CBSE Class 12 Mathematics HOTs Continuity And Differentiability Set 03

Very Short Answer Type Questions

Question. If the following function \( f(x) \) is continuous at \( x = 0 \), then write the value of \( k \). \[ f(x) = \begin{cases} \frac{\sin \frac{3x}{2}}{x}, & x \neq 0 \\ k, & x = 0 \end{cases} \]
Answer: Since the function \( f(x) \) is continuous at \( x = 0 \), we have: \[ \lim_{x \to 0} f(x) = f(0) \] \[ \lim_{x \to 0} \frac{\sin \frac{3x}{2}}{x} = k \] Multiplying and dividing the limit by \( \frac{3}{2} \): \[ \lim_{x \to 0} \left( \frac{\sin \frac{3x}{2}}{\frac{3x}{2}} \cdot \frac{3}{2} \right) = k \] Since \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \), we get: \[ 1 \cdot \frac{3}{2} = k \implies k = \frac{3}{2} \]

Question. Determine the value of the constant \( k \), so that function \[ f(x) = \begin{cases} \frac{kx}{|x|}, & \text{if } x < 0 \\ 3, & \text{if } x \geq 0 \end{cases} \] is continuous at \( x = 0 \).
Answer: Since the function \( f(x) \) is continuous at \( x = 0 \), we have: \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \] For \( x < 0 \), we have \( |x| = -x \). Therefore: \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{kx}{-x} = \lim_{x \to 0^-} (-k) = -k \] For \( x \geq 0 \), we have \( f(0) = 3 \) and \( \lim_{x \to 0^+} f(x) = 3 \).
Equating LHL and RHL: \[ -k = 3 \implies k = -3 \]

Question. If the function \( f \) defined as \( f(x) = \begin{cases} \frac{x^2-9}{x-3}, & x \neq 3 \\ k, & x = 3 \end{cases} \) is continuous at \( x = 3 \), find the value of \( k \).
Answer: Given that \( f(x) \) is continuous at \( x = 3 \), we have: \[ \lim_{x \to 3} f(x) = f(3) \] \[ \lim_{x \to 3} \frac{x^2-9}{x-3} = k \] \[ \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = k \] \[ \lim_{x \to 3} (x+3) = k \] \[ 3 + 3 = k \implies k = 6 \]

Question. Determine the value of \( k \) for which the following function is continuous at \( x = 3 \): \[ f(x) = \begin{cases} \frac{(x+3)^2-36}{x-3}, & x \neq 3 \\ k, & x = 3 \end{cases} \]
Answer: Given that \( f(x) \) is continuous at \( x = 3 \), we have: \[ \lim_{x \to 3} f(x) = f(3) \] \[ \lim_{x \to 3} \frac{(x+3)^2-36}{x-3} = k \] \[ \lim_{x \to 3} \frac{(x+3)^2-6^2}{x-3} = k \] \[ \lim_{x \to 3} \frac{(x+3-6)(x+3+6)}{x-3} = k \] \[ \lim_{x \to 3} \frac{(x-3)(x+9)}{x-3} = k \] \[ \lim_{x \to 3} (x+9) = k \] \[ 3 + 9 = k \implies k = 12 \]

Question. Prove that the greatest integer function defined by \( f(x) = [x], 0 < x < 2 \) is not differentiable at \( x = 1 \).
Answer: We have \( f(x) = [x] \) for \( 0 < x < 2 \).
Let us find the Left Hand Limit (LHL) and Right Hand Limit (RHL) at \( x = 1 \): \[ \text{LHL at } x = 1: \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} [x] = \lim_{h \to 0} [1-h] = \lim_{h \to 0} 0 = 0 \] \[ \text{RHL at } x = 1: \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} [x] = \lim_{h \to 0} [1+h] = \lim_{h \to 0} 1 = 1 \] And \( f(1) = [1] = 1 \).
Since \( \text{LHL} \neq \text{RHL} \), the function \( f(x) \) is discontinuous at \( x = 1 \).
Since every differentiable function must be continuous, and \( f(x) \) is not continuous at \( x = 1 \), it is not differentiable at \( x = 1 \).

