The maximum value of $f\left( x \right){\rm{ = }}\left| {\begin{array}{*{20}{c}} {{\rm{si}}{{\rm{n}}^{\rm{2}}}x}&{{\rm{1 + co}}{{\rm{s}}^{\rm{2}}}x}&{{\rm{cos2x}}}\\ {{\rm{1 + si}}{{\rm{n}}^{\rm{2}}}x}&{{\rm{co}}{{\rm{s}}^{\rm{2}}}x}&{{\rm{cos2x}}}\\ {{\rm{si}}{{\rm{n}}^{\rm{2}}}x}&{{\rm{co}}{{\rm{s}}^{\rm{2}}}x}&{{\rm{sin2x}}} \end{array}} \right|{\rm{,}}\,\,x \in R$ is

JEE MATHEMATICS MCQ: MATRICES

1 Easy

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Analyzing +4 | -1

The maximum value of

f\left( x \right)\mathrm{ = }\left| {\begin{array}{cccccccccccccccccccc} {\mathrm{si}{\mathrm{n}^\mathrm{2}}x}&{\mathrm{1 + co}{\mathrm{s}^\mathrm{2}}x}&{\mathrm{cos2x}}\\ {\mathrm{1 + si}{\mathrm{n}^\mathrm{2}}x}&{\mathrm{co}{\mathrm{s}^\mathrm{2}}x}&{\mathrm{cos2x}}\\ {\mathrm{si}{\mathrm{n}^\mathrm{2}}x}&{\mathrm{co}{\mathrm{s}^\mathrm{2}}x}&{\mathrm{sin2x}} \end{array}} \right|\mathrm{,}\,\,x \in R

View answer and explanation

Correct option: C

\sqrt 5

Option C (

\sqrt{5}

) is the correct answer because after simplifying the determinant and using trigonometric identities, the maximum value is found to be

\sqrt{5}

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\left|\begin{array}{ccc}a & b-c & b+c \\ c+a & b & c-a \\ a-b & a+b & c\end{array}\right|=0

a, b, c

a x+b y+c=0

{x_2} - {x_3} = 1,\; - {x_1} + 2{x_3} = - 2,\;{x_1} - 2,\;{x_1} - 2{x_2} = 3

2x + 3y + 1 = 0,3x + y - 2 = 0

ax + 2y - b = 0

\left|\begin{array}{ccc}1 & \left(\alpha^{2 x}-\alpha^{-2 x}\right)^2 & \left(\alpha^{2 x}+\alpha^{-2 x}\right)^2 \\ 1 & \left(\beta^{2 x}-\beta^{-2 x}\right)^2 & \left(\beta^{2 x}+\beta^{-2 x}\right)^2 \\ 1 & \left(\gamma^{2 x}-\gamma^{-2 x}\right)^2 & \left(\gamma^{2 x}+\gamma^{-2 x}\right)^2\end{array}\right|

\omega

\left| {\begin{array}{cccccccccccccccccccc} 1&\omega &{{\omega ^2}}\\ \omega &{{\omega ^2}}&1\\ {{\omega ^2}}&1&\omega \end{array}} \right|

A

2 \times 2

A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]

|A|