CBSE Class 12 Mathematics Continuity And Differentiability Worksheet Set 05

CBSE Class 12 Mathematics Continuity And Differentiability Worksheet 5. 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 \( y = \frac{x\sin^{-1}x}{\sqrt{1-x^2}} + \log\sqrt{1-x^2} \), show that \( \frac{dy}{dx} = \frac{sin^{-1}x}{(1-x^2)^{3/2}} \).
Answer: We have, \( y = \frac{x\sin^{-1}x}{\sqrt{1-x^2}} + \frac{1}{2}\log(1-x^2) \)
Diff w.r.t. x
\( \frac{dy}{dx} = \frac{\sqrt{1-x^2}\cdot\left(x\cdot\frac{1}{\sqrt{1-x^2}}+\sin^{-1}x\right)-x\sin^{-1}x\cdot\frac{1}{2\sqrt{1-x^2}}(-2x)}{(1-x^2)} + \frac{1}{2}\frac{1}{(1-x^2)}\cdot(-2x) \)
\( \Rightarrow \frac{dy}{dx} = \frac{x+\sqrt{1-x^2}\cdot\sin^{-1}x+\frac{x^2\sin^{-1}x}{\sqrt{1-x^2}}}{(1-x^2)} - \frac{x}{1-x^2} \)
\( \Rightarrow \frac{dy}{dx} = \frac{x\sqrt{1-x^2}+(1-x^2)\sin^{-1}x+x^2\sin^{-1}x}{\sqrt{1-x^2}(1-x^2)} - \frac{x}{1-x^2} \)
\( \Rightarrow \frac{dy}{dx} = \frac{x\sqrt{1-x^2}+\sin^{-1}x-x^2\sin^{-1}x+x^2\sin^{-1}x-x\sqrt{1-x^2}}{\sqrt{1-x^2}(1-x^2)} \)
\( \Rightarrow \frac{dy}{dx} = \frac{\sin^{-1}x}{(1-x^2)^{3/2}} \) Ans.

Diff. Of Infinite Series

Question. \( y = x^{x^{\dots \infty}} \). Find \( \frac{dy}{dx} \).
Answer: We have,
\( \Rightarrow y = x^y \)
taking log on both sides
\( \Rightarrow \log y = y\log x \)
Diff w.r.t. x
\( \frac{1}{y}\cdot\frac{dy}{dx} = y\cdot\frac{1}{x} + \log x\cdot\frac{dy}{dx} \)
\( \Rightarrow \frac{1}{y}\frac{dy}{dx} - \log x\cdot\frac{dy}{dx} = \frac{y}{x} \)
\( \Rightarrow \frac{dy}{dx} \left(\frac{1}{y} - \log x\right) = \frac{y}{x} \)
\( \Rightarrow \frac{dy}{dx} \left(\frac{1-y\log x}{y}\right) = \frac{y}{x} \)
\( \Rightarrow \frac{dy}{dx} = \frac{y^2}{x(1-y\log x)} \) Ans.

Question. \( y = \frac{\sin x}{1+\frac{\sin x}{1+\frac{\cos x}{1+\dots \infty}}} \). Find \( \frac{dy}{dx} \).
Answer: \( y = \frac{\sin x}{1+\frac{\cos x}{1+y}} \)
\( \Rightarrow y = \frac{(1+y)\sin x}{1+y+\cos x} \)
\( \Rightarrow y + y^2 + y\cos x = \sin x + y\sin x \)
then Diff w.r.t. 'x' (Do Yourself)
Ans. \( \frac{dy}{dx} = \frac{(1+y)\cos x+y\sin x}{1+2y+\cos x-\sin x} \)

Inverse Trigonometric Diff.

Question. \( y = \sin^{-1} \left(\frac{1-x^2}{1+x^2}\right) ; 0 < x < 1 \).
Answer: \( y = \sin^{-1} \left(\frac{1-x^2}{1+x^2}\right) \)
put \( x = \tan\theta \)
\( \Rightarrow y = \sin^{-1} \left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right) \)
\( \Rightarrow y = \sin^{-1}(\cos(2\theta)) \)
\( \Rightarrow y = \sin^{-1} \left(\sin \left(\frac{\pi}{2} - 2\theta\right)\right)  /> \( \Rightarrow y = \frac{\pi}{2} - 2\theta \quad \dots [0 < x < 1] \text{ (Conditions)} \)
\( [0 < \tan\theta < 1] \)
\( [0 < \theta < \frac{\pi}{4}] \)
\( \Rightarrow y = \frac{\pi}{2} - 2\tan^{-1}x \quad [0 < 2\theta < \frac{\pi}{2}] \)
Diff w.r.t. x \( \quad \left[\Rightarrow 0 < \frac{\pi}{2} - 2\theta < \frac{\pi}{2}\right] \)
\( \Rightarrow \frac{dy}{dx} = 0 - \frac{2}{1+x^2} = \frac{-2}{1+x^2} \) Ans.

