Q.

If f: be a continuous function satisfying 0π/2f(sin2x)sinxdx+α0π/4f(cos2x)cosxdx=0, then the value of α is          [2023]

1 -2  
2 3  
3 -3  
4 2  

Ans.

(1)

I=0π/4f(sin2x)sinxdx+π/4π/2f(sin2x)sinxdx+α0π/4f(cos2x)cosxdx=0 

Apply 0af(x)dx=0af(a-x)dx in the first part and put  x-π4=t in the second part, we get

I=0π/4f(cos2x)sin(π4-x)dx+0π/4f(cos2t)sin(π4+t)dt+α0π/4f(cos2x)cosxdx=0

=0π/4f(cos2x)[2sinπ4·cosx+αcosx]dx=0 

(α+2)0π/4f(cos2x)cosxdx=0    α=-2

[ In (0,π4), f(cos(2x)) and cosx are not zero.]