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 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)

$-3$

Given curve is $y=f\left(x\right)=x^2+bx-b$

$\Rightarrow f^{\prime }\left(x\right)=2x+b$

The equation of tangent at point (1, 1) is

$y-1={\left(\frac{dy}{dx}\right)}_{(1,1)}(x-1)$

$\Rightarrow y-1=(b+2)(x-1)$

$\Rightarrow (2+b)x-y=1+b$

$\Rightarrow \frac{x}{\left(\frac{1+b}{2+b}\right)}-\frac{y}{(1+b)}=1$

So, $OA=\frac{1+b}{2+b}$

and $OB=-(1+b)$

Now, area of $\Delta AOB=\frac{1}{2}\times \frac{(1+b)[-(1+b\left)\right]}{(2+b)}=2$

$\Rightarrow 4(2+b)+(1+b{)}^2=0$

$\Rightarrow b^2+6b+9=0$

$\Rightarrow (b+3{)}^2=0\Rightarrow b=-3$.

Hence, option C is correct.