CBSE Class 12 Mathematics Continuity and Differentiability Previous Year Questions Set 01

Here are the CBSE board Previous Year Questions on Chapter 5 Continuity and Differentiability for Class 12 Mathematics, from 2010 to 2020, covering continuity, differentiability and all methods of differentiation. Each question has a complete step-by-step solution with the marking split shown in brackets.

Continuity: Very Short Answer Questions (1 Mark)

Question 1. Determine the value of 'k' for which the following function is continuous at x = 3:
\( f(x) = \begin{cases} \frac{(x+3)^2 - 36}{x-3}, & x \neq 3 \\ k, & x = 3 \end{cases} \)   [CBSE 2017, 1M]
Answer:
\( f(x) = \begin{cases} \frac{(x+3)^2 - 36}{x-3}, & x \neq 3 \\ k, & x = 3 \end{cases} \)
Since the function is continuous at x = 3,
So \( \lim_{x \to 3} f(x) = f(3) \)
⇒ \( \lim_{x \to 3} \frac{(x+3)^2 - 36}{x - 3} = k \)
⇒ \( \lim_{x \to 3} \frac{(x+3)^2 - (6)^2}{x - 3} = k \)
⇒ \( \lim_{x \to 3} \frac{(x+9)(x-3)}{x - 3} = k \)
⇒ \( \lim_{x \to 3} (x + 9) = k \)
∴ k = 12   [1]

Question 2. The function f : R → R given by f(x) = −|x − 1| is   [CBSE 2020, 1M]
(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 at x = 1
Answer:
Ans. (C)
Given f : R → R, f(x) = −|x − 1|
Here f(x) is a modulus function.
Since a modulus function is always continuous in R but not differentiable at the point where its value is zero,
hence f(x) is continuous but not differentiable at x = 1.   [1]

Question 3. Find the value of k, so that the function \( f(x) = \begin{cases} kx^2 + 5 & \text{if } x \le 1 \\ 2 & \text{if } x \gt 1 \end{cases} \) is continuous at x = 1.   [CBSE 2020, 1M]
Answer:
Given that \( f(x) = \begin{cases} kx^2 + 5 & \text{if } x \le 1 \\ 2 & \text{if } x \gt 1 \end{cases} \)
f(x) is continuous at x = 1
so \( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^-} f(x) = f(1) \)
⇒ \( \lim_{x \to 1} 2 = \lim_{x \to 1} (kx^2 + 5) = k + 5 \)
⇒ 2 = k + 5 ⇒ k = −3   [1]

Continuity: Long Answer Questions-I (4 Marks)

Question 4. Discuss the continuity and differentiability of the function f(x) = |x| + |x − 1| in the interval (−1, 2).   [CBSE 2015, 4M]
Answer:
f(x) = |x| + |x − 1|
\( f(x) = \begin{cases} 1 - 2x, & -1 \lt x \le 0 \\ 1, & 0 \lt x \le 1 \\ 2x - 1, & 1 \lt x \lt 2 \end{cases} \)
The possible points of discontinuity and non-differentiability are x = 0 and x = 1.
(i) We have to check continuity at x = 0
\( \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 1 = 1 \)
\( \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-2x + 1) = 1 \)
f(0) = 1
So RHL = LHL = f(0)
∴ the function is continuous at x = 0   [1]
(ii) We have to check continuity at x = 1
\( \text{RHL} = \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x - 1) = 1 \)
\( \text{LHL} = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 1 = 1 \)
f(1) = 1
RHL = LHL = f(1)
So the function is continuous at x = 1   [1]
(iii) We have to check differentiability at x = 0
\( \text{RHD} = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h};\quad h \gt 0 \)
\( = \lim_{h \to 0} \frac{1 - 1}{h} = 0 \)
\( \text{LHD} = \lim_{h \to 0} \frac{f(0-h) - f(0)}{-h};\quad h \gt 0 \)
\( = \lim_{h \to 0} \frac{-2(-h) + 1 - 1}{-h} = -2 \)
RHD ≠ LHD
So f(x) is not differentiable at x = 0.   [1]
(iv) We have to check differentiability at x = 1
\( \text{RHD} = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h};\quad h \gt 0 \)
\( = \lim_{h \to 0} \frac{[2(1+h) - 1] - 1}{h} = 2 \)
\( \text{LHD} = \lim_{h \to 0} \frac{f(1-h) - f(1)}{-h};\quad h \gt 0 \)
\( = \lim_{h \to 0} \frac{1 - 1}{-h} = 0 \)
RHD ≠ LHD
So f(x) is not differentiable at x = 1   [1]
Hence, f(x) is continuous at every point of the interval (−1, 2), but it is not differentiable at x = 0 and x = 1.

