Question

The function $\mathrm{f}\left(\mathrm{x}\right)=|\mathrm{px}-\mathrm{q}|+\mathrm{r}\left|\mathrm{x}\right|,\mathrm{x}\in (-\infty ,\infty )$, where $\mathrm{p}>0,\mathrm{q}>0,\mathrm{r}>0$, assumes its minimum value only at one point if :

• A. $\mathrm{p}\ne \mathrm{q}$
• B. $\mathrm{r}\ne \mathrm{q}$
• C. $\mathrm{r}\ne \mathrm{p}$
• D. $\mathrm{p}=\mathrm{q}=\mathrm{r}$

Answer 1

The correct option is C

$\mathrm{r}\ne \mathrm{p}$

Explanation for correct option:

Given: $\mathrm{f}\left(\mathrm{x}\right)=|\mathrm{px}-\mathrm{q}|+\mathrm{r}\left|\mathrm{x}\right|,\mathrm{x}\in (-\infty ,\infty )$

$$\begin{gathered} =\left\{-\mathrm{px}+\mathrm{q}-\mathrm{rx},\mathrm{x}\le 0 \\ -\mathrm{px}+\mathrm{q}+\mathrm{rx},0<\mathrm{x}<\raisebox{1ex}{$\mathrm{q}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{p}$}\right. \\ \mathrm{px}-\mathrm{q}+\mathrm{rx},\raisebox{1ex}{$\mathrm{q}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{p}$}\right.<\mathrm{x}\right\} \end{gathered}$$

Thus, $\mathrm{f}$ has only one minimum point $\mathrm{x}=0$ when $\mathrm{r}\ne \mathrm{p}$

Explanation for incorrect options:

From the figure of the function we can see that;

when $\mathrm{p}=\mathrm{q}=\mathrm{r}$, $\mathrm{f}$ has infinite number of minimum points.

When $\mathrm{p}\ne \mathrm{q}$ and $\mathrm{r}\ne \mathrm{q}$, it does not affect the minimality of the function

Hence, the correct option is option(C) i.e. $\mathrm{r}\ne \mathrm{p}$.