The shaded figure given below is the solution set for the linear inequations. Choose the correct option.

$$\begin{aligned} & \text { If the function } \mathrm{f}(\mathrm{x})=1+\sin \frac{\pi}{2}, \quad-\infty<\mathrm{x} \leq 1 \\ & =\mathrm{ax}+\mathrm{b}, \quad 1<\mathrm{x}<3 \\ & =6 \tan \frac{x \pi}{12}, \quad 3 \leq x<6 \\ \end{aligned}$$

is continuous in $$(-\infty, 6)$$, then the values of $$\mathrm{a}$$ and $$\mathrm{b}$$ are respectively.

If $$\int \frac{x^3}{\sqrt{1+x^2}} d x=a\left(1+x^2\right)^{\frac{3}{2}}+b \sqrt{1+x^2}+c$$, then $$a+b=$$, (where $$c$$ is constant of integration)

$$\mathrm{A}^{-1}=\frac{-1}{2}\left[\begin{array}{cc}1 & -4 \\ -1 & 2\end{array}\right]$$, then $$2 A+I_2=\quad$$

where $$I_2$$ is a unit matrix of order 2