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Continuity — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsContinuity2 Marks
MCQ
The function defined by $f (x)=\left\{\begin{array}{cc}\left(x^2+ e ^{\frac{1}{2-x}}\right)^{-1} & , x \neq 2 \\ k & , x=2\end{array}\right.$, is continuous from right at the point $x=2$, then k is equal to
  • A
    $0$
  • ✓
    $\frac{1}{4}$
  • C
    $-\frac{1}{4}$
  • D
    4
✓

Answer

Correct option: B.
$\frac{1}{4}$
(B)
If $f (x)$ is continuous from right at $x=2$, then
$f (2)=\lim _{x \rightarrow 2^{+}} f (x)$
$\Rightarrow k =\lim _{ h \rightarrow 0} f (2+ h )$
$\Rightarrow k =\lim _{ h \rightarrow 0}\left[(2+ h )^2+ e ^{\frac{1}{2-(2+ h )}}\right]^{-1}$
$\Rightarrow k =\lim _{ h \rightarrow 0}\left[4+ h ^2+4 h+ e ^{\frac{-1}{h}}\right]^{-1}$
$\Rightarrow k =\left(4+0+0+ e ^{-\infty}\right)^{-1}$
$\Rightarrow k =\frac{1}{4}$

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