Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f(x)=|2x^2+5|x|-3|,x\in \mathbb{R}$. If $m$ and $n$ denote the number of points where $f$ is not continuous and not differentiable respectively, then $m+n$ is equal to :

Imagine you are standing before a function that seems to wear a mask of complexity: $f(x)=|2x^2+5|x|-3|$. At first glance, those absolute value bars might feel like a barrier, a wall between you and the solution.

But in mathematics, as in life, the most intimidating obstacles often hide the most elegant paths. Let us peel back that mask together.

The Transformation

The first step is to recognize a hidden symmetry. We see an $x^2$ term and an $|x|$ term. A fundamental property of real numbers is that $x^2=|x{|}^2$.

Why is this so powerful? Because it allows us to rewrite the entire function in terms of $|x|$. Suddenly, the function becomes:

By making this simple substitution, we have transformed a seemingly complex expression into a familiar quadratic form. We are no longer dealing with a mystery; we are dealing with a quadratic in disguise!

The Factorization

Let us define a temporary variable, $t=|x|$. Our expression inside the modulus is now $2t^2+5t-3$.

We need to factorize this. We look for two numbers that multiply to $-6$ (the product of $2$ and $-3$) and add to $5$. Those numbers are $6$ and $-1$.

So, we split the middle term: $2t^2+6t-t-3$. Factoring by grouping, we get $2t(t+3)-1(t+3)$, which simplifies to $(2t-1)(t+3)$.

Substituting $|x|$ back in, our function is now:

The Simplification

Here is where the magic happens. Look at the factor $(|x|+3)$. Since the absolute value of any real number is always non-negative, $|x|\ge 0$.

Therefore, $|x|+3$ is always at least $3$. It is strictly positive!

Because it is always positive, the outer modulus does not affect it. We can pull it out, leaving us with:

This is a much friendlier version of our original function.

The Hunt for Discontinuity

The question asks for $m$, the number of points of discontinuity. Since $|x|$ is continuous everywhere and our function is a product of continuous functions, $f(x)$ is continuous for all $x\in \mathbb{R}$.

There are no jumps, no holes, and no vertical asymptotes. Thus, $m=0$. The graph is a single, unbroken line.

The Hunt for Sharp Corners

Now, we search for $n$, the points of non-differentiability. These occur at 'sharp corners'—points where the slope changes abruptly. These corners happen where the expression inside the modulus becomes zero.

We have two cases:

1. $|x|=0$, which gives $x=0$.

2. $2|x|-1=0$, which means $|x|=\frac{1}{2}$, giving $x=\frac{1}{2}$ and $x=-\frac{1}{2}$.

We must verify these points. At $x=0$, the left-hand derivative is $5$ and the right-hand derivative is $-5$. They do not match, so $x=0$ is a point of non-differentiability.

At $x=\pm \frac{1}{2}$, the term $|2|x|-1|$ creates a V-shape, and since the other factor $(|x|+3)$ is non-zero at these points, the sharp corner persists. Thus, we have three points of non-differentiability: $0,\frac{1}{2},-\frac{1}{2}$.

So, $n=3$.

We have navigated the complexity and found our answers. With $m=0$ and $n=3$, the sum $m+n$ is simply $3$.

You have successfully decoded the function, analyzed its continuity, and rigorously tested its differentiability. This is the essence of JEE Advanced mathematics—not just calculating, but understanding the soul of the function. Keep this clarity with you as you tackle the next challenge!