Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f(x)$ be a continuous function defined for $1\le x\le 3$. If $f(x)$ takes rational values for all $x$ and $f(2)=10$, then $f(1.5)=.........$

Imagine you are standing on a bridge that spans from $x=1$ to $x=3$. You are holding a pen, and you are tasked with drawing a continuous line that represents the function $f(x)$.

The rules are strict: your pen can only touch rational coordinates. At $x=2$, you are anchored at the height $y=10$.

This problem, at first glance, feels like a riddle, but it is actually a profound test of your understanding of the topology of real numbers. Let us peel back the layers of this mystery together.

Many students instinctively try to imagine a function that wiggles around, perhaps oscillating wildly between rational values. They think, "Maybe it can just stay very close to $10$ but wiggle slightly?"

Let us test that intuition. Suppose our function $f(x)$ is not constant. If it is not constant, there must be two points, $x_1$ and $x_2$, where the function takes two different rational values, $y_1$ and $y_2$.

Because the function is continuous, the Intermediate Value Theorem (IVT) kicks in. The IVT is the law of the land for continuous functions; it dictates that if a function starts at $y_1$ and ends at $y_2$, it must pass through every single real number in the interval between $y_1$ and $y_2$.

The Dealbreaker

Here is where the physics of the problem collapses the possibility of a non-constant function. Between any two distinct real numbers $y_1$ and $y_2$, there exists an infinite sea of irrational numbers.

If our function were to move from $y_1$ to $y_2$, it would be forced to cross this sea. It would have to take on irrational values. But the problem explicitly forbids this!

The constraint $f(x)\in \mathbb{Q}$ acts as a cage. The function cannot escape into the irrational realm. Therefore, the only way to satisfy the condition of continuity while remaining strictly within the rational numbers is to never move at all.

The function cannot wiggle, it cannot curve, and it cannot slope. It must be perfectly flat.

Since the function must be constant, and we know that $f(2)=10$, the function must be:

This holds for all $x$ in the interval $[1,3]$. There is no room for variation.

When we ask for $f(1.5)$, we are simply looking at another point on this perfectly horizontal line. Thus, $f(1.5)=10$.

This problem teaches us that continuity is a powerful, rigid constraint. It binds the function to the topology of the real line, and when combined with a countable range like the rationals, it leaves only one possible outcome: total, unyielding constancy.

Keep this elegant logic in your toolkit; it is a classic JEE Advanced trap that rewards those who look past the algebra to the geometric soul of the function.