JEE PYQ: Quadratic Equation Question 31
Question 31 - 2020 (09 Jan Shift 1)
The number of real roots of the equation, $e^{4x} + e^{3x} - 4e^{2x} + e^x + 1 = 0$ is:
(1) 1
(2) 3
(3) 2
(4) 4
Show Answer
Answer: (1)
Solution
Let $e^x = t \in (0, \infty)$
Given equation: $t^4 + t^3 - 4t^2 + t + 1 = 0$
$\Rightarrow t^2 + t - 4 + \frac{1}{t} + \frac{1}{t^2} = 0$
$\Rightarrow \left(t^2 + \frac{1}{t^2}\right) + \left(t + \frac{1}{t}\right) - 4 = 0$
Let $t + \frac{1}{t} = y$
$(y^2 - 2) + y - 4 = 0 \Rightarrow y^2 + y - 6 = 0$
$y^2 + y - 6 = 0 \Rightarrow y = -3, 2$
$\Rightarrow y = 2 \Rightarrow t + \frac{1}{t} = 2$
$\Rightarrow e^x + e^{-x} = 2$
Hence, there only one solution of the given equation.
BETA
