Question

If $f\left(x\right)$ satisfies the relation $f\left(x\right)-\lambda \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sin x(\cos t\cdot f(t\left)\right)dt=\sin x,\lambda >2,$ then $f\left(x\right)$ decreases in which of the following interval?

• A. $(-\frac{3\pi }{2},\frac{3\pi }{2})$
• B. $(-\frac{\pi }{2},\frac{\pi }{2})$
• C. $(0,\pi )$
• D. $(\frac{\pi }{2},\frac{3\pi }{2})$

Answer 1

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

$f\left(x\right)-\lambda \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sin x(\cos t\cdot f(t\left)\right)dt=\sin x$
$\Rightarrow f\left(x\right)-\lambda \sin x\underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos t\cdot f\left(t\right)dt=\sin x$
$\Rightarrow f\left(x\right)-A\sin x=\sin x$
where, $A=\lambda \underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos t\cdot f\left(t\right)dt$
$f\left(x\right)=(A+1)\sin x\cdots \left(i\right)$
$\Rightarrow A=\lambda \underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos t(A+1)\sin tdt$
$\Rightarrow A=\frac{\lambda (A+1)}{2}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sin 2tdt$
$\Rightarrow A=\frac{\lambda (A+1)}{2}{\left[\frac{-\cos 2t}{2}\right]}_0^{\frac{\pi }{2}}$
$\Rightarrow A=\frac{\lambda (A+1)}{2}$
$\Rightarrow A=\frac{\lambda }{2-\lambda }$
Now, from equation $\left(i\right)$
$\Rightarrow f\left(x\right)=(\frac{\lambda }{2-\lambda }+1)\sin x$
$\Rightarrow f\left(x\right)=\left(\frac{2}{2-\lambda }\right)\sin x$
Since, $\lambda >2$ and $\sin x$ increases in $(-\frac{\pi }{2},\frac{\pi }{2})$
Hence, $f\left(x\right)$ decreases in $(-\frac{\pi }{2},\frac{\pi }{2})$