Question

The function $\frac{|x-1|}{x^2}$ is monotonically decreasing at the point

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

Answer 1

Answer: (A)

$f(x)=\{\begin{array}{ll}\frac{1-x}{x^2}& x<1,x\ne 0\\ \frac{x-1}{x^2},& x\ge 1\end{array}$
The given function is not differentiable at $x=1$

$f^{\prime }(x)=\{\begin{array}{llc}\frac{1}{x^2}-\frac{2}{x^3},& x<1,& x\ne 0\\ \frac{2}{x^3}-\frac{1}{x^2},& x>1& \end{array}$
Now $f^{\prime }(x)<0\Rightarrow \{\begin{array}{llc}\frac{x-2}{x^3}<0& \text{ given }& x<1\\ \frac{2-x}{x^3}<0& \text{ when }& x>1\end{array}$
$f(x)$ decreasing $\forall x\in (0,1)\cup (2,\infty )$ and $f(x)$ increases $\forall x\in (-\infty ,0)\cup (1,2)$
here $f(x)$ is decreasing at all points in $x\in (0,1)\cup (2,\infty )$ so will also be decreasing at $x=3$ at $x=1$ minima and at $x=2$ maxima