Animated Solution for Mathematics - Functions: Let $f$ be a one-one function with domain $\{x,y,z\}$ and range $\{1,2,3\}$. It is given that exactly one of the following statements is true and the remaining two are false $f(x)=1,f(y)\ne 1,f(z)\ne 2$ determine $f^{-1}(1)$.

Welcome, future engineer. Today, we are not just solving a math problem; we are embarking on a journey of logical deduction. In the high-stakes arena of JEE Advanced, you will often encounter problems that seem to be about functions or calculus, but are actually tests of your ability to organize information and handle constraints.

We are given a function $f:\{x,y,z\}\to \{1,2,3\}$ and a set of three statements, where exactly one is true. This is a classic logic puzzle, and to solve it, we must become detectives of the mathematical realm.

We have a domain set $D=\{x,y,z\}$ and a range set $R=\{1,2,3\}$. The function $f$ is a rule that assigns each element of $D$ to exactly one element of $R$.

There is a golden rule here: the function is one-one (injective). This means that no two elements in the domain can share the same output. If $f(x)=1$, then neither $y$ nor $z$ can map to $1$. This constraint is our most powerful tool.

We are told that exactly one of the following statements is true: 1. $f(x)=1$ 2. $f(y) eq 1$ 3. $f(z) eq 2$

Since we do not know which one is true, we must test each case. This is a systematic siege. We will assume one is true and see if the universe of the problem remains consistent. If we hit a contradiction—a violation of the one-one rule—we know our assumption was wrong.

Case 1: The False Start Let us assume Statement 1 is true. This means $f(x)=1$.

If this is the only true statement, then Statement 2 must be false. The negation of $f(y) eq 1$ is $f(y)=1$.

Now, look at the disaster we have created: both $x$ and $y$ are mapping to $1$. This violates the one-one property immediately. Case 1 is impossible.

Case 2: The Near Miss Let us assume Statement 2 is true. This means $f(y) eq 1$. Consequently, Statement 1 and Statement 3 must be false.

If Statement 1 is false, then $f(x) eq 1$. If Statement 3 is false, then $f(z)=2$.

Now, we have $f(z)=2$ and $f(x) eq 1$. Since $f(x)$ cannot be $1$ and cannot be $2$ (because $z$ took $2$), $f(x)$ must be $3$. This leaves $f(y)=1$. But wait! Our assumption for this case was that $f(y) eq 1$. We have arrived at a contradiction again. Case 2 is also impossible.

Case 3: The Breakthrough We are left with only one path: Case 3. Assume Statement 3 is true, meaning $f(z) eq 2$. This implies Statements 1 and 2 are false.

If Statement 1 is false, then $f(x) eq 1$. If Statement 2 is false, then $f(y)=1$.

Now, let us map this out. We know $f(y)=1$. We know $f(z) eq 2$. Since $1$ is already taken by $y$, $f(z)$ cannot be $1$. Since $f(z)$ cannot be $2$ (by our assumption) and cannot be $1$, $f(z)$ must be $3$. This leaves only one value for $x$: $f(x)=2$.

Let us verify our findings: $f(x)=2$, $f(y)=1$, and $f(z)=3$. This mapping is one-one, as every input has a unique output.

Checking the statements: 1. $f(x)=1$ is false (it is $2$). 2. $f(y) eq 1$ is false (it is $1$). 3. $f(z) eq 2$ is true (it is $3$).

Everything is consistent. The question asks for $f^{-1}(1)$, which is the input that maps to $1$. Looking at our valid mapping, we see that $f(y)=1$.

Therefore, the final answer is: $f^{-1}(1)=y$