◆ QA · Algebra

Functions & Graphs , formulas + CAT PYQs

Focused Algebra kit. The full chapter formula sheet (with explanations & basic examples) is tucked below; every CAT PYQ for Functions & Graphs is here.

47CAT PYQs
Algebrachapter
Formulas PYQs (47)↩ Full Algebra chapter

Algebra, formula sheet

Show the full Algebra formula sheet (explanations + basic examples)
1Polynomials & zeroes
  • Plain English: a polynomial is just a sum of x-powers; its "zeroes" are the x-values that make it equal 0.
  • A polynomial of degree n: aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀
  • k is a zero of p(x) if p(k) = 0. Zeroes are the x-coordinates where y = p(x) cuts the x-axis.
  • Max zeroes = degree: linear → 1, quadratic → 2, cubic → 3, degree n → n.
  • e.g. p(x) = x² − 9 has degree 2, so at most 2 zeroes: x = 3 and x = −3.
2Sum & product of roots
  • Plain English: you can read the sum and product of the roots straight off the coefficients, no need to solve.
  • Quadratic ax²+bx+c: α + β = −b/a, αβ = c/a
  • Cubic ax³+bx²+cx+d: α+β+γ = −b/a, αβ+βγ+γα = c/a, αβγ = −d/a
  • Build the equation: x² − (sum)x + (product) = 0.
  • e.g. x² − 5x + 6 = 0: sum = 5, product = 6 ⇒ roots 2 and 3 (2+3=5, 2×3=6).
3Discriminant & nature of roots
  • Plain English: the discriminant D is a single number that tells you how many real roots a quadratic has, before solving.
  • For ax²+bx+c (a≠0): D = b² − 4ac
  • D > 0 → two distinct real roots; D = 0 → equal real roots; D < 0 → no real roots (complex).
  • D a perfect square (a,b,c rational) → roots are rational.
  • Roots: x = (−b ± √D)/2a
  • e.g. x² + x + 1: D = 1 − 4 = −3 < 0 ⇒ no real roots.
4Sum of squares of roots (trick)
  • Plain English: you can get α²+β² from the sum and product alone, never solve for the roots first.
  • α² + β² = (α+β)² − 2αβ
  • To minimise a sum-of-squares-of-roots expression in a parameter, complete the square, minimum is at the vertex.
  • e.g. if α+β = 3 and αβ = 2, then α²+β² = 9 − 4 = 5.
5Algebraic identities
  • Plain English: memorised expand/factor templates that turn ugly expressions into products (or vice-versa) instantly.
  • (a±b)² = a² ± 2ab + b²
  • a² − b² = (a+b)(a−b)
  • (a+b+c)² = a²+b²+c² + 2(ab+bc+ca)
  • a³ ± b³ = (a±b)(a² ∓ ab + b²)
  • a³+b³+c³ − 3abc = (a+b+c)(a²+b²+c² − ab−bc−ca)
  • e.g. 97×103 = (100−3)(100+3) = 100² − 3² = 9991.
6Linear equations in two variables
  • Plain English: comparing the coefficient ratios tells you whether two lines cross once, never, or lie on top of each other.
  • a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0.
  • Unique solution (intersecting): a₁/a₂ ≠ b₁/b₂
  • No solution (parallel): a₁/a₂ = b₁/b₂ ≠ c₁/c₂
  • Infinite solutions (coincident): a₁/a₂ = b₁/b₂ = c₁/c₂
  • e.g. x+y=2 and 2x+2y=5: ratios 1/2 = 1/2 ≠ 2/5 ⇒ parallel, no solution.
7Inequalities, basic rules
  • Plain English: inequalities behave like equations, except multiplying or dividing by a negative reverses the arrow.
  • Adding/subtracting keeps direction; multiplying by a negative flips the sign.
  • If X > Y > 0 then 1/X < 1/Y.
  • For x > 0: x + 1/x ≥ 2 (equality at x = 1).
  • e.g. −2x > 6 ⇒ divide by −2 and flip ⇒ x < −3.
8Quadratic inequalities
  • Plain English: factor it, then "< 0" means between the roots and "> 0" means outside the roots.
  • (x−m)(x−n) < 0, n > m ⇒ m < x < n (between the roots).
  • (x−m)(x−n) > 0 ⇒ x < m or x > n (outside the roots).
  • Sign-of-product / wavy-curve method handles higher degree.
  • e.g. x² − 5x + 6 < 0 ⇒ (x−2)(x−3) < 0 ⇒ 2 < x < 3.
9Modulus (absolute value)
  • Plain English: |x| is the distance of x from 0, so it strips the sign and is never negative.
