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← JEE Main 2022 · 25 Jul · Shift 1
MathematicsSingle correct+4 / -1Question 4 of 90

JEE Main 2022 · 25 Jul · Shift 1, Mathematics question 4

The number of θ∈(0,4π)\theta\in(0,4\pi) for which the system of linear equations
3(sin⁡3θ)x−y+z=23(\sin3\theta)x-y+z=2
3(cos⁡2θ)x+4y+3z=33(\cos2\theta)x+4y+3z=3
6x+7y+7z=96x+7y+7z=9
has no solution, is :

Choose one option
Answer and worked solution

Official answer: B

Determinant. Expanding, Δ=21sin⁡3θ+42cos⁡2θ−42\Delta=21\sin3\theta+42\cos2\theta-42, zero when sin⁡3θ=2(1−cos⁡2θ)=4sin⁡2θ\sin3\theta=2(1-\cos2\theta)=4\sin^2\theta.

Solve. sin⁡θ (3−4sin⁡θ−4sin⁡2θ)=0\sin\theta\,(3-4\sin\theta-4\sin^2\theta)=0: sin⁡θ=0\sin\theta=0 (3 values in (0,4π)(0,4\pi)) or sin⁡θ=12\sin\theta=\frac12 (4 values). At each the system is inconsistent, so 7 values.

(B)

Topics:Systems of linear equationsTrigonometric equations
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