The triangle formed by the tangent to the curve $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+\mathrm{bx}-\mathrm{b}$ at the point $\left(1, 1\right)$ and the coordinates axes, lies in the first quadrant. If its area is 2, then the value of b is

$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+\mathrm{bx}-\mathrm{b}; \mathrm{f}\text{'}\left(\mathrm{x}\right)=2\mathrm{x}+\mathrm{b} \Rightarrow \mathrm{f}\text{'}\left(1\right)=\mathrm{b}+2$

Equation of tangent : $\mathrm{y}-1=(\mathrm{b}+2)(\mathrm{x}-1)$

Putting $\mathrm{x}=0\Rightarrow \mathrm{y}=1-\mathrm{b}-2=-\mathrm{b}-1>0\Rightarrow \mathrm{b}<-1$

Putting $\mathrm{y}=0\Rightarrow \mathrm{x}-1=-\frac{1}{\mathrm{b}+2}\Rightarrow \mathrm{x}=\frac{-1}{\mathrm{b}+2}+1$

$=\frac{\mathrm{b}+1}{\mathrm{b}+2}>0\Rightarrow \mathrm{b}<-2\text{ or }\mathrm{b}>-1$

Combining, two conditions $=\mathrm{b}<-2$

Now, $\frac{1}{2}|-\mathrm{b}-1|\left|\frac{\mathrm{b}+1}{\mathrm{b}+2}\right|=2;(\mathrm{b}+1{)}^2=4|\mathrm{b}+2|=-4\mathrm{b}-8$

$\Rightarrow (\mathrm{b}+3{)}^2=0\Rightarrow \mathrm{b}=-3$ follows the condition $\mathrm{b}<-2$