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

Answer: (D)

(B), (C), and (D) only

Answer 2

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.

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