Question

Let a function $f\left(x\right)=\underset{-\frac{\pi }{2}}{\overset{x}{\int }}(2{\sin }^{2}t+3\cos t)dt$ is defined in $[-\frac{\pi }{2},\frac{\pi }{2}].$ Then which of the following is/are correct

• A. $f_{min}=\pi -6$
• B. $f_{min}=0$
• C. $f_{max}=\pi +6$
• D. $f_{max}=6-\pi$

Answer 1

Answer: (B), (C)

$f_{max}=\pi +6$

Given : $f\left(x\right)=\underset{-\frac{\pi }{2}}{\overset{x}{\int }}(2{\sin }^{2}t+3\cos t)dt$
$\Rightarrow f^{\prime }\left(x\right)=2{\sin }^{2}x+3\cos x>0$ for $x\in [-\frac{\pi }{2},\frac{\pi }{2}]$
$\Rightarrow f\left(x\right)$ is strictly increasing .
So, $f_{max}=f\left(\frac{\pi }{2}\right)=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}(2{\sin }^{2}t+3\cos t)dt=2\underset{0}{\overset{\frac{\pi }{2}}{\int }}(2{\sin }^{2}t+3\cos t)dt=2\underset{0}{\overset{\frac{\pi }{2}}{\int }}(1-\cos 2t+3\cos t)dt=2{[t-\frac{\sin 2t}{2}+3\sin t]}_0^{\frac{\pi }{2}}=\pi +6$
and $f_{min}=f(-\frac{\pi }{2})=\underset{-\frac{\pi }{2}}{\overset{-\frac{\pi }{2}}{\int }}(2{\sin }^{2}t+3\cos t)dt=0$