Question:
Let m and M be respectively the minimum and maximum values of
\(\begin{vmatrix}\cos^2 x & 1+\sin^2 x & \sin 2x\\ 1+\cos^2 x & \sin^2 x & \sin 2x\\ \cos^2 x & \sin^2 x & 1+\sin 2x\end{vmatrix}\)
Then the ordered pair (m, M) is equal to :
Let m and M be respectively the minimum and maximum values of
\(\begin{vmatrix}\cos^2 x & 1+\sin^2 x & \sin 2x\\ 1+\cos^2 x & \sin^2 x & \sin 2x\\ \cos^2 x & \sin^2 x & 1+\sin 2x\end{vmatrix}\)
Then the ordered pair (m, M) is equal to :
\(\begin{vmatrix}\cos^2 x & 1+\sin^2 x & \sin 2x\\ 1+\cos^2 x & \sin^2 x & \sin 2x\\ \cos^2 x & \sin^2 x & 1+\sin 2x\end{vmatrix}\)
Then the ordered pair (m, M) is equal to :
Show Hint
Use \(R_1\to R_1-R_2\) and \(R_3\to R_3-R_2\) to simplify, then expand; the determinant reduces to \(-(2+\sin 2x)\).
Updated On: Nov 5, 2026
- \((-3, -1)\)
- \((-4, -1)\)
- \((1, 3)\)
- \((-3, 3)\)
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The Correct Option is A
Solution and Explanation
Let \(c=\cos^2x,\ s=\sin^2x,\ t=\sin 2x\), so \(c+s=1\).
Apply \(R_1\to R_1-R_2\) and \(R_3\to R_3-R_2\):
\(R_1=(-1,\ 1,\ 0)\), \(R_2=(1+c,\ s,\ t)\), \(R_3=(-1,\ 0,\ 1)\).
Expand along \(R_1\): \(\Delta=-1\,(s\cdot 1-t\cdot 0)-1\,\big((1+c)\cdot 1-t\cdot(-1)\big)=-s-(1+c+t)\).
\(\Delta=-(s+c)-1-t=-2-\sin 2x\).
Since \(-1\le \sin 2x\le 1\), \(\Delta\) lies between \(-2-1=-3\) (when \(\sin 2x=1\)) and \(-2+1=-1\) (when \(\sin 2x=-1\)).
So \(m=-3\) and \(M=-1\).
Answer: (1) \((-3, -1)\)
Apply \(R_1\to R_1-R_2\) and \(R_3\to R_3-R_2\):
\(R_1=(-1,\ 1,\ 0)\), \(R_2=(1+c,\ s,\ t)\), \(R_3=(-1,\ 0,\ 1)\).
Expand along \(R_1\): \(\Delta=-1\,(s\cdot 1-t\cdot 0)-1\,\big((1+c)\cdot 1-t\cdot(-1)\big)=-s-(1+c+t)\).
\(\Delta=-(s+c)-1-t=-2-\sin 2x\).
Since \(-1\le \sin 2x\le 1\), \(\Delta\) lies between \(-2-1=-3\) (when \(\sin 2x=1\)) and \(-2+1=-1\) (when \(\sin 2x=-1\)).
So \(m=-3\) and \(M=-1\).
Answer: (1) \((-3, -1)\)
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