JEE Main 201910 Jan, Morning shiftMathematicsMediumMatrices and Determinants245× since 2002Let d ∈\in∈ R, and A=[−24+d(sinθ)−21(sinθ)+2d5(2sinθ)−d(−sinθ)+2+2d],A = \left[ {\matrix{ { - 2} & {4 + d} & {\left( {\sin \theta } \right) - 2} \cr 1 & {\left( {\sin \theta } \right) + 2} & d \cr 5 & {\left( {2\sin \theta } \right) - d} & {\left( { - \sin \theta } \right) + 2 + 2d} \cr } } \right],A=−2154+d(sinθ)+2(2sinθ)−d(sinθ)−2d(−sinθ)+2+2d, θ∈[0,2π]\theta \in \left[ {0,2\pi } \right]θ∈[0,2π] If the minimum value of det(A) is 8, then a value of d is -AOption A. −-− 7BOption B. 2(2+2)2\left( {\sqrt 2 + 2} \right)2(2+2)COption C. −-− 5DOption D. 2(2+1)2\left( {\sqrt 2 + 1} \right)2(2+1)CheckSkip · show solution
