Question

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

• A. \(log\;e^3\)
• B. \(–log\;e^3\)
• C. \(log\;e^6\)
• D. \(–log\;e^6\)

Answer 1

The Correct Option is B

Given equation : \((e^{2x} – 4)(6e^{2x} – 5e^x + 1) = 0\)

\(⇒ e^{2x} – 4 = 0 \;or \;6e^{2x} – 5e^x + 1 = 0\)

\(⇒ e^{2x} = 4 \;or\; 6(e^x)^2 – 3e^x – 2e^x + 1 = 0\)

\(⇒ 2x = ln4 \;or (3e^x – 1)(2e^x – 1) = 0\)

\(⇒ x = In2 \;or\; e^x = \frac{1}{3}\; or\; e^x = \frac{1}{2}\)

else, \(x = ln\frac{1}{3}, -ln2\)

Sum of all real roots = \(ln2 – ln3 – ln2\)

= \(–ln3\)

Hence, the correct option is (B): \(–log\;e^3\)