JEE Main 2019MathematicsDeterminantsMediumMCQ

JEE Main 2019 — Determinants Question with Solution

JEE Main 2019 (10 Apr Shift 1)

Question

If Δ1=xsinθcosθ-sinθ-x1cosθ1x and Δ2=xsin2θcos2θ-sin2θ-x1cos2θ1x, x≠0; then for all θ∈0, π2 :

Choose an option

Show full solutionCorrect option: C
Correct answer
CΔ1+Δ2=-2x3

Step-by-step explanation

Δ1=xsinθ-sinθ-xcosθ1    cosθ1x
=x-x2-1-sinθ-xsinθ-cosθ+cosθ(-sinθ+xcosθ)
=-x3-x+xsin2θ+sinθcos⁡θ-sinθcosθ+xcos2θ
=-x3-x+xsin2⁡θ+cos2⁡θ
=-x3
Similarly: Δ2=-x3
∴Δ1+Δ2=-2x3

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About this question

This is a previous-year question from JEE Main 2019, covering the Determinants chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.