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 co-ordinate axes, lies in the first quadrant.If its area is $2$ sq.unit, then the value of $b$ is:

• A. $-3$ sq.unit
• B. $-2$ sq.unit
• C. $-1$ sq.unit
• D. $0$ sq.unit

Answer 1

The correct option is A $-3$ sq.unit

Equation of tangent at $(1,1)$ is
$y-1=(2+b)(x-1)$
For first quadrant $b<0$
$\therefore$ Area$=2$ (given)
$\frac{1}{2}\times \left(\frac{b+1}{b+2}\right)\times (-1-b)=2$
$\Rightarrow {(b+1)}^2+4b+8=0$
$\Rightarrow b^2+6b+9=0$
$\Rightarrow {(b+3)}^2=0$
$\therefore b=-3$