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If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) =

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Bunuel
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If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) = [#permalink] New post   23 Oct 2018, 21:56
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A
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Question Stats:

62% (01:33) correct 38% (01:46) wrong based on 298 sessions
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If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) =

A. 0.5
B. 1
C. 3
D. 4
E. 6
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D
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GMATinsight
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If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) = [#permalink] New post   Updated on: 23 Oct 2018, 22:39
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Bunuel wrote:
If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) =

A. 0.5
B. 1
C. 3
D. 4
E. 6


f(x) = 5 - 2x

f(3k) = 5-2(3k) = 5 - 6k

f(k+1) = 5-2(k+1) = 3 - 2k

f(3k) = f(k + 1)

i.e 5 - 6k = 3 - 2k

i.e. k = 0.5

i.e. f(k) = 5-2(0.5) = 4

Answer: Option D

Originally posted by GMATinsight on 23 Oct 2018, 22:09.
Last edited by GMATinsight on 23 Oct 2018, 22:39, edited 1 time in total.
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Re: If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) = [#permalink] New post   23 Oct 2018, 22:34
Hi GMATinsight

The question is asking for f(k) not for k value.

K is \(\frac{1}{2}\)

So f(k) = 5-2(\(\frac{1}{2}\))

f(k) = 5-1 => 4

OPTION : D
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If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) = [#permalink] New post   23 Oct 2018, 22:54
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The first thing to notice here is that we’re solving for \(f(k)\), not \(k\) itself. Make a note, circle it, underline it, do whatever you need to do to remind yourself what you’re actually solving for!

The next step is to see that we’re given \(f(3k) = f(k+1)\). Let’s simplify by “getting rid” of these functions and plug in the values provided, which are \(3k\) and \(k+1\).

Given that \(f(x) = 5 - 2x\), we get:

\(5-2(3k)\) = \(5-2(k+1)\)

Solving, we get:

\(5-6k = 5-2k-2\)

\(-4k = -2\)

\(k = 1/2\)

Choice A is a trap! Remember that we’re solving for \(f(k)\), not \(k\). Let’s plug our \(k\) value back into \(f(x)\).

\(f(1/2) = 5 - 2(1/2) = 4\)

Answer D.
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Re: If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) = [#permalink] New post   24 Oct 2018, 03:55
Expert Reply

Solution


Given:
    • f(x) = 5 – 2x
    • f(3k) = f(k + 1)

To find:
    • The value of f(k)

Approach and Working:
    • f(3k) = 5 – 2(3k) = 5 – 6k
    • f(k + 1) = 5 – 2(k + 1) = 3 – 2k
    • We are given, f(3k) = f(k + 1)
      o Implies, 5 – 6k = 3 – 2k
      o Thus, \(k = \frac{1}{2}\)
    • Therefore, f(k) = \(5 – 2(\frac{1}{2}) = 4\)

Hence, the correct answer is Option D.

Answer: D

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Re: If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) = [#permalink] New post   24 Oct 2018, 07:40
K is 1212

So f(k) = 5-2(1212)

f(k) = 5-1 => 4

D
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Re: If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) = [#permalink] New post   28 Oct 2018, 18:20
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Bunuel wrote:
If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) =

A. 0.5
B. 1
C. 3
D. 4
E. 6


We are given that f(x) = 5 - 2x. So f(3k) = 5 - 2(3k) = 5 - 6k and f(k + 1) = 5 - 2(k + 1) = 3 - 2k. Since f(3k) = f(k + 1), we equate the two expressions, and thus:

5 - 6k = 3 - 2k

2 = 4k

0.5 = k

Therefore, f(k) = f(0.5) = 5 - 2(0.5) = 4.

Answer: D
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If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) = [#permalink] New post   08 Mar 2020, 08:57
Bunuel wrote:
If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) =

A. 0.5
B. 1
C. 3
D. 4
E. 6


f(3k) = f(k+10) => 5-6k = 5-2(k+1) =>k = 1/2
f(1/2) = 4 (D)
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Re: If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) = [#permalink] New post   20 Apr 2023, 06:55
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