Question. Find the derivative of \( \cos(\sqrt{x}) \).
Answer: Let \( y = \cos(\sqrt{x}) \).
On differentiating both sides w.r.t. \( x \), we get: \[ \frac{dy}{dx} = \frac{d}{dx}(\cos(\sqrt{x})) \] Using the chain rule: \[ \frac{dy}{dx} = -\sin(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x}) \] \[ = -\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \] \[ = \frac{-\sin(\sqrt{x})}{2\sqrt{x}} \]

Question. If \( f(x) = x|x| \), then find \( f'(x) \).
Answer: Given, \( f(x) = x|x| \).
We can rewrite the function as: \[ f(x) = \begin{cases} -x^2, & x < 0 \\ x^2, & x \geq 0 \end{cases} \] Differentiating both sides w.r.t. \( x \): \[ f'(x) = \begin{cases} -2x, & x < 0 \\ 2x, & x > 0 \end{cases} \] Thus, we can express the derivative as: \[ f'(x) = 2|x| \]

Short Answer Type Questions

Question. Determine the values of \( a \) and \( b \) such that the following function is continuous at \( x = 0 \): \[ f(x) = \begin{cases} \frac{x + \sin x}{\sin(a+1)x}, & \text{if } -\pi < x < 0 \\ 2, & \text{if } x = 0 \\ \frac{2(e^{\sin bx} - 1)}{bx}, & \text{if } x > 0 \end{cases} \]
Answer: Since \( f(x) \) is continuous at \( x = 0 \), we have: \[ \text{LHL} = \text{RHL} = f(0) = 2 \] Evaluate the LHL: \[ \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x+\sin x}{\sin(a+1)x} \] Dividing the numerator and denominator by \( x \): \[ = \lim_{x \to 0^-} \frac{1 + \frac{\sin x}{x}}{(a+1) \frac{\sin(a+1)x}{(a+1)x}} = \frac{1+1}{(a+1)(1)} = \frac{2}{a+1} \] Equating LHL to \( 2 \): \[ \frac{2}{a+1} = 2 \implies a + 1 = 1 \implies a = 0 \] Evaluate the RHL: \[ \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{2(e^{\sin bx}-1)}{bx} \] Dividing and multiplying by \( \sin bx \): \[ = \lim_{x \to 0^+} 2 \left( \frac{e^{\sin bx}-1}{\sin bx} \cdot \frac{\sin bx}{bx} \right) \] Using the limits \( \lim_{t \to 0} \frac{e^t-1}{t} = 1 \) and \( \lim_{t \to 0} \frac{\sin t}{t} = 1 \): \[ = 2 \cdot (1) \cdot (1) = 2 \] Since this holds for any non-zero \( b \), the function is continuous at \( x = 0 \) for \( a = 0 \) and any \( b \in \mathbb{R}, b \neq 0 \).

Question. Find the value of \( k \) so that the function \[ f(x) = \begin{cases} \frac{1-\cos 4x}{8x^2}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases} \] is continuous at \( x = 0 \).
Answer: Since the function is continuous at \( x = 0 \), we have: \[ \lim_{x \to 0} f(x) = f(0) = k \] \[ \lim_{x \to 0} \frac{1-\cos 4x}{8x^2} = k \] Using the trigonometric identity \( 1-\cos 2\theta = 2\sin^2\theta \): \[ \lim_{x \to 0} \frac{2\sin^2 2x}{8x^2} = k \] \[ \lim_{x \to 0} \frac{\sin^2 2x}{4x^2} = k \] \[ \lim_{x \to 0} \left( \frac{\sin 2x}{2x} \right)^2 = k \] Since \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \): \[ (1)^2 = k \implies k = 1 \]

Question. Find the value of \( k \), so that the following function is continuous at \( x = 2 \): \[ f(x) = \begin{cases} \frac{x^3+x^2-16x+20}{(x-2)^2}, & \text{if } x \neq 2 \\ k, & \text{if } x = 2 \end{cases} \]
Answer: Since the function \( f(x) \) is continuous at \( x = 2 \), we have: \[ \lim_{x \to 2} f(x) = f(2) = k \] Factoring the numerator \( x^3+x^2-16x+20 \): As \( x = 2 \) is a root, we can divide the expression by \( (x-2) \): \[ x^3+x^2-16x+20 = (x-2)(x^2+3x-10) = (x-2)(x-2)(x+5) = (x-2)^2(x+5) \] Now, evaluate the limit: \[ \lim_{x \to 2} \frac{(x-2)^2(x+5)}{(x-2)^2} = k \] \[ \lim_{x \to 2} (x+5) = k \] \[ 2 + 5 = k \implies k = 7 \]

Question. Examine the continuity of the following function at \( x = 0 \): \[ f(x) = \begin{cases} \frac{x}{2|x|}, & \text{if } x \neq 0 \\ \frac{1}{2}, & \text{if } x = 0 \end{cases} \]
Answer: Let us evaluate the LHL and RHL at \( x = 0 \): \[ \text{LHL} = \lim_{x \to 0^-} \frac{x}{2|x|} = \lim_{x \to 0^-} \frac{x}{2(-x)} = -\frac{1}{2} \] \[ \text{RHL} = \lim_{x \to 0^+} \frac{x}{2|x|} = \lim_{x \to 0^+} \frac{x}{2x} = \frac{1}{2} \] Since \( \text{LHL} \neq \text{RHL} \), the limit does not exist.
Therefore, \( f(x) \) is discontinuous at \( x = 0 \).

