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If g(x)=max(y^(2)-xy)(0le yle1), then th...

If `g(x)=max(y^(2)-xy)(0le yle1)`, then the minimum value of g(x) (for real x) is

A

`(1)/(4)`

B

`3-sqrt3`

C

`3+sqrt8`

D

`(1)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
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`
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