JEE Main 2021 (Online) 31st August Evening ShiftQuestion 1
If $\alpha=\lim_{x\to\pi/4}\frac{\tan^{3}x-\tan x}{\cos\left(x+\frac{\pi}{4}\right)}$ and $\beta=\lim_{x\to0}(\cos x)^{\cot x}$ are the roots of the equation, $ax^{2}+bx-4=0$, then the ordered pair $(a,b)$ is :
Limits & Continuity
Solution
Let $x=\frac\pi4+h$: $\tan x-1\approx2h$, $\tan x(\tan x+1)\to2$, so numerator $\approx4h$; $\cos(\frac\pi2+h)=-\sin h\approx-h$. So $\alpha=-4$. $\ln\beta=\lim\cot x\ln\cos x=\lim\frac1x\cdot(-\frac{x^2}{2})=0$, so $\beta=1$. Roots $-4,1$ of $ax^2+bx-4=0$: product $-4/a=-4\Rightarrow a=1$; sum $-b/a=-3\Rightarrow b=3$. $(a,b)=(1,3)$. Option (C).