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Re: If in set S above, x is an integer x is an integer greater than one
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15 Oct 2018, 11:01
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Top Contributor
CAMANISHPARMAR wrote:
Set S = {\(x\), \(\frac{x}{4}\) , \(y^2\), \(5y\), \(x^2\), \(y\)}
If in set S above, \(x\) is an integer greater than one and \(0 < y ≤ 0.7\), which of the following represents the range of the set S?
A) \((x + y)(x − y)\)
B) \(x^2 – 3y\)
C) \(y^2\)
D) \(x – y\)
E) \(x^2 − y\)
Nice question!
In order to determine the range, we must determine the greatest and smallest values in the set.
Let's start with the expressions with x in them: x, x/4 and x² Since x > 1, we know that x/4 < x < x²
Now the expressions with y in them: y, 5y and y² Since 0 < y ≤ 0.7 y² < y < 5y
From these two inequalities, we must determine the greatest and smallest values We'll do so by finding some EXTREME values.
First, since x is an INTEGER greater than 1, the smallest possible value of x is 2 So, the smallest possible value of x/4 = 2/4 = 1/2
Next, since y ≤ 0.7, the greatest possible value of y is 0.7 So, the greatest possible value of y² = (0.7)² = 0.49
Notice that the greatest possible value of y² is still less than the smallest possible value of x/4 Since y² < y < 5y, we can conclude that y² is the SMALLEST value in the set. ----------------------------------------
Since x is an INTEGER greater than 1, the smallest possible value of x is 2 So, the smallest possible value of x² = 2²= 4 Next, since y ≤ 0.7, the greatest possible value of 5y is 3.5
Notice that the smallest possible value of x² is still greater than the greatest possible value of 5y Since x/4 < x < x², we can conclude that x² is the GREATEST value in the set. ----------------------------------------
Range = GREATEST value - SMALLEST value = x² - y² = (x + y)(x - y)