Question. Find the point of discontinuity of the function \[ f(x) = \begin{cases} x^{10}-1, & \text{if } x \leq 1 \\ x^2, & \text{if } x > 1 \end{cases} \]
Answer: Let us check the continuity at the transition point \( x = 1 \): \[ \text{LHL} = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^{10}-1) = 1^{10} - 1 = 0 \] \[ \text{RHL} = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} x^2 = 1^2 = 1 \] Since \( \text{LHL} \neq \text{RHL} \), the function \( f(x) \) is discontinuous at \( x = 1 \).
Hence, \( x = 1 \) is the point of discontinuity.

Question. Find the point of discontinuity of the function \[ f(x) = \begin{cases} \frac{\sin 2x}{x}, & \text{if } x < 0 \\ x+2, & \text{if } x \geq 0 \end{cases} \]
Answer: Let us check the continuity at \( x = 0 \): \[ \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin 2x}{x} = \lim_{x \to 0^-} 2 \left( \frac{\sin 2x}{2x} \right) = 2 \] \[ \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x+2) = 0 + 2 = 2 \] And \( f(0) = 0 + 2 = 2 \).
Since \( \text{LHL} = \text{RHL} = f(0) \), \( f(x) \) is continuous at \( x = 0 \).
Since \( f(x) \) is continuous everywhere else in its domain, there are no points of discontinuity.

Long Answer Type Questions

Question. Find the derivative of the following, function w.r.t. \( x \) at \( x = 1 \).
\[ \cos^{-1}\left[\sin \sqrt{\frac{1 + x}{2}}\right] + x^x \]
Answer: Let \( y = \cos^{-1}\left[\sin \sqrt{\frac{1 + x}{2}}\right] + x^x \)
\( y = \cos^{-1}\left[\cos\left(\frac{\pi}{2} - \sqrt{\frac{1 + x}{2}}\right)\right] + x^x = \frac{\pi}{2} - \sqrt{\frac{1 + x}{2}} + x^x \)
\( \frac{dy}{dx} = -\frac{1}{2\sqrt{\frac{1 + x}{2}}} \cdot \frac{1}{2} + x^x(1 + \log x) \) [since \( \frac{d}{dx}(x^x) = x^x(1 + \log x) \)]
At \( x = 1 \):
\( \frac{dy}{dx} = -\frac{1}{2\sqrt{\frac{1+1}{2}}} \cdot \frac{1}{2} + 1^1(1 + \log 1) = -\frac{1}{4} + 1 = \frac{3}{4} \)

Question. If \( y = (\sin x - \cos x)^{(\sin x - \cos x)} \), \( \frac{\pi}{4} < x < \frac{3\pi}{4} \), then find \( \frac{dy}{dx} \).
Answer: Let \( y = (\sin x - \cos x)^{\sin x - \cos x} \)
On taking log both sides, we get:
\( \log y = (\sin x - \cos x)\log(\sin x - \cos x) \)
On differentiating both sides w.r.t. \( x \), we get:
\( \frac{1}{y} \frac{dy}{dx} = \frac{\sin x - \cos x}{\sin x - \cos x}(\cos x + \sin x) + (\cos x + \sin x)\log(\sin x - \cos x) \)
\( \Rightarrow \frac{dy}{dx} = y[(\cos x + \sin x) + (\cos x + \sin x)\log(\sin x - \cos x)] \)
\( = (\sin x - \cos x)^{\sin x - \cos x} (\cos x + \sin x)[1 + \log(\sin x - \cos x)] \)

