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Bunuel
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Bunuel
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stne



\(|a+1|=|b-1|\)

This eqn only holds for certain combinations of \(a\) and \(b\)

i.e. if \(a=4\) then \(b\) has to be either \(6\) or \(-4\)

So \(a-b\) can be either \(-2\) or \(8\)

If \(a=-6\) then \(b\) can be \(6\) or \(-4\)

So \(a-b=-12\) or \(-2\)

Similarly we can verify by taking other numbers too

(1) \(ab>0\)

This implies \(a\) and \(b\) have to have the same sign

So if \(a\) is \(4 \) then \(b\) has to be \(6\) and \(a-b=-2 \)

If \(a=-6\) then \(b\) has to be \(-4\) and \(a-b=-2\)

(other values for a and b will give similar result)

SUFF.

(2) \(\frac{a}{b}≠ -1\)

Again from our pre-work we can see that we cannot have \(a=4\) and \(b=-4\) nor can we have \(a=-6\) and \(b=6\)

Thus the only other possibility is \(a=4\) and \(b=6\) or \(a=-6\) and \(b=-4 \)

Thus \(a-b=-2 \)

(other values for a and b will give similar result)

SUFF.

Ans D

Hope it helps.

But in statement 2, a and b can be 0 right?
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RiyaGangar


But in statement 2, a and b can be 0 right?
No, that is not possible.

0 divided by 0 is undefined.
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