Question

If a function $f\left(x\right)$ defined by $\left\{\begin{array}{l}a{e}^x+b{e}^{-x},-1\le x<1\\ c{x}^2,1\le x\le 3\\ a{x}^2+2cx,3<x\le 4\end{array}\right.$ be continuous for some $a,b,c\in Randf\text{'}\left(0\right)+f\text{'}\left(2\right)=e$, then the value of $a$ is:

• A. $\frac{1}{(e^2-3e+13)}$
• B. $\frac{e}{(e^2-3e-13)}$
• C. $\frac{e}{(e^2+3e+13)}$
• D. $\frac{e}{(e^2-3e+13)}$

Answer 1

The correct option is D

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

Explanation for correct option.

Step1. Finding value of $\text{'}c\text{'}$

Given,$f\left(x\right)$=$\left\{\begin{array}{l}a{e}^x+b{e}^{-x},-1\le x<1\\ c{x}^2,1\le x\le 3\\ a{x}^2+2cx,3<x\le 4\end{array}\right.$

$f\left(x\right)$ is continuous.

At $x=1,b=ce–a{e}^2$

At $x=3$

$9c=9a+6c$

$\Rightarrow$$c=3a$

Step2.Finding value of $‘a’$

Now,

$f\text{'}\left(0\right)+f\text{'}\left(2\right)=e$

$\Rightarrow$ $a-b+4c=e$

$\Rightarrow$$a-e(3a-ae)+4\times 3a=e$

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

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

$\mathbf{\therefore }\mathit{a}\mathbf{=}\frac{\mathbf{e}}{\mathbf{(}\mathbf{13}\mathbf{-}\mathbf{3}\mathbf{e}\mathbf{+}{\mathbf{e}}^{\mathbf{2}}\mathbf{)}}$

Hence, the correct option is $\mathbf{\left(}\mathit{D}\mathbf{\right)}$.