Question. \( y = \cos^{-1} \left(\frac{2x}{1+x^2}\right) ; -1 < x < 1 \).
Answer: \( y = \cos^{-1} \left(\frac{2x}{1+x^2}\right) \)
put \( x = \tan\theta \)
\( \Rightarrow y = \cos^{-1} \left(\frac{2\tan\theta}{1+\tan^2\theta}\right) \)
\( \Rightarrow y = \cos^{-1}(\sin(2\theta)) \)
\( \Rightarrow y = \cos^{-1}\left(\cos\left(\frac{\pi}{2}-2\theta\right)\right) \quad \dots -1 < x < 1 \text{ conditions} \)
\( \therefore y = \frac{\pi}{2} - 2\theta \quad \dots -1 < \tan\theta < 1 \)
\( \Rightarrow y = \frac{\pi}{2} - 2\tan^{-1}x \quad \dots -\frac{\pi}{4} < \theta < \frac{\pi}{4} \)
\( \Rightarrow \frac{dy}{dx} = 0 - \frac{2}{1+x^2} = \frac{-2}{1+x^2} \quad \dots -\frac{\pi}{2} < 2\theta < \frac{\pi}{2} \)
\( \frac{\pi}{2} > -2\theta > -\frac{\pi}{2} \)
\( \pi > \frac{\pi}{2} - 2\theta > 0 \)
\( \therefore \frac{\pi}{2} - 2\theta \in (0, \pi) \)

Question. \( y = \tan^{-1} \left(\frac{\sqrt{1+x^2}+1}{x}\right) \). Find \( \frac{dy}{dx} \).
Answer: \( y = \tan^{-1} \left(\frac{\sqrt{1+x^2}+1}{x}\right) \)
put \( x = \tan\theta \)
\( \Rightarrow y = \tan^{-1} \left(\frac{\sqrt{1+\tan^2\theta}+1}{\tan\theta}\right) \)
\( \Rightarrow y = \tan^{-1} \left(\frac{\sec\theta+1}{\tan\theta}\right) \)
\( \Rightarrow y = \tan^{-1} \left(\frac{\frac{1}{\cos\theta}+1}{\frac{\sin\theta}{\cos\theta}}\right) \)
\( \Rightarrow y = \tan^{-1} \left(\frac{1+\cos\theta}{\sin\theta}\right) \)
\( \Rightarrow y = \tan^{-1} \left(\frac{2\cos^2(\theta/2)}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right)  /> \( \Rightarrow y = \tan^{-1}\left(\cot\frac{\theta}{2}\right) \)
\( \Rightarrow y = \tan^{-1} \left(\tan \left(\frac{\pi}{2} - \frac{\theta}{2}\right)\right) \)
\( \Rightarrow y = \frac{\pi}{2} - \frac{\theta}{2} \)
\( \Rightarrow y = \frac{\pi}{2} - \frac{1}{2}\tan^{-1}x \)
Diff w.r.t. x
\( \Rightarrow \frac{dy}{dx} = 0 - \frac{1}{2}\cdot\frac{1}{1+x^2} = \frac{-1}{2(1+x^2)} \) Ans.

Question. \( y = \tan^{-1}\sqrt{\frac{1+\sin x}{1-\sin x}} \). Find \( \frac{dy}{dx} \).
Answer: \( y = \tan^{-1}\sqrt{\frac{1+\sin x}{1-\sin x}} \)
\( \Rightarrow y = \tan^{-1}\sqrt{\frac{1+\cos\left(\frac{\pi}{2}-x\right)}{1-\cos\left(\frac{\pi}{2}-x\right)}} \)
\( \Rightarrow y = \tan^{-1}\sqrt{\frac{2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2\sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}} \)
\( \Rightarrow y = \tan^{-1}\sqrt{\cot^2\left(\frac{\pi}{4}-\frac{x}{2}\right)} \)
\( \Rightarrow y = \tan^{-1}\left(\cot\left(\frac{\pi}{4}-\frac{x}{2}\right)\right) \)
\( \Rightarrow y = \tan^{-1}\left[\tan\left(\frac{\pi}{2}-\left(\frac{\pi}{4}-\frac{x}{2}\right)\right)\right] \)
\( \Rightarrow y = \frac{\pi}{2} - \frac{\pi}{4} + \frac{x}{2} \)
\( \Rightarrow y = \frac{\pi}{4} + \frac{x}{2}  /> Diff w.r.t. x
\( \Rightarrow \frac{dy}{dx} = \frac{1}{2} \) Ans.

