`f(x)=sinx+int_(-pi//2)^(pi//2)(sinx+tcosx)f(t)dt`
The range of `f(x)` is

Options

A. `[-(sqrt(3))/2,(sqrt(3))/2]`
B. `[-(sqrt(5))/3,(sqrt(5))/3]`
C. `[-(sqrt(5))/2,(sqrt(5))/2]`
D. none of these

Answer

B

Explanation

`f(x)=sinx+sin int_(-pi//2)^(pi//2) f(t)dt+cosx int_(-pi//2)^(pi//2) tf(t)dt`
`=sinx (1+int_(-pi//2)^(pi//2) f(t)dt)+cosx int_(-pi//2)^(pi//2) tf(t)dt`
`=Asinx+Bcosx`
Thus, `A=1+int_(-pi//2)^(pi//2) f(t)dt`
`=1+int_(-pi//2)^(pi//2) (Asint+Bcost)dt`
`=1+2Bint_(0)^(pi//2) cost dt`
`=A=1+2B` ............1
`B=int_(-pi//2)^(pi//2) tf(t)dt`
`=int_(-pi//2)^(pi//2)t(Asint+Bcost)dt`
`=2Aint_(0)^(pi//2) t sin tdt`
`=2A[-tcost+sint]_(0)^(pi//2)`
`:.B=2A`
From equations 1 and 2 we get
`A=-1//3, B=-2//3`
`:.f(x)=-1/3(sinx+2cosx)`
Thus, the range fo `f(x)` is `[-(sqrt(5))/3,(sqrt(5))/3]`
`f(x)=-1/2(sinx+2cosx)`
`=-(sqrt(5))/3 sin (x+tan^(-1)2)`
`=-(sqrt(5))/3cos(x-"tan"^(-1)1/2)`
`f(x)` in invertible if `-(pi)/2 le x=tan^(-1)2le(pi)/2`
or `-(pi)/2-tan^(-1)2 le xle (pi)/2 -tan^(-1) 2`
or `0lex-"tan"^(-1)1/2 le pi`
or `"tan"^(-1)1/2 le x le pi+"tan"^(-1)1/2`
or `pi le x-"tan"^(-1)1/2le 2pi`
or `x epsilon [pi+cot^(-1)2, 2pi +cot^(-1) 2]`
`int_(0)^(pi//2) f(x)dx=-1/3int_(0)^(pi//2) (sinx+2cosx)dx`
`=-1/3[-cosx+2sinx]_(0)^(pi//2) =-1`