The triangle formed by the tangent to the curve $$f\left( x \right) = {x^2} + bx - b$$     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

Solution :
$$\eqalign{ & {\text{Tangent}}\,{\text{to}}\,y = {x^2} + bx - b\,{\text{at}}\,\left( {1,1} \right)\,{\text{is}}\,y - 1 = \left( {2 + b} \right)\left( {x - 1} \right) \Rightarrow \left( {b + 2} \right)x - y = b + 1 \cr & x - {\text{intercept}} = \frac{{b + 1}}{{b + 2}}\,{\text{and}}\,y - {\text{intercept}} = - \left( {b + 1} \right) \cr & {\text{Given}}\,Ar\left( \Delta \right) = 2 \Rightarrow \frac{1}{2}\left( {\frac{{b + 1}}{{b + 2}}} \right)\left[ { - \left( {b + 1} \right)} \right] = 2 \cr & \Rightarrow {b^2} + 2b + 1 = - 4\left( {b + 2} \right) \Rightarrow {b^2} + 6b + 9 = 0 \cr & \Rightarrow {\left( {b + 3} \right)^2} = 0 \Rightarrow b = - 3 \cr} $$