If A= 1 sinθ 1 -sinθ 1 sinθ -1 -sinθ 1 , then for all θ∈(3π)/4,(5π)/4, detA lies in the interval :

Correct answer: (C) 3/2, 3. A= [ 1 sin θ 1 -sin θ 1 sin θ -1 -sin θ 1 ] ⇒ det(A)= (11+sin^2 θ )-sin θ(-sin θ+sin θ)+1 (sin^2 θ+1 ) =2+2 sin^2 θ ∵ θ ∈ ((3 π)/4, (5 π)/4 ) ∴ sin 2 θ ∈ [0, 1/2 ] ∴ det(A) ∈[2,3] which is a subset of (3/2, 3 ]

Q44 / 90 · JEE Main 2019 (12 Jan Shift 2) · Mathematics

If $A = [\begin{matrix} 1 & \sin θ & 1 \\ - \sin θ & 1 & \sin θ \\ - 1 & - \sin θ & 1 \end{matrix}]$ , then for all $θ \in (\frac{3 π}{4}, \frac{5 π}{4})$ , $\det (A)$ lies in the interval :

A
$(1, \frac{5}{2}]$
B
$[\frac{5}{2}, 4)$
Correct answer
C
$(\frac{3}{2}, 3]$
D
$(0, \frac{3}{2}]$

Solution

$A=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]$
$\Rightarrow \operatorname{det}(A)=\left(11+\sin ^2 \theta\right)-\sin \theta(-\sin \theta+\sin \theta)+1\left(\sin ^2 \theta+1\right)$
$=2+2 \sin ^2 \theta$
$\because \theta \in\left(\frac{3 \pi}{4}, \frac{5 \pi}{4}\right)$
$\therefore \sin 2 \theta \in\left[0, \frac{1}{2}\right]$
$\therefore \operatorname{det}(A) \in[2,3]$ which is a subset of $\left(\frac{3}{2}, 3\right]$

Hence, the correct answer is option: 'C'.

If A=1sinθ1-sinθ1sinθ-1-sinθ1, then for all θ∈3π4,5π4, detA lies in the interval :

Solution

Relevant Questions

If $A = [\begin{matrix} 1 & \sin θ & 1 \\ - \sin θ & 1 & \sin θ \\ - 1 & - \sin θ & 1 \end{matrix}]$ , then for all $θ \in (\frac{3 π}{4}, \frac{5 π}{4})$ , $\det (A)$ lies in the interval :