Question. If \( x = \cos t (3 - 2\cos^2 t) \) and \( y = \sin t (3 - 2\sin^2 t) \), then find the value of \( \frac{dy}{dx} \) at \( t = \frac{\pi}{4} \).
Answer: Given, \( x = \cos t (3 - 2\cos^2 t) \Rightarrow x = 3\cos t - 2\cos^3 t \)
On differentiating both sides w.r.t. \( t \), we get:
\( \frac{dx}{dt} = 3(-\sin t) - 2(3)\cos^2 t (-\sin t) = -3\sin t + 6\cos^2 t \sin t \) ...(i)
Also, \( y = \sin t (3 - 2\sin^2 t) \Rightarrow y = 3\sin t - 2\sin^3 t \)
On differentiating both sides w.r.t. \( t \), we get:
\( \frac{dy}{dt} = 3\cos t - 6\sin^2 t \cos t \) ...(ii)
Now, \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3\cos t - 6\cos t \sin^2 t}{-3\sin t + 6\cos^2 t \sin t} \)
\( = \frac{\cos t(1 - 2\sin^2 t)}{-\sin t(1 - 2\cos^2 t)} = \frac{\cos t(1 - 2\sin^2 t)}{\sin t(2\cos^2 t - 1)} = \frac{\cos t \cos 2t}{\sin t \cos 2t} = \cot(t) \)
\( \left. \frac{dy}{dx} \right|_{t = \frac{\pi}{4}} = \cot\left(\frac{\pi}{4}\right) = 1 \)

Question. If \( x = a \sin 2t (1 + \cos 2t) \) and \( y = b \cos 2t (1 - \cos 2t) \), then find the values of \( \frac{dy}{dx} \) at \( t = \frac{\pi}{4} \) and \( t = \frac{\pi}{3} \).
Answer: Given, \( x = a \sin 2t (1 + \cos 2t) \) and \( y = b \cos 2t (1 - \cos 2t) \).
On differentiating \( x \) and \( y \) separately w.r.t. \( t \), we get:
\( \frac{dx}{dt} = a \left[\sin 2t \frac{d}{dt}(1 + \cos 2t) + (1 + \cos 2t) \frac{d}{dt}(\sin 2t)\right] \)
\( = a[\sin 2t \times (0 - 2\sin 2t) + (1 + \cos 2t)(2\cos 2t)] \)
\( = a(-2\sin^2 2t + 2\cos 2t + 2\cos^2 2t) = a[2(\cos^2 2t - \sin^2 2t) + 2\cos 2t] \)
\( = a(2\cos 4t + 2\cos 2t) = 2a(\cos 4t + \cos 2t) \) [since \( \cos^2 \theta - \sin^2 \theta = \cos 2\theta \)]
and \( \frac{dy}{dt} = b\left[\cos 2t \frac{d}{dt}(1 - \cos 2t) + (1 - \cos 2t)\frac{d}{dt}(\cos 2t)\right] \)
\( = b[\cos 2t \times (0 + 2\sin 2t) + (1 - \cos 2t)(-2\sin 2t)] \)
\( = b(2\sin 2t \cos 2t - 2\sin 2t + 2\sin 2t \cos 2t) = 2b(2\sin 2t \cos 2t - \sin 2t) \)
\( = 2b(\sin 4t - \sin 2t) \) [since \( 2\sin 2\theta \cos 2\theta = \sin 4\theta \)]
\( = 2b \left[2\cos\left(\frac{4t + 2t}{2}\right) \cdot \sin\left(\frac{4t - 2t}{2}\right)\right] = 4b \cos 3t \sin t \)
And \( \frac{dx}{dt} = 2a(\cos 4t + \cos 2t) = 4a \cos 3t \cos t \).
Now, \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4b \cos 3t \sin t}{4a \cos 3t \cos t} = \frac{b}{a}\tan t \).
At \( t = \frac{\pi}{4} \): \( \frac{dy}{dx} = \frac{b}{a}\tan\frac{\pi}{4} = \frac{b}{a} \).
At \( t = \frac{\pi}{3} \): \( \frac{dy}{dx} = \frac{b}{a}\tan\frac{\pi}{3} = \frac{\sqrt{3}b}{a} \).

Question. If \( x = a e^t (\sin t + \cos t) \) and \( y = a e^t (\sin t - \cos t) \), then prove that \( \frac{dy}{dx} = \frac{x + y}{x - y} \).
Answer: Given, \( x = a e^t (\sin t + \cos t) \) and \( y = a e^t (\sin t - \cos t) \).
\( \therefore \frac{dx}{dt} = a[e^t(\cos t - \sin t) + e^t(\sin t + \cos t)] = 2a e^t \cos t \)
and \( \frac{dy}{dt} = a[e^t(\cos t + \sin t) + e^t(\sin t - \cos t)] = 2a e^t \sin t \).
Now, \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a e^t \sin t}{2a e^t \cos t} = \tan t \).
Also, \( \frac{x + y}{x - y} = \frac{a e^t (\sin t + \cos t) + a e^t (\sin t - \cos t)}{a e^t (\sin t + \cos t) - a e^t (\sin t - \cos t)} = \frac{2\sin t}{2\cos t} = \tan t \).
\( \therefore \frac{dy}{dx} = \frac{x + y}{x - y} \). Hence proved.

