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

-3

$f(x)=x^2+b x-b$

$\Rightarrow f'(x) = 2 x+b$

$\Rightarrow f'(1)=2+b$

∴ equation of tangent at (1, 1) is

$y-1=(2+b)(x-1) \Rightarrow (b+2) x-y-(b+1)=0$

∴ Length of x-intercept = $\frac{b+1}{b+2}$

Length of y-intercept = -(b+1)

∴  Area of $\Delta=-\frac{1}{2}\left(\frac{b+1}{b+2}\right)(b+1)=2$ (given)

$\Rightarrow b^2+6 b+9=0 \Rightarrow (b+3)^2=0 \Rightarrow b=-3$.