Question

Let \( f(\theta) = \begin{vmatrix} 1 & \cos \theta & -1 \\ -\sin \theta & 1 & -\cos \theta \\ -1 & \sin \theta & 1 \end{vmatrix} \). Suppose \( A \) and \( B \) are respectively the maximum and minimum values of \( f(\theta) \). Then \( (A, B) \) is equal to:

• A. \( (2, 1) \)
• B. \( (2, 0) \)
• C. \( (\sqrt{2}, 1) \)
• D. \( \left(2, \frac{1}{\sqrt{2}}\right) \)

Hint:

Simplify the expression before finding the maximum and minimum values. Utilize the range of trigonometric functions.

Answer 1

The Correct Option is B

Step 1: Calculate the determinant \( f(\theta) \).\n\[\nf(\theta) = 1(1 \cdot 1 - (-\cos \theta)(\sin \theta)) - \cos \theta((-\sin \theta)(1) - (-\cos \theta)(-1)) + (-1)((-\sin \theta)(\sin \theta) - (1)(-1))\n\]\n\[\nf(\theta) = (1 + \sin \theta \cos \theta) - \cos \theta(-\sin \theta - \cos \theta) - (-\sin^2 \theta + 1)\n\]\n\[\nf(\theta) = 1 + \sin \theta \cos \theta + \sin \theta \cos \theta + \cos^2 \theta + \sin^2 \theta - 1\n\]\n\[\nf(\theta) = 2 \sin \theta \cos \theta + (\sin^2 \theta + \cos^2 \theta) - 1 + 1 = \sin(2\theta) + 1\n\]\n\n
Step 2: Determine the maximum value \( A \).
\nThe maximum of \( \sin(2\theta) \) is 1.\n\[\nA = 1 + 1 = 2\n\]\n\n
Step 3: Determine the minimum value \( B \).
\nThe minimum of \( \sin(2\theta) \) is -1.\n\[\nB = 1 + (-1) = 0\n\]\n\n
Step 4: State \( (A, B) \).\n\[\n(A, B) = (2, 0)\n\]