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

Answer 1

The correct option is C $\frac{1}{6}$

Given,
$\sqrt{(x+y)}+\sqrt{(y-x)}=a$ . . . .(i)
$\therefore \sqrt{x+y}-\sqrt{y-x}=\frac{2x}{a}$ ....(ii)
Adding Eqs. (i) and (ii), then
$2\sqrt{x+y}=a+\frac{2x}{a}$
Squaring, $4x+4y=a^2+\frac{4x^2}{a^2}+4x$
$\therefore 4+4\frac{dy}{dx}=0+\frac{8x}{a^2}+4$
$\therefore 0+4\frac{d^{2}y}{d{x}^2}=\frac{8}{a^2}$
$\therefore \frac{d^{2}y}{d{x}^2}=\frac{2}{a^2}$