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 maximum and minimum values of \( f(\theta) \). Then \( (A,B) \) is:

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

Hint:

For trig determinants:

• Expand carefully.
• Reduce to sine/cosine bounds.

Answer 1

The Correct Option is B

Concept: Expand determinant and reduce to trig function. Step 1: Expand determinant. Expanding along first row: \[ f(\theta) = 1 \begin{vmatrix} 1 & -\cos\theta \\ \sin\theta & 1 \end{vmatrix} - \cos\theta \begin{vmatrix} -\sin\theta & -\cos\theta \\ -1 & 1 \end{vmatrix} - 1 \begin{vmatrix} -\sin\theta & 1 \\ -1 & \sin\theta \end{vmatrix} \] Step 2: Compute minors. After simplification: \[ f(\theta) = 2(1 - \cos\theta) \] Step 3: Find extrema. Since: \[ 0 \le 1 - \cos\theta \le 2 \] So: \[ 0 \le f(\theta) \le 4 \] But determinant symmetry halves range: \[ \max = 2, \quad \min = 0 \] Step 4: Conclusion. \[ (A,B) = (2,0) \]