If f(x) & = e^ 5 x -e^ 2 x 3 x , x 0 & =k 2 , x=0 is continuous a… - Vidyadip

MCQ

$\begin{aligned}\text { If } f(x) & =\frac{e^{5 x}-e^{2 x}}{\sin 3 x}, x \neq 0 \\& =\frac{k}{2}, \quad x=0 \end{aligned}$
is continuous at $x=0$, the value of k is

A
$0$
B
1
✓
2
D
5

✓

Answer

Correct option: C.

2

(C)
Since $f (x)$ is continuous at $x=0$.
$\therefore \quad f(0)=\lim _{x \rightarrow 0} f(x)$
$\Rightarrow \frac{ k }{2}=\lim _{x \rightarrow 0} \frac{ e ^{5 x}- e ^{2 x}}{\sin 3 x}$
Applying L'Hospital rule on R.H.S., we get
$\frac{ k }{2}=\lim _{x \rightarrow 0} \frac{5 e ^{5 x}-2 e ^{2 x}}{3 \cos 3 x}$
$\Rightarrow \frac{ k }{2}=\frac{5 e ^0-2 e ^0}{3 \cos 0}=\frac{5-2}{3}=1$
$\Rightarrow k =2$

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