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Solution :

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