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, then the value of 'b' is

• A. -3
• B. -2
• C. -1
• D. 0

Answer 1

The correct option is A -3

Given,

$f\left(x\right)=x^2+bx-b$

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

$f^{\prime }(x{)}_{(1,1)}=2(1)+b=2+b$

Equation of tangent at $(1,1)$

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

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

Therefore area of triangle $=|\frac{1}{2}\times \left(\frac{1+b}{2+b}\right)\times -(1+b\left)\right|=2sq.units$

$\therefore \frac{(1+b{)}^2}{2+b}=-4$

$1+b^2+2b=-8-4b$

$(b+3{)}^2=0$

$\therefore b=-3$