Question

The triangle formed by the tangent to the curve $f(x)=x^2+bx-b$ a t the point (1,1) and the coordinate axes, lies in the first quadrant. If its area is 2 sq units, then the value of b is

• A. 2
• B. 3
• C. -3
• D. 1

Answer 1

Answer: (C)

-3

Answer 2

Let $y=f(x)=x^2+bx-b$ The equation of the tangent at P (1, 1) to the curve 2y =$2x^2+2bx-2b$ is $y+1 = 2x.1 + 6(x+1) - 26$ $\therefore$ $y = (2+b)x-(1+b)$ Its meet the coordinate axes at $x_A=\frac{1+b}{2+b}$ and $y_b=-(1-b)$ $\therefore$ Area of $\Delta$ OAB = $\frac{1}{2}OA \times OB$ $-\frac{1}{2}\times\frac{(1+b)^2}{(2+b)}=2$ $\Rightarrow$ $(1+b)^2+4(2+b)=0 \Rightarrow b^2+6b+9=0$ $\Rightarrow (b+3)^2=0 \Rightarrow b=-3$