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Matrices and Determinants question

2019 · 10 Apr · Shift 1 · Q24
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  5. /2019 · 10 Apr · Shift 1 · Q24

Matrices and Determinants question

2019 · 10 Apr · Shift 1 · Q24

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If Δ1=∣xsin⁡θcos⁡θ−sin⁡θ−x1cos⁡θ1x∣{\Delta _1} = \left| {\begin{matrix} x & {\sin \theta } & {\cos \theta } \\ { - \sin \theta } & { - x} & 1 \\ {\cos \theta } & 1 & x \\ \end{matrix} } \right|Δ1​=​x−sinθcosθ​sinθ−x1​cosθ1x​​ and Δ2=∣xsin⁡2θcos⁡2θ−sin⁡2θ−x1cos⁡2θ1x∣{\Delta _2} = \left| {\begin{matrix} x & {\sin 2\theta } & {\cos 2\theta } \\ { - \sin 2\theta } & { - x} & 1 \\ {\cos 2\theta } & 1 & x \\ \end{matrix} } \right|Δ2​=​x−sin2θcos2θ​sin2θ−x1​cos2θ1x​​, xe0x e 0xe0; then for all θ∈(0,π2)\theta \in \left( {0,{\pi \over 2}} \right)θ∈(0,2π​) :
  1. A
    Δ1−Δ2{\Delta _1} - {\Delta _2}Δ1​−Δ2​ = x (cos 2 θ\thetaθ – cos 4 θ\thetaθ)
  2. B
    Δ1+Δ2{\Delta _1} + {\Delta _2}Δ1​+Δ2​ = - 2x3
  3. C
    Δ1+Δ2{\Delta _1} + {\Delta _2}Δ1​+Δ2​ = – 2(x3 + x –1)
  4. D
    Δ1−Δ2{\Delta _1} - {\Delta _2}Δ1​−Δ2​ = - 2x3
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