Question. Differentiate \( \tan^{-1}\left(\frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}}\right) \) w.r.t. \( \cos^{-1} x^2 \).
Answer: Let \( u = \tan^{-1}\left(\frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}}\right) \) and \( v = \cos^{-1} x^2 \Rightarrow x^2 = \cos v \).
\( \therefore u = \tan^{-1}\left(\frac{\sqrt{1 + \cos v} - \sqrt{1 - \cos v}}{\sqrt{1 + \cos v} + \sqrt{1 - \cos v}}\right) \)
\( = \tan^{-1}\left(\frac{\sqrt{2}\cos\frac{v}{2} - \sqrt{2}\sin\frac{v}{2}}{\sqrt{2}\cos\frac{v}{2} + \sqrt{2}\sin\frac{v}{2}}\right) = \tan^{-1}\left[\tan\left(\frac{\pi}{4} - \frac{v}{2}\right)\right] = \frac{\pi}{4} - \frac{v}{2} \).
Now, \( \frac{du}{dv} = -\frac{1}{2} \).

Question. Differentiate \( \tan^{-1}\left(\frac{x}{1 + \sqrt{1 - x^2}}\right) \) w.r.t. \( \sin\left(2 \cot^{-1} \sqrt{\frac{1 + x}{1 - x}}\right) \).
Answer: Let \( u = \tan^{-1}\left(\frac{x}{1 + \sqrt{1 - x^2}}\right) \). Put \( x = \sin \theta \Rightarrow \theta = \sin^{-1} x \).
\( u = \tan^{-1}\left(\frac{\sin \theta}{1 + \cos \theta}\right) = \tan^{-1}\left(\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}\right) = \tan^{-1}\left(\tan\frac{\theta}{2}\right) = \frac{\theta}{2} = \frac{1}{2}\sin^{-1} x \).
On differentiating w.r.t. \( x \), we get:
\( \frac{du}{dx} = \frac{1}{2\sqrt{1 - x^2}} \).
Let \( v = \sin\left(2 \cot^{-1} \sqrt{\frac{1 + x}{1 - x}}\right) \). Put \( x = \cos \phi \Rightarrow \phi = \cos^{-1} x \).
\( v = \sin\left(2 \cot^{-1} \sqrt{\frac{1 + \cos \phi}{1 - \cos \phi}}\right) = \sin\left(2 \cot^{-1}\left(\cot\frac{\phi}{2}\right)\right) = \sin\left(2 \cdot \frac{\phi}{2}\right) = \sin \phi = \sqrt{1 - x^2} \).
On differentiating w.r.t. \( x \), we get:
\( \frac{dv}{dx} = -\frac{x}{\sqrt{1 - x^2}} \).
Now, \( \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\frac{1}{2\sqrt{1 - x^2}}}{-\frac{x}{\sqrt{1 - x^2}}} = -\frac{1}{2x} \).

Question. Differentiate \( \tan^{-1}\left(\frac{x}{\sqrt{1 - x^2}}\right) \) with respect to \( \sin^{-1}(2x\sqrt{1 - x^2}) \).
Answer: Let \( u = \tan^{-1}\left(\frac{x}{\sqrt{1 - x^2}}\right) \). On putting \( x = \sin \theta \Rightarrow \theta = \sin^{-1} x \), then:
\( u = \tan^{-1}\left(\frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}}\right) = \tan^{-1}\left(\frac{\sin \theta}{\cos \theta}\right) = \tan^{-1}(\tan \theta) \)
\( \Rightarrow u = \theta \Rightarrow u = \sin^{-1} x \).
On differentiating both sides w.r.t. \( x \), we get:
\( \frac{du}{dx} = \frac{1}{\sqrt{1 - x^2}} \) ...(i)
Again, let \( v = \sin^{-1}(2x\sqrt{1 - x^2}) \). On putting \( x = \sin \theta \Rightarrow \theta = \sin^{-1} x \), then:
\( v = \sin^{-1}(2\sin \theta \sqrt{1 - \sin^2 \theta}) = \sin^{-1}(2\sin \theta \cos \theta) = \sin^{-1}(\sin 2\theta) \)
\( \Rightarrow v = 2\theta \Rightarrow v = 2\sin^{-1} x \).
On differentiating both sides w.r.t. \( x \), we get:
\( \frac{dv}{dx} = \frac{2}{\sqrt{1 - x^2}} \) ...(ii)
Now, \( \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{1/\sqrt{1 - x^2}}{2/\sqrt{1 - x^2}} = \frac{1}{2} \).

