JEE Main 2022 · 24 Jun · Shift 2 · Mathematics Q2

MathematicsSingle correct+4 / -1Question 2 of 90

The sum of all the real roots of the equation $\left(e^{2x}-4\right)\left(6e^{2x}-5e^x+1\right)=0$ is

\log_e3

-\log_e3

\log_e6

-\log_e6

Answer and worked solution

Official answer: B

First factor. $e^{2x}=4$ gives $x=\log_e2$ .

Second factor. With $t=e^x$ : $6t^2-5t+1=0$ , so $t=\frac12$ or $\frac13$ , i.e. $x=-\log_e2$ or $-\log_e3$ .

Sum. $\log_e2-\log_e2-\log_e3=-\log_e3$ .

(B)

Topics:Equations reducible to quadratic Logarithms

JEE Main 2022 · 24 Jun · Shift 2, Mathematics question 2