Let the function g(a, b) = f(a) + f(b).
For which function f below will g(a + b, a + b) = g(a, a) + g(b, b)?
let a = 1 and b = 2, a+b = 3
A) x+3,
2(a+b+3) = 2(a+3) + 2(b+3)
2(a+b+3) = 2(a+b+3) + 6 - incorrect
B) x^2
2(a+b)^2 = 2(a^2)+2(b^2) - Not always
C)|x|
2|a+b| = 2|a| + 2|b| - Not always
D)1/x
2/(a+b) = (2/a)+(2/b) = 2(a+b)/ab - Not always
E)x/4
2*(a+b)/4 = 2(a/4) + 2(b/4) = 2(a+b)/4 - Always Correct
Answer E
Functions & Algebra ||HELP||
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We can plug in values.dell2 wrote:Let the function g(a, b) = f(a) + f(b).
For which function f below will g(a + b, a + b) = g(a, a) + g(b, b)?
A: x+3
B: x^2
C: |x|
D: 1/x
E: x/4
OA : x/4
Since one of the answer choices is |x|, we should plug in at least one negative value.
Let a=-2 and b=3.
Then a+b = -2+3 = 1.
The question becomes:
For which function f below will g(1, 1) = g(-2, -2) + g(3, 3)?
Since g(a,b) = f(a) + f(b):
g(1,1) = f(1) + f(1) = 2f(1)
g(-2,-2) = f(-2) + f(-2) = 2f(-2)
g(3,3) = f(3) + f(3) = 2f(3)
Substituting these relationships into g(1,1) = g(-2,-2) + g(3,3), we get:
2f(1) = 2f(-2) + 2f(3)
f(1) = f(-2) + f(3).
Thus, the question now becomes:
For which function f below will f(1) = f(-2) + f(3)?
A: x+3
f(1) = 1+3 = 4.
f(-2) + f(3) = (-2+3) + (3+3) = 10.
Doesn't work. Eliminate A.
B: x²
f(1) = 1² = 1.
f(-2) + f(3) = (-2)² + 3² = 13.
Doesn't work. Eliminate B.
C: |x|
f(1) = |1| = 1.
f(-2) + f(3) = |-2| + |3| = 5.
Doesn't work. Eliminate C.
D: 1/x
f(1) = 1/1 = 1.
f(-2) + f(3) = -1/2 + 1/3 = -1/6.
Doesn't work. Eliminate D.
The correct answer is E.
E: x/4
f(1) = 1/4
f(-2) + f(3) = -2/4 + 3/4 = 1/4.
Success!
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It's a bit too involved.sana.noor wrote:Mitch is it Gmat type question?
Here are two problems from GMAT Prep that test the same basic concept:
https://www.beatthegmat.com/gmatprep-fun ... 66593.html
https://www.beatthegmat.com/for-which-of ... 65920.html
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
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