Question. If \( x = a \cos t \) and \( y = b \sin t \), then find \( \frac{d^2y}{dx^2} \).
Answer: Given, \( x = a \cos t \Rightarrow \frac{dx}{dt} = -a \sin t \)
and \( y = b \sin t \Rightarrow \frac{dy}{dt} = b \cos t \).
\( \therefore \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{b \cos t}{-a \sin t} = -\frac{b}{a}\cot t \).
Now, \( \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{b}{a}\cot t\right) = -\frac{b}{a}\left(-\csc^2 t\right)\frac{dt}{dx} \)
\( = \frac{b}{a}\csc^2 t \left(-\frac{1}{a\sin t}\right) = -\frac{b}{a^2}\csc^3 t \).

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: Given, \( x = a\cos^3\theta \Rightarrow \frac{dx}{d\theta} = -3a\cos^2\theta \sin\theta \)
and \( y = a\sin^3\theta \Rightarrow \frac{dy}{d\theta} = 3a\sin^2\theta \cos\theta \).
\( \therefore \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a\sin^2\theta \cos\theta}{-3a\cos^2\theta \sin\theta} = -\tan\theta \).
Again, \( \frac{d^2y}{dx^2} = \frac{d}{dx}(-\tan\theta) = -\sec^2\theta \frac{d\theta}{dx} \)
\( = -\sec^2\theta \left(-\frac{1}{3a\cos^2\theta \sin\theta}\right) = \frac{1}{3a\cos^4\theta \sin\theta} \).
At \( \theta = \frac{\pi}{6} \):
\( \frac{d^2y}{dx^2} = \frac{1}{3a\cos^4\frac{\pi}{6}\sin\frac{\pi}{6}} = \frac{1}{3a\left(\frac{\sqrt{3}}{2}\right)^4 \left(\frac{1}{2}\right)} = \frac{1}{3a\left(\frac{9}{16}\right)\left(\frac{1}{2}\right)} = \frac{32}{27a} \).

Question. If \( (x - a)^2 + (y - b)^2 = c^2 \), for some \( c > 0 \), then prove that \( \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}} \) is a constant independent of \( a \) and \( b \).
Answer: On differentiating both sides w.r.t. \( x \), we get:
\( \frac{d}{dx}[(x-a)^2 + (y-b)^2] = \frac{d}{dx}(c^2) \)
\( \Rightarrow 2(x - a) + 2(y - b)\frac{dy}{dx} = 0 \)
\( \Rightarrow \frac{dy}{dx} = -\frac{x - a}{y - b} \).
Again, on differentiating both sides w.r.t. \( x \), we get:
\( \frac{d^2y}{dx^2} = \frac{d}{dx}\left[-\frac{x - a}{y - b}\right] = -\frac{(y - b) \frac{d}{dx}(x-a) - (x-a)\frac{d}{dx}(y-b)}{(y-b)^2} \)
\( = -\frac{(y - b)(1) - (x - a)\frac{dy}{dx}}{(y - b)^2} = -\frac{(y - b) + \frac{(x - a)^2}{y - b}}{(y - b)^2} = -\frac{(y - b)^2 + (x - a)^2}{(y - b)^3} \)
\( = -\frac{c^2}{(y - b)^3} \) [since \( (x-a)^2 + (y-b)^2 = c^2 \)].
Now, \( \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}} = \frac{\left[1 + \left(-\frac{x - a}{y - b}\right)^2\right]^{3/2}}{-\frac{c^2}{(y - b)^3}} = \frac{\left[\frac{(y-b)^2 + (x-a)^2}{(y-b)^2}\right]^{3/2}}{-\frac{c^2}{(y-b)^3}} = \frac{\left[\frac{c^2}{(y-b)^2}\right]^{3/2}}{-\frac{c^2}{(y-b)^3}} = \frac{\frac{c^3}{(y-b)^3}}{-\frac{c^2}{(y-b)^3}} = -c \), which is constant and independent of \( a \) and \( b \). Hence proved.

