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Which of the following functions satisfies
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15 Mar 2021, 06:34
15 Mar 2021, 06:34
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Which of the following functions satisfies \(f(a+b)=f(a)f(b)\) for all positive numbers \(a, b\) ?
\(A. f(x)=x+1\)
\(B. f(x)=x^2+1\)
\(C. f(x)=\sqrt{x}\)
\(D. f(x)=\frac{1}{x}\)
\(E. f(x)=2^x\)
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
\(A. f(x)=x+1\)
\(B. f(x)=x^2+1\)
\(C. f(x)=\sqrt{x}\)
\(D. f(x)=\frac{1}{x}\)
\(E. f(x)=2^x\)
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
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Which of the following functions satisfies
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15 Mar 2021, 07:57
15 Mar 2021, 07:57
Carcass wrote:
Which of the following functions satisfies \(f(a+b)=f(a)f(b)\) for all positive numbers \(a, b\) ?
\(A. f(x)=x+1\)
\(B. f(x)=x^2+1\)
\(C. f(x)=\sqrt{x}\)
\(D. f(x)=\frac{1}{x}\)
\(E. f(x)=2^x\)
\(A. f(x)=x+1\)
\(B. f(x)=x^2+1\)
\(C. f(x)=\sqrt{x}\)
\(D. f(x)=\frac{1}{x}\)
\(E. f(x)=2^x\)
A. \(f(x)=x+1\)
\(f(a)=a+1\)
\(f(b)=b+1\)
\(f(a+b)=a+b+1\)
\(f(a+b)≠f(a)f(b)\)
B. \(f(x)=x^2+1\)
\(f(a)=a^2+1\)
\(f(b)=b^2+1\)
\(f(a+b)=(a+b)^2+1\)
\(f(a+b)≠f(a)f(b)\)
C. \(f(x)=\sqrt{x}\)
\(f(a)=\sqrt{a}\)
\(f(b)=\sqrt{b}\)
\(f(a+b)=\sqrt{(a+b)}\)
\(f(a+b)≠f(a)f(b)\)
D. \(f(x)=\frac{1}{x}\)
\(f(a)=\frac{1}{a}\)
\(f(b)=\frac{1}{b}\)
\(f(a+b)=\frac{1}{(a+b)}\)
\(f(a+b)≠f(a)f(b)\)
E.\( f(x)=2^x\)
\( f(a)=2^a\)
\( f(b)=2^b\)
\( f(a+b)=2^{(a+b)}\)
\(f(a+b)=f(a)f(b)\)
Hence, option E
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Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: Which of the following functions satisfies
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15 Mar 2021, 08:44
15 Mar 2021, 08:44
Carcass wrote:
Which of the following functions satisfies \(f(a+b)=f(a)f(b)\) for all positive numbers \(a, b\) ?
\(A. f(x)=x+1\)
\(B. f(x)=x^2+1\)
\(C. f(x)=\sqrt{x}\)
\(D. f(x)=\frac{1}{x}\)
\(E. f(x)=2^x\)
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
\(A. f(x)=x+1\)
\(B. f(x)=x^2+1\)
\(C. f(x)=\sqrt{x}\)
\(D. f(x)=\frac{1}{x}\)
\(E. f(x)=2^x\)
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
Upon scanning the answer choices, we might recognize that answer choice E, f(x) = 2^x, involves a variable exponent and that the given information that f(a+b) = f (a)f(b) looks A LOT like the Product Law: (k^a)(k^b) = k^(a+b)
So, let's check E first.
E. f(x) = 2^x
If f(x) = 2^x, then f(a) = 2^a, f(b) = 2^b and f(a+b) = 2^(a+b)
We get: f(a)f(b) = (2^a)(2^b)
= 2^(a+b) [apply Product Law]
= f(a+b)
PERFECT!!!
Answer: E
Cheers,
Brent
