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JEE Main 20248 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2024 · 8 April · Evening Shift · Mathematics · Limits

If α= _ x arrow 0^ + ( e^ x -e^ x x - x ) and β= _ x arrow 0 (1+...

If α=lim⁡x→0+(etan⁡x−extan⁡x−x)\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}aretherootsofthequadraticequationare the roots of the quadratic equationa x^2+b x-\sqrt{\mathrm{e}}=0,then, then12 \log _{\mathrm{e}}(\mathrm{a}+\mathrm{b}) is equal to__________

Correct answer

0

Step-by-step solution

α=lim⁡x→0+ex(etan⁡x−x−1)tan⁡x−x=1β=lim⁡x→0(1+sin⁡x)12cot⁡x=e1/2x2−(1+e)+e=0ax2+bx−e=0 On comparing a=−1,b=e+112ln⁡(a+b)=12×12=6\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}
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