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If 0 < p < 1, which of the following has the least value?

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Senior Manager
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If 0 < p < 1, which of the following has the least value?  [#permalink]

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New post 22 Feb 2017, 20:56
3
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A
B
C
D
E

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  35% (medium)

Question Stats:

66% (01:07) correct 34% (01:28) wrong based on 191 sessions

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If 0 < p < 1, which of the following has the least value?


A. \(\frac{1}{p^2}\)

B. \(\frac{1}{\sqrt{p}}\)

C. \(\frac{1}{p^2+1}\)

D. \(\frac{1}{\sqrt{p+1}}\)

E. \(\frac{1}{(p+1)^2}\)
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Re: If 0 < p < 1, which of the following has the least value?  [#permalink]

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New post 22 Feb 2017, 23:04
1
E.

Did the question by taking an example.

Let p=.5(1/2)

A. 1/p^2 = 1/(1/4) = 4

B. 1/sqr of p = 1/sqr(1/2) = √2 = 1.414

C. 1/(p^2+1) = 1/(1/4 +1 ) = 1/(5/4) = 4/5 = .8

D. 1/ sqr of (p+1) = 1/sqr(3/2) = √2/√3 = 1.414/1.732 ~ 14/17 ~ .8

E. 1/(p+1)^2 = 1/(3/2)^2 = 1/(9/4) = 4/9 ~ .4


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Re: If 0 < p < 1, which of the following has the least value?  [#permalink]

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New post 07 Dec 2017, 00:52
vikasp99 wrote:
If 0 < p < 1, which of the following has the least value?

A. 1/p^2
B. 1/sqr of p
C. 1/(p^2+1)
D. 1/ sqr of (p+1)
E. 1/(p+1)^2


Bunuel
How to mathematically compare C and D??
Please help
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Re: If 0 < p < 1, which of the following has the least value?  [#permalink]

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New post 18 Jul 2018, 22:09
Since there was a root involved in denominator in choices I took p as 1/4..it helped to get more clear values for quick visual comparison.so C was 1/17 and D was 1/sqrt(5).

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If 0 < p < 1, which of the following has the least value?  [#permalink]

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New post 18 Jul 2018, 22:35
2
1
vikasp99 wrote:
If 0 < p < 1, which of the following has the least value?


A. \(\frac{1}{p^2}\)

B. \(\frac{1}{\sqrt{p}}\)

C. \(\frac{1}{p^2+1}\)

D. \(\frac{1}{\sqrt{p+1}}\)

E. \(\frac{1}{(p+1)^2}\)


since all are fractions with 1 in numerator, the fraction with largest denominator will be the smallest
Also p is a positive fraction<1, so square p^2 will be smaller than \(\sqrt{p}\)...
\(\sqrt{p}>p>p^2\)...
so
A. \(p^2\)
B \(\sqrt{p}\)
C. \(p^2+1\)
D. \(\sqrt{p+1}\)
E. \((p+1)^2=p^2+1+2p > p^2+1\)... hence E > D

we should compare only C,D and E as only these are >1 and E>D
so lets compare only D and E

\(\sqrt{p+1}\)...here p+1>1, so square root will be less than original, hence\(\sqrt{p+1}<p+1......D<p+1\)
But any number >1 should have its square > original .. hence \((p+1)^2>p+1........E>p+1\)

so \(D<p+1<E\)

so denominator in E is the greatest and the fraction given in E is the least

E
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If 0 < p < 1, which of the following has the least value? &nbs [#permalink] 18 Jul 2018, 22:35
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