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JEE Main 201910 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2019 · 10 April · Morning Shift · Mathematics · Determinants

If Δ _ 1 = | cases x & sinθ & cosθ - sinθ & - x & 1 cosθ & 1 & x...

If {\Delta}_{1} = \left|\begin{cases}x & sin\theta & cos\theta \\ - sin\theta & - x & 1 \\ cos\theta & 1 & x\end{cases}\right| and {\Delta}_{2} = \left|\begin{cases}x & \sin 2\theta & \cos 2\theta \\ - \sin 2\theta & - x & 1 \\ \cos 2\theta & 1 & x\end{cases}\right|, x≠0;x \ne 0; then for all θ∈(0,π2)\theta \in \left(0, \dfrac{\pi}{2}\right) :

Options

  1. AΔ1+Δ2=−2(x3+x−1){\Delta}_{1} + {\Delta}_{2} = - 2 ( {x}^{3} + x - 1 )
  2. BΔ1−Δ2=x(cos⁡2θ−cos⁡4θ){\Delta}_{1} - {\Delta}_{2} = x ( \cos 2\theta - \cos 4\theta )
  3. CΔ1+Δ2=−2x3{\Delta}_{1} + {\Delta}_{2} = - 2{x}^{3}
  4. DΔ1−Δ2=−2x3{\Delta}_{1} - {\Delta}_{2} = - 2{x}^{3}

Correct answer

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

Step-by-step solution

{\Delta}_{1} = \left|\begin{cases}x & sin\theta \\ - sin\theta & - x \\ cos\theta & 1\end{cases}cosθ1x\begin{array}{ll}cos\theta \\ 1 \\ x\end{array}\right|<br/><br />= x\left(- {x}^{2} - 1\right) - sin\theta\left(- xsin\theta - cos\theta\right) + cos\theta ( - sin\theta + xcos\theta )<br/><br />= - {x}^{3} - x + xsi{n}^{2}\theta + sin\theta cos \theta - sin\theta cos\theta + xco{s}^{2}\theta<br/><br />= - {x}^{3} - x + x\left(\sin^{2} \theta + \cos^{2} \theta\right)<br/><br />= - {x}^{3}<br/>Similarly:<br />Similarly:{\Delta}_{2} = - {x}^{3}<br/><br />∴ {\Delta}_{1} + {\Delta}_{2} = - 2{x}^{3}$
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