CBSE Class 12 Mathematics HOTs Continuity And Differentiability Set 04

Question. If \( x = a \sec^3 \theta \), \( y = a \tan^3 \theta \), find \( \frac{d^2y}{dx^2} \) at \( \theta = \frac{\pi}{4} \).
Answer: Here \( x = a \sec^3 \theta \)
\(\Rightarrow \frac{dx}{d\theta} = a \cdot 3\sec^2 \theta \cdot \sec \theta \tan \theta = 3a \sec^3 \theta \tan \theta \)
and \( y = a \tan^3 \theta \)
\(\Rightarrow \frac{dy}{d\theta} = a \cdot 3\tan^2 \theta \cdot \sec^2 \theta \)
\(\therefore \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a\tan^2\theta \sec^2\theta}{3a\sec^3\theta \tan\theta} = \frac{\tan\theta}{\sec\theta} = \sin\theta \)
On differentiating w.r.t. \( x \), we get
\( \frac{d^2y}{dx^2} = \cos\theta \cdot \frac{d\theta}{dx} = \frac{\cos\theta}{3a \sec^3\theta \tan\theta} = \frac{1}{3a} \cos^4 \theta \cdot \cot \theta \)
\(\therefore \left.\frac{d^2y}{dx^2}\right|_{\theta = \frac{\pi}{4}} = \frac{1}{3a} \cos^4 \frac{\pi}{4} \cdot \cot \frac{\pi}{4} = \frac{1}{3a} \left(\frac{1}{\sqrt{2}}\right)^4 \cdot 1 = \frac{1}{3a} \cdot \frac{1}{4} \cdot 1 = \frac{1}{12a} \)

Question. If \( y = A e^{mx} + B e^{nx} \), show that \( \frac{d^2y}{dx^2} - (m+n)\frac{dy}{dx} + mny = 0 \).
Answer: Given \( y = A e^{mx} + B e^{nx} \)
Differentiating w.r.t. \( x \), we get
\( \frac{dy}{dx} = m A e^{mx} + n B e^{nx} \)
\(\Rightarrow \frac{d^2y}{dx^2} = m^2 A e^{mx} + n^2 B e^{nx} \)
Now, L.H.S. \( = \frac{d^2y}{dx^2} - (m+n)\frac{dy}{dx} + mny \)
\( = m^2 A e^{mx} + n^2 B e^{nx} - (m+n)(m A e^{mx} + n B e^{nx}) + mn(A e^{mx} + B e^{nx}) \)
\( = A e^{mx}[m^2 - (m+n)m + mn] + B e^{nx}[n^2 - (m+n)n + mn] \)
\( = A e^{mx} \times 0 + B e^{nx} \times 0 = 0 = \text{R.H.S.} \)

Question. If \( x = a(\cos t + t \sin t) \) and \( y = a(\sin t - t \cos t) \), then find the value of \( \frac{d^2y}{dx^2} \) at \( t = \frac{\pi}{4} \).
Answer: Here, \( x = a(\cos t + t \sin t) \)
\(\Rightarrow \frac{dx}{dt} = a[-\sin t + \sin t + t \cos t] = at \cos t \) ...(1)
and \( y = a(\sin t - t \cos t) \)
\(\Rightarrow \frac{dy}{dt} = a[\cos t - (\cos t - t \sin t)] = at \sin t \)
\(\dots \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{at \sin t}{at \cos t} = \tan t \)
\(\Rightarrow \frac{d^2y}{dx^2} = \sec^2 t \cdot \frac{dt}{dx} = \sec^2 t \cdot \frac{1}{at \cos t} \) [Using (1)]
\( = \frac{1}{at \cos^3 t} \)
\(\therefore \left.\frac{d^2y}{dx^2}\right|_{t = \frac{\pi}{4}} = \frac{1}{a\left(\frac{\pi}{4}\right)\cos^3 \frac{\pi}{4}} = \frac{4}{\pi a} \cdot (\sqrt{2})^3 = \frac{8\sqrt{2}}{\pi a} \)

