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# If f is the function defined by f(x) = 27x for x ≥ 0 and f(x) = x^4 fo

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Joined: 02 Sep 2009
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If f is the function defined by f(x) = 27x for x ≥ 0 and f(x) = x^4 fo  [#permalink]

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24 Oct 2018, 05:10
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45% (medium)

Question Stats:

59% (01:10) correct 41% (01:11) wrong based on 111 sessions

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If f is the function defined by f(x) = 27x for x ≥ 0 and f(x) = x^4 for x < 0, what is the value of f (k)?

(1) |k| = 3
(2) k < 0

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If f is the function defined by f(x) = 27x for x ≥ 0 and f(x) = x^4 fo  [#permalink]

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Updated on: 24 Oct 2018, 06:14
Bunuel wrote:
If f is the function defined by f(x) = 27x for x ≥ 0 and f(x) = x^4 for x < 0, what is the value of f (k)?

(1) |k| = 3
(2) k < 0

Statement 1) satisfies both functions because k can be -3 or 3.

$$27*3 = 81$$

$$(-3)^4 = 81$$

Sufficient.

Statement 2) k < 0 but we don't know if k = -1 or -2 or any other value.

Insufficient.

I originally made a careless mistake of not substituting in the equation and choose answer choice C.

Thanks to nkin for pointing it out!

Originally posted by Salsanousi on 24 Oct 2018, 05:46.
Last edited by Salsanousi on 24 Oct 2018, 06:14, edited 2 times in total.
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Re: If f is the function defined by f(x) = 27x for x ≥ 0 and f(x) = x^4 fo  [#permalink]

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24 Oct 2018, 06:05
1
Stmt 1:

k = +3 or -3,

If k = 3 then, f(k) = 27*3 = 81
If k = -3, then, f(k) = (-3)^4 = 81

Since we get the same value of f(k) for both values of k, Stmt 1 is sufficient.

Eliminate B, C, E

Stmt 2:

k < 0

Hence f(k) = k^4

But if k = -1, f(k) = 1, if k = -2, f(k) = 16, and so on. As we get different values, for different values of k, Stmt 2, is insufficient.

Hence, option A
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Re: If f is the function defined by f(x) = 27x for x ≥ 0 and f(x) = x^4 fo  [#permalink]

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24 Oct 2018, 06:16
Statement 1:

k = +3 or -3
So f(k) for k = +3 is 27(3) = 81
f(k) for k = -3 is $$(-3)^4$$ = 81

In both the cases, we are getting a unique answer. So statement 1 is sufficient.

Statement 2:

Option : A
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Re: If f is the function defined by f(x) = 27x for x ≥ 0 and f(x) = x^4 fo   [#permalink] 24 Oct 2018, 06:16
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