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 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: Find the equation of the tangent

Given equation of the curve is $f\left(x\right)=x^2+bx-b$

The equation of the tangent of the curve at point $\left(x_0,y_0\right)$ is given by

$\left(y-y_0\right)=m\left(x-x_0\right)$ where $m$ is the slope of the curve.

The slope of the curve $f\left(x\right)=x^2+bx-b$ is

$$\begin{gathered} m=\frac{dy}{dx} \\ =\frac{df\left(x\right)}{dx} \\ =2x+b \end{gathered}$$

At $(1,1)$, $m=2+b$

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

$$\begin{gathered} \left(y-1\right)=\left(2+b\right)\left(x-1\right) \\ \Rightarrow \frac{y}{2+b}-x=\frac{1}{2+b}-1 \\ \Rightarrow \frac{y}{2+b}-x=\frac{-1-b}{2+b} \\ \Rightarrow \frac{x}{{\displaystyle \frac{b+1}{2+b}}}-\frac{y}{1+b}=1 \end{gathered}$$

Step-2: Find the value of $b$

We know that the line $\frac{x}{a}+\frac{y}{b}=1$ intercepts the $x$ and $y$ axis at $a$ and $b$ respectively

Therefore the tangent intercept the $x-$ axis at $\frac{b+1}{b+2}$ and the $y-$ axis at $-\left(1+b\right)$

So the height of the triangle is $-\left(1+b\right)$ and the length of the base is $\frac{b+1}{b+2}$

Given that the area of the triangle is $2$

$\therefore$$\frac{1}{2}\times$ (base) $\times$ (height) $=2$

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

Hence option $\left(C\right)$ i.e. $-3$ is correct