If $\mathrm{A}=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]$; then for all $\theta \in\left(\frac{3 \pi}{4}, \frac{5 \pi}{4}\right)$, $\operatorname{det}(\mathrm{A})$ lies in the interval :

(D) $\left(\frac{3}{2}, 3\right]$

(A) $\left(1, \frac{5}{2}\right]$

(B) $\left[\frac{5}{2}, 4\right)$

(C) $\left(0, \frac{3}{2}\right]$

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Determinants

A = 1 - sin θ - 1 sin θ 1 - sin θ 1 sin θ 1

; then for all

θ \in (\frac{3 π}{4}, \frac{5 π}{4})

det (A)

lies in the interval :

Answer: (D)

Expanding the determinant,

det A = 1 (1 + sin^{2} θ) - sin θ (- sin θ + sin θ) + 1 (sin^{2} θ + 1) = 2 (1 + sin^{2} θ)

. On

θ \in (\frac{3 π}{4}, \frac{5 π}{4})

sin θ

runs between

\pm \frac{1}{2}

, touching

0

θ = π

, so

sin^{2} θ \in [0, \frac{1}{2})

and

det A = 2 (1 + sin^{2} θ) \in [2, 3)

, a range that lies inside

(\frac{3}{2}, 3]

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option