Animated Solution for Mathematics - Basic Mathematics: The function $f(x)=|px-q|+r|x|,x\in (-\infty ,\infty )$ where $p>0,q>0,r>0$ assumes its minimum value only on one point if

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to dissect a problem that seems simple on the surface but hides a beautiful geometric truth.

We are looking at the function $f(x)=|px-q|+r|x|$, where $p,q,r>0$. At first glance, it is just a sum of two absolute values.

In the world of JEE, we don't just solve equations; we visualize them. Imagine this function as a path you are walking on. The points where the path changes direction are the secrets to unlocking this problem.

To understand the shape of our function, we must identify the points where the behavior changes. These are the critical points where the arguments inside the modulus signs become zero.

For the first term, $|px-q|$, the argument is zero when $px-q=0$, which gives us:

For the second term, $r|x|$, the argument is zero when $x=0$. Since $p,q>0$, we know that $\frac{q}{p}$ is a positive value. Thus, our number line is divided into three distinct regions: $(-\infty ,0]$, $[0,\frac{q}{p}]$, and $[\frac{q}{p},\infty )$.

Let us analyze the function in each region. In the first region, where $x\le 0$, both $px-q$ and $x$ are negative.

The function becomes:

The slope here is $-(p+r)$, which is strictly negative. The function is sliding downhill.

Now, let us jump to the third region, where $x\ge \frac{q}{p}$. Here, both expressions are positive, so:

The slope is $(p+r)$, which is strictly positive. The function is climbing uphill. This confirms that the minimum must lie somewhere between $x=0$ and $x=\frac{q}{p}$.

This is where the magic happens. In the middle interval, $0<x<\frac{q}{p}$, the term $x$ is positive, but $px-q$ is still negative.

Substituting these into our function, we get:

The slope of this middle segment is $m=r-p$. This single value, $r-p$, is the gatekeeper of our solution.

If $r-p=0$, the slope is zero, and the function is a horizontal line in this region. If the function is horizontal, every point in the interval $[0,\frac{q}{p}]$ is a minimum point, leading to infinitely many minima.

Since the problem requires a unique minimum, we must have $r - p eq 0$, which means $r eq p$.

If $r>p$, the slope is positive, and the minimum is at $x=0$. If $r<p$, the slope is negative, and the minimum is at $x=\frac{q}{p}$. In both these cases, we get a unique minimum.

Keep this visual approach in your toolkit, and you will never fear piecewise functions again!