kakz wrote:If yz does not equal zero. Is 0<y<1?
(1)y<1/y
(2)y=z^2
Statement 1: y < 1/y.
The CRITICAL POINTS are where y = 1/y or where the inequality is undefined.
y = 1/y when y=1 or y=-1.
y < 1/y is undefined when y=0.
When y is any other value, y < 1/y or y > 1/y.
Thus, there are 4 ranges to consider: y < -1, -1<y<0, 0<y<1, and y>1.
To determine the range of y, test one value to the left and right of each critical point.
If y = -2, then the inequality becomes -2 < -1/2.
This works. Thus, y<-1 is part of the range.
If y = -1/2, then the inequality becomes -1/2 < -2.
Doesn't work. Thus, -1<y<0 is not part of the range.
If y = 1/2, then the inequality becomes 1/2 < 2.
This works. Thus, 0<y<1 is part of the range.
If y = 2, then the inequality becomes 2 < 1/2.
Doesn't work. Thus, y>1 is not part of the range.
Thus, it's possible that y<-1 or that 0<y<1.
INSUFFICIENT.
Statement 2: y=z².
Since the square of a number cannot be negative, and it is given that yz≠0, we know that y>0.
Thus, it's possible that y=1/2 (in which case 0<y<1) or that y=2 (in which case y>1).
INSUFFICIENT.
Statements 1 and 2 combined:
The only range that satisfies both statements is 0<y<1.
SUFFICIENT.
The correct answer is
C.
An algebraic way to determine the CRITICAL POINTS of y<1/y is to multiply by y².
Since y²≥0, we don't have to worry about changing the direction of the inequality:
y²(y) < y²(1/y)
y³ < y
y³ - y < 0
y(y² - 1) < 0
y(y+1)(y-1) < 0.
The CRITICAL POINTS are where y(y+1)((y-1) = 0:
y=-1, y=0 and y=1.
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