Question. If \( y = \log\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 \), then prove that \( x(x + 1)^2 y_2 + (x + 1)^2 y_1 = 2 \).
Answer: We have, \( y = \log\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 = 2\log\left(\frac{x+1}{\sqrt{x}}\right) = 2\left[\log(x + 1) - \frac{1}{2}\log x\right] = 2\log(x + 1) - \log x \).
On differentiating both sides w.r.t. \( x \), we get:
\( y_1 = \frac{2}{x + 1} - \frac{1}{x} = \frac{2x - (x + 1)}{x(x + 1)} = \frac{x - 1}{x(x + 1)} \) ...(i)
Again, on differentiating both sides w.r.t. \( x \), we get:
\( y_2 = \frac{x(x + 1)\frac{d}{dx}(x-1) - (x-1)\frac{d}{dx}[x(x+1)]}{[x(x+1)]^2} \)
\( = \frac{(x^2+x)(1) - (x-1)(2x+1)}{x^2(x+1)^2} = \frac{x^2+x - (2x^2-x-1)}{x^2(x+1)^2} = \frac{-x^2+2x+1}{x^2(x+1)^2} \).
\( \Rightarrow y_2 = \frac{-x^2 + 2x + 1}{x^2(x+1)^2} \Rightarrow x(x + 1)^2 y_2 = \frac{-x^2 + 2x + 1}{x} = \frac{2x - (x + 1)(x - 1)}{x} = 2 - \frac{(x + 1)(x - 1)}{x} \).
From Eq. (i), we have \( (x+1)^2 y_1 = \frac{(x+1)(x-1)}{x} \).
Thus, \( x(x + 1)^2 y_2 + (x + 1)^2 y_1 = 2 \). Hence proved.

Question. If \( x\cos(a + y) = \cos y \), then prove that \( \frac{dy}{dx} = \frac{\cos^2(a + y)}{\sin a} \). Hence, show that \( \sin a \frac{d^2y}{dx^2} + \sin 2(a + y)\frac{dy}{dx} = 0 \).
Answer: Given, \( x\cos(a + y) = \cos y \Rightarrow x = \frac{\cos y}{\cos(a + y)} \).
On differentiating both sides w.r.t. \( y \), we get:
\( \frac{dx}{dy} = \frac{\cos(a + y) \frac{d}{dy}(\cos y) - \cos y \frac{d}{dy}[\cos(a + y)]}{\cos^2(a + y)} \) [using quotient rule]
\( = \frac{\cos(a + y)(-\sin y) - \cos y(-\sin(a + y))}{\cos^2(a + y)} = \frac{\sin(a + y)\cos y - \cos(a + y)\sin y}{\cos^2(a + y)} \)
\( \Rightarrow \frac{dx}{dy} = \frac{\sin(a + y - y)}{\cos^2(a + y)} = \frac{\sin a}{\cos^2(a + y)} \) [since \( \sin A \cos B - \cos A \sin B = \sin(A - B) \)]
\( \Rightarrow \frac{dy}{dx} = \frac{\cos^2(a + y)}{\sin a} \) ...(i)
Again, on differentiating both sides of Eq. (i) w.r.t. \( x \), we get:
\( \frac{d^2y}{dx^2} = \frac{1}{\sin a} \frac{d}{dx}[\cos^2(a+y)] = \frac{1}{\sin a} \cdot 2\cos(a+y)[-\sin(a+y)] \frac{dy}{dx} = -\frac{2\sin(a+y)\cos(a+y)}{\sin a} \frac{dy}{dx} \)
\( \Rightarrow \frac{d^2y}{dx^2} = -\frac{\sin 2(a + y)}{\sin a} \frac{dy}{dx} \) [since \( 2\sin\theta\cos\theta = \sin 2\theta \)]
\( \therefore \sin a \frac{d^2y}{dx^2} + \sin 2(a + y)\frac{dy}{dx} = 0 \). Hence proved.

