Let `f(x) = |(2cos^2x, sin2x, -sinx), (sin2x, 2 sin^2x, cosx), (sinx, -cosx,0)|`, then the value of `int_0^(pi//2){f(x) + f'(x)} dx` is

Options

A. `pi`
B. `pi//2`
C. `2pi`
D. `3pi//2`

Answer

A

Explanation

Applying `C_(1) to C_(1) -2 sin xC_(3) " and " C_(2) to C_(2) +2 cos x C_(3) ` we get
`f(x)= |{:(2,,0,,-sin x),(0,,2,,cos x),(sin x ,,-cos x ,,0):}|`
`=2 cos^(2) x+ 2sin^(2) x=2`
`:. F(x) =0`
`:. int_(0)^(pi//2) [f(x)+ f'(x) =dx = int_(0)^(pi//2) 2dx =pi`