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STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
MCQ
If  $\begin{gathered} f(x)\, = \,\left\{ \begin{gathered}
  x\left( {\frac{{{e^{1/x}} - {e^{ - 1/x}}}}{{{e^{1/x}} + {e^{ - 1/x}}}}} \right)\,,\,\,x \ne 0 \hfill \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,x\, = \,0\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ 
\end{gathered}  \right. \hfill \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ \hfill \\ 
\end{gathered}$ then correct statement is
  • A
    $f$ is continuous at all points except $x = 0$
  • ✓
    $f$ is continuous at every point but not differentiable
  • C
    $f$ differentiable at every point
  • D
    $f$ is differentiable only at the origin
✓

Answer

Correct option: B.
$f$ is continuous at every point but not differentiable
b
$A+x=0;$

${\rm{LHD}} = \mathop {\lim }\limits_{h \to 0} \frac{{{\rm{f}}(0 - {\rm{h}}) - {\rm{f}}(0)}}{{ - {\rm{h}}}}$

$ = \mathop {\lim }\limits_{h \to 0} \frac{{ - h\left( {\frac{{{e^{ - 1/h}} - {e^{1/h}}}}{{{e^{ - 1/h}} + {e^{1/h}}}}} \right)}}{{ - h}}$

$ = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{1/h}}}}{{{e^{1/h}}}}\left( {\frac{{{e^{ - 2/h}} - 1}}{{{e^{ - 2h}} + 1}}} \right)$

$\frac{{{e^{ - \infty }} - 1}}{{{e^{ - \infty }} + 1}} =  - 1$

${\rm{RHD}} = \mathop {\lim }\limits_{h \to 0} \frac{{{\rm{f}}(0 + {\rm{h}}) - {\rm{f}}(0)}}{{\rm{h}}}$

$ = \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {\frac{{{e^{1/h}} - {e^{ - 1/h}}}}{{{e^{1/h}} + {e^{ - 1/h}}}}} \right) - 0}}{h} = 1$

Not differentiable but continuous

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