Let f : R $$\to$$ (0, $$\infty$$) and g : R $$\to$$ R.

f(x) > 0 $$\forall$$x $$\in$$ R

It is given that f'(2) = 0, g(2) = 0, f''(2) $$\ne$$ 0 and g'(2) $$\ne$$ 0.

It is also given that

$$\mathop {\lim }\limits_{x \to 2} {{f(x)g(x)} \over {f'(x)g'(x)}} = 1$$ $$\left( {{0 \over 0}} \right)$$

Applying L' Hospital rule, we get

$$\mathop {\lim }\limits_{x \to 2} {{f'(x)g(x) + g'(x)f(x)} \over {f''(x)g'(x) + f'(x)g''(x)}} = 1$$

For finite limit, we get

$${{f'(2)g(2) + g'(2)f(2)} \over {f''(2)g'(2) + f'(2)g''(2)}} = 1$$

$${{g'(2)f(2)} \over {f''(2)g'(2)}} = 1$$

$${{f(2)} \over {f''(2)}} = 1 \Rightarrow f''(2) = f(2) > 0$$ and f'(2) = 0 which means that f(x) has local minima at x = 2.

Hence, option (A) is correct.

$$f(2) - f''(2) = 0$$

Therefore, we can say that $$f(x) - f''(x) = 0$$ has at least one solution in x $$\in$$ R.

Hence, option (D) is correct.