t}\) × \(\frac{d t}{d x}\)
⇒ (1 – sin² t) \(\frac{d^2 y}{d x^2}\) = [- 4 cos t sin 2t + 2 cos 2t sin t] \(\frac{1}{\cos t}\)
= – 4 sin 2t + 2 \(\frac{\cos 2 t}{\cos t}\) sin t
⇒ (1 – x²) \(\frac{d^2 y}{d x^2}\) = – 4y + x \(\frac{d y}{d x}\) [using (1) and (2)]
⇒ (1 – x²) \(\frac{d^2 y}{d x^2}\) – x \(\frac{d y}{d x}\) + 4y = 0.

Question 26.
Verify Rolle’s theorem for the following functions and find point (or points) in the interval where derivative is zero :
f(x) = 2 cos 2 (x – \(\frac{\pi}{4}\)) in [0, π]
Solution:
Given f(x) = 2 cos 2 (x – \(\frac{\pi}{4}\))
= 2 cos (2x – \(\frac{\pi}{2}\))
= 2 cos {- (\(\frac{\pi}{2}\) – 2x)}
= 2 cos (\(\frac{\pi}{2}\) – 2x)
[∵ cos (- θ) = cos θ]
⇒ f(x) = 2 sin 2x ………..(1)
Since, sin 2x is continuous and differentiable everywhere.
∴ f(x) is continuous in [0, π] and f is derivable in (0, π).
also, f(0) = 2 × 0 = 0;
f(π) = 2 sin 2π
= 2 × 0 = 0
∴ f(0) = f(π)
Thus, all the three conditions of Rolle’s theorem are satisfied for function f in [0, π].
Then ∃ atleast one real number c ∈ (0, π)
s.t. f’ (c) = 0
Diff. eqn. (1) w.r.t. x; we have
f’(x) = 4 cos 2x
Now f’ (c) = 0
⇒ 4 cos 2c = 0
⇒ cos 2c = 0
⇒ 2c = (2n + 1) \(\frac{\pi}{2}\), ∀ n ∈ I
⇒ c = (2n + 1) \(\frac{\pi}{4}\), ∀ n ∈ I
∴ c = \(\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \ldots,-\frac{\pi}{4}, \ldots\)
But c ∈ (0, π)
∴ c = \(\frac{\pi}{4}\), \(\frac{3 \pi}{4}\)
Thus, ∃ teo real numbers \(\frac{\pi}{4}\) and \(\frac{3 \pi}{4}\)
s.t. f'(\(\frac{\pi}{4}\)) = 0 = f'(\(\frac{3 \pi}{4}\)).

Question 27.
Verify Lagrange’s mean value theorem for the following function. Also find c of this theorem : f(x) = log x, 1 ≤ x ≤ 2e.
Solution:
Given f(x) = log x, 1 ≤ x ≤ 2e
Since log x is continuous in its domain.
∴ f(x) is continuous in [1, 2e].
Further f’ (x) = \(\frac{1}{x}\), which is defined ∀ x ∈ (1, 2e)
∴ f is derivable in (1, 2e).
Thus, both conditions of lagrange’s mean value theorem are satisfied for f in [1, 2e].
Then ∃ atleast one real number c ∈ (1, 2e) such that
f'(c) = \(\frac{f(2 e)-f(1)}{2 e-1}\) ………….(1)
Diff. eqn. (1) w.r.t. x ; we have
f'(x) = \(\frac{1}{x}\) ;
f(1) = log 1 = 0 ;
f(e) = log 2e
∴ from (2) ; we have
\(\frac{1}{c}=\frac{\log 2 e-0}{2 e-1}\)
⇒ c = \(\frac{2 e-1}{\log 2 e}\)
⇒ c = \(\frac{2 e-1}{\log 2+\log e}\)
= \(\frac{2 e-1}{1+\log 2}\) ∈ (1, 2e)
[∵ e = 2.7183 and log 2 = 0.3010]
[Since 2e – 1 < 2e
⇒ \(\frac{2 e-1}{1+\log 2}\) < \(\frac{2 e}{1+\log 2}\) < 2e and 2e – 1 > 2
⇒ \(\frac{2 e-1}{1+\log 2}\) > \(\frac{2}{1+\log 2}\) > 1]
Thus, ∃ a real number c = \(\frac{2 e-1}{1+\log 2}\) ∈ (1, 2e)
s.t. f'(c) = \(\frac{f(2 e)-f(1)}{2 e-1}\)
Thus, lagrange’s theorem is verified and c = \(\frac{2 e-1}{1+\log 2}\).

Continuous practice using ML Aggarwal Class 12 ISC Solutions Chapter 5 Continuity and Differentiability Ex 5.14 can lead to a stronger grasp of mathematical concepts.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.14

Question 1.
Examine the continuity the function f(x) = \(\left\{\begin{array}{cc}
|x| & , \quad x \leq 0 \\
x & , 0<x<1 \\ 2-x & , 1 \leq x \leq 2 \\ 3 x-5, & x>2
\end{array}\right.\) at each of the points 0, 1, 2.
Solution:
Continuity at x = 0:
L.H.L. = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) |x|
= \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) – x = 0

Continuity at x = 1:
\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) x = 1
\(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) 2 – x
= 2 – 1 = 1
also f(1) = 2 – 1 = 1
∴ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = f(1)
∴ f is continuous at x = 1.

Continuity at x = 2:
L.H.L. = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) 2 – x
= 2 – 2
= 0
R.H.L. = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) 3x – 5
= 6 – 5 = 1
∴ \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) ≠ \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x)
Thus, f(x) is discontinuous at x = 2.

Question 2.
Determine the constant k so that the function f(x) = may be continuous.
Solution:

Also f(2) = k
Now, the function f(x) is continuous at x = 2.
if \(\underset{x \rightarrow 2}{\mathrm{Lt}}\) f(2) = if \(\frac{1}{8}\) = k.

Question 3.
Examine for continuity and differentiability, each of the following functions:
(i) f(x) = \(\left\{\begin{array}{cc}
x \sin \frac{1}{x}, & x<0 \\ 0 & , x \geq 0 \end{array}\right.\) at x = 0 (ii) f(x) = \(\left\{\begin{array}{cc} |x| \sin \frac{1}{x}, & x>0 \\
0, & x \leq 0
\end{array}\right.\) at x = 0
(iii) f(x) = \(\left\{\begin{array}{cc}
(x-c)^2 \cos \frac{1}{x-c}, & x \neq c \\
0 & , x=c
\end{array}\right.\) at x = c.
Solution:
(i) Continuity at x = 0:
\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(0) = 0
\(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) x sin \(\frac{1}{x}\)
Let g(x) = x
and h(x) = sin \(\frac{1}{x}\)
\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) g(x) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) x = 0
and sin \(\frac{1}{x}\) is bounded in the deleted ngd of 0.
[∵ – 1 ≤ sin t ≤ 1 ∀ t ∈ R]
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) g(x) h(x) = 0
⇒ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) x sin \(\frac{1}{x}\) = 0
⇒ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = 0
∴ \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = 0,
also f(0) = 0
Thus \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x0 = f(0)
Hence, f is continuous at x = 0.

Differentiability at x = 0:
Lf'(0) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) \(\frac{f(x)-f(0)}{x-0}\)

∴ f(x) is not differentiable at x = 0.

(ii) Continuity at x = 0:
\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) |x| sin \(\frac{1}{x}\)
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x sin \(\frac{1}{x}\)
Let g(x) = x and
h(x) = sin \(\frac{1}{x}\)
\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) g(x) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) x = 0
and sin \(\frac{1}{x}\) is boundede in the deleted ngd of 0.
[∵|sin t| ≤ 1 ∀ t ∈ R]
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) g(x) h(x) = 0
⇒ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) x sin \(\frac{1}{x}\) = 0
⇒ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = 0
⇒ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = 0
\(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) 0 = 0
and f(0) = 0
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = f(0)
Thus f(x) is continuous at x = 0

Differentiability at x = 0:
Rf'(0) = \(\ {Lt}_{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}\)
= \({Lt}_{x \rightarrow 0^{+}} \frac{x \sin \frac{1}{x}-0}{x}\)
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) sin \(\frac{1}{x}\), which does not exists.
[since sin \(\frac{1}{x}\) be a real number oscillating between – 1 and 1]
If x = \(\frac{1}{2 n \pi+\frac{\pi}{2}}\), n ∈ N, n → ∞ ⇒ x → 0
∴ sin \(\frac{1}{x}\) = sin (2nπ + \(\frac{\pi}{2}\))
= sin \(\frac{\pi}{2}\) = 1
If x = \(\frac{1}{2 n \pi-\frac{\pi}{2}}\), n ∈ N, n → ∞ ⇒ x → 0
∴ sin \(\frac{1}{x}\) = sin (2nπ – \(\frac{\pi}{2}\))
= sin (- \(\frac{\pi}{2}\))
= – 1
∴ Rf'(0) does not exists.
Thus f is not differentiable at x = 0.

(iii) Continuity at x = c:
Let g(x) = (x – c)² ;
h(x) = cos \(\frac{1}{x-c}\)
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) g(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) (x – c)² = 0
and h(x) = cos \(\frac{1}{x-c}\) is bounded in the ddeleted ngd of c.
[∵ – 1 ≤ cos \(\frac{1}{x}\) ≤ 1 ∀ x ∈ R]
∴ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) g(x) h(x) = 0
⇒ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) (x – c)² cos \(\frac{1}{x-c}\) = 0
⇒ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = 0 ;
also f(c) = 0
∴ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = f(c)
Thus f is continuous at x = c.

Differentiability at x = c:
Lf'(c) = \(\ {Lt}_{x \rightarrow c^{-}} \frac{f(x)-f(c)}{x-c}\)
= \(\mathrm{Lt}_{h \rightarrow 0^{+}} \frac{f(c-h)-f(c)}{c-h-c}\)
= \(\ {Lt}_{h \rightarrow 0^{+}} \frac{(c-h-c)^2 \cos \left(\frac{-1}{h}\right)-0}{-h}\)
[put x = c – h as x → c^– ⇒ h → 0⁺]
= \(\ {Lt}_{h \rightarrow 0^{+}} \frac{h^2 \cos \frac{1}{h}}{h}\)
= \(\underset{h \rightarrow 0^{+}}{\mathrm{Lt}}\) h cos \(\frac{1}{h}\) = 0
[Here g(h) = h
and \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) g(h) = \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) h = 0
and f(h) = cos \(\frac{1}{h}\) is bounded in the deleted ngd of 0.
∴ \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) g(h) f(h) = \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) h cos \(\frac{1}{h}\) = 0]
Rf'(c) = \(\ e{Lt}_{x \rightarrow c^{+}} \frac{f(x)-f(c)}{x-c}\)
= \(\ {Lt}_{h \rightarrow 0^{+}} \frac{f(c+h)-f(c)}{c+h-c}\)
[put x = c + h as x → c⁺
⇒ h → 0⁺]
= \(\ {Lt}_{h \rightarrow 0^{+}} \frac{(c+h-c)^2 \cos \frac{1}{c+h-c}-0}{c+h-c}\)
= \(\ {Lt}_{h \rightarrow 0^{+}} \frac{h^2 \cos \frac{1}{h}-0}{h}\)
= \(\underset{h \rightarrow 0^{+}}{\mathrm{Lt}}\) h cos \(\frac{1}{h}\) = 0
∴ Lf'(c) = R f'(c)
Thus f(x) is also differentiable at x = c.
Hence f(x) is continuous as well as differentiable at x = c.

Question 4.
Prove that the derivative of an odd function is always an even function.
Solution:
Let f(x) be an odd function
∴ f(- x) = – f(x) …………….(1)
Diff. eqn. (1) both sides w.r.t. x, we have
f'(- x) \(\frac{d}{d x}\) (- x) = – \(\frac{d}{d x}\) f(x)
⇒ f'(- x) (- 1) = – f'(x)
⇒ f'(- x) = f'(x)
Thus f'(x) be an even function.
Hence, the deerivative of an add function is always an even function.

Question 5.
If 2 f(x) + 3 f(- x) = x² + x + 1, find f'(1).
Solution:
Given 2 f(x) + 3 f(- x) = x² + x + 1 ……………..(1)
Changing x to – x in eqn. (1) ; we have
2 f(- x) + 3 f(x) = x² – x + 1 ………….(2)
Multiply eqn. (1) by 2 and eqn. (2) by 3 and subtracting ; we have
– 5 f(x) = 2x² + 2x + 2 – 3x² + 3x – 3
⇒ – 5 f(x) = – x² + 5x – 1
⇒ f(x) = \(\frac{1}{5}\) [x² – 5x + 1]
Diff. both sides w.r.t. x ; we have
f'(x) = \(\frac{1}{5}\) (2x – 5)
⇒ f'(1) = \(\frac{1}{5}\) (2 – 5)
= – \(\frac{3}{5}\)

Question 6.
Differentiate the following functions w.r.t. x :
(i) tan^-1 \(\left(\frac{\sqrt{x}(3-x)}{1-3 x}\right)\)
(ii) tan^-1 \(\left(\frac{\left(3 a^2 x-x^3\right)}{a\left(a^2-3 x^2\right)}\right)\)
Solution:
(i) Let y = tan^-1 \(\left(\frac{\sqrt{x}(3-x)}{1-3 x}\right)\)
put √x = tan θ.
⇒ θ = tan^-1 x
Thus from (1) ; we have
y = tan^-1 \(\left\{\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right\}\)
⇒ y = tan^-1 (tan 3θ) = 3θ = tan^-1 √x
Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}=\frac{3}{1+(\sqrt{x})^2} \frac{d}{d x} \sqrt{x}\)
= \(\frac{3}{(1+x)} \frac{1}{2 \sqrt{x}}\)
= \(\frac{3}{2 \sqrt{x}(1+x)}\)

(ii) Let y = tan^-1 \(\left(\frac{\left(3 a^2 x-x^3\right)}{a\left(a^2-3 x^2\right)}\right)\)
⇒ y = tan^-1 \(\left\{\frac{\frac{3 x}{a}-\left(\frac{x}{a}\right)^3}{1-3\left(\frac{x}{a}\right)^2}\right\}\) ……(1)
put \(\frac{x}{a}\) = tan θ
⇒ θ = tan^-1 \(\frac{x}{a}\)
∴ from (1) ;
y = tan^-1 \(\left\{\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right\}\)
⇒ y = tan^-1 (tan 3θ)
⇒ y = 3θ
= 3 tan^-1 \(\frac{x}{a}\)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=3 \cdot \frac{1}{1+\left(\frac{x}{a}\right)^2} \cdot \frac{1}{a}\)
= \(\frac{3 a}{a^2+x^2}\)

Question 7.
Differentiate tan^-1 \(\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)\) w.r.t. tan^-1 ax.
Solution:
Let y = tan^-1 \(\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)\) ………..(1)
and z = tan^-1 ax ………..(2)
So we want to diff. y w.r.t. z i.e. to find \(\frac{d y}{d z}\).
put ax = tan θ
i.e. θ = tan^-1 ax
∴ from (1) ;
y = tan^-1 \(\left(\frac{\sec \theta-1}{\tan \theta}\right)\)

Question 8.
If y = \(\frac{\sin x}{1+\frac{\cos x}{1+\frac{\sin x}{1+\frac{\cos x}{1+\ldots \infty}}}}\), prove that \(\frac{d y}{d x}=\frac{(1+y) \cos x+y \sin x}{1+2 y+\cos x-\sin x}\).
Solution:
Given y = \(\frac{\sin x}{1+\frac{\cos x}{1+y}}\)
⇒ y = \(\frac{(1+y) \sin x}{1+y+\cos x}\)
⇒ y + y² + y cos x = (1 + y) sin x ……….(1)
Diff. eqn. (1) both sides w.r.t. x, we have
\(\frac{d y}{d x}\) + 2y \(\frac{d y}{d x}\) + cos x \(\frac{d y}{d x}\) + y (- sin x) = (1 + y) cos x + sin x \(\frac{d y}{d x}\)
⇒ (1 + 2y + cos x – sin x) \(\frac{d y}{d x}\) = (1 + y) cos x + y sin x
⇒ \(\frac{d y}{d x}=\frac{(1+y) \cos x+y \sin x}{1+2 y+\cos x-\sin x}\).

Question 9.
If y = (log_{cos x} sin x) (log_{sin x} cos x)^-1, prove that \(\frac{d y}{d x}\)_π/4 = – 8 log₂ e.
Solution:

Question 10.
Differentiate x^x sin^-1 √x w.r.t. x.
Solution:
Let y = x^x sin^-1 √x
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = x^x \(\frac{1}{\sqrt{1-(\sqrt{x})^2}} \frac{1}{2 \sqrt{x}}\) + sin^-1 √x \(\frac{d}{d x}\) x^x
= \(\frac{x^x}{\sqrt{1-x}} \frac{1}{2 \sqrt{x}}\) + sin^-1 √x \(\frac{d}{d x}\) e^{log xx}
= \(\frac{x^x}{2 \sqrt{x-x^2}}\) + sin^-1 √x \(\frac{d}{d x}\) e^{log xx}
= \(\frac{x^x}{2 \sqrt{x-x^2}}\) + sin^-1 √x e^{x log x} [x × \(\frac{1}{x}\) + log x . 1]
= \(\frac{x^x}{2 \sqrt{x-x^2}}\) + sin^-1 √x e^{log xx} [1 + log x]
= \(\frac{x^x}{2 \sqrt{x-x^2}}\) + sin^-1 √x . x^x (1 + log x).

Question 11.
If x = sin t and y = sin 2t, prove that
(i) (1 – x²) (\(\frac{d y}{d x}\))² = 4 (1 – y²)
(ii) (1 – x²) \(\frac{d^2 y}{d x^2}\) – x \(\frac{d y}{d x}\) + 4y = 0.
Solution:
(i) Given x = sin t ………..(1)
and y = sin 2t ………..(2)
Diff. eqn. (1) and (2) w.r.t. t ; we have
\(\frac{d x}{d t}\) = cos t ;
\(\frac{d y}{d t}\) = 2 cos 2t
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{2 \cos 2 t}{\cos t}\) ………..(3)
On squaring both sides of eqn. (3) ; we have
cos² t (\(\frac{d y}{d x}\))² = 4 cos² 2t
⇒ (1 – sin² t) (\(\frac{d y}{d x}\))² = 4 [1 – sin² 2t]
⇒ (1 – x²) (\(\frac{d y}{d x}\))² = 4 [1 – y²] ………..(4)
Diff. eqn. (4) w.r.t. x ; we have
(1 – x²) 2 \(\frac{d y}{d x}\) \(\frac{d^2 y}{d x^2}\) + (\(\frac{d y}{d x}\))² (- 2x) = – 8y \(\frac{d y}{d x}\)
Dividing throughout by 2 \(\frac{d y}{d x}\) ≠ 0 ; we get
(1 – x²) \(\frac{d^2 y}{d x^2}\) – x \(\frac{d y}{d x}\) = – 4y
⇒ (1 – x²) \(\frac{d^2 y}{d x^2}\) – x \(\frac{d y}{d x}\) + 4y = 0.

