4-2.Quadratic Equations and Inequations
hard

समीकरण $\left(e^{2 x}-4\right)\left(6 e^{2 x}-5 e^x+1\right)=0$के सभी वास्तविक मूलों का योगफल होगा

A

$\log _{ c } 3$

B

$-\log _{e} 3$

C

$\log _{ e } 6$

D

$-\log _{e} 6$

(JEE MAIN-2022)

Solution

$\left( e ^{2 x}-4\right)\left(6 e ^{2 x }-3 e ^{ x }-2 e ^{ x }+1\right)=0$

$\left( e ^{2 x }-4\right)\left(3 e ^{ x }-1\right)\left(2 e ^{ x }-1\right)=0$

$e ^{2 x }=4 \text { or }^{ x }=\frac{1}{3} \text { or }^{ x }=\frac{1}{2}$

$\Rightarrow \text { sum of real roots }=\frac{1}{2} \ln 4+\ln \frac{1}{3}+\ln \frac{1}{2}$

$=-\ln 3$

Standard 11
Mathematics

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