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If Δ _1=| … | and Δ _2=| … |, x≠ 0; then for all θ ∈ (0, π 2) :

JEE Main 2019 — Mathematics Algebra

2019mcqmedium

If Δ1=∣xsinθcosθ−sinθ−x1cosθ1x∣{\Delta }_{1}=|\begin{matrix}x & sin\theta & cos\theta \\ -sin\theta & -x & 1 \\ cos\theta & 1 & x\end{matrix}|Δ1​=∣x−sinθcosθ​sinθ−x1​cosθ1x​∣ and Δ2=∣xsin2θcos2θ−sin2θ−x1cos2θ1x∣,{\Delta }_{2}=|\begin{matrix}x & sin2\theta & cos2\theta \\ -sin2\theta & -x & 1 \\ cos2\theta & 1 & x\end{matrix}|,Δ2​=∣x−sin2θcos2θ​sin2θ−x1​cos2θ1x​∣, x≠0;x\neq 0;x=0; then for all θ∈(0,π2)\theta \in (0,\frac{\pi }{2})θ∈(0,2π​) :

Official previous-year question

Held on 10 Apr 2019 · Verified 6 Jul 2026.

Options

  1. A

    Δ1+Δ2=−2(x3+x−1){\Delta }_{1}+{\Delta }_{2}=-2({x}^{3}+x-1)Δ1​+Δ2​=−2(x3+x−1)

  2. B

    Δ1−Δ2=x(cos2θ−cos4θ){\Delta }_{1}-{\Delta }_{2}=x(cos2\theta -cos4\theta )Δ1​−Δ2​=x(cos2θ−cos4θ)

  3. C

    Δ1+Δ2=−2x3{\Delta }_{1}+{\Delta }_{2}=-2{x}^{3}Δ1​+Δ2​=−2x3

  4. D

    Δ1−Δ2=−2x3{\Delta }_{1}-{\Delta }_{2}=-2{x}^{3}Δ1​−Δ2​=−2x3

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