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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 $f (x) =$  $\left[ {\begin{array}{*{20}{c}}  {\begin{array}{*{20}{c}}  {\frac{{\begin{array}{*{20}{c}}  {{2^x}}&{ + {2^{3 - x}}}&{ - 6} \end{array}}}{{\begin{array}{*{20}{c}}  {\sqrt {{2^{ - x}}} }&{ - {2^{1 - x}}} \end{array}}}}&{if}&{x > 2} \end{array}} \\   {} \\   {} \\   {} \\   {\begin{array}{*{20}{c}}   {\frac{{\begin{array}{*{20}{c}}  {{x^2}}&{ - 4} \end{array}}}{{x - \sqrt {3x - 2} }}}&{if}&{x < 2}  \end{array}} \end{array}} \right.$  then
  • A
    $f (2) = 8 \Rightarrow\,\, f$ is continuous at $x = 2$
  • B
    $f (2) = 16 \,\,\Rightarrow\,\, f$ is continuous at $x = 2$
  • ✓
    $f (2^-) \ne f (2^+)\,\, \Rightarrow\,\, f$ is discontinuous
  • D
    $f$ has a removable discontinuity at $x = 2$
✓

Answer

Correct option: C.
$f (2^-) \ne f (2^+)\,\, \Rightarrow\,\, f$ is discontinuous
c
$f (2^+) = 8 ; f (2^-) = 16 $

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