JEE Corner — Relations and Functions for JEE Main and Advanced

Why This Section Exists

The textbook chapter gives you the definitions; JEE tests how fast you can compute with them. The high-yield toolkit this section adds:

1. Counting formulas — relations, functions, and functions with conditions;
2. The domain playbook — the complete checklist for square roots, rational functions and their combinations;
3. Greatest integer and fractional part — the properties of $[x]$ and $\{x\}$ that power a whole family of JEE one-liners;
4. Modulus equations and inequalities — $|x| = a$, $|x| < a$, $|x| > a$ solved as intervals;
5. A preview of one-one and onto — the Class 12 vocabulary that JEE assumes early.

The worked examples and practice questions here are modelled on the JEE Main and Advanced pattern and difficulty; they are practice questions in the exam style, not reproductions of specific past papers.

[Board Note] CBSE students: counting functions, $\{x\}$ and the modulus-inequality machinery sit beyond the Class 11 Board syllabus. Treat this section as a JEE investment — the domain playbook alone will repay it in the calculus chapters.

Counting Relations and Functions

Let $n(A) = m$ and $n(B) = n$.

• Relations from A to B: subsets of $A \times B$, so $2^{mn}$.
• Functions from A to B: each of the m inputs independently picks one of n images: $n^m$.
• Functions with a fixed value (say $f(a_1) = b_1$ forced): the rest choose freely: $n^{m-1}$.
• One-one functions (distinct inputs, distinct images; needs $m \leq n$): $n(n-1)(n-2)\cdots(n-m+1)$ — images chosen without repetition.
• Non-empty relations: $2^{mn} - 1$; relations that are NOT functions: $2^{mn} - n^m$.

Worked contrast for m = 3, n = 2: relations $2^6 = 64$; functions $2^3 = 8$; one-one functions: impossible to squeeze 3 distinct images into 2 targets — zero.

Key Point: Read the exponent from the DOMAIN: $n^m$ has the domain size upstairs. Muddling $n^m$ with $m^n$ (or with $2^{mn}$) is the single most common error in this topic.

[JEE Tip] "How many functions from A to B are not one-one?" = total minus one-one = $n^m - n(n-1)\cdots(n-m+1)$. Compute the two counts separately; never try to count non-injective maps directly.

The Domain Playbook

Every JEE domain question decomposes into these atomic rules, applied together and intersected:

1. Polynomial: domain $\mathbb{R}$ — no restriction.
2. $\frac{1}{g(x)}$: exclude the roots of g.
3. $\sqrt{g(x)}$: require $g(x) \geq 0$.
4. $\frac{1}{\sqrt{g(x)}}$: require $g(x) > 0$ — strict, because the root sits in a denominator.
5. $[x]$-based pieces: $\frac{1}{[x]}$ needs $[x] \neq 0$, i.e. x outside $[0, 1)$.

Quadratic inequalities power rules 3 and 4: for an upward parabola with roots $\alpha < \beta$, we get $g \geq 0$ outside the roots ($(-\infty, \alpha] \cup [\beta, \infty)$) and $g \leq 0$ between them ($[\alpha, \beta]$).

Modulus equations and inequalities (a > 0)

• $|x| = a \iff x = \pm a$
• $|x| < a \iff -a < x < a$ — a single interval
• $|x| > a \iff x < -a$ or $x > a$ — two rays
• $|x - c| < a \iff c - a < x < c + a$ — an interval centred at c

These convert modulus conditions into interval language instantly. Example: the domain of $\sqrt{1 - |x|}$ needs $|x| \leq 1$: domain $[-1, 1]$.

The range playbook (first tools)

• Solve for x in terms of y; the legal y form the range (used for $\frac{x^2}{1 + x^2}$, $\frac{1}{x - 3}$, etc.).
• Complete the square for quadratics: range = [vertex value, $\infty$) upward.
• Bound chains for compositions: $x^2 \geq 0 \Rightarrow x^2 + 2 \geq 2 \Rightarrow \frac{1}{x^2 + 2} \in \left(0, \frac{1}{2}\right]$.

[JEE Tip] For $y = \frac{ax + b}{cx + d}$ (c $\neq$ 0), the range is $\mathbb{R} - \left\{\frac{a}{c}\right\}$ — solve for x and the only illegal output is the one that kills the x-term. Quote it; don't re-derive it in the exam.

Greatest Integer and Fractional Part — the JEE Pair

Define the fractional part $\{x\} = x - [x]$. Then every real splits as x = [x] + {x}: integer part plus leftover.

