Solved Examples

How to Use This Section

Here are 32 fully worked problems covering the entire chapter — ordered pairs and Cartesian products, relations with their domains and ranges, the function test, evaluation and graphs of standard functions, domain-range computations, and the algebra of real functions — arranged in a deliberate easy → medium → hard progression.

One suggestion that multiplies the value of every problem: attempt each yourself before reading the solution. Cover the solution, commit to an answer, then compare. Problems 1-10 warm up on products and relations, 11-22 build the function muscles, and 23-32 finish with domain-range and proof-style questions at Board and JEE difficulty.

Solved Examples

Example 1. Find x and y if: (i) $(2x, x + y) = (6, 5)$ (ii) $(x - y, x + y) = (4, 10)$.

Solution.

(i) Step 1 — split into coordinate equations. $2x = 6$ and $x + y = 5$.

Step 2 — solve in order. $x = 3$; substitute: $3 + y = 5$, so $y = 2$.

(ii) Step 3 — a genuine simultaneous system. $x - y = 4$ and $x + y = 10$.

Step 4 — add the equations. $2x = 14 \Rightarrow x = 7$; then $y = 10 - 7 = 3$.

Step 5 — verify. $(7 - 3, 7 + 3) = (4, 10)$ ✓.

Takeaway: coordinates may interlock — treat the two coordinate equations as a small simultaneous system and solve by the usual elimination.

Example 2. A = {x : x $\in \mathbb{N}$, x < 3}, B = {x : x is a letter of the word FEE}. Find $A \times B$ and $n(A \times B)$.

Solution.

Step 1 — roster-convert both factors FIRST. A: naturals below 3 = {1, 2}. B: letters of FEE = F, E, E → drop the repeat → {F, E}.

Step 2 — multiply. $A \times B$ = {(1, F), (1, E), (2, F), (2, E)}.

Step 3 — count. $n(A \times B) = 2 \times 2 = 4$ ✓ matches the listing.

Takeaway: hidden set sizes control everything — roster-convert before multiplying, and remember repeated letters collapse.

Example 3. For A = {1, 2, 3} and B = {2, 3, 4}, list the pairs of $A \times B$ with equal coordinates.

Solution.

Step 1 — translate the condition. A pair (a, a) lies in $A \times B$ exactly when $a \in A$ AND $a \in B$ — that is, $a \in A \cap B$.

Step 2 — compute the intersection. $A \cap B = \{2, 3\}$.

Step 3 — list the diagonal pairs. (2, 2) and (3, 3).

Takeaway: diagonal pairs of $A \times B$ correspond one-to-one with elements of $A \cap B$ — a tiny identity that shortcuts several MCQ types.

Example 4. For A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6}, verify that $A \times (B \cap C) = (A \times B) \cap (A \times C)$.

Solution.

Step 1 — left side. $B \cap C$: the sets {1, 2, 3, 4} and {5, 6} share nothing, so $B \cap C = \phi$, and $A \times \phi = \phi$.

Step 2 — right side. Every pair of $A \times B$ has second entry in B; every pair of $A \times C$ has second entry in C. A common pair would need its second entry in BOTH — impossible since $B \cap C = \phi$. So the intersection is $\phi$.

Step 3 — compare. Both sides equal $\phi$ ✓ — the identity survives its degenerate case.

Takeaway: verifying an identity on an empty-overlap example stress-tests understanding — the equality holds with room to spare.

Example 5. With A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6}, D = {5, 6, 7, 8}, verify that $A \times C$ is a subset of $B \times D$.

Solution.

Step 1 — list the smaller product. $A \times C$ = {(1,5), (1,6), (2,5), (2,6)}.

Step 2 — test each pair against $B \times D$. First entries 1, 2 all lie in B ($A \subset B$); second entries 5, 6 all lie in D ($C \subset D$). All four pairs qualify ✓.

Step 3 — extract the principle. $A \subset B$ and $C \subset D$ together give $A \times C \subset B \times D$ — factorwise containment implies product containment.

Takeaway: the general principle spares you the listing whenever the factor containments are visible.

Example 6. An arrow diagram from P = {5, 6, 7} sends each number to its remainder on division by 3. Write the relation in roster form; give domain and range.

