If $A^{-1}=\alpha A+\beta I$, where $I$ is an identity matrix of order $2 \times 2$, then $\alpha+\beta$ equals....................
$ a+b=3, b+d=7,(3-b)(7-b)-b^2=1 $
$ 21-10 b=1 \rightarrow b=2, a=1, d=5$
$A=\left[\begin{array}{ll}1 & 2 \\ 2 & 5\end{array}\right], A^{-1}=\left[\begin{array}{cc}5 & -2 \\ -2 & 1\end{array}\right]$
$\mathrm{A}^{-1}=\alpha \mathrm{A}+\beta \mathrm{I}$
$\left[\begin{array}{cc}5 & -2 \\ -2 & 1\end{array}\right]=\left[\begin{array}{cc}\alpha+\beta & 2 \alpha \\ 2 \alpha & 5 \alpha+\beta\end{array}\right]$
$\alpha=-1, \beta=6 \rightarrow \alpha+\beta=5$
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{\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