Question 5. If \( f(x) = \begin{cases} \frac{\sin (a+1)x + 2\sin x}{x}, & x \lt 0 \\ 2, & x = 0 \\ \frac{\sqrt{1 + bx} - 1}{x}, & x \gt 0 \end{cases} \) is continuous at x = 0, then find the values of a and b.   [CBSE 2016, 4M]
Answer:
f(x) is continuous at x = 0.
So RHL = LHL = f(0)
\( \text{RHL} = \lim_{x \to 0^+} \frac{\sqrt{1+bx} - 1}{x} \)
\( = \lim_{x \to 0^+} \frac{\sqrt{1+bx} - 1}{x} \times \frac{\sqrt{1+bx} + 1}{\sqrt{1+bx} + 1} \)
\( = \lim_{x \to 0^+} \frac{(1+bx) - 1}{x\left[\sqrt{1+bx} + 1\right]} = \frac{b}{2} \)   [1½]
\( \text{LHL} = \lim_{x \to 0^-} \left[\frac{\sin(a+1)x}{x} + \frac{2\sin x}{x}\right] \)
⇒ \( \lim_{x \to 0^-} (a+1)\frac{\sin(a+1)x}{(a+1)x} + 2\lim_{x \to 0^-} \frac{\sin x}{x} \)
= (a + 1) + 2 = a + 3   [1½]
f(0) = 2
∴ \( \frac{b}{2} = a + 3 = 2 \)   [½]
a = −1 and b = 4   [½]

Differentiability and Differentiation: Very Short Answer Questions (1 Mark)

Question 6. If \( y = \cos\left(\sqrt{3x}\right) \), then find \( \frac{dy}{dx} \).   [CBSE 2019, 1M]
Answer:
\( y = \cos\left(\sqrt{3x}\right) \)
⇒ \( \frac{dy}{dx} = -\sin\left(\sqrt{3x}\right) \times \frac{1}{2\sqrt{3x}} \times 3 = -\frac{\sqrt{3}\,\sin\left(\sqrt{3x}\right)}{2\sqrt{x}} \)   [1]

Question 7. Differentiate sec²(x²) with respect to x².   [CBSE 2020, 1M]
OR
If y = f(x²) and \( f'(x) = e^{\sqrt{x}} \), then find \( \frac{dy}{dx} \).
Answer:
Let u = sec²(x²) and v = x²
Now \( \frac{du}{dv} = \frac{du}{dx} \times \frac{dx}{dv} \)
\( \frac{du}{dv} = 2\sec x^2 \cdot \sec x^2 \tan x^2 \times 2x \times \frac{1}{2x} \)
\( \frac{du}{dv} = 2\sec^2 x^2 \tan x^2 \)
so, \( \frac{d(\sec^2 x^2)}{d(x^2)} = 2\sec^2 x^2 \tan x^2 \)   [1]
OR
Given y = f(x²) and \( f'(x) = e^{\sqrt{x}} \)
Now \( \frac{dy}{dx} = f'(x^2) \times 2x \)
\( \frac{dy}{dx} = e^{\sqrt{x^2}} \cdot 2x = 2x\,e^{|x|} \)   (= 2x e^x for x ≥ 0)   [1]

Differentiability and Differentiation: Short Answer Questions (2 Marks)

Question 8. Differentiate \( \tan^{-1}\left(\frac{1 + \cos x}{\sin x}\right) \) with respect to x.   [CBSE 2018, 2M]
Answer:
Let \( y = \tan^{-1}\left[\frac{1 + \cos x}{\sin x}\right] \)
⇒ \( y = \tan^{-1}\left[\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}\right] \)
⇒ \( y = \tan^{-1}\left[\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}}\right] = \tan^{-1}\left[\cot\frac{x}{2}\right] \)   [1]
⇒ \( y = \tan^{-1}\left[\tan\left(\frac{\pi}{2} - \frac{x}{2}\right)\right] \)
⇒ \( y = \frac{\pi}{2} - \frac{x}{2} \)   Differentiating w.r.t. x,
⇒ \( \frac{dy}{dx} = 0 - \frac{1}{2} = -\frac{1}{2} \)   [1]

Question 9. If \( f(x) = \sqrt{\frac{\sec x - 1}{\sec x + 1}} \), find \( f'\left(\frac{\pi}{3}\right) \).   [CBSE 2020, 2M]
OR
Find f′(x) if f(x) = (tan x)^{tan x}.
Answer:
Given \( f(x) = \sqrt{\frac{\sec x - 1}{\sec x + 1}} \)
\( f(x) = \sqrt{\frac{1 - \cos x}{1 + \cos x}} \)
\( f(x) = \sqrt{\frac{2\sin^2 x/2}{2\cos^2 x/2}} \)
\( f(x) = \left|\tan\frac{x}{2}\right| \)   (∵ x lies in the 1st quadrant)
f(x) = tan x/2   Differentiating w.r.t. x,
\( f'(x) = \sec^2\frac{x}{2} \times \frac{1}{2} \)   [1]
\( f'\left(\frac{\pi}{3}\right) = \sec^2\frac{\pi}{6} \times \frac{1}{2} \)
\( f'\left(\frac{\pi}{3}\right) = \frac{4}{3} \times \frac{1}{2} = \frac{2}{3} \)   [1]
OR
Given f(x) = (tan x)^{tan x}
Taking log on both sides,
log f(x) = log (tan x)^{tan x}
log f(x) = tan x log tan x
Differentiating w.r.t. x,
\( \frac{1}{f(x)} \cdot f'(x) = \tan x \times \frac{1}{\tan x} \times \sec^2 x + \log \tan x \cdot \sec^2 x \)
\( \frac{f'(x)}{f(x)} = \sec^2 x + \sec^2 x \log \tan x \)   [1]
f′(x) = f(x) · sec²x (1 + log tan x)
f′(x) = (tan x)^{tan x} · sec²x (1 + log tan x)   [1]

