Paragraph: Let f(x)= 2 x+x on [(-π)/(2), (π)/(2)] Question: What is the greatest value of f(x) ?

Correct answer: B. 3 2 +(π)/(12) f(x)= 2 x+x on [-(π)/(2), (π)/(2)]f(x)=-2 2 x+1 for maxima \& minima f(x)=0 aligned & 2 x=(1)/(2) & x=(π)/(12) or (5 π)/(12)( Not in interval) of [(-π)/(2), (π)/(2)]. aligned So, maximum value of f(x) occurs at x=(π)/(12) ie f((π)/(12))= (π)/(6)+(π)/(12)= 3 2 +(π)/(12)

NDA 2024

Exam: NDA
Year: 2024
Subject: Mathematics
Chapter: Application of Derivatives

NDA 2024 · Mathematics · Application of Derivatives

Paragraph: Let f(x)= 2 x+x on [(-π)/(2), (π)/(2)] Question: What is...

Paragraph: Let

f(x)=\cos 2 x+x

on \left[\frac{-\pi}{2}, \frac{\pi}{2}\right]
Question: What is the greatest value of

f(x)

Options

A $\frac{\sqrt{3}}{2}-\frac{\pi}{12}$
B $\frac{\sqrt{3}}{2}+\frac{\pi}{12}$
C $\frac{\sqrt{3}}{2}+\frac{\pi}{9}$
D $\frac{\sqrt{3}}{2}+\frac{\pi}{6}$

Correct answer

B. $\frac{\sqrt{3}}{2}+\frac{\pi}{12}$

Step-by-step solution

f(x)=\cos 2 x+x

on \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]f(x)=-2 \sin 2 x+1

for maxima \& minima

f(x)=0

\begin{aligned} & \sin 2 x=\frac{1}{2} \\ & x=\frac{\pi}{12} \text{ or } \frac{5 \pi}{12}\left(\text{ Not in interval) of }\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\right. \end{aligned}

So, maximum value of

f(x)

occurs at

x=\frac{\pi}{12}

ie

f\left(\frac{\pi}{12}\right)=\cos \frac{\pi}{6}+\frac{\pi}{12}=\frac{\sqrt{3}}{2}+\frac{\pi}{12}$

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