Celestial09 wrote:

Hello Experts!

Kindly share your views and methods on this question.

Thanks

Celestial

Hi

Celestial09,

The question is a classic example where students miss out on the concept that \(|x| => 0\) and not \(|x| > 0\). Let's analyze the question statement first to see what exactly we need to find.

Analyze the given information in the questionThe question asks us if \(a^2*b > 0\). We know that \(a^2 => 0\). To answer the question, we need to know for sure if \(a\) and \(b\) are not equal to 0 along with the sign of \(b\). Please note that sign of \(a\) does not make any difference as \(a^2\) will never be negative. With this understanding, let's evaluate the statements now.

Analyze statement-I independentlySt-I tells us that \(|a| = b\). Since \(|a| => 0\), it means that \(b => 0\). Using this information, \(|a| = b = 0\) or \(|a| = b > 0\). Since the statement does not tell us for sure if \(a\) and \(b\) are not equal to 0 we can't say if \(a^2 * b > 0\).

Analyze statement-II independentlySt-II tells us that \(ab < 0\) i.e. \(a\), \(b\) have the opposite signs. So,

i. If \(a > 0\), then \(b < 0\). This would mean \(a^2 * b < 0\)

ii. If \(a < 0\), then \(b > 0\). This would mean that \(a^2 * b > 0\) as \(a^2\) is positive irrespective of the sign of \(a\).

Thus, the statement does not tell us for sure if \(a^2 * b > 0\).

However, the statement tells us about an important nature of both \(a\) and \(b\). Since product of \(a\) and \(b\) is not equal to 0, we can say that neither of \(a\) or \(b\) is equal to zero. This nature of \(a\) and \(b\) would be helpful when we combine the analysis from st-I & II

Combine analysis from st-I & IISt-I tells us that \(|a| = b => 0\) and st-II tells us that \(b < 0\) or \(b > 0\) and \(a\) not equal to 0.

Combining both the statements, we can say that \(|a| = b > 0\). Since none of \(a\) and \(b\) is equal to \(0\) and \(b > 0\), we can safely say that \(a^2 > 0\) and \(b > 0\) i.e. \(a^2* b > 0\).

Hence combining both the statements is sufficient to answer our question.

Hope its clear!

Regards

Harsh

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