MCQ
If $f(x) = \left\{ {\begin{array}{*{20}{c}}{\frac{{1 - \sin x}}{{\pi - 2x}},}&{x \ne \frac{\pi }{2}}\\{\,\,\,\,\,\,\,\,\,\,\,\,\,\lambda \,,}&{x = \frac{\pi }{2}}\end{array}} \right.$, be continuous at $x = \pi /2,$ then value of $\lambda $ is
  • A
    $-1$
  • B
    $1$
  • $0$
  • D
    $2$

Answer

Correct option: C.
$0$
c
(c) $f(x)$ is continuous at $x = \frac{\pi }{2}$, then

$\mathop {\lim }\limits_{x \to \pi /2} f(x) = f(0)$ or $\lambda = \mathop {\lim }\limits_{x \to \pi /2} \frac{{1 - \sin x}}{{\pi - 2x}}$, $\left( {\frac{0}{0}{\rm{ form}}} \right)$

Applying $L -$ Hospital’s rule,

$\lambda = \mathop {\lim }\limits_{x \to \pi /2} \frac{{ - \cos x}}{{ - 2}}$ 

==> $\lambda = \mathop {\lim }\limits_{x \to \pi /2} \frac{{\cos x}}{2} = 0.$

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