If a function $f(x)$ defined by


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 :

  1. (1) $\frac{1}{e^{2}-3 e+13}$

  2. (2) $\frac{e}{e^{2}-3 e-13}$

  3. (3) $\frac{e}{e^{2}+3 e+13}$

  4. (4) $\frac{e}{e^{2}-3 e+13}$

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)


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

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

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

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

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