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Bunuel
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Bunuel
If \(f^2 g < 0\), which of the following must be true?


(A) \(f < 0\)

(B) \(g < 0\)

(C) \(fg < 0\)

(D) \(fg > 0\)

(E) \(f^2 < 0\)


Given

\(f^2 g < 0\)

As \(f^2\) can never be negative , g must be negative.

The best answer is B.
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Bunuel
If \(f^2 g < 0\), which of the following must be true?


(A) \(f < 0\)

(B) \(g < 0\)

(C) \(fg < 0\)

(D) \(fg > 0\)

(E) \(f^2 < 0\)

\(f^2 g < 0\)
\(f^2\) is always +ve.
Hence g<0

Answer B
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Bunuel
If \(f^2 g < 0\), which of the following must be true?


(A) \(f < 0\)

(B) \(g < 0\)

(C) \(fg < 0\)

(D) \(fg > 0\)

(E) \(f^2 < 0\)


B, since f^2 will always be positive
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Given that \(f^2 g < 0\)

We know that Square of a number is always >= 0
=> \(f^2\) >= 0

So, for \(f^2 g < 0\), g has to be < 0
As product of two numbers is negative only one one is positive and other is negative.
=> g < 0

So, Answer will be B
Hope it helps!

Watch the following video to learn the Basics of Inequalities

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