carcass wrote:

If \(x < y < z\) but \(x^2 > y^2 > z^2 > 0\), which of the following must be positive?

A.\(x^3\) \(y^4 z^5\)

B. \(x^3 y^5 z^4\)

C. \(x^4 y^3 z^5\)

D. \(x^4 y^5 z^3\)

E. \(x^5 y^4 z^3\)

Think of the cases in which '\(x < y < z\) but \(x^2 > y^2 > z^2 > 0\)' happens.

A simple case I can think of is all negative numbers: -5 < -4 < -3 but 25 > 16 > 9 > 0

Another thing that comes to mind is that z can be positive as long as its absolute value remains low: -5 < -4 < 3 but 25 > 16 > 9 > 0

We need to find the option that must stay positive:

A.\(x^3\) \(y^4 z^5\)

Will be negative in this case: -5 < -4 < 3

i.e. x negative, y negative, z positive

B. \(x^3 y^5 z^4\)

Will be positive in both the cases.

C. \(x^4 y^3 z^5\)

Will be negative in this case: -5 < -4 < 3

i.e. x negative, y negative, z positive

D. \(x^4 y^5 z^3\)

Will be negative in this case: -5 < -4 < 3

i.e. x negative, y negative, z positive

E. \(x^5 y^4 z^3\)

Will be negative in this case: -5 < -4 < 3

i.e. x negative, y negative, z positive

Notice that for an expression to stay positive, we need the power of both x and y to be either even or both to be odd since x and y are both negative. Also, we need the power of z to be even so that it doesn't affect the sign of the expression. Only (B) satisfies these conditions.

We don't need to consider any other numbers since we have already rejected 4 options using these numbers. The fifth must be positive in all cases.

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Karishma

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