If α=lim x rightarrow 0 + (frac e √(tan x) -e √(x) √(tan x)-√(x) ) and β=lim x rightarrow 0 (1+sin x) 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 .

Mathematics · Limits, Continuity and Differentiability

JEE Main 2024 — 8 April, Shift 2 — Question 20

If $\alpha=\lim _{x \rightarrow 0^{+}}\left(\frac{e^{\sqrt{\tan x}}-e^{\sqrt{x}}}{\sqrt{\tan x}-\sqrt{x}}\right)$ and $\beta=\lim _{x \rightarrow 0}(1+\sin x)^{\frac{1}{2} \cot x}$ are the roots of the quadratic equation

$\mathrm{ax}^{2}+\mathrm{bx}-\sqrt{\mathrm{e}}=0$ , then 12 $\log _{e}(a+b)$ is equal to $\qquad$ .

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

$\alpha=\lim _{x \rightarrow 0^{+}} e^{\sqrt{x}} \frac{\left(e^{\sqrt{\tan x}-\sqrt{x}}-1\right)}{\sqrt{\tan x}-\sqrt{x}}$ $=1$

$\beta=\lim _{x \rightarrow 0}(1+\sin x)^{\frac{1}{2} \cot x}$ $=\mathrm{e}^{1 / 2}$

$x^{2}-(1+\sqrt{e})+\sqrt{e}=0$

$a x^{2}+b x-\sqrt{e}=0$

On comparing

$\mathrm{a}=-1, \mathrm{~b}=\sqrt{\mathrm{e}}+1$

$12 \ln (\mathrm{a}+\mathrm{b})=12 \times \frac{1}{2}=6$

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Exam: JEE Main 2024
Paper: 8 April, Shift 2
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Indeterminate forms & its solving methods

For\ a,b 0,\ \text{let} f(x)= begincases tan big((a+1)xbig)+btan x/x, & x<0, \\[10pt] 3, & x=0, \\[10pt] dfracsqrtax+b^2x^2-√(ax)b√(a)…

Answer: 6

Step-by-step solution

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