Let $$f$$ be a twice differentiable function defined on $$R$$ such that $$f(0) = 1$$, $$f'(0) = 2$$ and $$f'(x) \neq 0$$ for all $$x \in R$$. If $$\begin{vmatrix} f(x) & f'(x) \\ f'(x) & f''(x) \end{vmatrix} = 0$$, for all $$x \in R$$, then the value of $$f(1)$$ lies in the interval

We are given that $$f$$ is twice differentiable on $$\mathbb{R}$$ with $$f(0) = 1$$, $$f'(0) = 2$$, $$f'(x) \neq 0$$ for all $$x$$, and $$\begin{vmatrix} f(x) & f'(x) \\ f'(x) & f''(x) \end{vmatrix} = 0$$ for all $$x$$.

Expanding the determinant gives $$f(x) \cdot f''(x) - [f'(x)]^2 = 0$$, which can be rewritten as $$\frac{f(x) \cdot f''(x) - [f'(x)]^2}{[f(x)]^2} = 0$$.

This is exactly $$\frac{d}{dx}\left(\frac{f'(x)}{f(x)}\right) = 0$$. Therefore, $$\frac{f'(x)}{f(x)} = k$$ for some constant $$k$$.

Using the initial conditions, $$k = \frac{f'(0)}{f(0)} = \frac{2}{1} = 2$$. So $$\frac{f'(x)}{f(x)} = 2$$, which gives $$\frac{d}{dx}[\ln|f(x)|] = 2$$.

Integrating, $$\ln|f(x)| = 2x + C_1$$, so $$f(x) = Ce^{2x}$$. Using $$f(0) = 1$$, we get $$C = 1$$, and therefore $$f(x) = e^{2x}$$.

Now, $$f(1) = e^2 \approx 7.389$$, which lies in the interval $$(6, 9)$$.

Therefore, the value of $$f(1)$$ lies in the interval $$(6, 9)$$.