Question 12.
If y^1/m + y^-1/m = 2x, prove that (x² – 1) y₂ + xy₁ = m²y.
Solution:
Given y^1/n + y^-1/n = 2x
⇒ y^1/n + \(\frac{1}{y^{1 / n}}\) = 2x
⇒ (y^1/n)² – 2xy^1/n + 1 = 0 ; …………(1)
putting y^1/n = t in eqn. (1) ; we get
⇒ t² – 2xt + 1 = 0
⇒ t = \(\frac{2 x \pm \sqrt{4 x^2-4}}{2}\)
⇒ t = x ± \(\sqrt{x^2-1}\)

Case I:
When y^1/n = x + \(\sqrt{x^2-1}\)
⇒ y = (x + \(\sqrt{x^2-1}\))ⁿ …………(2)
Diff. eqn. (2) w.r.t. x, we have
∴ y₁ = n (x + \(\sqrt{x^2-1}\))^n-1 \(\frac{d}{d x}\) [x + \(\sqrt{x^2-1}\)]
∴ y₁ = n (x + \(\sqrt{x^2-1}\))^n-1 \(\left[1+\frac{x}{\sqrt{x^2-1}}\right]\)
= \(\frac{n\left(x+\sqrt{x^2-1}\right)^{n-1}\left[x+\sqrt{x^2-1}\right]}{\sqrt{x^2-1}}\)
⇒ \(\sqrt{x^2-1}\) y₁ = ny ………..(3) [using (2)]
Again diff. both sides of eqn. (3) w.r.t. x, we have
⇒ \(\sqrt{x^2-1}\) y₂ + \(\frac{y_1}{\sqrt{x^2-1}}\) × x = ny₁
⇒ (x² – 1) y₂ + xy₁ = n²y [using (3)]

Case – II:
When y^1/n = x – \(\sqrt{x^2-1}\)
⇒ y = [x – \(\sqrt{x^2-1}\)]ⁿ ………..(4)
Diff. eqn. (4) w.r.t. x; we have
∴ y₁ = n (x – \(\sqrt{x^2-1}\))^n-1 \(\left(1-\frac{x}{\sqrt{x^2-1}}\right)\)
⇒ \(\sqrt{x^2-1}\) y₁ = – ny
Again diff. eqn. (5) w.r.t. x ; we have
⇒ \(\sqrt{x^2-1}\) y₂ + xy₁ = n²y [using eqn. (5)]

Question 13.
If x = e^t sin t and y = e^t cos t, then prove that (x + y)² \(\frac{d^2 y}{d x^2}\) = 2 (x \(\frac{d y}{d x}\) – y).
Solution:
Given, x = e^t sin t …………(1)
y = e^t cos t ………..(2)
Diff. eqn. (1) and (2) both sides w.r.t. t ; we have
\(\frac{d x}{d t}\) = e^t cos t + e^t sin t
= e^t (cos t + sin t) ………….(3)
and \(\frac{d x}{d t}\) = e^t (- sin t) + cos t e^t
= e^t (cos t – sin t)
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{e^t(\cos t-\sin t)}{e^t(\cos t+\sin t)}\)
= \(\frac{\cos t-\sin t}{\cos t+\sin t}\) ………….(4)
Diff. eqn. (4) both sides w.r.t. x, we have

Question 14.
Using Rolle’s theorem, show that between any two distinct real roots of the equation ax³ + bx² + cx + d = 0, there is atleast one real root of the equation 3ax²+ 2bx + c = 0.
Solution:
Let f(x) = ax³ + bx² + cx + d
Let α and β be the distinct real rootsof the equation
f(x) = 0 and let α < β.
∴ f (α) = 0 = (β)
Since f(x) be a polynomial function and hence f(x) be continuous and differentiable ∀ x ∈ R.
Thus f(x) is continuous in [α, β].
f(x) is derivable in (α, β).
Also f(α) = f(β)
Thus, all the three conditions of Rolle’s theorem are satisfied so 3 atleast one real number y ∈ (α, β)
s.t f’ (γ) = 0
Diff. eqn. (1) both sides w.r.t. x, we have
∴f’(x) = 3ax² + 2bx + c
Now f’(y) = 0
⇒ 3ay² + 2by + c = 0
⇒ γ be the root of the equation 3ax² + 2bx + c = 0
Hence, between any two distinct real roots of the equation f(x)= 0, there is atleast one real root of the equation 3ax² + 2bx + c = 0.

Question 15.
Discuss the applicability of Rolle’s theorem for the following functions in the indicated intervals :
(i) f(x) = \(\begin{cases}x+3, & x \leq 2 \\ 7-x, & x>2\end{cases}\) in [- 3, 7]
(ii) f(x) = \(\left\{\begin{array}{cc}
x^2+1, & 0 \leq x \leq 1 \\
3-x, & 1<x \leq 2
\end{array}\right.\) in [0, 2] (NCERT Exampler)
Solution:
(i) Given, f(x) = \(\begin{cases}x+3, & x \leq 2 \\ 7-x, & x>2\end{cases}\)
\(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) 7 – x
= 7 – 2 = 5
\(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) x + 3
= 2 + 3 = 5
also f(2) = 2 + 3 = 5
∴ \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = f(2)
Thus f(x) is continuous at x = 2.

Differentiability at x = 2:
Lf'(2) = \(\ {Lt}_{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2}\)
= \(\ {Lt}_{x \rightarrow 2^{-}} \frac{(x+3)-(2+3)}{x-2}\)
= \(\ {Lt}_{x \rightarrow 2^{-}} \frac{x-2}{x-2}\) = 1

and Rf'(2) = \(\ {Lt}_{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2}\)
= \(\ {Lt}_{x \rightarrow 2^{+}} \frac{7-x-(2+3)}{x-2}\)
= \(\ {Lt}_{x \rightarrow 2^{+}} \frac{2-x}{x-2}\) = – 1
∴ Lf'(2) ≠ Rf'(2)
Hence f(x) is not derivable at x = 2 ∈ (- 3, 7)
∴ f(x) is not derivable in (- 3, 7)
So condition (ii) of Rolle’s theorem is not satisfied.
Hence Rolle’s theorem is not applicable to the function f(x) in [- 3, 7]

(ii) Given f(x) = \(\left\{\begin{array}{cc}
x^2+1 ; & 0 \leq x \leq 1 \\
3-x & ; 1 \end{array}\right.\) in [0, 2]

Derivability at x = 1:
Rf'(1) = \(\ {Lt}_{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{+}} \frac{3-x-(1+1)}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{+}} \frac{3-x-2}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{+}} \frac{1-x}{x-1}\) = – 1

Lf'(1) = \(\ {Lt}_{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{\left(x^2+1\right)-(1+1)}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{x^2-1}{x-1}\)
= \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) x + 1
= 1 + 1 = 2
∴ Lf'(1) ≠ Rf'(1)
Thus, f(x) is not derivable at x I E (0, 2)
Hence fis not derivable in (0, 2)
So condition (ii) of RoBe’s theorem is not satisfied.
Hence Rolle’s theorem is not applicable to the function f(x) in [0, 2].

Question 16.
Discuss the applicability of Lagrange’s mean value theorem for the following functions in the indicated intervals.
(i) f(x) = \(\left\{\begin{array}{rll}
2+x^3 & , \text { if } & x \leq 1 \\
3 x & \text {, if } & x>1
\end{array}\right.\) on [- 1, 2]
(ii) f(x) = \(\begin{cases}\frac{1}{x}, & x \neq 0 \\ 0, & x=0\end{cases}\) in [- 1, 1]
(iii) f(x) = \(\left\{\begin{array}{cl}
3 x+2 & , x<2 \\ 14-3 x & , 2 \leq x \end{array}\right.\) in [- 2, 6]
(iv) f(x) = \(\left\{\begin{array}{cc} x \sin \frac{1}{x}, & x \neq 0 \\ 0 & , x=0 \end{array}\right.\) in [- 1, 1]
Solution:
(i) Given \(\left\{\begin{array}{rll} 2+x^3 & , \text { if } & x \leq 1 \\ 3 x & \text {, if } & x>1
\end{array}\right.\) on [- 1, 2]
Clearly f(x) is polynomial in x,
also \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) 3x
= \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) 3 (1 + h) = 3
and \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) (2 + x³)
= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) [2 + (1 – h)³]
= 2 + (1 – 0)³ = 3
∴ \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = f(1)
i.e. f(x) is continuous at x = 1 and hence f(x) is continuous in [- 1, 2]

at x = 1:

∴ Lf’(1) = Rf’(1)
∴ f(x) is differentiable at x = 1
Hencef is derivable on (- 1, 2).
Further f(- 1) = 2 + (- 1)³ = 1
and f(2) = 3 × 2 = 6
Thus all the two conditions of L.M.V. theorem are satisfied
∴ ∃ atleast our real number c ∈ (- 1, 2).
s.t. f'(c) = \(\frac{f(b)-f(a)}{b-a}\)
Also f'(c) = \(\left\{\begin{array}{cc}
3 x^2, & x \leq 1 \\
3, & x>1
\end{array}\right.\)

When x ≤ 1 :
i.e. 3c² = \(\frac{6-1}{2-(-1)}=\frac{5}{3}\)
c = \(\pm \frac{\sqrt{5}}{3}\)
but c = – \(\frac{\sqrt{5}}{3}\) ∈ (- 1, 2)

When x > 1 ; f'(c) = 3
i.e. 3 = \(\frac{6-1}{2-(-1)}\)
⇒ 3 = \(\frac{5}{3}\) which is impossible
Hence L.M.V. theorem is applicable and c = \(\frac{\sqrt{5}}{3}\).

(ii) as x → 0^–, \(\frac{1}{x}\) → ∞
as x → 0⁺, \(\frac{1}{x}\) → ∞
Thus f(x) = \(\frac{1}{x}\) oscillating between – ∞ to + ∞.
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) does not exist.
Thus f(x) is discontinuous at x = 0 ∈ [- 1, 1]
Hence lagrange mean value theorem is not applicable.

(iii) Given f(x) = \(\begin{cases}3 x+2 & ; \quad x<2 \\ 14-3 x & ; \quad 2 \leq x\end{cases}\) in [- 2, 6]
L.H. Limit = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) 3x + 2
= 6 + 2 = 8
R.H.Limit = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) (14 – 3x)
= 14 – 6 = 8
and f(2) = 8
∴ f is continuous at x = 2.

Differentiability at x = 2:

∴ Lf'(2) ≠ Rf'(2)
Thus f(x) is not derivable at x = 2 ∈ (- 2, 6)
Hence f(x) is not derivable in (- 2, 6).
Thus lagrange’s mean value theorem is not applicable for f(x) in [- 2, 6].

(iv) f'(0) = \(\ {Lt}_{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{x \sin \frac{1}{x}-0}{x-0}\)
= \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) sin \(\frac{1}{x}\),
which does not exists
[Since sin \(\frac{1}{x}\) be a real number oscillating between – 1 and 1]
at x = \(\frac{1}{2 n \pi+\frac{\pi}{2}}\), n ∈ N, n → ∞
⇒ x → 0
∴ sin \(\frac{1}{x}\) = sin (2nπ + \(\frac{\pi}{2}\))
= siin \(\frac{\pi}{2}\) = 1
at x = \(\frac{1}{2 n \pi+\frac{\pi}{2}}\), n ∈ N, n is large
⇒ x → 0
∴ sin \(\frac{1}{x}\) = sin (2nπ – \(\frac{\pi}{2}\))
= sin \(\frac{-\pi}{2}\) = – 1
Thus f(x)is not derivable at x ∈ (- 1, 1)
Hencef(x) is not derivable in (- 1, 1).
Thus the condition (ii) of lagrange’s mean values is not satisfied.
Thus, the lagrange’s mean value theorem is not applicable for the function f(x) in [- 1, 1].

Question 17.
Using Lagrange’s Mean Value Theorem, show that | sin α – sin β | ≤ |α – β|.
Solution:
Two cases arise.

Case – I:
When α = β
∴ | sin α – sin β | = | sin α – sin α |
= |0| = 0
|α – β| = |α – α|
= |0| = 0
Thus, | sin α – sin β | = |α – β|

Case – II:
When α ≠ β, let α < β
Let f(x) = sin x ………….(1)
which is continuous and differentiable everywhere.
Thus f(x) is continuous in [α, β] and f(x) is derivable in (α, β).
So both the conditions of lagrange’s mean value theorem are satisfied.
Then ∃ atleast one real number c ∈ (α, β)
s.t. f'(c) = \(\frac{f(\beta)-f(\alpha)}{\beta-\alpha}\) …………..(2)
Diff. eqn. (1) w.r.t. x ; we have
f'(x) = cos x
Now f'(c) = cos c ;
f(β) = sin β ;
f(α) = sin α
∴ from (2) ;
\(\frac{\sin \beta-\sin \alpha}{\beta-\alpha}\) = cos c
⇒ \(\frac{\sin \beta-\sin \alpha}{\beta-\alpha}\) = |cos c| ≤ 1
[∵ – 1 ≤ cos x ≤ 1 ∀ x ∈ R]
⇒ |sin α – sin β| ≤ |β – α|
= |- (α – β)|
= |α – β|
[∵ |- x| = |x| ∀ x ∈ R]
So, combining two cases, we have |sin α – sin β| ≤ |α – β|.

Effective ML Aggarwal Class 12 Solutions ISC Chapter 5 Continuity and Differentiability Ex 5.13 can help bridge the gap between theory and application.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.13

Verify Lagrange’s mean value theorem for the following (1 to 4) functions in the given intervals. Also find ‘c’ of this theorem :

Question 1.
(i) f(x) = x² in [2, 4] (NCERT)
(ii) f(x) = x² – 4x – 3 in [1, 4] (NCERT)
(iii) f(x) = x + \(\frac {1}{x}\) in [1, 3]. [ISC 2019]
Solution:
(i) Given f(x) = x² is continuous on [2, 4]
and derivable on (2, 4) as f(x) is polynomial in x which is continuous and differentiable everywhere.
∴ f’(x) = 2x
and f(2) = 4;
f (4) = 16
Thus both the conditions of L.M.V are satisfied.
∴ ∃ atleast one real number c ∈ (2, 4) s.t.
f'(c) = \(\frac{f(b)-f(a)}{b-a}\)
⇒ 2c = \(\frac{16-4}{4-2}\)
Hence L.M.V theorem is verified and c = 3.

(ii) Given f(x) = x² – 4x – 3
∴ f(x) is continuous in [1, 4]
[∵ f(x) is polynomial in x]
also f’(x) = 2x – 4 exists ∀ x ∈ (1, 4)
∴ f(x) is derivable in (1, 4).
Now, f( 1) = – 6
and f(4) = 3
∴ Both the conditions of L.M.V. theorem are satiafied.
∴ ∃ atleast one real no. c ∈ (1, 4)
s.t. f'(c) = \(\frac{f(4)-f(1)}{4-1}\)
2c – 4 = \(\frac{-3-(-6)}{4-1}\) = 1
c = \(\frac{5}{2}\) ∈ (1, 4).
∴ L.M.V theorem is verified and c = \(\frac{5}{2}\).

(iii) Given f(x) = x + \(\frac {1}{x}\)
∴ f'(x) = 1 – \(\) which exists in [1, 3]
Thus, f(x) is continuous in [1, 3] and derivable in (1, 3)
Now f(1) = 1 + \(\frac{1}{1}\) = 2 ;
f(3) = 3 + \(\frac{1}{3}\)
= \(\frac{10}{3}\)
Therefore both conditions of Lagrange’s mean value theorem are satisfied so ∃ atleast one real number c ∈ (1, 3) such that
\(\frac{f(3)-f(1)}{3-1}\) = f'(c)
⇒ \(\frac{\frac{10}{3}-2}{2}=1-\frac{1}{c^2}\)
⇒ \(\frac{2}{3}=1-\frac{1}{c^2}\)
⇒ \(\frac{1}{c^2}=1-\frac{2}{3}=\frac{1}{3}\)
⇒ c² = 3
⇒ c = ± √3
But c ∈ (1, 3)
∴ c = √3 ∈ (1, 3)
Thus ∃ c = √3 ∈ (1, 3) s.t.
f'(c) = \(\frac{f(3)-f(c)}{3-1}\)
Hence, Lagrange’s mean value theorem is verified.

Question 1 (old).
(iii) f(x) = x² – 2x + 4 on [1, 5]
Solution:
Given f(x) = x² – 2x + 4
since polynomial function is everywhere continuous and differentiable.
Thus f (x) is continuous on [1, 5] and differentiable on (1, 5).
So both conditions of L.M.V theorem are satisfied.
So ∃ atleast one real number c ∈ (1, 5) such that
f'(c) = \(\frac{f(5)-f(1)}{5-1}\) ……….(1)
since f(x) = x² – 2x + 4
∴ f’(x) = 2x – 2
f(1) = 1 – 2 + 4 = 3 ;
f(5) = 25 – 10 + 4 = 19
∴ from (1) ;
2c – 2 = \(\frac{19-3}{4}\)
= \(\frac{16}{4}\) = 4
Thus c = 3 ∈ (1, 5) s.t.
f’ (c) = \(\frac{f(5)-f(1)}{5-1}\)
Thus, L.M.V Theorem is verified.

Question 2.
(i) f(x) = x³ – 2x² – x + 3 on [0, 1] (NCERT Exampler)
(ii) f(x) = (x – 4) (x – 6) (x – 8) on [4, 8].
Solution:
(i) Given f(x) = x³ – 2x² – x + 3 in [0, 1]
Since f(x) is polynomial in x.
∴ f(x) is continuous in [0, 1] and derivable in (0, 1).
Now f'(x) = 3x² – 4x – 1
and f'(0) = 3
and f(1) = 1 – 2 – 1 + 3 = 1
∴ Both the conditions of L.M.V. theorem are satisfied.
∴ ∃ atleast one real number c ∈ (0, 1) s.t.
f'(c) = \(\frac{f(b)-f(a)}{b-a}\)
i.e. 3c² – 4c – 1 = \(\frac{1-3}{1-0}\)
= – 2
⇒ 3c² – 4c + 1 = 0
⇒ c = 1, \(\frac{1}{3}\)
but c = 1 ∉ (0, 1)
∴ c = \(\frac{1}{3}\) ∈ (0, 1)
Hence L.M.V. theorem is applicable and c = \(\frac{1}{3}\).

(ii) Given f(x) = (x – 4) (x – 6) (x – 8) …………..(1)
Clcarly f(x) be a polynomial function hence continuous and differentiable eveiy here.
∴ f(x) is continuous in [4, 8] and derivable in (4, 8).
Thus, both conditions of Lagrange’s mean value Theorem are satisfied.
So ∃ exists at least one real number c ∈ (4, 8)
s.t f'(c) = \(\frac{f(8)-f(4)}{8-4}\) …………(2)
Duff. (1) both sides w.r.t. x, we have
Since f(x) = x³ – 18x² + 104x – 192
f’(x) = 3x² – 36x + 104
also, f(8) = 0; f(4) = 0
∴ from (2) ;
3c² – 36c + 104 = \(\frac{0-0}{4}\) = 0
⇒ c = \(\frac{36 \pm \sqrt{1296-1248}}{6}\)
= \(\frac{36 \pm \sqrt{48}}{6}\)
⇒ c = \(\frac{36 \pm 4 \sqrt{3}}{6}\)
= 6 ± \(\frac{2}{3}\) √3.

Question 3.
(i) f(x) = sin x in [0, \(\frac{\pi}{2}\)]
(ii) f(x) = x – 2 sin x in [- π, π]
Solution:
(i) Given f(x) = sin x …………….. (1)
Clearly f (x) be continuous and differentiable in its domain.
∴ f be continuous in [0, \(\frac{\pi}{2}\)] and derivable in (o, \(\frac{\pi}{2}\)).
Thus both conditions of Lagrange’s mean value theorem are satisfied for function f in [0, \(\frac{\pi}{2}\)]
Then ∃ atleast one real number c ∈ (0, \(\frac{\pi}{2}\))
s.t. f'(c) = \(\frac{f\left(\frac{\pi}{2}\right)-f(0)}{\frac{\pi}{2}-0}\) ………..(2)
Hence f'(x) = cos x ;
f(\(\frac{\pi}{2}\)) = 1 ;
f(0) = 0
∴ from (2) ;
cos c = \(\frac{1-0}{\frac{\pi}{2}-0}=\frac{2}{\pi}\)
⇒ c = cos^-1 \(\left(\frac{2}{\pi}\right)\)
So ∃ a real number c = cos^-1 \(\frac{2}{\pi}\) ∈ (0, \(\frac{\pi}{2}\))
s.t. f'(c) = \(\frac{f\left(\frac{\pi}{2}\right)-f(0)}{\frac{\pi}{2}-0}\)
Hence lagrange’s mean value Theorem is verified
and c = cos^-1 \(\left(\frac{2}{\pi}\right)\)
[When 0 < c < \(\frac{\pi}{2}\)
⇒ 0 < cos c < 1
⇒ 0 < \(\frac{2}{\pi}\) < 1, which is true]

(ii) Given f(x) = x – 2 sin x ……………(1)
Clearly polynomial function and trigonometric function are continuous and derivable everywhere.
Thus f(x) is continuous in [- π, π] and derivable on (- π, π).
So both conditions of lagrange’s mean value are satisfied.
So ∃ atleast one real number e E (- π, π).
s.t. f'(c) = \(\frac{f(\pi)-f(-\pi)}{\pi-(-\pi)}\) ……….(2)
Diff. w.r.t. x ; we get
f'(x) = 1 – 2 cos x ;
f(π) = π – 2 sin π
= π – 0 = π
f(- π) = – π – 2 sin (- π) = – π
∴ from (2) ;
1 – 2 cos c = \(\frac{\pi-(-\pi)}{\pi-(-\pi)}\) = 1
⇒ – 2 cos c = 0
⇒ cos c = 0
⇒ c = (2n + 1) \(\frac{\pi}{2}\) ∀ n ∈ I
But c ∈ (- π, π)
∴ c = \(\frac{\pi}{2}\), – \(\frac{\pi}{2}\)
Thus ∃ two real numbers \(\frac{\pi}{2}\) and – \(\frac{\pi}{2}\)
s.t. f'(\(\frac{\pi}{2}\)) = \(\frac{f(\pi)-f(-\pi)}{\pi-(-\pi)}\)
and f'(- \(\frac{\pi}{2}\)) = \(\frac{f(\pi)-f(-\pi)}{\pi-(-\pi)}\)
∴ Lagrange mean value theorem is verified and c = ± \(\frac{\pi}{2}\).

