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Re: If 0 < p < 1, which of the following has the least value?
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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).

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

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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