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

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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