Question:

Let the function $\phi(x)$ be defined by the determinant: $$ \phi(x) = \begin{vmatrix} 2 & 3 & -1 \\ \log_e(1+x^2) & e^x & \sin x \\ \cos x & \tan x & \sin x \end{vmatrix} $$ for $x \geq 0$. Step 1: Evaluate $\phi(x)$ at $x=0$. Substitute $x=0$ into the determinant: $$ \phi(0) = \begin{vmatrix} 2 & 3 & -1 \\ \log_e(1+0^2) & e^0 & \sin 0 \\ \cos 0 & \tan 0 & \sin 0 \end{vmatrix} $$ $$ \phi(0) = \begin{vmatrix} 2 & 3 & -1 \\ \log_e(1) & 1 & 0 \\ 1 & 0 & 0 \end{vmatrix} $$ $$ \phi(0) = \begin{vmatrix} 2 & 3 & -1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{vmatrix} $$ Step 2: Calculate $\phi(0)$. Expand the determinant along the second row: $$ \phi(0) = -0 \cdot \begin{vmatrix} 3 & -1 \\ 0 & 0 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & -1 \\ 1 & 0 \end{vmatrix} - 0 \cdot \begin{vmatrix} 2 & 3 \\ 1 & 0 \end{vmatrix} $$ $$ \phi(0) = 1 \cdot (2 \cdot 0 - (-1) \cdot 1) $$ $$ \phi(0) = 1 \cdot (0 + 1) = 1 $$ Since $\phi(0) = 1 \neq 0$, it follows that $\phi(x)$ is not identically zero. Furthermore, $\phi(x)$ is not divisible by $x$, as divisibility by $x$ would imply $\phi(0)=0$. Step 3: Calculate $\phi'(x)$. The derivative of a determinant is the sum of determinants obtained by differentiating each row (or column) in turn. $$ \phi'(x) = \begin{vmatrix} \frac{d}{dx}(2) & \frac{d}{dx}(3) & \frac{d}{dx}(-1) \\ \log_e(1+x^2) & e^x & \sin x \\ \cos x & \tan x & \sin x \end{vmatrix} + \begin{vmatrix} 2 & 3 & -1 \\ \frac{d}{dx}(\log_e(1+x^2)) & \frac{d}{dx}(e^x) & \frac{d}{dx}(\sin x) \\ \cos x & \tan x & \sin x \end{vmatrix} + \begin{vmatrix} 2 & 3 & -1 \\ \log_e(1+x^2) & e^x & \sin x \\ \frac{d}{dx}(\cos x) & \frac{d}{dx}(\tan x) & \frac{d}{dx}(\sin x) \end{vmatrix} $$ $$ \phi'(x) = \begin{vmatrix} 0 & 0 & 0 \\ \log_e(1+x^2) & e^x & \sin x \\ \cos x & \tan x & \sin x \end{vmatrix} + \begin{vmatrix} 2 & 3 & -1 \\ \frac{2x}{1+x^2} & e^x & \cos x \\ \cos x & \tan x & \sin x \end{vmatrix} + \begin{vmatrix} 2 & 3 & -1 \\ \log_e(1+x^2) & e^x & \sin x \\ -\sin x & \sec^2 x & \cos x \end{vmatrix} $$ The first determinant is zero because its first row consists entirely of zeros. $$ \phi'(x) = \begin{vmatrix} 2 & 3 & -1 \\ \frac{2x}{1+x^2} & e^x & \cos x \\ \cos x & \tan x & \sin x \end{vmatrix} + \begin{vmatrix} 2 & 3 & -1 \\ \log_e(1+x^2) & e^x & \sin x \\ -\sin x & \sec^2 x & \cos x \end{vmatrix} $$ Step 4: Evaluate $\phi'(x)$ at $x=0$. Substitute $x=0$ into the expression for $\phi'(x)$: $$ \phi'(0) = \begin{vmatrix} 2 & 3 & -1 \\ \frac{2(0)}{1+0^2} & e^0 & \cos 0 \\ \cos 0 & \tan 0 & \sin 0 \end{vmatrix} + \begin{vmatrix} 2 & 3 & -1 \\ \log_e(1+0^2) & e^0 & \sin 0 \\ -\sin 0 & \sec^2 0 & \cos 0 \end{vmatrix} $$ $$ \phi'(0) = \begin{vmatrix} 2 & 3 & -1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{vmatrix} + \begin{vmatrix} 2 & 3 & -1 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{vmatrix} $$ Calculate the first determinant: Expand along the third row. $$ 1 \cdot \begin{vmatrix} 3 & -1 \\ 1 & 1 \end{vmatrix} - 0 + 0 = 1(3 \cdot 1 - (-1) \cdot 1) = 1(3+1) = 4 $$ Calculate the second determinant: Expand along the second row. $$ -0 + 1 \cdot \begin{vmatrix} 2 & -1 \\ 0 & 1 \end{vmatrix} - 0 = 1(2 \cdot 1 - (-1) \cdot 0) = 1(2-0) = 2 $$ Therefore, $$ \phi'(0) = 4 + 2 = 6 $$ Since $\phi'(0) = 6 \neq 0$, it follows that $\phi'(x)$ is not identically zero. Conclusion: The value of $\phi(0)$ is $1$. This demonstrates that $\phi(x)$ is not identically zero and is not divisible by $x$. The value of $\phi'(0)$ is $6$. This demonstrates that $\phi'(x)$ is not identically zero.