Question

The slope of the normal at the point with abscissa $x=-2$ of the graph of the function $f\left(x\right)=|(x^2-|x|)|$, is?

• A. $\frac{1}{3}$
• B. $-\frac{1}{3}$
• C. $\frac{1}{6}$
• D. $-\frac{1}{6}$

Answer 1

The correct option is A $\frac{1}{3}$

Given that $f\left(x\right)=|x(x-1)|$

$f\left(x\right)=\{\begin{array}{cc}x(x+1)& ,\forall x<0\\ -x(x-1),& 0⩽x\le 1\\ x(x-1),& x>1\end{array}$

$f(-2)=-2(-3)=6$

$\therefore$ The point is (-2,6)

Now $,\frac{dy}{dx}=(x+1)+x=2x+1$

${\left(\frac{dy}{dx}\right)}_{x=-2}=-4+1=$-3$

slope of tangent $m=-3$

$\therefore$ slope of Normal $=\frac{-1}{m}=\frac{1}{3}$

Hence, $\left(A\right)$ is the correct option.