Question 4.
(i) f(x) = 2 sin x + sin 2x in [0, π]
(ii) f(x) = sin x – sin 2x on [0, π]
Solution:
(i) Given f(x) = 2 sin x + sin 2x
Since every trigonometric function is continuous and differentiable in its domain.
The sum of two continuous and differentiable functions ¡s continuous and differentiable.
(i) f(x) is continuous in [0, π]
(ii) f(x) is duff. in (0, π)
Thus both conditions of Lagrange’s mean value theorem are satisfied
so ∃ atleast one real no. c ∈ (0, π)
s.t \(\frac{f(\pi)-f(0)}{\pi-0}\) = f’(c) ………….(1)
Here f(π) = 2 sin π + sin 2π = 0
and f(0) = 0 + 0 = 0
and f’(x) = 2 cos x + 2 cos 2x
∴ from (1) ;
\(\frac{0}{\pi}\) = 2 (cos c + cos 2c)
⇒ cos 2c + cos c = 0
⇒ 2 cos \(\frac{3 c}{2}\) cos \(\frac{c}{2}\) = 0
⇒ cos \(\frac{3 c}{2}\) = 0 or
cos \(\frac{c}{2}\) = 0
⇒ \(\frac{3 c}{2}=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \ldots\)
or \(\frac{c}{2}=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \ldots\)
⇒ c = \(\frac{\pi}{3}, \pi, \frac{5 \pi}{3}, \ldots\)
or c = π, 3π, 5π, ……………
But c ∈ (0, π)
∴ c = \(\frac{\pi}{3}\)
Thus there exists atleast one c = \(\frac{\pi}{3}\) ∈ (0, π)
s.t. f'(c) = \(\frac{f(\pi)-f(0)}{\pi-0}\)

(ii) f(x) = sin x – sin 2x on [0, π]
Since sine function is continuous and derivable everywhere
∴ f is continuous in [0, 2π] and derivable in (0, 2π)
∴ Both the conditions of L.M.V are satisfied.
∴ ∃ atleast one real no c ∈ (0, 2π) s.t.
f'(c) = \(\frac{f(b)-f(a)}{b-a}\)
Now f’(c) = cos x – 2 cos 2x ;
f(x) = 0 ;
f(2π) = 0
i.e. cos c – 2 cos 2c = \(\frac{f(2 \pi)-f(0)}{2 \pi-0}\)
i.e. cos c – 2 cos 2c = \(\frac{0-0}{2 \pi}\) = 0
⇒ cos c – 2 cos 2c = 0
⇒ cos c – 2 (2 cos² c – 1) = 0
⇒ 4 cos² c – cos c – 2 = 0
∴ cos c = \(\frac{1 \pm \sqrt{33}}{8}\)
∴ c = cos^-1 \(\left(\frac{1 \pm \sqrt{33}}{8}\right)\) ∈ [0, 2π]
[Since 0 < c < 2π
⇒ – 1 < cos c < 1]
Thus L.M.V. is applicable and
c = cos^-1 \(\left(\frac{1 \pm \sqrt{33}}{8}\right)\)

Question 5.
(i) f(x) = px² + qx + r, p ≠ 0, on [0, 1]
(ii) f(x) = x on [a, b]
(iii) f(x) = (x – 1)^2/3 on [1, 2].
Solution:
(i) Given f(x) = px² + qx + r, p ≠ 0
Sincef(x) be a polynomial function. So it
is continuous and derivable everywhere.
Thus f(x) is continuous in [0, 1] and derivable in (0, 1).
Therefore, both conditions of lagrange’s mean value Theorem are satisfied.
So ∃ atleast one real number c ∈ (0, 1)
s.t. f'(c) = \(\frac{f(1)-f(0)}{1-0}\) ………..(2)
Diff. (1) w.r.t. x, we have
f'(x) = 2px + q ;
f(1) = p + q + r ;
f(0) = r
∴ from (2) ; we have
2pc + q = \(\frac{p+q+r-r}{1}\)
= p + q
⇒ 2pc = p
⇒ c = \(\frac{1}{2}\) ∈ (0, 1) [∵ p ≠ 0]
Thus, ∃ a real number c = \(\frac{1}{2}\) ∈ (0, 1)
such that f'(\(\frac{1}{2}\)) = \(\frac{f(1)-f(0)}{1-0}\)
Hence lagrange’s mean value theorem is verified and c = \(\frac{1}{2}\) .

(ii) Given f(x) = x on [a, b]
Clearly f(x) is polynomial in x.
∴ f is continuous n [a, b] and derivable in (a, b)
s.t. f'(c) = \(\frac{f(b)-f(a)}{b-a}\)
⇒ 1 = \(\frac{b-a}{b-a}\)
⇒ 1 = 1 which is true.
Hence L.M.V. is applicable and ∃ one real number c ∈ (a, b) s.t.
f(c) = \(\frac{f(b)-f(a)}{b-a}\)

(iii) f(x) = (x – 1)^2/3
∴ f'(x) = \(\frac{2}{3(x-1)^{1 / 3}}\)
which does not exist at x = 1 and 1 ∉ (1, 2)
∴ f is derivable in (1, 2) and hence continuous in [1, 2]
also f(1) = 0
and f(2) = 1
∴ Both the conditions of L.M.V. theorem are satisfied.
∴ ∃ atleast one real number c ∈ (1, 2)
s.t. f'(c) = \(\frac{f(2)-f(1)}{2-1}\)
\(\frac{2}{3(c-1)^{1 / 3}}=\frac{1-0}{1}\)
= 1
⇒ (c – 1)^2/3 = \(\frac{2}{3}\)
⇒ c – 1 = \(\frac{8}{27}\)
⇒ c = \(\frac{35}{27}\) ∈ (1, 2)
Thus L.M.V. theorem ia applicable and c = \(\frac{35}{27}\).

Question 6.
Show that the function f(x) = x² – 6x + 1 on [1, 3] satisfies Lagrange’s mean value theorem. Also find the coordinates of a point at which the tangent to the curve represented by the above function is parallel to the chord joining A (1, – 4) and B (3, – 8). (ISC 2004)
(ii) Use Lagrange’s mean valuc theorem to deterniine a point P on the curve y = \(\sqrt{x-2}\) defined in the interval [2, 3] where the tangent ¡s parallel to the chord joining the end points on the curve. (ISC 2008)
Solution:
(i) Given y = f(x)
= x² – 6x + 1
Here we discuss the applicability of L.M.V theorem in [1, 3]
Since f(x) is polynomial in x
∴ it ¡s continuous in [1, 3]
also, f’ (x) = 2x – 6 which is exists ∀ x ∈ (1, 3)
∴ f(x) is derivable in (1, 3)
Now,
f(1) = 1- 6 + 1 = – 4
and f(3) = 9 – 18 + 1 = – 8
Thus all the conditions of L.M.V theorem are satisfied
∴ ∃ atleast one real number c ∈ (1, 3)
s.t. f'(c) = \(\frac{f(3)-f(1)}{3-1}\)
⇒ 2c – 6 = \(\frac{-8-(-4)}{2}\)
= – 2
⇒ 2c = 4
⇒ c = 2 ∈ (1, 3)
i.e.when x = 2
then y = 4 – 12 + 1 = – 7
Hence the required point is (2, – 7).
Thus, there exists a point (2, – 7) on the given curve y = – 6x + 1 where the tangent ¡s parallel to the chord joining the points (1, – 4) and (3, – 8).

(ii) Given y = f(x) = \(\sqrt{x-2}\) ………….(1)
Here we use the lagrange’s mean value theorem for [2, 3]
∴ f'(x) = \(\frac{1}{2 \sqrt{x-2}}\)
and f(2) = 0
and f(3) = 1
Clearly f(x) exists ∀ x ∈ (2, 3)
i.e. f(x) is desirable in (2, 3) and hence continuous in [2, 3].
∴ by L.M.V theorem ∃ atleast one real number x ∈ (2, 3).
s.t. f'(x) = \(\frac{f(3)-f(2)}{3-2}\)
⇒ \(\frac{1}{2 \sqrt{x-2}}=\frac{1-0}{1}\) = 1
⇒ 2 \(\sqrt{x-2}\) = 1
⇒ (x- 2) = \(\frac{1}{4}\)
⇒ x = \(\frac{9}{4}\) ∈ (2, 3)
∴ From (1) ;
y = \(\sqrt{\frac{9}{4}-2}=\frac{1}{2}\)
Thus, ∃ a point on the given curve where the tangent is parallel to the chord joining the points on the curve.

Question 7.
What can you say about the applicability of Lagrange’s mean value theorem for the following functions in the indicated intervals ?
(i) f(x) = x^1/3 in [- 1, 2]
(ii) f(x) = |x| in [- 2, 3]
(iii) f(x) = 3 – (2 – x)^2/3 in [0, 3]
Solution:
(i) Given f(x) = x^1/3
∴ f’(x) = \(\frac{1}{3} \frac{1}{x^{2 / 3}}\) does not exists at x = 0 ∈ (- 1, 1)
:. f(x) is not derivable at (- 1, 1)
Hence L.M.V theorem is not applicable.

(ii) Given f(x) = |x|
Clearly f(x) be continuous at x = 0
∴ f be continuous in [- 2, 3].
Thus, derivative of f(x) does not exists at x = 0 ∈ (- 2, 3).
Hence f(x) is not derivable on (- 2, 3).
Hence condition (ii) of lagrange’s mean value is not satisfied byf in [- 2, 3]
Thus, lagrange mean value theorem is not applicable to function fin [- 2, 3].

(ii) Given f(x) = 3 – (2 – x)^2/3 ; x ∈ [0, 3]
∴ f'(x) = – \(\frac{2}{3}\) (2 – x)^-1/3 (- 1)
= \(\frac{2}{3(2-x)^{1 / 3}}\), x ≠ 2
Thus, deerivative of f(x) does not exist at x = 2 ∈ (0, 3)
∴ f is not derivable at x = 2 ∈ (0, 3)
Thus f(x) is not derivable in (0, 3).
Hence lagrange’s mean value theorem is not applicable to function f in [0, 3].

Students can cross-reference their work with ML Aggarwal Class 12 Solutions Chapter 5 Continuity and Differentiability Ex 5.12 to ensure accuracy.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.12

Verify Rolle’s theorem for the following (1 to 5) functions and find point (or points) in the interval where derivative is zero :

Question 1.
(i) f(x) = x² – 5x + 6 in [1, 4]
(ii) = x² + 2x – 8, x ∈ [- 4, 2]
(iii) f(x) = x³ – 3x in [- √3, 0] (NCERT)
Solution:
(i) Given f(x) = x² – 5x + 6 …………(1)
Since f(x) be a polynomial function and hence continuous everywhere and
∴ f(x) be continuous in [1, 4].
Also f(x) being a polynomial function so it is differentiable everywhere.
∴ f(x) is derivable in (1, 4).
Also, f(1) = 1² – 5 + 6 = 2 ;
f(4) = 16 – 20 + 6 = 2
f(1) = f(4)
Thus, all the three conditions of Rolle’s Theorem are satisfied.
So ∃ atleast one real number
c ∈ (1, 4) s.t f’(c) = 0
Diff. (1) w.r.t. x, we have
f’(x) = 2x – 5
Now f'(c) = 0
⇒ 2c – 5 = 0.
⇒ c = \(\frac{5}{2}\) ∈ (1, 4)
So there exists \(\frac{5}{2}\) ∈ (1, 4) s.t. f'(\(\frac{5}{2}\)) = 0
Hence Rolle’s theorem is verified and c = \(\frac{5}{2}\).

(ii) Given, f(x) = x² + 2x – 8.
Since f(x) is polynomial in x so it is continuous in [- 4, 2]
also f’(x) = 2x + 2 exists ∀ x ∈ (- 4, 2)
∴ f is derivable in (- 4, 2).
also, f(- 4) = 16 – 8 – 8 = 0
and f(2) = 4 + 4 – 8 = 0
∴ f (- 4) = f(2)
Thus all the three conditions of Roile’s theorem are satisfied.
∴ ∃ one real number c ∈ (- 4, 2) s.t.f’ (c) = 0
i.e. 2c + 2 = 0
⇒ c = – 1 ∈ (- 4, 2)
Hence Rolle’s theorem is verified.

(iii) Given f(x) = x³ – 3x ………( 1)
Clearly f(x) be a polynomial in x.
∴ f (x) be contriuous in [- √3, 0].
Also f(x) be derivable in (- √3, 0).
Since f’(x) = 3x² – 3 which exists in (- √3, 0).
f(- √3) = (- √3)³ – 3 (- √3)
= – 3√3 + 3√3 = 0
f(0) = 0
Thus, f(- √3) = f(0)
Thus, all the three conditions of Rolle’s theorem are satisfied.
So ∃ atleast one real number c ∈ (- √3, 0) s.t. f’(c) = 0
Diff. eqn. (1) both sides w.r.t. x, we have
f’(x) = 3x² – 3
Now f’(c) = 0
⇒ 3c²– 3 = 0
⇒ c = ± 1
Clearly c = 1 ∉ (- √3, 0)
So there exists c = – 1 ∈ (- √3, 0)
s.t. f’(c) = 0
Hence Roll&s theorem is verified and c = – 1.

Question 1 (old).
(ii) f(x) = x² + 2 on [- 2, 2] (NCERT)
Solution:
We have f(x) = x² + 2
Since f(x) is polynormal in x
∴ it is continuous in [- 2, 2]
also f'(x) = 2x exists ∀ x ∈ (2, 2)
∴ f is derivable in (- 2, 2).
also f(2) = 2² + 2 = 6
and f(- 2) = (- 2)² + 2 = 6
∴ f(2) = f(- 2)
Hence all the conditions of Rolle’s theorem are satisfied
∴ ∃ atleast one number c ∈ (- 2, 2) s.t. f'(c) = 0
i.e. 2c = 0
⇒ c = 0 ∈ (- 2, 2)
Hence Rolle’s theorem is verified.

Question 2.
(i) f(x) = x³ + 3x² – 24x – 80 in [- 4, 5]
(ii) f(x) = x (x – 1)² in O, 1 (NCERT Excmplar)
(iii) f(x) = (x – 1) (x – 2) (x – 3) in [1, 3]
(iv) f(x) = \(\sqrt{4-x^2}\) in [- 2, 2] (NCERT Exemplar)
Solution:
(i) Given f (x) = x³ + 3x² – 24x – 80 in [- 4, 5]
Since f(x) is polynomial in x
∴ Continuous everywhere and differentiable.
Hence f is continuous in [- 4, 5] and derivable in (- 4, 5).
Also, f(- 4) = – 64 – 48 + 96 – 80 = 0
and f(5) = 125 + 75 – 120 – 80 = 0
∴ All the three conditions of rolle’s theorem are satisfied.
∴ ∃ atleast one real no. c ∈ (- 4, 5) s.t. f’(c) = 0
i.e. 3c² + 6c – 24 = 0
⇒ c² + 2c – 8 = 0
i.e. c = \(\frac{-2 \pm 6}{2}\)
= 2, – 4
but c = – 4 ∉ (- 4, 5)
∴ c = 2 ∈ (- 4, 5)
Hence Rolle’s theorem is verified and e 2.

(ii) Given f(x) = x (x – 1)² on [0, 1]
and f(x) = x (x² – 2x + 1)
Since f(x) ¡s polynomial in x
∴ continuous in [0, 1] and derivable in (0, 1) as f’(x) = 3x² – 4x + 1 exists ∀ x ∈ (0, 1)
also, f(0) = 0 = f(1)
∴ all the three conditions of RoBe’s theorem are satisfied.
∴ ∃ atleast one real number c ∈ (0, 1) s.t. f’(c) = 0
i.e. 3c² – 4c + 1 = 0
⇒ c = 1, \(\frac{1}{3}\)
but c = 1 ∉ (0, 1)
∴ c = \(\frac{1}{3}\) ∈ (0, 1)
Hence Rolle’s theorem is verified and c = \(\frac{1}{3}\).

(iii) Given f(x) = (x- 1) (x – 2) (x -3)
= (x – 1) (x² – 5x + 6)
= (x³ – 6x² + 11x – 6)
f(x) = x³ – 6x² + 11x – 6
Since f(x) is polynomial in x
∴ it is continuous and derivable everywhere.
∴ f(x) is continuous in [1, 3] and derivable in (1, 3)
Also f(1) = 0 = f(3)
All the three conditions of Rolle’s theorem are satisfied.
∴∃ atleast one real number c ∈ (1, 3) s.t. f’(c) = 0
i.e. 3c² – 12c + 11 = 0
⇒ c = \(\frac{12 \pm \sqrt{12}}{6} \)
= \(\frac{6 \pm \sqrt{3}}{3}\)
= 2 + \(\frac{1}{\sqrt{3}}\), 2 – \(\frac{1}{\sqrt{3}}\)
∴ c = \(\frac{6 \pm \sqrt{3}}{3}\) ∈ (1, 3)
Hence there are more than one c ∈ (1, 3) s.t.f’ (c) = 0.

(iv) Given f(x) = \(\sqrt{4-x^2}\) …………..(1)
Clearly D_F = [- 2, 2].
So f(x) is clearly continuous in its domain i.e. [- 2, 2]
Also,f’ (x) = \(\frac{1}{2 \sqrt{4-x^2}}\) (- 2x)
= \(\frac{-x}{\sqrt{4-x^2}}\)
Clearly f’ (x) exists in (- 2, 2).
So f(x) is clearly derivable in (- 2, 2).
Also f(- 2) – 0 = F (2)
So all the three conditions of Rolle’s theorem are satisfied.
So 3 atleast one real number c ∈ (- 2, 2) s.t f’ (C) = 0
Now, f’ (c) = 0
\(\frac{-c}{\sqrt{4-c^2}}\) = 0
c = 0 ∈ (- 2, 2)
So there exists a real number c ∈ (- 2, 2) s.t f’(c) = 0
Hence Rolle’s Theorem verified and c = 0.

Question 3.
(i) f(x) = cos 2x in [- \(\frac{\pi}{4}\), \(\frac{\pi}{4}\)]
(ii) f(x) = sin x – 1 in [\(\frac{\pi}{2}\), \(\frac{5 \pi}{2}\)]
Solution:
(i) Given f(x) = cos 2x in [- \(\frac{\pi}{4}\), \(\frac{\pi}{4}\)]
Since cosine function is continuous and derivable everywhere in its domain.
∴ f(x) is cintinuous in [- \(\frac{\pi}{4}\), \(\frac{\pi}{4}\)] and derivable in (- \(\frac{\pi}{4}\), \(\frac{\pi}{4}\))
also f(- \(\frac{\pi}{4}\)) = 0 = f(\(\frac{\pi}{4}\))
∴ all the three conditions of Roll’s theorem satisfied.
∃ atleast one real number c ∈ (- \(\frac{\pi}{4}\), \(\frac{\pi}{4}\))
s.t. f'(c) = 0 i.e. – 2 sin 2c = 0
⇒ sin 2c = 0
⇒ c = 0 ∈ (- \(\frac{\pi}{4}\), \(\frac{\pi}{4}\))
Hence Roll’s theorem is verified and c = 0.

(ii) Since sine function and constant functions are continuous and differentiable everywhere.
∴ f(x) is continuous in [\(\frac{\pi}{2}\), \(\frac{5 \pi}{2}\)]
and derivable in (\(\frac{\pi}{2}\), \(\frac{5 \pi}{2}\)).
Also f(\(\frac{\pi}{2}\)) = 1 – 1 = o
and f(\(\frac{5 \pi}{2}\)) = sin \(\frac{5 \pi}{2}\) – 1
= 1 – 1 = 0
∴ All the three conditions of Rolle’s theorem are satisfied.
∴ ∃ atleast one real no c ∈ (\(\frac{\pi}{2}\), \(\frac{5 \pi}{2}\))
s.t. f'(c) = 0
i.e. cos c = 0
⇒ c = nπ + \(\frac{\pi}{2}\) ∀ n ∈ I
Thus, c = \(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\), \(\frac{5 \pi}{2}\)
But c = – \(\frac{\pi}{2}\), \(\frac{\pi}{2}\) ∉ (\(\frac{\pi}{2}\), \(\frac{5 \pi}{2}\))
∴ c = \(\frac{3 \pi}{2}\) [\(\frac{\pi}{2}\), \(\frac{5 \pi}{2}\)]
Hence Rolls theorem is verified and
c = \(\frac{3 \pi}{2}\).

