JEE Main 202329 Jan, Morning shiftMathematicsMediumIntegral Calculus369× since 2002Let f(x)=x+aπ2−4sinx+bπ2−4cosx,x∈Rf(x) = x + {a \over {{\pi ^2} - 4}}\sin x + {b \over {{\pi ^2} - 4}}\cos x,x \in Rf(x)=x+π2−4asinx+π2−4bcosx,x∈R be a function which satisfies f(x)=x+∫0π/2sin(x+y)f(y)dyf(x) = x + \int\limits_0^{\pi /2} {\sin (x + y)f(y)dy}f(x)=x+0∫π/2sin(x+y)f(y)dy. then (a+b)(a+b)(a+b) is equal toAOption A. −2π(π+2)- 2\pi (\pi + 2)−2π(π+2)BOption B. −π(π−2)- \pi (\pi - 2)−π(π−2)COption C. −π(π+2)- \pi (\pi + 2)−π(π+2)DOption D. −2π(π−2)- 2\pi (\pi - 2)−2π(π−2)CheckSkip · show solution
