Matrices & Determinants
Determinants
Grade Class 12
Question:
Let $f(x) = \begin{vmatrix} 1+\sin^2 x & \cos^2 x & \sin 2x \\ \sin^2 x & 1+\cos^2 x & \sin 2x \\ \sin^2 x & \cos^2 x & 1+\sin 2x \end{vmatrix}, x \in \left[\frac{\pi}{6}, \frac{\pi}{3}\right]$. If $\alpha$ and $\beta$ respectively are the maximum and the minimum values of $f$, then
(1) $\beta^2 - 2\sqrt{\alpha} = \frac{19}{4}$
(2) $\beta^2 + 2\sqrt{\alpha} = \frac{19}{4}$
(3) $\alpha^2 - \beta^2 = 4\sqrt{3}$
(4) $\alpha^2 + \beta^2 = \frac{9}{2}$
Step-by-Step Solution
Key Concept: Apply row operations to simplify the determinant. R1 -> R1 - R3 and R2 -> R2 - R3. This will simplify the determinant to a form where it can be easily evaluated as a function of sin(2x).
Applying $R_1 \to R_1 - R_3$ and $R_2 \to R_2 - R_3$, we get $f(x) = \begin{vmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ \sin^2 x & \cos^2 x & 1+\sin 2x \end{vmatrix} = 1(1+\sin 2x + \cos^2 x) - 0 + (-1)(0 - \sin^2 x) = 1 + \sin 2x + \cos^2 x + \sin^2 x = 2 + \sin 2x$. Since $x \in [\pi/6, \pi/3]$, $2x \in [\pi/3, 2\pi/3]$. Thus, $\sin 2x \in [\sqrt{3}/2, 1]$. The maximum value $\alpha = 2+1 = 3$ and the minimum value $\beta = 2 + \sqrt{3}/2$. Checking option (1): $\beta^2 - 2\sqrt{\alpha} = (2 + \sqrt{3}/2)^2 - 2\sqrt{3} = 4 + 3/4 + 2\sqrt{3} - 2\sqrt{3} = 19/4$.
Correct Answer: (1)