Question 4.
(i) f(x) = sin 2x in [0, \(\frac{\pi}{2}\)] (NCERT Exampler)
(ii) f(x) = sin 3x in [0, π].
Solution:
(i) Given f(x) = sin 2x in [0, \(\frac{\pi}{2}\)]
Since sin function is continuous and derivable everywhere
∴ f(x) is continuous in [0, \(\frac{\pi}{2}\)] and derivable in (0, \(\frac{\pi}{2}\))
Also, f(0) = 0 = f(\(\frac{\pi}{2}\))
∴ All the three conditions of Roll’s theorem are sarisfied.
∴ ∃ atleast one real no. c ∈ (0, \(\frac{\pi}{2}\)) s.t. f'(c) = 0
i.e. 2 cos 2c = 0
⇒ cos 2c = 0
⇒ 2c = \(\frac{\pi}{2}\)
⇒ c = \(\frac{\pi}{4}\) ∈ (0, \(\frac{\pi}{2}\))
Hence Roll’s theorem is verified and c = \(\frac{\pi}{4}\).

(ii) Given f(x) = sin 3x
since sin 3x trigonometric sine function, is continuous and differentiable everywhere.
∴ f(x) is continuous on [0, π] and f (x) is differentiable on (0, π).
f(0) = 0 ; f(π) = sin 3π = 0
∴ f(0) = f(π)
Thus, all the three conditions of Rolle’s theorem arc satisfied.
Now we want to show that ∃ atleast one real number c ∈ (0, π) s.t. f’ (c) = 0
we have, f(x) = sin 3x
∴ f’ (x) = 3 cos 3x
Now f’(c) = 0
⇒ 3 cos 3c = 0
⇒ 3c = \(\frac{\pi}{2}\)
⇒ c = \(\frac{\pi}{6}\)
i.e. c = π/6 ∈ (0, π) s.t.f’ (c) = 0
Hence Rolle’s theorem verified.

Question 5.
(i) f(x) = e^x sin x on [0, π] (ISC 2005)
(ii) f(x) = e^x cos x on [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
(iii) f(x) = e^-x sin x in [0, π]
(iv) f(x) = e^2x (sin 2x – cos 2x) in [\(\frac{\pi}{8}\), \(\frac{5 \pi}{8}\)]
Solution:
(i) Given f(x) = e^x sinx
since, e^x and sin x are dif1erentiable and continuous everywhere.
Therefore, product of two functions
i.e. f(x) = e^x sin x is continuous on [0, π] and differentiable on (0, π).
Now f(0) = 0;
f(π) = e^π × 0 = 0
∴ f(0) = f(c)
So, all the three conditions of Rolle’s theorem are satisfied.
Now we want to show that, ∃ atleast one real number
c ∈ (0, π) such that f’ (c) = 0.
We have f (x) = e^x sin x
∴ f'(x) = e^x cos x + sin x e^x
= e^x (cos x + sin x)
Now f’ (c) = 0
⇒ e^c (cos c + sin c) = 0
⇒ cos c + sin c = 0 [∵ e^c > 0]
⇒ tan c = – 1
= tan \(\frac{3 \pi}{4}\)
⇒ c = \(\frac{3 \pi}{4}\) ∈ (0, π) such that f’ (c) = 0
Hence Rolle’s theorem verified.

(ii) Given f(x) = e^x cos x in [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
Since cosine and exponential function are continuous everywhere.
Also product of two continuous functions is continuous.
∴ f (x) is continuous in [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)] and derivable in (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)).
Also f’(x) = e^x cos x – e^x sin x
Also, f(- \(\frac{\pi}{2}\)) = 0
= f(\(\frac{\pi}{2}\))
Therefore all the three conditions of Rolle’s theorem are satisfied.
∴∃ atleast are real numbers c ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
s.t. f’(c) = 0, i.e. e^c (cos c – sin c) = 0
⇒ cos c – sin c = 0
[∵ e^c ≠ 0]
⇒ tan c = 1
⇒ c = \(\frac{\pi}{4}\) ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
∴ Rolle’s theorem is verified.

(iii) Given f(x) = e^-x sin x
Clearly e^-x is continuous and derivable everywhere and sin x is also continuous and differentiable in its domain.
Also product of two continuous functions is continuous.
Thus f(x) be continuous in [0, π]
and f(x) be derivable in (0, π).
Also f(0) = e^-0 sin 0 = 0;
f(π) = e^-π sin π
= e^-π × 0 = 0
∴ f(0) = f(π)
Thus, all the three conditions of Rolle’s theorem are satisfied.
So ∃ atleast one real number c ∈ (0, π) such that f’ (c) = 0
Diff. eqn. (1) both sides w.r.t. x, we have
⇒ f'(x) = e^-x cos x + sin x e^{– x} (- 1)
⇒ f’ (x) = e^{– x} (cos x – sin x)
Now f’ (c) = 0
⇒ e^{– c} (cos c – sin c) = 0
⇒ cos c – sin c = 0 (∵ e^{– c} > 0)
⇒ tan c = 1
= tan \(\frac{\pi}{4}\)
⇒ c = nπ + \(\frac{\pi}{4}\) ∀ n ∈ I
but c ∈ (0, π)
∴ c = \(\frac{\pi}{4}\)
Thus there exists \(\frac{\pi}{4}\) ∈ (0, π) s.t. f'(c) = 0
i.e. f'(\(\frac{\pi}{4}\)) = 0
Hence Rolle’s theorem is verified and c = \(\frac{\pi}{4}\).

(iv) Given f(x) = e^2x (sin 2x – cos 2x)
Clearly exponential function is differentiable everywhere and sine, cosine functions are differentiable everywhere in its domain,
f’(x) = e^2x (2 cos 2x + 2 sin 2x) + (sin 2x – cos 2x) e^2x . 2
= e^2x (4 sin 2x) which exists ∀ x ∈ R
Thus f(x) is continuous in [\(\frac{\pi}{8}\), \(\frac{5 \pi}{8}\)] and derivable in (\(\frac{\pi}{8}\), \(\frac{5 \pi}{8}\))
f(\(\frac{\pi}{8}\)) = \(e^{\frac{\pi}{4}\left(\sin \frac{\pi}{4}-\cos \frac{\pi}{4}\right)}\)
= 0

f(\(\frac{5 \pi}{8}\)) = \(e^{\frac{5 \pi}{4}}\left(\sin \frac{5 \pi}{4}-\cos \frac{5 \pi}{4}\right)\)
= \(e^{\frac{5 \pi}{4}}\left[\sin \left(\pi+\frac{\pi}{4}\right)-\cos \left(\pi+\frac{\pi}{4}\right)\right]\)
= \(e^{\frac{5 \pi}{4}}\left[-\sin \frac{\pi}{4}+\cos \frac{\pi}{4}\right]\)
= 0

Thus all the three conditions of Rolle’s Theorem are satisfied so atleast one real number C ∈ s.t.f'(c) = 0
⇒ 4e^2c sin 2c = 0
⇒ sin 2c = 0 [∵ e^2c > 0]
⇒ 2c = 0 π, 2π, 3π, ………….
⇒ c = 0, \(\frac{\pi}{2}\), π, \(\frac{3 \pi}{2}\), ………….
Clearly c = \(\frac{\pi}{2}\) ∈ (\(\frac{\pi}{8}\), \(\frac{5 \pi}{8}\)).
∴ Rolle’s theorem verified.

Question 5 (old).
(iii) f(x) = e^2x (sin 2x – cos 2x) in [\(\frac{\pi}{8}\), \(\frac{5 \pi}{8}\)]. (ISC 2008)
Solution:
Given f(x) = e^2x sin x
Clearly e^x is continuous and derivable everywhere and sin x is also continuous and differentiable in its domain.
Also product of two continuous functions is Continuous.
Thus f(x) be continuous in [0, π]
and f(x) be derivable in (0, π).
Also f(0) = e^-0 sin 0 = 0;
f(π) = e^{– π} sin π
= e^{– π} × 0 = 0
∴ f(0) = f( π)
Thus, all the three conditions of Rolle’s theorem are satisfied.
So ∃ atleast are real number c ∈ (0, π) such that f’ (c) = 0
Diff. eqn. (1) both sides w.r.t. x, we have
f’(x) = e^-x cos x + sin x e^-x (- 1)
⇒ f'(x) = e^-x (cos x – sin x)
Now f'(c) = 0
⇒ e^{– c} (cos c – sin c) = 0
⇒ cos c – sin c = 0 (∵ e^{– c} > 0)
⇒ tan c = 1
= tan \(\frac{\pi}{4}\)
⇒ c = nπ + \(\frac{\pi}{4}\) ∀ n ∈ I
but c ∈ (0, π)
∴ c = \(\frac{\pi}{4}\)
Thus there exists \(\frac{\pi}{4}\) ∈ (0, π) s.t. f'(c) = 0
i.e. f'(\(\frac{\pi}{4}\)) = 0
Hence Rolle’s theorem is verified and c = \(\frac{\pi}{4}\).

Question 6.
Apply Rolle’s theorem to find point (or points) on the following curves where the tangent is parallel to x-axis :
(i) y = x² in [- 2, 2]
(ii) y = – 1 + cos x on (0, 2π) (NCERT Exemplar)
Solution:
(i) Given y = x² in [- 2, 2]
Since f (x) is polynomial in x
∴ it is continuous and derivable everywhere
∴ f(x) is continuous in [- 2, 2] and derivable in (- 2, 2).
Also f(- 2) = 4 = f(2)
Hence all the three conditions of Rolle’s theorem are satisfied.
∴ ∃ atleast one real no. c ∈ (- 2, 2) s.t.f’(c) = 0
i.e. c = 0 ∈ (- 2, 2).
When x = 0,
y = 0²= 0
∴ (0, 0) be the required point at which tangent is || to x-axis.
[By Geometrical interpretation of Rolle’s Theorem]

(ii) Let f(x) = y = – 1 + cos x
Clearly f(x) is continuous in [0, 2π] and
derivable in (0, 2π).
Since curve and constant functions are continuous and everywhere differentiable
Now f(0) = – 1 + 1 = 0;
f(2π) = – 1 + cos 2π
= – 1 + 1 = 0
∴ f(0) = f (2π)
Thus, all the three conditions of Rolle’s theorem are satisfied.
So ∃ atleast one real number c ∈ (0, 2π) s.t.f’ (c) = 0
Diff. eqn. (1) w.r.t. x, we get
f’ (x) = – sin x
Now f’(c) = 0
⇒ – sin c = 0
But π ∈ (0, 2π)
∴ c = π ∈ (0, 2π)
So there exists a real number c = π ∈ (0, 2π)
s.t f’ (π) = 0 i.e. at which the tangent is || to x-axis
When x = π
∴ from (1);
y = – 1 + cos π
= – 1 – 1
= – 2
Hence, the required point on given curve at which the tangent is parallel to x-axis be (π, – 2).

Question 7.
Verify the conditions of Rolle’s Theorem for the following function :
f(x) = log (x² + 2) – log 3 on [- 1, 1].
Find a point in the given interval where the tangent to the curve is parallel to x axis. (ISC 2016)
Solution:
Let f(x) = log (x² + 2) – log 3
Clearly f(x) is continuous in [- 1, 1].
Now f'(x) = \(\frac{2 x}{x^2+2}\) exists ∀ x ∈ (- 1, 1)
∴ f is derivable in (- 1, 1).
further f(- 1) = log 3 – log 3 = 0
and f(1) = log 3 – log 3 = 0
∴ f(- 1) = f (1) = 0
Thus all the three conditions of Rolle’s theorem are satisfied.
∴∃ atleast one real number c ∈ (- 1, 1) s.t.f’(c) = 0
i.e. \(\frac{2 c}{c^2+2}\) = 0
⇒ c = 0 ∈ (- 1, 1)
Hence Rolle’s theorem is verified.

Question 8.
If Rolle’s theorem holds for the function f(x) = x³ + ax² + bx in [1, 2] at the point x = \(\frac{4}{3}\), then find the values of a and b. (ISC 2009)
Solution:
Given f(x) = x³ + ax² + bx …………..(1)
Since it is given that, Rolle’s theorem holds for the function f(x) in [1, 2].
∴ f(1) = f(2)
⇒ 1 + a + b = 8 + 4a + 2b
⇒ 3a + b = – 7 ………….(2)
Also ∃ atleast one rai number x = c ∈ (1, 2) s.t f’(c) = 0
Also it is given that, Rolle’s theorem holds for f(x) is [1, 2] at the point x = \(\frac{4}{3}\).
∴ f'(\(\frac{4}{3}\)) = 0
Diff. (1) w.r.t. x; we have
f’(x) = 3x² + 2ax + b
Now f'(\(\frac{4}{3}\) = 0
0 = 3 × \(\frac{16}{9}\) + 2a × \(\frac{4}{3}\) + b
8a + 3b = – 16 …………(3)
On solving eqn. (2) and eqn. (3); we have a = – 5 ; b = 8.

Question 9.
Examine if Rolle’s theorcm is applicable to the function f(x) = |x| for x ∈ [- 2, 2] What can you say about the converse of Rolle’s theorem ? (NCERT)
Solution:
at x = 2
\(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) [x] = 2 and
\(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) [x]
= (2 – 1) = 1
∴ f(x) is not continuous at x = 2 ∈ [- 2, 2]
Hence Rolle’s theorem is not applicable.

Question 10.
What can you say about the applicability of Rolle’s theorem for the following functions in the indicated intervals ?
(i) f(x) = x^1/3 in [- 1, 1]
(ii) f(x) = x^2/3 in [- 2, 2]
(iii) f(x) = 2 + (x – 1)^2/3 in [0, 2]
(iv) f(x) = 1 + |x – 2| in [0, 4]
(v) f(x) = tan x in [0, π]
(vi) f(x) = sec x in [0, 2π]
(vii) f(x) = \(\frac{x(x-2)}{x-1}\) in [0, 2]
(viii) f(x) = x² + 1 in [- 1, 2]
Solution:
(i) Given f(x) = x^1/3
∴ f'(x) = \(\frac{1}{3} \frac{1}{x^{2 / 3}}\) does not exists at x = 0 ∈ (- 1, 1)
∴ function f(x) is not derivable in (- 1, 1)
Hence Rolle’s theorem is not applicable.

(ii) Given f(x) = x^2/3
∴ f'(x) = \(\frac{2}{3}\) x^-1/3
= \(\frac{2}{3 x^{1 / 3}}\)
which does not exists at x = 0 ∈ (- 1, 1)
∴ function f(x) is not derivable in (- 1, 1)
Hence Rolle’s theorem is not verified.

(iii) Given f(x) = 2 + (x -1)^2/3 in [0, 2]
∴ f'(x) = \(\frac{2}{3}\) (x – 1)^-1/3
= \(\frac{2}{3(x-1)^{1 / 3}}\)
which does not exists at x = 1.
∴ function f(x) is not derivable in x = 1 ∈ (0, 2).
Hence Rolle’s theorem is not applicable.

(iv) Given f(x) = 1 + |x – 2|
Clearly f(x) is continuous in [0, 4]

Differentiability at x = 2
Lf'(2) = \(\ {Lt}_{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2}\)

[∵ as x → 2⁺
⇒ x > 2
⇒ x – 2 > 0
∴ |x – 2| = x – 2]
∴ Lf'(2) ≠ Rf'(2)
Thus f is not derivable or differentiable at x = 2.
Also, f'(x) = \(\frac{x-2}{|x-2|}\) ; x ≠ 2
Thus derivable of f(x) does not exists at x = 2.
∴ f(x) is not derivable at x = 2 ∈ (0, 4)
Hence f(x) is not derivable in (0, 4).
So condition (ii) of Rolle’s theorem is not satisfied.
Thus Rolle’s theorem is not applicable for f(x) = 1 + |x – 2| in [0, 4].

(v) Given f(x) = tan x in [0, 4]
Since f'(x) = sec² x does not exists at x = \(\frac{\pi}{2}\)
Thus f(x) is not differentiable at x = \(\frac{\pi}{2}\) ∈ [0, π]
∴ f is not derivable in (0, π)
Hence Rolle’s theorem is not applicable.

(vi) Given, f(x) = sec x
Clearly at x = \(\frac{\pi}{2}\) ∈ [0, 2π]
f(\(\frac{\pi}{2}\)) = sec \(\frac{\pi}{2}\) which does not exists
as \(\underset{x \rightarrow \frac{\pi^{-}}{2}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow \frac{\pi^{-}}{2}}{\mathrm{Lt}}\) sec x → + ∞
and \(\underset{x \rightarrow \frac{\pi^{+}}{2}}{\mathrm{Lt}}\) f(x) = latex]\underset{x \rightarrow \frac{\pi^{+}}{2}}{\mathrm{Lt}}[/latex] sec x → – ∞
Hence f(x) is continuous at x = \(\frac{\pi}{2}\) ∈ [0, 2π]
∴ f(x) is discontinuous in [0, 2π]
Hence the condition (i) of Roll’s theorem is not satisfied.
Thus, Rolle’s theorem is not appicable to functionf(x) = sec x in [0, 2π].

(vii) We have ;
f(x) = \(\frac{x(x-2)}{x-1}\)
Now f'(x) = \(\frac{(x-1)(2 x-2)-\left(x^2-2 x\right)}{(x-1)^2}\)
= \(\frac{2\left(x^2-2 x+1\right)-\left(x^2-2 x\right)}{(x-1)^2}\)
= \(\frac{x^2-2 x+2}{(x-1)^2}\)
Clearly f’( x) does not exists at x = 1 ∈ (0, 2)
Hence Rolle’s theorem is not applicable.

(viii) Given f(x) = x² + 1
Clearly f(x) be a polynomial function
∴ f(x) be continuous in [- 1, 2] and derivable in (- 1, 2).
But f(- 1) = 1 + 1 = 2;
f(2) = 2²+ 1 = 5
∴ f(- 1) ≠ f(2)
So condition (iii) of Rolle’s theorem is not satisfied.
Hence Rolle’s theorem is not applicable to function f(x) = 1 + x² in [- 1, 2].

Access to comprehensive Class 12 ISC Maths Solutions Chapter 5 Continuity and Differentiability Ex 5.11 Solutions encourages independent learning.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.11

Question 1.
Find the second order derivatives of the following functions:

(i) x²⁰ (NCERT)
(ii) x² + 3x + 2 (NCERT)
(iii) x³ – 5x² + 3x + 4
(iv) x³ + tan x (NCERT)
(v) log x (NCERT)
(vi) tan^-1 x (NCERT)
Solution:
(i) Let y = x²⁰ ;
Diff. w.r.t. x, we have
\(\frac{d y}{d x}\) = 20 x¹⁹ ;
Again Diff. both sides w.r.t. x, we get
∴ \(\frac{d^2 y}{d x^2}\) = 380 x¹⁸

(ii) Let y = x² + 3x + 2
\(\frac{d y}{d x}\) = 2x + 3
and \(\frac{d^2 y}{d x^2}\) = 2

(iii) Let y = x³ – 5x² + 3x + 4
Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = 3x² – 10x + 3
Diff. again w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = 6x – 10

(iv) Let y = x³ + tan x
∴ \(\frac{d y}{d x}\) = 3x² + sec² x
and \(\frac{d^2 y}{d x^2}\) = 6x + 2 sec² x tan x

(v) Let y = log x,
Diff. both sides w.r.t. x, we have
∴ \(\frac{d y}{d x}\) = \(\frac{1}{x}\)
and \(\frac{d^2 y}{d x^2}\) = – \(\frac{1}{x^2}\) ; x ≠ 0.

(vi) Let y = tan^-1 x
Diff. both sides w.r.t. x, we have
∴ \(\frac{d y}{d x}\) = \(\frac{1}{1+x^2}\)
Diff. again both sides w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = – \(\frac{1}{\left(1+x^2\right)^2} \times 2 x\)
= \(\frac{-2 x}{\left(1+x^2\right)^2}\)

Question 2.
(i) If y = log (x – 2), x > 2, find \(\frac{d^2 y}{d x^2}\).
(ii) If y = cot x, find \(\frac{d^2 y}{d x^2}\) at x = \(\frac{\pi}{4}\).
(iii) If x = at² and y = 2at, then find \(\frac{d^2 y}{d x^2}\).
Solution:
(i) Given y = log (x – 2), x > 2
Differentiating w.r.t. x ; we have
\(\frac{d y}{d x}=\frac{1}{x-2}\) ;
again differentiating w.r.t. x, we have
\(\frac{d^2 y}{d x^2}=-\frac{1}{(x-2)^2}\).