Solution.

Step 1 — compute each remainder. $5 = 3 + 2 \to 2$; $6 = 2 \times 3 \to 0$; $7 = 2 \times 3 + 1 \to 1$.

Step 2 — roster form. R = {(5, 2), (6, 0), (7, 1)}.

Step 3 — domain and range. Domain = {5, 6, 7}; range = {0, 1, 2}.

Step 4 — bonus observation. Each input has exactly one image — division leaves ONE remainder — so this relation is in fact a function.

Takeaway: modular arithmetic rules are single-valued, so remainder relations are always functions.

Example 7. On A = {1, 2, 3, 4}, let R = {(x, y) : x, y $\in$ A, x < y}. Write R in roster form and find its domain and range.

Solution.

Step 1 — list increasing pairs, row by row. x = 1: (1,2), (1,3), (1,4). x = 2: (2,3), (2,4). x = 3: (3,4). x = 4: nothing exceeds 4 inside A.

Step 2 — count. $3 + 2 + 1 = 6$ pairs ✓ (equivalently $\binom{4}{2}$: choose two distinct values, order them increasingly).

Step 3 — domain and range. Domain = {1, 2, 3} — the maximum 4 sends no arrow. Range = {2, 3, 4} — the minimum 1 receives none.

Takeaway: strict-order relations always lose the maximum from the domain and the minimum from the range.

Example 8. A has 2 elements, B has 3. Find (i) the number of relations from A to B; (ii) from B to A; (iii) on A.

Solution.

Step 1 — the rule. Relations from X to Y are subsets of $X \times Y$: count $= 2^{n(X)n(Y)}$.

(i) Step 2. $n(A \times B) = 2 \times 3 = 6$: $2^6 = 64$.

(ii) Step 3. $n(B \times A) = 3 \times 2 = 6$: also $2^6 = 64$ — different pairs, same count.

(iii) Step 4. $n(A \times A) = 4$: $2^4 = 16$.

Takeaway: only the SIZE of the product matters — reversing direction changes the pairs but never the count.

Example 9. Which are functions? Give reasons. (i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)} (ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)} (iii) {(1,3), (1,5), (2,5)}.

Solution.

Step 1 — the test. Scan first coordinates for repeats with different partners.

(i) Step 2. First elements 2, 5, 8, 11, 14, 17 — all distinct. Function (all inputs share the output 1 — a constant-like function). Domain {2, 5, 8, 11, 14, 17}, range {1}.

(ii) Step 3. First elements 2, 4, …, 14 — distinct. Function; domain {2, 4, 6, 8, 10, 12, 14}, range {1, …, 7}.

(iii) Step 4. The input 1 carries two images, 3 and 5. Not a function.

Takeaway: state domain and range only AFTER the function test passes — for a non-function the question is void.

Example 10. If $f(x) = \frac{x - 1}{x + 1}$ ($x \neq -1$), find (i) f(0) (ii) f(1) (iii) $f\left(\frac{1}{2}\right)$ (iv) an x with f(x) = 0.

Solution.

(i) Step 1. $f(0) = \frac{0 - 1}{0 + 1} = \frac{-1}{1} = -1$.

(ii) Step 2. $f(1) = \frac{0}{2} = 0$.

(iii) Step 3. $f\left(\frac{1}{2}\right) = \frac{\frac{1}{2} - 1}{\frac{1}{2} + 1} = \frac{-\frac{1}{2}}{\frac{3}{2}} = -\frac{1}{3}$.

(iv) Step 4. A fraction vanishes exactly when its numerator does (and the denominator does not): $x - 1 = 0$, so x = 1 — consistent with (ii).

Takeaway: the numerator carries the roots, the denominator carries the exclusions — keep their roles separate.

Example 11. Write $f(x) = |2x - 3|$ as a piecewise function and evaluate f(0), f(1.5), f(5).

Solution.

Step 1 — find the corner. The inside vanishes at $2x - 3 = 0$, i.e. $x = \frac{3}{2}$ — split the line there.

Step 2 — write the pieces. For $x \geq \frac{3}{2}$ the inside is non-negative: $f(x) = 2x - 3$. For $x < \frac{3}{2}$ it is negative, so the modulus flips it: $f(x) = 3 - 2x$.

