If A=1sin⁡θ1−sin⁡θ1sin⁡θ−1−sin⁡θ1; then for all θ∈3π4,5π4,det⁡(A) lies in the interval

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

Solution:

Here, $\det (A) = |\begin{array}{ccc} 1 & \sin θ & 1 \\ - \sin θ & 1 & \sin θ \\ - 1 & - \sin θ & 1 \end{array}|$

$\begin{array}{l} 1 (1 + \sin^{2} θ) - \sin θ (0) + 1 (\sin^{2} θ + 1) = 2 (1 + \sin^{2} θ) \\ ∵ θ \in (\frac{3 π}{4}, \frac{5 π}{4}) \Rightarrow \sin θ \in (- \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) \\ \Rightarrow \sin^{2} θ \in [0, \frac{1}{2}) ∴ \det (A) \in [2, 3) \end{array}$

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

Solution:

Related content