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If f(x)=x^2+2b c+2c^2a n dg(x)=-x^2-2c x...

If `f(x)=x^2+2b c+2c^2a n dg(x)=-x^2-2c x+b^2` are such that min `f(x)> m a xf(x)` , ten he relation between `ba n dc` is a. no relation b. `0

A

No real value of b and c

B

`0 le clt b sqrt(2)`

C

`|c|lt|b|sqrt(2)`

D

`|c|ge|b|sqrt(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
Given `f(x) = x ^(2) + 2bx + 2c^(2) and g(x) = - x ^(2) - 2 cx + b ^(2)`
Then `f (x) ` is minimum and `g(x)` is maximum at
`(x = (-b)/(4a) and f(x) = (-D)/( 4a))`, respectively.
` therefore min f(x) = (- (4 b^(2) - 8c^(2) ))/( 4) = ( 2c^(2) - b ^(2))`
and `max g(x) = - ( 4c^(2) + 4 b^(2))/( 4(-1)) = (b ^(2) + c ^(2))`
Now, ` minf(x) gt max g (x)`
`rArr " " 2c ^(2) - b ^(2) gt b ^(2) + c ^(2)`
`rArr " "c ^(2) gt 2 b ^(2)`
`rArr " " |c| gt sqrt 2 |b|`
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