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, lies in the first quadrant. If the area is 2 then the value of b is
A. -1
B. 3
C. -3
D. 1

Hint:
Formula Used: Area of triangle = $\dfrac{1}{2}\times base\times height$

Complete step by step Solution:
Given that
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, lies in the first quadrant
To find the value of b
Here $f\left( x \right)={{x}^{2}}+bx-b$
Slop m is given by
$m=\dfrac{dy}{dx}$
By finding a slop of given equation $f\left( x \right)={{x}^{2}}+bx-b$
\[\begin{align}
  & m=f\left( x \right)\dfrac{dy}{dx} \\
 & m=\left( {{x}^{2}}+bx-b \right)\dfrac{dy}{dx} \\
 & m=2x+b \\
\end{align}\]
At point (1,1)
$m=2+b$
The equation of the tangent at a point whose slop is $m=2+b$
Here the equation of circle $\begin{align}
  & \left( y-Y \right)=m\left( x-X \right) \\
 & \\
\end{align}$
$\begin{align}
  & \left( y-Y \right)=m\left( x-X \right) \\
 & \\
\end{align}$
Where (X, Y) is the fixed coordinates whose value is (1,1)
$\begin{align}
  & \left( y-Y \right)=m\left( x-X \right) \\
 & \\
\end{align}$
$\begin{align}
  & \left( y-1 \right)=\left( 2+b \right)\left( x-1 \right) \\
 & \left( y-1 \right)=2x-2+bx-b \\
 & y=\left( 2+b \right)x-\left( 1+b \right) \\
 & \\
 & \\
\end{align}$
By dividing (1+b) 0n both sides we get
\[\begin{align}
  & y=\left( 2+b \right)x-\left( 1+b \right) \\
 & \dfrac{y}{\left( 1+b \right)}=\dfrac{\left( 2+b \right)x}{\left( 1+b \right)}-1 \\
 & \\
\end{align}\]$\left( \dfrac{1+b}{2+b} \right)$
Here when we put x=o we get y =$-\left( 1+b \right)$
Or when we put y=0 we get x=$\left( \dfrac{1+b}{2+b} \right)$
Here x represents base and y represents the height
Area of triangle = $\dfrac{1}{2}\times base\times height$
\[\begin{align}
  & 2=\dfrac{1}{2}\times base\times height \\
 & 4=base\times height \\
 & 4=\left( \dfrac{1+b}{2+b} \right)\times -\left( 1+b \right) \\
 & 4\left( 2+b \right)={{\left( 1+b \right)}^{2}} \\
 & 8+4b=-\left( 1+{{b}^{2}}+2b \right) \\
 & 8+4b=-1-{{b}^{2}}-2b \\
 & ={{b}^{2}}+6x+9 \\
 & ={{\left( b+3 \right)}^{2}} \\
 & b=-3 \\
\end{align}\]

Therefore, the correct option is (C).

Note: while doing the question you should revise the formula if you forgot the formula and used the wrong formula you get the wrong answer