If a function f(x) defined by f(x)=begin{case | vedclass

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(D) Given $f(x)$ is continuous at $x=1$ and $x=3$.
For continuity at $x=1$: $\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{+}} f(x) \Right

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

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(D) Given $f(x)$ is continuous at $x=1$ and $x=3$.
For continuity at $x=1$: $\lim_{x ightarrow 1^{-}} f(x) = \lim_{x ightarrow 1^{+}} f(x) \Rightarrow a e + b e^{-1} = c(1)^2 \Rightarrow a e + b/e = c \Rightarrow b = c e - a e^2 \quad (1)$.
For continuity at $x=3$: $\lim_{x ightarrow 3^{-}} f(x) = \lim_{x ightarrow 3^{+}} f(x) \Rightarrow c(3)^2 = a(3)^2 + 2c(3) \Rightarrow 9c = 9a + 6c \Rightarrow 3c = 9a \Rightarrow c = 3a \quad (2)$.
Substitute $(2)$ into $(1)$: $b = (3a)e - a e^2 = a(3e - e^2) \quad (3)$.
Now,$f^{\prime}(x) = \begin{cases} a e^x - b e^{-x}, & -1 < x < 1 \ 2cx, & 1 < x < 3 \ 2ax + 2c, & 3 < x < 4 \end{cases}$.
Given $f^{\prime}(0) + f^{\prime}(2) = e$.
$f^{\prime}(0) = a e^0 - b e^0 = a - b$.
$f^{\prime}(2) = 2c(2) = 4c$.
So,$a - b + 4c = e$.
Substitute $b = 3ae - ae^2$ and $c = 3a$ into the equation:
$a - (3ae - ae^2) + 4(3a) = e$.
$a - 3ae + ae^2 + 12a = e$.
$a(e^2 - 3e + 13) = e$.
Therefore,$a = \frac{e}{e^2 - 3e + 13}$.

If a function $f(x)$ defined by $f(x)=\begin{cases} a e^{x}+b e^{-x}, & -1 \leq x<1 \\ c x^{2}, & 1 \leq x \leq 3 \\ a x^{2}+2 c x, & 3 < x \leq 4 \end{cases}$ is continuous for some $a, b, c \in R$ and $f'(0)+f'(2)=e$,then the value of $a$ is:

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