Question. If \( x = a\left(\cos t + \log \tan \frac{t}{2}\right) \), \( y = a \sin t \), evaluate \( \frac{d^2y}{dx^2} \) at \( t = \frac{\pi}{3} \).
Answer: Refer to answer 85. We get \( \frac{dy}{dx} = \tan t \)
Again differentiating w.r.t. \( x \), we get
\( \frac{d^2y}{dx^2} = \sec^2 t \cdot \frac{dt}{dx} = \frac{\sec^2 t}{a \cos^2 t / \sin t} = \frac{\sin t}{a \cos^4 t} \)
\(\therefore \left.\frac{d^2y}{dx^2}\right|_{t = \frac{\pi}{3}} = \frac{\sin \frac{\pi}{3}}{a \cos^4 \frac{\pi}{3}} = \frac{\sqrt{3}/2}{a(1/2)^4} = \frac{8\sqrt{3}}{a} \)

Question. If \( y = \log\left[x + \sqrt{x^2+a^2}\right] \), show that \( (x^2+a^2)\frac{d^2y}{dx^2} + x\frac{dy}{dx} = 0 \).
Answer: Given that \( y = \log\left(x + \sqrt{x^2+a^2}\right) \) ...(1)
Differentiating (1) w.r.t. 'x' on both sides, we get
\( \frac{dy}{dx} = \frac{1}{x + \sqrt{x^2+a^2}} \cdot \frac{d}{dx}\left(x + \sqrt{x^2+a^2}\right) \)
\( = \frac{1}{x + \sqrt{x^2+a^2}} \cdot \left[1 + \frac{1}{2\sqrt{x^2+a^2}} \cdot 2x\right] \)
\( = \frac{1}{x + \sqrt{x^2+a^2}} \cdot \frac{(\sqrt{x^2+a^2} + x)}{\sqrt{x^2+a^2}} \)
\(\Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{x^2+a^2}} \Rightarrow \sqrt{x^2+a^2} \frac{dy}{dx} = 1 \) ...(2)
Again differentiating (2) on both sides w.r.t. x, we get
\( \sqrt{x^2+a^2} \frac{d^2y}{dx^2} + \frac{2x}{2\sqrt{x^2+a^2}} \frac{dy}{dx} = 0 \)
\(\Rightarrow (x^2+a^2)\frac{d^2y}{dx^2} + x\frac{dy}{dx} = 0 \)

Question. If \( x = a \cos^3 \theta \) and \( y = a \sin^3 \theta \), then find the value of \( \frac{d^2y}{dx^2} \) at \( \theta = \frac{\pi}{6} \).
Answer: Here \( x = a \cos^3 \theta \), \( y = a \sin^3 \theta \)
\(\Rightarrow \frac{dx}{d\theta} = 3a \cos^2 \theta \cdot (-\sin\theta) \) and \( \frac{dy}{d\theta} = 3a \sin^2 \theta \cdot \cos\theta \)
\(\Rightarrow \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\theta} = -\tan\theta \)
Differentiating w.r.t. \( x \), we get
\( \frac{d^2y}{dx^2} = -\sec^2\theta \frac{d\theta}{dx} = \frac{-\sec^2\theta}{-3a\cos^2\theta\sin\theta} = \frac{1}{3a\cos^4\theta\sin\theta} \)
\(\therefore \left.\frac{d^2y}{dx^2}\right|_{\theta = \frac{\pi}{6}} = \frac{1}{3a \cos^4 \frac{\pi}{6} \cdot \sin \frac{\pi}{6}} = \frac{1}{3a \cdot \left(\frac{\sqrt{3}}{2}\right)^4 \cdot \frac{1}{2}} = \frac{32}{27a} \)

