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Question

The fraction exceeding its pth power by the greatest number possible, where p 2, is


A

(1p)1p1

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B

(1p)(p1)

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C

p1p1

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D

None of these

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Solution

The correct option is A

(1p)1p1


Let y=xxp, where x is the fraction.
dydx=1pxp1
For maximum or minimum dydx=0
1pxp1=0 x=(1p)1(p1)Now, d2ydx2=p(p1)xp2 [d2ydx2]x=(1p)1p1=p(p1)(1p)p2(p1)<0
y is maximum at x=(1p)1(p1)


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