Let f be a twice differentiable function on R . If f ^ ' (0) = 4...

Given,

$f\left(x\right) + {∫}_{0}^{x}\left(x - t\right){f}^{'}\left(t\right)dt =$\left({e}^{2x} + {e}^{- 2x}\right)\cos 2x + \dfrac{2x}{a} ⋯ \left(i\right)$</p><p>Putting$x = 0$both side we get,$f\left(0\right) = 2 ⋯ \left(ii\right)$</p><p>On differentiating equation \left(i\right)$ w.r.t. $x$ we get:

${f}^{'}\left(x\right) + {∫}_{0}^{x}{f}^{'}\left(t\right)dt + x{f}^{'}\left(x\right) - x{f}^{'}\left(x\right) = 2\left({e}^{2x} - {e}^{- 2x}\right)\cos 2x - 2\left({e}^{2x} + {e}^{- 2x}\right)\sin 2x + \dfrac{2}{a}$

$⇒ {f}^{'}\left(x\right) + f\left(x\right) - f\left(0\right) = 2\left({e}^{2x} - {e}^{- 2x}\right)\cos 2x - 2\left({e}^{2x} + {e}^{- 2x}\right)\sin 2x + \dfrac{2}{a}$

Replace $x$ by $0$ we get:

$⇒ 4 = \dfrac{2}{a} ⇒ a = \dfrac{1}{2}$

Now putting the value of $a$ in${\left(2a + 1\right)}^{5} \cdot {a}^{2}$

We get, ${2}^{5} \cdot \dfrac{1}{{2}^{2}} = {2}^{3} = 8$