Animated Solution for Mathematics - Functions: Let $\{x\}$ and $[x]$ denotes the fractional and integral part of a real number $x$ respectively. Solve $4\{x\}=x+[x]$.

Welcome, fellow traveler on the road to JEE Advanced. Today, we are going to demystify a problem that often trips up even the brightest minds. We are looking at the equation $4\{x\}=x+[x]$.

At first glance, it looks like a simple algebraic expression, but it is actually a beautiful interplay between the discrete world of integers and the continuous world of real numbers.

The Key to the Kingdom

Before we dive into the algebra, let us pause and appreciate the structure of any real number $x$. Every number you have ever encountered can be split into two parts: its 'floor' or integer part, denoted as $[x]$, and its 'remainder' or fractional part, denoted as $\{x\}$.

This is the fundamental identity: $x=[x]+\{x\}$.

Think of this as a coordinate system for numbers. If you are standing at $\mathrm{1.666...}$, your integer part is $1$, and your fractional part is $\mathrm{0.666...}$. By substituting this identity into our original equation, we transform a confusing expression into a clear, solvable relationship.

Watch what happens when we replace $x$ in $4\{x\}=x+[x]$:

Now, let us group our terms. We have two $[x]$ terms on the right side, giving us $2[x]$.

Our equation now reads $4\{x\}=2[x]+\{x\}$. If we subtract $\{x\}$ from both sides, we arrive at a much cleaner, more elegant form:

This is the heart of the problem. We have successfully isolated the fractional part on one side and the integer part on the other.

But here is where the 'JEE magic' happens. We cannot just solve this like a standard linear equation because $\{x\}$ is not just any variable. It is bound by the ironclad definition $0\le \{x\}<1$.

Narrowing the Possibilities

Since we know $3\{x\}=2[x]$, we can use our constraint on $\{x\}$ to create a boundary for $[x]$. If $0\le \{x\}<1$, then multiplying by $3$ gives us $0\le 3\{x\}<3$.

Because $3\{x\}$ is equal to $2[x]$, we can substitute it directly:

Dividing by $2$, we find that $0\le [x]<1.5$. This is a massive breakthrough!

Since $[x]$ must be an integer, the only possible values for $[x]$ are $0$ and $1$. We have effectively reduced an infinite number line down to just two specific cases.

Now, we simply test our candidates.

Case 1: If $[x]=0$, then $3\{x\}=2(0)$, which means $\{x\}=0$. Combining these, $x=[x]+\{x\}=0+0=0$. This is our first solution.

Case 2: If $[x]=1$, then $3\{x\}=2(1)$, which means $\{x\}=\frac{2}{3}$. Combining these, $x=[x]+\{x\}=1+\frac{2}{3}=\frac{5}{3}$. This is our second solution.

Look at what we have achieved. We started with a daunting equation involving functions that seem to 'break' standard algebra, and through the power of the fundamental identity and careful constraint analysis, we found the exact values.

The solutions are $x=0$ and $x=\frac{5}{3}$.

Remember, in JEE Advanced, the math is rarely about brute force. It is about understanding the 'soul' of the functions you are working with. Keep practicing, keep questioning, and most importantly, keep finding the beauty in the logic.