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

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Joined: 02 Sep 2009
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If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) =  [#permalink]

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23 Oct 2018, 21:56
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45% (medium)

Question Stats:

62% (01:48) correct 38% (01:51) wrong based on 68 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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If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) =  [#permalink]

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Updated on: 23 Oct 2018, 22:39
1
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
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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]

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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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Joined: 23 Oct 2018
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If f(x) = 5 - 2x and f(3k) = f(k + 1), then f(k) =  [#permalink]

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23 Oct 2018, 22:54
1
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]

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24 Oct 2018, 03:55

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]

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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]

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