InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
If `g(x)=max(y^(2)-xy)(0le yle1)`, then the minimum value of g(x) (for real x) isA. `(1)/(4)`B. `3-sqrt3`C. `3+sqrt8`D. `(1)/(2)` |
|
Answer» Correct Answer - B `y^(2)-xy=(y-(x)/(2))^(2)-(x^(2))/(4) rArr y^(2)-xy` is decreasing for `yle (x)/(2)` and increasing for `yge(x)/(2)` Thus the largest value of `y^(2)-xy` must be at `y=0,(x)/(2)` or 1. The values are `0,(x^(2))/(4),1-x` for `x in (0,1)` And for `x in (0,1)g(x)=max((x^(2))/(4),1-x)` Also `x^(2)+4x=4=0 rArr x = -2 pm sqrt8` `rArr" "g(x)=1-x " for "x le sqrt8-2 and (x^(2))/(4)` for `x ge sqrt8-2` `rArr" g(x) is minimum at "x=sqrt8-2` and minimum value is `3-sqrt8` |
|