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JEE Main 2003
Exam
JEE Main
Year
2003
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2003 · Mathematics · Continuity and Differentiability

If f(x)= cases x e^ -((1)/(|x|)+(1)/(x)) , & x ≠ 0 then f(x) is 0 &...

If f(x)=f(x)=\begin{cases}x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}, & x \neq 0 \text{ then } f(x) \text{ is } \\ 0 & , x=0\end{cases}$$

Options

  1. Adiscontinuous every where
  2. Bcontinuous as well as differentiable for all xx
  3. Ccontinuous for all xx but not differentiable at x=0x=0
  4. Dneither differentiable nor continuous at x=0x=0

Correct answer

C. continuous for all xx but not differentiable at x=0x=0

Step-by-step solution

f(0)=0;f(x)=xe−(1∣x∣+1x)f(0)=0 ; f(x)=x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)} R.H.L. lim⁡h→0(0+h)e−2/h=lim⁡h→0he2/h=0\lim _{h \rightarrow 0}(0+h) \mathrm{e}^{-2 / h}=\lim _{h \rightarrow 0} \frac{h}{\mathrm{e}^{2 / h}}=0 L.H.L Lim⁡h→0(0−h)e−(1h−1h)=0\operatorname{Lim}_{h \rightarrow 0}(0-h) e^{-\left(\frac{1}{h}-\frac{1}{h}\right)}=0 Therefore, f(x)f(x) is continuous R.H.D. lim⁡h→0(0+h)e−(1h+1h)−he−(1h+1h)h=0\lim _{h \rightarrow 0} \frac{(0+h) e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}-h e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}}{h}=0 L.H.D. Lim⁡h→0(0−h)e−(1h−1h)−he−(1h+1h)−h=1\operatorname{Lim}_{h \rightarrow 0} \frac{(0-h) \mathrm{e}^{-\left(\frac{1}{h}-\frac{1}{h}\right)}-h e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}}{-h}=1 Therefore, L.H.D. ≠\neq R.H.D. f(x)\mathrm{f}(\mathrm{x}) is not differentiable at x=0\mathrm{x}=0.
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