JEE Main 2022 (Online) 25th June Evening ShiftQuestion 1
$\lim_{x\to\frac{\pi}{2}}\left(\tan^{2}x\left(\left(2\sin^{2}x+3\sin x+4\right)^{\frac{1}{2}}-\left(\sin^{2}x+6\sin x+2\right)^{\frac{1}{2}}\right)\right)$ is equal to
Limits & Continuity
Solution
Concept. Put $s=\sin x\to1$; rationalise and write $\tan^2x$ using $s$. Formulas used. (1) $\sqrt A-\sqrt B=\frac{A-B}{\sqrt A+\sqrt B}$(2) $\tan^2x=\frac{s^2}{1-s^2}=\frac{s^2}{(1-s)(1+s)}$Step 1 - $A-B$. $(2s^2+3s+4)-(s^2+6s+2)=s^2-3s+2=(s-1)(s-2)$. Step 2 - Denominator of (1). At $s=1$: $\sqrt9+\sqrt9=6$. Step 3 - Product. By (2): $$\frac{s^2}{(1-s)(1+s)}\cdot\frac{(s-1)(s-2)}{6}=\frac{-s^2(s-2)}{6(1+s)}$$ Step 4. At $s=1$: $\frac{-1\cdot(-1)}{6\cdot2}=\frac1{12}$. Answer: A