Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let the function $f(x)=(x^2+1)|x^2-ax+2|+\cos |x|$ be not differentiable at the two points $x=\alpha =2$ and $x=\beta$. Then the distance of the point $(\alpha ,\beta )$ from the line $12x+5y+10=0$ is equal to:

Unmasking the Beast

When you first look at a function like $f(x)=(x^2+1)|x^2-ax+2|+\cos |x|$, it is natural to feel a surge of intimidation. It looks like a chaotic mix of polynomials, absolute values, and trigonometry.

But in the world of JEE Advanced, we don't fear functions; we dissect them. We are going to peel back the layers of this expression to reveal the elegant geometry hiding underneath.

The Detective Work

First, let's identify our suspects. We have three components here. The term $(x^2+1)$ is a simple polynomial—it is smooth, continuous, and differentiable everywhere.

Then we have $\cos |x|$. Remember, cosine is an even function, meaning $\cos (-x)=\cos (x)$. Therefore, $\cos |x|$ is just $\cos x$, which is also perfectly smooth.

So, where does the non-differentiability come from? It must be the absolute value term: $|x^2-ax+2|$.

In calculus, an absolute value function $|g(x)|$ creates a 'sharp corner' or a cusp exactly where $g(x)=0$. This is where the graph 'bounces' off the x-axis, and the derivative fails to exist.

The problem gives us a gift: it tells us one of these points is at $x=\alpha =2$. This means that when we plug $x=2$ into our quadratic expression, the result must be zero.

Solving for the Unknown

Now, we act like detectives. We set up our equation: $x^2-ax+2=0$ at $x=2$.

Substituting the value, we get $(2{)}^2-a(2)+2=0$. This simplifies beautifully to $4-2a+2=0$, or $6-2a=0$. Solving for $a$, we find $a=3$.

With $a$ unlocked, our quadratic is no longer a mystery; it is $x^2-3x+2$. To find the second point of non-differentiability, $\beta$, we simply need the other root of this quadratic.

We factorize: $x^2-3x+2=(x-2)(x-1)=0$. The roots are $x=2$ and $x=1$.

Since we already knew $\alpha =2$, our second point must be $\beta =1$. We have successfully mapped our point of interest to $P(2,1)$.

The Geometric Finale

We have arrived at the final stage of our journey. We have a point $P(2,1)$ and a line $12x+5y+10=0$. The problem asks for the perpendicular distance from this point to the line.

This is a classic application of the distance formula from coordinate geometry:

Here, $A=12$, $B=5$, and $C=10$. Plugging in our coordinates $(2,1)$, the numerator becomes:

The denominator is the magnitude of the normal vector:

Finally, we calculate the distance:

Look at that! The complexity of the original function melted away, leaving us with a clean, integer result.

This is the beauty of mathematics—no matter how intimidating the problem looks, if you break it down into its fundamental components, the path to the solution becomes clear. You have mastered the function, and you have mastered the geometry. The final answer is 3.