Question

Slope of tangent to the curve $y^3=3a{x}^2+6x+b$ at $(1,1)$ is $2,$ then the absolute value of $a+b$ is

Answer 1

Given, $y^3=3a{x}^2+6x+b$
Differentiating w.r.t. $x$
$\Rightarrow 3y^2\frac{dy}{dx}=6ax+6$
$\Rightarrow {\frac{dy}{dx}|}_{(1,1)}=\frac{6(a+1)}{3}=2(a+1)\Rightarrow 2(a+1)=2$
$\Rightarrow a=0$
and $(1,1)$ lies on the curve $y^3=3a{x}^2+6x+b$
$\Rightarrow 1=3a+6+b$
$\Rightarrow b=1-3a-6$
$\Rightarrow b=1+0-6=-5$
$\therefore |a+b|=5$