If a function f(x) defined by
$f(x)=\left\{\begin{array}{l}a e^{x}+b e^{-x},-1 \leq x<1 \\ c x^{2}, 1 \leq x \leq 3 \\ a x^{2}+2 c x, 3 $a, b, c \in \mathbf{R}$ and $f^{\prime}(0)+f^{\prime}(2)=e$, then the value of $a$ is :
Correct Option: , 4
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), $a-3 a e+a e^{2}+12 a=e$ $\Rightarrow 13 a-3 a e+a e^{2}=e$ $\Rightarrow a=\frac{e}{e^{2}-3 e+13}$