We have to ascertain whether |x| < 1.VJesus12 wrote: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.
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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