Animated Solution for Mathematics - Limits, Continuity and Differentiability: The number of points, where the function $f:\mathbb{R}\to \mathbb{R}$, $f(x)=|x-1|\cos |x-2|\sin |x-1|+(x-3)|x^2-5x+4|$, is NOT differentiable, is :

When you first look at the function: it is natural to feel a spike of anxiety. There are modulus bars everywhere!

But in the world of JEE Advanced, intimidation is just a mask for a beautiful, structured reality. Let us peel back the layers of this function together.

The first step in any complex problem is to simplify the landscape. Look at the quadratic term: $|x^2-5x+4|$. This is a classic quadratic that begs to be factorized as $(x-1)(x-4)$.

So, our function becomes:

Notice something? The term $|x-1|$ is a common factor! By factoring it out, we get:

This is a massive breakthrough. We have transformed a sprawling expression into a product of a simple modulus and a more complex function. Let us call the expression inside the square brackets $g(x)$. Now, our function is simply $f(x)=|x-1|\cdot g(x)$.

Where can a function fail to be differentiable? Only where the 'sharp corners' exist. These corners occur where the arguments inside the modulus bars are zero.

Looking at our function, we have three primary suspects: $x=1$, $x=2$, and $x=4$. These are the points where the geometry of the function might break. We need to investigate each one with the precision of a surgeon.

Before we test these points, let us recall a powerful tool in our JEE toolkit. For a function $F(x)=|x-a|\cdot g(x)$, the function is differentiable at $x=a$ if and only if $g(a)=0$.

Why? Because if $g(a)=0$, the function $g(x)$ effectively 'flattens' the sharp cusp of the modulus, making the transition smooth. If $g(a) eq 0$, the cusp remains, and the derivative does not exist. This is our litmus test.

Let us test our suspects one by one. First, consider $x=1$. We have our function in the form $f(x)=|x-1|\cdot g(x)$, where $g(x)=\cos |x-2|\sin |x-1|+(x-3)|x-4|$.

To check differentiability at $x=1$, we evaluate $g(1)$:

Since $g(1)=-6$, which is clearly not zero, the function is NOT differentiable at $x=1$. The sharp corner survives!

Next, let us look at $x=2$. Here, the term $\cos |x-2|$ appears. Many students panic here, thinking the modulus creates a cusp.

But remember your trigonometry! The cosine function is even, meaning $\cos (-\theta )=\cos (\theta )$. Therefore, $\cos |x-2|=\cos (x-2)$. This is a smooth, continuous, and differentiable function. There is no cusp here, and the function is differentiable at $x=2$.

Finally, let us check $x=4$. We look at the second term of our original function: $(x-3)|x-1||x-4|$. We can rewrite this as $|x-4|\cdot [(x-3)|x-1|]$. Let $h(x)=(x-3)|x-1|$.

For the function to be differentiable at $x=4$, we need $h(4)=0$. Let us calculate:

Since $h(4) = 3 eq 0$, the function is NOT differentiable at $x=4$.

We have systematically tested our suspects. At $x=1$, the function fails. At $x=2$, it passes with flying colors. At $x=4$, it fails again.

Thus, there are exactly two points of non-differentiability. You see? When you break a problem down into its fundamental components—factorization, critical points, and the Golden Rule—even the most intimidating functions reveal their secrets.