Good Question!

BITSAT 2015 Sample Questions › 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

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7 Answers

30

Ashish Kothari ·2011-07-02 23:23:58

Any hints ? :P

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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

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

qwerty's hint should finish off the problem

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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

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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

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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)

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rishabh ·2011-07-03 06:56:24

thanks for the correction...silly me

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