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Consider the following functions: $\quad[2015-I I]$
1- $\quad f(x)=\left\{\begin{array}{lll}\frac{1}{x} & \text { if } & x \neq 0 \\ 0 & \text { if } & x=0\end{array}\right.$
2- $\quad f(x)=\left\{\begin{array}{cll}2 x+5 & \text { if } & x>0 \\ x^{2}+2 x+5 & \text { if } & x \leq 0\end{array}\right.$
Which of the above functions is/are derivable at $\mathrm{x}=0$ ?
Options:
1- $\quad f(x)=\left\{\begin{array}{lll}\frac{1}{x} & \text { if } & x \neq 0 \\ 0 & \text { if } & x=0\end{array}\right.$
2- $\quad f(x)=\left\{\begin{array}{cll}2 x+5 & \text { if } & x>0 \\ x^{2}+2 x+5 & \text { if } & x \leq 0\end{array}\right.$
Which of the above functions is/are derivable at $\mathrm{x}=0$ ?
Solution:
2578 Upvotes
Verified Answer
The correct answer is:
2 only
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}\frac{1}{\mathrm{x}} & \mathrm{x} \neq 0 \\ 0 & \mathrm{x}=0\end{array}\right.$
as there is a discontinuity at $\mathrm{x}=0$, so function is not differentiable at $\mathrm{x}=0$
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}2 \mathrm{x}+5 & \mathrm{x}>0 \\ \mathrm{x}^{2}+2 \mathrm{x}+5 \quad \mathrm{x} \leq 0\end{array}\right.$
$\mathrm{f}(0)=5$
$\mathrm{LHD}=\lim _{\mathrm{x} \rightarrow 0^{-}} \frac{\mathrm{f}(\mathrm{x})-\mathrm{f}(0)}{\mathrm{x}-0}$
$=\lim _{\mathrm{x} \rightarrow 0} \frac{\mathrm{x}^{2}+2 \mathrm{x}+5-5}{\mathrm{x}}=\lim _{\mathrm{x} \rightarrow 0} \mathrm{x}+2=2$
$\mathrm{RHD}=\lim _{\mathrm{x} \rightarrow 0^{+}} \frac{\mathrm{f}(\mathrm{x})-\mathrm{f}(0)}{\mathrm{x}-0}$
$=\lim _{\mathrm{x} \rightarrow 0} \frac{2 \mathrm{x}+5-5}{\mathrm{x}}=2$
$\therefore$ It is differentiable at $\mathrm{x}=0$
$\therefore$ Only $(2)$ is differentiable at $\mathrm{x}=0$
as there is a discontinuity at $\mathrm{x}=0$, so function is not differentiable at $\mathrm{x}=0$
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}2 \mathrm{x}+5 & \mathrm{x}>0 \\ \mathrm{x}^{2}+2 \mathrm{x}+5 \quad \mathrm{x} \leq 0\end{array}\right.$
$\mathrm{f}(0)=5$
$\mathrm{LHD}=\lim _{\mathrm{x} \rightarrow 0^{-}} \frac{\mathrm{f}(\mathrm{x})-\mathrm{f}(0)}{\mathrm{x}-0}$
$=\lim _{\mathrm{x} \rightarrow 0} \frac{\mathrm{x}^{2}+2 \mathrm{x}+5-5}{\mathrm{x}}=\lim _{\mathrm{x} \rightarrow 0} \mathrm{x}+2=2$
$\mathrm{RHD}=\lim _{\mathrm{x} \rightarrow 0^{+}} \frac{\mathrm{f}(\mathrm{x})-\mathrm{f}(0)}{\mathrm{x}-0}$
$=\lim _{\mathrm{x} \rightarrow 0} \frac{2 \mathrm{x}+5-5}{\mathrm{x}}=2$
$\therefore$ It is differentiable at $\mathrm{x}=0$
$\therefore$ Only $(2)$ is differentiable at $\mathrm{x}=0$
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