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

3 years ago

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Answers : (3)

                                        

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).

3 years ago
                                        

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

3 years ago
                                        

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).

3 years ago

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