Functions Question 16
Question: The value of $ \underset{x\to \frac{\pi }{2}}{\mathop{\lim }},\frac{\int_{\pi /2}^{x}{t,dt}}{\sin (2x-\pi )} $ is
[MP PET 1998]
Options:
A) $ \infty $
B) $ \frac{\pi }{2} $
C) $ \frac{\pi }{4} $
D) $ \frac{\pi }{8} $
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Answer:
Correct Answer: C
Solution:
$ y=\underset{x\to \pi /2}{\mathop{\lim }},\frac{\int_{\pi /2}^{x}{t,.,dt}}{\sin ,(2x-\pi )}\Rightarrow y=\underset{x\to \pi /2}{\mathop{\lim }},\frac{[ \frac{t^{2}}{2} ]_{\pi /2}^{x}}{\sin ,(2x-\pi )} $
$ y=\underset{x\to \pi /2}{\mathop{\lim }},\frac{( \frac{x^{2}}{2}-\frac{{{\pi }^{2}}}{8} )}{\sin ,(2x-\pi )}\Rightarrow y=\underset{x\to \pi /2}{\mathop{\lim }},\frac{1}{8}\frac{(4x^{2}-{{\pi }^{2}})}{\sin ,(2x-\pi )} $
$ y=\underset{x\to \pi /2}{\mathop{\lim }},\frac{1}{8}\frac{(2x-\pi )(2x+\pi )}{\sin ,(2x-\pi )} $
$ y=\frac{1}{8}\frac{\underset{x\to \pi /2}{\mathop{\lim }}(2x+\pi )}{\underset{x\to \pi /2}{\mathop{\lim }},\frac{\sin ,(2x-\pi )}{,(2x-\pi )}} $ ,
$ ( \because ,\underset{\theta \to 0}{\mathop{\lim }},\frac{\theta }{\sin \theta }=1 ) $
$ y=\frac{1}{8}\times 2\pi =\frac{\pi }{4} $ .
BETA
