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)

Given curve is $y=f\left(x\right)=x^2+bx-b$
On differentiating w.r.t. $x$, we get
$f^{\prime }\left(x\right)=2x+b$
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 $\Delta AOB=\frac{1}{2}\times \frac{\left(1+b\right)\left[-\left(1+b\right)\right]}{\left(2+b\right)}=2$
$\Rightarrow 4\left(2+b\right)+{\left(1+b\right)}^2=0$
$\Rightarrow 8+4b+1+b^2+2b=0$
$\Rightarrow b^2+6b+9=0$
$\Rightarrow {\left(b+3\right)}^2\Rightarrow b=-3$