MCQ
$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\int_{\pi /2}^x {t\,dt} }}{{\sin (2x - \pi )}} =$
- A$\infty $
- B$\frac{\pi }{2}$
- ✓$\frac{\pi }{4}$
- D$\frac{\pi }{8}$
$\Rightarrow \,\,y = \mathop {\lim }\limits_{x \to \pi /2} \,\,\frac{{\left[ {\frac{{{t^2}}}{2}} \right]_{\pi /2}^x}}{{\sin \,(2x - \pi )}}$
$y = \mathop {\lim }\limits_{x \to \pi /2} \,\,\frac{{\left( {\frac{{{x^2}}}{2} - \frac{{{\pi ^2}}}{8}} \right)}}{{\sin \,(2x - \pi )}}\,\, $
$\Rightarrow \,\,y = \mathop {\lim }\limits_{x \to \pi /2} \,\,\frac{1}{8}\frac{{(4{x^2} - {\pi ^2})}}{{\sin \,(2x - \pi )}}\,\,$
$y = \mathop {\lim }\limits_{x \to \pi /2} \,\,\frac{1}{8}\frac{{(2x - \pi )\,\,(2x + \pi )}}{{\sin \,(2x - \pi )}}$
$y = \frac{1}{8}\,\,\frac{{\mathop {\lim }\limits_{x \to \pi /2} \,(2x + \pi )}}{{\mathop {\lim }\limits_{x \to \pi /2} \,\,\frac{{\sin \,(2x - \pi )}}{{\,(2x - \pi )}}}}$,
$\left( {\because \,\,\,\mathop {\lim }\limits_{\theta \to 0} \,\,\frac{\theta }{{\sin \theta }} = 1} \right)$
$y = \frac{1}{8} \times 2\pi = \frac{\pi }{4}$.
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