The number of real roots of the equation e6x – e4x – 2e3x

Engineering

Mathematics

Monotonicity

Question

The number of real roots of the equation

e^6x – e^4x – 2e^3x – 12e^2x + e^x +1= 0 is

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

e^6x – e^4x – 2e^3x –12e^2x + e^x + 1 = 0

⇒ (e^3x – 1)2 – e^x(e^3x – 1) – 12e^2x = 0

⇒ m² – mn – 12n² = 0 where m = e^3x – 1, n = e^x

⇒ (m–4n) (m+3n) = 0

⇒ m = 4n or m + 3n = 0

⇒ e^3x –1 = 4e^x or (e^3x–1) . e^x = 0

Case-I e^3x–1 = 4e^x

Case –II (e^3x–1) e^x = 0

⇒ e^3x = 1

⇒ x = 0 (one solution)

$∴$ Total 2 roots

Please subscribe our Youtube channel to unlock this solution.