Question. If \( y = x \log\left(\frac{x}{a+bx}\right) \), then prove that \( x^3\frac{d^2y}{dx^2} = \left(x\frac{dy}{dx} - y\right)^2 \).
Answer: Here, \( y = x \log\left(\frac{x}{a+bx}\right) \) ... (1)
\(\Rightarrow y = x[\log x - \log(a+bx)] = x \log x - x \log(a+bx) \)
\(\Rightarrow \frac{dy}{dx} = x \cdot \frac{1}{x} + 1 \cdot \log x - \left[1 \cdot \log(a+bx) + x \cdot \frac{1}{a+bx} \cdot b\right] \)
\( = 1 - \frac{bx}{a+bx} + \log x - \log(a+bx) \)
\(\Rightarrow \frac{dy}{dx} = \frac{a}{a+bx} + \log\left(\frac{x}{a+bx}\right) \) ...(2)
\(\Rightarrow \frac{dy}{dx} = \frac{a}{a+bx} + \frac{y}{x} \) [Using (1)]
Again differentiating (2) w.r.t. x, we get
\( \frac{d^2y}{dx^2} = a \cdot (-1)(a+bx)^{-2} \cdot b + \frac{\left(x\frac{dy}{dx} - y\right)}{x^2} \)
\( = \frac{-ab}{(a+bx)^2} + \frac{a}{x(a+bx)} = \frac{-abx + a(a+bx)}{x(a+bx)^2} = \frac{a^2}{x(a+bx)^2} \)
Now, R.H.S. \( = \left(x \frac{dy}{dx} - y\right)^2 = \left\{x \cdot \left[\frac{a}{a+bx} + \frac{y}{x}\right] - y\right\}^2 = \left(\frac{ax}{a+bx}\right)^2 \)
and L.H.S. \( = x^3 \frac{d^2y}{dx^2} = x^3 \cdot \frac{a^2}{x(a+bx)^2} = \frac{a^2x^2}{(a+bx)^2} = \left(\frac{ax}{a+bx}\right)^2 = \text{R.H.S.} \)

Question. If \( x = \tan\left(\frac{1}{a} \log y\right) \), then show that \( (1+x^2)\frac{d^2y}{dx^2} + (2x-a)\frac{dy}{dx} = 0 \).
Answer: We have, \( x = \tan\left(\frac{1}{a} \log y\right) \)
\(\Rightarrow \frac{1}{a} \log y = \tan^{-1} x \)
Differentiating w.r.t. x, we get
\( \frac{1}{ay} \frac{dy}{dx} = \frac{1}{1+x^2} \Rightarrow (1+x^2)\frac{dy}{dx} = ay \)
Again differentiating w.r.t. x, we get
\( (1+x^2)\frac{d^2y}{dx^2} + 2x \frac{dy}{dx} = a \frac{dy}{dx} \)
\(\Rightarrow (1+x^2)\frac{d^2y}{dx^2} + (2x-a)\frac{dy}{dx} = 0 \)

Question. If \( x = \cos \theta \) and \( y = \sin^3 \theta \), then prove that \( y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 3\sin^2\theta(5\cos^2\theta - 1) \).
Answer: Given \( x = \cos \theta \) and \( y = \sin^3 \theta \)
\(\Rightarrow \frac{dx}{d\theta} = -\sin\theta \) and \( \frac{dy}{d\theta} = 3\sin^2\theta \cos\theta \)
\(\therefore \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3\sin^2\theta\cos\theta}{-\sin\theta} = -3\sin\theta\cos\theta \)
Differentiating w.r.t. x, we get
\( \frac{d^2y}{dx^2} = \frac{d}{d\theta}(-3\sin\theta\cos\theta) \cdot \frac{d\theta}{dx} = -3[\cos^2\theta - \sin^2\theta] \cdot \left(\frac{-1}{\sin\theta}\right) = \frac{3(\cos^2\theta - \sin^2\theta)}{\sin\theta} \)
Now, L.H.S. \( = y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 \)
\( = \sin^3\theta \left[\frac{3(\cos^2\theta - \sin^2\theta)}{\sin\theta}\right] + (-3\sin\theta\cos\theta)^2 \)
\( = 3\sin^2\theta(\cos^2\theta - \sin^2\theta) + 9\sin^2\theta\cos^2\theta \)
\( = 3\sin^2\theta(\cos^2\theta - \sin^2\theta + 3\cos^2\theta) \)
\( = 3\sin^2\theta(4\cos^2\theta - \sin^2\theta) \)
\( = 3\sin^2\theta(4\cos^2\theta - 1 + \cos^2\theta) \)
\( = 3\sin^2\theta(5\cos^2\theta - 1) = \text{R.H.S.} \)

