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If the function f(x) = x^2 + 1 and f(y+1)^(1/2) = 1, what is the value
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04 May 2022, 05:36
04 May 2022, 05:36
Expert Reply
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Question Stats:
91% (00:36) correct
8% (00:56) wrong based on 23 sessions
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If the function \(f(x) =x^2+1\) and \(f(\sqrt{y+1})=1\), what is the value \(y\)?
(A) -4
(B) -1
(C) 1
(D) 3
(E) 5
(A) -4
(B) -1
(C) 1
(D) 3
(E) 5
Part of the project: GRE Quantitative Comparison & Problem Solving Challenge
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GRE Math Essentials for a TOP Quant Score - Q170!!! (2021)
GRE Geometry Formulas for a Q170
GRE - Math Book
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials for a TOP Quant Score - Q170!!! (2021)
GRE Geometry Formulas for a Q170
GRE - Math Book
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If the function f(x) = x^2 + 1 and f(y+1)^(1/2) = 1, what is the value
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04 May 2022, 06:07
04 May 2022, 06:07
1
Carcass wrote:
If the function \(f(x) =x^2+1\) and \(f(\sqrt{y+1})=1\), what is the value \(y\)?
(A) -4
(B) -1
(C) 1
(D) 3
(E) 5
(A) -4
(B) -1
(C) 1
(D) 3
(E) 5
APPROACH #1: Number sense
Given: \(f(x) =x^2+1\)
In other words, \(f(something) = something^2+1\)
So, if \(f(something) = 1\), we can be certain that \(something^2 = 0\)
Similarly, since we're told \(f(\sqrt{y+1})=1\), we can be certain that \((\sqrt{y+1})^2 = 0\), which means \(y = -1\)
Answer: B
APPROACH #2: Algebra
If \(f(x) =x^2+1\), then \(f(\sqrt{y+1})= (\sqrt{y+1})^2 + 1\)
Since we're told \(f(\sqrt{y+1})=1\), we can write: \(f(\sqrt{y+1})= (\sqrt{y+1})^2 + 1 = 1\)
Take: \((\sqrt{y+1})^2 + 1 = 1\)
Subtract \(1\) from both sides: \((\sqrt{y+1})^2 = 0\)
Simplify: \(y+1 = 0\)
Solve: \(y = -1\)
Answer: B
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Status:GRE Quant Tutor
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Re: If the function f(x) = x^2 + 1 and f(y+1)^(1/2) = 1, what is the value
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30 Jun 2022, 22:52
30 Jun 2022, 22:52
Given that \(f(x) =x^2+1\) and \(f(\sqrt{y+1})=1\) and we need to find the value of y
To find \(f(\sqrt{y+1})\) we need to compare what is inside the bracket in \(f(\sqrt{y+1})\) and f(x)
=> We need to substitute x with \(\sqrt{y+1}\) in \(f(\sqrt{y+1})=1\) to get the value of \(f(\sqrt{y+1})\)
=> \(f(\sqrt{y+1})\) = \((\sqrt{y+1})^2 + 1\) = y+1 + 1 = y + 2 = 1 (given)
=> y = 1-2 = -1
So, Answer will be B
Hope it helps!
Watch the following video to learn the Basics of Functions and Custom Characters
To find \(f(\sqrt{y+1})\) we need to compare what is inside the bracket in \(f(\sqrt{y+1})\) and f(x)
=> We need to substitute x with \(\sqrt{y+1}\) in \(f(\sqrt{y+1})=1\) to get the value of \(f(\sqrt{y+1})\)
=> \(f(\sqrt{y+1})\) = \((\sqrt{y+1})^2 + 1\) = y+1 + 1 = y + 2 = 1 (given)
=> y = 1-2 = -1
So, Answer will be B
Hope it helps!
Watch the following video to learn the Basics of Functions and Custom Characters
