Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please solve R.D.Sharma class 12 chapter Differentiation exercise 10.3 question 1 maths textbook solution.

Please solve R.D.Sharma class 12 chapter Differentiation exercise 10.3 question 1 maths textbook solution.

Answers (1)

Answer :  \frac{d y}{d x}=\frac{2}{\sqrt{1-x^{2}}}
Hint:

\begin {array} {l}\cos ^{-1}(\cos \theta)=\theta\ \ if \ \ \theta \varepsilon[0, \pi]\\\\ \cos ^{-1}(\cos \theta)=-\theta \ \ if \ \ \theta \varepsilon[-\pi, 0] \end{array}
Given:

\cos ^{-1}\left\{2 x \sqrt{1-x^{2}}\right\}, \frac{1}{\sqrt{2}}<x<1

Solution:

\begin {array}{ll} Lets\: y=\cos ^{-1}\left\{2 x \sqrt{1-x^{2}}\right\} \\\\ Lets \: \mathrm{x}=\cos \theta \Rightarrow \theta=\cos ^{-1} \mathrm{x} \\\\ Now \ \ y=\cos ^{-1}\left\{2 \cos \theta \sqrt{1-\cos ^{2} \theta}\right\}\end{array}

\begin {array}{ll} Using \therefore \cos ^{2} \theta+\sin ^{2} \theta=1,2 \sin \theta \cos \theta=\sin 2 \theta\\\\ \mathrm{y}=\cos ^{-1}\left\{2 \cos \theta \sqrt{\sin ^{2} \theta}\right\}\\\\ \end{array}

\begin {array}{ll} y=\cos ^{-1}(2 \cos \theta \sin \theta]\\\\ y=\cos ^{-1}(\sin 2 \theta)\\\\ \therefore \cos \left(\frac{\pi}{2}-\theta\right) \Rightarrow \sin \theta\\\\ y=\cos ^{-1}\left(\cos \left(\frac{\pi}{2}-2 \theta\right)\right)\\\\ \end{array}

Now by considering the limits,

\begin{array}{l} \frac{1}{\sqrt{2}}<x<1\\\\ \Rightarrow \frac{1}{\sqrt{2}}<\cos \theta<1\\\\ \Rightarrow 0<\theta<\frac{\pi}{4} \ \ \ \ \ \ \ \ \ \ \therefore\left[\cos 90^{\circ}=0, \cos 0^{\circ}=1\right]\\\\ \Rightarrow 0<2 \theta<\frac{\pi}{2}\\\\ \Rightarrow 0>-2 \theta>-\frac{\pi}{2}\\\\ \Rightarrow \frac{\pi}{2}>\frac{\pi}{2}-2 \theta>\frac{\pi}{2}-\frac{\pi}{2} \end{array}

Therefore,

\begin{array}{l} y=\cos ^{-1}\left(\cos \left(\frac{\pi}{2}-2 \theta\right)\right)\\\\ y=\cos ^{-1}\left(\cos \left(\frac{\pi}{2}-2 \theta\right)\right)\\\\ y=\left(\frac{\pi}{2}-2 \theta\right) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array}                                                                                                    \left\{\cos ^{-1}(\cos \theta)=\theta\right\}

y=\frac{\pi}{2}-2 \cos ^{-1} x

\text { if } \theta \varepsilon[0, \pi]
Differentiating with Respect to x, we get

\begin{array}{l} \frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{2}-2 \cos ^{-1} x\right)\\\\ \therefore \frac{d}{d x}\left(\cos ^{-1} x\right) \Rightarrow \frac{-1}{\sqrt{1-x^{2}}}\\\\ \therefore \frac{\mathrm{d}}{\mathrm{dx}} \text { (constant) }=0\\\\ \frac{d y}{d x}=0-2\left(\frac{-1}{\sqrt{1-x^{2}}}\right)\\\\ \frac{d y}{d x}=\frac{2}{\sqrt{1-x^{2}}} \end{array}

 

Posted by

infoexpert22

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question