Differentiability and Differentiation: Long Answer Questions-I (4 Marks)

Question 10. If y = cosec⁻¹ x, x > 1, then show that \( x(x^2 - 1)\frac{d^2y}{dx^2} + (2x^2 - 1)\frac{dy}{dx} = 0 \).   [CBSE 2010, 4M]
Answer:
y = cosec⁻¹ x
Differentiating both sides w.r.t. x,
\( \frac{dy}{dx} = -\frac{1}{x\sqrt{x^2 - 1}} \)   (squaring on both sides)   [1]
⇒ \( x^2(x^2 - 1)\left(\frac{dy}{dx}\right)^2 = 1 \)   [1]
Again differentiating w.r.t. x,
\( 2x^2(x^2 - 1)\left(\frac{dy}{dx}\right)\frac{d^2y}{dx^2} + (4x^3 - 2x)\left(\frac{dy}{dx}\right)^2 = 0 \)   [1]
\( 2x\frac{dy}{dx}\left[x(x^2 - 1)\frac{d^2y}{dx^2} + (2x^2 - 1)\frac{dy}{dx}\right] = 0 \)
\( x(x^2 - 1)\frac{d^2y}{dx^2} + (2x^2 - 1)\frac{dy}{dx} = 0 \)   \( \left(\because x \neq 0,\ \frac{dy}{dx} \neq 0\right) \)   [1]

Question 11. If x = a(cos t + t sin t) and y = a(sin t − t cos t), \( 0 \lt t \lt \frac{\pi}{2} \), find \( \frac{d^2x}{dt^2},\ \frac{d^2y}{dt^2} \) and \( \frac{d^2y}{dx^2} \).   [CBSE 2012, 4M]
Answer:
x = a(cos t + t sin t), y = a(sin t − t cos t), 0 < t < π/2
\( \frac{dx}{dt} = a[-\sin t + \sin t + t\cos t] \)
\( \frac{dx}{dt} = a\,(t\cos t) \)   .........(i)   [½]
\( \frac{d^2x}{dt^2} = a[\cos t - t\sin t] \)   [½]
and y = a(sin t − t cos t)
\( \frac{dy}{dt} = a(\cos t - \cos t + t\sin t) \)
\( \frac{dy}{dt} = a\,t\sin t \)   .........(ii)   [½]
\( \frac{d^2y}{dt^2} = a(\sin t + t\cos t) \)   [½]
\( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{at\sin t}{at\cos t} = \tan t \)   Using (i) and (ii)   [1]
\( \frac{d^2y}{dx^2} = \sec^2 t \times \frac{1}{(dx/dt)} \) ⇒ \( \frac{d^2y}{dx^2} = \sec^2 t \times \frac{1}{at\cos t} = \frac{1}{at\cos^3 t} \)   [1]

Question 12. If y^x = e^{y − x}, then prove that \( \frac{dy}{dx} = \frac{(1 + \log y)^2}{\log y} \).   [CBSE 2013, 4M]
Answer:
y^x = e^{y − x} ⇒ e^{x log y} = e^{y − x}
x log y = y − x
x(log y + 1) = y
\( x = \frac{y}{1 + \log y} \)   [1]
Differentiating w.r.t. y,
\( \frac{dx}{dy} = \frac{(1 + \log y)\cdot 1 - y \times \frac{1}{y}}{(1 + \log y)^2} \)   [1½]
⇒ \( \frac{dx}{dy} = \frac{\log y}{(1 + \log y)^2} \)
⇒ \( \frac{dy}{dx} = \frac{(1 + \log y)^2}{(\log y)} \)   [1½]

