Question

The triangle formed by the tangent to the curve $f\left(x\right)=x^2+bx-b$ at the point $(1,1)$ and the coordinates axes, lies in the first quardant. 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 $-3$

Let $y=f\left(x\right)=x^2+bx-b$
The equation of the tangent at $P(1,1)$ to the curve $2y={2x}^2+2bx-2b$
is $y+1=2x\times 1+b(x+1)-2b\Rightarrow y=x(2+b)-(1+b)$
Its meet the coordinate axes at
$x_A=\frac{1+b}{2+b}$ and $y_B=-(1+b)$
Therefore $△OAB=\frac{1}{2}\times OA\times OB$
$=-\frac{1}{2}\times \frac{{(1+b)}^2}{2+b}=2\Rightarrow {(1+b)}^2+4(2+b)=0\Rightarrow b^2+6b+9=0\Rightarrow {(b+3)}^2=0\Rightarrow b=-3$