Properties of $[x]$:

• $[x] \leq x < [x] + 1$ and $x - 1 < [x] \leq x$
• $[x + n] = [x] + n$ for any integer n (integers slide out)
• $[x] + [-x] = 0$ if x is an integer, $-1$ otherwise
• $[x] \geq n \iff x \geq n$; $[x] < n \iff x < n$ (for integer n)

Properties of $\{x\}$:

• $0 \leq \{x\} < 1$ always; $\{x\} = 0 \iff x \in \mathbb{Z}$
• $\{x + n\} = \{x\}$ for integer n — periodic with period 1
• $\{-x\} = 1 - \{x\}$ for non-integer x. Example: $\{-2.3\} = 0.7$, NOT $-0.3$ or 0.3.

Solving with the bracket: to solve $[x] = n$, translate to $n \leq x < n + 1$. To solve equations like $[x]^2 - 5[x] + 6 = 0$: factor as a quadratic in $[x]$ — $[x] = 2$ or 3 — then translate: $x \in [2, 3) \cup [3, 4) = [2, 4)$.

One-one and onto — a preview

A function $f : A \to B$ is one-one (injective) if distinct inputs get distinct images, and onto (surjective) if range = codomain. These are Class 12 Chapter 1 material, but JEE Main quotes the words freely — recognise them now: $f(x) = 2x + 1$ on $\mathbb{R}$ is both; $f(x) = x^2$ on $\mathbb{R}$ is neither (f($-1$) = f(1), and negatives are never hit).

[JEE Tip] $\{x\} = x - [x]$ makes any equation mixing x and $[x]$ solvable: substitute x = n + t with $n = [x] \in \mathbb{Z}$, $t = \{x\} \in [0, 1)$, and solve for n and t separately — the integer and fractional unknowns decouple.

Solved Examples

Example 1. A = {1, 2, 3, 4}, B = {a, b, c}. Count (i) relations from A to B (ii) functions from A to B (iii) functions with f(1) = a.

Solution.

(i) Step 1. Relations are subsets of $A \times B$: $n(A \times B) = 4 \times 3 = 12$, so $2^{12} = 4096$ relations.

(ii) Step 2. A function lets each of the 4 inputs pick one of 3 images independently: $3 \times 3 \times 3 \times 3 = 3^4 = 81$. The DOMAIN size sits in the exponent.

(iii) Step 3. Fix f(1) = a — no choice there; inputs 2, 3, 4 still choose freely: $3^3 = 27$.

Takeaway: three different formulas from one setup — identify WHICH object (relation / function / conditioned function) is asked before computing.

Example 2. How many one-one functions exist from a set with 3 elements to a set with 5 elements?

Solution.

Step 1 — first input. 5 possible images.

Step 2 — second input. Must avoid the first image: 4 choices.

Step 3 — third input. Must avoid both: 3 choices.

Step 4 — multiply. $5 \times 4 \times 3 = 60$ one-one functions.

Takeaway: one-one = images without repetition = a falling product $n(n-1)(n-2)\cdots$; if the domain outnumbered the codomain the count would be 0.

Example 3. Find the domain of $f(x) = \frac{1}{\sqrt{x^2 - 3x + 2}}$.

Solution.

Step 1 — pick the right rule. The root sits in a DENOMINATOR: need $x^2 - 3x + 2 > 0$, strictly.

Step 2 — factor. Product 2, sum $-3$: $(x - 1)(x - 2) > 0$.

Step 3 — sign chart. Upward parabola, roots 1 and 2: positive OUTSIDE the roots.

Step 4 — domain. $(-\infty, 1) \cup (2, \infty)$ — endpoints excluded by strictness.

Takeaway: the only difference from a plain $\sqrt{g}$ domain is strict inequality — it costs exactly the two endpoints.

Example 4. Find the domain of $f(x) = \sqrt{1 - |x - 2|}$.

Solution.

Step 1 — radicand condition. $1 - |x - 2| \geq 0$, i.e. $|x - 2| \leq 1$.

Step 2 — translate the modulus inequality. $|x - c| \leq a$ is the closed interval of radius a centred at c: $2 - 1 \leq x \leq 2 + 1$.

Step 3 — domain. $[1, 3]$.

Takeaway: $|x - c| \leq a \iff c - a \leq x \leq c + a$ — one line, no case-splitting.

Example 5. Find the range of $f(x) = \frac{2x + 3}{x - 1}$, $x \neq 1$.

Solution.

