If the function f(x)= l (e^ k x -1 ) k x 4 x^2 , x 0 16, x=0 . is… - Vidyadip

MCQ

If the function $f(x)=\left\{\begin{array}{l}\frac{\left(e^{k x}-1\right) \tan k x}{4 x^2}, x \neq 0 \\ 16, x=0\end{array}\right.$ is continuous at $x=0$, then $k=$

A
$\pm \frac{1}{8}$
B
$\pm 4$
C
$\pm 2$
✓
$\pm 8$

✓

Answer

Correct option: D.

$\pm 8$

Since, $f(x)$ is continuous at $x=0$.
$\therefore \lim _{x \rightarrow 0} f(x)=f(0)$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\left(e^{k x}-1\right) \tan k x}{4 x^2}=16$
$\Rightarrow \frac{1}{4} \lim _{x \rightarrow 0} \frac{\left(e^{k x}-1\right)}{k x} \times \frac{\tan k x}{k x} \times k^2=16$
$\Rightarrow \frac{k^2}{4} \times 1 \times 1=16$
$\Rightarrow k^2=64 $
$\Rightarrow k= \pm 8$

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