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The functions f(x) and g(x) are defined by f(x) = x^2 - 1 and g(x) = 1

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Bunuel
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The functions f(x) and g(x) are defined by f(x) = x^2 - 1 and g(x) = 1 [#permalink] New post   17 Sep 2018, 01:55
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The functions \(f(x)\) and \(g(x)\) are defined by \(f(x) = x^2 – 1\) and \(g(x) = 1 – 2x\). Given that \(f(g(k)) = 3\), which of the following could be the value of k?



A. 1/2

B. \(\frac{√3}{2}\)

C. 1

D. 3/2

E. -1
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D
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u1983
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The functions f(x) and g(x) are defined by f(x) = x^2 - 1 and g(x) = 1 [#permalink] New post   Updated on: 17 Sep 2018, 18:53
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Bunuel wrote:
The functions \(f(x)\) and \(g(x)\) are defined by \(f(x) = x^2 – 1\) and \(g(x) = 1 – 2x\). Given that \(f(g(k)) = 3\), which of the following could be the value of k?

A. 1/2
B. \(\frac{√3}{2}\)
C. 1
D. 3/2
E. -1


\(f(g(k)) = 3\)
or \(f(1-2k) = 3\)
or \((1-2k)^2 -1 = 3\)
or \((1-2k)^2 = 4\)
or \(|1-2k| =2\)
or \(k= -1/2\) or \(3/2\)............Ans D

Originally posted by u1983 on 17 Sep 2018, 08:15.
Last edited by u1983 on 17 Sep 2018, 18:53, edited 1 time in total.
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BrentGMATPrepNow
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Re: The functions f(x) and g(x) are defined by f(x) = x^2 - 1 and g(x) = 1 [#permalink] New post   17 Sep 2018, 10:08
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Bunuel wrote:
The functions \(f(x)\) and \(g(x)\) are defined by \(f(x) = x^2 – 1\) and \(g(x) = 1 – 2x\). Given that \(f(g(k)) = 3\), which of the following could be the value of k?



A. 1/2

B. \(\frac{√3}{2}\)

C. 1

D. 3/2

E. -1


Looks like a great candidate for the something method (see video below)

We're told that f(g(k)) = 3

Let's let g(k) = something
So, we have f(something) = 3

f(x) = x² - 1
So, f(something) = something² - 1 = 3
If something² - 1 = 3, then...
something² = 4
This tells us that EITHER something = 2 OR something = -2
In other words, g(k) = 2 OR g(k) = -2


Let's test each case.

case 1) g(k)= 2
g(x) = 1 – 2x
So, g(k) = 1 – 2k
This means: 1 - 2k = 2
Subtract 1 from both sides: -2k = 1
Solve: k = -1/2
Check the answer choices.....NOT THERE!
Looks like case 2 will yield the correct answer then.

case 2) g(k)= -2
This means: 1 - 2k = -2
Subtract 1 from both sides: -2k = -3
Solve: k = (-3/(-2) = 3/2
Check the answer choices.....
Answer: D

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Re: The functions f(x) and g(x) are defined by f(x) = x^2 - 1 and g(x) = 1 [#permalink] New post   17 Sep 2018, 13:52
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f(g(k)) = 3

f(x) = x²-1 = 3 -> x² = 4 -> x = 2 or -2

If x = 2
g(x) = 1 - 2*x = 2 -> x = -1/2

If x = -2
g(x) = 1 - 2*k = -2 -> x = 3/2

Answer D
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The functions f(x) and g(x) are defined by f(x) = x^2 - 1 and g(x) = 1 [#permalink] New post   06 Jul 2020, 12:27
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\(g(k)=1-2k\)
\(f(g(k))=g(k)^2-1=3\)
\(](1-2k)^2-1=3\)
\(4k^2-4k-3=0]\)
\((2k-3)(2k+1)=0\)
\((2k-3)=0\) or \((2k+1)=0\)
\(k=\frac{3}{2}\) or \(k=-\frac{1}{2}\)

Answer D
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Re: The functions f(x) and g(x) are defined by f(x) = x^2 - 1 and g(x) = 1 [#permalink] New post   14 Jun 2020, 11:50
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Bunuel wrote:
The functions \(f(x)\) and \(g(x)\) are defined by \(f(x) = x^2 – 1\) and \(g(x) = 1 – 2x\). Given that \(f(g(k)) = 3\), which of the following could be the value of k?



A. 1/2

B. \(\frac{√3}{2}\)

C. 1

D. 3/2

E. -1


Solution


    • \(f(g(k)) = (g(k))^2 – 1 = 3 \)
    \(⟹ (g(k))^2 = 4\)
    \(⟹(1 – 2k )^2 = 4\)
    \(⟹ (1-2k)^2 – 2^2 = 0\)
    \(⟹(1-2k -2)(1-2k + 2) = 0\)
    \(⟹(-1 – 2k) (3-2k) = 0\)
    \(⟹ k = -\frac{1}{2}\) or \(\frac{3}{2}\)
However, -1/2 is not in option,hence, \(k = \frac{3}{2}\)
Thus, the correct answer is Option D.
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Re: The functions f(x) and g(x) are defined by f(x) = x^2 - 1 and g(x) = 1 [#permalink] New post   02 Nov 2021, 09:44
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Re: The functions f(x) and g(x) are defined by f(x) = x^2 - 1 and g(x) = 1 [#permalink]
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