Step 3 — evaluate. $f(0) = 3 - 0 = 3$ (left piece); $f(1.5) = 2(1.5) - 3 = 0$ (the corner); $f(5) = 10 - 3 = 7$ (right piece).

Takeaway: every modulus is secretly a two-piece linear function split at the zero of its inside — write the split once and evaluation becomes mechanical.

Example 12. Evaluate sgn($x^2 + 1$) for any real x, and sgn($x^2 - 4$) at x = 1 and x = 3.

Solution.

Step 1 — sign of $x^2 + 1$. Since $x^2 \geq 0$, we get $x^2 + 1 \geq 1 > 0$ for EVERY real x: the signum is constantly 1.

Step 2 — sign of $x^2 - 4$ at x = 1. $1 - 4 = -3 < 0$: sgn $= -1$.

Step 3 — at x = 3. $9 - 4 = 5 > 0$: sgn $= 1$.

Takeaway: signum questions are pure sign-analysis — decide the sign of the inside expression, output $1$, $-1$ or $0$ accordingly.

Example 13. Compute (i) $[2.7] + [-2.7]$ (ii) $[x] + [-x]$ for x not an integer (iii) $[3.4] - [-3.4]$.

Solution.

(i) Step 1. $[2.7] = 2$; $[-2.7] = -3$ (drop DOWN below $-2.7$). Sum: $2 + (-3) = -1$.

(ii) Step 2 — the general pattern. For non-integer x with $n < x < n+1$: $[x] = n$ and $-(n+1) < -x < -n$ gives $[-x] = -(n + 1)$. Sum: $n - n - 1 = -1$ — always. (For integer x both floors are exact and the sum is 0.)

(iii) Step 3. $[3.4] = 3$; $[-3.4] = -4$. Difference: $3 - (-4) = 7$.

Takeaway: $[x] + [-x]$ equals 0 for integers and $-1$ otherwise — a one-line result worth memorising.

Example 14. f = {(1, 1), (2, 3), (0, $-1$), ($-1$, $-3$)} is a linear function from $\mathbb{Z}$ to $\mathbb{Z}$. Find f(x).

Solution.

Step 1 — set the form. $f(x) = mx + c$.

Step 2 — the x = 0 pair first. (0, $-1$) gives $c = -1$ immediately.

Step 3 — one more pair. (1, 1): $m + c = 1$, so $m = 2$.

Step 4 — verify on the unused pairs. $f(2) = 4 - 1 = 3$ ✓; $f(-1) = -2 - 1 = -3$ ✓. Hence $f(x) = 2x - 1$.

Takeaway: the pair with x = 0 hands you c for free — always spend it first, then verify with leftovers.

Example 15. f(x) = $x^2$ on $\{-4, -3, ..., 4\}$. Tabulate f, and state the domain and range of the full real function $y = x^2$.

Solution.

Step 1 — tabulate. x: $-4, -3, -2, -1, 0, 1, 2, 3, 4$ → f(x): 16, 9, 4, 1, 0, 1, 4, 9, 16.

Step 2 — note the symmetry. $f(-x) = f(x)$: the table mirrors around x = 0, exactly as the parabola mirrors about the y-axis.

Step 3 — the real function. Domain: every real can be squared → $\mathbb{R}$. Range: squares are $\geq 0$ and every $y \geq 0$ equals $(\sqrt{y})^2$ → $[0, \infty)$.

Takeaway: the symmetric table is the numerical shadow of the parabola's mirror symmetry.

Example 16. For $f(x) = \frac{1}{x}$, tabulate f at x = $-2, -1.5, -1, -0.5, 0.25, 0.5, 1, 1.5, 2$ and state its domain and range.

Solution.

Step 1 — tabulate. $f(-2) = -0.5$; $f(-1.5) \approx -0.67$; $f(-1) = -1$; $f(-0.5) = -2$; $f(0.25) = 4$; $f(0.5) = 2$; $f(1) = 1$; $f(1.5) \approx 0.67$; $f(2) = 0.5$.