Step 1 — solve for x in terms of y. $y(x - 1) = 2x + 3 \Rightarrow xy - y = 2x + 3 \Rightarrow x(y - 2) = y + 3 \Rightarrow x = \frac{y + 3}{y - 2}$.

Step 2 — legality check. The expression exists for every $y \neq 2$; at y = 2 the equation collapses to the impossible $0 = 5$.

Step 3 — range. $\mathbb{R} - \{2\}$.

Step 4 — the shortcut confirmed. For $\frac{ax + b}{cx + d}$ the forbidden output is $\frac{a}{c} = \frac{2}{1} = 2$ — the horizontal asymptote.

Takeaway: quote the $\frac{a}{c}$ shortcut in exams; re-derive it only when a problem modifies the template.

Example 6. Find the range of $f(x) = \frac{1}{2 - \sin x}$, using only the fact that $\sin x \in [-1, 1]$.

Solution.

Step 1 — bound the denominator. $\sin x \in [-1, 1] \Rightarrow -\sin x \in [-1, 1] \Rightarrow 2 - \sin x \in [1, 3]$ — never zero, so f is defined for ALL real x.

Step 2 — reciprocate the positive interval. Taking reciprocals of $[1, 3]$ swaps the endpoints: $f(x) \in \left[\frac{1}{3}, 1\right]$.

Step 3 — check attainment. $\sin x = 1$ gives $f = 1$ ✓; $\sin x = -1$ gives $f = \frac{1}{3}$ ✓ — both endpoints attained.

Step 4 — range. $\left[\frac{1}{3}, 1\right]$.

Takeaway: chain bounds through each operation, and remember reciprocation of a positive interval swaps its endpoints.

Example 7. Solve $[x]^2 - 5[x] + 6 = 0$ for real x.

Solution.

Step 1 — treat the bracket as an unknown. Set $u = [x]$: $u^2 - 5u + 6 = (u - 2)(u - 3) = 0$, so $u = 2$ or $u = 3$.

Step 2 — unpack each integer value. $[x] = 2 \iff 2 \leq x < 3$; $[x] = 3 \iff 3 \leq x < 4$.

Step 3 — union. $[2, 3) \cup [3, 4) = [2, 4)$.

Takeaway: solve for the bracket as an ordinary unknown FIRST, then translate each integer value into its half-open interval — the answer is a run of reals, not two isolated points.

Example 8. Compute (i) $\{3.8\}$ (ii) $\{-1.4\}$ (iii) $\{5\}$ (iv) $[2.3] + \{2.3\} + [-2.3] + \{-2.3\}$.

Solution.

(i) Step 1. $\{3.8\} = 3.8 - [3.8] = 3.8 - 3 = 0.8$.

(ii) Step 2 — the negative case, by definition. $[-1.4] = -2$ (drop DOWN), so $\{-1.4\} = -1.4 - (-2) = 0.6$. Never $-0.4$: the fractional part is always in $[0, 1)$.

(iii) Step 3. Integers have zero fractional part: $\{5\} = 0$.

(iv) Step 4. $2 + 0.3 + (-3) + 0.7 = 0$ — inevitable, since $[u] + \{u\} = u$ makes the four terms rebuild $2.3 + (-2.3) = 0$.

Takeaway: for negatives compute $\{x\} = x - [x]$ and trust the definition over intuition.

Example 9. Solve $x + [x] = 4.6$, if a solution exists.

Solution.

Step 1 — decouple integer and fractional parts. Substitute $x = n + t$ with $n = [x] \in \mathbb{Z}$ and $t = \{x\} \in [0, 1)$: the equation becomes $n + t + n = 4.6$, i.e. $2n + t = 4.6$.

Step 2 — pin down n. $t = 4.6 - 2n$ must lie in $[0, 1)$: $0 \leq 4.6 - 2n < 1 \iff 3.6 < 2n \leq 4.6 \iff 1.8 < n \leq 2.3$. The only integer is n = 2.

Step 3 — recover t and x. $t = 4.6 - 4 = 0.6$, so $x = 2.6$. Check: $2.6 + [2.6] = 2.6 + 2 = 4.6$ ✓.

Takeaway: the $n + t$ substitution turns any x-and-bracket equation into a one-line integer hunt.

Example 10. Find the domain of $f(x) = \frac{1}{\sqrt{|x| - x}}$.

Solution.

Step 1 — the strict condition. Root in denominator: $|x| - x > 0$, i.e. $|x| > x$.

Step 2 — case x $\geq$ 0. $|x| = x$, so the condition reads $x > x$ — false everywhere here.

