Q55 / 90 · JEE Main 2024 (08 Apr Shift 2) · Mathematics
If $\alpha=\lim _{x \rightarrow 0^{+}}\left(\frac{\mathrm{e}^{\sqrt{\tan x}}-\mathrm{e}^{\sqrt{x}}}{\sqrt{\tan x}-\sqrt{x}}\right)$ and $\beta=\lim _{x \rightarrow 0}(1+\sin x)^{\frac{1}{2} \cot x}$ are the roots of the quadratic equation $a x^2+b x-\sqrt{\mathrm{e}}=0$, then $12 \log _{\mathrm{e}}(\mathrm{a}+\mathrm{b})$ is equal to__________
Correct answer: 6
Solution
$\begin{aligned} & \alpha=\lim _{x \rightarrow 0^{+}} e^{\sqrt{x}} \frac{\left(e^{\sqrt{\tan x}-\sqrt{x}}-1\right)}{\sqrt{\tan x}-\sqrt{x}} \\ & =1 \\ & \beta=\lim _{x \rightarrow 0}(1+\sin x)^{\frac{1}{2} \cot x} \\ & =e^{1 / 2} \\ & x^2-(1+\sqrt{e})+\sqrt{e}=0 \\ & a x^2+b x-\sqrt{e}=0 \\ & \text { On comparing } \\ & a=-1, b=\sqrt{e}+1 \\ & 12 \ln (a+b)=12 \times \frac{1}{2}=6\end{aligned}$
Hence, the correct answer is 6.