JEE Main 2024 (Online) 8th April Evening ShiftQuestion 1
If $\alpha=\lim_{x\to0^{+}}\left(\frac{e^{\sqrt{\tan x}}-e^{\sqrt{x}}}{\sqrt{\tan x}-\sqrt{x}}\right)$ and $\beta=\lim_{x\to0}(1+\sin x)^{\frac{1}{2}\cot x}$ are the roots of the quadratic equation $ax^{2}+bx-\sqrt{e}=0$, then $12\log_e(a+b)$ is equal to $\_\_\_\_\_\_$ .
Limits & Continuity
Answer$6$
Solution
$\alpha$: with $u=\sqrt{\tan x},v=\sqrt x\to0$, $\frac{e^u-e^v}{u-v}\to e^0=1$. $\beta=e^{\lim\frac12\cot x\sin x}=e^{1/2}=\sqrt e$. Product of roots $=-\frac{\sqrt e}{a}=\sqrt e\Rightarrow a=-1$. Sum $=-\frac ba=b=1+\sqrt e$. So $a+b=\sqrt e$ and $12\ln\sqrt e=6$.