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JEE Main 20209 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
9 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Quadratic Equation

JEE Main 2020 · 9 January · Morning Shift · Mathematics · Quadratic Equation

The number of real roots of the equation, e ^ 4x + e ^ 3x - 4 e ^...

The number of real roots of the equation, e4x+e3x−4e2x+ex+1=0{e}^{4x} + {e}^{3x} - 4{e}^{2x} + {e}^{x} + 1 = 0 is:

Options

  1. A11
  2. B33
  3. C22
  4. D44

Correct answer

A. 11

Step-by-step solution

Let ex=t∈(0,∞){e}^{x} = t \in \left(0, \infty\right)
Given equation
t4+t3−4t2+t+1=0{t}^{4} + {t}^{3} - 4{t}^{2} + t + 1 = 0
t2+t−4+1t+1t2=0{t}^{2} + t - 4 + \dfrac{1}{t} + \dfrac{1}{{t}^{2}} = 0
\left({t}^{2} + \dfrac{1}{{t}^{2}}\right) + \left(t + \dfrac{1}{t}\right) - 4 = 0<br/>Let<br />Lett + \dfrac{1}{t} = \alpha<br/>(α2−2)+α−4=0<br />\left({\alpha}^{2} - 2\right) + \alpha - 4 = 0
α2+α−6=0{\alpha}^{2} + \alpha - 6 = 0
α2+α−6=0{\alpha}^{2} + \alpha - 6 = 0
α=−3,2⇒α=2⇒ex+e−x=2\alpha = - 3, 2 ⇒ \alpha = 2 ⇒ {e}^{x} + {e}^{- x} = 2
x=0x = 0 is the only solution.
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