Is ∣x∣<1?

[VJesus12]

Is ∣x∣<1? $$1)\ \ \ x=\frac{1}{3+y^2}$$ $$2)\ \ \ y=-2$$ The OA is the option A.

Could someone tell me how can I show that the first statement is sufficient? Thanks in advance.

[Jay@ManhattanReview]

Is ∣x∣<1? $$1)\ \ \ x=\frac{1}{3+y^2}$$ $$2)\ \ \ y=-2$$ The OA is the option A.

Could someone tell me how can I show that the first statement is sufficient? Thanks in advance.

We have to ascertain whether |x| < 1.

The minimum value of |x| is 0, thus, we have to determine whether 0 ≤ |x| < 1.

Let's take each statement one by one.

1. x = 1/(3 + y^2)

Though the value of y is not given, we must not dismiss Statement 1 only because of this reason.

Note that the minimum value of y^2 is 0 as y^2 is a non-negative number. Whether we decrease or increase the value of y, the value of y^2 would increase, thereby increasing the value of the denominator of x, i.e., (3 + y^2).

As the value of (3 + y^2) increases, the value of 1/(3 + y^2) decreases.

This implies that 0 < 1/(3 + y^2) < 1/3.

=> |x| < 1. Sufficient.

2. y = -2

We do not have any information on the relationship of y with x. Insufficient.

The correct answer: A

Hope this helps!

-Jay
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