`f(x)` satisfies the relation `f(x)-lamda int_(0)^(pi//2)sinxcostf(t)dt=sinx` If `f(x)=2` has the least one real root, then

Correct Answer - D
`f(x)-lamda int_(0)^(pi//2)sinx cost f (t) dt=sinx`
or `f(x)-lamda sinx int_(0)^(pi//2) cost f(t)dt=sinx`
or `f(x)=Asinx=sinx`
or `f(x)=(A+1)sinx` where
`A=lamdaint_(0)^(pi//2) cost f(t)dt`
or `A=lamda int_(0)^(pi//2) cos (A+1)sin dt`
`=(lamda(A+1))/2int_(0)^(pi//2) sin 2tdt`
`=(lamda(A+1))/2[(-cos 2t)/2]_(0)^(pi//2)`
`=(lamda(A+1))/2`
`:. A=(lamda)/(2-lamda)`
`:.f(x)=((lamda)/(2-lamda)+1)sinx=(2/(2-lamda))sinx`
`(2/(2-lamda))sinx=2`
or `sinx=(2-lamda)`
or `|2-lamda|le1`
or `-1le lamda-2le 1`
or `1 le lamda le 3`
`int_(0)^(pi//2) f(x)dx=3`
or `int_(0)^(pi//2) 2/(2-lamda) sinxdx=3`
or `-[2/(2-lamda) cosx]_(0)^(pi//2) =3`
or `2/(2-lamda)=3`