Calculus - RIDDLER's RIDDLE #1

1

Akash shrivastava ·2009-04-02 06:19:57

answer is 0
and any one can solve by using log

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1

ANKIT MAHATO ·2009-04-02 07:15:39

the book is wrong .. thankx dude .. i wanted to check if my method was correct

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1

Akand ·2009-04-02 07:12:15

dude if it was 1/e .....then y wud d heck will they give such huge powers?????
i think its e^-2/5....

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1

ANKIT MAHATO ·2009-04-02 06:59:46

i got the same answer but it was give 1/e as the answer .. .

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1

Akand ·2009-04-02 06:58:49

wasnt thinkin while i was integrating hehehe [3][3][3]

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1

Akand ·2009-04-02 06:58:14

ya sorry.................corrected

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1

ANKIT MAHATO ·2009-04-02 06:57:07

₀¹âˆ«log x dx is not 0

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1

ANKIT MAHATO ·2009-04-02 06:56:05

no dude ... it has an e term ..

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1

Akand ·2009-04-02 06:55:50

y=(n!/nⁿ)^{{(2n4+1)/(5n5+1)}}
logy={(2n⁴+1)/(5n⁵+1)}log(n!/nⁿ)
logy=(2/n+1/n⁵)/(5+1/n5){log(1/n)+log(2/n)...log(n/n)
logy=2/5(1/n)Î£log(r/n)
logy=2/5âˆ«logxdx
.....
logy=-2/5
y=e^-2/5

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1

Akand ·2009-04-02 06:53:58

is d ans 1.??????

if yes ill post d soln

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1

ANKIT MAHATO ·2009-04-02 06:34:02

no u r not correct ... but ur method is close ...

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1

karthik singaraju ·2009-04-02 02:41:18

....do not try to riddle me or else u will be caught in a riddle which will make u think that RIDDLE is himself a very big riddle ! .. I have no mass or volume no begining or end you see me everyday ... what am i ?

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