(ii) Given y = cot x
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = – cosec² x
Differentiating both sides w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = – 2 cosec x \(\frac{d}{d x}\) (cosec x)
= 2 cosec² x cot x
at x = \(\frac{\pi}{4}\),
\(\frac{d^2 y}{d x^2}\) = (2√2)² × 1 = 4

(iii) Given x = at² ………….(1)
and y = 2at …………(2)
diff. eqns. (1) and (2) w.r.t. t ; we have
\(\frac{d x}{d t}\) = 2at ;
\(\frac{d y}{d t}\) = 2a
∴ \(\frac{d y}{d x}\) = \(\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{2 a}{2 a t}\)
= \(\frac{1}{t}\)
Diff. both sides w.r.t. x ; we have
\(\frac{d^2 y}{d x^2}=\frac{d}{d t}\left(\frac{1}{t}\right) \frac{d t}{d x}\)
= \(-\frac{1}{t^2} \frac{1}{2 a t}\)
= – \(\frac{1}{2 a t^3}\)

Question 3.
Find the second order derivative of the following functions:
(i) sin^-1 x (NCERT)
(ii) x cos x (NCERT)
(iii) x sin 2x
(iv) e^x sin 5x (NCERT)
(v) e^2x sin 3x
(vi) log (log x) (NCERT)
(vii) \(\frac{\log x}{x}\)
(viii) x² log |cos x|
(ix) \(\frac{2 x+1}{2 x+3}\)
Solution:
(i) Let y = sin^-1 x ;
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}=\frac{1}{\sqrt{1-x^2}}\)
Again differentiating w.r.t. x ; we have
\(\frac{d^2 y}{d x^2}\) = – \(\frac{1}{2}\) (1 – x²)^{–\(\frac{1}{2}\)-1} (- 2x)
= \(\frac{x}{\left(1-x^2\right)^{3 / 2}}\)

(ii) Let y = x cos x ;
Diff. both sides w.r.t. x ; we get
\(\frac{d y}{d x}\) = x \(\frac{d}{d x}\) (cos x) + cos x \(\frac{d}{d x}\) (x)
= – x sin x + cos x
Again differentiating w.r.t. x ; we get
\(\frac{d^2 y}{d x^2}\) = – \(\frac{d}{d x}\) (x sin x) + \(\frac{d}{d x}\) cos x
= – [x cos x + sin x . 1] – sin x
= – x cos x – 2 sin x

(iii) Let y = x sin 2x ;
Diff. both sides w.r.t. x ; we get
\(\frac{d y}{d x}\) = 2x cos 2x + sin 2x
Again differentiating w.r.t. x ; we have
\(\frac{d^2 y}{d x^2}\) = 2 [- 2x sin 2x + cos 2x] + 2 cos 2x
∴ \(\frac{d^2 y}{d x^2}\) = – 4x sin 2x + 4 cos 2x

(iv) Let y = e^x sin 5x ;
Differentiate both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = e^x (5 cos 5x) + sin 5x e^x
\(\frac{d^2 y}{d x^2}\) = – 25 sin 5x e^x + 5 cos 5x e^x + 5 cos 5x e^x + sin 5x e^x
∴ \(\frac{d^2 y}{d x^2}\) = 2e^x [5 cos 5x – 12 sin 5x]

(v) Let y = e^2x sin 3x
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = e^2x (3 cos 3x) + sin 3x . e^2x . 2
= e^2x (3 cos 3x + 2 sin 3x)
Again differentiating w.r.t. x ; we have
\(\frac{d^2 y}{d x^2}\) = e^2x (- 9 sin 3x + 6 cos 3x) + (3 cos 3x + 2 sin 3x) 2e^2x
= e^2x (- 5 sin 3x + 12 cos 3x)

(vi) Let y = log (log x)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{1}{\log x} \cdot \frac{1}{x}\)
Again differentiating w.r.t. x ; we have
\(\frac{d^2 y}{d x^2}\) = \(\frac{d}{d x}\) (x log x)^-1
= (- 1) (x log x)^-2 \(\frac{d}{d x}\) (x log x)
= \(\frac{-1}{(x \log x)^2}\) [x . \(\frac{1}{x}\) + log x . 1]
= \(\frac{-(1+\log x)}{(x \log x)^2}\)

(vii) Given y = \(\frac{log x}{x}\) ;
Diff. both sides w.r.t. x, we have

(viii) Let y = x² log |cos x|
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = 2x log |cos x| + \(\frac{x^2}{|\cos x|} \frac{\cos x}{|\cos x|}\) (- sin x)
= 2x log |cos x| + x² (- tan x)
[∵ |cos x| . |cos x| = |cos² x|
= |cos x|²
= cos² x]
Again differentiating w.r.t. x, we get
\(\frac{d^2 y}{d x^2}\) = \(\frac{2 x}{|\cos x|} \frac{\cos x}{|\cos x|}\) (- sin x) + log |cos x| . 2 – (x² sec² x + 2x tan x)
= – 4x tan x – x² sec² x + 2 log |cos x|

(ix) Let y = \(\frac{2 x+1}{2 x+3}\)
= \(\frac{2 x+3-2}{2 x+3}\)
= 1 – \(\frac{2}{2 x+3}\)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=0+\frac{2}{(2 x+3)^2} \times 2\)
= \(\frac{4}{(2 x+3)^2}\)
Again differentiating w.r.t. x, we get
\(\frac{d^2 y}{d x^2}\) = \(\frac{-4 \times 2}{(2 x+3)^3} \times 2\)
= \(\frac{16}{(2 x+3)^3}\)

Question 4.
Find the second derivatives of the following functions:
(i) sec ax
(ii) cot (1 – 2x) (NCERT)
(iii) sin 3x cos 5x
(iv) sin³ x (NCERT)
(v) cos (2x² – 1)
(vi) \(\sqrt{1-x^2}\). (NCERT)
Solution:
(i) Let y = sec ax
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = a sec ax tan ax
On differentiating again w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = a [a sec ax sec² ax + a tan ax sec ax tan ax]
= a² [sec³ ax + sec ax (sec² ax – 1)]
= a² sec ax (2 sec² ax – 1)

(ii) Let y = cot (1 – 2x)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = – cosec² (1 – 2x) (- 2)
Again diff. both sides w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = 2 × 2 cosec (1 – 2x) {- cosec (1 – 2x) cot (1 – 2x)} (- 2)
= 8 cot (1 – 2x) cosec² (1 – 2x)

(iii) Let y = sin 3x cos 5x
⇒ y = \(\frac{1}{2}\) (2 sin 3x cos 5x)
⇒ y = \(\frac{1}{2}\) [sin 8x + sin (- 2x)]
⇒ y = \(\frac{1}{2}\) [sin 8x – sin 2x]
diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{1}{2}\) [8 cos 8x – 2 cos x]
Again differentiating w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = \(\frac{1}{2}\) [- 64 sin 8x + 4 sin 2x]
= – 32 sin 8x + 2 sin 2x

(iv) Let y = sin³ x
= \(\frac{3 \sin x-\sin 3 x}{4}\)
[∵ sin 3x = 3 sin x – 4 sin³ x]
On differentiating w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = \(\frac{1}{4}\) [- 3 sin x + 9 sin 3x]
= \(\frac{3}{4}\) [3 sin 3x – sin x]

(v) Let y = cos (2x² – 1) ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = – sin (2x² – 1) (4x)
Again diff. both sides w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = – 4 [sin (2x² – 1) + x cos (2x² – 1) 4x]
= – 4 [sin (2x² – 1) + 4x² cos (2x² – 1)]

(vi) Let y = \(\sqrt{1-x^2}\) ;
Diff. both sides w.r.t. x, we have

Question 5.
(i) If y = cos^-1 x, find \(\frac{d^2 y}{d x^2}\) interms of y alone. (NCERT)
(ii) Find \(\frac{d^2 y}{d x^2}\) when y = log \(\left(\frac{x^2}{e^x}\right)\).
Solution:
(i) Given y = cos^-1 x ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=-\frac{1}{\sqrt{1-x^2}}\)
∴ \(\frac{d^2 y}{d x^2}\) = – (- \(\frac{1}{2}\)) (1 – x²)^-3/2
= \(\frac{-x}{\left(1-x^2\right)^{3 / 2}}\)
= – \(\frac{\cos y}{\sin ^3 y}\)
= – cot y cosec² y
[Since y = cos^-1 x
⇒ x = cos y, y ∈ [0, π]]

(ii) Given y = log \(\left(\frac{x^2}{e^2}\right)\)
= log x² – log e²
⇒ y = 2 log x – 2 log e ……….(1)
∴ \(\frac{d y}{d x}=\frac{2}{x}\)
again differentiating w.r.t. x, we get
\(\frac{d^2 y}{d x^2}=\frac{-2}{x^2}\).

Question 6.
(i) If y = cot x, prove that \(\frac{d^2 y}{d x^2}\) + 2 y \(\frac{d y}{d x}\) = 0.
(ii) If y = 5 cos x – 3 sin x, prove that \(\frac{d^2 y}{d x^2}\) + y = 0. (NCERT)
Solution:
(i) Given y = cot x ………..(1)
Diff. (1) w.r.t. x ; we get
\(\frac{d y}{d x}\) = – cosec² x
Diff. again eqn. (2) w.r.t. x ; we get
\(\frac{d^2 y}{d x^2}\) = + 2 cosec² x cot x
\(\frac{d^2 y}{d x^2}\) = 2 (- \(\frac{d y}{d x}\)) y
[using (1) and (2)]
⇒ \(\frac{d^2 y}{d x^2}\) + 2y \(\frac{d y}{d x}\) = 0

(ii) Given y = 5 cos x – 3 sin x ……….(1)
Diff. (1) w.r.t. x ; we get
\(\frac{d y}{d x}\) = – 5 sin x – 3 cos x ……..(2)
Diff. (2) w.r.t. x ; we get
\(\frac{d^2 y}{d x^2}\) = – 5 cos x + 3 sin x = – y [using (1)]
∴ \(\frac{d^2 y}{d x^2}\) + y = 0 [Hence proved]

Question 7.
(i) If y = x + tan x, prove that cos² x . \(\frac{d^2 y}{d x^2}\) – 2y + 2x = 0.
(ii) If y = tan x + sec x, prove that \(\frac{d^2 y}{d x^2}\) = \(\frac{\cos x}{(1-\sin x)^2}\). (NCERT Exampler)
Solution:
(i) Given y = x + tan x ………..(1)
Differentiate eqn. (1) w.r.t. x, we get
\(\frac{d y}{d x}\) = 1 + sec² x
= 2 + tan² x ………..(2)
Differentiate (2) w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = 2 tan x (sec² x)
⇒ cos² x \(\frac{d^2 y}{d x^2}\) = 2 tan x
⇒ cos² x \(\frac{d^2 y}{d x^2}\) = 2 (y – x) [using eqn. (1)]
⇒ cos² x \(\frac{d^2 y}{d x^2}\) – 2y + 2x = 0 [Hence proved]

(ii) Let y = sec x + tan x …………(1)
Differentiate eqn. (1) both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = sec x tan x + sec² x
= sec x . y ……….(2) [using (1)]
Again diff. both sides w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = sec x \(\frac{d y}{d x}\) + y sec x tan x
= sec x ((sec x) y) + y sec x tan x [using (2)]
⇒ \(\frac{d^2 y}{d x^2}\) = y sec x [sec x + tan x]
= y² sec x [using (1)]
⇒ cos x \(\frac{d^2 y}{d x^2}\) = y² [Hence proved]

Question 8.
(i) If y = sec x – tan x, prove that cos x . \(\frac{d^2 y}{d x^2}\) = y².
(ii) If y = x + cot x, prove that sin² x . \(\frac{d^2 y}{d x^2}\) – 2y + 2x = 0
Solution:
(i) Given y = sec x – tan x ………..(1)
Differentiate eqn. (1) both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = sec x tan x – sec² x
= – sec x (sec x – tan x)
= – (sec x) y …………….(2) [using (1)]
Diff. eqn. (2) w.r.t. x, we get
\(\frac{d^2 y}{d x^2}\) = – [sec x \(\frac{d y}{d x}\) + y sec x tan x]
= – [- y sec² x + y sec tan x] [using (2)]
⇒ \(\frac{d^2 y}{d x^2}\) = + (sec x) y [sec x – tan x]
= y² sec x
⇒ cos x \(\frac{d^2 y}{d x^2}\) = y² [hence proved]

(ii) Given y = x + cot x
Diff. (1) w.r.t. x, we have
\(\frac{d y}{d x}\) = 1 – cosec² x
again diff. both sides w.r.t. x, we get
⇒ \(\frac{d^2 y}{d x^2}\) = – 2 cosec x (- cosec x cot x)
= 2 cosec² x cot x
⇒ sin² x \(\frac{d^2 y}{d x^2}\) = 2 cot x
= 2 (y – x)
∴ sin² x \(\frac{d^2 y}{d x^2}\) – 2y + 2x = 0.

Question 9.
(i) If y = ae^mx + be^-mx, prove that y₂ – m² y = 0.
(ii) If y = ae^2x + be^-x, prove that \(\frac{d^2 y}{d x^2}\) – \(\frac{d y}{d x}\) – 2y = 0
Solution:
(i) Given y = ae^mx + be^-mx ………..(1)
Diff. both sides of eqn. (1) w.r.t. x, we have
\(\frac{d y}{d x}\) = ae^mx . m + be^{– mx} . (- m)
Diff. again both sides w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = m² ae^mx + bm² e^-mx
= m² [ae^mx + be^-mx]
= m² y
⇒ \(\frac{d^2 y}{d x^2}\) – m²y = 0
i.e. y₂ – m² y = 0

(ii) Given y = ae^2x + be^-x
Diff. both sides of eqn. (1) w.r.t. x, we have
\(\frac{d y}{d x}\) = 2ae^2x – be^-x
Differentiate eqn. (2) both sides w.r.t. x ; we have
\(\frac{d^2 y}{d x^2}\) = 4ae^2x + be^-x
L.H.S. = \(\frac{d^2 y}{d x^2}\) – \(\frac{d y}{d x}\)
= 4ae^2x + be^-x – (2ae^2x – be^-x) – 2 (ae^2x + be^-x)
= e^2x (4a – 2a – 2a) + e^-x (b + b – 2b)
= 0 . e^2x + 0 . e^-x
= 0
= R.H.S.
Thus \(\frac{d^2 y}{d x^2}\) – \(\frac{d y}{d x}\) – 2y = 0.

Question 9 (old).
(ii) If y = 500 e^7x + 600 e^-7x, prove that \(\frac{d^2 y}{d x^2}\) = 49 y. (NCERT)
Solution:
Given y = 500 e^7x + 600 e^-7x …………….(1)
Diff. (1) w.r.t. x , we get
\(\frac{d y}{d x}\) = 3500 e^7x – 4200 e^-7x
Again differentiating both sides w.r.t. x, we get
\(\frac{d^2 y}{d x^2}\) = 3500 × 7 e^7x + 4200 × 7e ^-7x
= 49 [500 e^7x + 600 e^-7x]
= 49 y [using eqm. (1)]

Question 10.
If y = A cos nx + B sin nx, show that \(\frac{d^2 y}{d x^2}\) + n² y = 0
Solution:
Given y = A cos nx + B sin nx …………….(1)
Diff. both sides of eqn. (1) w.r.t. x, we have
\(\frac{d y}{d x}\) = – A sin nx + Bn cos nx
Again diff. w.r.t. x ; we get
\(\frac{d^2 y}{d x^2}\) = – An² cos nx – Bn² sin nx
= – n² y [using (1)]
∴ \(\frac{d^2 y}{d x^2}\) + n² y = 0 [Hence proved]

Question 10 (old).
(i) If y = A sin x + B cos x, prove that \(\frac{d^2 y}{d x^2}\) + y = 0 (NCERT)
Solution:
Given y = A sin x + B cos x …………(1)
Diff. (1) w.r.t. x, we get
\(\frac{d y}{d x}\) = A cos x – B sin x ……….(2)
Diff. (2) w.r.t. x ; we get
\(\frac{d^2 y}{d x^2}\) = – A sin x – B cos x = – y
∴ \(\frac{d^2 y}{d x^2}\) + y = 0.

Question 11.
(i) If y = tan^-1 x, prove that (1 + x²) \(\frac{d^2 y}{d x^2}\) + 2x \(\frac{d y}{d x}\) = 0
(ii) If y = sin^-1 x, prove that (1 – x²) \(\frac{d^2 y}{d x^2}\) – x \(\frac{d y}{d x}\) = 0. (NCERT)
Solution:
(i) Given, y = tan^-1 x ………..(1)
\(\frac{d y}{d x}\) = \(\frac{1}{1+x^2}\)
⇒ (1 + x²) \(\frac{d y}{d x}\) = 1
again differentiate (1) w.r.t. x, we get
(1 + x²) \(\frac{d^2 y}{d x^2}\) + 2x \(\frac{d y}{d x}\) = 0

(ii) Given y = sin^-1 x ;
Diff. w.r.t. x, we get
\(\frac{d y}{d x}=\frac{1}{\sqrt{1-x^2}}\)
\(\sqrt{1-x^2} \frac{d y}{d x}\) = 1 ;
Again diff. both sides w.r.t. x ; we have
\(\sqrt{1-x^2} \frac{d^2 y}{d x^2}\) + \(\frac{d y}{d x} \frac{1}{2}\) (1 – x²)^-1/2 (- 2x) = 0
⇒ \(\sqrt{1-x^2} \frac{d^2 y}{d x^2}-\frac{x}{\sqrt{1-x^2}} \frac{d y}{d x}\) = 0
⇒ (1 – x²) \(\frac{d^2 y}{d x^2}\) – x . \(\frac{d y}{d x}\) = 0 [Hence proved].

Question 12.
(i) If y = sin (log x), prove that x² \(\frac{d^2 y}{d x^2}\) + x \(\frac{d y}{d x}\) + y = 0
(ii) If y = 2 cos (log x) + 3 sin (log x), prove that x² \(\frac{d^2 y}{d x^2}\) + x \(\frac{d y}{d x}\) + y = 0
Solution:
(i) Given y = sin (log x) ………..(1)
Diff. (1) both sides w.r.t. x ; we get
\(\frac{d y}{d x}\) = cos (log x) . \(\frac{1}{x}\)
⇒ x \(\frac{d y}{d x}\) = cos (log x)
Again differentiating both sides w.r.t. x ; we get
x \(\frac{d^2 y}{d x^2}\) + \(\frac{d y}{d x}\) . 1 = – sin (log x) . \(\frac{1}{x}\)
⇒ x² \(\frac{d^2 y}{d x^2}\) + x \(\frac{d y}{d x}\) = – y [using eqn. (1)]
⇒ x² \(\frac{d^2 y}{d x^2}\) + x \(\frac{d y}{d x}\) + y = 0

(ii) Given y = 2 cos (log x) + 3 sin (log x) ……..(1)
diff. eqn. (1) w.r.t. x ; we have
\(\frac{d y}{d x}\) = – 2 sin (log x) . \(\frac{1}{x}\) + 3 cos (log x) . \(\frac{1}{x}\)
⇒ x \(\frac{d y}{d x}\) = – 2 sin (log x) + 3 cos (log x) ……….(2)
Diff. eqn. (2) both sides w.r.t. x ;
x \(\frac{d^2 y}{d x^2}\) + \(\frac{d y}{d x}\) . 1 = – 2 cos (log x) . \(\frac{1}{x}\) – 3 sin (log x) .\(\frac{1}{x}\)
⇒ x² \(\frac{d^2 y}{d x^2}\) + x \(\frac{d y}{d x}\) = – [2 cos (log x) + 3 sin (log x)] [using (1)]
= – y
⇒ x² \(\frac{d^2 y}{d x^2}\) + x \(\frac{d y}{d x}\) + y = 0.

Question 12 (old).
If y = 3 cos (log x) + 4 sin (log x), prove that x² \(\frac{d^2 y}{d x^2}\) + x \(\frac{d y}{d x}\) + y = 0.
Solution:
Given y = 3 cos (log x) + 4 sin (log x) ………….(1)
Differentiate both sides w.r.t. x ; we get
\(\frac{d y}{d x}\) = y₁
= – 3 sin (log x) . \(\frac{1}{x}\) + 4 cos (log x) . \(\frac{1}{x}\)
∴ xy₁ = – 3 sin (log x) + 4 cos (log x)
Differentiate (2) both sides w.r.t. x, we get
xy₂ + y₁ = – 3 cos (log x) . \(\frac{1}{x}\) – 4 sin (log x) . \(\frac{1}{x}\)
⇒ x (xy₂ + y₁) = – y [using (1)]
⇒ x²y₂ + xy₁ + y = 0 [Hence proved]

Question 13.
If y = x cos x, prove that x² \(\frac{d^2 y}{d x^2}\) – 2x \(\frac{d y}{d x}\) + (2 + x²) y = 0
Solution:
Given y = x cos x ………….(1)
Diff. (1) both sides w.r.t. x ; we get
\(\frac{d y}{d x}\) = cos x – x sin x ………..(2)
Diff. eqn. (2) w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = – sin x – [sin x + x cos x]
= – x cos x – 2 sin x ……….(3)
∴ L.H.S. = x² \(\frac{d^2 y}{d x^2}\) – 2x \(\frac{d y}{d x}\) + (2 + x²) y
= x² [- x cos x – 2 sin x] – 2x [cos x – x sin x] + (2 + x²) x cos x [using (1), (2) and (3)]
= – x³ cos x – 2x² sin x – 2x cos x + 2x² sin x + 2x cos x + x³ cos x
= 0
= R.H.S.
Thus, x² \(\frac{d^2 y}{d x^2}\) – 2x \(\frac{d y}{d x}\) + (2 + x²) y = 0.

