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Continuity — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsContinuity1 Mark
MCQ
If the fucnction $\text{f(x)}=\begin{cases}(\cos\text{x})^{\frac{1}{\text{x}}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ is continuouse at $x = 0,$ then the value of $k$ is:
  • A
    $0$
  • ✓
    $1$
  • C
    $-1$
  • D
    $e$
✓

Answer

Correct option: B.
$1$
$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}(\cos\text{x})^{\frac{1}{\text{x}}}$
$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}(1+\cos \text{x}-1)^{\frac{1}{\text{x}}}$
$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}\Big(1-2\sin^2\frac{\text{x}}{2}\Big)^\frac{1}{\text{x}}$
$\text{f(0)}=\lim\limits\Big(1-2\sin^2\frac{\text{x}}{1}\Big)^{\frac{1}{-2\sin^2\frac{\text{x}}{2}}\times\frac{-2\sin^2\frac{\text{x}}{2}}{\text{x}}}$
$\text{f(0)}= \lim\limits_{\text{x}\rightarrow0}\text{e}^{\frac{-2\sin^2\frac{\text{x}}{2}}{\text{x}}}$
$\text{f(0)}=\lim \limits_{\text{x}\rightarrow0}\text{e}^{\frac{-2\sin\frac{\text{x}}{2}}{\text{x}}\times\sin\frac{\text{x}}{2}}$
$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}\text{e}^{\frac{-2\sin\frac{\text {x}}{2}}{\frac{\text{x}}{2}}\times\frac{1}{2}\sin\frac{\text{x}}{2}}$
$\text{f(0)}=\lim\limits_{\text{x}\rightarrow0}\text{e}^{1\times\sin\frac{1}{2}}=\text{e}^0=1$
$\Rightarrow\text{k}=1$

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