Question

If $\Delta = \begin{vmatrix} 1 & 3\cos\theta & 1 \\ \sin\theta & 1 & 3\cos\theta \\ 1 & \sin\theta & 1 \end{vmatrix}$, then maximum value of $\Delta$ is:

• A. 1
• B. 4
• C. 9
• D. 10

Answer 1

Answer: (D)

10

Answer 2

To find the maximum value of $\Delta$, we first expand the determinant:

$\Delta = 1(1 - 3\sin\theta\cos\theta) - 3\cos\theta(\sin\theta - 3\cos\theta) + 1(\sin^2\theta - 1)$ $\Delta = 1 - 3\sin\theta\cos\theta - 3\sin\theta\cos\theta + 9\cos^2\theta + \sin^2\theta - 1$ $\Delta = \sin^2\theta - 6\sin\theta\cos\theta + 9\cos^2\theta$ $\Delta = (\sin\theta - 3\cos\theta)^2$

To maximize $\Delta$, we need to maximize $|\sin\theta - 3\cos\theta|$. The maximum value of $A\sin\theta + B\cos\theta$ is $\sqrt{A^2 + B^2}$. In this case, $A = 1$ and $B = -3$, so the maximum value of $\sin\theta - 3\cos\theta$ is $\sqrt{1^2 + (-3)^2} = \sqrt{10}$.

Therefore, the maximum value of $\Delta$ is $(\sqrt{10})^2 = 10$.