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Question Number 10776 by j.masanja06@gmail.com last updated on 24/Feb/17

if   D = determinant ((1,(3sinθ),1),((sinθ),1,(3cosθ)),(1,(sinθ),1)) the maxi  mum value of D is?

$$\mathrm{if}\:\:\:\mathrm{D}\:=\begin{vmatrix}{\mathrm{1}}&{\mathrm{3sin}\theta}&{\mathrm{1}}\\{\mathrm{sin}\theta}&{\mathrm{1}}&{\mathrm{3cos}\theta}\\{\mathrm{1}}&{\mathrm{sin}\theta}&{\mathrm{1}}\end{vmatrix}\:\mathrm{the}\:\mathrm{maxi} \\ $$$$\mathrm{mum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{D}\:\mathrm{is}? \\ $$

Answered by sandy_suhendra last updated on 24/Feb/17

D=1+9 sinθ cosθ+sinθ cosθ−(1+3 sinθ cosθ+3 sinθ cosθ)           =4 sinθ cosθ=2 sin 2θ  because maximum for sin 2θ=1  so D maximum=2

$$\mathrm{D}=\mathrm{1}+\mathrm{9}\:\mathrm{sin}\theta\:\mathrm{cos}\theta+\mathrm{sin}\theta\:\mathrm{cos}\theta−\left(\mathrm{1}+\mathrm{3}\:\mathrm{sin}\theta\:\mathrm{cos}\theta+\mathrm{3}\:\mathrm{sin}\theta\:\mathrm{cos}\theta\right)\:\:\:\:\: \\ $$$$\:\:\:\:=\mathrm{4}\:\mathrm{sin}\theta\:\mathrm{cos}\theta=\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\theta \\ $$$$\mathrm{because}\:\mathrm{maximum}\:\mathrm{for}\:\mathrm{sin}\:\mathrm{2}\theta=\mathrm{1} \\ $$$$\mathrm{so}\:\mathrm{D}\:\mathrm{maximum}=\mathrm{2} \\ $$

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