_ x arrow (π)/(2) ([1- ((x)/(2))][1- x])/([1+ ((x)/(2))][π-2 x^3]) is

Correct answer: D. (1)/(32) _ x arrow (π)/(2) ( ((π)/(4)-(x)/(2)) ·(1- x))/((π-2 x)^3) Let x=(π)/(2)+y ; y arrow 0 = _ y arrow 0 (- (-(y)/(2)) ·(1- y))/((-2 y)^3)= _ y arrow 0 (- (y)/(2) 2 ^2 (y)/(2))/((-8) · (y^3)/(8) · 8) = _ y arrow 0 (1)/(32) ( (y)/(2))/(((y)/(2))) · [( y / 2)/(y / 2)]^2=(1)/(32)

JEE Main 2003

Exam: JEE Main
Year: 2003
Subject: Mathematics
Chapter: Limits

JEE Main 2003 · Mathematics · Limits

_ x arrow (π)/(2) ([1- ((x)/(2))][1- x])/([1+ ((x)/(2))][π-2 x^3])...

\lim _{x \rightarrow \frac{\pi}{2}} \frac{\left[1-\tan \left(\frac{x}{2}\right)\right][1-\sin x]}{\left[1+\tan \left(\frac{x}{2}\right)\right]\left[\pi-2 x^3\right]}

Options

A $\infty$
B $\frac{1}{8}$
C0
D $\frac{1}{32}$

Correct answer

D. $\frac{1}{32}$

Step-by-step solution

\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{4}-\frac{x}{2}\right) \cdot(1-\sin x)}{(\pi-2 x)^3}

Let

\mathrm{x}=\frac{\pi}{2}+\mathrm{y} ; \mathrm{y} \rightarrow 0

=\lim _{y \rightarrow 0} \frac{-\tan \left(-\frac{y}{2}\right) \cdot(1-\cos y)}{(-2 y)^3}=\lim _{y \rightarrow 0} \frac{-\tan \frac{y}{2} 2 \sin ^2 \frac{y}{2}}{(-8) \cdot \frac{y^3}{8} \cdot 8}

=\lim _{y \rightarrow 0} \frac{1}{32} \frac{\tan \frac{y}{2}}{\left(\frac{y}{2}\right)}

\cdot

\left[\frac{\sin y / 2}{y / 2}\right]^2=\frac{1}{32}

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