Question 13. Differentiate the following with respect to x: \( \sin^{-1}\left(\frac{2^{x+1}\cdot 3^x}{1 + (36)^x}\right) \).   [CBSE 2013, 4M]
Answer:
Let \( y = \sin^{-1}\left(\frac{2^{x+1}\cdot 3^x}{1 + (36)^x}\right) \)
\( y = \sin^{-1}\left(\frac{2\cdot (6)^x}{1 + (6^x)^2}\right) \)   [1]
Put 6^x = tan θ; θ = tan⁻¹ 6^x
\( y = \sin^{-1}\left(\frac{2\tan\theta}{1 + \tan^2\theta}\right) \)
y = sin⁻¹(sin 2θ) ⇒ y = 2θ   [1]
y = 2 tan⁻¹(6^x)   (Now differentiating w.r.t. x)
\( \frac{dy}{dx} = \frac{2}{1 + (36)^x} \times 6^x \log_e 6 \)   [1]
\( = \left(\frac{2\cdot 2^x\cdot 3^x}{1 + 36^x}\right)\log_e 6 = \left(\frac{2^{x+1}\cdot 3^x}{1 + 36^x}\right)\cdot\log_e 6 \)   \( \left(\because \frac{d}{dx}a^x = a^x\log_e a\right) \)   [1]
Note: sin⁻¹(sin 2θ) = 2θ needs 2θ ≤ π/2, i.e. 6^x ≤ 1 (x ≤ 0). For x > 0 the derivative is the negative of this value. The answer above is the one given in the CBSE marking scheme.

Question 14. If x sin(a + y) + sin a cos(a + y) = 0, prove that \( \frac{dy}{dx} = \frac{\sin^2(a + y)}{\sin a} \).   [CBSE 2013, 4M]
Answer:
x sin(a + y) + sin a cos(a + y) = 0
\( x = -\frac{\sin a\cos(a + y)}{\sin(a + y)} = -\sin a\cot(a + y) \)   [1]
Differentiating w.r.t. y,
\( \frac{dx}{dy} = \left[\sin a\,\text{cosec}^2(a + y)\right] \)   [1]
\( \frac{dx}{dy} = \frac{\sin a}{\sin^2(a + y)} \)   [1]
\( \frac{dy}{dx} = \frac{\sin^2(a + y)}{\sin a} \)   [1]

Question 15. If y = Pe^ax + Qe^bx, show that \( \frac{d^2y}{dx^2} - (a + b)\frac{dy}{dx} + aby = 0 \).   [CBSE 2014, 4M]
Answer:
y = Pe^ax + Qe^bx   ...(i)
Differentiating w.r.t. x,
\( \frac{dy}{dx} = aPe^{ax} + bQe^{bx} \)   [½]
\( \frac{dy}{dx} = a[y - Qe^{bx}] + bQe^{bx} \)   [From eq. (i)]   [1]
\( \frac{dy}{dx} = ay - aQe^{bx} + bQe^{bx} \)
\( \frac{dy}{dx} = ay - (a - b)Qe^{bx} \)   ...(ii)   [1]
Again differentiating w.r.t. x,
\( \frac{d^2y}{dx^2} = a\frac{dy}{dx} - (a - b)Qbe^{bx} \)
\( \frac{d^2y}{dx^2} = \left[a\frac{dy}{dx} + b\frac{dy}{dx} - aby\right] \)   [From eq. (ii)]   [1]
\( \frac{d^2y}{dx^2} - (a + b)\frac{dy}{dx} + aby = 0 \)   [½]

Question 16. If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 − cos 2t), show that at \( t = \frac{\pi}{4},\ \left(\frac{dy}{dx}\right) = \frac{b}{a} \).   [CBSE 2014, 4M]
Answer:
x = a sin 2t (1 + cos 2t)
\( x = a\sin 2t + \frac{a}{2}\,2\sin 2t\cos 2t = a\sin 2t + \frac{a}{2}\sin 4t \)
Differentiating w.r.t. t,
\( \frac{dx}{dt} = 2a\cos 2t + \frac{4a}{2}\cos 4t \)
\( \frac{dx}{dt} = 2a(\cos 2t + \cos 4t) \)   ...........(i)   [1]
Now y = b cos 2t (1 − cos 2t)
y = b cos 2t − b cos² 2t
\( y = b\cos 2t - b\left(\frac{1 + \cos 4t}{2}\right) \)   \( \left(\because \cos^2\theta = \frac{1 + \cos 2\theta}{2}\right) \)
\( y = b\cos 2t - \frac{b}{2} - \frac{b}{2}\cos 4t \)
Differentiating w.r.t. t,
⇒ \( \frac{dy}{dt} = -2b\sin 2t + \frac{4b}{2}\sin 4t \)
⇒ \( \frac{dy}{dt} = 2b(\sin 4t - \sin 2t) \)   ......(ii)   [1]
Now \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2b(\sin 4t - \sin 2t)}{2a(\cos 2t + \cos 4t)} \)   [1]
\( \left(\frac{dy}{dx}\right)_{t = \frac{\pi}{4}} = \frac{b}{a}\left[\frac{\sin\pi - \sin\frac{\pi}{2}}{\cos\frac{\pi}{2} + \cos\pi}\right] \)
\( \left(\frac{dy}{dx}\right)_{t = \frac{\pi}{4}} = \frac{b}{a} \)   [1]

