Question

If a function $f(x)$ defined by

$f(x)=\{\begin{array}{ll}a{e}^x+b{e}^{-x},& -1\le x<1\\ c{x}^2,& 1\le x\le 3\\ a{x}^2+2cx,& 3<x\le 4\end{array}$

be continuous for some $a,b,c\in R$ and $f^{\prime }(0)+f^{\prime }(2)=e,$ then the value of of a is

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

Answer 1

Answer: (D)

$f(x)=<br/><br/>\{\begin{array}{ll}<br/><br/>a{e}^x+b{e}^{-x},& -1\le x<1\\ <br/><br/>c{x}^2,& 1\le x\le 3\\ <br/><br/>a{x}^2+2cx,& 3<x\le 4<br/><br/>\end{array}$
For continuity at $x=1$
$\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right)$
$\Rightarrow ae+b{e}^{-1}=c$
$\Rightarrow b=ce-a{e}^2\dots (1)$
For continuity at $x=3$
$\underset{x\to 3^{-}}{\lim }f\left(x\right)=\underset{x\to 3^{+}}{\lim }f\left(x\right)$
$\Rightarrow 9c=9a+6c$
$\Rightarrow c=3a\dots (2)$
$f^{\prime }(0)+f^{\prime }(2)=e$
${\left(a{e}^x-b{e}^x\right)}_{x=0}+(2cx{)}_{x=2}=e$
$\Rightarrow a-b+4c=e\dots$(3)
From $(1),(2)\&(3)$
$a-3ae+a{e}^2+12a=e$
$\Rightarrow a\left(e^2+13-3e\right)=e$
$\Rightarrow a=\frac{e}{e^2-3e+13}$