Step 2 — read the trend. Small inputs produce huge outputs and vice versa — the two branches of the hyperbola hug the axes without touching them.

Step 3 — domain and range. Domain: all reals except the forbidden 0 → $\mathbb{R} - \{0\}$. Range: $\frac{1}{x} = y$ solves to $x = \frac{1}{y}$ for any $y \neq 0$, and $y = 0$ is impossible → $\mathbb{R} - \{0\}$.

Takeaway: the reciprocal's domain and range are the same punctured line — 0 is unreachable from either side.

Example 17. Find the domain of $f(x) = \sqrt{x^2 - 5x + 6}$.

Solution.

Step 1 — set up the radicand condition. Need $x^2 - 5x + 6 \geq 0$.

Step 2 — factor. Product 6, sum $-5$: $(x - 2)(x - 3) \geq 0$.

Step 3 — sign chart. An upward parabola with roots 2 and 3 is negative BETWEEN its roots and non-negative outside: $x \leq 2$ or $x \geq 3$.

Step 4 — domain. $(-\infty, 2] \cup [3, \infty)$, endpoints included ($\sqrt{0}$ is fine).

Takeaway: root-of-quadratic domains are sign-chart problems — keep the outside of the roots for an upward parabola, the inside for a downward one.

Example 18. Find the range of $f(x) = x^2 - 4x + 7$ on $\mathbb{R}$.

Solution.

Step 1 — complete the square. $x^2 - 4x + 7 = (x^2 - 4x + 4) + 3 = (x - 2)^2 + 3$.

Step 2 — bound. $(x - 2)^2 \geq 0$ with equality exactly at x = 2, so $f(x) \geq 3$, and the minimum 3 IS attained.

Step 3 — coverage upward. As x moves away from 2 the square grows without bound, hitting every value above 3.

Step 4 — range. $[3, \infty)$, closed at 3.

Takeaway: completing the square converts any quadratic's range into a one-line read: vertex value up to infinity (or down from it for a downward parabola).

Example 19. Find the range of $f(x) = \frac{1}{x - 3}$, $x \neq 3$.

Solution.

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

Step 2 — solvability. The expression exists for every $y \neq 0$; and $y = 0$ is impossible (a reciprocal never vanishes).

Step 3 — range. $\mathbb{R} - \{0\}$.

Takeaway: \"solve for x, ask which y are legal\" — the universal range method; shifting the denominator moved the domain but not the range.

Example 20. Find the domain of $f(x) = \frac{\sqrt{4 - x^2}}{x - 1}$.

Solution.

Step 1 — root condition. $4 - x^2 \geq 0 \iff x^2 \leq 4 \iff -2 \leq x \leq 2$: the interval $[-2, 2]$.

Step 2 — denominator condition. $x - 1 \neq 0$: delete x = 1.

Step 3 — combine. $[-2, 2] - \{1\} = [-2, 1) \cup (1, 2]$. Both endpoints survive — the root sits in the NUMERATOR, so its zero is harmless.

Takeaway: stack every condition, intersect, delete — and present the answer as a union of intervals showing the hole.

Example 21. Define $f : \mathbb{N} \to \mathbb{N}$ by f(n) = $\frac{n}{2}$ if n is even, and f(n) = $\frac{n + 1}{2}$ if n is odd. Compute f(1), f(2), f(7), f(10), and show f is a function but not one-one.

Solution.

Step 1 — evaluate by matching parity. f(1): odd rule, $\frac{2}{2} = 1$. f(2): even rule, $\frac{2}{2} = 1$. f(7): odd rule, $\frac{8}{2} = 4$. f(10): even rule, 5.

Step 2 — function check. Every natural number is even or odd, never both — the cases are exclusive and exhaustive, so each input gets exactly one image ✓.

Step 3 — shared image. $f(1) = f(2) = 1$: two different inputs, one output — allowed for a function, but it shows f is not one-one (injectivity is a Class 12 refinement).

Takeaway: piecewise-by-parity definitions are genuine functions when the cases are exclusive and exhaustive; sharing outputs is never a violation.

Example 22. f(x) = x + 1, g(x) = 2x $-$ 3. Find (f + g)(x), (f $-$ g)(x), $\left(\frac{f}{g}\right)(x)$, and evaluate each at x = 2.

