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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
Let $f\,(x)\, = \,\left\{ {\begin{array}{*{20}{c}}
{ - 1\,,\,\,\,\, - 2\, \le x\, < \,0}\\
{{x^2} - 1,\,\,\,0,\, \le \,x\, \le 2}
\end{array}} \right.$ and $g\,(x)\, = \,\left| {f\,(x)\,} \right|\, + \,f\,(\,\left| x \right|\,),$ Then, in the interval $(-2\,,2),\,g$ is
  • A
    differentiable at all points
  • B
    not continuous
  • C
    not differentiable at two points
  • ✓
    not differentiable at one point
✓

Answer

Correct option: D.
not differentiable at one point
d
$\left| {f\left( x \right)} \right| = \left\{ {\begin{array}{*{20}{c}}
1&{ - 2 \le x < 0}\\
{\left| {{x^2} - 1} \right|}&{0 \le x \le 2}
\end{array}} \right.$

$f\left( {\left| x \right|} \right) = \left\{ {\begin{array}{*{20}{c}}
{ - 1}&{ - 2 \le \left| x \right| < 0}\\
{{x^2} - 1}&{0 \le \left| x \right| \le 2}
\end{array}} \right.$

$ \Rightarrow f\left( {\left| x \right|} \right) = \left\{ {\begin{array}{*{20}{c}}
{{x^2} - 1}&{ - 2 \le x \le 2}
\end{array}} \right\}$

$g\left( x \right) = \left| {f\left( x \right)} \right| + f\left( {\left| x \right|} \right):\left\{ {\begin{array}{*{20}{c}}
{1 + {x^2} - 1}&{ - 2 \le x < 0}\\
{\left| {{x^2} - 1} \right| + {x^2} - 1}&{0 \le x \le 2}
\end{array}} \right.$

$g\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{{x^2}}&{ - 2 < x < 0}\\
0&{0 \le x < 1}\\
{2\left( {{x^2} - 1} \right)}&{1 \le x < 2}
\end{array}} \right.$

$\therefore g\left( x \right)$ is not differentiable at $x=1$

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