JEE Main 2023 · 31 January · evening · Single correct · +4 / −1
The equation e⁴ˣ+8 e³ˣ+13 e²ˣ-8 eˣ+1=0, x ∈ ℝ has :
- (a) two solutions and both are negative
- (b) two solutions and only one of them is negative
- (c) four solutions two of which are negative
- (d) no solution
Answer and solution
Answer (a) two solutions and both are negative
Solution
Maths sir · 5 steps
Put t = eˣ, divide by t², and the quartic folds into a quadratic in t - 1/t.
- Substitute. Let t = eˣ, so t > 0. The equation becomes t⁴ + 8t³ + 13t² - 8t + 1 = 0. The coefficients 1, 8, 13, -8, 1 pair up from the ends, so divide by t².
(t² + 1/t²) + 8(t - 1/t) + 13 = 0
- Reduce to quadratic. Put u = t - 1/t. Then t² + 1/t² = u² + 2, so u² + 8u + 15 = 0, which gives (u+3)(u+5) = 0. So u = -3 or u = -5.
- Back to t. From t² + 3t - 1 = 0, the only positive root is t = (-3+√13)/2 ≈ 0.30. From t² + 5t - 1 = 0, the only positive root is t = (-5+√29)/2 ≈ 0.19. The other two roots are negative, and eˣ can never be negative, so they are rejected.
- Sign of x. Both valid values satisfy t - 1/t < 0, so t < 1. Then x = ln t < 0 for each of them.
- So. There are exactly two real solutions, and both are negative. That is option (a).
The idea to keep. When the coefficients read the same from both ends (up to sign), divide by the middle power and substitute t ± 1/t. With t = eˣ, keep only the roots with t > 0.
Answer from the NTA final answer key. Solution by PrepWiser.