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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 19 Maths Textbook Solution.

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3  Question 19 Maths Textbook Solution.

Answers (1)

Answer:\frac{dy}{dx}=\frac{-1}{2\sqrt{1-x^{2}}}

Hint:

\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constants })=0 \\ &\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1} \end{aligned}

Given:

\sin ^{-1}\left\{\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right\}, 0<x<1

Solution:

Let

y=\sin ^{-1}\left\{\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right\}

Let,x=\cos 2\theta

\begin{aligned} &\text { Now, } \sin ^{-1}\left\{\frac{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}{2}\right\} \\ &\text { Using } 1-2 \sin ^{2} \theta=\cos 2 \theta \\ &2 \cos ^{2} \theta-1=\cos 2 \theta \\ &y=\sin ^{-1}\left\{\frac{\sqrt{2 \cos ^{2} \theta}+\sqrt{2 \sin ^{2} \theta}}{2}\right\} \end{aligned}

Now,

\begin{aligned} &\mathrm{y}=\sin ^{-1}\left\{\frac{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}{2}\right\} \\ &\mathrm{y}=\sin ^{-1}\left\{ \cos \theta \frac{1}{\sqrt{2}}+\sin \theta \frac{1}{\sqrt{2}}\right\} \\ &\mathrm{y}=\sin ^{-1}\left\{\sin \theta \cos \left(\frac{\mathrm{\pi}}{4}\right)+\cos \theta \sin \left(\frac{\mathrm{\pi}}{4}\right)\right\} \\ &\therefore \sin \left(\frac{\mathrm{\pi}}{4}\right)=\frac{1}{\sqrt{2}} \& \cos \left(\frac{\mathrm{\pi}}{4}\right)=\frac{1}{\sqrt{2}} \\ &\mathrm{U} \sin \mathrm{sin}(\mathrm{A}+\mathrm{B})=\sin \mathrm{A} \cos \mathrm{B}+\cos \mathrm{Asin} \mathrm{B} \\ &\mathrm{y}=\sin ^{-1}\left\{\sin \left(\theta+\frac{\mathrm{\pi}}{4}\right)\right\} \end{aligned}

Considering the limits,

\begin{aligned} &0<\mathrm{x}<1 \\ &0<\cos 2 \theta<1 \\ &0<2 \theta<\frac{\mathrm{\pi}}{2} \\ &0<\theta<\frac{\mathrm{\pi}}{4} \\ &0+\frac{\mathrm{\pi}}{4}<\left(\theta+\frac{1}{4}\right)<\frac{\mathrm{\pi}}{2}+\frac{\pi}{4} \\ &\frac{\mathrm{\pi}}{4}<\left(\theta+\frac{\pi}{4}\right)<\frac{\mathrm{\pi}}{2} \end{aligned}

Now,y=\sin ^{-1}                                                                            \left \{ \sin \left ( \theta +\frac{\pi}{4} \right ) \right \}

y=\theta +\frac{\pi }{4}                                                                                                       \sin ^{-1}\left ( \sin \theta \right )=\theta ,\: if\: \theta \varepsilon \left [ -\frac{\pi}{2},\frac{\pi}{2} \right ]

    y=\frac{1}{2} \cos ^{-1} x+\frac{\pi}{4}                                                                                                                            \left \{ since\: x=\cos 2\theta \right \}

\frac{d y}{d x}=\frac{d}{d x}\left(\frac{1}{2} \cos ^{-1} x+\frac{\pi}{4}\right)                                                                                

\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constant })=0 \ \\ &\frac{d}{\mathrm{dx}}\left(\cos ^{-1} \mathrm{x}\right)=\frac{-1}{\sqrt{1-\mathrm{x}^{2}}} \\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left(\frac{-1}{\sqrt{1-\mathrm{x}^{2}}}\right)+0 \\ &\frac{\mathrm{dy}}{\mathrm{dx}} \Rightarrow \frac{-1}{2 \sqrt{1-\mathrm{x}^{2}}} \end{aligned}

 

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