Solution.

Step 1 — the three combinations. $(f + g)(x) = 3x - 2$; $(f - g)(x) = (x + 1) - (2x - 3) = -x + 4$; $\left(\frac{f}{g}\right)(x) = \frac{x + 1}{2x - 3}$ with $x \neq \frac{3}{2}$.

Step 2 — evaluate the formulas at 2. $3(2) - 2 = 4$; $-2 + 4 = 2$; $\frac{3}{1} = 3$.

Step 3 — cross-check pointwise. $f(2) = 3$, $g(2) = 1$: sum $3 + 1 = 4$ ✓, difference $3 - 1 = 2$ ✓, quotient $\frac{3}{1} = 3$ ✓.

Takeaway: evaluating both ways — formula first vs pointwise — is a free, complete check of the algebra.

Example 23. $f(x) = \sqrt{x + 2}$ and $g(x) = \sqrt{2 - x}$. Find the domains of fg and $\frac{f}{g}$, and compute (fg)(1).

Solution.

Step 1 — ingredient domains. dom f: $x + 2 \geq 0 \to [-2, \infty)$. dom g: $2 - x \geq 0 \to (-\infty, 2]$. Intersection: $[-2, 2]$.

Step 2 — the product. dom(fg) = $[-2, 2]$, and $(fg)(x) = \sqrt{(x + 2)(2 - x)} = \sqrt{4 - x^2}$.

Step 3 — evaluate. $(fg)(1) = \sqrt{(3)(1)} = \sqrt{3}$.

Step 4 — the quotient. Additionally delete the zeros of g: $g(x) = 0$ at x = 2. dom$\left(\frac{f}{g}\right) = [-2, 2)$.

Takeaway: the quotient loses exactly the zero of the denominator — here one endpoint — from the product's domain.

Example 24. For $f(x) = x^2$, simplify $\frac{f(x + h) - f(x)}{h}$ ($h \neq 0$).

Solution.

Step 1 — expand the shifted value. $f(x + h) = (x + h)^2 = x^2 + 2xh + h^2$.

Step 2 — subtract. $f(x + h) - f(x) = 2xh + h^2 = h(2x + h)$.

Step 3 — divide by h. $\frac{h(2x + h)}{h} = 2x + h$ (legal since $h \neq 0$).

Takeaway: as h shrinks, $2x + h \to 2x$ — you have computed the derivative of $x^2$ a year early; factoring h out before cancelling is the key move.

Example 25. Are f(x) = $\frac{x^2 - 1}{x - 1}$ and g(x) = x + 1 the same function?

Solution.

Step 1 — simplify f where legal. For $x \neq 1$: $f(x) = \frac{(x - 1)(x + 1)}{x - 1} = x + 1$ — the values agree with g everywhere BOTH are defined.

Step 2 — compare domains. dom f = $\mathbb{R} - \{1\}$ (the original formula divides by zero at 1); dom g = $\mathbb{R}$.

Step 3 — verdict. Equal functions require equal domains AND equal values; the domains differ, so $f \neq g$ — f is g with one point punched out.

Takeaway: a function is the PACKAGE (domain, rule); cancelling a common factor silently changes the domain — a favourite exam trap.

Example 26. A function's graph is the upper semicircle of radius 2 centred at the origin. Find the function, its domain and its range.

Solution.

Step 1 — start from the circle. $x^2 + y^2 = 4$; the upper half keeps $y \geq 0$.

Step 2 — solve for y. $y = \sqrt{4 - x^2}$ — the non-negative root exactly matches the upper half.

Step 3 — domain. Radicand $\geq 0$: $[-2, 2]$.

Step 4 — range. The height runs from 0 (at $x = \pm 2$) up to the radius 2 (at x = 0): $[0, 2]$.

Takeaway: \"upper semicircle of radius r\" and $\sqrt{r^2 - x^2}$ are the same object — translate freely between picture and formula.

Example 27. A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16}, f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}. Are these true? (i) f is a relation from A to B. (ii) f is a function from A to B.

Solution.

