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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY1 Mark
MCQ
If $\text{f(x)}=\text{a}|\sin\text{x}|+\text{be}^{|\text{x}|}+\text{c|x|}^3$and if f(x) is differentiable at x = 0, then:
  • A
    $\text{a}=\text{b}=\text{c}=0$
  • ✓
    $\text{a}=0,\text{b}=0;\text{c}\in\text{R}$
  • C
    $\text{b}=\text{c}=0,\text{a}\in\text{R}$
  • D
    $\text{c}=0,\text{a}=0,\text{b}\in\text{R}$
✓

Answer

Correct option: B.
$\text{a}=0,\text{b}=0;\text{c}\in\text{R}$
We have,

$\text{f(x)}=\text{a}|\sin\text{x}|+\text{be}^{|\text{x}|}+\text{c|x|}^3$

$=\begin{cases}\text{a}\sin\text{x}+\text{bx}^\text{x}+\text{cx}^3 & 0<\text{x}<\frac{\pi}{2}\\-\text{a}\sin\text{x}+\text{be}^{-\text{x}}-\text{cx}^3 & -\frac{\pi}{2}<\text{x}<0\end{cases}$

Here, f(x) is differentiable at x = 0

Therefore, (LHL at x = 0) = (RHL at x = 0)

$\Rightarrow\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$

$\Rightarrow\lim\limits_{\text{x}\rightarrow0^{-}}\frac{-\text{a}\sin\text{x}+\text{be}^{-\text{x}}-\text{cx}^3-\text{b}}{\text{x}}=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{a}\sin\text{x}+\text{be}^{\text{x}}-\text{cx}^3-\text{b}}{\text{x}}$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{-\text{a}\sin(0-\text{h})+\text{be}^{-(0-\text{h)}}-\text{c}(0-\text{h})^3-\text{b}}{0-\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}\sin(0+\text{h})+\text{be}^{(0+\text{h)}}+\text{c}(0+\text{h})^3-\text{b}}{0+\text{h}}$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}\sin\text{h}+\text{be}^{\text{h}}+\text{ch}^3-\text{b}}{-\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}\sin\text{h}+\text{be}^{\text{h}}+\text{ch}^3-\text{b}}{\text{h}} $

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}\cos\text{h}+\text{be}^{\text{h}}+3\text{ch}^2}{-1}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}\cos\text{h}+\text{be}^{\text{h}}+3\text{ch}^2}{1} $ (By L'Hospital rule)

$\Rightarrow-(\text{a}+\text{b})=\text{a}+\text{b}$

$\Rightarrow-2(\text{a}+\text{b})=0$

$\Rightarrow\text{a}+\text{b}=0$

This is true for all value of c

$\therefore\text{c}\in\text{R}$

In the given option (b) satisfies a + b = 0 and $\text{c}\in\text{R.}$

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