If = lim x 0 + ( e tan x - e x tan x - x ) and = lim x 0 ( 1 + sin x ) 1 2 cot x are the roots of the quadratic equation a x 2 + b x - e = 0 , then 12 log e ( a + b ) is equal to _______ . [2024]

Question

If $α = \lim_{x \to 0^{+}} (\frac{e^{\sqrt{\tan x}} - e^{\sqrt{x}}}{\sqrt{\tan x} - \sqrt{x}})$ and $β = \lim_{x \to 0} {(1 + \sin x)}^{\frac{1}{2} \cot x}$ are the roots of the quadratic equation $a x^{2} + b x - \sqrt{e} = 0,$ then $12 \log_{e} (a + b)$ is equal to _______ . [2024]

Smart Achievers · Accepted Answer

(6)

$α = \lim_{x \to 0^{+}} \frac{e^{\sqrt{\tan x}} - e^{\sqrt{x}}}{(\sqrt{\tan x} - \sqrt{x})}$

$= \lim_{x \to 0^{+}} \frac{e^{\sqrt{x}} (e^{\sqrt{\tan x} - \sqrt{x}} - 1)}{(\sqrt{\tan x} - \sqrt{x})} = 1$ $[∵ \lim_{x \to 0} \frac{e^{x} - 1}{x} = 1]$

$β = \lim_{x \to 0} {(1 + \sin x)}^{\frac{1}{2} \cot x} = \lim_{x \to 0} e^{(\sin x) (\frac{1}{2} \cot x)}$

$= \lim_{x \to 0} e^{\frac{1}{2} \cos x} = e^{1 / 2} = \sqrt{e}$

Since, $α, β$ be the roots of the quadratic equation, $a x^{2} + b x - \sqrt{e} = 0$

$∴$ Products of roots $= \frac{- \sqrt{e}}{a} = \sqrt{e} \Rightarrow a = - 1$

and sum of roots $= \frac{- b}{a} = 1 + \sqrt{e} \Rightarrow b = \sqrt{e} + 1$

Hence, $12 \log_{e} (a + b) = 12 \log_{e} (\sqrt{e} + 1 - 1)$

$= 12 \log_{e} (e^{1 / 2}) = 6$

Solution

Q.
If $α = \lim_{x \to 0^{+}} (\frac{e^{\sqrt{\tan x}} - e^{\sqrt{x}}}{\sqrt{\tan x} - \sqrt{x}})$ and $β = \lim_{x \to 0} {(1 + \sin x)}^{\frac{1}{2} \cot x}$ are the roots of the quadratic equation $a x^{2} + b x - \sqrt{e} = 0,$ then $12 \log_{e} (a + b)$ is equal to _______ . [2024]

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Solution

Q. If α=limx→0+(etanx-extanx-x) and β=limx→0(1+sinx)12cotx are the roots of the quadratic equation ax2+bx-e=0, then 12loge(a+b) is equal to _______ . [2024]

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Q.
If $α = \lim_{x \to 0^{+}} (\frac{e^{\sqrt{\tan x}} - e^{\sqrt{x}}}{\sqrt{\tan x} - \sqrt{x}})$ and $β = \lim_{x \to 0} {(1 + \sin x)}^{\frac{1}{2} \cot x}$ are the roots of the quadratic equation $a x^{2} + b x - \sqrt{e} = 0,$ then $12 \log_{e} (a + b)$ is equal to _______ . [2024]