If the minimum and the maximum values of the function f:π4, π2→R defined by fθ=−sin2θ−1−sin2θ1−cos2θ−1−cos2θ11210−2are m and M respectively, then order pair (m, M) is equal to

fθ=−sin2θ−1−sin2θ1−cos2θ−1−cos2θ11210−2 R1→R1+R2+R3=1170−cos2θ−1−cos2θ11210−2 R3→R3-R1=1170−cos2θ−1−cos2θ113−2C1→C1-C1-C3 =4700−1−cos2θ103−2=42+2cos2θ−3=4cos2θ2θ∈π2,π⇒cos2θ∈−1,0⇒4cos2θ∈−4,0, m=-4 ,M=0

If the minimum and the maximum values of the function f:π4, π2→R defined by fθ=−sin2θ−1−sin2θ1−cos2θ−1−cos2θ11210−2are m and M respectively, then order pair (m, M) is equal to

fθ=−sin2θ−1−sin2θ1−cos2θ−1−cos2θ11210−2 R1→R1+R2+R3=1170−cos2θ−1−cos2θ11210−2 R3→R3-R1=1170−cos2θ−1−cos2θ113−2C1→C1-C1-C3 =4700−1−cos2θ103−2=42+2cos2θ−3=4cos2θ2θ∈π2,π⇒cos2θ∈−1,0⇒4cos2θ∈−4,0, m=-4 ,M=0

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If the minimum and the maximum values of the function $f : [\frac{π}{4}, \frac{π}{2}] \to R$ defined by $f (θ) = |\begin{matrix} - \sin^{2} θ & - 1 - \sin^{2} θ & 1 \\ - \cos^{2} θ & - 1 - \cos^{2} θ & 1 \\ 12 & 10 & - 2 \end{matrix}|$ are m and M respectively, then order pair (m, M) is equal to

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$(- 4, 0)$

$(- 4, 4)$

$(0, 4)$

$(0, 2 \sqrt{2})$

answer is A.

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Detailed Solution

$\begin{matrix} f (θ) = |\begin{matrix} - \sin^{2} θ & - 1 - \sin^{2} θ & 1 \\ - \cos^{2} θ & - 1 - \cos^{2} θ & 1 \\ 12 & 10 & - 2 \end{matrix}| \\ R_{1} \to R_{1} + R_{2} + R_{3} \\ = |\begin{matrix} 11 & 7 & 0 \\ - \cos^{2} θ & - 1 - \cos^{2} θ & 1 \\ 12 & 10 & - 2 \end{matrix}| \\ R_{3} \to R_{3} - R_{1} \\ = |\begin{matrix} 11 & 7 & 0 \\ - \cos^{2} θ & - 1 - \cos^{2} θ & 1 \\ 1 & 3 & - 2 \end{matrix}| \\ C_{1} \to C_{1} - C_{1} - C_{3} \\ = |\begin{matrix} 4 & 7 & 0 \\ 0 & - 1 - \cos^{2} θ & 1 \\ 0 & 3 & - 2 \end{matrix}| \\ = 4 (2 + 2 \cos^{2} θ - 3) \\ = 4 \cos 2 θ \end{matrix}$