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# Let f (x) = |x| and g (x) = |x3|, then (a) f (x) and g (x) both are continuous at x = 0 (b) f (x) and g (x) both are differentiable at x = 0 (c) f (x) is differentiable but g (x) is not differentiable at x = 0 (d) f (x) and g (x) both are not differentiable at x = 0

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## Option (a) f (x) and g (x) both are continuous at x = 0 Given: $f\left(x\right)=\left|x\right|,g\left(x\right)=\left|{x}^{3}\right|$ We know $\left|x\right|$ is continuous at x=0 but not differentiable at x = 0 as (LHD at x = 0) ≠ (RHD at x = 0). Now, for the function $g\left(x\right)=\left|{x}^{3}\right|=\left\{\begin{array}{l}{x}^{3},x\ge 0\\ -{x}^{3},x<0\end{array}\right\$ Continuity at x = 0: (LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}g\left(x\right)=\underset{h\to 0}{\mathrm{lim}}g\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}-\left(-{h}^{3}\right)=\underset{h\to 0}{\mathrm{lim}}{h}^{3}=0.$ (RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}{h}^{3}=0.$ and $g\left(0\right)=0.$ Thus, $\underset{x\to {0}^{-}}{\mathrm{lim}}g\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}g\left(x\right)=g\left(0\right)$. Hence, $g\left(x\right)$ is continuous at x = 0. Differentiability at x = 0: (LHD at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(0-h\right)-f\left(0\right)}{0-h-0}=\underset{h\to 0}{\mathrm{lim}}\frac{{h}^{3}-0}{-h}=0.$ (RHD at x = 0) = $\underset{x\to {c}^{+}}{\mathrm{lim}}\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(0+h\right)-f\left(0\right)}{0+h-0}=\underset{h\to 0}{\mathrm{lim}}\frac{{h}^{3}-0}{h}=\underset{h\to 0}{\mathrm{lim}}\frac{{h}^{3}}{h}=0$ Thus, (LHD at x = 0) = (RHD at x = 0). Hence, the function $g\left(x\right)$ is differentiable at x = 0.  Suggest Corrections  0      Similar questions  Related Videos   Differentiating One Function wrt Other
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