Solution:
$\begin{vmatrix}1&\sin\theta&1\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix}$
$=\begin{vmatrix}1&\sin\theta&2\\-\sin\theta&1&\sin\theta\\-1&-\sin\theta&1\end{vmatrix}$ [Applying C3 → C3 + C1]
$=2\times\begin{vmatrix}-\sin\theta&1\\-1&-\sin\theta\end{vmatrix}$ [Expanding along C3]
$=2(\sin^2\theta+1)$
Given, $0\leq\theta\leq2\pi$
$-1\leq\sin\theta\leq1$
$0\leq\sin^2\theta\leq1$
$|\text{A}|=2(\sin^2\theta+1)$
$|\text{A}|=2\times1=2$ $[\theta=0]$
$|\text{A}|=2\times2=4$ $[\theta=2\pi]$
$\text{Det (A)}\in[2,4]$
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If $f(x) = {\rm{ }}\left\{ {\begin{array}{*{20}{c}}
{\frac{{\left( {{e^x} - 1} \right)^2}}{{\sin {\mkern 1mu} \left( {\frac{x}{k}} \right){\mkern 1mu} \log {\mkern 1mu} \left( {1 + \frac{x}{4}} \right)}}{\mkern 1mu} ,{\mkern 1mu} x \ne 0}\\
{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 12{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} ,x{\mkern 1mu} {\mkern 1mu} = 0{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} }
\end{array}} \right.$
is a continuous function then the value of $k$ is