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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
The function $f(x) = \left[ {\begin{array}{*{20}{c}}  {\left| {x\, - \,3} \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}&{,\,\,\,\,x\, \geqslant \,1} \\   {\left( {\tfrac{{{x^2}}}{4}} \right)\, - \,\left( {\tfrac{{3\,x}}{2}} \right)\, + \,\left( {\tfrac{{13}}{4}} \right)}&{,\,\,\,\,x\, < \,1} \end{array}} \right.$ is :
  • A
    continuous at $x = 1$
  • B
    diff. at $x = 1$
  • C
    continuous at $x = 3$
  • ✓
    All of the above
✓

Answer

Correct option: D.
All of the above
d
$f (x) = \left[ \begin{gathered}  x - 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,if\,\,x \geqslant \,3 \hfill \\   3 - x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,if\,\,\,\,1 \leqslant x < 3 \hfill \\  \frac{{{x^2}}}{4}\, - \,\frac{{3x}}{2}\, + \frac{{13}}{4}\,\,if\,x\, < 1 \hfill \\ \end{gathered}  \right.$
$f ‘(1^+) =$ $\mathop {Limit}\limits_{h \to 0} \frac{{f(1 + h) - f(1)}}{h}\,\,$

= $\mathop {Limit}\limits_{h \to 0} \frac{{3 - (1 + h) - 2}}{h}\,\, $

$= \, - 1$
$f ‘(1^-) =$ $\mathop {Limit}\limits_{h \to 0} \frac{{\frac{{{{(1 - h)}^2}}}{4}\,\, - \,\,\frac{3}{2}\,(1 - h)\, + \frac{{13}}{4}\, - 2}}{{ - \,\,h}}$

=$\mathop {Limit}\limits_{h \to 0} \frac{{{{(1 - h)}^2}\, - 6(1 - h)\, + 5}}{{ - 4h}}\,$
=$\mathop {Limit}\limits_{h \to 0} \frac{{{h^2} - 2h\, + 6h}}{{ - 4h}}\,\,$ 

$= - 1$
$\Rightarrow\,\, f$  is continuous at $x =1 $

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