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27 May 2021, 08:30
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E
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65% (02:00) correct
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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. \(f(x)=\frac{x}{5}\)
27 May 2021, 08:45
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Bunuel
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
General Discussion
16 Dec 2021, 06:16
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We need to find a function f(x) out of the answer choices which satisfies the condition \(f(y−z)=f(y)−f(z)\) for all possible values of y and z
Let's take option by option and check
A. \(f(x) = x^2\)
To find f(y-z) we need to substitute x= y-z in \(f(x) = x^2\)
We get \(f(y-z) = (y-z)^2\)
Similarly, \(f(y) = y^2\)
And \(f(z) = z^2\)
Clearly, \(f(y-z) ≠ f(y)−f(z)\)
As, \((y-z)^2 ≠ y^2 - z^2\)
B. \(f(x) = x + (x − 1)^2\)
\(f(y-z) = y-z + (y-z − 1)^2\)
\(f(y) = y + (y − 1)^2\)
\(f(z) = z + (z − 1)^2\)
\(f(y) - f(z) = y + (y − 1)^2 - z + (z − 1)^2\) ≠\(y-z + (y-z − 1)^2\)
C. \(f(x)=x−1\)
\(f(y-z)=y-z−1\)
\(f(y)=y−1\)
\(f(z)=z−1\)
\(f(y) - f(z) = y−1 - (z-1) = y -z ≠ y-z−1\)
D. \(f(x)=\frac{5}{x}\)
\(f(y-z)=\frac{5}{(y-z)}\)
\(f(y)=\frac{5}{y}\)
\(f(z)=\frac{5}{z}\)
\(f(y) - f(z) = \frac{5}{y} - \frac{5}{z} = \frac{5z - 5y}{yz} ≠\frac{5}{(y-z)}\)
E. \(f(x)=\frac{x}{5}\)
\(f(y-z)=\frac{y-z}{5}\)
\(f(y)=\frac{y}{5}\)
\(f(z)=\frac{z}{5}\)
\(f(y) - f(z) = \frac{y}{5} - \frac{z}{5} = \frac{y-z}{5}\)
So, Answer is E
Hope it helps!
Watch the following video to learn the Basics of Functions and Custom Characters
Let's take option by option and check
A. \(f(x) = x^2\)
To find f(y-z) we need to substitute x= y-z in \(f(x) = x^2\)
We get \(f(y-z) = (y-z)^2\)
Similarly, \(f(y) = y^2\)
And \(f(z) = z^2\)
Clearly, \(f(y-z) ≠ f(y)−f(z)\)
As, \((y-z)^2 ≠ y^2 - z^2\)
B. \(f(x) = x + (x − 1)^2\)
\(f(y-z) = y-z + (y-z − 1)^2\)
\(f(y) = y + (y − 1)^2\)
\(f(z) = z + (z − 1)^2\)
\(f(y) - f(z) = y + (y − 1)^2 - z + (z − 1)^2\) ≠\(y-z + (y-z − 1)^2\)
C. \(f(x)=x−1\)
\(f(y-z)=y-z−1\)
\(f(y)=y−1\)
\(f(z)=z−1\)
\(f(y) - f(z) = y−1 - (z-1) = y -z ≠ y-z−1\)
D. \(f(x)=\frac{5}{x}\)
\(f(y-z)=\frac{5}{(y-z)}\)
\(f(y)=\frac{5}{y}\)
\(f(z)=\frac{5}{z}\)
\(f(y) - f(z) = \frac{5}{y} - \frac{5}{z} = \frac{5z - 5y}{yz} ≠\frac{5}{(y-z)}\)
E. \(f(x)=\frac{x}{5}\)
\(f(y-z)=\frac{y-z}{5}\)
\(f(y)=\frac{y}{5}\)
\(f(z)=\frac{z}{5}\)
\(f(y) - f(z) = \frac{y}{5} - \frac{z}{5} = \frac{y-z}{5}\)
So, Answer is E
Hope it helps!
Watch the following video to learn the Basics of Functions and Custom Characters
