1&2&3 \\
0&5&4 \\
0&3&2
\end{array}} \right]$ and $A^3 -8A^2 + \alpha A + \beta I = O$ then ordered pair $(\alpha , \beta)$ is
- ✓$(5, 2)$
- B$(5, -2)$
- C$(-5, 2)$
- D$(2, 5)$
$\Rightarrow \lambda^{3}-8 \lambda^{2}+5 \lambda+2=0$
$\Rightarrow A^{3}-8 A^{2}+5 A+2 I=0$
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The solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\frac{2\text{xy}}{1+\text{x}^2}=\frac{1}{(1+\text{x}^2)^2}$ is:
$\text{y}(1+\text{x}^2)=\text{C}+\tan^{-1}\text{x}$
$\frac{\text{y}}{1+\text{x}^2}=\text{C}+\tan^{-1}\text{x}$
$\text{y}\log(1+\text{x}^2)=\text{C}+\tan^{-1}\text{x}$
$\text{y}(1+\text{x}^2)=\text{C}+\sin^{-1}\text{x}$
$\mathrm{F}(\mathrm{x})=\int\limits_{1}^{\mathrm{x}} \mathrm{t}^{2} \mathrm{g}(\mathrm{t}) \mathrm{dt},$ where $\mathrm{g}(\mathrm{t})=\int\limits_{1}^{\mathrm{t}} \mathrm{f}(\mathrm{u}) \mathrm{du}$
Then for the function $\mathrm{F}$, the point $\mathrm{x}=1$ is