Question

A differentiable function $f$ satisfies $f\left(x\right)=\underset{0}{\overset{x}{\int }}\left(f\right(t)\cos t-\cos (t-x\left)\right)dt$. Which of the following hold(s) good?

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

Answer 1

Answer: (A), (B), (C)

The correct options are
A $f\left(x\right)$ has the minimum value $1-e$
B $f\left(x\right)$ has the maximum value $1-e^{-1}$
C $f^{\prime \prime }\left(\frac{\pi }{2}\right)=e$

$f\left(x\right)=\underset{0}{\overset{x}{\int }}\left(f\right(t)\cos t-\cos (t-x\left)\right)dt=\underset{0}{\overset{x}{\int }}f\left(t\right)\cos tdt-\underset{0}{\overset{x}{\int }}\cos (-t)dt[\because \underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx]$
$\Rightarrow f\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)\cos tdt-\sin x$
Differentiating both sides w.r.t. $x$, we get
$f^{\prime }\left(x\right)=f\left(x\right)\cos x-\cos x$

Let $f\left(x\right)=y;f^{\prime }\left(x\right)=\frac{dy}{dx}$
Then, $\frac{dy}{dx}-y\cos x=-\cos x$
$I.F.=e^{-\int \cos xdx}=e^{-\sin x}$
Hence, $y\cdot e^{-\sin x}=\int -e^{-\sin x}\cos xdx$
$\Rightarrow y\cdot e^{-\sin x}=C+e^{-\sin x}$
$\Rightarrow y=C{e}^{\sin x}+1$
If $x=0,$ then $y=0$ (from the given relation)
$\Rightarrow C=-1$
Hence, $f\left(x\right)=1-e^{\sin x}$

Now, minimum value $=(1-e)$ when $x=\frac{\pi }{2}$
and maximum value $=(1-e^{-1})$ when $x=\frac{-\pi }{2}$

$f^{\prime }\left(x\right)=-e^{\sin x}\cos x$
$f^{\prime }\left(0\right)=-1$
$f^{\prime \prime }\left(x\right)=-[{\cos }^{2}x\cdot e^{\sin x}-e^{\sin x}\cdot \sin x]f^{\prime \prime }\left(\frac{\pi }{2}\right)=e$