If |b|>=2 and x = |a|*b, then which of the following is always true?

I don't think you need to plug in any values to check the inequality.
I don't understand your approach.Can you please elaborate a bit more.

If you consider both the values for the first option then it will not be true always
if \(a>0\) the inequality will become
\(a-a*b^2\geq 0\) will be -ve (for some finite values of \(a\),excluding \(0\)).
and when \(a<0\),the inequality will become
\(-a+a*b^2\geq 0\) will be +ve (for some finite values of \(a\),excluding \(0\)).

So this will not be always true.Thats why I asked the question

|b| >= 2. This means that b^2 >= 4. On the number line you will get two roots. (2,-2) both inclusive. Hence b >= 2 and b <=-2

x = |a| b
x can be +ve or -ve. since b can be +ve or -ve.
Since b can be +ve or -ve. xb can be +ve or -ve.

Ultimately you will have to plugin "smart values" to determine the values of a-xb and a+xb.

This is definitely a tough question.

You are right about everything.But we have to look at the expression \(x*b\) which will become \(|a|b^2\) after plugging value of \(x\),so \(b^2\) is always going to be +ve. It doesn't depend upon the value of \(b\) at all.

Inequality is always \(a - |a|*b^2\)

Also \(|a|*b^2\) is always positive irrespective of the sign of a or b (except when a=b=0, of course)

If a is negative than we are subtracting a positive term from a negative term, so it is always negative

If a is positive than we are subtracting from 'a' a positive term that is 4 times 'a', so again always negative.

Remember that sign of 'a' doesnt change the fact that the expression we are evaluating is still a - |a|*b^2


Therefore, I don't think plugging in is required in this case. We are told that \(b^2 >=4\)and \(x = |a|* b\)

Since question is asking us about the signs of \(a-b*x\) and \(a+b*x\), we need to work scenarios for both of them

\(a-b*x\) reduces to \(a-b^2*|a|\) Now \(b^2 is >=4\) and \(|a|\)is always positive so we are always subtracting a higher number than 'a' from 'a' (irrespective of whether 'a' is negative or positive) so \(a-b^2*|a|\)will always be <=0


\(a+b*x\) reduces to \(a+b^2*|a|\) Now \(b^2 is >=4\) and \(|a|\) is always positive so we are always adding a higher number than 'a' to 'a' (irrespective of whether 'a' is negative or positive) so \(a+b^2*|a|\)will always be >=0

Hence, E is correct option.

That's an awesome interpretation ! Guys thank you so much ! beyondgmatscore I love your quant skills.

\(a + x*b\)
\(= a + |a|b^2\)

as \(b^2>=4\) so,

\(= a + |a|4\) ..... the value of this expression will always be +ve because 4|a|, a bigger +ve term, is added in a smaller term.
\(a - |a|4\) ..... this will always be -ve because 4|a|, a bigger +ve term, is subtracted from a smaller term.

therefore,

\(a + x*b >= 0\)
\(a - x*b <= 0\)

E. A could be 0, positive or negative. but the x=|a| * b tells you that x will be the same sign as b. so if b is negative x will be negative and likewise if b is positive so will x. Looking at the choices both B and D hold true.

There are two expressions

a + x*b = a + |a|b^2 = a (1 +/- b^2)
when a<0 = a(1 - b^2) > 0
when a>=0 = a(1 + b^2) >=0

a - x*b = a - |a|b^2 = a (1 -/+ b^2)
when a<0 =a(1+b^2)< 0
when a>0 =a(1-b^2)< 0

So its must be E

E
There are basically two equations to solve (considering B^2 >= 4)
?a - (+ab^2) .... always negative doesn't matter what ?a is...
?a + (+ab^2) .... always positive doesn't matter what ?a is...
Thus B & D = E!

this question is really tough and i don't understand the explanation given above. Requesting Math expert to provide easy and logical explanation by dissecting every answer choices.

Hi Shapla,

The answer choices show two different expressions
\(a−x∗b\)
\(a+x∗b\)

and the question asks whether it is true that one of the expressions is always positive or always negative, or a combination of two options (answer choice E).

The question stem tells us that \(|b|\geq{2}\), and that \(x=|a|*b\). If we plug in that definition for x into each of the two expressions, then we get:

\(a-|a|*b*b\) = \(a-|a|*b^{2}\)
\(a+|a|*b*b\) = \(a+|a|*b^{2}\)

And since we know that \(b^{2}\geq{4}\) (because \(|b|\geq{2}\)), then we can plug in the value of 4 for \(b^{2}\) and the expressions become:

\(a-4*|a|\)
\(a+4*|a|\)

You might ask "But how can we assume the value of 4 for \(b^{2}\), when it could be anything greater than 4 too?"

If you look at the expressions, if we make \(b^{2}\) bigger than 4, it won't affect the sign of the expression. Note that \(4*|a|\) will always be positive, and always have a larger magnitude than \(a\). So, now when we look at the expressions, specifically the sign of the expressions, regardless of the sign of \(a\), we can see that:

\(a-4*|a|\) will always be \(\leq{0}\)
\(a+4*|a|\) will always be \(\geq{0}\)

Therefore the answer is E

I hope that cleared up your doubts.

thanks davedekoos

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