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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
$f(x) = \left\{ {\begin{array}{*{20}{c}}
  {\left[ {\cos \pi x} \right];\,\,\,\,x \leqslant 1} \\ 
  {2\left\{ x \right\} - 1\,\,\,\,x > 1} 
\end{array}} \right.$  comment on the derivability at $x =1,$ where $[ \ ]$ denotes greatest integer function $and\, \{ \}$ denotes fractional part function
  • A
    $f ' (1^-) = 0$
  • B
    $f ' (1^+) = 2$
  • C
    Not diff. at $x = 1$
  • ✓
    All of them
✓

Answer

Correct option: D.
All of them
d
${f^\prime }({1^ - }) = \mathop {\lim }\limits_{h \to 0} \frac{{f(1 - h) - f(1)}}{{ - h}}$

$ = \mathop {\lim }\limits_{h \to 0} \frac{{[\cos (\pi  - \pi h)] + 1}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \frac{{ - 1 + 1}}{{ - h}} = 0$

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

$ = \mathop {\lim }\limits_{h \to 0} \frac{{2h}}{h} = 2$

Hence $f(x)$ is not differentiable at $x=1$

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