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27 Dec 2019, 02:12
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
57% (02:16) correct
43%
(02:32)
wrong
based on 297
sessions
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f(x) = ax + b. a and b are both positive numbers. What is the value of f(1)?
1) f(f(x)) = 4x + 5
2) f(f(1)) = 4
Posted from my mobile device
1) f(f(x)) = 4x + 5
2) f(f(1)) = 4
Posted from my mobile device
27 Dec 2019, 13:40
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f(1)= a+b
We need to figure out a+b
Statement 1-
f(x)= ax+b
\(f(f(x)= f(ax+b)= a(ax+b)+b\)
\(f(f(x)== a^2x+ab+b\)
\(a^2x+ab+b= 4x+5\)
hence; \(a^2=4\); and (ab+b)=5. Also, a,b>0.
a=2 and b= 5/3
Sufficient
Statement 2-
\(f(f(x)== a^2x+ab+b\)
\(f(f(1)== a^2+ab+b\)=4
We have 2 unknowns and 1 equation.
Insufficient
We need to figure out a+b
Statement 1-
f(x)= ax+b
\(f(f(x)= f(ax+b)= a(ax+b)+b\)
\(f(f(x)== a^2x+ab+b\)
\(a^2x+ab+b= 4x+5\)
hence; \(a^2=4\); and (ab+b)=5. Also, a,b>0.
a=2 and b= 5/3
Sufficient
Statement 2-
\(f(f(x)== a^2x+ab+b\)
\(f(f(1)== a^2+ab+b\)=4
We have 2 unknowns and 1 equation.
Insufficient
MIJI
29 Dec 2019, 19:43
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MIJI
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.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
Condition 1)
\(f(f(x)) = f(ax+b) = a(ax+b)+b = a^2x + ab + b = 4x + 5\).
Then we have \(a^2 = 4\), which means \(a = 2\) or \(a = -2\).
Since \(a\) is a positive number, we have \(a = 2\).
Then we have \(ab + b = 2b + b = 3b = 5\) or \(b = \frac{5}{3}\) and \(f(1) = a + b = \frac{11}{3}\).
Since condition 1) yields a unique solution, it is sufficient.
Condition 2)
\(f(f(1)) = f(a+b) = a(a+b) + b = a^2 + ab + b = 4\).
If \(a = 1\) and \(b = \frac{3}{2}\), then we have \(f(1)=a+b=\frac{5}{2}\).
If \(a = 2\) and \(b = 3\), then we have \(f(1)=a+b=5\).
Since condition 2) does not yield a unique solution, it is not sufficient
Therefore, A is the answer.
If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
