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Let f:R to R be given by f(x+y)=f(x)-f(y...

Let `f:R to R` be given by `f(x+y)=f(x)-f(y)+2xy+1"for all "x,y in R` If f(x) is everywhere differentiable and `f'(0)=1`, then f'(x)=

A

2x+1

B

2x-1

C

x+1

D

x-1

Text Solution

Verified by Experts

The correct Answer is:
B
We have
`f(x+y)=f(x)-f(y)+2xy+1"or all "x,y in R..........(i)`
Putting x=y=0, we get
`f(0)=f(0)-f(0)+0+1 Rightarrow f(0)=1`
Now,
`f'(x)=underset(h to 0)lim (f(x+h)-f(x))/(h)`
`Rightarrow f'(x)=underset(h to 0)lim (f(x)-f(h)+2xh+1-f(x))/(h)`
`Rightarrow f'(x)=underset(h to 0)lim {2x-(f(h))/(h)1}`
`Rightarrow f'(x)=2x-underset(h to 0)lim (f(h)-f(0))/(h)=2x-f'(0)=2x-1`
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