Question

\(\text{ If } \Delta = \begin{vmatrix} 1 & \cos x & 1 \\ -\cos x & 1 & \cos x \\ -1 & -\cos x & 1 \\ \end{vmatrix} , \text{ then:}\)
\((A)\space \Delta = 2(1 - \cos^2 x)\)
\((B)\space \Delta = 2(2 - \sin^2 x)\)
(C) Minimum value of ∆ is 2
(D) Maximum value of \( \Delta \) is 4

• A. (A), (C), and (D) only
• B. (A), (B), and (C) only
• C. (A), (B), (C), and (D)
• D. (B), (C), and (D) only

Answer 1

The Correct Option is D

The determinant \( \Delta \) is given as:

\[\Delta = \begin{vmatrix} 1 & \cos x & 1 \\ -\cos x & 1 & \cos x \\ -1 & -\cos x & 1 \end{vmatrix}.\]

Expand the determinant:

\[ \Delta = 1 \cdot \begin{vmatrix} 1 & \cos x \\ -\cos x & 1 \end{vmatrix} - \cos x \cdot \begin{vmatrix} -\cos x & \cos x \\ -1 & 1 \end{vmatrix} +1 \cdot \begin{vmatrix} -\cos x & 1 \\ -1 & -\cos x \end{vmatrix}.\]

Compute each minor:

\[ \begin{vmatrix} 1 & \cos x \\ -\cos x & 1 \end{vmatrix} = 1 - \cos^2 x, \quad \begin{vmatrix} -\cos x & \cos x \\ -1 & 1 \end{vmatrix} = \cos x -\cos x = 0, \quad \begin{vmatrix} -\cos x & 1 \\ -1 & -\cos x \end{vmatrix} = \cos^2 x - 1. \]

Substitute back:

\[ \Delta = (1 - \cos^2 x) + 0 + (\cos^2 x - 1) = 2 - \sin^2 x.\]

Simplify:

\[\Delta = 2(2 - \sin^2 x).\]

This matches option (B).

The minimum value of \( \Delta \) occurs when \( \sin^2 x = 1 \), giving:

\[\Delta_{\min} = 2(2 - 1) = 2.\]

The maximum value of \( \Delta \) occurs when \( \sin^2 x = 0 \), giving:

\[\Delta_{\max} = 2(2 - 0) = 4.\]

Thus, options (B), (C), and (D) are correct.

\(\boxed{(B), (C), \text{ and } (D) \text{ only}}\).