Question 17. If x = cos t (3 − 2cos²t) and y = sin t (3 − 2sin²t), find the value of \( \frac{dy}{dx} \) at \( t = \frac{\pi}{4} \).   [CBSE 2014, 4M]
Answer:
x = 3cos t − 2cos³t
Differentiating w.r.t. t,
\( \frac{dx}{dt} = -3\sin t + 2 \times 3\cos^2 t\sin t \)
= 3sin t (2cos²t − 1)
= 3sin t · cos 2t   ...(i)   [1]
y = sin t (3 − 2sin²t)
y = 3sin t − 2sin³t
Differentiating w.r.t. t,
\( \frac{dy}{dt} = 3\cos t - 2 \times 3\sin^2 t\cos t \)
= 3cos t (1 − 2sin²t)
= 3cos t · cos 2t   ...(ii)   [1]
Dividing equation (ii) by (i),
\( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3\cos t\cdot\cos 2t}{3\sin t\cdot\cos 2t} = \cot t \)   [1]
⇒ \( \left(\frac{dy}{dx}\right)_{t = \frac{\pi}{4}} = \cot\frac{\pi}{4} = 1 \)   [1]

Question 18. If x = a(cos 2t + 2t sin 2t) and y = a(sin 2t − 2t cos 2t), then find \( \frac{d^2y}{dx^2} \).   [CBSE 2015, 4M]
Answer:
x = a(cos 2t + 2t sin 2t)
Differentiating w.r.t. t,
\( \frac{dx}{dt} = a[-2\sin 2t + 2(2t\cos 2t + \sin 2t)] \)
\( \frac{dx}{dt} = 4at\cos 2t \)   .........(1)   [1]
y = a(sin 2t − 2t cos 2t)
Differentiating w.r.t. t,
\( \frac{dy}{dt} = a\left[2\cos 2t - 2\left(-2t\sin 2t + \cos 2t\right)\right] \)
\( \frac{dy}{dt} = 4at\sin 2t \)   ..........(2)   [1]
∵ \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4at\sin 2t}{4at\cos 2t} \)
\( \frac{dy}{dx} = \tan 2t \)   [1]
\( \frac{d^2y}{dx^2} = 2\sec^2 2t \times \frac{dt}{dx} \)
\( \frac{d^2y}{dx^2} = 2\sec^2 2t \times \frac{1}{4at\cos 2t} \)
\( \frac{d^2y}{dx^2} = \frac{1}{2at} \times \sec^3 2t \)   [1]

Question 19. If (ax + b)e^y/x = x, then show that \( x^3\left(\frac{d^2y}{dx^2}\right) = \left(x\frac{dy}{dx} - y\right)^2 \).   [CBSE 2015, 4M]
Answer:
(ax + b)e^y/x = x
\( e^{y/x} = \frac{x}{ax + b} \)   ……(i)
Taking log on both sides,
\( \frac{y}{x} = \left[\log x - \log(ax + b)\right] \)   [½]
Differentiating w.r.t. x,
\( \frac{x\cdot\frac{dy}{dx} - y}{x^2} = \frac{1}{x} - \frac{a}{ax + b} \)   [1]
\( \frac{x\cdot\frac{dy}{dx} - y}{x^2} = \frac{ax + b - ax}{x(ax + b)} \)   [½]
\( x\frac{dy}{dx} - y = \frac{bx}{(ax + b)} \)
\( x\frac{dy}{dx} - y = be^{y/x} \)   .........(ii)   [Using eq. (i)]   [½]
Differentiating w.r.t. x,
\( x\cdot\frac{d^2y}{dx^2} + \frac{dy}{dx} - \frac{dy}{dx} = \frac{be^{y/x}\left(x\cdot\frac{dy}{dx} - y\right)}{x^2} \)
\( x^3\left(\frac{d^2y}{dx^2}\right) = \left(x\frac{dy}{dx} - y\right)^2 \)   [Using eq. (ii)]   [1]

