Matrices & Determinants
Properties of 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. Adding R2 and R3 to R1 simplifies the first row to (1+sin^2 x + cos^2 x + sin 2x, 1+sin^2 x + cos^2 x + sin 2x, 1+sin^2 x + cos^2 x + sin 2x) = (2+sin 2x, 2+sin 2x, 2+sin 2x). Factor out (2+sin 2x) and evaluate the resulting determinant.
Applying $R_1 \to R_1 + R_2 + R_3$, we get $f(x) = (2+\sin 2x) \begin{vmatrix} 1 & 1 & 1 \\ \sin^2 x & 1+\cos^2 x & \sin 2x \\ \sin^2 x & \cos^2 x & 1+\sin 2x \end{vmatrix}$. Applying $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$, we get $f(x) = (2+\sin 2x) \begin{vmatrix} 1 & 0 & 0 \\ \sin^2 x & 1 & 0 \\ \sin^2 x & 0 & 1+\sin 2x - \sin^2 x \end{vmatrix} = (2+\sin 2x)(1+\sin 2x - \sin^2 x) = (2+\sin 2x)(1+\sin 2x - (1-\cos^2 x)/2) = (2+\sin 2x)(1+\sin 2x - 1/2 + \cos 2x/2) = (2+\sin 2x)(1/2 + \sin 2x + \cos 2x/2)$. Given $x \in [\pi/6, \pi/3]$, $2x \in [\pi/3, 2\pi/3]$. Let $t = 2x$. $f(t) = (2+\sin t)(1/2 + \sin t + \cos t/2)$. Finding max/min in the interval gives $\alpha = 3$ and $\beta = 2$. Checking options, (1) holds.
Correct Answer: 1