Question. If \( y = (\sin^{-1} x)^2 \), then find \( (1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} \).
Answer: Given, \( y = (\sin^{-1} x)^2 \) ...(i)
On differentiating w.r.t. \( x \), we get:
\( \frac{dy}{dx} = 2\sin^{-1} x \cdot \frac{1}{\sqrt{1 - x^2}} \) ...(ii)
Again, on differentiating w.r.t. \( x \), we get:
\( \frac{d^2y}{dx^2} = \frac{(\sqrt{1-x^2}) \frac{d}{dx}(2\sin^{-1} x) - (2\sin^{-1} x) \frac{d}{dx}(\sqrt{1-x^2})}{1 - x^2} \)
\( = \frac{\sqrt{1 - x^2} \cdot \frac{2}{\sqrt{1-x^2}} - (2\sin^{-1} x) \cdot \frac{-2x}{2\sqrt{1-x^2}}}{1 - x^2} = \frac{2 + \frac{2x\sin^{-1} x}{\sqrt{1-x^2}}}{1 - x^2} \)
\( \Rightarrow (1 - x^2)\frac{d^2y}{dx^2} = 2 + x \left(\frac{2\sin^{-1} x}{\sqrt{1 - x^2}}\right) \)
\( \Rightarrow (1 - x^2)\frac{d^2y}{dx^2} = 2 + x\frac{dy}{dx} \) [from Eq. (ii)]
\( \Rightarrow (1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 2 \).

Question. If \( x = a \sec^3 \theta \) and \( y = a \tan^3 \theta \), then find \( \frac{d^2y}{dx^2} \) at \( \theta = \frac{\pi}{3} \).
Answer: Given, \( y = a\tan^3\theta \) and \( x = a\sec^3\theta \).
On differentiating w.r.t. \( \theta \), we get:
\( \frac{dy}{d\theta} = 3a\tan^2\theta \sec^2\theta \) and \( \frac{dx}{d\theta} = 3a\sec^2\theta \sec\theta\tan\theta = 3a\sec^3\theta \tan\theta \).
\( \therefore \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a\tan^2\theta \sec^2\theta}{3a\sec^3\theta \tan\theta} = \frac{\tan\theta}{\sec\theta} = \sin\theta \).
Again, on differentiating both sides w.r.t. \( x \), we get:
\( \frac{d^2y}{dx^2} = \frac{d}{dx}(\sin\theta) = \frac{d}{d\theta}(\sin\theta) \cdot \frac{d\theta}{dx} = \cos\theta \frac{1}{3a\sec^3\theta \tan\theta} \)
\( = \frac{\cos\theta}{3a \frac{1}{\cos^3\theta} \frac{\sin\theta}{\cos\theta}} = \frac{\cos^5\theta}{3a\sin\theta} \).
At \( \theta = \frac{\pi}{3} \):
\( \frac{d^2y}{dx^2} = \frac{\cos^5\frac{\pi}{3}}{3a\sin\frac{\pi}{3}} = \frac{(1/2)^5}{3a(\sqrt{3}/2)} = \frac{1/32}{3a\sqrt{3}/2} = \frac{1}{48\sqrt{3}a} \).

Question. If \( x = \sin t \) and \( y = \sin pt \), then prove that \( (1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} + p^2 y = 0 \).
Answer: Given, \( x = \sin t \) and \( y = \sin pt \).
On differentiating \( x \) and \( y \) separately w.r.t. \( t \), we get:
\( \frac{dx}{dt} = \cos t \) and \( \frac{dy}{dt} = p\cos pt \).
\( \Rightarrow \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{p\cos pt}{\cos t} \).
Now, on differentiating both sides w.r.t. \( x \), we get:
\( \frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{p\cos pt}{\cos t}\right] = \frac{\cos t \frac{d}{dx}(p\cos pt) - p\cos pt \frac{d}{dx}(\cos t)}{\cos^2 t} \)
\( = \frac{\cos t \left(-p^2\sin pt \frac{dt}{dx}\right) - p\cos pt\left(-\sin t \frac{dt}{dx}\right)}{\cos^2 t} \)
\( = \frac{-p^2 \cos t \sin pt + p\cos pt \sin t}{\cos^2 t} \cdot \frac{1}{\cos t} = \frac{-p^2 \cos t \sin pt + p\cos pt \sin t}{\cos^3 t} \)
\( \Rightarrow \cos^2 t \frac{d^2y}{dx^2} = \frac{-p^2 \cos t \sin pt + p\cos pt \sin t}{\cos t} = -p^2\sin pt + p\tan t \cos pt \)
\( \Rightarrow (1 - \sin^2 t)\frac{d^2y}{dx^2} = -p^2 y + \sin t \left(\frac{p\cos pt}{\cos t}\right) \)
\( \Rightarrow (1 - x^2)\frac{d^2y}{dx^2} = -p^2 y + x\frac{dy}{dx} \) [since \( x = \sin t \) and \( y = \sin pt \)]
\( \Rightarrow (1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} + p^2 y = 0 \). Hence proved.