Question 14.
(i) If y = (cos^-1 x)², prove that (1 – x²) y₂ – xy₁ = 2.
(ii) If y = (tan^-1 x)², prove that (x² + 1) \(\frac{d^2 y}{d x^2}\) + 2x (x² + 1) \(\frac{d y}{d x}\) = 2.
(iii) If y = (sec^-1 x)², x > 1, show that x² (x² – 1) \(\frac{d^2 y}{d x^2}\) + (2x³ – x) \(\frac{d y}{d x}\) – 2 = 0
Solution:
(i) Given y = (cos^-1 x)²
Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = y₁
= 2 cos^-1 x \(\left(\frac{-1}{\sqrt{1-x^2}}\right)\)
⇒ \(\sqrt{1-x^2}\) y₁ = – 2 cos^-1 x …………(1)
Diff. eqn. (1) both sides w.r.t. x, we get
\(\sqrt{1-x^2}\) y₂ + y₁ \(\frac{1}{2}\) (1 – x²)^{–\(\frac{1}{2}\)} (- 2x)
= + \(\frac{2}{\sqrt{1-x^2}}\)
⇒ \(\sqrt{1-x^2}\) y₂ – \(\frac{x y_1}{\sqrt{1-x^2}}\) = \(\frac{2}{\sqrt{1-x^2}}\)
multiplying both sides by \(\sqrt{1-x^2}\) ; we have
(1 – x²) y₂ – xy₁ = 2

(ii) Given, y = (tan^-1 x)² ;
Differentiate both sides w.r.t. x, we get
y₁ = 2 tan^-1 x . \(\frac{1}{1+x^2}\)
⇒ (1 + x²) y₁ = 2 tan^-1 x ……….(1)
agaion differentiate both sides w.r.t. x, we get
(1 + x²) y₂ + 2x (1 + x²) y₁ = 2.

(iii) Given y = (sec^-1 x)²
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}\) = 2 sec^-1 x \(\frac{d}{d x}\) sec^-1 x

Question 15.
(i) If y = log (x + \(\sqrt{x^2+1}\)), prove that (x² + 1) \(\frac{d^2 y}{d x^2}\) + x \(\frac{d y}{d x}\) = 0.
(ii) If y = log (x + \(\sqrt{x^2+a^2}\)), prove that (x² + a²) y₂ + xy₁ = 0
Solution:
(i) Given y = log (x + \(\sqrt{x^2+1}\)) ………..(1)
Diff. (1) w.r.t. x, we get
\(\frac{d y}{d x}=\frac{1}{x+\sqrt{x^2+1}}\left[1+\frac{1}{2}\left(x^2+1\right)^{-1 / 2} 2 x\right]\)
= \(\frac{1}{x+\sqrt{x^2+1}}\left[\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}\right]\)
= \(\frac{1}{\sqrt{x^2+1}}\)
⇒ \(\sqrt{x^2+1} \frac{d y}{d x}\) ……….(2)
diff. eqn. (2) w.r.t. x, we get
\(\sqrt{x^2+1} \frac{d^2 y}{d x^2}+\frac{d y}{d x} \cdot \frac{1}{2}\) (x² + 1)^-1/2 . 2x = 0
\(\sqrt{x^2+1} \frac{d^2 y}{d x^2}+\frac{x}{\sqrt{x^2+1}} \frac{d y}{d x}\) = 0 ;
Multiplying both sides by \(\sqrt{x^2+1}\) ; we have
⇒ (x² + 1) \(\frac{d^2 y}{d x^2}\) + x \(\frac{d y}{d x}\) = 0 [Hence proved]

(ii) Given y = log (x + \(\sqrt{x^2+a^2}\))
Diff. both sides w.r.t. x, we get

Multiplying throughout by \(\sqrt{x^2+a^2}\) ; we have
(x² + a²) y₂ + xy₁ = 0

Question 16.
If y = cos (sin x), prove that \(\frac{d^2 y}{d x^2}\) + tan x \(\frac{d y}{d x}\) + y cos² x = 0.
Solution:
Given y = cos (sin x) ………..(1)
Diff. both sides of eqn. (1) w.r.t. x, we have
\(\frac{d y}{d x}\) = – sin (sin x) . cos x
again diff. both sides w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = – [sin (sin x) (- sin x) + cos² x cos (sin x)]
⇒ \(\frac{d^2 y}{d x^2}\) = sin x sin (sin x) – cos² x cos (sin x)
L.H.S. = \(\frac{d^2 y}{d x^2}\) + tan x \(\frac{d y}{d x}\) + y cos² x
= sin x sin (sin x) – cos² x cos (sin x) + tan x {- cos x sin (sin x)} + cos² cos (sin x)
= sin x sin (sin x) – cos² x cos (sin x) – sin x sin (sin x) + cos² cos (sin x)
= 0
= R.H.S.

Question 17.
(i) If y = e^{tan-1 x}, prove that (1 + x²) \(\frac{d^2 y}{d x^2}\) + (2x – 1) \(\frac{d y}{d x}\) = 0.
(ii) If x = tan (\(\frac{1}{a}\) log y), then prove that (1 + x²) \(\frac{d^2 y}{d x^2}\) + (2x – a) \(\frac{d y}{d x}\) = 0.
(iii) If y = e^{a cos-1 x} – 1 < x < 1, prove that (1 – x²) \(\frac{d^2 y}{d x^2}\) – x \(\frac{d y}{d x}\) – a²y = 0.
Solution:
(i) Given y = e^{tan-1 x} …….(1)
diff. both sides of eqn. (1) w.r.t. x ; we have
\(\frac{d y}{d x}\) = e^{tan-1 x} \(\frac{d}{d x}\) tan^-1 x
\(\frac{d y}{d x}\) = e^{tan-1 x} \(\frac{1}{1+x^2}\)
⇒ (1 + x²) \(\frac{d y}{d x}\) = y [using eqn. (1)]
Diff. again both sides w.r.t. x ; we have
(1 + x²) \(\frac{d^2 y}{d x^2}\) + \(\frac{d}{d x}\) (2x) = \(\frac{d y}{d x}\)
⇒ (1 + x²) \(\frac{d^2 y}{d x^2}\) + (2x – 1) \(\frac{d y}{d x}\) = 0

(ii) Given x = tan (\(\frac{1}{a}\) log y) ……..(1)
⇒ a tan^-1 x = log y
⇒ y = e^a tan^-1 x
Diff. eqn. (1) w.r.t. x, we get
y₁ = e^a tan^-1 x \(\left(\frac{a}{1+x^2}\right)\)
⇒ (1 + x²) y₁ = ay …………(2) [using (1)]
Diff. eqn. (2) w.r.t. x, we get
(1 + x²) y₂ + 2xy₁ = ay₁
⇒ (1 + x²) + (2x – a) y₁ = 0 [Hence Proved]

(iii) Given y = e^{a cos-1 x} ………..(1)
Diff. (1) w.r.t. x, we get
y₁ = e^{a cos-1 x} \(\left(\frac{-a}{\sqrt{1-x^2}}\right)\)
⇒ \(\sqrt{1-x^2}\) y₁ = – ay
Diff. eqn. (2) w.r.t. x, we get
⇒ \(\sqrt{1-x^2}\) y₂ + y . \(\frac{1}{2}\) (1 – x²)^-1/2 (-2x) = ay₁
⇒ (1 – x²) – xy₁ = a² y [usin eqn. (2)]
[Hence Proved].

Question 18.
If y = log \(\left(\frac{x}{a+b x}\right)^x\), prove that x³ \(\frac{d^2 y}{d x^2}\) = (x \(\frac{d y}{d x}\) – y)²
Solution:
Given y = x log_e \(\left(\frac{x}{a+b x}\right)\)
Diff. both sides w.r.t. x ; we get

Question 19.
If √x + √y = √a, find \(\frac{d^2 y}{d x^2}\) at x = a.
Solution:
Given, √x + √y = √a ……….(1)
Diff. eqn. (1) both sides w.r.t. x, we have
\(\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=-\frac{\sqrt{y}}{\sqrt{x}}\)
diff. again w.r.t. x ; we get
\(\frac{d^2 y}{d x^2}\) = – \(\left[\frac{1}{\sqrt{x}}\left(+\frac{1}{2}\right) y^{-\frac{1}{2}} \frac{d y}{d x}+\sqrt{y}\left(-\frac{1}{2}\right) x^{-\frac{3}{2}}\right]\)
= – \(\left[\frac{1}{2 \sqrt{x}} \frac{1}{\sqrt{y}}\left(-\frac{\sqrt{y}}{\sqrt{x}}\right)-\frac{\sqrt{y}}{2 x^{3 / 2}}\right]\)
at x = a
∴ from eqn. (1) ; y = 0
Thus, (\(\frac{d^2 y}{d x^2}\))_x=a = – \(\left[-\frac{1}{2 a}-\frac{0}{2 a^{3 / 2}}\right]=\frac{1}{2 a}\).

Question 20.
If x^myⁿ = (x + y)^m+n, then prove that
(i) \(\frac{d y}{d x}=\frac{y}{x}\)
(ii) \(\frac{d^2 y}{d x^2}\) = 0 (NCERT Exampler)
Solution:
Given x^myⁿ = (x + y)^m+n
Tsaking logarithm on both sides ; we have
m log x + n log y = (m + n) log (x + y)
[∵ log a^b = b log a
and log (ab) = log a + log b]
Diff. both sides w.r.t. x ; we have

Question 21.
If x = a (θ – sin θ), y = a (1 + cos θ), find \(\frac{d^2 y}{d x^2}\).
Solution:
Given x = a (θ – sin θ)
and y = a (1 + cos θ)
∴ \(\frac{d x}{d \theta}\) = a (1 – cos θ)
and \(\frac{d y}{d \theta}\) = – a sin θ
Thus, \(\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}\)
= \(\frac{-a \sin \theta}{a(1-\cos \theta)}\)
= \(\frac{-2 \sin \theta / 2 \cos \theta / 2}{2 \sin ^2 \theta / 2}\)
= – cot \(\frac{\theta}{2}\)

Question 21 (old).
If x cos (a + y) = cos y, prove that sin a \(\frac{d^2 y}{d x^2}\) + sin 2 (a + y) \(\frac{d y}{d x}\) = 0.
Solution:
Given x cos (a + y) = cos y
⇒ x = \(\frac{\cos y}{\cos (a+y)}\) …………..(1)
Diff. eqn. (1) w.r.t. y ; we have

Question 22.
If x = log t and y = \(\frac{1}{t}\), prove that \(\frac{d^2 y}{d x^2}\) + \(\frac{d y}{d x}\) = 0.
Solution:
Given x = log t ………(1)
and y = \(\frac{1}{t}\) ……………(2)
Diff. eqn. (1) and (2) w.r.t. t, we have
\(\frac{d x}{d t}=\frac{1}{t}\) ;
\(\frac{d y}{d t}=-\frac{1}{t^2}\)
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{-\frac{1}{t^2}}{\frac{1}{t}}=-\frac{1}{t}\)
= – y [using (2)]
Diff. again both sides w.r.t. x, we have
\(\frac{d^2 y}{d x^2}=-\frac{d y}{d x}\)
⇒ \(\frac{d^2 y}{d x}+\frac{d y}{d x}\) = 0.

Question 23.
If x = a sin pt and y = b cos pt, find the value of \(\frac{d^2 y}{d x^2}\) at t = 0.
Solution:
Given x = a sin pt ………..(1)
and y = b cos pt …………(2)
Diff. eqn. (1) and (2) w.r.t. t, we have
\(\frac{d x}{d t}\) = ap cos pt ……….(3)
\(\frac{d y}{d t}\) = – bp sin pt ………..(4)
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{-b p \sin p t}{a p \cos p t}\)
= – \(\frac{b}{a}\) tan pt ……….(5) [using (3) and (4)]
Diff. eqn. (5) w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = \(\frac{d}{d x}\) (- \(\frac{b}{a}\) tan pt)
= – \(\frac{b}{a}\) sec² pt . p. \(\frac{d t}{d x}\)
= \(-\frac{b p}{a} \sec ^2 p t \times \frac{1}{a p \cos p t}\) [using (3)]
= – \(\frac{b}{a^2}\) (sec³ pt)
∴ at t = 0,
\(\frac{d^2 y}{d x^2}=-\frac{b}{a^2} \times 1\)
= \(-\frac{b}{a^2}\)

Question 24.
If x = a sec³ θ and y = a tan³ θ, find \(\frac{d^2 y}{d x^2}\).
Solution:
Given x = a sec³ θ ………….(1)
y = a tan³ θ …………(2)
Diff. eqn (1) and (2) w.r.t. θ ; we have
\(\frac{d x}{d \theta}\) = 3a sec² θ (sec θ tan θ)
= 3a sec³ θ tan θ
and \(\frac{d x}{d \theta}\) = 3a tan² θ sec² θ
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}\)
= \(\frac{3 a \tan ^2 \theta \sec ^2 \theta}{3 a \sec ^3 \theta \tan \theta}\)
= \(\frac{\tan \theta}{\sec \theta}\)
= sin θ
diff. both sides w.r.t. x ;
\(\frac{d^2 y}{d x^2}\) = cos θ \(\frac{d \theta}{d x}\)
= cos θ \(\frac{1}{3 a \sec ^3 \theta \tan \theta}\)
= \(\frac{1}{3 a \sec ^4 \theta \tan \theta}\)
at θ = \(\frac{\pi}{3}\) ;
\(\frac{d^2 y}{d x^2}\) = \(\frac{1}{3 a(2)^4 \sqrt{3}}
= [latex]\frac{1}{48 \sqrt{3} a}\)

Question 25.
If x = a (1 + cos t), y = a (t + sin t), find \(\frac{d^2 y}{d x^2}\) at t = \(\frac{\pi}{2}\).
Solution:
Given x = a (1 + cos t)
and y = a (t + sin t)
∴ \(\frac{d x}{d t}\) = – a sin t ;
\(\frac{d y}{d t}\) = a (1 + cos t)
Thus, \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}\)
= \(\frac{a(1+\cos t)}{-a \sin t}\)
∴ \(\frac{d y}{d x}=\frac{2 \cos ^2 t / 2}{-2 \sin t / 2 \cos t / 2}\)
= – cot \(\frac{t}{2}\)
Again diff. w.r.t. x ; we get
\(\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(-\cot \frac{t}{2}\right)\)
= cosec² \(\frac{t}{2} \cdot \frac{1}{2} \frac{d t}{d x}\)
= \(\frac{1}{2} \ {cosec}^2 \frac{t}{2}\left(\frac{1}{-a \sin t}\right)\)
at t = \(\frac{\pi}{2}\),
\(\frac{d^2 y}{d x^2}=\frac{1}{2} \ {cosec}^2 \frac{\pi}{4}\left(\frac{1}{-a \sin \pi / 2}\right)\)
= \(-\frac{1}{2 a} \times(\sqrt{2})^2\)
= \(-\frac{1}{a}\)

Question 26.
If x = cos t + log (tan \(\frac{t}{2}\)), y = sin t, then find \(\frac{d^2 y}{d x^2}\) and \(\frac{d^2 y}{d x^2}\) at t = \(\frac{\pi}{4}\).
Solution:
Given x = cos t + log (tan \(\frac{t}{2}\)) ………..(1)
and y = sin t ……….(2)
diff. eqn. (1) and (2) w.r.t. x ; we have

Students appreciate clear and concise ISC Maths Class 12 Solutions Chapter 5 Continuity and Differentiability Ex 5.10 that guide them through exercises.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.10

Find \(\frac{d y}{d x}\) in the following (1 to 6) questions when :

Question 1.
(i) x = at², y = 2 at (NCERT)
(ii) x = 2at², y = at⁴ (NCERT)
Solution:
(i) Given x =at² …………(1)
and y = 2at
Differentiate (1) and (2) w.r.t. ‘t’, we have
\(\frac{d x}{d t}\) = 2at
and \(\frac{d y}{d t}\) = 2a
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{2 a}{2 a t}=\frac{1}{t}\), t ≠ 0

(ii) Given, x = 2at² ………….(1)
and y = at⁴ ………….(2)
Differentiate (1) and (2) w.r.t. ‘t’, we have
\(\frac{d x}{d t}\) = 4at
and \(\frac{d y}{d t}\) = 4at³
∴ \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}\)
= \(\frac{4 a t^3}{4 a t}\)
= t²

Question 2.
(i) x = a cos θ, y = b sin θ (NCERT)
(ii) x = a cos θ, y = a sin θ (NCERT)
Solution:
(i) Given, x = a cos θ ;
y = b sin θ
Differentiating both sides w.r.t. θ ; we get
∴ \(\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}\)
= \(\frac{b \cos \theta}{-a \sin \theta}\)
= – \(\frac{b}{a}\) cot θ

(ii) Given, x = a cos θ ………..(1)
and y = a sin θ ………….(2)
Diff. (1) and (2) w.r.t. ‘θ’, we have
\(\frac{d x}{d \theta}\) = – a sin θ
and \(\frac{d y}{d \theta}\) = a cos θ
∴ \(\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}\)
= \(\frac{a \cos \theta}{-a \sin \theta}\)
= – cot θ.

Question 3.
(i) x = a sec θ, y = b tan θ (NCERT)
(ii) x = sin t, y = cos 2t (NCERT)
Solution:
(i) Given x = a sec θ
and y = b tan θ
Diff. both given eqn. w.r.t. θ, we have
∴ \(\frac{d x}{d \theta}\) = a sec θ tan θ
and \(\frac{d y}{d \theta}\) = b sec² θ
Thus \(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}\)
= \(\frac{b \sec ^2 \theta}{a \sec \theta \tan \theta}\)
= \(\frac{b}{a}\) cosec θ

(ii) Given x = sin t
and y = cos 2t ………..(2)
Diff. eqn. (1) and eqn. (2) w.r.t. t, we have
\(\frac{d x}{d t}\) = cos t
\(\frac{d y}{d t}\) = – sin 2t . 2
Thus, \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{-2 \sin 2 t}{\cos t}\)
= \(\frac{-4 \sin t \cos t}{\cos t}\)
= – 4 sin t

Question 4.
(i) x = 4t, y = (NCERT)
(ii) x = t + \(\frac{1}{t}\), y = t – \(\frac{1}{t}\) (NCERT Exampler)
(iii) x = a (t – sin t), y = a (1 + cos t)
Solution:
(i) Given x = 4t ……….(1)
and y = 4/t ………..(2)
Diff. (1) and (2) w.r.t. ‘t’, we have
\(\frac{d x}{d t}\) = 4
and \(\frac{d y}{d t}\) = – 4/t²
∴ \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}\)
= \(\frac{-\frac{4}{t^2}}{4}\)
= – \(\frac{1}{t^2}\)

(ii) Given x = (t + \(\frac{1}{t}\))
and y = (t – \(\frac{1}{t}\)) ……………(1)
Differentiating both sides of eqn. (1) w.r.t. x ; we have
\(\frac{d x}{d t}=\left(1-\frac{1}{t^2}\right)\) ;
\(\frac{d y}{d t}=\left(1+\frac{1}{t^2}\right)\)
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{\left(t^2+1\right)}{\left(t^2-1\right)}\)
= \(\frac{\left(t+\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)}=\frac{x}{y}\) [using eqn. (1)]

(iii) x = a (t – sin t)
and y = a (1 + cos t)
∴ \(\frac{d x}{d t}\) = a (1 – cos t)
and \(\frac{d y}{d t}\) = – a sin t
Thus \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}\)
= \(\frac{-a \sin t}{a(1-\cos t)}\)
= \(\frac{-2 a \sin \frac{t}{2} \cos \frac{t}{2}}{2 a \sin ^2 \frac{t}{2}}\)
= – cot \(\frac{t}{2}\)

Question 5.
(i) x = a cos³ t, y = b sin t
(ii) x = a (1 – sin t), y = a (1 + cos t)
Solution:
Given x = a cos³ t ………..(1)
and y = b sin³ t ………..(2)
Diff. eqn. (1) and eqn. (2) w.r.t. t, we have
\(\frac{d x}{d t}\) = 3a cos² t (- sin t)
and \(\frac{d y}{d t}\) = 3b sin² t (cos t)
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{3 b \sin ^2 t \cos t}{3 a \cos ^2 t \sin t}\)
= – \(\frac{b}{a}\) tan t

(ii) Given x = a (1 – sin t) ……….(1)
and y = a (1 + cos t) ………..(2)
Diff. both eqns. w.r.t. t, we have
\(\frac{d x}{d t}\) = – a cos t ;
\(\frac{d y}{d t}\) = – a sin t
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{-a \sin t}{-a \cos t}\)
= tan t.

