Question

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 of a is

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

Answer 1

Answer: (D)

$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} $
For continuity at $x=1$
$\displaystyle\lim_{x\to1^{-}} f\left(x\right) = \displaystyle\lim_{x\to1^{+}} f\left(x\right)$
$\Rightarrow a e+b e^{-1}=c$
$\Rightarrow b = ce - ae^2 \dots (1)$
For continuity at $x=3$
$\displaystyle\lim_{x\to3^{-}} f\left(x\right) = \displaystyle\lim_{x\to3^{+}} f\left(x\right)$
$\Rightarrow 9 c=9 a+6 c$
$\Rightarrow c =3 a \ldots(2)$
$f'(0)+f^{\prime}(2)=e$
$\left(a e^{x}-b e^{x}\right)_{x=0}+(2 c x)_{x=2}=e$
$\Rightarrow a-b+4 c=e \dots$(3)
From $(1),(2) \&(3)$
$a-3 a e+a e^{2}+12 a=e$
$\Rightarrow a\left(e^{2}+13-3 e\right)=e$
$\Rightarrow a=\frac{e}{e^{2}-3 e+13}$