Question

If $f\left(x\right)$ satisfies the relation $f\left(x\right)-\lambda \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sin x(\cos t\cdot f(t\left)\right)dt=\sin x$ and $f\left(x\right)=2$ has at least one real root, then

• A. $\lambda \in [1,4]-\left\{2\right\}$
• B. $\lambda \in [-1,2]$
• C. $\lambda \in [0,1]$
• D. $\lambda \in [1,3]-\left\{2\right\}$

Answer 1

The correct option is D $\lambda \in [1,3]-\left\{2\right\}$

$f\left(x\right)-\lambda \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sin x(\cos t\cdot f(t\left)\right)dt=\sin x$
$\Rightarrow f\left(x\right)-\lambda \sin x\underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos t\cdot f\left(t\right)dt=\sin x$
$\Rightarrow f\left(x\right)-A\sin x=\sin x$
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 $\left(i\right)$
$\Rightarrow f\left(x\right)=(\frac{\lambda }{2-\lambda }+1)\sin x$
$\Rightarrow f\left(x\right)=\left(\frac{2}{2-\lambda }\right)\sin x$
$f\left(x\right)=2$ has atleast one root
$\Rightarrow \left(\frac{2}{2-\lambda }\right)\sin x=2$
$\Rightarrow \sin x=(2-\lambda )$
$\Rightarrow |2-\lambda |\le 1$
$\Rightarrow -1\le \lambda -2\le 1$
$\Rightarrow 1\le \lambda \le 3$ and $\lambda \ne 2$