abhasjha wrote:IS (x+1)/(x-3) <0?
(1) -1<x<1
(2) X^2-4<0.
Determine the CRITICAL POINTS:
The values of x where (x+1)/(x-3) is equal to 0 or is undefined.
(x+1)/(x-3) = 0 when x=-1.
(x+1)/(x-3) is undefined when x=3.
Thus, the critical points are x=-1 and x=3.
To determine where (x+1)/(x-3) < 0, test one value to the left and one value to the right of each critical point.
x<-1:
If x=-2, then (x+1)/(x-3) = (-2+1)/(-2-3) = 1/5.
Since (x+1)/(x-3) > 0 when x=-2, x<-1 is not a valid range.
-1<x<3:
If x=0, then (x+1)/(x-3) = (0+1)/(0-3) = -1/3.
Since (x+1)/(x-3) < 0 when x=0, -1<x<3 is a valid range.
x>3:
If x=4, then (x+1)/(x-3) = (4+1)/(4-3) = 5.
Since (x+1)/(x-3) > 0 when x=4, x>3 is not a valid range.
Thus, (x+1)/(x-3) < 0 when -1<x<3.
Question stem, rephrased:
Is -1<x<3?
Statement 1: -1<x<1
Thus, -1<x<3.
SUFFICIENT.
Statement 2: x²<4
If x=1, then -1<x<3.
If x=-1.5, then it is not true that -1<x<3.
INSUFFICIENT.
The correct answer is
A.
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