JEE Main 2019 — Math solution

JEE Main2019Math

If {\Delta _1} = \left| {\begin{array}{*{20}{c}} x{\sin \theta }{\cos \theta } \\ { - \sin \theta }{ - x}1 \\ {\cos \theta }1x \end{array}} \right| and {\Delta _2} = \left| {\begin{array}{*{20}{c}} x{\sin 2\theta }{\cos 2\theta } \\ { - \sin 2\theta }{ - x}1 \\ {\cos 2\theta }1x \end{array}} \right|,\,x \ne 0; then

for all $θ \in (0, \frac{π}{2})$

A
$Δ$ ₁ + $Δ$ ₂ = – 2x³
B
$Δ$ ₁ – $Δ$ ₂ = – 2x³
C
$Δ$ ₁ + $Δ$ ₂ = – 2(x³ + x – 1)
D
$Δ$ ₁ – $Δ$ ₂ = x(cos2 $θ$ – cos4 $θ$ )

Solution

{\Delta _1} = \left| {\begin{array}{*{20}{c}}x\\{ - \sin \theta }\\{\cos \theta }\\{}\end{array}\,\,\,\,\,\begin{array}{*{20}{c}}{\sin \theta }\\{ - x}\\1\\{}\end{array}\,\,\,\,\,\begin{array}{*{20}{c}}{\cos \theta }\\1\\x\\{}\end{array}} \right|

$= x (- x^{2} - 1) - sin θ (- x sin θ - cos θ)$

$+ cos θ (- sin θ + x cos θ)$

$= - x^{3} - x + x sin^{2} θ + sin θ cos θ - cos θ sin θ$

$+ x cos^{2} θ$

$= - x^{3} - x + x = - x^{3}$

$S imi l a r l y Δ_{1} = - x^{3}$

$Δ_{1} + Δ_{2} = - 2 x^{3}$

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