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LEt F: R->R is a differntiable function ...

LEt `F: R->R `is a differntiable function `f(x+2y) =f(x) + f(2y) +4xy` for all `x,y in R`

A

`f'(1)=f'(0)+1`

B

`f'(1)=f'(0)-1`

C

`f'(0)=f'(1)+2`

D

`f'(0)=f'(1)-2`

Text Solution

Verified by Experts

The correct Answer is:
D
We have
`f(x+2y)=f(x)+f(2y)+4xy" for all "x,y inR`
Putting x=y=0, we get f(0)=0
Now, `f(x+2y)=f(x)+f(2y)+4xy`
`Rightarrow (f(x+2y)-f(x))/(2y)=2x+(f(2y))/(2y)`
`Rightarrow underset(y to 0)lim (f(x+2y)-f(x))/(2y)=underset( y to 0)lim {2x+(f(2y)-f(0))/(2y)}`
`Rightarrow f'(x)=2x+f'(0)"for all x"`
`Rightarrow f'(1)=2+f'(0)" "["Replacing x by 1"]`
`Rightarrow f'(0)=f'(1)-2`
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