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Re: If -2 < x < 1/2 and 7 < y 10, what [#permalink]
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chetan2u wrote:
If \(-2 < {x} \le {\frac{1}{2}}\) and \(7 < {y} \le 10\), what is the least value of \(x^2*y\) possible?

A. \(2\)
B. \(\frac{7}{4}\)
C. \(0\)
D. \(-8\)
E. \(-14\)



Irrespective of the range of x, the variable x^2 will always non-negative. If we take into consideration the value of x, which is between -2 and 1/2, we can very easily conclude that the least value of x^2 can be 0, when x = 0.

Since we need to find the least value of x^2 * y, we can easily say it has to be zero, since 0 * anything is always 0.

The correct answer is Option C.
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If -2 < x < 1/2 and 7 < y 10, what [#permalink]
chetan2u wrote:
If \(-2 < {x} \le {\frac{1}{2}}\) and \(7 < {y} \le 10\), what is the least value of \(x^2*y\) possible?

A. \(2\)
B. \(\frac{7}{4}\)
C. \(0\)
D. \(-8\)
E. \(-14\)



From the asked question, we can conclude that whatever be the value of "x", it can't be negative as x has even exponent.

"y" has a positive range. Therefore eliminate (D) & (E) as answer will always be positive or zero.

Least value of x will be "0" as all the other values (negative or positive) will give positive result for x^2 .

Take any value for "y" from the specified range,

Take x = 0 , y = 8

Least value of x^2 * y will always be "0"

(C)
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If -2 < x < 1/2 and 7 < y 10, what [#permalink]
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