Question

The triangle formed by the tangent to the curve $f(x)=x^{2}+b x-b$ 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 $y=f(x)=x^{2}+b x-b$
On differentiating w. r. t $x$, we get
$\frac{d y}{d x}=2 x +b$
The equation of the tagent at $(1,1)$ is
$y-1=\left(\frac{d y}{d x}\right)_{(1,1)}(x-1)$
$\Rightarrow y-1=(b+2)(x-1)$
$\Rightarrow (2+b) x-y=1+b$
$\Rightarrow \frac{x}{(1+b) /(2+b)}-\frac{y}{(1+b)}=1$
So, $O A=\frac{1+b}{2+b}$ and $O B=-(1+b)$

Now, area of $\triangle A O B$
$=\frac{1}{2} \frac{(1+b)}{(2+b)}[-(1+b)]=2$ (given)
$\Rightarrow 4(2+b)+(1+b)^{2}=0$
$\Rightarrow 8+4 b+1+b^{2}+2 b=0$
$\Rightarrow b^{2}+6 b+9=0$
$\Rightarrow (b+3)^{2}=0$
$\Rightarrow b=-3$