જો f(x) = * 20 c 1 - x - 2x , & x 2 , , , , , , , , , , , , , ,, … - Vidyadip

MCQ

જો $f(x) = \left\{ {\begin{array}{*{20}{c}}{\frac{{1 - \sin x}}{{\pi - 2x}},}&{x \ne \frac{\pi }{2}}\\{\,\,\,\,\,\,\,\,\,\,\,\,\,\lambda \,,}&{x = \frac{\pi }{2}}\end{array}} \right.$ એ $x = \pi /2$ આગળ સતત હોય તો $\lambda $ ની કિમત મેળવો.

A
$-1$
B
$1$
✓
$0$
D
$2$

✓

Answer

Correct option: C.

$0$

(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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