Question

Let $-\pi \le \theta \le \pi$ and
$\Delta \left(\theta \right)=\left|\begin{array}{ccc}1& \sin \theta & 1\\ -\sin \theta & 1& \sin \theta \\ -1& -\sin \theta & 1\end{array}\right|$
If l is the length of interval of $\theta$ for which $\left[\Delta \left(\theta \right)\right]=2$ (where [x] denotes the greatest integer $\le$x). Find $\frac{4l}{\pi }$.

Answer 1

$△\left(\theta \right)=\left|\begin{array}{ccc}1& \sin \theta & 1\\ -\sin \theta & 1& \sin \theta \\ -1& -\sin \theta & 1\end{array}\right|$

$=1(1+{\sin }^2\theta )-\sin \theta \left(0\right)+1(1+{\sin }^2\theta )$

$(△\left(\theta \right)=2)$

$=2+2{\sin }^2\theta$

$2<2+2{\sin }^2\theta <3$

$=2{\sin }^2\theta ϵ[0,1]$

$\Rightarrow {\sin }^2\theta ϵ[0,\frac{1}{2}]$

$\Rightarrow \theta ϵ[0,\frac{\pi }{4}]$

$l=\frac{\pi }{4}$

Now,

$\frac{4l}{\pi }=\frac{4\times \frac{\pi }{4}}{\pi }=1$

Correct answer $1$