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Continuity — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsContinuity2 Marks
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} \lim _{x \rightarrow 0} \frac{e^{k x}-1}{k x} \times \lim _{x \rightarrow 0} \frac{\tan k x}{k x}=16 $
$\Rightarrow \frac{k^2}{4} \times 1 \times 1=16$
$ \Rightarrow k^2=64 $
$\Rightarrow k= \pm 8$

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