vjsharma25 wrote:
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
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.