Question

The triangle formed by the tangent to the curve $f\left(x\right)=x^2+bx-b$ at the point $\left(1,1\right)$ and the co-ordinate axes, lies 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

Step 1: Solve for the equation of tangent

Given that triangle formed by the tangent to the curve $f\left(x\right)=x^2+bx-b$ at the point $\left(1,1\right)$ and the co-ordinate axes, lies in the first quadrant and its area is $2$
Given function $f\left(x\right)=x^2+bx-b$
We know that equation of line passing through $\left(x_1,y_1\right)$ with slope $m$ is $y-y_1=m\left(x-x_1\right)$

Here $\frac{df\left(x\right)}{dx}$ is slope. Equation of tangent at $\left(1,1\right)$ is
$$\begin{gathered} \Rightarrow y-y_1={\left(\frac{df\left(x\right)}{dx}\right)}_{(x_1,y_1)}\left(x-x_1\right) \\ \Rightarrow y-1={\left(\frac{d(x^2+bx-b)}{dx}\right)}_{(1,1)}\left(x-1\right) \\ \Rightarrow y-1=(2+b)(x-1) \\ \Rightarrow 1+b=(2+b)x-y \\ \Rightarrow \frac{x}{\left({\displaystyle \frac{1+b}{2+b}}\right)}-\frac{y}{1+b}=1 \end{gathered}$$

Step 2: Solve for the required value
Intercept form of straight line is $\frac{x}{a}+\frac{y}{b}=1$
Area of triangle formed by the intercept form of line with coordinate axis $=\frac{1}{2}\times a\times b$

Therefore according to the given data area of the required triangle,

$$\begin{gathered} \Rightarrow \frac{1}{2}\times \left(\frac{1+b}{2+b}\right)\left[-\left(1+b\right)\right]=2 \\ \Rightarrow 4(2+b)+{(1+b)}^2=0 \\ \Rightarrow 8+4b+b^2+2b+1=0\mathbf{[}\mathbf{\because }{\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{a}\mathbf{b}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{]} \\ \Rightarrow b^2+6b+9=0 \\ \Rightarrow {(b+3)}^2=0 \\ \Rightarrow b=-3 \end{gathered}$$

Hence option(C) is correct.