Question

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 lie in the first quadrant. If its area is $2$, then the value of $b$ is

• A. $-1$
• B. $3$
• C. $-3$
• D. $1$

Answer 1

The correct option is C

$-3$

Explanation for correct option :

Let $y=f\left(x\right)$

Given $f\left(x\right)=x^2+bx-b$

$$\begin{gathered} m=\frac{dy}{dx} \\ \Rightarrow m=\frac{d}{dx}\left(x^2+bx-b\right) \\ \Rightarrow m=2x+b \end{gathered}$$

The equation of the tangent at $(1,1)$

$\begin{array}{rcl}y-y_1& =& m\left(x-x_1\right)\\ & \Rightarrow & y-1=\left(b+2\right)\left(x-1\right)\\ & \Rightarrow & y-1=bx-b+2x-2\\ & \Rightarrow & y-x\left(b+2\right)+\left(b+1\right)=0\end{array}$

We know that Area of triangle $=\frac{1}{2}\times \left(\mathrm{base}\right)\times \left(\mathrm{height}\right)$

$\begin{array}{rcl}& \Rightarrow & 2=\frac{1}{2}\times \frac{b+1}{b+2}\times \left(-\left(b+1\right)\right)\\ & \Rightarrow & 4b+8=-b^2-2b-1\\ & \Rightarrow & b^2+6b+9=0\\ & \Rightarrow & {\left(b+3\right)}^2=0\\ & \Rightarrow & b=-3\end{array}$

Hence $\mathrm{Option}\left(\mathrm{C}\right)$ is the answer.