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Differentiability — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiability1 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, $(\text{LHL}$ at $x = 0) = (\text{RHL}$ 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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