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QJEEM24Ma576c: JEE Main 2024 Mathematics - Class 12 JEEM24Ma576c Solution: If α = lim x → 0 + ( e tan ⁡ x − e x tan ⁡ x − x ) and β = lim…

If α=limx→0+(etan⁡x−extan⁡x−x) and β=limx→0(1+sin⁡x)12cot⁡x are the roots of the quadratic equation ax2+bx−e=0\mathrm{a} x^2+\mathrm{b} x-\sqrt{\mathrm{e}}=0 , then 12log⁡e(a+b)12 \log _{\mathrm{e}}(\mathrm{a}+\mathrm{b}) is equal to _________.

Step-by-Step Verified Solution

αamp;=lim⁡x→0+etan⁡x−ex(tan⁡x−x)amp;=lim⁡x→0ex(etan⁡x−x−1)(tan⁡x−x)=1βamp;=lim⁡x→0(1+sin⁡x)12cot⁡x=lim⁡x→0e(sin⁡x)(12cot⁡x)amp;=lim⁡x→0e12cos⁡x=e1/2\begin{aligned}\alpha & =\lim _{x \rightarrow 0^{+}} \frac{e^{\sqrt{\tan x}}-e^{\sqrt{x}}}{(\sqrt{\tan x}-\sqrt{x})} \\& =\lim _{x \rightarrow 0} \frac{e^{\sqrt{x}}\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}=\lim _{x \rightarrow 0} e^{(\sin x)\left(\frac{1}{2} \cot x\right)} \\& =\lim _{x \rightarrow 0} e^{\frac{1}{2} \cos x}=e^{1 / 2}\end{aligned}

amp; Product of roots =e=−ea⇒a=−1amp; Sum of roots =−ba=1+eamp;=b=e+1amp;⇒12ln⁡(a+b)=12ln⁡(e+1−1)=12ln⁡(e1/2)=6\begin{aligned}& \text { Product of roots }=\sqrt{e}=\frac{-\sqrt{e}}{a} \Rightarrow a=-1 \\& \text { Sum of roots }=\frac{-b}{a}=1+\sqrt{e} \\& \qquad=b=\sqrt{e}+1 \\& \Rightarrow 12 \ln (a+b)=12 \ln (\sqrt{e}+1-1)=12 \ln \left(e^{1 / 2}\right)=6\end{aligned}

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