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

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|.