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(B) Given the determinant: $\left| \begin{array}{cc} f'(x) & f(x) \ f''(x) & f'(x) \end{array} ight| = 0$This implies $(f'(x))^2 - f(x)f''(x) = 0$.

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(B) Given the determinant: $\left| \begin{array}{cc} f'(x) & f(x) \ f''(x) & f'(x) \end{array} ight| = 0$This implies $(f'(x))^2 - f(x)f''(x) = 0$.Dividing by $(f'(x))^2$,we get $\frac{f'(x)f'(x) - f(x)f''(x)}{(f'(x))^2} = 0$,which is the derivative of the quotient $\frac{f(x)}{f'(x)}$.Thus,$\frac{d}{dx} \left( \frac{f(x)}{f'(x)} ight) = 0$,which means $\frac{f(x)}{f'(x)} = C$ (a constant).Using the initial conditions $f(0) = 1$ and $f'(0) = 2$,we find $C = \frac{f(0)}{f'(0)} = \frac{1}{2}$.So,$\frac{f(x)}{f'(x)} = \frac{1}{2} \Rightarrow \frac{f'(x)}{f(x)} = 2$.Integrating both sides,$\ln|f(x)| = 2x + k$. Since $f(0) = 1$,we get $k = 0$,so $f(x) = e^{2x}$.We need to find the number of solutions for $e^{2x} = x^2$.Let $g(x) = e^{2x} - x^2$. $g'(x) = 2e^{2x} - 2x = 2(e^{2x} - x)$. Since $e^{2x} > x$ for all real $x$,$g'(x) > 0$,so $g(x)$ is strictly increasing.As $x 	o -\infty$,$g(x) 	o \infty$. As $x 	o 0$,$g(0) = 1$. As $x 	o -\infty$,$e^{2x} 	o 0$ and $x^2 	o \infty$. By the Intermediate Value Theorem,there is exactly one solution.

Suppose $\left| \begin{array}{cc} f'(x) & f(x) \\ f''(x) & f'(x) \end{array} \right| = 0$ where $f(x)$ is a continuously differentiable function with $f'(x) \ne 0$ and satisfies $f(0) = 1$ and $f'(0) = 2$. Then the number of solution$(s)$ of the equation $f(x) = x^2$ is equal to:

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