Let m and M be respectively the minimum and maximum values of cos2x1+sin2xsin2x1+cos2xsin2xsin2xcos2xsin2x1+sin2x Then the ordered pair (m,M) is equal to:

Δ=cos2x1+sin2xsin2x1+cos2xsin2xsin2xcos2xsin2x1+sin2x by C1→C1+C2Δ=21+sin2xsin2x2sin2xsin2x1sin2x1+sin2x by R1→R1-R2 and R2→R2-R3Δ=01010-11sin2x1+sin2x⇒Δ=(-1)[(1+sin2x)-(-1)]=-(2+sin2x) (by expanding along R1 ) Now M=max(Δ)=-1 m=min(Δ)=-3∴(m,M)=(-3,-1)

Let m and M be respectively the minimum and maximum values of cos2x1+sin2xsin2x1+cos2xsin2xsin2xcos2xsin2x1+sin2x Then the ordered pair (m,M) is equal to:

Δ=cos2x1+sin2xsin2x1+cos2xsin2xsin2xcos2xsin2x1+sin2x by C1→C1+C2Δ=21+sin2xsin2x2sin2xsin2x1sin2x1+sin2x by R1→R1-R2 and R2→R2-R3Δ=01010-11sin2x1+sin2x⇒Δ=(-1)[(1+sin2x)-(-1)]=-(2+sin2x) (by expanding along R1 ) Now M=max(Δ)=-1 m=min(Δ)=-3∴(m,M)=(-3,-1)

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Let m and M be respectively the minimum and maximum values of $|\begin{matrix} \cos^{2} x & 1 + \sin^{2} x & \sin 2 x \\ 1 + \cos^{2} x & \sin^{2} x & \sin 2 x \\ \cos^{2} x & \sin^{2} x & 1 + \sin 2 x \end{matrix}|$ $Then the ordered pair (m, M) is equal to:$

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$(- 3, 3)$

$(1, 3)$

$(- 3, - 1)$

$(- 4, - 1)$

answer is C.

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Detailed Solution

$\begin{array}{l} Δ = |\begin{matrix} \cos^{2} x & 1 + \sin^{2} x & \sin 2 x \\ 1 + \cos^{2} x & \sin^{2} x & \sin 2 x \\ \cos^{2} x & \sin^{2} x & 1 + \sin 2 x \end{matrix}| \\ by C_{1} \to C_{1} + C_{2} \\ Δ = |\begin{matrix} 2 & 1 + \sin^{2} x & \sin 2 x \\ 2 & \sin^{2} x & \sin 2 x \\ 1 & \sin^{2} x & 1 + \sin 2 x \end{matrix}| \end{array}$

$\begin{array}{l} by R_{1} \to R_{1} - R_{2} and R_{2} \to R_{2} - R_{3} \\ Δ = |\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & - 1 \\ 1 & \sin^{2} x & 1 + \sin 2 x \end{matrix}| \\ \Rightarrow Δ = (- 1) [(1 + \sin 2 x) - (- 1)] = - (2 + \sin 2 x) \\ (by expanding along R_{1}) \\ Now \{\begin{cases} M = \max (Δ) = - 1 \\ m = \min (Δ) = - 3 \end{cases} \\ ∴ (m, M) = (- 3, - 1) \end{array}$