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a is the integer portion of 11+72 and b is the decimal portion of 11-72. What is the value of 1a-2+2-1b?

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a is the integer portion of 11+72 and b is the decimal portion of 11-72. What is the value of 1a-2+2-1b?

by Max@Math Revolution » Mon Mar 23, 2020 7:52 am

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[GMAT math practice question]

a is the integer portion of \(\sqrt{11+\sqrt{72}}\) and b is the decimal portion of \(\sqrt{11-\sqrt{72}}\) . What is the value of 1/a-2+ \(\sqrt{2}\) -1/b?

A. \(-\sqrt{2}\)
B. \(-\sqrt{3}\)
C. -1
D. -1- \(\sqrt{5}\)
E. -2
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Re: a is the integer portion of 11+72 and b is the decimal portion of 11-72. What is the value of 1a-2+2-1b?

by Max@Math Revolution » Wed Mar 25, 2020 5:48 am
=>

Remember that
\(\sqrt{a+b+2\sqrt{ab}}=\sqrt{a}+\sqrt{b}\) and \(\sqrt{a+b-2\sqrt{ab}}=\sqrt{a}-\sqrt{b}\left(a>b\right)\)

\(\sqrt{11+\sqrt{72}}=\sqrt{11+\sqrt{4\cdot18}}\) \(=\sqrt{11+2\sqrt{18}}=\sqrt{\left(9+2\right)+2\sqrt{9\cdot2}}\) \(=\sqrt{9}+\sqrt{2}=3+\sqrt{2}\)
Then, we have a = 4 since 1 < √2 < 2 or 4 < 3 + √2 < 5.

\(\sqrt{11-\sqrt{72}}\)
\(=\sqrt{11-\sqrt{4\cdot18}}=\sqrt{11-2\sqrt{18}}\) \(=\sqrt{\left(9+2\right)-2\sqrt{9\cdot2}}=\sqrt{9}-\sqrt{2}=3-\sqrt{2}\)
We have b = 3 - √2 - 1 = 2-√2 since 1 < 3 - √2 < 2.

\(\frac{1}{a-2+\sqrt{2}}-\frac{1}{b}\)
\(=\frac{1}{4-2+\sqrt{2}}-\frac{1}{2-\sqrt{2}}\)
\(=\frac{1}{2+\sqrt{2}}-\frac{1}{2-\sqrt{2}}\)
\(=\frac{2-\sqrt{2}}{\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}-\frac{2+\sqrt{2}}{\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)}\)
\(=\frac{2-\sqrt{2}}{4-2}-\frac{2+\sqrt{2}}{4-2}=\frac{2-\sqrt{2}-\left(2+\sqrt{2}\right)}{2}\)
\(=\frac{-2\sqrt{2}}{2}=-\sqrt{2}\)
Therefore, A is the answer.
Answer: A
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