Step 3 — case x < 0. $|x| = -x$, and $-x > x$ holds for all negative x (indeed $|x| - x = -2x > 0$).

Step 4 — domain. $(-\infty, 0)$.

Takeaway: memorise the split — $|x| - x$ is 0 on the non-negatives and $-2x > 0$ on the negatives; exams reuse it constantly.

Example 11. Find the domains of (i) $f(x) = \sqrt{x} + \sqrt{-x}$ and (ii) $g(x) = \sqrt{x} + \frac{1}{\sqrt{-x}}$.

Solution.

(i) Step 1. The two roots demand $x \geq 0$ AND $-x \geq 0$ (i.e. $x \leq 0$) simultaneously.

Step 2. The only number satisfying both is x = 0: domain = {0}, a single point (where $f(0) = 0$).

(ii) Step 3. Now the second condition is STRICT ($-x > 0$, root in denominator), i.e. $x < 0$ — which contradicts $x \geq 0$ outright.

Step 4. No x survives: domain = $\phi$ — g is defined nowhere.

Takeaway: domains can legally collapse to a single point or to nothing — intersect honestly rather than assuming an interval must survive.

Example 12. Classify each as one-one and/or onto: (i) $f : \mathbb{R} \to \mathbb{R}$, f(x) = 2x + 1 (ii) $f : \mathbb{R} \to \mathbb{R}$, f(x) = $x^2$ (iii) $f : \mathbb{R} \to [0, \infty)$, f(x) = $x^2$.

Solution.

(i) Step 1 — one-one test. $2a + 1 = 2b + 1 \Rightarrow 2a = 2b \Rightarrow a = b$: distinct inputs cannot collide. One-one ✓.

Step 2 — onto test. Any target y is hit by $x = \frac{y - 1}{2}$: range = codomain = $\mathbb{R}$. Onto ✓. Both.

(ii) Step 3. $f(-1) = 1 = f(1)$: not one-one. Range $[0, \infty) \neq \mathbb{R}$ (negatives never hit): not onto. Neither.

(iii) Step 4. Same collision kills one-one; but the DECLARED codomain is now $[0, \infty)$ = the range, so every codomain element is hit. Onto but not one-one.

Takeaway: onto depends on the declared codomain — (ii) and (iii) share the same rule and domain, yet differ in surjectivity; the property belongs to the whole triple (domain, codomain, rule).

Example 13. How many functions from A = {1, 2, 3} to B = {x, y, z, w} are NOT one-one?

Solution.

Step 1 — total functions. $4^3 = 64$.

Step 2 — one-one functions. Falling product: $4 \times 3 \times 2 = 24$.

Step 3 — subtract. Not one-one: $64 - 24 = 40$.

Takeaway: complement counting — counting collisions directly is a combinatorial mess; total minus injective is two lines.

Example 14. Find the range of $f(x) = [x] + [-x]$ over $\mathbb{R}$, and of $g(x) = \{x\} + \{-x\}$.

Solution.

Step 1 — f at integers. Both brackets are exact: $n + (-n) = 0$.

Step 2 — f at non-integers. Write $x = n + t$, $0 < t < 1$: $[x] = n$, $[-x] = -n - 1$, sum $= -1$. Range of f = {$-1$, 0}.

Step 3 — g at integers. Both fractional parts vanish: 0.

Step 4 — g at non-integers. $\{-x\} = 1 - \{x\}$, so the sum is $\{x\} + 1 - \{x\} = 1$. Range of g = {0, 1}.

Takeaway: both ranges are two-point sets — derive them once from the identities $[x] + [-x] = -1$ and $\{-x\} = 1 - \{x\}$ (non-integer case), then quote.

Example 15. Find the domain of $f(x) = \sqrt{4 - x} + \frac{1}{\sqrt{x^2 - 1}}$.

Solution.

Step 1 — first term. $4 - x \geq 0 \iff x \leq 4$: the ray $(-\infty, 4]$.

Step 2 — second term (strict). Root in denominator: $x^2 - 1 > 0 \iff x < -1$ or $x > 1$.

Step 3 — intersect. $(-\infty, 4] \cap \big((-\infty, -1) \cup (1, \infty)\big) = (-\infty, -1) \cup (1, 4]$.

Step 4 — audit the boundary points. $x = 4$: first term $\sqrt{0}$ fine, second term defined ✓ — included. $x = \pm 1$: second term divides by zero — excluded.

Takeaway: the full playbook in one problem — root rule, strict root-in-denominator rule, quadratic sign chart, intersection — the exact anatomy of an exam domain question.