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