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Let f(x+y)=f(x)+f(y)+2x y-1 for all real...

Let `f(x+y)=f(x)+f(y)+2x y-1` for all real `xa n dy` and `f(x)` be a differentiable function. If `f^(prime)(0)=cosalpha,` then prove that `f(x)>0AAx in Rdot`

Text Solution

Verified by Experts

We have
`f'(x)=underset(hrarr0)lim(f(x+h)-f(x))/(h)`
`=underset(hrarr0)lim(2x+(f(h)-1)/(h))`
Now substituting x= y 0 in the given functional relation, we get
`f(0)=f(0)+f(0)+0-1 or f(0)=1`
`therefore" "f'(x)=2x+underset(hrarr0)lim(f(h)-f(0))/(h)=2x+f'(0)`
Integrating, we get `f(x) = x^(2)+x cos alpha + C.`
Since f(0)=1,
1=C
`" or "f(x)=x^(2)+ cos alpha + 1`.
It is a quadratic in x with discriminant
`D=cose^(2)alpha -4 lt 0`
and coefficient of `x^(2)=1gt0`. Therefore
Alternative method :
`f(x+y)=f(x)+f(y)+2xy-1" (1)"`
Differentiate w.r.t. x keeping as constant
`f'(x+y)=f'(x)+2y`
Put = 0 and y = x. Then
`f'(x)=f'(0)+2x`
`=cos alpha +2x`
`therefore" "f(x)= x cos alpha +x^(2)+c" (2)"`
`Put x=y =0 in (1). Then f(0) = f(0)+f(0)-1 or f(0)=1`.
Then from (2), we get `f(x) = x^(2)+(cos alpha)x + 1.`
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