Let `f(x)=x^(2)+xg'(1)+g''(2)` and `g(x)=f(1)x^(2)+xf'(x)+f''(x)` then `f(g(1))` is equal to

f(x)=x^(2)+xg'(1)+g''(2)and g(x)=f(1)x^(2)+xf'(x)+f'(x). The value of g(0) is

f(x)=x^(2)+xg'(1)+g''(2)and g(x)=f(1)x^(2)+xf'(x)+f'(x). The value of f(3) is

Let f(x)=x^(2)+xg^(2)(1)+g'(2) and g(x)=f(1)*x^(2)+xf'(x)+f''(x) then find f(x) and g(x)

Let f (x) and g (x) be two differentiable functions, defined as: f (x)=x ^(2) +xg'(1)+g'' (2) and g (x)= f (1) x^(2) +x f' (x)+ f''(x). The value of f (1) +g (-1) is:

Let f (x) and g (x) be two differentiable functions, defined as: f (x)^(2)=x ^(2) +xg'(1)+g'' (2) and g (x)= f(1)x^(2) +x f' (x)+ f''(x). The number of integers in the domain of the function F(x)= sqrt(-(f(x))/(g (x)))+sqrt(3-x) is:

f(x)=x^(2)+xg'(1)+g''(2)and g(x)=f(1)x^(2)+xf'(x)+f'(x). The domain of the function sqrt((f(x))/(g(x))) is

Let f(x)=3x^(2)+4xg'(1)+g''(2) and, g(x)=2x^(2)+3xf'(2)+f''(3)" for all "x in R. Then,