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Geometry Level 2

∣ sec ⁡ 2 x 1 1 cos ⁡ 2 x cos ⁡ 2 x csc ⁡ 2 x 1 cos ⁡ 2 x cot ⁡ 2 x ∣ \begin{vmatrix} \sec ^{ 2 }{ x } & 1 & 1 \\ \cos ^{ 2 }{ x } & \cos ^{ 2 }{ x } & \csc ^{ 2 }{ x } \\ 1 & \cos ^{ 2 }{ x } & \cot ^{ 2 }{ x } \end{vmatrix}

The expression above represents a determinant of a matrix.

If the range of the expression above is ( a , b ) (a,b) , submit your answer as the product a ⋅ b a\cdot b .


The answer is 0.

Vishnu Kadiri by Vishnu Kadiri , 19, India

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1 solution

Vishnu Kadiri
Mar 4, 2019

f ( x ) = ∣ sec ⁡ 2 x 1 1 cos ⁡ 2 x cos ⁡ 2 x csc ⁡ 2 x 1 cos ⁡ 2 x cot ⁡ 2 x ∣ = cos ⁡ 2 x ∣ sec ⁡ 2 x sec ⁡ 2 x 1 cos ⁡ 2 x 1 csc ⁡ 2 x 1 1 cot ⁡ 2 x ∣ = cos ⁡ 2 x cot ⁡ 2 x ∣ sec ⁡ 2 x sec ⁡ 2 x tan ⁡ 2 x cos ⁡ 2 x 1 sec ⁡ 2 x 1 1 1 ∣ = cot ⁡ 2 x ∣ 1 1 sin ⁡ 2 x cos ⁡ 2 x 1 sec ⁡ 2 x 1 1 1 ∣ = cot ⁡ 2 x ∣ 0 cos ⁡ 2 x sin ⁡ 2 x − sin ⁡ 2 x − tan ⁡ 2 x sec ⁡ 2 x 0 0 1 ∣ = cot ⁡ 2 x sin ⁡ 2 x cos ⁡ 2 x = cos ⁡ 4 x f\left( x \right) =\begin{vmatrix} \sec ^{ 2 }{ x } & 1 & 1 \\ \cos ^{ 2 }{ x } & \cos ^{ 2 }{ x } & \csc ^{ 2 }{ x } \\ 1 & \cos ^{ 2 }{ x } & \cot ^{ 2 }{ x } \end{vmatrix}=\cos ^{ 2 }{ x } \begin{vmatrix} \sec ^{ 2 }{ x } & \sec ^{ 2 }{ x } & 1 \\ \cos ^{ 2 }{ x } & 1 & \csc ^{ 2 }{ x } \\ 1 & 1 & \cot ^{ 2 }{ x } \end{vmatrix}=\cos ^{ 2 }{ x } \cot ^{ 2 }{ x } \begin{vmatrix} \sec ^{ 2 }{ x } & \sec ^{ 2 }{ x } & \tan ^{ 2 }{ x } \\ \cos ^{ 2 }{ x } & 1 & \sec ^{ 2 }{ x } \\ 1 & 1 & 1 \end{vmatrix}=\cot ^{ 2 }{ x } \begin{vmatrix} 1 & 1 & \sin ^{ 2 }{ x } \\ \cos ^{ 2 }{ x } & 1 & \sec ^{ 2 }{ x } \\ 1 & 1 & 1 \end{vmatrix}=\cot ^{ 2 }{ x } \begin{vmatrix} 0 & \cos ^{ 2 }{ x } & \sin ^{ 2 }{ x } \\ -\sin ^{ 2 }{ x } & -\tan ^{ 2 }{ x } & \sec ^{ 2 }{ x } \\ 0 & 0 & 1 \end{vmatrix}=\cot ^{ 2 }{ x } \sin ^{ 2 }{ x } \cos ^{ 2 }{ x } =\cos ^{ 4 }{ x } Thus, least upper bound is 1 and greatest lower bound is 0.

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