Let m and M be respectively the minimum and maximum values of the determinant
$\begin{vmatrix}cos^2x & 1+sin^2x& sin 2x\\1+cos^2x& sin^2x& sin 2x\\cos^2x& sin^2x& 1+sin 2x\end{vmatrix}$
Then the ordered pair (m, M) is equal to:
Answer & explanation
Correct answer: option 3
The correct answer is option (3) : (-3, -1)
Let $Δ=\begin{vmatrix}cos^2x & 1+sin^2x& sin 2x\\1+cos^2x& sin^2x& sin 2x\\cos^2x& sin^2x& 1+sin 2x\end{vmatrix}$. Then
Applying $C_1→C_1+C_2, $ we obtain
$Δ=\begin{vmatrix}2 & 1+sin^2x& sin 2x\\2 & sin^2x & sin 2x\\2 & sin^2x& 1+sin 2x\end{vmatrix}$.
Applying $R_1→R_1-2R_3, R_2 → R_2-2R_3$, we obtain
$Δ=\begin{vmatrix}0 & cos^2x& -(2+sin 2x)\\0 & -sin^2x & -(2+sin 2x)\\1 & sin^2x& 1+sin 2x\end{vmatrix}$
Applying $R_1→R_1-R_2,R_3→R_3+R_2, $ we obtain
$⇒Δ= \begin{vmatrix}0 & 1 & 0\\0 & -sin^2x & -(2+sin 2x)\\1 &0 & -1\end{vmatrix}= -(2+sin2x)$
Now, $-1≤sin\, 2x ≤1$
$⇒-1≤sin\, 2x ≤ 1 ⇒-3≤-2-sin\, 2x ≤ -1, ⇒ -3≤ Δ≤ -1$
Hence, $m=-3$ and $M=-1.$