If a function $f(x)$ defined by

Since, function $f(x)$ is continuous at $x=1,3$

$\therefore f(1)=f\left(1^{+}\right)$

$\Rightarrow a e+b e^{-1}=c$.....(1)

$f(3)=f\left(3^{+}\right)$

$\Rightarrow 9 c=9 a+6 c \Rightarrow c=3 a$.....(2)

From (1) and (2),

$b=a e(3-e)$......(3)

$f^{\prime}(x)=\left[\begin{array}{cc}a e^{x}-b e^{-x} & -1

$f^{\prime}(0)=a-b, f^{\prime}(2)=4 c$

Given, $f^{\prime}(0)+f^{\prime}(2)=e$

$a-b+4 c=e$...........(4)

From eqs. (i), (ii), (iii) and (iv),

$a-3 a e+a e^{2}+12 a=e$

$\Rightarrow 13 a-3 a e+a e^{2}=e$

$\Rightarrow a=\frac{e}{e^{2}-3 e+13}$