Question

The number of critical points of $f(x)=\frac{|x-1|}{x^2}$ is

• A. 1
• B. 2
• C. 3
• D. none of these.

Answer 1

Answer: (C)

$f\left(x\right)=\frac{|x-1|}{x^2}=\frac{x-1}{x^2},x\ge 1$
$=\frac{1-x}{x^2},x<1$
Clearly $f(x)$ is not diff. at $x=0$ and $x=1$.
$\therefore$ by def. there are two of the critical points.
For points other than these two, we have
$f^{\prime }\left(x\right)=\frac{x^2\cdot 1-\left(x-1\right)2x}{x^4}$
i.e, $\frac{-x+2}{x^3},x>1$
$=\frac{x^2\left(-1\right)-\left(1-x\right)2x}{x^4}$
i.e., $\frac{x-2}{x^3}$, $x<1$
$f^{\prime }\left(x\right)=0$ at $x=2$
$\therefore x=2$ is also a critical point. Hence $f(x)$ has three critical points i.e. $(0,1,2)$.