Bunuel
For all integers \(n,\) the function \(f\) is defined by \(f(n)=(a)^{\frac{6}{n}}\) where \(a\) is a constant. What is the value of \(f(1)\)?
(1) \(f(2)=64\)
(2) \(f(3)=16\)
Given: \(f(n)=(a)^{\frac{6}{n}}\) where \(a\) is a constant. Target question: What is the value of \(f(1)\)? Statement 1: \(f(2)=64\) Plug \(n = 2\) into the given function to get: \(f(2)=(a)^{\frac{6}{2}} = 64\)
Simplify: \((a)^{3} = 64\)
So, it must be the case that \(a = 4\), which means our function is defined as follows: \(f(n)=(4)^{\frac{6}{n}}\)
So,
\(f(1)=(4)^{\frac{6}{1}} = 4^6 = 4096\) [
On test day, I would never waste time actually calculating 4^6, since all we need to do is determine whether there's only one answer to the target question]
Since we can answer the
target question with certainty, statement 1 is SUFFICIENT
Statement 2: \(f(3)=16\)Plug \(n = 3\) into the given function to get: \(f(2)=(a)^{\frac{6}{3}} = 16\)
Simplify: \((a)^{2} = 16\)
There are two possible solutions to this equation: \(a = 4\) and \(a = -4\)
Important: Some students will assume that since we have two possible values for \(a\), statement 2 must be insufficient. However, the target question doesn't ask us for the value of \(a\); it asks us for the value of \(f(1)\). So let's see what happens for each possible a-value. Case a: \(a = 4\). In this case,
\(f(1)=(4)^{\frac{6}{1}} = 4^6 = 4096\) Case b: \(a = -4\). In this case,
\(f(1)=(-4)^{\frac{6}{1}} = (-4)^6 = 4096\) Aha!! We get the same answer to the target question in both cases, which means it must be the case that
\(f(1)=4096\) Since we can answer the
target question with certainty, statement 2 is SUFFICIENT
Answer: D
Cheers,
Brent