Animated Solution for Mathematics - Limits, Continuity and Differentiability: If a function $f(x)$ defined by $f(x)=\{\begin{array}{ll}a{e}^x+b{e}^{-x},& -1<x<1\\ c{x}^2& ,1\le x\le 3\\ a{x}^2+2cx,& 3<x<4\end{array}$ be continuous for some $a,b,c\in R$ and $f^{\prime }(0)+f^{\prime }(2)=e$, then the value of $a$ is :

Welcome, future engineers. Today, we are going to dissect a problem that, at first glance, looks like a daunting maze of variables and piecewise definitions. In the world of JEE Advanced, a problem like this is not a test of your ability to memorize formulas; it is a test of your ability to organize chaos.

We are looking at a function $f(x)$ defined in three distinct regions. It is a bridge built in three parts, and our job is to ensure that the bridge is perfectly connected.

Imagine you are standing on the number line. From $x=-1$ to $x=1$, the function behaves like an exponential curve, $f(x)=a{e}^x+b{e}^{-x}$. As you cross the threshold of $x=1$, the function transforms into a parabola, $f(x)=c{x}^2$. Finally, after passing $x=3$, it shifts into another quadratic form, $f(x)=a{x}^2+2cx$.

For the function to be continuous, there can be no jumps. The 'handshake' at the transition points must be perfect.

At $x=1$, the left-hand limit must equal the right-hand limit:

Substituting $x=1$, we get the elegant relationship:

Now, we move to the second junction, $x=3$. The left-hand limit from the parabola $c{x}^2$ must meet the right-hand limit from the quadratic $a{x}^2+2cx$:

Substituting $x=3$, we get $9c=9a+6c$. A quick subtraction of $6c$ from both sides gives us $3c=9a$, or simply:

We have successfully reduced the complexity. We now know that $c$ is just three times $a$. Let's use this to simplify our first equation by substituting $c=3a$ into equation (1):

Our goal is to isolate $b$ so we can express everything in terms of $a$. Rearranging the terms, we find:

Multiplying by $e$, we arrive at:

We have expressed both $b$ and $c$ as functions of $a$. The entire function is now effectively controlled by a single variable.

The problem introduces a condition involving the derivative: $f^{\prime }(0)+f^{\prime }(2)=e$. To solve this, we must differentiate our piecewise function.

For $x\in (-1,1)$, $f^{\prime }(x)=a{e}^x-b{e}^{-x}$. For $x\in (1,3)$, $f^{\prime }(x)=2cx$. For $x\in (3,4)$, $f^{\prime }(x)=2ax+2c$.

We need $f^{\prime }(0)$ and $f^{\prime }(2)$. Since $0$ falls in the first interval, $f^{\prime }(0)=a{e}^0-b{e}^{-0}=a-b$. Since $2$ falls in the second interval, $f^{\prime }(2)=2c(2)=4c$.

Substituting these into our condition $f^{\prime }(0)+f^{\prime }(2)=e$, we get:

This is the moment of truth. We substitute our expressions for $b$ and $c$ (in terms of $a$) into equation (4):

Expanding this, we get:

Combining the like terms ($a+12a=13a$), we obtain:

Factoring out $a$:

And finally, we isolate $a$:

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