Question. If \( y = \sin^{-1} x \), show that \( (1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 0 \).
Answer: Given \( y = \sin^{-1} x \)
Differentiating w.r.t. x, we get
\( \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} \)
Again differentiating w.r.t. x, we get
\( \frac{d^2y}{dx^2} = -\frac{1}{2}(1-x^2)^{-3/2} \cdot (-2x) = \frac{x}{(1-x^2)^{3/2}} \)
\(\Rightarrow \frac{d^2y}{dx^2} = \frac{x}{(1-x^2)\sqrt{1-x^2}} \)
\(\Rightarrow (1-x^2)\frac{d^2y}{dx^2} = \frac{x}{\sqrt{1-x^2}} \)
\(\Rightarrow (1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 0 \)

Question. If \( y = (\tan^{-1}x)^2 \), show that \( (x^2+1)^2\frac{d^2y}{dx^2} + 2x(x^2+1)\frac{dy}{dx} = 2 \).
Answer: \( y = (\tan^{-1}x)^2 \)
Differentiating w.r.t. x, we get
\( \frac{dy}{dx} = 2\tan^{-1}x \cdot \frac{d}{dx}(\tan^{-1}x) = \frac{2\tan^{-1}x}{1+x^2} \)
\(\Rightarrow (x^2+1)\frac{dy}{dx} = 2\tan^{-1}x \)
Again differentiating w.r.t. x, we get
\( (x^2+1)\frac{d^2y}{dx^2} + 2x\frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} \)
\(\Rightarrow (x^2+1)^2\frac{d^2y}{dx^2} + 2x(x^2+1)\frac{dy}{dx} = 2 \)

Question. If \( x = a(\cos t + t \sin t) \) and \( y = a(\sin t - t \cos t) \), \( 0 < t < \frac{\pi}{2} \), find \( \frac{d^2x}{dt^2} \), \( \frac{d^2y}{dt^2} \) and \( \frac{d^2y}{dx^2} \).
Answer: Refer to answer 95.
Since, we have \( \frac{dx}{dt} = at \cos t \)
\(\therefore \frac{d^2x}{dt^2} = a[-t\sin t + \cos t] = -at\sin t + a\cos t \)
and \( \frac{dy}{dt} = at \sin t \Rightarrow \frac{d^2y}{dt^2} = a[t\cos t + \sin t] \)

Question. If \( x = a(\theta - \sin\theta) \), \( y = a(1 + \cos\theta) \), find \( \frac{d^2y}{dx^2} \).
Answer: Refer to answer 84.
We have, \( \frac{dy}{dx} = \frac{-\sin\theta}{1-\cos\theta} = \frac{-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}} = -\cot\frac{\theta}{2} \)
\(\Rightarrow \frac{dy}{dx} = -\cot\frac{\theta}{2} \Rightarrow \frac{d^2y}{dx^2} = \frac{d}{d\theta}\left(-\cot\frac{\theta}{2}\right) \frac{d\theta}{dx} = \frac{1}{2}\text{cosec}^2\frac{\theta}{2} \cdot \frac{1}{a(1-\cos\theta)} \)
\( = \frac{1}{2}\text{cosec}^2\frac{\theta}{2} \cdot \frac{1}{2a\sin^2\frac{\theta}{2}} = \frac{\text{cosec}^4\frac{\theta}{2}}{4a} \)

Question. If \( x = a(\theta + \sin\theta) \) and \( y = a(1 - \cos\theta) \), find \( \frac{d^2y}{dx^2} \).
Answer: Refer to answer 109.

