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← JEE Main 2022 · 28 Jun · Shift 1
MathematicsInteger+4 / -1Question 22 of 90

JEE Main 2022 · 28 Jun · Shift 1, Mathematics question 22

The number of real solutions of the equation e4x+4e3x−58e2x+4ex+1=0e^{4x}+4e^{3x}-58e^{2x}+4e^x+1=0 is ______.

Answer and worked solution

Official answer: 2

Divide by e2xe^{2x}. With t=ex+e−x≥2t=e^x+e^{-x}\ge2: t2−2+4t−58=0t^2-2+4t-58=0, so t=6t=6 or −10-10.

t=6t=6 gives two values of xx.

Answer: 2