Q.Let $f: R \to R$ be defined by $f(x)=\begin{cases} 1 & \text{if } x \in Q \\ -1 & \text{if } x \notin Q \end{cases}$, then $f\left(\dfrac{1}{2}\right)+f(\pi)+f(\sqrt{2})=$

Function Equality — When Are Two Functions the Same?

You write f(x) = x * x; your friend writes g(x) = x^2. Same function? Intuitively yes — for every input they give the same output.

But what if your friend's program only works for positive numbers? Then for $x = -3$ your function gives $9$ but theirs crashes. Now they're not the same function, even though the formulas look alike.

That's the core idea: two functions are equal only when they agree on every input they're supposed to handle.

The Precise Definition

Two functions $f$ and $g$ are equal ($f = g$) if and only if:

1. They have the same domain $D$.
2. For every $x$ in $D$, $f(x) = g(x)$.

If either condition fails, the functions are different.

Why This Matters

Two expressions can look identical after simplification yet have different domains — that's the exam trap.

Equal: $f(x) = \sqrt{x^2}$ and $g(x) = |x|$ — both have domain $\mathbb{R}$, and $\sqrt{x^2} = |x|$ for every $x$. So $f = g$.

Not equal: $f(x) = \frac{x-1}{x-1}$ and $g(x) = 1$ — $f$ has domain $\mathbb{R} \setminus \{1\}$, $g$ has domain $\mathbb{R}$. Different domains → $f \neq g$, even though $f(x) = 1$ wherever $f$ is defined.

A Quick Test

1. $f(x) = \sqrt{x} \cdot \sqrt{x}$, $g(x) = x$ — domain of $f$ is $x \geq 0$, domain of $g$ is all reals. Not equal (different domains).
2. $f(x) = \sin^2 x + \cos^2 x$, $g(x) = 1$ — both domain $\mathbb{R}$, equal for every $x$. Equal. …