Question. If \( y = \text{cosec}^{-1} x \), \( x > 1 \), then show that \( x(x^2 - 1)\frac{d^2y}{dx^2} + (2x^2 - 1)\frac{dy}{dx} = 0 \).
Answer: We have, \( y = \text{cosec}^{-1} x \)
\( \frac{dy}{dx} = \frac{-1}{x\sqrt{x^2-1}} \Rightarrow x\sqrt{x^2-1}\frac{dy}{dx} = -1 \)
Differentiating w.r.t. x, we get
\( x\sqrt{x^2-1}\frac{d^2y}{dx^2} + \left[x \cdot \frac{2x}{2\sqrt{x^2-1}} + \sqrt{x^2-1}\right]\frac{dy}{dx} = 0 \)
\(\Rightarrow x\sqrt{x^2-1}\frac{d^2y}{dx^2} + \left[\frac{x^2+x^2-1}{\sqrt{x^2-1}}\right]\frac{dy}{dx} = 0 \)
\(\Rightarrow x(x^2-1)\frac{d^2y}{dx^2} + (2x^2-1)\frac{dy}{dx} = 0 \)

Question. If \( y = (\cot^{-1} x)^2 \), then show that \( (x^2 + 1)^2\frac{d^2y}{dx^2} + 2x(x^2 + 1)\frac{dy}{dx} = 2 \).
Answer: We have, \( y = (\cot^{-1}x)^2 \)
\(\Rightarrow \frac{dy}{dx} = 2(\cot^{-1}x) \times \frac{-1}{1+x^2} \)
\(\Rightarrow (x^2+1)\frac{dy}{dx} = -2\cot^{-1}x \)
\(\Rightarrow (x^2+1)^2\left(\frac{dy}{dx}\right)^2 = 4(\cot^{-1}x)^2 \)
\(\Rightarrow (x^2+1)^2\left(\frac{dy}{dx}\right)^2 = 4y \)
Differentiating w.r.t. x, we get
\( (x^2+1)^2 \left[2\frac{dy}{dx} \frac{d^2y}{dx^2}\right] + 2(x^2+1)(2x)\left(\frac{dy}{dx}\right)^2 = 4\frac{dy}{dx} \)
\(\therefore (x^2+1)^2 \frac{d^2y}{dx^2} + 2x(x^2+1)\frac{dy}{dx} = 2 \)

Question. If \( y = \frac{\sin^{-1}x}{\sqrt{1-x^2}} \), then show that \( (1-x^2)\frac{d^2y}{dx^2} - 3x\frac{dy}{dx} - y = 0 \).
Answer: We have, \( y = \frac{\sin^{-1}x}{\sqrt{1-x^2}} \)
\(\Rightarrow \sqrt{1-x^2} \cdot y = \sin^{-1}x \)
Differentiating w.r.t. x, we get
\( \sqrt{1-x^2}\frac{dy}{dx} - \frac{xy}{\sqrt{1-x^2}} = \frac{1}{\sqrt{1-x^2}} \)
\(\Rightarrow (1-x^2)\frac{dy}{dx} - xy = 1 \)
Again differentiating w.r.t. x, we get
\( (1-x^2)\frac{d^2y}{dx^2} - 2x \frac{dy}{dx} - x \frac{dy}{dx} - y = 0 \)
\(\Rightarrow (1-x^2)\frac{d^2y}{dx^2} - 3x\frac{dy}{dx} - y = 0 \)

