Question

The triangle formed by the tangent to the curve $f\left(x\right)=x^2+bx-b$ at the point $(1,10)$ 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

The correct option is C $-3$

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

$\Rightarrow b=-3$