Question
$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\int_{\pi /2}^x {t\,dt} }}{{\sin (2x - \pi )}} =$
$\Rightarrow \,\,y = \mathop {\lim }\limits_{x \to \pi /2} \,\,\frac{{\left[ {\frac{{{t^2}}}{2}} \right]_{\pi /2}^x}}{{\sin \,(2x - \pi )}}$
$y = \mathop {\lim }\limits_{x \to \pi /2} \,\,\frac{{\left( {\frac{{{x^2}}}{2} - \frac{{{\pi ^2}}}{8}} \right)}}{{\sin \,(2x - \pi )}}\,\,$
$\Rightarrow \,\,y = \mathop {\lim }\limits_{x \to \pi /2} \,\,\frac{1}{8}\frac{{(4{x^2} - {\pi ^2})}}{{\sin \,(2x - \pi )}}\,\,$
$y = \mathop {\lim }\limits_{x \to \pi /2} \,\,\frac{1}{8}\frac{{(2x - \pi )\,\,(2x + \pi )}}{{\sin \,(2x - \pi )}}$
$y = \frac{1}{8}\,\,\frac{{\mathop {\lim }\limits_{x \to \pi /2} \,(2x + \pi )}}{{\mathop {\lim }\limits_{x \to \pi /2} \,\,\frac{{\sin \,(2x - \pi )}}{{\,(2x - \pi )}}}}$,
$\left( \because \,\,\,\underset{\theta \to 0}{\mathop{\lim }}\,\,\,\frac{\theta }{\sin \theta }=1 \right)$
$y = \frac{1}{8} \times 2\pi = \frac{\pi }{4}$.
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$S_1=\{z \in C:|z|<4\}, S_2=\left\{z \in C: \operatorname{Im}\left[\frac{z-1+\sqrt{3} i}{1-\sqrt{3} i}\right] > 0\right\}$ तथा
$S_3:\{z \in C: \operatorname{Re} z>0\} .$
$1.$ $S$ का क्षेत्रफल $=$
$(A)$ $\frac{10 \pi}{3}$ $(B)$ $\frac{20 \pi}{3}$ $(C)$ $\frac{16 \pi}{3}$ $(D)$ $\frac{32 \pi}{3}$
$2.$ $\min _{z \in S}|1-3 i-z|=$
$(A)$ $\frac{2-\sqrt{3}}{2}$ $(B)$ $\frac{2+\sqrt{3}}{2}$ $(C)$ $\frac{3-\sqrt{3}}{2}$$(D)$ $\frac{3+\sqrt{3}}{2}$
इस प्रश्न के उतर दीजिये $1$ ओर $2.$