Explanation

(c)
Given curve is $y = f(x) = x^{2} + bx - b$
On differentiating, we get $\frac{dy}{dx} = 2x + b$
The equation of the tangent at (1, 1) is
$y - 1 = \left( \frac{dy}{dx} \right)_{(1,\ 1)} = (x - 1)$
$\Rightarrow y - 1 = (b + 2)(x - 1)$
$\Rightarrow \ \ (2 + b)x - y = 1 + b$
$\Rightarrow \ \frac{x}{\frac{(1 + b)}{(2 + b)}} - \frac{y}{1 + b} = 1$
So, OA$= \frac{1 + b}{2 + b}$
and $OB = - (1 + b)$
[image:6_image14.emf]
Now, area of $\mathrm{\Delta}\ AOB = \frac{1}{2}.\frac{1 + b}{2 + b}.\left\lbrack - (1 + b) \right\rbrack = 2$ (given)
$\Rightarrow \ \ \ 4(2 + b) + (1 + b)^{2} = 0$
$\Rightarrow \ \ \ 8 + 4b + 1 + b^{2} + 2b = 0$
$\Rightarrow \ \ \ b^{2} + 6b + 9 = 0$
$\Rightarrow \ \ (b + 3)^{2} = 0\ \ \Rightarrow b = - 3$