Question

$f\left(x\right)$ satisfies the relation $f\left(x\right)-\lambda \underset{0}{\overset{\pi /2}{\int }}\sin x\cos tf\left(t\right)dt=\sin x.$

If $\lambda >2,$ then $f\left(x\right)$ decreases in which of the following interval?

• A. $(0,\pi )$
• B. $(\frac{\pi }{2},\frac{3\pi }{2})$
• C. $(-\frac{\pi }{2},\frac{\pi }{2})$
• D. none of these

Answer 1

The correct option is C $(-\frac{\pi }{2},\frac{\pi }{2})$

$f\left(x\right)-\lambda \underset{0}{\overset{\pi /2}{\int }}\sin x\cos tf\left(t\right)dt=\sin x\Rightarrow f\left(x\right)-\lambda \sin x\underset{0}{\overset{\pi /2}{\int }}\cos tf\left(t\right)dt=\sin x$
$\Rightarrow f\left(x\right)-A\sin x=\sin x$
where, $A=\lambda \underset{0}{\overset{\pi /2}{\int }}\cos tf\left(t\right)dt$

$\Rightarrow f\left(x\right)=(A+1)\sin x$
$\Rightarrow f\left(t\right)=(A+1)\sin t$

$\therefore A=\lambda \underset{0}{\overset{\pi /2}{\int }}(A+1)\cos t\sin tdt\Rightarrow A=\frac{\lambda (A+1)}{2}\underset{0}{\overset{\pi /2}{\int }}\sin 2tdt\Rightarrow A=\frac{\lambda (A+1)}{2}{[-\frac{\cos 2t}{2}]}_0^{\pi /2}\Rightarrow A=\frac{\lambda (A+1)}{2}\Rightarrow A=\frac{\lambda }{2-\lambda }$
So,
$f\left(x\right)=(\frac{\lambda }{2-\lambda }+1)\sin x\Rightarrow f\left(x\right)=\left(\frac{2}{2-\lambda }\right)\sin x$

Now, $\lambda >2$
$\Rightarrow f\left(x\right)=-k\sin x,(k>0)$
Hence, $f\left(x\right)$ is decreasing when $\sin x$ is increasing.
From the given options, $\sin x$ is increasing in $(-\frac{\pi }{2},\frac{\pi }{2})$.

Therefore, $f\left(x\right)$ is decreasing in $(-\frac{\pi }{2},\frac{\pi }{2})$.