The triangle formed by the tangent to the curve f(x) = x² + 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

-3

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

∴ $f'(1)=2+b$

∴ Equation of tangent at (1, 1) is

y – 1 = (2 + b) (x – 1)

⇒ (b + 2) x – y – (b + 1) = 0

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

Length of y-intercept = – (b + 1)

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

⇒ b² + 6b + 9 = 0 ⇒ b = -3

Hence (3) is the correct answer.