CBSE Class 12 Mathematics Continuity And Differentiability Worksheet Set 03

CBSE Class 12 Mathematics Continuity And Differentiability Worksheet 3. The Continuity And Differentiability questions in the worksheets have been specifically designed by best mathematics teachers so that the students can practise them to clear their Continuity And Differentiability concepts and get better marks in class 12 mathematics tests and examinations. Students can free download these Continuity And Differentiability worksheets in pdf and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the Continuity And Differentiability chapter and other subjects too. Use them for better understanding of the subjects.

Question. If \( y = \cos^{-1}x \), find \( \frac{d^2y}{dx^2} \) in terms of \( y \) alone.
Answer: We have, \( y = \cos^{-1}x \ldots\ldots(1) \)
Diff w.r.t x
\( \frac{dy}{dx} = \frac{-1}{\sqrt{1-x^2}} \)
\( \Rightarrow \sqrt{1-x^2}\frac{dy}{dx} = -1 \quad \text{(cross mult.)} \)
\( \frac{dy}{dx} = \frac{-1}{\sqrt{1-x^2}} \)
Diff again w.r.t x (using Quotient rule)
\( \Rightarrow \frac{d^2y}{dx^2} = \frac{\sqrt{1-x^2}(0) - (-1)\frac{1}{2\sqrt{1-x^2}}(-2x)}{(1-x^2)} \)
\( \Rightarrow \frac{d^2y}{dx^2} = \frac{-\frac{x}{\sqrt{1-x^2}}}{(1-x^2)} \)
\( \Rightarrow \frac{d^2y}{dx^2} = \frac{-x}{(1-x^2)\sqrt{1-x^2}} \)
from (1) \( x = \cos y \), put in \( \frac{d^2y}{dx^2} \)
\( \Rightarrow \frac{d^2y}{dx^2} = \frac{-\cos y}{(1-\cos^2 y)\sqrt{1-\cos^2 y}} \)
\( \Rightarrow \frac{-\cos y}{\sin^2 y \cdot \sin y} \)
\( \frac{d^2y}{dx^2} = -\cot y \cdot \csc^2 y \) Ans.

Question. If \( y = Ae^{mx} + Be^{nx} \), show that \( \frac{d^2y}{dx^2} - (m + n)\frac{dy}{dx} + mny = 0 \).
Answer: We have, \( y = Ae^{mx} + Be^{nx} \ldots\ldots(1) \)
Diff w.r.t x ,
\( \frac{dy}{dx} = mAe^{mx} + nBe^{nx} \ldots\ldots(2) \)
Diff again w.r.t x
\( \frac{d^2y}{dx^2} = m^2Ae^{mx} + n^2Be^{nx} \ldots\ldots(3) \)
Taking LHS
\( \frac{d^2y}{dx^2} - (m + n)\frac{dy}{dx} + mny \ Bir \)
Substituting the value of \( \frac{d^2y}{dx^2}, \frac{dy}{dx} \) & \( y \ldots\ldots\{\text{from (1), (2) \& (3)\} \)
\( m^2Ae^{mx} + n^2Be^{nx} - (m + n)(mAe^{mx} + nBe^{nx}) + mn(Ae^{mx} + Be^{nx}) \)
\( \Rightarrow m^2Ae^{mx} + n^2Be^{nx} - m^2Ae^{mx} - mnBe^{nx} - mnAe^{mx} - n^2Be^{nx} + mnAe^{mx} + mnBe^{nx} = 0 \) RHS. (Proved)

Question. If \( y = \tan x + \sec x \), show that \( \frac{d^2y}{dx^2} = \frac{\cos x}{(1-\sin x)^2} \).
Answer: We have \( y = \tan x + \sec x \)
Diff w.r.t x
\( \frac{dy}{dx} = \sec^2 x + \sec x\tan x \)
\( \Rightarrow \frac{dy}{dx} = \frac{1+\sin x}{\cos^2 x} \)
\( \Rightarrow \frac{dy}{dx} = \frac{1+\sin x}{1-\sin^2 x} \)
\( \Rightarrow \frac{dy}{dx} = \frac{1+\sin x}{(1-\sin x)(1+\sin x)} \)
\( \Rightarrow \frac{dy}{dx} = \frac{1}{1-\sin x} \)
Diff again w.r.t. x (using Quotient rule)
\( \Rightarrow \frac{d^2y}{dx^2} = \frac{(1-\sin x)(0) - 1(-\cos x)}{(1-\sin x)^2} \)
\( \Rightarrow \frac{d^2y}{dx^2} = \frac{\cos x}{(1-\sin x)^2} \) (Proved)

