A triangle is formed by the tangents to the curve \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}+\mathrm{bx}-\mathrm{b}\) at the point \((1,1)\) and the coordinate axes, lies in the first Quadrant. If the area is 2, then value of \(b\) is : (a) \(-1\) (b) 3 (c) \(-3\) (d) 1

We'll find the equation of tangent by using the derivative of the curve and the point-slope form of a linear equation. The derivative of the given curve function is: \(f'(x) = 2x + b\) Now we'll need to find the slope of the tangent line at the point (1, 1): m = \(f'(1) = 2(1) + b = 2 + b\) Now, we'll use the point-slope form of the linear equation to write the tangent line equation: \(y - 1 = (2 + b)(x - 1)\)

To find the x-intercept, we let y = 0: \(0 - 1 = (2 + b)(x - 1)\) Solving for x, we have: \(x = \frac{b + 1}{b + 2}\) To find the y-intercept, let x = 0: \(y - 1 = (2 + b)(0 - 1)\) Solving for y, we have: \(y = 1 - 2 - b = -1 - b\)

We now have the following vertices for the triangle: A (0, 0) - the origin. B \(\left(\frac{b + 1}{b + 2}, 0\right)\) - the x-intercept. C (0, -1 - b) - the y-intercept.

We can calculate the area of the triangle using the vertices coordinates: Area = \(\frac{1}{2}\) × base × height Area = \(\frac{1}{2}\) × \(\frac{b + 1}{b + 2}\) × (-1 - b)

Equate the area to 2: \(\frac{1}{2} \times \frac{b + 1}{b + 2} \times (-1 - b) = 2\) \( b(1 + b) = -2(b + 2) \) Expand and rearrange the equation: \(b^2 + b = -2b - 4\) \(b^{2} + 3b + 4 = 0\) Now, we'll solve for the values of \(\(b\)\): \(b = \frac{-3 \pm \sqrt{(-3)^{2}-4(4)}}{2}\) \(b = \frac{-3 \pm \sqrt{9 - 16}}{2}\) \(b = \frac{-3 \pm \sqrt{-7}}{2}\) There is no real solution for the equation, so we should recheck our reasoning about the triangle in the first quadrant. Upon rereading the problem, we see the area of the triangle is 2 and not the value of b. Thus, our equation for the area of the triangle should be: \(\frac{1}{2} \times \frac{b + 1}{b + 2} \times (-1 - b) = -2\) \( -2b(b + 1) = -4(b + 2) \) Now solving for b: \( 2b(b + 1) = 4(b + 2) \) \(2b^{2} + 2b = 4b + 8\) \(2b^{2} - 2b - 8 = 0\) Factoring the quadratic equation: \(2(b^{2} - b - 4) = 0\) Solving for b: \(b = \frac{1 \pm \sqrt{(-1)^{2}+4(4)}}{1}\) \(b = \frac{1 \pm \sqrt{17}}{1}\) However, since the triangle lies in the first quadrant, only the negative value will be valid: \(b = -1 + \sqrt{17}\) So, the value of b is not among the given options (a), (b), (c), or (d).