Good Question!

BITSAT 2015 Coaching › Calculus questions › Good Question!

i believe we have not discussed this before -

prove that e^pi > pi ^e.

It is not as easy as it luks !

#Mathematics #Calculus

UP 0 DOWN 0 0 7
Post on FacebookPost anonymously?

7 Answers

30

Ashish Kothari ·2011-07-02 23:23:58

Any hints ? :P

UP 0 DOWN 0
Post on FacebookPost anonymously?

23

qwerty ·2011-07-03 00:24:17

hint : consider the function x^(1/x) ..analyse where it is increasing ...decreasing ..its maxima ..etc etc

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

aditya ravichandran ·2011-07-03 01:49:56

qwerty's hint should finish off the problem

UP 0 DOWN 0
Post on FacebookPost anonymously?

21

Shubhodip ·2011-07-03 01:58:14

consider f(x) = x^1/x. Verify that f_max = f(e) = e^1/e
hence e^1/e>pi^1/pi implying e^pi> pi^e

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

rishabh ·2011-07-03 02:36:13

let y = x^1/x .

dy/dx = x^{(1/x - 2)} + x^1/x*{lnx}(-1/x²)
which is < 0 for x > 1

Hence y is decreasing for x > 1

=> y(pi) < y(e)

(pi)^1/pi < e^1/e

raise both sides to the power e*pi

=> pi^e < e^pi

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

aditya ravichandran ·2011-07-03 03:25:47

Shubodip solution is right

@Rishabh
f'(x)=0
when ln(x)=1
i.e
at x=e

so f(max)=f(e)

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

rishabh ·2011-07-03 06:56:24

thanks for the correction...silly me

UP 0 DOWN 0
Post on FacebookPost anonymously?

Your Answer

Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω