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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
If $\text{f(x)}=\begin{cases}\text{x}\sin\frac{\pi}{2}(\text{x}+1),&\text{x}\leq0\\\frac{\tan\text{x}-\sin\text{x}}{\text{x}^3},&\text{x} > 0\end{cases}$ is continuous at $x = 0,$ then a equals :
  • ✓
    $\frac{1}{2}$
  • B
    $\frac{1}{3}$
  • C
    $\frac{1}{4}$
  • D
    $\frac{1}{6}$
✓

Answer

Correct option: A.
$\frac{1}{2}$
Given, $\text{f(x)}=\begin{cases}\text{x}\sin\frac{\pi}{2}(\text{x}+1),&\text{x}\leq0\\\frac{\tan\text{x}-\sin\text{x}}{\text{x}^3},&\text{x} > 0\end{cases}$
We have,
$(\text{LHL}$ at ${x}=0)=\lim\limits_{\text{x}\rightarrow0^{-}}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(0-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim\limits_{\text{h}\rightarrow0}\text{a}\sin\Big(\frac{\pi}{2}(-\text{h+1})\Big)=\text{a}\sin\Big(\frac{\pi}{2}\Big)=\text{a}$
$(\text{RHL}$ at ${x}=0)=\lim\limits_{\text{x}\rightarrow0^+}\text{f(x)}\\=\lim\limits_{\text{h}\rightarrow0}\text{f}(0+\text{h})=\lim\limits_{\text{h}\rightarrow0}\text{f(h)}=\lim\limits\frac{\tan\text{h}-\sin\text{h}}{\text{h}^3}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\sin\text{h}}{\cos\text{h} }-\sin\text{h}}{\text{h}^3}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\sin\text{h}}{\cos\text{h}}(1-\cos\text{h})}{\text{h}^3}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(1-\cos\text{h})\tan\text{h}}{\text{h}^3}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\sin^2\frac{\text{h}}{2}\tan\text{h}}{4\times\frac{\text{h}^2}{4}\times\text{h}}$
$=\frac{2}{4}\lim\limits_{\text{h}\rightarrow0}\frac{\sin^2\frac{\text{h}}{2}\tan\text{h}}{\frac{\text{h}^2}{4}\times\text{h}}$
$=\frac{1}{2}\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\sin\frac{\text{h}}{2}}{\frac{\text{h}}{2}}\bigg)^2\times\lim\limits_{\text{h}\rightarrow0}\frac{\tan\text{h}}{\text{h}}$
$=\frac{1}{2}\times1\times1$
$=\frac{1}{2}$
If $f(x)$ is continuous at $x = 0,$ then
$\lim\limits_{\text{x}\rightarrow0^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0^+}\text{f(x)}$
$\Rightarrow\text{a}=\frac{1}{2}$

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