JEE Main 2024 · 6 April · evening · Single correct · +4 / −1
If the function f(x)=(1/x)²ˣ ; x>0 attains the maximum value at x=1/e then :
- (a) e^π<πᵉ
- (b) e^(2 π)<(2 π)ᵉ
- (c) (2 e)^π>π⁽²ᵉ⁾
- (d) e^π>πᵉ
Answer and solution
Answer (d) e^π>πᵉ
Solution
Maths sir · 4 steps
The maximum is at x = 1/e, so compare it with x = 1/π and clear the fractional powers.
- Use the maximum. The largest value of f is at x = 1/e, so every other positive x gives a smaller value. Take x = 1/π, which is not 1/e. Then f(1/e) > f(1/π).
- Evaluate both. f(1/e) = e^(2/e) and f(1/π) = π^(2/π).
e^(2/e) > π^(2/π) ⇒ (e^(2/e))^(eπ/2) > (π^(2/π))^(eπ/2) ⇒ e^π > πᵉ
- Power is safe. Both sides are positive and eπ/2 is positive. Raising both sides to a positive power keeps the inequality the same way round.
- So. e^π > πᵉ, which is option (d).
The idea to keep. When a question tells you where a function is maximum, compare the maximum value with the value at a smartly chosen point. Then remove the fractional powers by raising both sides to a positive power.
Answer from the NTA final answer key. Solution by PrepWiser.