Verified Solution • 2025–26 Session moderate

QJEEM20Ma99e6: JEE Main 2020 Mathematics - Class 11 JEEM20Ma99e6 Solution: The number of real roots of the equation, e 4x + e 3x – 4e 2x + …

The number of real roots of the equation,
e4x + e3x – 4e2x + ex + 1 = 0 is :

Multiple Choice Options

  • A 1
  • B 2
  • C 3
  • D 4

Step-by-Step Verified Solution

e4x + e3x – 4e2x + ex + 1 = 0

Dividing by e2x, we get

e2x + ex - 4 + 1ex{1 \over {{e^x}}} + 1e2x{1 \over {{e^{2x}}}} = 0

⇒\Rightarrow (e2x+1e2x)+(ex+1ex)\left( {{e^{2x}} + {1 \over {{e^{2x}}}}} \right) + \left( {{e^x} + {1 \over {{e^x}}}} \right) - 4 = 0

⇒\Rightarrow (ex+1ex)2−2{\left( {{e^x} + {1 \over {{e^x}}}} \right)^2} - 2 + (ex+1ex)\left( {{e^x} + {1 \over {{e^x}}}} \right) - 4 = 0

Let ex+1ex=z{{e^x} + {1 \over {{e^x}}} = z}

(z2 – 2) + (z) – 4 = 0

⇒\Rightarrow z2 + z – 6 = 0

⇒\Rightarrow z = –3, 2

∴\therefore ex+1ex=2{{e^x} + {1 \over {{e^x}}} = 2}

⇒\Rightarrow (ex – 1)2 = 0 ⇒\Rightarrow x = 0.

∴\therefore Number of real roots = 1

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