Question 20. If x cos(a + y) = cos y, then prove that \( \frac{dy}{dx} = \frac{\cos^2(a + y)}{\sin a} \). Hence, show that \( \sin a\frac{d^2y}{dx^2} + \sin 2(a + y)\frac{dy}{dx} = 0 \).   [CBSE 2016, 4M]
OR
Find \( \frac{dy}{dx} \), if \( y = \sin^{-1}\left[\frac{6x - 4\sqrt{1 - 4x^2}}{5}\right] \).
Answer:
x cos(a + y) = cos y
\( x = \frac{\cos y}{\cos(a + y)} \)
Differentiating both sides w.r.t. y,
⇒ \( \frac{dx}{dy} = \frac{-\cos(a + y)\sin y + \cos y\sin(a + y)}{\cos^2(a + y)} \)
⇒ \( \frac{dx}{dy} = \frac{\sin a}{\cos^2(a + y)} \)
⇒ \( \frac{dy}{dx} = \frac{\cos^2(a + y)}{\sin a} \)
Differentiating both sides w.r.t. x,
\( \frac{d^2y}{dx^2} = \frac{-2\cos(a + y)\sin(a + y)}{\sin a}\cdot\frac{dy}{dx} \)   [2]
\( \frac{d^2y}{dx^2} = \frac{-\sin 2(a + y)}{\sin a}\cdot\frac{dy}{dx} \)
\( \sin a\frac{d^2y}{dx^2} + \sin 2(a + y)\frac{dy}{dx} = 0 \)   [2]
OR
\( y = \sin^{-1}\left[\frac{6x - 4\sqrt{1 - 4x^2}}{5}\right] \)
\( y = \sin^{-1}\left[\frac{3}{5}\cdot 2x - \frac{4}{5}\sqrt{1 - (2x)^2}\right] \)
Put 2x = sin θ and \( \frac{3}{5} = \cos\phi \), i.e. \( \frac{4}{5} = \sin\phi \)
⇒ \( \theta = \sin^{-1} 2x \) and \( \phi = \cos^{-1}\frac{3}{5} \) or \( \sin^{-1}\frac{4}{5} \)   [1]
∴ \( y = \sin^{-1}\left[\sin\theta\cos\phi - \cos\theta\sin\phi\right] \)
\( y = \sin^{-1}[\sin(\theta - \phi)] \)
\( y = \theta - \phi \)   [1]
\( y = \sin^{-1} 2x - \sin^{-1}\frac{4}{5} \)   [1]
Differentiating w.r.t. x,
\( \frac{dy}{dx} = \frac{1}{\sqrt{1 - 4x^2}}\cdot 2 - 0 \)
\( \frac{dy}{dx} = \frac{2}{\sqrt{1 - 4x^2}} \)   [1]

Question 21. If x^y + y^x = a^b, then find \( \frac{dy}{dx} \).   [CBSE 2017, 4M]
OR
If e^y(x + 1) = 1, then show that \( \frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2 \).
Answer:
x^y + y^x = a^b
Let u = x^y and v = y^x
u + v = a^b
Differentiating both sides w.r.t. x,
\( \frac{du}{dx} + \frac{dv}{dx} = 0 \)   .......(i)   [1]
∵ u = x^y
Taking log on both sides,
log u = y log x
Differentiating w.r.t. x,
\( \frac{1}{u}\cdot\frac{du}{dx} = \frac{y}{x} + \log x\cdot\frac{dy}{dx} \)
\( \frac{du}{dx} = u\left[\frac{y}{x} + \log x\cdot\frac{dy}{dx}\right] \)
\( \frac{du}{dx} = x^y\left[\frac{y}{x} + \log x\cdot\frac{dy}{dx}\right] \)
\( \frac{du}{dx} = x^{y-1}y + x^y\log x\cdot\frac{dy}{dx} \)   .......(ii)   [1]
∵ v = y^x
Taking log on both sides,
log v = x log y
Differentiating w.r.t. x,
\( \frac{1}{v}\cdot\frac{dv}{dx} = \frac{x}{y}\cdot\frac{dy}{dx} + \log y \)
\( \frac{dv}{dx} = v\left[\frac{x}{y}\cdot\frac{dy}{dx} + \log y\right] \)
\( \frac{dv}{dx} = y^x\left[\frac{x}{y}\cdot\frac{dy}{dx} + \log y\right] \)
\( \frac{dv}{dx} = xy^{x-1}\frac{dy}{dx} + y^x\log y \)   ........(iii)   [1]
From equations (i), (ii) and (iii),
⇒ \( yx^{y-1} + x^y\log x\frac{dy}{dx} + xy^{x-1}\frac{dy}{dx} + y^x\log y = 0 \)
⇒ \( \frac{dy}{dx}\left(x^y\log x + xy^{x-1}\right) = -\left(y^x\log y + yx^{y-1}\right) \)
⇒ \( \frac{dy}{dx} = -\left(\frac{y^x\log y + yx^{y-1}}{x^y\log x + xy^{x-1}}\right) \)   [1]
OR
∵ e^y(x + 1) = 1
\( e^y = \frac{1}{x + 1} \)   .......(i)
Differentiating w.r.t. x,
\( e^y\cdot\frac{dy}{dx} = -\frac{1}{(x + 1)^2} \)   [1½]
\( \frac{1}{(x + 1)}\cdot\frac{dy}{dx} = -\frac{1}{(x + 1)^2} \)   From equation (i)
\( \frac{dy}{dx} = -\frac{1}{(x + 1)} \)   .......(ii)   [1]
\( \frac{d^2y}{dx^2} = \frac{1}{(x + 1)^2} \)   [1]
\( \frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2 \)   From equation (ii)   [½]

