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[GMAT math practice question]
(function) \(f(x)\) is a function. What is the value of \(f(2006)\)?
\(1) f(11)=11\)
\(2) f(x+3)=\frac{(f(x)-1)}{(f(x)+1)}\)
(function) \(f(x)\) is a function. What is the value of \(f(2006)\)?
\(1) f(11)=11\)
\(2) f(x+3)=\frac{(f(x)-1)}{(f(x)+1)}\)
15 Oct 2019, 00:33
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[GMAT math practice question]
(number properties) \(m\) and \(n\) are integers. What is the value \((-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}\)?
\(1) m = n + 1\)
\(2) m = 3\)
(number properties) \(m\) and \(n\) are integers. What is the value \((-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}\)?
\(1) m = n + 1\)
\(2) m = 3\)
Updated on: 13 Jul 2021, 00:26
Originally posted by MathRevolution on 16 Oct 2019, 02:51.
Last edited by MathRevolution on 13 Jul 2021, 00:26, edited 1 time in total.
Last edited by MathRevolution on 13 Jul 2021, 00:26, edited 1 time in total.
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MathRevolution
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
Since we have many variables to determine a function and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
Since we have \(f(x+3) = \frac{(f(x)-1)}{(f(x)+1)}\) and \(f(11)=11\), we have \(f(14)=\frac{(f(11)-1)}{(f(11)+1)}=\frac{10}{12}=\frac{5}{6}\) when we substitute \(11\) for \(x.\)
We have \(f(17) = \frac{(f(14)-1)}{(f(14)+1)} = ((\frac{5}{6})-1)/(\frac{5}{6})+1) = (-(\frac{1}{6}))/(\frac{11}{6}) = \frac{-1}{11}\) when we substitute \(14\) for \(x.\)
We have \(f(20) = \frac{(f(17)-1)}{(f(17)+1)} = (-(\frac{1}{11})-1)/(-(\frac{1}{11})+1) = (-(\frac{12}{11}))/(\frac{10}{11}) = \frac{-12}{10} = \frac{-6}{5}\), when we substitute \(17\) for \(x.\)
We have \(f(23) = \frac{(f(20)-1)}{(f(20)+1)} = (-(\frac{6}{5})-1)/(-(\frac{6}{5})+1) = (-(\frac{11}{5}))/(-(\frac{1}{5}))=11\), when we substitute \(20\) for \(x.\)
Then we have the following patterns.
\(f(11) = f(23) = f(35) = … = f(12k-1) = 11\)
\(f(14) = f(26) = f(38) = … = f(12k+2) = \frac{5}{6}\)
\(f(17) = f(29) = f(41) = … = f(12k+5) = \frac{-1}{11}\)
\(f(20) = f(32) = f(44) = … = f(12k+8) = \frac{-6}{5}\)
So, \(f(2006) = f(12*167+2) = \frac{5}{6}.\)
Therefore, C is the answer.
Answer: C
In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions when the answer is A, B, C, or D.
