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

290 Points
```										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,
```
5 years ago
17 Points
```										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
```
5 years ago
290 Points
```										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,
```
5 years ago
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