- A$2A$
- B$4A$
- C$6A$
- ✓None of these
$I = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$,then, ${A^2} + 9I = \,\left[ {\begin{array}{*{20}{c}}{15}&{11}&7\\{ - 11}&{13}&{ - 11}\\7&{11}&{21}\end{array}} \right]$.
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$\frac{d y}{d x}=1+x e^{y-x},-\sqrt{2}\,<\,x\,<\,\sqrt{2}, y (0)=0$ then, the minimum value of $y(x)$ , $\mathrm{x} \in(-\sqrt{2}, \sqrt{2})$ is equal to:
$12\pi\ \text{cm}^{3}/\text{sec}.$
$180\pi\ \text{cm}^{3}/\text{sec}.$
$225\pi\ \text{cm}^{3}/\text{sec}.$
$3\pi\ \text{cm}^{3}/\text{sec}.$
$I=\int_{0}^{10} \frac{[x] e^{[x]}}{e^{x-1}} d x,$
where $[ x ]$ denotes the greatest integer less than or equal to $x$. Then the value of $I$ is equal to: