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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
The function defined by $f(x)\, = \,\left\{ {\begin{array}{*{20}{c}}{{{\left( {{x^2} + {e^{\frac{1}{{2 - x}}}}} \right)}^{ - 1}}}&,&{x \ne 2}\\k&,&{x = 2}\end{array}} \right.$, is continuous from right at the point $x = 2$, then $k$ is equal to
  • A
    $0$
  • ✓
    $\frac{1}{4}$
  • C
    -1/4
  • D
    None of these
✓

Answer

Correct option: B.
$\frac{1}{4}$
b
(b) $f(x) = {\left[ {{x^2} + {e^{\frac{1}{{2 - x}}}}} \right]^{ - 1}}$ and $f(2) = k$

If $f(x)$ is continuous from right at $x = 2$ then $\mathop {\lim }\limits_{x \to {2^ + }} f(x) = f(2) = k$

==> $\mathop {\lim }\limits_{x \to {2^ + }} {\left[ {{x^2} + {e^{\frac{1}{{2 - x}}}}} \right]^{ - 1}} = k$

==> $k = \mathop {\lim }\limits_{h \to 0} f(2 + h)$

==> $k = \mathop {\lim }\limits_{h \to 0} \,{\left[ {{{(2 + h)}^2} + {e^{\frac{1}{{2 - (2 + h)}}}}} \right]^{\, - 1}}$

==> $k = \mathop {\lim }\limits_{h \to 0} \,{\left[ {\,4 + {h^2} + 4h + {e^{ - 1/h}}\,} \right]^{\, - 1}}$

==> $k = {[4 + 0 + 0 + {e^{ - \infty }}]^{\, - 1}}$

==> $k = \frac{1}{4}$.

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