JEE Main 2020 (Online) 9th January Morning ShiftQuestion 1
The number of real roots of the equation, $e^{4x}+e^{3x}-4e^{2x}+e^x+1=0$ is :
Differentiation and its Application
Solution
Concept. With $t=e^x>0$, 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. $t^4+t^3-4t^2+t+1=0$. By (1): $$u^2-2+u-4=0\ \Longrightarrow\ (u+3)(u-2)=0.$$ Step 2 - Valid $u$. $u\ge2$, so $u=2$, which gives $t=1$ (double root). Step 3 - Count. $t=1$ means $x=0$: exactly one real root. Answer: (A)