1. Given piecewise function

We are told that $f(x)$ is continuous and

$f'(0)+f'(2)=e.$

We need to find $a$.

2. Use continuity at $x=1$

For continuity at $x=1$,

$\lim_{x\to 1^-}f(x)=f(1).$

From the first branch,

$\lim_{x\to 1^-}f(x)=ae+b e^{-1}.$

From the second branch,

$f(1)=c(1)^2=c.$

So,

$ae+\frac{b}{e}=c \qquad ...(1)$

3. Use continuity at $x=3$

For continuity at $x=3$,

$f(3)=\lim_{x\to 3^+}f(x).$

From the second branch,

$f(3)=c(3)^2=9c.$

From the third branch,

$\lim_{x\to 3^+}f(x)=a(3)^2+2c(3)=9a+6c.$

Hence,

$9c=9a+6c$

$3c=9a$

$c=3a. \qquad ...(2)$

4. Find derivatives needed

At $x=0$

Since $0\in[-1,1)$, use the first branch:

$f(x)=ae^x+be^{-x}$

Differentiate:

$f'(x)=ae^x-be^{-x}$

Thus,

$f'(0)=a-b. \qquad ...(3)$

At $x=2$

Since $2\in[1,3]$, use the second branch:

$f(x)=cx^2$

Differentiate:

$f'(x)=2cx$

Thus,

$f'(2)=4c. \qquad ...(4)$

Given

$f'(0)+f'(2)=e,$

so using (3) and (4),

$a-b+4c=e. \qquad ...(5)$

5. Substitute $c=3a$ into continuity equation

From (1):

$ae+\frac{b}{e}=c=3a.$

Multiply by $e$:

$ae^2+b=3ae$

$b=3ae-ae^2=a(3e-e^2). \qquad ...(6)$

6. Use derivative condition

From (5):

$a-b+4c=e$

Using $c=3a$,

$a-b+12a=e$

$13a-b=e. \qquad ...(7)$

Now substitute $b=a(3e-e^2)$ from (6):

$13a-a(3e-e^2)=e$

$a\big(13-3e+e^2\big)=e$

$a(e^2-3e+13)=e$

Therefore,

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

7. Match with options

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

This is Option C.

8. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.