Question

A differentiable function $f$ satisfies the equation $f\left(x\right)=\underset{0}{\overset{x}{\int }}\left(f\right(t)\cos t-\cos (t-x\left)\right)dt$, then the correct option(s) is/are

• A. $f^{\prime }\left(0\right)=-1$
• B. $f\left(x\right)$ has a maximum value $1-e$
• C. $f\left(x\right)$ has a minimum value $1-e^{-1}$
• D. $f^{\prime \prime }\left(\frac{\pi }{2}\right)=e$

Answer 1

Answer: (A), (D)

The correct options are
A $f^{\prime }\left(0\right)=-1$
D $f^{\prime \prime }\left(\frac{\pi }{2}\right)=e$

$f\left(x\right)=\underset{0}{\overset{x}{\int }}\left(f\right(t)\cos t-\cos (t-x\left)\right)dt\Rightarrow f\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)\cos tdt-\underset{0}{\overset{x}{\int }}\cos (t-x)dt\Rightarrow f\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)\cos tdt-{\left[\sin \right(t-x\left)\right]}_0^x\Rightarrow f\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)\cos tdt-\sin x\Rightarrow f^{\prime }\left(x\right)=f\left(x\right)\cos x-\cos x$

We know that $f\left(0\right)=0$
$\Rightarrow f^{\prime }\left(0\right)=-1$

$f^{\prime }\left(x\right)=f\left(x\right)\cos x-\cos x$
Let $y=f\left(x\right)$
$\Rightarrow \frac{dy}{dx}=\cos x(y-1)\Rightarrow \int \frac{dy}{y-1}=\int \cos xdx\Rightarrow \ln |y-1|=\sin x+c\Rightarrow |y-1|=C{e}^{\sin x},C=e^c$
$\Rightarrow y-1=\pm C{e}^{\sin x},C=e^c$

$y\left(0\right)=0$
$\therefore f\left(x\right)=1-e^{\sin x}\Rightarrow f^{\prime }\left(x\right)=-e^{\sin x}\cos x$

For maxima and minima,
$e^{\sin x}\cos x=0\Rightarrow \cos x=0\Rightarrow x=\frac{(2n+1)\pi }{2}$

$f^{\prime \prime }\left(x\right)=-e^{\sin x}{\cos }^{2}x+e^{\sin x}\sin x$
At $x=\frac{\pi }{2}$
$f^{\prime \prime }\left(\frac{\pi }{2}\right)=e>0$
So, minimum value of $f$ is,
$f\left(\frac{\pi }{2}\right)=1-e$

$f^{\prime \prime }\left(\frac{3\pi }{2}\right)=-e^{-1}<0$
So, maximum value of $f$ is,
$f\left(\frac{3\pi }{2}\right)=1-e^{-1}$