Question

Let $f$ be a non-negative function defined on the interval $[0,\frac{\pi }{2}]$
If $\underset{0}{\overset{x}{\int }}\left(f^{{}^{\prime }}\right(t)-\sin 2t)dt=\underset{x}{\overset{0}{\int }}f\left(t\right)\tan tdt$
and $f\left(0\right)=1$, assume $y=f\left(x\right)$ then

• A. $f\left(0\right)=1$ is the maximum value of $f$
• B. $f\left(0\right)=1$ is the minimum value of $f$
• C. $\underset{0}{\overset{\frac{\pi }{2}}{\int }}ydx=3-\frac{\pi }{2}$
• D. $f\left(\frac{\pi }{3}\right)=1$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $f\left(0\right)=1$ is the minimum value of $f$
C $\underset{0}{\overset{\frac{\pi }{2}}{\int }}ydx=3-\frac{\pi }{2}$
D $f\left(\frac{\pi }{3}\right)=1$

$\underset{0}{\overset{x}{\int }}\left(f^{{}^{\prime }}\right(t)-\sin 2t)dt=\underset{x}{\overset{0}{\int }}f\left(t\right)\tan tdt$
Use Newton Leibnitz theorem to differentiate RHS & LHS
$f^{\prime }\left(x\right)-\sin 2x=-f\left(x\right)\tan x$
$\Rightarrow f^{{}^{\prime }}\left(x\right)+f\left(x\right)\tan x=\sin 2x$
$\Rightarrow \frac{dy}{dx}+y\tan x=\sin 2x$
Integrating factor,
$=e^{\int \tan xdx}=\sec x$
$\Rightarrow y\sec x=\int \sin 2x\sec xdx$
$\Rightarrow \frac{y}{\cos x}=-2\cos x+c$

Given $f\left(0\right)=1$
$\Rightarrow 1=-2+c\Rightarrow c=3$
So,
$f\left(x\right)=3\cos x-2{\cos }^{2}x$
$\Rightarrow f\left(\frac{\pi }{3}\right)=\frac{3}{2}-2\times \frac{1}{4}=\frac{3}{2}-\frac{1}{2}=1$
$\Rightarrow \underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(x\right)dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}(3\cos x-\cos 2x-1)dx$
$={[3\sin x-\frac{\sin 2x}{2}-x]}_0^{\frac{\pi }{2}}$
$=3-\frac{\pi }{2}$

$f\left(x\right)=3\cos x-2{\cos }^{2}x$
$f^{\prime }\left(x\right)=\sin x(-3+4\cos x)$
At $x=0\Rightarrow f^{\prime }\left(0\right)=0$

$f^{\prime \prime }\left(x\right)=-3\cos x+4\cos 2x$
$f^{\prime \prime }\left(0\right)=1>0$
$x=0$ point of local minima and the minima will be,
$f\left(0\right)=1$