The triangle formed by the tangent to the cur | vedclass

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(C) Given the curve $f(x) = x^2 + bx - b$. Since the point $(1, 1)$ lies on the curve,we have $1 = 1^2 + b(1) - b$,which is $1 = 1$,confirming the poi

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$-3$

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(C) Given the curve $f(x) = x^2 + bx - b$. Since the point $(1, 1)$ lies on the curve,we have $1 = 1^2 + b(1) - b$,which is $1 = 1$,confirming the point is on the curve for any $b$.First,find the derivative: $\frac{dy}{dx} = 2x + b$.At the point $(1, 1)$,the slope of the tangent is $m = 2(1) + b = 2 + b$.The equation of the tangent line at $(1, 1)$ is $y - 1 = (2 + b)(x - 1)$.Simplifying,$y - 1 = (2 + b)x - (2 + b)$,which gives $(2 + b)x - y = 1 + b$.To find the intercepts,set $y = 0$ to get $x = \frac{1 + b}{2 + b}$ (point $A$),and set $x = 0$ to get $y = -(1 + b)$ (point $B$).Since the triangle lies in the first quadrant,the intercepts must be positive: $\frac{1 + b}{2 + b} > 0$ and $-(1 + b) > 0$.This implies $1 + b The area of the triangle is $\frac{1}{2} 	imes |OA| 	imes |OB| = \frac{1}{2} 	imes \left| \frac{1 + b}{2 + b} ight| 	imes |-(1 + b)| = 2$.Since $b  0$ and $-(1 + b) > 0$.Thus,$\frac{1}{2} \cdot \frac{1 + b}{2 + b} \cdot (-(1 + b)) = 2$.$-(1 + b)^2 = 4(2 + b) \Rightarrow -(1 + 2b + b^2) = 8 + 4b \Rightarrow b^2 + 6b + 9 = 0$.$(b + 3)^2 = 0$,so $b = -3$. This satisfies $b < -2$.

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

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