(i) Step 1 — test membership of each pair in $A \times B$. First entries 1, 2, 3, 4, 2 all lie in A ✓; second entries 5, 9, 1, 5, 11 all lie in B ✓. So $f \subset A \times B$: a relation ✓.

(ii) Step 2 — function test. Scan first coordinates: 2 appears TWICE, with images 9 and 11. One input, two images — not a function.

Takeaway: every function is a relation, but not conversely — this relation fails by exactly one repeated first element.

Example 28. A = {1, 2, 3}, B = {a, b}. Count (i) the relations from A to B, (ii) the functions from A to B.

Solution.

(i) Step 1. Relations are subsets of $A \times B$: $n(A \times B) = 6$, so $2^6 = 64$ relations.

(ii) Step 2. A function makes each of the 3 inputs pick exactly ONE of the 2 images, independently: $2 \times 2 \times 2 = 2^3 = 8$ functions.

Step 3 — compare. Only 8 of the 64 relations pass the one-image test — functions are rare among relations.

Takeaway: relations count as $2^{pq}$, functions as $q^p$ — never swap the two formulas.

Example 29. Find the range of f(x) = $[x] - x$ for real x, and of g(x) = $x - [x]$.

Solution.

Step 1 — recognise g. $g(x) = x - [x]$ is the fractional part $\{x\}$: it is 0 at integers and climbs toward (never reaching) 1 between them.

Step 2 — range of g. $[0, 1)$ — 0 attained, 1 approached only.

Step 3 — f is the negative. $f(x) = [x] - x = -g(x)$: the range mirrors through 0 to $(-1, 0]$ — now $-1$ is open and 0 closed.

Takeaway: negating a function reflects its range about 0 AND swaps which endpoint is open — track both effects.

Example 30. Find the domain of $f(x) = \frac{1}{\sqrt{x - 2}}$.

Solution.

Step 1 — root condition. $x - 2 \geq 0$, i.e. $x \geq 2$.

Step 2 — denominator condition. $\sqrt{x - 2} \neq 0$, i.e. $x \neq 2$.

Step 3 — combine. The two conditions together give the STRICT inequality $x > 2$: domain = $(2, \infty)$.

Takeaway: a root in a denominator upgrades $\geq$ to $>$ — the boundary point falls to the division rule.

Example 31. For $f(x) = \begin{cases} x^2, & x < 0 \\ x, & 0 \leq x \leq 1 \\ \frac{1}{x}, & x > 1 \end{cases}$, evaluate $f(-2)$, $f\left(\frac{1}{2}\right)$, f(1), f(4), and state the range of the middle piece.

Solution.

Step 1 — match each input to its interval FIRST. $-2 < 0$: first rule. $\frac{1}{2} \in [0, 1]$: middle rule. $1 \in [0, 1]$: middle rule (the boundary belongs to the middle piece). $4 > 1$: last rule.

Step 2 — evaluate. $f(-2) = (-2)^2 = 4$; $f\left(\frac{1}{2}\right) = \frac{1}{2}$; $f(1) = 1$; $f(4) = \frac{1}{4}$.

Step 3 — middle-piece range. The identity rule on $[0, 1]$ outputs exactly $[0, 1]$.

Takeaway: locate the input's interval BEFORE touching any formula — boundary ownership (here 0 and 1 belong to the middle) decides close calls.

Example 32. Show that the range of $f(x) = \frac{1}{1 + x^2}$ (x real) is $(0, 1]$.

Solution.

Step 1 — upper bound, attained. $1 + x^2 \geq 1$ gives $f(x) \leq 1$, with equality exactly at x = 0 — the maximum 1 is attained.

Step 2 — positivity, never zero. Numerator 1 and denominator $1 + x^2$ are positive, so $f(x) > 0$ for all x; as $|x|$ grows, f(x) shrinks toward 0 without reaching it.

Step 3 — everything between occurs. For any $0 < y \leq 1$, solve: $1 + x^2 = \frac{1}{y}$ gives $x^2 = \frac{1}{y} - 1 \geq 0$ — a real solution exists.

Step 4 — conclude. Range = $(0, 1]$. ∎

Takeaway: a complete range proof has three parts — bound it, attain the attained end, and solve for x to certify everything in between.