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

Solution :
$$f\left( x \right) = {x^2} + bx - b\,;\,f'\left( x \right) = 2x + b\, \Rightarrow f'\left( 1 \right) = b + 2$$
Equation of tangent $$y - 1 = \left( {b + 2} \right)\left( {x - 1} \right)$$
Putting $$x = 0$$
$$ \Rightarrow y = 1 - b - 2 = - b - 1 > 0 \Rightarrow b < - 1$$
Putting $$y = 0$$
$$\eqalign{ & \Rightarrow x - 1 = - \frac{1}{{b + 2}} \cr & \Rightarrow x = \frac{{ - 1}}{{b + 2}} + 1 \cr & \Rightarrow x = \frac{{b + 1}}{{b + 2}} > 0 \cr & \Rightarrow b < - 2{\text{ or }}b > - 1 \cr} $$
Combining, the two conditions $$ = b < - 2$$
$$\eqalign{ & {\text{Now, }} \cr & \frac{1}{2}\left| { - b - 1} \right|\left| {\frac{{b + 1}}{{b + 2}}} \right| = 2 \cr & \Rightarrow {\left( {b + 1} \right)^2} = 4\left| {b + 2} \right| \cr & \Rightarrow {\left( {b + 1} \right)^2} = - 4b - 8 \cr & \Rightarrow {\left( {b + 3} \right)^2} = 0 \cr & \Rightarrow b = - 3\,\,{\text{follows the condition}}\,b < - 2 \cr} $$