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Which of the following functions f(x) satisfies the condition f(y −z)
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11 Jan 2022, 07:32
11 Jan 2022, 07:32
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Which of the following functions f(x) satisfies the condition \(f(y−z)=f(y)−f(z)\) for all possible values of y and z?
A. \(f(x) = x^2\)
B. \(f(x) = x + (x − 1)^2\)
C. \(f(x)=x−1\)
D. \(f(x)=\frac{5}{x}\)
E. \(f(x)=\frac{x}{5}\)
A. \(f(x) = x^2\)
B. \(f(x) = x + (x − 1)^2\)
C. \(f(x)=x−1\)
D. \(f(x)=\frac{5}{x}\)
E. \(f(x)=\frac{x}{5}\)
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Re: Which of the following functions f(x) satisfies the condition f(y −z)
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11 Jan 2022, 07:46
11 Jan 2022, 07:46
Carcass wrote:
Which of the following functions f(x) satisfies the condition \(f(y−z)=f(y)−f(z)\) for all possible values of y and z?
A. \(f(x) = x^2\)
B. \(f(x) = x + (x − 1)^2\)
C. \(f(x)=x−1\)
D. \(f(x)=\frac{5}{x}\)
E. \(f(x)=\frac{x}{5}\)
A. \(f(x) = x^2\)
B. \(f(x) = x + (x − 1)^2\)
C. \(f(x)=x−1\)
D. \(f(x)=\frac{5}{x}\)
E. \(f(x)=\frac{x}{5}\)
Let's test a set of values: \(y = 3\) and \(z = 1\)
If the given condition is true, then it must be the case that \(f(3 − 1) = f(3) − f(1)\)
When we simplify the first function, we get: \(f(2) = f(3) − f(1)\)
So for each answer choice, let's test whether \(f(2) = f(3) − f(1)\)
A. \(f(x) = x^2\)
\(f(2) = f(3) − f(1)\) becomes \(2^2 = 3^2 − 1^2\)
Simplify: \(4 = 9 − 1\)
NOT TRUE.
Eliminate A
B. \(f(x) = x + (x − 1)^2\)
\(f(2) = f(3) − f(1)\) becomes \(2 + (2-1)^2 = (3 + (3-1)^2) − (1 + (2-1)^2)\)
Simplify: \(3 = 7 − 2\)
NOT TRUE.
Eliminate B
C. \(f(x)=x−1\)
\(f(2) = f(3) − f(1)\) becomes \(2 - 1 = (3 - 1) − (1 - 1)\)
Simplify: \(1 = 2 − 0\)
NOT TRUE.
Eliminate C
D. \(f(x)=\frac{5}{x}\)
\(f(2) = f(3) − f(1)\) becomes \(\frac{5}{2} = \frac{5}{3} - \frac{5}{1}\)
Simplify: \(\frac{5}{2} = -\frac{10}{3}\)
NOT TRUE.
Eliminate D
By the process of elimination, the correct answer must be E.
But let's check to make sure.
E. \(f(x)=\frac{x}{5}\)
\(f(2) = f(3) − f(1)\) becomes \(\frac{2}{5} = \frac{3}{5} - \frac{1}{5}\)
WORKS!
Answer: E
gmatclubot
Re: Which of the following functions f(x) satisfies the condition f(y −z) [#permalink]
11 Jan 2022, 07:46
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