MHTCET 15 May 2024 Shift 1 Paper

Concept:Simplify the trigonometric ratio using the identity $\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}} = \tan\left(\frac{\pi}{4}-\frac{x}{2}\right)$ and then substitute $\theta = \pi - 2x$ to use standard limit formulas.Explanation:Let $l = \lim_{x \to \frac{\pi}{2}} \frac{\left(1-\tan \frac{x}{2}\right)(1-\sin x)}{\left(1+\tan \frac{x}{2}\right)(\pi-2x)^3}$.Write the limit as:$$l = \lim_{x \to \frac{\pi}{2}} \left[\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right] \cdot \frac{1-\sin x}{(\pi-2x)^3}.$$Using the identity $\frac{1-\tan A}{1+\tan A} = \tan\left(\frac{\pi}{4}-A\right)$, with $A=\frac{x}{2}$, we get:$$\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}} = \tan\left(\frac{\pi}{4}-\frac{x}{2}\right).$$So,$$l = \lim_{x \to \frac{\pi}{2}} \frac{\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)(1-\sin x)}{(\pi-2x)^3}.$$Now substitute $\theta = \pi - 2x$. Then $x = \frac{\pi}{2} - \frac{\theta}{2}$, and as $x \to \frac{\pi}{2}$, we have $\theta \to 0$.Also, $\frac{\pi}{4}-\frac{x}{2} = \frac{\pi}{4} - \frac{1}{2}\left(\frac{\pi}{2}-\frac{\theta}{2}\right) = \frac{\theta}{4}$.And $1-\sin x = 1-\sin\left(\frac{\pi}{2}-\frac{\theta}{2}\right) = 1-\cos\frac{\theta}{2}$.Thus,$$l = \lim_{\theta \to 0} \frac{\tan\frac{\theta}{4}\left(1-\cos\frac{\theta}{2}\right)}{\theta^3}.$$Use $1-\cos\frac{\theta}{2} = 2\sin^2\frac{\theta}{4}$:$$l = \lim_{\theta \to 0} \frac{\tan\frac{\theta}{4} \cdot 2\sin^2\frac{\theta}{4}}{\theta^3}.$$Rewrite to apply standard limits $\lim_{u\to0}\frac{\tan u}{u}=1$ and $\lim_{u\to0}\frac{\sin u}{u}=1$:$$l = \lim_{\theta \to 0} \frac{\tan\frac{\theta}{4}}{\frac{\theta}{4}} \cdot \frac{2\sin^2\frac{\theta}{4}}{\frac{\theta^3}{64}} \cdot \frac{1}{?}$$Better: multiply and divide conveniently:$$l = \lim_{\theta \to 0} \frac{\tan\frac{\theta}{4}}{\frac{\theta}{4}} \cdot \frac{1}{4} \cdot \frac{2\sin^2\frac{\theta}{4}}{\theta^2} \cdot \frac{1}{\theta}$$Actually, let's do carefully:$$l = \lim_{\theta \to 0} \frac{\tan\frac{\theta}{4}}{\frac{\theta}{4}} \cdot \frac{2\sin^2\frac{\theta}{4}}{\left(\frac{\theta}{4}\right)^2} \cdot \frac{1}{32}.$$Because $\theta^3 = \left(\frac{\theta}{4}\right)^3 \cdot 64$, but we have $\tan$ and $\sin^2$: Let's verify: Denominator $\theta^3 = \left(\frac{\theta}{4}\right)^3 \cdot 64$. Numerator has $\tan\frac{\theta}{4} \cdot 2\sin^2\frac{\theta}{4}$. So denominator can be written as $64 (\frac{\theta}{4})^3$. Thus factor: $\frac{1}{64} \cdot \frac{\tan\frac{\theta}{4}}{\frac{\theta}{4}} \cdot \frac{2\sin^2\frac{\theta}{4}}{(\frac{\theta}{4})^2}$. But $2\sin^2 u / u^2 = 2$ times $(\sin u/u)^2$. So factor becomes $\frac{2}{64} = \frac{1}{32}$. Hence:$$l = \frac{1}{32} \lim_{\theta \to 0} \left[\frac{\tan\frac{\theta}{4}}{\frac{\theta}{4}} \cdot \left(\frac{\sin\frac{\theta}{4}}{\frac{\theta}{4}}\right)^2\right].$$Using the standard limits, both factors tend to 1:$$l = \frac{1}{32} \cdot 1 \cdot 1^2 = \frac{1}{32}.$$Answer:$\frac{1}{32}$, which corresponds to option B.