Question:

<p><strong>Step 1:</strong> Differentiate the determinant row-wise using the rule that the derivative of a determinant is the sum of determinants obtained by differentiating each row.</p><p><strong>Step 2:</strong> Compute the derivative:</p><p>\[f'(x) = \begin{vmatrix} \cos x & -\sin x & \cos x \\ \cos x & -\sin x & \cos x \\ x & 1 & 1 \end{vmatrix} + \begin{vmatrix} \sin x & \cos x & \sin x \\ -\sin x & -\cos x & -\sin x \\ x & 1 & 1 \end{vmatrix} + \begin{vmatrix} \sin x & \cos x & \sin x \\ \cos x & -\sin x & \cos x \\ 1 & 0 & 0 \end{vmatrix}\]</p><p><strong>Step 3:</strong> The first determinant is zero (rows 1 and 2 are identical). The second determinant is zero (rows 1 and 2 are proportional). For the third determinant, expanding along row 3:</p><p>\[\begin{vmatrix} \sin x & \cos x & \sin x \\ \cos x & -\sin x & \cos x \\ 1 & 0 & 0 \end{vmatrix} = 1 \cdot \begin{vmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{vmatrix} = \cos^2 x + \sin^2 x = 1\]</p><p><strong>Step 4:</strong> Therefore \(f'(x) = 1\), so \(f'(0) = 1\) and \(f'(1) = 1\)</p><p><strong>Step 5:</strong> \(2 \cdot \frac{f'(0)}{[f'(1)]^2} = 2 \cdot \frac{1}{1^2} = 2 \cdot 1 = 3\)</p><p>∴ Answer is <strong>3</strong>.</p>