Question

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

Answer 1

Answer: 8

$f^{\prime }(0)=4$
$f(x)+\underset{0}{\overset{x}{\int }}(x-t)f^{\prime }(t)dt=\left(e^{2x}+e^{-2x}\right)\cos 2x+\frac{2}{a}x$
Put $x=0:f(0)=2$
$f^{\prime }(x)+x\left(f^{\prime }(x)\right)+\underset{0}{\overset{x}{\int }}{f}^{\prime }(t)dt-x{f}^{\prime }(x)$
$=\left(e^{2x}+e^{-2x}\right)(-2\sin 2x)$
$+\cos 2x\left(2e^{2x}-2e^{-2x}\right)+\frac{2}{a}$
$\Rightarrow f^{\prime }(x)+f(x)-2=\left(e^{2x}+e^{-2x}\right)(-2\sin 2x)$
$+\cos 2x\left(2e^{2x}-2e^{-2x}\right)+\frac{2}{a}$
Put $x=0$
$4+2-2=0+(2-2)+2/a$
$\Rightarrow a=\frac{1}{2}$
$(2a+1{)}^5{a}^2=2^5\cdot \frac{1}{2^2}=8$