The concept being tested here is odd exponents preserve the sign of the base while even exponents do not.
For example
\((-2)^6 =+64\)
\((2)^6 = +64\)
\((-2)^6 =+64 = (2)^6 = +64\)
\((-2)^5 = -32\)
\((2)^5 = +32\)
\((-2)^5 = -32 \neq (2)^5 = +32\)
Now, if all three, \(a\),\(b\),\(c\) are positive, then the inequality is easily satisfied.
But it can also be satisfied if \(a\) and \(c\) are negative. Explanation below.
Now since \(a^3b^4c^7 > 0\), we can conclude that the sign of \(b\) is immaterial to satisfying the inequality. However, both \(a\) and \(c\) have to be negative so that \(a^3\) and \(c^7\) are negative and when multiplied together become positive to satisfy the inequality.
Therefore, \(a\) is negative, \(c\) is negative and \(b\) can be positive or negative.
A. ab : \(a=-ve\), \(b=+ve\)
or \(b=-ve\), thus \(ab=+ve\)
or \(-ve\), thus \(ab\) could be negative
B. abc : \(a=-ve\), \(b=+ve\)
or \(-ve\),\(c=-ve\), thus \(abc = +ve\)
or \(-ve\), thus \(abc\) could be positive
C. ac : \(a=-ve\) and \(c=-ve\), thus \(ac=+ve\), thus \(ac\)
must be positive
Therefore,
C is the correct answer.
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