If a function $f(x)$ defined by
$f(x)= \begin{cases}a e^{x}+b e^{-x}, & -1 \leq x<1 \\ c x^{2} & , \quad 1 \leq x \leq 3 \\ a x^{2}+2 c x, & 3 be continuous for some $a, b, c \in R$ and $\mathrm{f}^{\prime}(0)+\mathrm{f}^{\prime}(2)=\mathrm{e}$, then the value of of $\mathrm{a}$ is :
Correct Option: , 4
$f(x)= \begin{cases}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 For continuity at $x=1$ $\operatorname{Lim}_{x \rightarrow 1^{-}} f(x)=\operatorname{Lim}_{x \rightarrow 1^{+}} f(x)$ $\Rightarrow \mathrm{ae}+\mathrm{be}^{-1}=\mathrm{c} \Rightarrow \mathrm{b}=\mathrm{ce}-\mathrm{ae}^{2}$ ......(1) For continuity at $x=3$ $\operatorname{Lim}_{x \rightarrow 3^{-}} f(x)=\operatorname{Lim}_{x \rightarrow 3^{+}} f(x)$ $\Rightarrow 9 \mathrm{c}=9 \mathrm{a}+6 \mathrm{c}$ $\Rightarrow c=3 a$.............(2) $f^{\prime}(0)+f^{\prime}(2)=e$ $\left(a e^{x}-b e^{x}\right)_{x}=0+(2 c x)_{x=2}=e$ $\Rightarrow a-b+4 c=e$.......(3) From (1), (2) & (3) $a-3 a e+a e^{2}+12 a=e$ $\Rightarrow a\left(e^{2}+13-3 e\right)=e$ $\Rightarrow a=\frac{e}{e^{2}-3 e+13}$