Question 5 (old).
(i) x = a sin³ t, y = a cos³ t (ISC 2004)
Solution:
Given x = a sin³ t
and y = a cos³ t
Diff. eqn. (1) and eqn. (2) w.r.t. ‘t’, we have
\(\frac{d x}{d t}\) = 3a sin² t cos t ;
\(\frac{d y}{d t}\) = 3a cos² t (- sin t)
Thus, \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{-3 a \cos ^2 t \sin t}{3 a \sin ^2 t \cos t}\)
= – cos t

Question 6.
(i) x = a (θ + sin θ), y = a (1 – cos θ)
(ii) x = a (1 – cos θ), y = (θ + sin θ)
Solution:
(i) Given x = a (θ + sin θ) ………..(1)
and y = a (1 – cos θ) …………..(2)
Diff. eqn. (1) and (2) w.r.t. θ, we have
\(\frac{d x}{d \theta}\) = a (1 + cos θ)
\(\frac{d x}{d \theta}\) = a sin θ
Thus, \(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}\)
= \(\frac{a \sin \theta}{a(1+\cos \theta)}\)
= \(\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}}\)
= tan \(\frac{\theta}{2}\)

(ii) Given x = a (1 – cos θ) ;
y = a (θ + sin θ)
Diff. both sides w.r.t. θ ; we get
\(\frac{d x}{d \theta}\) = a sin θ;
\(\frac{d x}{d \theta}\) = a (1 + cos θ)
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}\)
= \(\frac{a(1+\cos \theta)}{a \sin \theta}\)
= \(\frac{1+\cos \theta}{sin \theta}\)
= \(\frac{2 \cos ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\)
= 2 cot \(\frac{\theta}{2}\)

Question 7.
(i) x = e^t (sin t + cos t), y = e^t (sin t – cos t)
(ii) x = cos θ – cos 2θ, y = sin θ – sin 2θ (NCERT)
Solution:
(i) Given x = e^t (sin t + cos t)
and y = e^t (sin t – cos t)
Diff. both eqn.’s w.r.t. t, we have
\(\frac{d x}{d t}\) = e^t (cos t – sin t) + (sin t + cos t) e^t
= e^t (2 cos t)
and \(\frac{d y}{d t}\) = e^t (cos t + sin t) + (sin t – cos t) e^t
= e^t (2 sin t)
∴ \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}\)
= \(\frac{2 e^t \sin t}{2 e^t \cos t}\)
= tan t

(ii) Given x = cos θ – cos 2θ ……….(1)
and y = sin θ – sin 2θ ………..(2)
Diff. eqn. (1) and eqn. (2) w.r.t. θ, we have
\(\frac{d x}{d \theta}\) = – sin θ + 2 sin 2θ
and \(\frac{d y}{d \theta}\) = cos θ – 2 cos 2θ
∴ \(\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}\)
= \(\frac{\cos \theta-2 \cos 2 \theta}{-2 \sin \theta+2 \sin 2 \theta}\)

Question 8.
(i) x = a (cos θ + cos 2θ), y = b (sin θ + sin 2θ)
(ii) x = \(\frac{1+\log t}{t^2}\), y = \(\frac{3+ 2\log t}{t^2}\) (NCERT Exampler)
Solution:
Given x = a (cos θ + cos 2θ) …………(1)
and y = b (sin θ + sin 2θ) ………..(2)
Diff. eqn. (1) and eqn. (2) w.r.t. θ, we have
\(\frac{d x}{d \theta}\) = a (- sin θ – 2 sin 2θ) ……….(3)
and \(\frac{d y}{d \theta}\) = b (cos θ + 2 cos 2θ) …………..(4)
∴ \(\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}\)
= \(\frac{b(\cos \theta+2 \cos 2 \theta)}{a(-\sin \theta-2 \sin 2 \theta)}\)
= \(-\frac{b(\cos \theta+2 \cos 2 \theta)}{a(\sin \theta+2 \sin 2 \theta)}\)
[using eqn. (3) and eqn. (4)]

(iv) Given x = \(\frac{1+\log t}{t^2}\) ……….(1)
and y = \(\frac{3+ 2\log t}{t^2}\) …………..(2)
Differentiate eqn. (1) and (2) both sides w.r.t. ‘t’ ; we have

Question 9.
x = 3 cos θ – 2 cos³ θ, y = 3 sin θ – 2 sin³ θ
Solution:
Given x = 3 cos θ – 2 cos³ θ
and y = 3 sin θ – 2 sin³ θ
Diff. both given eqn’s w.r.t. θ, we have
\(\frac{d x}{d \theta}\) = – 3 sin θ – 6 cos² θ (- sin θ)
= – 3 sin θ + 6 cos² θ sin θ ………..(1)
and \(\frac{d y}{d \theta}\) = 3 cos θ – 6 sin² θ cos ² …………….(2)
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}\)
= \(\frac{3 \cos \theta-6 \sin ^2 \theta \cos \theta}{-3 \sin \theta+6 \cos ^2 \theta \sin \theta}\)
[using (1) and (2)]
= \(\frac{3 \cos \theta\left[1-2 \sin ^2 \theta\right]}{-3 \sin \theta\left[1-2 \cos ^2 \theta\right]}\)
= \(\frac{3 \cos \theta\left(1-2 \sin ^2 \theta\right)}{-3 \sin \theta\left[1-2\left(1-\sin ^2 \theta\right)\right]}\)
= \(\frac{3 \cos \theta\left(1-2 \sin ^2 \theta\right)}{3 \sin \theta\left(1-2 \sin ^2 \theta\right)}\)
= cos θ.

Question 10.
(i) If x = 2 cos t – cos 2t and y = 2 sin t – sin 2t, find \(\frac{d y}{d x}\) at t = \(\frac{\pi}{4}\).
(ii) If x = 3 sin t – sin 3t, y = 3 cos t – cos 3t, find \(\frac{d y}{d x}\) at t = \(\frac{\pi}{3}\). (NCERT Exampler)
Solution:
(i) Given x = 2 cos t – cos 2t
and y = 2 sin t – sin 2t
Diff. both sides w.r.t. ‘t’ ; we have
\(\frac{d x}{d t}\) = – 2 sin t + 2 sin 2t ;
\(\frac{d y}{d t}\) = 2 cos t – 2 cos 2t
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{2(\cos t-\cos 2 t)}{2(\sin 2 t-\sin t)}\)
at t = \(\frac{\pi}{4}\);
\(\frac{d y}{d x}=\frac{\cos \left(\frac{\pi}{4}\right)-\cos \frac{\pi}{2}}{\sin \frac{\pi}{2}-\sin \frac{\pi}{4}}\)
= \(\frac{\frac{1}{\sqrt{2}}}{1-\frac{1}{\sqrt{2}}}\)
= \(\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}\)
= √2 + 1

(ii) Given x = 3 sin t – sin 2t
= 3 sin t – (3 sin t – 4 sin³ t)
⇒ x = 4 sin³ t …….(1)
and y = 3 cos t – cos 3t ………..(2)
Diff. eqn. (1) and (2) w.r.t. t ; we get
\(\frac{d x}{d t}\) = 12 sin² t cos t
\(\frac{d y}{d t}\) = – 3 sin t + 3 sin 3t
∴ \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}\)
= \(\frac{-\sin t+\sin 3 t}{4 \sin ^2 t \cos t}\)
at t = \(\frac{\pi}{3}\) ;
\(\frac{d y}{d x}=\frac{-\frac{\sqrt{3}}{2}+0}{4 \times \frac{3}{4} \times \frac{1}{2}}\)
= \(\frac{-\sqrt{3} \times 2}{2 \times 3}\)
= \(-\frac{1}{\sqrt{3}}\)

Question 11.
(i) If x = \(\frac{2 b t}{1+t^2}\) and y = \(\frac{a\left(1-t^2\right)}{1+t^2}\), find \(\frac{d y}{d x}\) at t = 2.
(ii) If x = ae^θ (sin θ – cos θ) and y = ae^θ (sin θ + cos θ), find \(\frac{d y}{d x}\) at θ = \(\frac{\pi}{4}\).
Solution:
Given x = \(\frac{2 b t}{1+t^2}\)
and y = \(\frac{a\left(1-t^2\right)}{1+t^2}\)
Diff. both equations w.r.t. t ; we have

(ii) Given x = a e^θ (sin θ – cos θ) ……………..(1)
and y = a e^θ (sin θ + cos θ)
diff. eqn. (1) w.r.t. θ ; we have
\(\frac{d x}{d \theta}\) = a e^θ (sin θ – cos θ) + a e^θ (cos θ + sin θ)
= a e^θ (sin θ – cos θ + cos θ + sin θ)
= 2ae^θ sin θ
and y = ae^θ (sin θ + cos θ) ………….(2)
diff. eqn. (2) w.r.t. θ ; we have
\(\frac{d y}{d \theta}\) = ae^θ (sin θ + cos θ) + ae^θ (cos θ – sin θ)
= ae^θ (sin θ + cos θ + cos θ – sin θ)
= 2ae^θ cos θ
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}\)
= \(\frac{2 a e^\theta \cos \theta}{2 a e^\theta \sin \theta}\)
= cot θ
at θ = \(\frac{\pi}{4}\) ;
\(\frac{d y}{d x}\) = cot \(\frac{\pi}{4}\) = 1.

Question 12.
(i) \(\frac{x^2}{1-x^2}\) w.r.t. x²
(ii) sin x² w.r.t. x³ (ISC 2009)
(iii) cot³ (2x + 1) w.r.t. x² + 1
(iv) sin² x w.r.t. e^{cos x}
Solution:
(i) Let y = \(\frac{x^2}{1-x^2}\)
and z = x²
So we want to differentiate y w.r.t. i.e. we want to find \(\frac{d y}{d z}\)
Diff. given eqns w.r.t. x, we have
\(\frac{d y}{d x}=\frac{\left(1-x^2\right) 2 x-x^2(-2 x)}{\left(1-x^2\right)^2}\)
= \(\frac{2 x}{\left(1-x^2\right)^2}\)
and \(\frac{d z}{d x}\) = 2x
∴ \(\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}\)
= \(\frac{2 x}{\frac{\left(1-x^2\right)^2}{2 x}}\)
= \(\frac{1}{\left(1-x^2\right)^2}\)

(ii) Let y = sin x²
and z = x³
So we want to diff. y w.r.t. z i.e. to find \(\frac{d y}{d z}\)
Diff. both eqn’s w.r.t. x : we have
\(\frac{d y}{d x}\) = cos x² . 2x ;
\(\frac{d z}{d x}\) = 3x²
∴ \(\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}\)
= \(\frac{2 x \cos x^2}{3 x^2}\)
= \(\frac{2}{3 x}\) cos x²

(iii) Let y = cot³ (2x + 1)
and z = x² + 1
So, we want to diff. y w.r.t. z i.e. to find \(\frac{d y}{d z}\).
Diff. given eqn’s both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = 3 cot² (2x + 1) {- cosec² (2x + 1)} . 2
and \(\frac{d z}{d x}\) = 2x
∴ \(\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}\)
= \(\frac{-6 \cot ^2(2 x+1) \ {cosec}^2(2 x+1)}{2 x}\)
= \(\frac{-3 \cot ^2(2 x+1) \ {cosec}^2(2 x+1)}{x}\)

(iv) Let y = sin² x ………….(1)
and z = e^{cos x}
We want to diff. y w.r.t. z i.e. to find \(\frac{d y}{d z}\)
diff. eqn. (1) and eqn. (2) w.r.t. x ;
\(\frac{d y}{d x}\) = 2 sin x cos x ;
\(\frac{d z}{d x}\) = e^{cos x} (- sin x)
∴ \(\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}\)
= \(\frac{2 \sin x \cos x}{e^{\cos x}(-\sin x)}\)
= – \(\frac{2 \cos x}{2^{\cos x}}\)

Question 8 (old).
(iv) log (sin x) w.r.t. \(\sqrt{cos x}\)
Solution:
Given, y = log (sin x)
and z = \(\sqrt{cos x}\)
i.e. we want diff. y w.r.t. z i.e. to find \(\frac{d y}{d z}\)
Now, diff. both eqn’s w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{1}{sin x}\) cos x
= cot x
and \(\frac{d z}{d x}\) = \(\frac{1}{2}\) (cos x)^{\(-\frac{1}{2}\)} .(- sin x)
= – \(\frac{\sin x}{2 \sqrt{\cos x}}\)
∴ \(\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}\)
= \(\frac{2 \cot x}{-\sin x}\) \(\sqrt{cos x}\)
= – 2 cot x cosec x \(\sqrt{cos x}\)

Question 13.
(i) Differentiate sin^-1 \(\left(\frac{2 x}{1+x^2}\right)\) w.r.t. tan^-1 x.
(ii) Differentiate tan^-1 \(\left(\frac{2 x}{1-x^2}\right)\) w.r.t. tan^-1 x.
Solution:
(i) Let y = sin^-1 \(\left(\frac{2 x}{1+x^2}\right)\) ……….(1)
and z = tan^-1 x
i.e. we want to find \(\frac{d y}{d z}\)
put x = tan θ in eqn. (1) ; we have
y = sin^-1 \(\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\)
= sin^-1 (sin 2θ)
= 2θ
= 2 tan^-1 x
∴ \(\frac{d y}{d x}\) = \(\frac{2}{1+x^2}\)
and \(\frac{d z}{d x}\) = \(\frac{1}{1+x^2}\)
Thus, \(\frac{d y}{d z}=\frac{d y / d x}{d z / d x}\)
= \(\frac{2 / 1+x^2}{1 / 1+x^2}\)
= 2

(ii) Let y = tan^-1 \(\left(\frac{2 x}{1-x^2}\right)\) …………….(1)
and z = tan^-1 x ………(2)
i.e. we want to find \(\frac{d y}{d z}\)
put x = tan θ
⇒ θ = tan^-1 x in eqn. (1) ; we have
∴ y = tan^-1 \(\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)\)
⇒ y = tan^-1 (tan 2θ)
= 2θ
= 2 tan^-1 x
Thus \(\frac{d y}{d x}\) = \(\frac{2}{1+x^2}\)
and \(\frac{d z}{d x}\) = \(\frac{1}{1+x^2}\)
∴ \(\frac{d y}{d x}\) = 2.

Question 14.
Differentiate cos^-1 \(\left(\frac{1}{\sqrt{1+t^2}}\right)\) w.r.t. sin^-1 \(\left(\frac{t}{\sqrt{1+t^2}}\right)\).
Solution:
Let y = cos^-1 \(\left(\frac{1}{\sqrt{1+t^2}}\right)\) ………..(1)
and z = sin^-1 \(\left(\frac{t}{\sqrt{1+t^2}}\right)\)
putting t = tan θ
i.e. θ = tan^-1 t in eqn (1), we have
y = cos^-1 \(\left(\frac{1}{\sqrt{1+\tan ^2 \theta}}\right)\)
= cos^-1 (cos θ)
⇒ y = θ = tan^-1 t
Diff. w.r.t. t ; we have
∴ \(\frac{d y}{d t}=\frac{1}{1+t^2}\) ………..(3)
putting t = tan Φ i.e. Φ = tan^-1 t, in eqn. (2) ; we have
z = sin^-1 \(\left(\frac{\tan \phi}{\sec \phi}\right)\)
= sin^-1 (sin Φ) = Φ
⇒ z = tan^-1 t
Diff. both sides w.r.t. t, we get
∴ \(\frac{d z}{d t}=\frac{1}{1+t^2}\) ……………..(4)
Now, we want to diff. y w.r.t. z to find \(\frac{d y}{d z}\).
∴ \(\frac{d y}{d z}=\frac{\frac{d y}{d t}}{\frac{d z}{d t}}\)
= \(\frac{\frac{1}{1+t^2}}{\frac{1}{1+t^2}}\)
= 1 [using (3) and (4)]

Question 15.
(i) Differentiate sin^-1 \(\left(\frac{2 x}{1+x^2}\right)\) w.r.t. cos^-1 \(\left(\frac{1-x^2}{1+x^2}\right)\).
(ii) Differentiate tan^-1 \(\left(\frac{3 x-x^3}{1-3 x^2}\right)\) w.r.t. tan^-1 \(\frac{2 x}{1-x^2}\).
Solution:
(i) Let y = sin^-1 \(\left(\frac{2 x}{1+x^2}\right)\) …………(1)
and z = cos^-1 \(\left(\frac{1-x^2}{1+x^2}\right)\) …………(2)
i.e. we want to diff. y w.r.t. z to find \(\frac{d y}{d z}\)
To simplify eqn. (1) ;
we put x = tan θ
i.e. θ = tan^-1 x
y = sin^-1 \(1\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\)
= sin^-1 (sin 2θ)
⇒ y = 2θ
= 2 tan^-1 x
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{2}{1+x^2}\) …………….(3)
To simplify eqn. (2)
we put x = tan Φ
i.e. Φ = tan^-1 x
∴ z = cos^-1 \(\left(\frac{1-\tan ^2 \phi}{1+\tan ^2 \phi}\right)\)
= cos^-1 (cos 2Φ)
⇒ z = 2Φ = 2 tan^-1 x
Diff. both sides w.r.t. x, we have
\(\frac{d z}{d x}=\frac{2}{1+x^2}\) ………………(4)
∴ \(\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}\)
= \(\frac{\frac{2}{1+x^2}}{\frac{2}{1+x^2}}\)
= 1 [using eqn. (3) and eqn. (4)]

(ii) Let y = tan^-1 \(\left(\frac{3 x-x^3}{1-3 x^2}\right)\) …………….(1)
and z = tan^-1 \(\frac{2 x}{1-x^2}\) ……………(2)
i.e. we want to diff. y w.r.t. z i.e. to find \(\frac{d y}{d z}\).
Putting x = tan θ
⇒ θ = tan^-1 x in eqn. (1)
and x = tan Φ
⇒ Φ = tan^-1 x in eqn. (2) ; we have
∴ y = tan^-1 \(\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right)\)
On differentiating both eqn.’s w.r.t. x, we have
∴ \(\frac{d y}{d x}=\frac{3}{1+x^2}\)
and \(\frac{d z}{d x}=\frac{2}{1+x^2}\)
Thus, \(\frac{d y}{d z}=\frac{d y / d x}{d z / d x}\)
= \(\frac{3}{1+x^2} \times \frac{1+x^2}{2}\)
= \(\frac{3}{2}\)

Question 16.
(i) Differentiate tan^-1 \(\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) w.r.t. tan^-1 x.
(ii) Differentiate tan^-1 \(\left(\frac{\sqrt{1-x^2}}{x}\right)\) w.r.t. cos^-1 (2x \(\sqrt{1-x^2}\)).
Solution:
(i) Let y = tan^-1 \(\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) …………..(1)
and z = tan^-1 x …………(2)
Now we want to find \(\frac{d y}{d z}\)
put x = tan θ in eqn. (1) ; we have
y = tan^-1 \(\left(\frac{1-\cos \theta}{\sin \theta}\right)\)
= tan^-1 \(\left(\frac{2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)\)
= tan^-1 (tan \(\frac{\theta}{2}\))
⇒ y = \(\frac{\theta}{2}\)
= \(\frac{1}{2}\) tan^-1 x
and z = tan^-1 x
Diff. both given eqn’s w.r.t. x, we have
⇒ \(\frac{d y}{d x}=\frac{1}{2\left(1+x^2\right)}\)
and \(\frac{d z}{d x}=\frac{1}{1+x^2}\)
∴ \(\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}\)
= \(\frac{\frac{1}{2\left(1+x^2\right)}}{\frac{1}{1+x^2}}\)
= \(\frac{1}{2}\)

(ii) Let y = tan^-1 \(\left(\frac{\sqrt{1-x^2}}{x}\right)\) ………..(1)
and z = cos^-1 (2x\(\sqrt{1-x^2}\)) ………………(2)
Put x =sin θ
i.e. θ = sin^-1 x in eqn. (1) ; we have
y = tan^-1 \(\left(\frac{\sqrt{1-\sin ^2 \theta}}{\sin \theta}\right)\)
= tan^-1 (cot θ)
⇒ y = tan^-1 (tan (\(\frac{\pi}{2}\) – θ))
⇒ y = \(\frac{\pi}{2}\) – θ
= \(\frac{\pi}{2}\) – sin^-1 x
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=-\frac{1}{\sqrt{1-x^2}}\) ………..(3)
putting x = sin Φ i.e. Φ = sin^-1 x in eqn. (2) ; we have
∴ z = cos^-1 (2 sin Φ \(\sqrt{1-\sin ^2 \phi}\))
= cos^-1 (2 sin Φ cos Φ)
z = cos^-1 (sin 2Φ)
= cos^-1 (cos (\(\frac{\pi}{2}\) – 2Φ))
z = \(\frac{\pi}{2}\) – 2Φ
= \(\frac{\pi}{2}\) – sin^-1 x
Diff. both sides w.r.t. x ; we have
\(\frac{d z}{d x}=-\frac{2}{\sqrt{1-x^2}}\) ……………..(4)
We want to diff. y w.r.t. z i.e. to find \(\frac{d y}{d z}\)
∴ \(\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}\)
= \(\frac{-\frac{1}{\sqrt{1-x^2}}}{\frac{-2}{\sqrt{1-x^2}}}\)
= \(\frac{1}{2}\)

Question 17.
Prove that the derivative of tan^-1 \(\left(\frac{x}{1+\sqrt{1-x^2}}\right)\) w.r.t. sin^-1 x is independent of x.
Solution:
Let y = tan^-1 \(\left(\frac{x}{1+\sqrt{1-x^2}}\right)\) ………..(1)
and z = sin^-1 x
Now we want to diff. y w.r.t. z
Put x = sin θ
∴ θ = sin^-1 x in eqn. (1) ; we have
y = tan^-1 \(\left(\frac{\sin \theta}{1+\cos \theta}\right)\)
= tan^-1 \(\left(\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}}\right)\)
⇒ y = tan^-1 (tan \(\frac{\theta}{2}\))
= \(\frac{\theta}{2}\)
= \(\frac{1}{2}\) sin^-1 x
Diff. both sides w.r.t. x, we have
Thus \(\frac{d y}{d x}\) = \(\frac{1}{2} \frac{1}{\sqrt{1-x^2}}\)
Diff. eqn. (2) w.r.t. x
\(\frac{d z}{d x}=\frac{1}{\sqrt{1-x^2}}\)
∴ \(\frac{d y}{d z}=\frac{d y / d x}{d z / d x}\)
= \(\frac{1}{2 \sqrt{1-x^2}} \times \sqrt{1-x^2}\)
= \(\frac{1}{}2\)
∴ \(\frac{d y}{d z}\) is independent of x.

