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

Hint:

When working with determinants, especially for \( 3 \times 3 \) matrices, it’s essential to break the determinant into smaller \( 2 \times 2 \) minors for each element in the first row. After expanding, simplify the expressions and evaluate each minor. Additionally, be mindful of trigonometric identities like \( \cos^2 x + \sin^2 x = 1 \), which can help simplify the result. Also, remember to check the bounds for maximizing or minimizing the function (in this case, based on \( \sin^2 x \)).

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.

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

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:

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

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

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

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