MY CART (5)

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
                   

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

3 years ago

Share

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

Post Your Answer

More Questions On Differential Calculus

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!!
Click Here for details
Blue circles r doubts
 
 
Dear Aakanksha, For such kind of problems to be done in a short time it is better to subsitute some value in place of x.If you substitute 0,in f(x) you get it to be 5/4 and g(5/4)=1. Hence...
 
erra akhil 9 days ago
 
by solving f(x) you will get it as constant functn whose value is 5/4 then u can easily find G(F(x))
 
Ajay Kotwal 9 days ago
let f defined on [0,1] be twice differentiable such that |f’’(x)|
 
 
plz post ur complete question so that we can solve
 
Phani 9 days ago
 
Complete the question properly
 
erra akhil 9 days ago
How d n /dx n becomes d n-1 /dx n-1 …...please explain
 
 
Hello student, Please find answer to your question For example, Apply this
  img
Jitender Singh 11 months ago
 
Thanx …........you are awesome sir …....you helped me whenever I required …..this website is awesome ….....
 
milind 11 months ago
A particle prjected from origin moves in x-y plane with a velocity (vect.) v = 3i + 6xj where i and j are the unit vectors along x and y axix respectively. Find the equation of the path...
 
 
Dear Sunrit The path equation is y = x 2 S x = x = ∫v x .dt = 3∫dt = 3t S y = y = ∫v y .dt = ∫6x dt = ∫6(3t) dt = ∫18tdt = 9t 2 Thus y = x 2
 
shashi K Sharma 10 months ago
if 5 C n , 24 C n and 6 C n are in H.P. then n is equal to?
 
 
as it is observed from the question that maximum vakue of n can be 5 but on solving we get value of n is 0 there is no other value of n except 0 approve if useful
 
ng29 2 months ago
 
of course it is wrong if x,y,z arein hp then 1/x,1/y,1/z are in ap so on expanding the factorial we coclude that only possible value is n=0 approve if useful
 
ng29 2 months ago
 
ans. given in the paper is n=2 if it is wrong ...plz could u send me ur solution
 
mansi dabriwal 2 months ago
integrate 1/(a+bcosx)^2 dx
 
 
Ans:Hello student, please find answer to your question Now, simply use the standard form & partial farction
  img
Jitender Singh 11 months ago
View all Questions »