# If a function f(x) defined by

Question:

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}$is : 1.$\frac{e}{e^{2}-3 e-13}$2.$\frac{e}{e^{2}+3 e+13}$3.$\frac{1}{\mathrm{e}^{2}-3 \mathrm{e}+13}$4.$\frac{\mathrm{e}}{\mathrm{e}^{2}-3 \mathrm{e}+13}$Correct Option: , 4 Solution:$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)

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

$\Rightarrow a\left(e^{2}+13-3 e\right)=e$

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