Question 22. If y = sin(sin x), prove that \( \frac{d^2y}{dx^2} + \tan x\frac{dy}{dx} + y\cos^2 x = 0 \).   [CBSE 2018, 4M]
Answer:
Given y = sin(sin x)   ...(i)
Differentiating w.r.t. x,
\( \frac{dy}{dx} = \cos(\sin x)\cdot\cos x \)   ...(ii)   [1]
Again differentiating w.r.t. x,
\( \frac{d^2y}{dx^2} = \cos(\sin x)\frac{d}{dx}(\cos x) + \cos x\frac{d}{dx}\cos(\sin x) \)
\( \frac{d^2y}{dx^2} = -\sin x\cdot\cos(\sin x) + \cos x\cdot\{-\sin(\sin x)\cdot\cos x\} \)   [1]
\( \frac{d^2y}{dx^2} = \frac{-\sin x}{\cos x}\cdot\cos x\cdot\cos(\sin x) - \cos^2 x\cdot\sin(\sin x) \)   [1]
\( \frac{d^2y}{dx^2} = \frac{-\sin x}{\cos x}\cdot\frac{dy}{dx} - y\cos^2 x \)   [Using (i) & (ii)]
\( \frac{d^2y}{dx^2} + \tan x\frac{dy}{dx} + y\cos^2 x = 0 \)   [1]

Question 23. If (x² + y²)² = xy, find \( \frac{dy}{dx} \).   [CBSE 2018, 4M]
OR
If x = a(2θ − sin 2θ) and y = a(1 − cos 2θ), find \( \frac{dy}{dx} \) when \( \theta = \frac{\pi}{3} \).
Answer:
Given (x² + y²)² = xy
x⁴ + y⁴ + 2x²y² = xy
Differentiating both sides w.r.t. x,
\( 4x^3 + 4y^3\frac{dy}{dx} + 2\left[2xy^2 + x^2\cdot 2y\frac{dy}{dx}\right] = x\frac{dy}{dx} + y \)   [2]
\( \frac{dy}{dx}(4y^3 + 4x^2y - x) = (y - 4x^3 - 4xy^2) \)
⇒ \( \frac{dy}{dx} = \frac{y - 4x^3 - 4xy^2}{4y^3 + 4x^2y - x} \)   [2]
OR
x = a(2θ − sin 2θ) and y = a(1 − cos 2θ)
∵ x = a(2θ − sin 2θ)
Differentiating w.r.t. θ,
\( \frac{dx}{d\theta} = a(2 - \cos 2\theta \times 2) \)
⇒ \( \frac{dx}{d\theta} = 2a(1 - \cos 2\theta) \)
⇒ \( \frac{dx}{d\theta} = 4a\sin^2\theta \)   ......(i)   [1]
∵ y = a(1 − cos 2θ)
Differentiating w.r.t. θ,
\( \frac{dy}{d\theta} = 2a\sin 2\theta \)
⇒ \( \frac{dy}{d\theta} = 4a\sin\theta\cos\theta \)   .......(ii)   [1]
∵ \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \)
\( \frac{dy}{dx} = \frac{4a\sin\theta\cos\theta}{4a\sin^2\theta} \)   [using (i) & (ii)]   [1]
\( \frac{dy}{dx} = \frac{\cos\theta}{\sin\theta} \)
\( \left(\frac{dy}{dx}\right)_{\theta = \frac{\pi}{3}} = \frac{\cos\left(\frac{\pi}{3}\right)}{\sin\left(\frac{\pi}{3}\right)} = \frac{1}{\sqrt{3}} \)   [1]

Question 24. If x = ae^t(sin t + cos t) and y = ae^t(sin t − cos t), then prove that \( \frac{dy}{dx} = \frac{x + y}{x - y} \).   [CBSE 2019, 4M]
OR
Differentiate x^{sin x} + (sin x)^{cos x} with respect to x.   [CBSE 2009, 2019, 4M]
Answer:
x = ae^t(sin t + cos t)
Differentiating both sides w.r.t. t,
∴ \( \frac{dx}{dt} = a\left[e^t(\cos t - \sin t) + (\sin t + \cos t)e^t\right] \)
= ae^t(sin t + cos t) − ae^t(sin t − cos t)
⇒ \( \frac{dx}{dt} = x - y \)   ............(i)   [1½]
∵ y = ae^t(sin t − cos t)
Differentiating both sides w.r.t. t,
\( \frac{dy}{dt} = a\left[e^t(\cos t + \sin t) + (\sin t - \cos t)e^t\right] \)
= ae^t(sin t + cos t) + ae^t(sin t − cos t)
\( \frac{dy}{dt} = x + y \)   ............(ii)   [1½]
∵ \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{x + y}{x - y} \)   [using (i) & (ii)]   [1]
OR
Let y = x^{sin x} + (sin x)^{cos x}
y = u + v (let)
Differentiating w.r.t. x,
⇒ \( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \)   .........(i)   [1]
∵ u = x^{sin x}
⇒ log u = sin x · log x
\( \frac{1}{u}\frac{du}{dx} = \sin x\cdot\frac{1}{x} + \log x\cdot\cos x \)
⇒ \( \frac{du}{dx} = x^{\sin x}\left[\frac{\sin x}{x} + \log x\cdot\cos x\right] \)   ........(ii)   [1]
∵ v = (sin x)^{cos x}
⇒ log v = cos x · log (sin x)
⇒ \( \frac{1}{v}\frac{dv}{dx} = \cos x\cdot\frac{1}{\sin x}\cdot\cos x + \log(\sin x)\cdot(-\sin x) \)
⇒ \( \frac{dv}{dx} = (\sin x)^{\cos x}\left[\cos x\cdot\cot x - \sin x\log(\sin x)\right] \)   ..........(iii)   [1]
From equation (i),
\( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \)
\( = x^{\sin x}\left[\frac{\sin x}{x} + \log x\cdot\cos x\right] + (\sin x)^{\cos x}\left[\cos x\cdot\cot x - \sin x\cdot\log(\sin x)\right] \)   [using (i), (ii) & (iii)]   [1]

