Question

The number of real roots of the equation 

e^4x + 2e^3x – e^x – 6 = 0 is : 

(1) 2

(2) 4

(3) 1

(4) 0

Answer 1

Correct answer is: (3) 1

Let e^x = t > 0 

ƒ(t) = t⁴ + 2t³ – t – 6 = 0 

ƒ'(t) = 4t³ + 6t² – 1

ƒ(0) = –6, ƒ(1) = –4, ƒ(2) = 24

\(\Rightarrow\) Number of real roots = 1