Let f (x) = [ * 20 c * 20 c * 20 c 2^x & + 2^ 3 - x & - 6 * 20 c … - Vidyadip

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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