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by sana.noor » Tue Oct 08, 2013 5:24 am
The function f(x) is defined as,

f(x) = x2 {read it as x squared} for |x|<=1
f(x) = 1/x2 {read it as 1 upon x squared} for |x|>1

Is -0.9<a<0.9?

(1) f(-a)=1/f(b)
(2) a=1/b
OA is A
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by GMATGuruNY » Tue Oct 08, 2013 6:31 am
sana.noor wrote:The function f(x) is defined as,

f(x) = x2 {read it as x squared} for |x|<=1
f(x) = 1/x2 {read it as 1 upon x squared} for |x|>1

Is -0.9<a<0.9?

(1) f(-a)=1/f(b)
(2) a=1/b
OA is A
To determine how the value of f(x) is constrained, test cases BETWEEN -0.9 and 0.9 and BEYOND this range:

Case 1: If x=0 -- in which case |x|≤1 -- then f(x) = x², so f(0) = 0² = 0.
Case 2: If x=±1/2 -- in which case |x|≤1 -- then f(x) = x², so f(±1/2) = (±1/2)² = 1/4.
Case 3: If x=±1 -- in which case |x|≤1 -- then (x) = x², so f(±1) = (±1)² = 1.
Case 4: If x=±2 -- in which case |x|>1 -- then f(x) = 1/x², so f(±2) = 1/(±2)² = 1/4.

The cases above illustrate that there are only THREE OPTIONS for f(x):
f(x) = 0.
f(x) = a POSITIVE fraction.
f(x) = 1.

Statement 1: f(-a) = 1/f(b)
f(-a) * f(b) = 1.
Given the 3 options for f(x), it is possible that f(-a) * f(b) = 1 only if f(-a) = 1 and f(b) = 1.
Since f(-a) = 1, we know that -a=±1, as shown in Case 3 above.
Thus, a=±1.
Result:
It is NOT true that -0.9 < a < 0.9.
SUFFICIENT.

Statement 2: a = 1/b
ab = 1.
It's possible that a=b=1, in which case it is NOT true that -0.9 < a < 0.9.
It's possible that a=1/2 and b=2, in which case -0.9 < a < 0.9.
INSUFFICIENT.

The correct answer is A.
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