Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain solution RD Sharma class 12 chapter Maxima and Minima exercise Multiple choice question question 24 maths

Explain solution RD Sharma class 12 chapter Maxima and Minima exercise Multiple choice question, question 24 maths

Answers (1)

Answer: option (a) Minimum at x=\frac{\pi}{2}

Hint: For local maxima or minima, we must have f'(x)=0.

Given: f(x)=1+2\sin x+3\cos^2x

Solution:

We have,

f(x)=1+2\sin x+3\cos^2x

f'(x)=2\cos x-6\cos x\sin x

f'(x)=2\cos x(1-3 \sin x)              (i)

For maxima and minima f'(x)=0

\begin{aligned} &\Rightarrow 2 \cos x(1-3 \sin x)=0 \\ &\Rightarrow 2 \cos x=0 \underset{\text { or }} \Rightarrow(1-3 \sin x)=0 \end{aligned}

\Rightarrow \cos x=0\Rightarrow \sin x =\frac{1}{3}

\Rightarrow x=\frac{\pi}{2}                                    \Rightarrow x=\sin ^{-1}\left ( \frac{1}{3} \right )

Now,

\begin{aligned} &f^{\prime}(x)=2 \cos x(-3 \cos x)+(1-3 \sin x)(-2 \sin x) \\ &=-6 \cos ^{2} x-2 \sin x+6 \sin ^{2} x \\ &=-6\left(\cos ^{2} x-\sin ^{2} x\right)-2 \sin x \\ &-6 \cos 2 x-2 \sin x \quad\left[\because \cos ^{2} x-\sin ^{2} x=\cos 2 x\right] \end{aligned}

\begin{aligned} &\text { at } x=\frac{\pi}{2}, f^{\prime \prime}\left(\frac{\pi}{2}\right) \\ &=-6 \cos \pi-2 \sin \frac{\pi}{2} \\ &=-6(-1)-2 \times 1 \quad\left[\because \cos \pi=-1, \sin \frac{\pi}{2}=1\right] \\ &-2+6=4>0 \end{aligned}

So, x=\frac{\pi}{2} is local minima

\begin{aligned} &\text { at } x=\sin ^{-1}\left(\frac{1}{3}\right) \\ &f^{\prime \prime}(x)=f^{\prime \prime}\left(\sin ^{-1}\left(\frac{1}{3}\right)\right) \\ &=-2 \sin \left(\sin ^{-1}\left(\frac{1}{3}\right)\right)-6\left(1-2 \sin ^{2} x\right) \quad\left[\because \cos 2 x=1-2 \sin ^{2} x\right] \end{aligned}

\begin{aligned} &=-2 \times \frac{1}{3}-6+12(\sin x)^{2} \\ &=\frac{-2}{3}-6+12\left(\sin \left(\sin ^{-1}\left(\frac{1}{3}\right)\right)\right)^{2} \\ &=\frac{-2}{3}-6+12 \times\left(\frac{1}{3}\right)^{2}=\frac{-2}{3}-6+12 \times \frac{1}{9} \end{aligned}

\begin{aligned} &=\frac{-2}{3}-6-\frac{4}{3}=\frac{-2-18-4}{3} \\ &=\frac{-24}{3}=-8 \\ &f^{\prime \prime}(x)<0 \end{aligned}

So, x=\sin ^{-1} \left (\frac{1}{3} \right ) is local maxima

Posted by

infoexpert24

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