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

We have the piece‐wise definition

$$f(x)=\begin{cases} ae^{x}+be^{-x}, & -1\le x<1\\[4pt] cx^{2}, & 1\le x\le3\\[4pt] ax^{2}+2cx, & 3<x\le4 \end{cases}$$

Because the question says that $$f(x)$$ is continuous everywhere in its domain, the left-hand and right-hand limits at the points where the rule changes must be equal to the actual function value. Thus we must impose continuity at $$x=1$$ and at $$x=3$$.

At $$x=1$$, the value coming from the first branch is

$$f(1^-)=ae^{1}+be^{-1}=ae+ \frac{b}{e}.$$

The value given by the second branch is

$$f(1^+)=c\,(1)^2=c.$$

Equating these two for continuity, we get

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

At $$x=3$$, the value from the second branch is

$$f(3^-)=c\,(3)^2=9c,$$

while the value from the third branch is

$$f(3^+)=a\,(3)^2+2c\,(3)=9a+6c.$$

Equality of these two expressions gives

$$9c=9a+6c.$$

Simplifying,

$$9c-6c=9a\;\;\Longrightarrow\;\;3c=9a\;\;\Longrightarrow\;\;c=3a\qquad\qquad (2).$$

Now we substitute $$c=3a$$ into equation (1):

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

Isolating $$b$$, we write

$$\frac{b}{e}=3a-ae\;\;\Longrightarrow\;\;b=e\bigl(3a-ae\bigr)=ae(3-e)\qquad\qquad (3).$$

Next we turn to the derivative. For each interval we differentiate the corresponding expression.

• For $$-1<x<1$$ we have $$f(x)=ae^{x}+be^{-x}$$, so

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

• For $$1<x<3$$ we have $$f(x)=cx^{2}$$, so

$$f'(x)=2cx.$$

• For $$3<x\le4$$ we have $$f(x)=ax^{2}+2cx$$, so

$$f'(x)=2ax+2c.$$

Therefore

$$f'(0)=ae^{0}-be^{0}=a-b,$$

because $$e^{0}=1.$$

Also, $$2$$ lies in the interval $$1\le x\le3,$$ so we use the middle derivative:

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

The problem tells us that

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

Substituting the expressions just found,

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

Now we insert the relations (2) and (3): $$c=3a$$ and $$b=ae(3-e).$$

Thus

$$(a-ae(3-e))+4(3a)=e.$$

First combine like terms:

$$a-ae(3-e)+12a=e.$$

Notice that $$a+12a=13a,$$ so

$$13a-ae(3-e)=e.$$

Factor out $$a$$ from the left‐hand side:

$$a\bigl(13-e(3-e)\bigr)=e.$$

Now compute the bracket:

$$e(3-e)=3e-e^{2},$$

so

$$13-e(3-e)=13-(3e-e^{2})=13-3e+e^{2}=e^{2}-3e+13.$$

Hence we have

$$a\,(e^{2}-3e+13)=e.$$

Finally, solving for $$a$$ gives

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

This value matches Option D.

Hence, the correct answer is Option D.