Question. \( y = \cot^{-1} \left\{ \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right\} \). Show that \( \frac{dy}{dx} \) is independent of \( x \).
Answer: We have, \( y = \cot^{-1} \left\{ \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right\} \)
\( \Rightarrow y = \tan^{-1} \left\{ \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}} \right\} \quad \dots \left\{\tan^{-1}\left(\frac{1}{x}\right) = \cot^{-1}x\right\} \)
Divide by \( \sqrt{1+\sin x} \)
\( \Rightarrow y = \tan^{-1} \left\{ \frac{1-\sqrt{\frac{1-\sin x}{1+\sin x}}}{1+\sqrt{\frac{1-\sin x}{1+\sin x}}} \right\} \)
\( \Rightarrow y = \tan^{-1}(1) - \tan^{-1}\sqrt{\frac{1-\sin x}{1+\sin x}} \quad \dots \left\{\tan^{-1}\left(\frac{x-y}{1+xy}\right) = \tan^{-1}x - \tan^{-1}y\right\} \)
\( \Rightarrow y = \frac{\pi}{4} - \tan^{-1}\sqrt{\frac{1-\cos\left(\frac{\pi}{2}-x\right)}{1+\cos\left(\frac{\pi}{2}-x\right)}} \)
\( \Rightarrow y = \frac{\pi}{4} - \tan^{-1}\sqrt{\frac{2\sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}} \)
\( \Rightarrow y = \frac{\pi}{4} - \tan^{-1}\left(\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)\right) \)
\( \Rightarrow y = \frac{\pi}{4} - \left(\frac{\pi}{4}-\frac{x}{2}\right) \)
\( \Rightarrow y = \frac{x}{2} \)
Diff w.r.t. x
\( \frac{dy}{dx} = \frac{1}{2} \) clearly it is independent of x. Ans.

Question. \( y = \sin^{-1} \left(\frac{2x}{1+x^2}\right) ; x \in (1, \infty) \). Find \( \frac{dy}{dx} \).
Answer: \( y = \sin^{-1} \left(\frac{2x}{1+x^2}\right) \)
Put \( x = \tan\theta \)
\( \Rightarrow y = \sin^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right) \quad \dots \text{(Conditions)} \)
\( \Rightarrow y = \sin^{-1}(\sin(2\theta)) \quad \dots 1 < x < \infty \)
\( \Rightarrow y = \sin^{-1}(\sin(\pi - 2\theta)) \quad \dots 1 < \tan\theta < \infty /> \( \therefore y = \pi - 2\theta \quad \dots \frac{\pi}{4} < \theta < \frac{\pi}{2} \)
\( \Rightarrow y = \pi - 2\tan^{-1}x \quad \dots \frac{\pi}{2} < 2\theta < \pi \)
Diff w.r.t x \( \quad \dots -\frac{\pi}{2} > -2\theta > -\pi \)
\( \Rightarrow \frac{dy}{dx} = 0 - 2\cdot\frac{1}{1+x^2} \quad \dots \frac{\pi}{2} > \pi-2\theta > 0 \)
\( \Rightarrow \frac{dy}{dx} = \frac{-2}{1+x^2} \) Ans. \( \quad (\pi-2\theta) \in \left(0, \frac{\pi}{2}\right) \)

Question. \( y = \tan^{-1} \left\{ \frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}} \right\} ; -1 < x < 1 \). Find \( \frac{dy}{dx} \).
Answer: We have, \( y = \tan^{-1} \left\{ \frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}} \right\} \)
Divide by \( \sqrt{1+x^2} \)
\( \Rightarrow y = \tan^{-1} \left\{ \frac{1+\sqrt{\frac{1-x^2}{1+x^2}}}{1-\sqrt{\frac{1-x^2}{1+x^2}}} \right\} \)
\( \Rightarrow y = \tan^{-1}(1) + \tan^{-1}\left(\sqrt{\frac{1-x^2}{1+x^2}}\right) \quad \dots \left\{\tan^{-1}\left(\frac{x+y}{1-xy}\right) = \tan^{-1}x + \tan^{-1}y\right\} \)
put \( x^2 = \cos(2\theta) \)
\( \Rightarrow y = \frac{\pi}{4} + \tan^{-1}\sqrt{\frac{1-\cos(2\theta)}{1+\cos(2\theta)}} \)
\( \Rightarrow y = \frac{\pi}{4} + \tan^{-1}\sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}} \)
\( \Rightarrow y = \frac{\pi}{4} + \tan^{-1}(\tan\theta) \)
\( \therefore y = \frac{\pi}{4} + \theta \quad \dots -1 < x < 1 \)
\( \Rightarrow y = \frac{\pi}{4} + \frac{1}{2}\cos^{-1}(x^2) \quad \dots 0 < x^2 < 1 \)
Diff w.r.t. \( x \quad \dots 0 < \cos(2\theta) < 1 \)
\( \Rightarrow \frac{dy}{dx} = 0 - \frac{1}{2}\cdot\frac{1}{\sqrt{1-x^4}}\cdot(2x) \quad \dots 0 < 2\theta < \frac{\pi}{2} \)
\( \Rightarrow \frac{dy}{dx} = \frac{-x}{\sqrt{1-x^4}} \) (Ans) \( \quad \dots 0 < \theta < \frac{\pi}{4} \)

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