Question. If \( (x - a)^2 + (y - b)^2 = c^2 \ldots\ldots(1) \), Show that \( \frac{\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}} \) is a constant and independent of a and b.
Answer: We have, \( (x - a)^2 + (y - b)^2 = c^2 \ldots\ldots(1) \)
Diff w.r.t. x
\( 2(x - a) + 2(y - b)\frac{dy}{dx} = 0 \)
\( \Rightarrow (x - a) + (y - b)\frac{dy}{dx} = 0 \)
\( \Rightarrow \frac{dy}{dx} = \frac{-(x - a)}{(y - b)} \ldots\ldots(2) \)
Diff again w.r.t. x (Quotient rule on RHS)
\( \Rightarrow \frac{d^2y}{dx^2} = -\left[\frac{(y-b)(1) - (x-a)\frac{dy}{dx}}{(y-b)^2}\right] \)
\( \Rightarrow \frac{d^2y}{dx^2} = -\left[\frac{(y-b) + (x-a)\left(\frac{x-a}{y-b}\right)}{(y-b)^2}\right] \ldots\ldots\{\text{from eq. (2)\} \)
\( \Rightarrow \frac{d^2y}{dx^2} = -\left[\frac{(y-b)^2 + (x-a)^2}{(y-b)^3}\right] \)
\( \Rightarrow \frac{d^2y}{dx^2} = \frac{-(c^2)}{(y-b)^3} \ldots\ldots\{\text{From eq. (1)\} \)
Consider \( \frac{\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\left(\frac{d^2y}{dx^2}\right)} \)
\( = \frac{\left(1 + \frac{(x-a)^2}{(y-b)^2}\right)^{3/2}}{\frac{-c^2}{(y-b)^3}} \)
\( = \frac{\left(\frac{(x-a)^2 + (y-b)^2}{(y-b)^2}\right)^{3/2}}{\frac{-c^2}{(y-b)^3}} \)
\( = \frac{\left[\frac{c^2}{(y-b)^2}\right]^{3/2}}{\frac{-c^2}{(y-b)^3}} \)
\( = \frac{\frac{c^3}{(y-b)^3}}{\frac{-c^2}{(y-b)^3}} \)
\( = \frac{c^3}{-c^2} = -c \) which is a constant and independent of a and b. (Proved)

Question. If \( x = a(\theta - \sin\theta) \) and \( y = a(1 + \cos\theta) \) find \( \frac{d^2y}{dx^2} \).
Answer: We have ,
\( x = a(\theta - \sin\theta) \qquad y = a(1 + \cos\theta) \)
Diff w.r.t. \( \theta \qquad \) Diff w.r.t \( \theta \)
\( \frac{dx}{d\theta} = a(1 - \cos\theta) \ldots\ldots(1) \qquad \frac{dy}{d\theta} = a(-\sin\theta) \)
Now \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \)
\( \Rightarrow \frac{dy}{dx} = \frac{-a\sin\theta}{a(1-\cos\theta)} \)
\( \Rightarrow \frac{dy}{dx} = \frac{-\sin\theta}{1-\cos\theta} = \frac{-2\sin(\theta/2)\cdot\cos(\theta/2)}{2\sin^2(\theta/2)} \)
\( \Rightarrow \frac{dy}{dx} = -\cot(\theta/2) \)
Now Diff w.r.t x
\( \frac{d^2y}{dx^2} = -\left[-\csc^2\left(\frac{\theta}{2}\right)\right] \cdot \frac{1}{2} \cdot \frac{d\theta}{dx} \quad \{\text{main step}\} \)
\( \Rightarrow \frac{d^2y}{dx^2} = \frac{1}{2}\csc^2\left(\frac{\theta}{2}\right) \cdot \frac{1}{a(1-\cos\theta)} \ldots\ldots\{\text{from eq. (1)\} \)
\( \Rightarrow \frac{d^2y}{dx^2} = \frac{1}{2a} \cdot \csc^2\left(\frac{\theta}{2}\right) \cdot \frac{1}{2\sin^2(\theta/2)} \)
\( \Rightarrow \frac{d^2y}{dx^2} = \frac{1}{4a}\csc^4\left(\frac{\theta}{2}\right) \) Ans.

