If `f(pi) = 2 and int_0^pi (f(x)+f^('')(x)) sinx dx=5`, then `f(0)` is equal to (it is given that `f(x)` is continuous in `[0,pi])`. (a) 7 (b) 3 (c) 5 (d) 1

Options

A. 7
B. 3
C. 5
D. 1

Answer

B

Explanation

`int_(0)^(pi) [f(x)+f''(x)]sinx dx`
`=int_(0)^(pi)f(x)sinx dx+int_(0)^(pi) f''(x)sin x dx`
`=(f(x)(-cosx))_(0)^(pi)+int_(0)^(pi)f'(x)cosxdx`
`+sinxf'(x)|_(0)^(pi)-int_(0)^(pi) cos xf'(x)dx`
`=f(pi)+f(0)=5` (given)
`:. f(0)=5-f(pi)=5-2=3`