Question 25. If (x − a)² + (y − b)² = c², for some c > 0, 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.   [CBSE 2019, 4M]
Answer:
Given, (x − a)² + (y − b)² = c²   ……(i)
Differentiating w.r.t. x,
\( 2(x - a) + 2(y - b)\frac{dy}{dx} = 0 \)
⇒ \( \frac{dy}{dx} = -\frac{(x - a)}{(y - b)} \)   ……(ii)   [1]
Again, differentiating w.r.t. x,
\( \frac{d^2y}{dx^2} = -\left[\frac{(y - b)(1) - (x - a)\frac{dy}{dx}}{(y - b)^2}\right] \)   [1]
\( \frac{d^2y}{dx^2} = -\left[\frac{(y - b) + (x - a)\frac{(x - a)}{(y - b)}}{(y - b)^2}\right] \)   [using (ii)]
\( = -\left[\frac{(x - a)^2 + (y - b)^2}{(y - b)^3}\right] \)
\( = -\frac{c^2}{(y - b)^3} \)   ..... (iii)   [using (i)]   [1]
Now, \( \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}} = \frac{\left[1 + \frac{(x - a)^2}{(y - b)^2}\right]^{3/2}}{\frac{-c^2}{(y - b)^3}} \)   {using (ii) and (iii)}
\( = \frac{\left[\frac{(x - a)^2 + (y - b)^2}{(y - b)^2}\right]^{3/2}}{-c^2/(y - b)^3} \)
\( = \frac{c^3}{-c^2} = -c \);   [using (i)]
which is a constant independent of a and b.   [1]

Question 26. If \( \tan^{-1}\left(\frac{y}{x}\right) = \log\sqrt{x^2 + y^2} \), prove that \( \frac{dy}{dx} = \frac{x + y}{x - y} \).   [CBSE 2020, 4M]
OR
If \( y = e^{a\cos^{-1}x},\ -1 \lt x \lt 1 \), then show that \( (1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2y = 0 \).
Answer:
Given \( \tan^{-1}\left(\frac{y}{x}\right) = \log\sqrt{x^2 + y^2} \)
Differentiating w.r.t. x,
\( \frac{1}{\left(1 + \frac{y^2}{x^2}\right)} \times \left(\frac{x\frac{dy}{dx} - y}{x^2}\right) = \frac{1}{\sqrt{x^2 + y^2}} \times \frac{\left(2x + 2y\frac{dy}{dx}\right)}{2\sqrt{x^2 + y^2}} \)   [1]
⇒ \( \frac{x^2}{(x^2 + y^2)} \times \left(x\frac{dy}{dx} - y\right) \times \frac{1}{x^2} = \frac{1}{(x^2 + y^2)}\left(x + y\frac{dy}{dx}\right) \)   [1]
⇒ \( x\frac{dy}{dx} - y = x + y\frac{dy}{dx} \)   [1]
⇒ \( (x - y)\frac{dy}{dx} = x + y \)
⇒ \( \frac{dy}{dx} = \frac{x + y}{x - y} \)   [1]
OR
Given \( y = e^{a\cos^{-1}x} \)   ...(1)
Differentiating w.r.t. x,
\( \frac{dy}{dx} = -e^{a\cos^{-1}x}\cdot a \times \frac{1}{\sqrt{1 - x^2}} \)   ..(2)   [1]
\( \sqrt{1 - x^2}\cdot\frac{dy}{dx} = -ay \)   (from (1))
⇒ \( (1 - x^2)\left(\frac{dy}{dx}\right)^2 = a^2y^2 \)   [1]
Again differentiating w.r.t. x,
\( 2(1 - x^2)\left(\frac{dy}{dx}\right)\left(\frac{d^2y}{dx^2}\right) - 2x\left(\frac{dy}{dx}\right)^2 = 2a^2y\frac{dy}{dx} \)   [1]
⇒ \( 2\left(\frac{dy}{dx}\right)\left[(1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2y\right] = 0 \)
⇒ \( (1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2y = 0 \)   \( \left(\because \frac{dy}{dx} \neq 0, \text{ as } y \text{ is not constant}\right) \)   [1]