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If f is the function defined by f(x) = 2x for x ≥ 0 and f(x) = x^2 for

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If f is the function defined by f(x) = 2x for x ≥ 0 and f(x) = x^2 for  [#permalink]

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New post 07 Nov 2017, 22:59
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

67% (01:00) correct 33% (00:57) wrong based on 85 sessions

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Re: If f is the function defined by f(x) = 2x for x ≥ 0 and f(x) = x^2 for  [#permalink]

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New post 07 Nov 2017, 23:10
Bunuel wrote:
If f is the function defined by f(x) = 2x for x ≥ 0 and f(x) = x^2 for x < 0, what is the value of f(c) ?

(1) |c| = 2

(2) c < 0


(1) c an be 2 then f(c) will be 2*2 =4 or c can be -2 then f(c) will be (-2)^2 = 4..sufficient

(2) not sufficient as c^2 will give different results

A
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Re: If f is the function defined by f(x) = 2x for x ≥ 0 and f(x) = x^2 for  [#permalink]

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New post 08 Nov 2017, 02:55
Luckisnoexcuse wrote:
Bunuel wrote:
If f is the function defined by f(x) = 2x for x ≥ 0 and f(x) = x^2 for x < 0, what is the value of f(c) ?

(1) |c| = 2

(2) c < 0


(1) c an be 2 then f(c) will be 2*2 =4 or c can be -2 then f(c) will be (-2)^2 = 4..sufficient

(2) not sufficient as c^2 will give different results

A


Answer is correct. Explanation is not.

1. Here it means, any positive or negative value of c is 2.
So \(f(c) = c^2\) when c>0
f(c) = 4
Sufficient.

2. Here c<0. So f(c) = 2c.
when c = -1, f(c) = -2
when c = -2, f(c) = -4
Insufficient.

A.
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If f is the function defined by f(x) = 2x for x ≥ 0 and f(x) = x^2 for  [#permalink]

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New post 08 Nov 2017, 23:20
1
f(x) = 2x ----- x ≥ 0
f(x) = x^2 ----x<0
f(c)= ?

(1) |c| = 2
this implies c=+2 or c= -2

lets consider c=+2
f(c)= 2*2 = 4

lets consider c=-2
f(c)= (-2)^2 = 4
In any case, we are getting a unique value i.e. 4
Hence Statement 1 is sufficient.

(2) c < 0
This statement gives us an idea about the sign of c, but it does not tell anything about the value of c.
Hence It is not possible to calculate f(c)
Hence, Statement 2 is Insufficient.

Answer is A.

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If f is the function defined by f(x) = 2x for x ≥ 0 and f(x) = x^2 for &nbs [#permalink] 08 Nov 2017, 23:20
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