if g(x)=max|y^2-xy| (0≤y≤1) then the minimum value of g(x) for real x

Hi Debdutta,

As y is always positive, let us see what happens in these cases

1. x>0, in this case we are subtracting a positive term from y^2 and hence possibility of minimum

2. x<0, in this case we are subtracting a negative term from y^2 and hence a possibility for maximum.

So we say g(x) = max|y^2 - xy|, x>0 for g(x) = minimum.

Now say y lies in (0,1] ie y is not equal to zero. In this case there will always exist a positive x where g(x) is always > 0.

And hence the minimum value will be possibile for x=y, ie the minimum value will be 0.

In case y=0, then g(x) = 0 for all x, and in this case also minimum value is 0.

Hence minimum value of g(x) is clearly 0.

Hope it helps.

Regards,

Ashwin (IIT Madras).

sorry mr ashwin by mistake i approved ur ans but the ans is 3-8^1/2 i think u r not considering the word max before mod

Hi Debdutta,

3-√8 is definitely not the right answer. You could have checked that on your own.

I have considered the word MAX in the question.

Clearly 0 is lesser than 3-√8, as 3-√8 is a positive number.

So minimum value has to be 0.

If somewhere you have seen the answer to this question as 3-√8, then it is wrong.

(Or else you have not mentioned some condition in the question)

Thanks.

Regards,

Ashwin (IIT Madras).