Question 18.
Differentiate x^x w.r.t. x log x.
Solution:
Let y = x^x ………..(1)
and z = x log x ………..(2)
We want to diff. y w.r.t. z i.e. to find \(\frac{d y}{d z}\)
Taking logarithm on both sides of eqn. (1) we have
log y = x log x ;
diff. both sides w.r.t. x ; we have
\(\frac{1}{y} \frac{d y}{d x}\) = x × \(\frac{1}{x}\) + log x . 1
⇒ \(\frac{d y}{d x}\) = x^x [1 + log x] ………..(3)
Diff. eqn. (2) w.r.t. x ; we have
\(\frac{d z}{d x}\) = x × \(\frac{1}{x}\) + log x . 1
= 1 + log x
∴ \(\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}\)
= \(\frac{x^x(1+\log x)}{1+\log x}\)
= x^x.

Utilizing Understanding ISC Mathematics Class 12 Solutions Chapter 5 Continuity and Differentiability Ex 5.8 as a study aid can enhance exam preparation.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.8

Differentiate the following (1 to 6) functions w.r.t. x:

Question 1.
(i) e^x + 3 sin x
(ii) 10^x + \(\frac{1}{3}\) e^x – 2 log x
Solution:
(i) Let y = e^x + 3 sin x ;
Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = e^x + 3 cos x

(ii) Let y = 10^x + \(\frac{1}{3}\) e^x – 2 log x
Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = 10^x log 10 + \(\frac{1}{3}\) e^x – \(\frac{2}{x}\)

Question 2.
(i) e^-x (NCERT)
(ii) sin (log x), x > 0 (NCERT)
Solution:
(i) Let y = e^-x ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = – e^-x

(ii) Let y = sin (log x), x > 0
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = cos (log x) \(\frac{d}{d x}\) log x
= \(\frac{\cos (\log x)}{x}\).

Question 3.
(i) e^{cos x} (NCERT)
(ii) e^{sin-1 x} (NCERT)
Solution:
(i) Let y = e^{cos x}
On differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = e^{cos x} \(\frac{d}{d x}\) (cos x)
= e^{cos x} (- sin x)

(ii) Let y = e^{sin-1 x}
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = e^{sin-1 x} \(\frac{d}{d x}\) (sin^-1 x)
= e^{sin-1 x} \(\frac{1}{\sqrt{1-x^2}}\)

Question 4.
(i) e^x3 (NCERT)
(ii) \(\sqrt{e^{\sqrt{x}}}\), x > 0
Solution:
(i) Let y = e^x3 ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = e^x3 \(\frac{d}{d x}\) x³
= e^x3 . 3x²

(ii) Let y = \(\sqrt{e^{\sqrt{x}}}\), x > 0
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{d}{d x}\left(e^{\sqrt{x}}\right)^{1 / 2}\)
= \(\frac{1}{2}\left(e^{\sqrt{x}}\right)^{\frac{1}{2}-1} \frac{d}{d x} e^{\sqrt{x}}\)
= \(\frac{1}{2 \sqrt{e^{\sqrt{x}}}} e^{\sqrt{x}} \frac{1}{2 \sqrt{x}}\)
= \(\frac{e^{\sqrt{x}}}{4 \sqrt{x} \sqrt{e^{\sqrt{x}}}}\)

Question 5.
(i) log (log x), x > 1 (NCERT)
(ii) log₇ (log x), x > 1 (NCERT)
Solution:
(i) Let y = log (log x), x > 1
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) log (log x)
= \(\frac{1}{\log x}\) \(\frac{d}{d x}\) (log x)
= \(\frac{1}{x \log x}\)

(ii) Let y = log₇ (log x)
Differentiating both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\log (\log x)}{\log 7}\right)\)
= \(\frac{1}{\log 7} \frac{1}{\log x} \frac{1}{x}\)
= \(\frac{1}{x \log x \log 7}\)

Question 6.
(i) log (cos e^x) (NCERT)
(ii) tan^-1 (e^{sin x})
Solution:
(i) Let y = log (cos e^x) ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{1}{\cos e^x}\) \(\frac{d}{d x}\) cos e^x
= \(\frac{1}{\cos e^x}\) (- sin e^x) \(\frac{d}{d x}\) e^x
= – e^x tan (e^x)

(ii) Let y = tan^-1 (e^{sin x})
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = \(\frac{1}{1+\left(e^{\sin x}\right)^2} \frac{d}{d x}\) e^{sin x}
= \(\frac{1}{1+e^{2 \sin x}}\) e^{sin x} . cos x

Question 7.
Is the function f(x) = log (2x – 1) derivable at x = 0?
Solution:
(i) Given f(x) = log (2x – 1)
for domain of f : f(x) must be a real number
⇒ log (2x – 1) must be a real number
⇒ 2x – 1 > 0
⇒ x > \(\frac{1}{2}\)
∴ D_f = (\(\frac{1}{2}\), ∞).

Question 8.
If f(x) = log (log x), find f'(e).
Solution:
Given f(x) = log (log x) ;
Diff. both sides w.r.t. x, we have
f'(x) = \(\frac{1}{\log x} \cdot \frac{1}{x}\)
∴ f'(e) = \(\frac{1}{\log e} \cdot \frac{1}{e}\)
= \(\frac{1}{e \log e}\)
= \(\frac{1}{e \times 1}=\frac{1}{e}\)

Question 9.
If y = log (tan x), show that \(\frac{d y}{d x}\) = 2 cosec 2x.
Solution:
Let y = log (tan x) ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=-\frac{1}{\tan x}\) sec² x
= \(\frac{1}{\frac{\sin x}{\cos x} \times \cos ^2 x}\)
= \(\frac{2}{\sin 2 x}\)
\(\frac{d y}{d x}\) = 2 cosec 2x.

Question 10.
If f(1) = 4 and f'(1) = 2, find the value of the derivative of log f(e^x) w.r.t. x at x = 0.
Solution:
Let y = log f(e^x)
Differentiating w.r.t. x, we have
\(\frac{d y}{d x}=\frac{1}{f\left(e^x\right)} \frac{d}{d x}\) f(e^x)
= \(\frac{f^{\prime}\left(e^x\right)}{f\left(e^x\right)}\) × e^x
at x = 0,
\(\frac{d y}{d x}=\frac{f^{\prime}(1)}{f(1)}\) × e⁰
= \(\frac{2}{4}=\frac{1}{2}\)
[Since f'(1) = 2 and f(1) = 4]

Question 11.
If f(x) = e^x g(x), g(0) = 2 and g'(0) = 1, then find f'(0).
Solution:
Given f(x) = e^x g(x)
Diff. both sides w.r.t. x, we have
f'(x) = e^x g'(x) + g(x) e^x …………….(1)
putting x = 0 in eqn. (1) ; we have
f'(0) = e⁰ g'(0) + g(0) e⁰
= g'(0) + g(0)
= 1 + 2 = 3
[∵ g(0) = 2 and g'(0) = 1]

Differentiate the following functions w.r.t. x :

Question 12.
(i) \(\frac{\mathcal{e}^x}{\sin x}\)
(ii) \(\frac{e^x}{1+\sin x}\).
Solution:
(i) Let y = \(\frac{\mathcal{e}^x}{\sin x}\)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{\sin x \frac{d}{d x} e^x \frac{d}{d x} \sin x}{\sin ^2 x}\)
[using quotient rule]
= \(\frac{(\sin x) e^x-e^x \cos x}{\sin ^2 x}\)
= \(\frac{e^x(\sin x-\cos x)}{\sin ^2 x}\)

(ii) Given y = \(\frac{e^x}{1+\sin x}\) ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{(1+\sin x) e^x-e^x \cos x}{(1+\sin x)^2}\)
= \(\frac{e^x(1+\sin x-\cos x)}{(1+\sin x)^2}\)

Question 13.
(i) 2x tan^-1 x – log (1 + x²)
(ii) \(\frac{\cos x}{\log x}\), x > 0 (NCERT)
Solution:
(i) Let y = 2x tan^-1 x – log (1 + x²)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = 2 [x . \(\frac{1}{1+x^2}\) + tan^-1 x . 1] – \(\frac{2 x}{1+x^2}\)
= 2 tan^-1 x

(ii) Let y = \(\frac{\cos x}{\log x}\), x > 0
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{\log x \frac{d}{d x} \cos x-\cos x \frac{d}{d x} \log x}{(\log x)^2}\) [quotient rule]
= \(\frac{-\sin x \log x-\frac{\cos x}{x}}{(\log x)^2}\)
= \(\frac{-x \sin x \log x+\cos x}{x(\log x)^2}\)

Question 14.
(i) e^{sec x2} + 3 cos^-1 x (NCERT)
(ii) e^x + e^x2 + ………….. + e^x5.
Solution:
(i) Let y = e^{sec x2} + 3 cos^-1 x
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = e^{sec x2} \(\frac{d}{d x}\) sec² x + \(\frac{d}{d x}\) 3 cos^-1 x
= e^{sec x2} (2 sec x) \(\frac{d}{d x}\) sec x + 3 \(\left(\frac{-1}{\sqrt{1-x^2}}\right)\)
= 2 sec² x tan x e^{sec2 x} – \(\frac{3}{\sqrt{1-x^2}}\)

(ii) Let y = e^x + e^x2 + ………….. + e^x5
i.e. y = e^x + e^x2 + e^x3 + e^x4 + e^x5
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = e^x + e^x2 \(\frac{d}{d x}\) x² + e^x3 \(\frac{d}{d x}\) x³ + e^x4 \(\frac{d}{d x}\) x⁴ + e^x5 \(\frac{d}{d x}\) x⁵
= e^x + 2x e^x2 + 3x² e^x3 + 4x³ e^x4 + 5x⁴ e^x5

Question 15.
(i) \(\frac{5^x \log x}{x^2+1}\)
(ii) cos (log x + e^x). (NCERT)
Solution:
(i) Let y = \(\frac{5^x \log x}{x^2+1}\)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{\left(x^2+1\right) \frac{d}{d x}\left(5^x \log x\right)-\left(5^x \log x\right) \frac{d}{d x}\left(x^2+1\right)}{\left(x^2+1\right)^2}\)
= \(\frac{\left(x^2+1\right)\left(\frac{5^x}{x}+\log x \cdot 5^x \log 5\right)-\left(5^x \log x\right) 2 x}{\left(x^2+1\right)^2}\)
= \(\frac{5^x\left[\left(x^2+1\right)(1+x \log 5 \cdot \log x)-2 x^2 \log x\right]}{x\left(x^2+1\right)^2}\)

(ii) Let y = cos (log x + e^x)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = – sin (log x + e^x) . \(\frac{d}{d x}\) (log x + e^x)
= – sin (log x + e^x) (\(\frac{1}{x}\) + e^x)
∴ \(\frac{d y}{d x}\) = – \(\frac{1}{x}\) sin (log x + e^x) (1 + x e^x).

Question 16.
(i) log \(\left(\frac{x+\sqrt{x^2-a^2}}{x-\sqrt{x^2-a^2}}\right)\)
(ii) log \(\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\)
Solution:
(i) Let y = log \(\left(\frac{x+\sqrt{x^2-a^2}}{x-\sqrt{x^2-a^2}}\right)\)
Diff. both sides w.r.t. x, we have

(ii) y = log \(\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\)
= \(\frac{1}{2}\) [log (1 – cos x) – log (1 + cos x)]
[∵ log a^b = b log a ;
log \(\frac{a}{b}\) = log a – log b]
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{1}{2}\left[\begin{array}{c}
\frac{1}{1-\cos x}(+\sin x) \\
-\frac{1}{1+\cos x}(-\sin x)
\end{array}\right]\)
= \(\frac{1}{2}\left[\frac{\sin x(1+\cos x+1-\cos x)}{(1-\cos x)(1+\cos x)}\right]\)
= \(\frac{\sin x}{1-\cos ^2 x}\)
= \(\frac{\sin x}{\sin ^2 x}\)
= cosec x.

Question 17.
(i) e^{cot-1 x2}
(ii) x \(\sqrt{1+x^2}\) + log (x + \(\sqrt{x^2+1}\)).
Solution:
(i) Let y = e^{cot-1 x2} ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = e^{cot-1 x2} \(\frac{d}{d x}\) cot^-1 x²
= e^{cot-1 x2} \(\left\{\frac{-1}{1+\left(x^2\right)^2} \times 2 x\right\}\)
= – \(\frac{2 x e^{\cot ^{-1} x^2}}{1+x^4}\)

(ii) Let y = x \(\sqrt{1+x^2}\) + log (x + \(\sqrt{x^2+1}\))
Diff. both sides w.r.t. x, we get

Question 18.
(i) If y = \(\frac{\log \left(x+\sqrt{x^2+1}\right)}{\sqrt{x^2+1}}\), prove that (x² + 1) \(\frac{d y}{d x}\) + xy = 1.
(ii) If y = e^{2 log x + 3x}, prove that \(\frac{d y}{d x}\) = x (2 + 3x) e^3x.
Solution:
(i) Given y \(\sqrt{x^2+1}\) = log (\(\sqrt{x^2+1}\) + x)
Differentiating both sides w.r.t. x ; we get

⇒ xy + (x² + 1) \(\frac{d y}{d x}\) = 1
⇒ (x² + 1) \(\frac{d y}{d x}\) + xy = 1.

(ii) Let y = e^{2 log x + 3x}
= e^{2 log x} . e^3x
= e^{log x2} e^3x
⇒ y = x² e^3x . 3 + e^3x . 2x
= e^3x (2 + 3x) x.

Question 19.
Differentiate tan^-1 \(\left(\frac{2^{x+1}}{1-4^x}\right)\) w.r.t. x.
Solution:
Let y = tan^-1 \(\left(\frac{2^{x+1}}{1-4^x}\right)\)
= tan^-1 \(\left(\frac{2 \times 2^x}{1-\left(2^x\right)^2}\right)\)
put 2^x = tan θ
i.e. θ = tan^-1 (2x)
Then y = tan^-1 \(\left(\frac{2 \tan \theta}{1-\tan ^2 \theta}\right)\)
⇒ y = tan^-1 (tan 2θ) = 2θ
⇒ y = 2 tan^-1 (2^x) ;
Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}=\frac{2}{1+\left(2^x\right)^2} \frac{d}{d x}\) (2^x)
= \(\frac{2}{1+2^{2 x}}\) 2^x log 2
= \(\frac{2^{x+1} \log 2}{1+4^x}\).

Question 20.
Find \(\frac{d y}{d x}\) when
(i) xy + xe^-y + ye^x = x²
(ii) e^x-y = log \(\left(\frac{x}{y}\right)\)
Solution:
(i) Given, xy + xe^-y + ye^x = x² …………..(1)
Diff. eqn. (1) w.r.t. x, taking y as a function of x, we have
x \(\frac{d}{d x}\) + y . 1 + x e^-y (- \(\frac{d y}{d x}\)) + e^-y + y e^x + e^x \(\frac{d y}{d x}\) = 2x
⇒ [x – xe^-y + e^x] \(\frac{d y}{d x}\) = 2x – y e^x – e^-y – y
∴ \(\frac{d y}{d x}\) = \(\frac{2 x-y e^x-e^{-y}-y}{x+e^x-x e^{-y}}\)

(ii) Given e^x-y = log \(\left(\frac{x}{y}\right)\)
Diff. eqn. (1) both sides w.r.t. x, we have
\(e^{x-y}\left[1-\frac{d y}{d x}\right]=\frac{1}{x}-\frac{1}{y} \frac{d y}{d x}\)
⇒ \(\left[\frac{1}{y}-e^{x-y}\right] \frac{d y}{d x}=\frac{1}{x}-e^{x-y}\)
⇒ \(\frac{\left[1-y e^{x-y}\right]}{y} \frac{d y}{d x}=\left[\frac{1-x e^{x-y}}{x}\right]\)
⇒ \(\frac{d y}{d x}=\frac{y\left(1-x e^{x-y}\right)}{x\left(1-y e^{x-y}\right)}\).

Question 21.
If log (x² + y² = 2 tan^-1 \(\frac{y}{x}\), show that \(\frac{d y}{d x}=\frac{x+y}{x-y}\).
Solution:
Given, log (x² + y² = 2 tan^-1 \(\frac{y}{x}\)
diff. both sides w.r.t. x ; we have
\(\frac{1}{x^2+y^2}\left[2 x+2 y \frac{d y}{d x}\right]=\frac{2}{1+\frac{y^2}{x^2}}\left[\frac{x \frac{d y}{d x}-y \cdot 1}{x^2}\right]\)
⇒ \(\frac{1}{x^2+y^2}\left[x+y \frac{d y}{d x}\right]=\frac{1}{x^2+y^2}\left[x \frac{d y}{d x}-y\right]\)
⇒ (y – x) \(\frac{d y}{d x}\) = – x – y
⇒ \(\frac{d y}{d x}\) = \(\frac{x+y}{x-y}\).

Question 22.
If y log x = x – y, prove that \(\frac{d y}{d x}\) = \(\frac{\log x}{(1+\log x)^2}\).
Solution:
Given y log x = x – y
⇒ y (1 + log x) = x
⇒ y = \(\frac{x}{1+\log x}\) ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{(1+\log x) \frac{d}{d x} x-x \frac{d}{d x}(1+\log x)}{(1+\log x)^2}\)
\(\frac{d y}{d x}=\frac{(1+\log x) \cdot 1-x\left(0+\frac{1}{x}\right)}{(1+\log x)^2}\)
= \(\frac{1+\log x-1}{(1+\log x)^2}\)
⇒ \(\frac{d y}{d x}=\frac{\log x}{(1+\log x)^2}\).

Question 23.
Differentiate the following functions w.r.t. x :
(i) log_x (2x – 3)
(ii) log_{cos x} sin x.
Solution:
(i) Let y = log_x (2x – 3)
= \(\frac{\log (2 x-3)}{\log x}\)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{\left[(\log x) \frac{d}{d x} \log (2 x-3)-\log (2 x-3) \frac{d}{d x} \log x\right]}{(\log x)^2}\)
= \(\frac{\left[\log x \cdot \frac{2}{2 x-3}-\frac{\log (2 x-3)}{x}\right]}{(\log x)^2}\)
= \(\left[\frac{2 x \log x-(2 x-3) \log (2 x-3)}{(2 x-3) x(\log x)^2}\right]\).

(ii) Let y = log_{cos x} sin x
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{\log \cos x \frac{d}{d x}(\log \sin x)-\log \sin x \frac{d}{d x} \log \cos x}{[\log \cos x]^2}\)
= \(\frac{\log \cos x \cdot \frac{\cos x}{\sin x}-(\log \sin x) \cdot\left(-\frac{\sin x}{\cos x}\right)}{(\log \cos x)^2}\)
= \(\frac{\cot x \log \cos x+\tan x \log \sin x}{(\log \cos x)^2}\).