Question. If \( x = a(\cos\theta + \theta\sin\theta) \) & \( y = a(\sin\theta - \theta\cos\theta) \). Find \( \frac{d^2y}{dx^2} \) at \( \theta = \frac{\pi}{3} \).
Answer: We have, \( x = a(\cos\theta + \theta\sin\theta) \)
Diff w.r.t \( \theta \)
\( \frac{dx}{d\theta} = a[-\sin\theta + \theta\cdot\cos\theta + \sin\theta] \)
\( \frac{dx}{d\theta} = a\theta\cos\theta \ldots\ldots(1) \)
\( y = a(\sin\theta - \theta\cos\theta) \)
Diff w.r.t. \( \theta \)
\( \frac{dy}{d\theta} = a(\cos\theta - (-\theta\sin\theta + \cos\theta)) \)
\( \Rightarrow \frac{dy}{d\theta} = a\theta\sin\theta \)
Now \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \)
\( \Rightarrow \frac{dy}{dx} = \frac{a\theta\sin\theta}{a\theta\cos\theta} = \tan\theta \)
Diff now w.r.t. 'x'
\( \frac{d^2y}{dx^2} = \sec^2\theta \cdot \frac{d\theta}{dx} \)
\( \Rightarrow \frac{d^2y}{dx^2} = \sec^2\theta \cdot \frac{1}{a\theta\cos\theta} = \frac{1}{a\theta} \cdot \sec^3\theta \)
\( \Rightarrow \left(\frac{d^2y}{dx^2}\right)_{\theta=\pi/3} = \frac{1}{a \cdot \frac{\pi}{3}}\sec^3(\pi/3) \)
\( \Rightarrow \left(\frac{d^2y}{dx^2}\right)_{\theta=\pi/3} = \frac{3}{a\pi}(2)^3 \ldots\ldots\{\cos\frac{\pi}{3} = 1/2 \therefore \sec\pi/3 = 2\} \)
\( \Rightarrow \left(\frac{d^2y}{dx^2}\right)_{\theta=\pi/3} = \frac{24}{a\pi} \) Ans.

Question. Find A and B so that \( y = A\sin(3x) + B\cos(3x) \) satisfies the equation \( \frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 3y = 10\cos(3x) \).
Answer: We have, \( y = A\sin(3x) + B\cos(3x) \)
Diff w.r.t. x
\( \frac{dy}{dx} = 3A\cdot\cos(3x) - 3B\sin(3x) \)
Diff again w.r.t. x
\( \frac{d^2y}{dx^2} = -9A\sin(3x) - 9B\cos(3x) \)
given equation :
\( \frac{d^2y}{dx^2} + \frac{4dy}{dx} + 3y = 10\cos(3x) \)
subst. The values of \( \frac{d^2y}{dx^2}, \frac{dy}{dx}, y \)
\( -9A\sin(3x) - 9B\cos(3x) + 4(3A\cos(3x) - 3B\sin(3x)) + 3(A\sin(3x) + B\cos(3x)) = 10\cos(3x) \)
\( \Rightarrow -9A\sin(3x) - 9B\cos(3x) + 12A\cos(3x) - 12B\sin(3x) + 3(A\sin(3x) + B\cos(3x)) = 10\cos(3x) \)
\( \Rightarrow \sin(3x)[-9A - 12B + 3A] + \cos(3x)[-9B + 12A + 3B] = 10\cos(3x) \)
\( \Rightarrow \sin(3x)(-6A - 12B) + \cos(3x)(-6B + 12A) = 10\cos(3x) \)
equating the coefficients of \( \sin(3x) \) and \( \cos(3x) \) on both sides.
We have, \( -6A - 12B = 0 \)
and \( -6B + 12A = 10 \)
solving these two equations we get
\( A = \frac{2}{3}, B = \frac{-1}{3} \) Ans.

