Question

Let $f$ be a twice differentiable function on $R$.
If $f ^{\prime}(0)=4$ and $f(x)+\int\limits_0^x(x-t) f^{\prime}(t) d t=\left(e^{2 x}+e^{-2 x}\right) \cos 2 x+\frac{2}{a} x,$
then $(2 a+1)^5 a^2$ is equal to ______.

Answer 1

The correct answer is 8
f′(0)=4
f(x)+0∫x​(x−t)f′(t)dt=(e2x+e−2x)cos2x+a2​x
Put x=0:f(0)=2
f′(x)+x(f′(x))+0∫x​f′(t)dt−xf′(x)
=(e2x+e−2x)(−2sin2x)
+cos2x(2e2x−2e−2x)+a2​
⇒f′(x)+f(x)−2=(e2x+e−2x)(−2sin2x)
+cos2x(2e2x−2e−2x)+a2​
Put x=0
4+2−2=0+(2−2)+2/a
⇒a=21​
(2a+1)5a2=25⋅221​=8