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BITSAT 2017 Slot 2 Solved Paper

Section: Mathematics

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Question : 133 of 150

Marks: +1, -0

The maximum value of

f(x)=x+\sin 2x, x \in [0,2\pi]

is

$\frac{\pi}{2}$
$2\pi$
$\frac{3\pi}{4}$
$\frac{3\pi}{2}$

Solution:

Given,

f(x)=x+\sin 2x

\Rightarrow f'(x)=1+2 \cos 2x

For maximum or minimum value,

f'(x)=0

\Rightarrow 1+2 \cos 2x=0 \Rightarrow \cos 2x=-\frac{1}{2}

\Rightarrow 2x=2n\pi \pm \frac{2\pi}{3}

\Rightarrow x=n\pi \pm \frac{\pi}{3}

\Rightarrow x=\frac{4\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{3}

Find

f(0), f\left(\frac{2\pi}{3}\right), f\left(\frac{4\pi}{3}\right), f\left(\frac{5\pi}{3}\right), f(2\pi)

\Rightarrow f(0)=0, f\left(\frac{2\pi}{3}\right)=\frac{2\pi}{3}-0.8

\Rightarrow f\left(\frac{4\pi}{3}\right)=\frac{4\pi}{3}+0.8, f\left(\frac{5\pi}{3}\right)

=\frac{5\pi}{3}-0.8

\Rightarrow f(2\pi)=2\pi+0=2\pi

\therefore

Maximum value of

f(x)=2\pi

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