- A$x - y$
- B$x + y$
- ✓$\frac{{x - y}}{{1 + xy}}$
- D$\frac{{x + y}}{{1 - xy}}$
$ \Rightarrow \,\,{\tan ^{ - 1}}\,\left( {\frac{{x - y}}{{1 + xy}}} \right) = {\tan ^{ - 1}}A$.
Hence $A = \frac{{x - y}}{{1 + xy}}$.
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($A$) $\mathrm{M}$ is invertible
($B$) There exists a nonzero column matrix $\left(\begin{array}{l}a_1 \\ a_2 \\ a_3\end{array}\right)$ such that $M\left(\begin{array}{l}a_1 \\ a_2 \\ a_3\end{array}\right)=\left(\begin{array}{l}-a_1 \\ -a_2 \\ -a_3\end{array}\right)$
($C$) The set $\left\{\mathrm{X} \in \mathbb{R}^3: \mathrm{MX}=\mathbf{0} \neq \neq 0\right\}$, where $\mathbf{0}=\left(\begin{array}{l}0 \\ 0 \\ 0\end{array}\right)$
($D$) The matrix $(M-2 I)$ is invertible, where $I$ is the $3 \times 3$ identity matrix
$\int_0^1 {(1 + {{\cos }^8}x)(a{x^2} + bx + c)\,dx} = \int_0^2 {(1 + {{\cos }^8}x)(a{x^2} + bx + c)\,dx} $
Then the quadratic equation $a{x^2} + bx + c = 0$ has