A=1−sinθ−1sinθ1−sinθ1sinθ1
⇒det(A)=(11+sin2θ)−sinθ(−sinθ+sinθ)+1(sin2θ+1)
=2+2sin2θ
∵θ∈(43π,45π)
∴sin2θ∈[0,21]
∴det(A)∈[2,3] which is a subset of (23,3]
JEE Main 2019 — Mathematics Algebra
If A=[1−sinθ−1sinθ1−sinθ1sinθ1], then for all θ∈(43π,45π), det(A) lies in the interval :
Held on 12 Jan 2019 · Verified 6 Jul 2026.
(1,25]
[25,4)
(23,3]
(0,23]
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