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JEE Main & Advanced Mathematics Functions Question Bank Critical Thinking

  • question_answer
    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]

    A) \[\infty \]

    B) \[\frac{\pi }{2}\]

    C) \[\frac{\pi }{4}\]

    D) \[\frac{\pi }{8}\]

    Correct Answer: C

    Solution :

    • \[y=\underset{x\to \pi /2}{\mathop{\lim }}\,\,\,\frac{\int_{\pi /2}^{x}{t\,.\,dt}}{\sin \,(2x-\pi )}\,\,\Rightarrow \,\,y=\underset{x\to \pi /2}{\mathop{\lim }}\,\,\,\frac{\left[ \frac{{{t}^{2}}}{2} \right]_{\pi /2}^{x}}{\sin \,(2x-\pi )}\]           
    • \[y=\underset{x\to \pi /2}{\mathop{\lim }}\,\,\,\frac{\left( \frac{{{x}^{2}}}{2}-\frac{{{\pi }^{2}}}{8} \right)}{\sin \,(2x-\pi )}\,\,\Rightarrow \,\,y=\underset{x\to \pi /2}{\mathop{\lim }}\,\,\,\frac{1}{8}\frac{(4{{x}^{2}}-{{\pi }^{2}})}{\sin \,(2x-\pi )}\,\,\]           
    • \[y=\underset{x\to \pi /2}{\mathop{\lim }}\,\,\,\frac{1}{8}\frac{(2x-\pi )\,\,(2x+\pi )}{\sin \,(2x-\pi )}\]           
    • \[y=\frac{1}{8}\,\,\frac{\underset{x\to \pi /2}{\mathop{\lim }}\,\,(2x+\pi )}{\underset{x\to \pi /2}{\mathop{\lim }}\,\,\,\frac{\sin \,(2x-\pi )}{\,(2x-\pi )}}\],      
    • \[\left( \because \,\,\,\underset{\theta \to 0}{\mathop{\lim }}\,\,\,\frac{\theta }{\sin \theta }=1 \right)\]           
    • \[y=\frac{1}{8}\times 2\pi =\frac{\pi }{4}\].


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