Question. If \( x = a\cos\theta + b\sin\theta \) and \( y = a\sin\theta - b\cos\theta \), show that \( y^2 \frac{d^2y}{dx^2} - x \frac{dy}{dx} + y = 0 \).
Answer: We have, \( x = a\cos\theta + b\sin\theta \qquad y = a\sin\theta - b\cos\theta \)
Diff w.r.t \( \theta \)
\( \frac{dx}{d\theta} = -a\sin\theta + b\cos\theta \qquad \frac{dy}{d\theta} = a\cos\theta + b\sin\theta \)
\( \frac{dx}{d\theta} = -(a\sin\theta - b\cos\theta) \qquad \frac{dy}{d\theta} = x \)
\( \frac{dx}{d\theta} = -y \)
Now \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = -\frac{x}{y} \)
\( \Rightarrow y \frac{dy}{dx} = -x \ldots\ldots(1) \)
Diff again w.r.t x (product rule on LHS)
\( \Rightarrow y\cdot\frac{d^2y}{dx^2} + \frac{dy}{dx}\cdot\frac{dy}{dx} = -1 \)
\( \Rightarrow y\frac{d^2y}{dx^2} + \frac{dy}{dx}\left(\frac{-x}{y}\right) = -1 \ldots\ldots\{\text{from eq. (1)\} \)
\( \Rightarrow y^2\frac{d^2y}{dx^2} - x\frac{dy}{dx} = -y \)
\( \Rightarrow y^2\frac{d^2y}{dx^2} - x\frac{dy}{dx} + y = 0 \) (Proved)

Implicit Functions And General Differentiation

Question. If \( \log(x^2 + y^2) = 2\tan^{-1}\left(\frac{y}{x}\right) \), show that \( \frac{dy}{dx} = \frac{x+y}{x-y} \).
Answer: We have, \( \log(x^2 + y^2) = 2\tan^{-1}\left(\frac{y}{x}\right) \)
Diff. w.r.t. x
\( \frac{1}{x^2+y^2} \cdot \left(2x + 2y\frac{dy}{dx}\right) = 2 \cdot \frac{1}{1+\frac{y^2}{x^2}} \cdot \left[\frac{x\cdot\frac{dy}{dx}-y(1)}{x^2}\right] \)
\( \Rightarrow \frac{2}{x^2+y^2}\left(x + y\frac{dy}{dx}\right) = 2 \cdot \frac{x^2}{x^2+y^2}\left(\frac{xdy/dx-y}{x^2}\right) \)
\( \Rightarrow x + y\frac{dy}{dx} = x\frac{dy}{dx} - y \)
\( \Rightarrow x + y = x\frac{dy}{dx} - y\frac{dy}{dx} \)
\( \Rightarrow x + y = (x - y)\frac{dy}{dx} \)
\( \Rightarrow \frac{dy}{dx} = \frac{x+y}{x-y} \) (Proved)

Question. If \( x\sqrt{1+y} + y\sqrt{1+x} = 0 \), show that \( \frac{dy}{dx} = -\frac{1}{(1+x)^2} \).
Answer: We have, \( x\sqrt{1+y} + y\sqrt{1+x} = 0 \)
\( \Rightarrow x\sqrt{1+y} = -y\sqrt{1+x} \)
squaring
\( \Rightarrow x^2(1 + y) = y^2(1 + x) \)
\( \Rightarrow x^2 + x^2y = y^2 + y^2x \)
\( \Rightarrow x^2 - y^2 = y^2x - x^2y \)
\( \Rightarrow (x + y)(x - y) = yx(y - x) \)
\( \Rightarrow (x + y)(x - y) = -yx(x - y) \)
\( \Rightarrow x + y = -yx \)
\( \Rightarrow y + yx = -x \)
\( \Rightarrow y(1 + x) = -x \)
\( \Rightarrow y = -\frac{x}{1+x} \)
Diff w.r.t. x (Quotient rule)
\( \Rightarrow \frac{dy}{dx} = \frac{(1+x)(-1) - (-x)(1)}{(1+x)^2} \)
\( \Rightarrow \frac{dy}{dx} = \frac{-1-x+x}{(1+x)^2} \)
\( \Rightarrow \frac{dy}{dx} = -\frac{1}{(1+x)^2} \) (Proved)

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