Question

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

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