JEE Main 2021 (Online) 27th July Evening ShiftQuestion 1
The number of real roots of the equation $e^{4x}-e^{3x}-4e^{2x}-e^x+1=0$ is equal to ________.
Differentiation and its Application
Answer$2$
Solution
Concept. With $t=e^x>0$ the equation is a reciprocal quartic; divide by $t^2$ and put $u=t+\frac1t\ge2$. Formulas used. (1) $t^2+\frac{1}{t^2}=u^2-2$Step 1 - Substitute. $t^4-t^3-4t^2-t+1=0$. Dividing by $t^2$ and using (1): $$u^2-2-u-4=0\ \Longrightarrow\ (u-3)(u+2)=0.$$ Step 2 - Valid $u$. $u\ge2$, so $u=3$. Step 3 - Values of $t$. $t+\frac1t=3$ gives $t=\frac{3\pm\sqrt5}{2}$, both positive. Each gives one $x=\ln t$. Answer: 2