Question

The triangle formed by the tangent to the curve $f (x) = x2 + bx - b$ at the point $(1,1)$ and the coordinate 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

Answer: (C)

-3

Answer 2

Given curve is $y = f\left(x\right)=x^{2} + bx-b$
On differentiating w.r.t. $x$, we get
$f'\left(x\right) =2x+b$
The equation of tangent at point (1, 1) is
$y-1=\left(\frac{dy}{dx}\right)_{\left(1,1\right)} \left(x-1\right)$
$\Rightarrow y -1=\left(b+2\right)\left(x-1\right)$
$\Rightarrow \left(2+b\right)x-y=1+b$
$\Rightarrow \frac{x}{\left(\frac{1+b}{2+b}\right)} - \frac{y}{\left(1+b\right)} = 1$

So, $OA = \frac{1+b}{2+b}$
and $OB =-\left(1+b\right)$
Now, area of $\Delta AOB = \frac{1}{2}\times \frac{\left(1+b\right)\left[-\left(1+b\right)\right]}{\left(2+b\right) } = 2$
$\Rightarrow 4 \left(2 + b\right) + \left(1 + b\right)^{2} = 0$
$\Rightarrow 8+4 b+1+b^{2}+2 b=0$
$\Rightarrow b^{2} + 6b + 9 = 0$
$\Rightarrow \left(b+3\right)^{2} \Rightarrow b=-3$