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

Correct answer: A. - left(1+(π)/(2)) f(x)= 2 x+x on [-(π)/(2), (π)/(2)] Find the value of f(x) at boundary of interval i.e at x=(-π)/(2) and x=(π)/(2) aligned & f((-π)/(2))=-1+(-π)/(2) & f((π)/(2))=-1+(π)/(2) aligned So, minimum value of f(x)^2=- (1+(π)/(2))

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 least value of

f(x)

Options

A $- \l left(1+\frac{\pi}{2})$
B $- \left(\frac{1}{2}+\frac{\pi}{2}\right)$
C $- \l left(1+\frac{\pi}{4})$
D $-2 \left(\frac{1}{2}-\frac{\pi}{4}\right)$

Correct answer

A. $- \l left(1+\frac{\pi}{2})$

Step-by-step solution

f(x)=\cos 2 x+x

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

Find the value of

f(x)

at boundary of interval i.e at

\mathrm{x}=\frac{-\pi}{2} and \mathrm{x}=\frac{\pi}{2}

\begin{aligned} & f\left(\frac{-\pi}{2}\right)=-1+\frac{-\pi}{2} \\ & f\left(\frac{\pi}{2}\right)=-1+\frac{\pi}{2} \end{aligned}

So, minimum value of

\mathrm{f}(\mathrm{x})^2=- \l\left(1+\frac{\pi}{2}\right)$

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