Question. If \( y = e^x(\sin x + \cos x) \), then prove that \( \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0 \).
Answer: \( \frac{dy}{dx} = e^x(\sin x + \cos x) + e^x(\cos x - \sin x) = 2e^x \cos x \)
Again differentiating w.r.t. x, we get
\( \frac{d^2y}{dx^2} = 2e^x \cos x - 2e^x \sin x = 2e^x(\cos x - \sin x) \)
L.H.S. \( = \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y \)
\( = 2e^x(\cos x - \sin x) - 4e^x\cos x + 2[e^x(\sin x + \cos x)] \)
\( = 2e^x[\cos x - \sin x - 2\cos x + \sin x + \cos x] \)
\( = 2e^x \times 0 = 0 = \text{R.H.S.} \)

Question. If \( y = e^x \sin x \), then prove that \( \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0 \).
Answer: We have, \( y = e^x \sin x \)
\(\Rightarrow \frac{dy}{dx} = e^x \cos x + e^x \sin x \)
\(\Rightarrow \frac{dy}{dx} = e^x \cos x + y \) ...(1)
Again differentiating w.r.t. x, we get
\( \frac{d^2y}{dx^2} = e^x \cos x - e^x \sin x + \frac{dy}{dx} \)
\(\Rightarrow \frac{d^2y}{dx^2} = \left(\frac{dy}{dx} - y\right) - y + \frac{dy}{dx} \) [From (1)]
\(\Rightarrow \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0 \)

Question. If \( y = \sin(\log x) \), then prove that \( x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} + y = 0 \).
Answer: We have, \( y = \sin(\log x) \)
Differentiating w.r.t. x, we get
\( \frac{dy}{dx} = \cos(\log x) \times \frac{1}{x} = \frac{\cos(\log x)}{x} \)
\(\Rightarrow x \frac{dy}{dx} = \cos(\log x) \)
Again differentiating w.r.t. x, we get
\( x \frac{d^2y}{dx^2} + \frac{dy}{dx} = \frac{-\sin(\log x)}{x} \)
\(\Rightarrow x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = 0 \)

Question. If \( y = x + \tan x \), then prove that \( \cos^2 x \frac{d^2y}{dx^2} - 2y + 2x = 0 \).
Answer: We have, \( y = x + \tan x \)
\(\Rightarrow \frac{dy}{dx} = 1 + \sec^2 x \)
\(\Rightarrow \frac{d^2y}{dx^2} = 2\sec x \cdot \sec x \tan x \)
\(\Rightarrow \frac{d^2y}{dx^2} = \frac{2\tan x}{\cos^2 x} \Rightarrow \cos^2 x \cdot \frac{d^2y}{dx^2} = 2\tan x = 2(y-x) \)
\(\Rightarrow \cos^2 x \frac{d^2y}{dx^2} - 2y + 2x = 0 \)

Question. Verify Rolle’s theorem for the function \( f(x) = x^2 - 4x + 3 \) on \( [1, 3] \).
Answer: We have, \( f(x) = x^2 - 4x + 3 \)
(i) \( f(x) \) being a polynomial function is continuous in \( [1, 3] \)
(ii) \( f(x) \) being a polynomial function is differentiable in \( (1, 3) \)
(iii) \( f(3) = 3^2 - 4(3) + 3 = 0 \) and \( f(1) = 1^2 - 4(1) + 3 = 0 \). Thus \( f(1) = f(3) \)
Thus, all the conditions of Rolle's theorem are satisfied, so there exists atleast one point \( c \in (1, 3) \) such that \( f'(c) = 0 \)
\( f'(x) = 2x - 4 \Rightarrow f'(c) = 2c - 4 \)
\(\therefore f'(c) = 2c - 4 = 0 \Rightarrow c = 2 \in (1, 3) \)
Hence, the Rolle's theorem is verified.