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

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Senior Manager
Joined: 02 Jan 2017
Posts: 293
If 0 < p < 1, which of the following has the least value?  [#permalink]

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22 Feb 2017, 20:56
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35% (medium)

Question Stats:

65% (01:26) correct 35% (01:37) wrong based on 218 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. $$\frac{1}{(p+1)^2}$$
Manager
Joined: 15 Mar 2014
Posts: 140
Location: Singapore
Concentration: Technology, General Management
GPA: 3.5
WE: Operations (Telecommunications)
Re: If 0 < p < 1, which of the following has the least value?  [#permalink]

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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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Joined: 22 Nov 2016
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Re: If 0 < p < 1, which of the following has the least value?  [#permalink]

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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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Joined: 02 Aug 2009
Posts: 7764
If 0 < p < 1, which of the following has the least value?  [#permalink]

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18 Jul 2018, 22:35
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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?   [#permalink] 18 Jul 2018, 22:35
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