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26 Apr 2020, 21:09
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D
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Difficulty:
85%
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Question Stats:
53% (02:21) correct
47%
(02:25)
wrong
based on 148
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Consider 3 numbers \(X = a^{a}b^{b}c^{c}\), \(Y = a^{a}b^{c}c^{b}\) and \(Z = a^{b}b^{c}c^{a}\), where 0 < a < b < c < 1.
Which of the following options is the correct order?
A. \(Z<X<Y\)
B. \(Y<Z<X\)
C. \(Y<X<Z\)
D. \(Z<Y<X\)
E. \(X<Z<Y\)
Which of the following options is the correct order?
A. \(Z<X<Y\)
B. \(Y<Z<X\)
C. \(Y<X<Z\)
D. \(Z<Y<X\)
E. \(X<Z<Y\)
26 Apr 2020, 22:27
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nick1816
Logically
You can say that the larger number should be with larger power, smallest with smaller and so on...
So X does exactly the same, and Z totally opposite, so they are in extreme corners...X>Y>Z
D
Substitute
Take a=1/10, b=2/10 and c=3/10
\(X = a^{a}b^{b}c^{c}=(\frac{1}{10})^{\frac{1}{10}}*(\frac{2}{10})^{\frac{2}{10}}*(\frac{3}{10})^{\frac{3}{10}}=((\frac{1}{10})^1*(\frac{2}{10})^2*(\frac{3}{10})^3)^{\frac{1}{10}}=(\frac{1*2*2*3*3*3}{10^6})^\frac{1}{10}=\sqrt[10]{\frac{108}{10^6}}\),
\(Y = a^{a}b^{c}c^{b}=(\frac{1}{10})^{\frac{1}{10}}*(\frac{2}{10})^{\frac{3}{10}}*(\frac{3}{10})^{\frac{2}{10}}=((\frac{1}{10})^1*(\frac{2}{10})^3*(\frac{3}{10})^2)^{\frac{1}{10}}=(\frac{1*2*2*2*3*3}{10^6})^\frac{1}{10}=\sqrt[10]{\frac{72}{10^6}}\) and
\(Z = a^{b}b^{c}c^{a}=(\frac{1}{10})^{\frac{2}{10}}*(\frac{2}{10})^{\frac{3}{10}}*(\frac{3}{10})^{\frac{1}{10}}=((\frac{1}{10})^2*(\frac{2}{10})^3*(\frac{3}{10})^1)^{\frac{1}{10}}=(\frac{1*1*2*2*2*3}{10^6})^\frac{1}{10}=\sqrt[10]{\frac{24}{10^6}}\),
X>Y>Z
D
27 Apr 2020, 21:07
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\(\frac{X}{Y} = \frac{a^{a}b^{b}c^{c}}{ a^{a}b^{c}c^{b}}\)
\(\frac{X}{Y} = \frac{b^{b} * c^{c}}{ b^{c}c^{b}} = (\frac{c}{b})^{c-b}\)
As c>b>0, \((\frac{c}{b})^{c-b}> 1 \)
\(\frac{X}{Y}> 1\) or \(X>Y\)........(1)
\(\frac{Y}{Z} = \frac{a^{a}b^{c}c^{b}}{ a^{b}b^{c}c^{a}}\)
\(\frac{X}{Y} = \frac{a^{a} * c^{b}}{ a^{b}c^{a}} = (\frac{c}{a})^{b-a}\)
As c>b>a>0, \((\frac{c}{a})^{b-a}> 1 \)
\(\frac{Y}{Z}> 1\) or \(Y>Z\)........(2)
From (1) and (2)
\(X>Y>Z\)
\(\frac{X}{Y} = \frac{b^{b} * c^{c}}{ b^{c}c^{b}} = (\frac{c}{b})^{c-b}\)
As c>b>0, \((\frac{c}{b})^{c-b}> 1 \)
\(\frac{X}{Y}> 1\) or \(X>Y\)........(1)
\(\frac{Y}{Z} = \frac{a^{a}b^{c}c^{b}}{ a^{b}b^{c}c^{a}}\)
\(\frac{X}{Y} = \frac{a^{a} * c^{b}}{ a^{b}c^{a}} = (\frac{c}{a})^{b-a}\)
As c>b>a>0, \((\frac{c}{a})^{b-a}> 1 \)
\(\frac{Y}{Z}> 1\) or \(Y>Z\)........(2)
From (1) and (2)
\(X>Y>Z\)