  • |x| = max(x, −x); −|x| ≤ x ≤ |x|.
  • |a+b| ≤ |a|+|b| and |a|−|b| ≤ |a−b|; |ab| = |a||b|.
  • |x| ≤ k ⇒ −k ≤ x ≤ k. |x| ≥ k ⇒ x ≥ k or x ≤ −k.
  • |f| + |g| = |f+g| only when f, g have the same sign.
  • e.g. |x| ≤ 3 ⇒ −3 ≤ x ≤ 3; |x − 4| = 2 ⇒ x = 6 or x = 2.
10AM-GM-HM inequality
  • Plain English: for positive numbers the plain average is always ≥ the geometric average, the go-to tool for "find the minimum".
  • For positive reals: AM ≥ GM ≥ HM, equality when all equal.
  • Two numbers: AM = (a+b)/2, GM = √(ab), HM = 2ab/(a+b).
  • AM × HM = GM²
  • If a₁a₂…aₙ = 1 then a₁+a₂+…+aₙ ≥ n.
  • e.g. for a = 2, b = 8: AM = 5 ≥ GM = √16 = 4. ✓
11Maxima & minima of a quadratic
  • Plain English: a parabola's turning point is at x = −b/2a; that's where the min (opens up) or max (opens down) lives.
  • ax²+bx+c: vertex at x = −b/2a; extreme value = (4ac − b²)/4a = −D/4a.
  • a > 0 → opens up → minimum; a < 0 → opens down → maximum.
  • min/max of max-of-two / min-of-two lines occurs where the two graphs intersect.
  • e.g. x² − 6x + 5: vertex at x = 3, minimum value = 9 − 18 + 5 = −4.
12Functions, domain, range, even/odd
  • Plain English: domain is what you may feed in, range is what comes out; even/odd describe the graph's symmetry.
  • Domain = allowed inputs; range = resulting outputs.
  • Even: f(−x) = f(x) (graph symmetric about y-axis), e.g. x², |x|.
  • Odd: f(−x) = −f(x) (symmetric about origin), e.g. x³, 1/x.
  • Inverse exists only if f is one-to-one.
  • e.g. f(x) = x³ is odd: f(−2) = −8 = −f(2). ✓
13Functional equations
  • Plain English: the form of a functional rule reveals the function, "turns + into ×" means exponential, etc.
  • f(x+y) = f(x)·f(y) ⇒ exponential type, f(x) = aˣ.
  • f(xy) = f(x)·f(y) ⇒ power/multiplicative; f(1) = 1.
  • If f(a+x) = f(a−x), the graph is symmetric about x = a; roots pair around a (sum of 4 roots = 4a).
  • e.g. f(x+y) = f(x)f(y) with f(1) = 3 ⇒ f(2) = f(1)² = 9.
14Graph shifting
  • Plain English: changes outside f() move the graph vertically; changes inside f() move it horizontally (and oppositely).
  • f(x)+c → shift up c; f(x)−c → shift down c.
  • f(x+c) → shift left c; f(x−c) → shift right c.
  • −f(x) → reflect in x-axis; f(−x) → reflect in y-axis.
  • e.g. y = (x−2)² is y = x² shifted 2 units right.
15Logarithm, definition
  • Plain English: log_b x just asks "what power of b gives x?", it's the inverse of raising to a power.
  • y = log_b x ⇔ x = bʸ (b > 0, b ≠ 1, x > 0).
  • log_a a = 1; log_a 1 = 0; a^(log_a m) = m.
  • e.g. log₂8 = 3 because 2³ = 8.
16Logarithm laws
  • Plain English: logs turn multiplication into addition, division into subtraction, and powers into multipliers.
  • log_a(xy) = log_a x + log_a y
  • log_a(x/y) = log_a x − log_a y
  • log_a(xᵐ) = m·log_a x
  • log_(aⁿ)(xᵐ) = (m/n)·log_a x
  • Change of base: log_a x = (log x)/(log a); log_a x = 1/log_x a
  • e.g. log₂40 = log₂(8×5) = log₂8 + log₂5 = 3 + log₂5.
17Indices (laws of exponents)
  • Plain English: same base, add exponents when multiplying, subtract when dividing, multiply when raising a power to a power.
  • pᵐ·pⁿ = pᵐ⁺ⁿ; pᵐ/pⁿ = pᵐ⁻ⁿ; (pᵐ)ⁿ = pᵐⁿ
  • pⁿ·qⁿ = (pq)ⁿ; (p/q)ⁿ = pⁿ/qⁿ
  • p⁻ⁿ = 1/pⁿ; p⁰ = 1; p^(1/n) = ⁿ√p
  • e.g. 2³·2⁴ = 2⁷ = 128; 8^(2/3) = (∛8)² = 2² = 4.
18Surds & rationalisation
  • Plain English: a surd is an unresolved root like √2; "rationalising" clears it from a denominator using the conjugate.
  • √(ab) = √a·√b; √(a/b) = √a/√b.
  • Rationalise a/(b+√c) by multiplying top & bottom by the conjugate (b−√c).
  • If a+√b is a root of a rational quadratic, so is its conjugate a−√b.
  • e.g. 1/(√3 − 1) × (√3 + 1)/(√3 + 1) = (√3 + 1)/2.
19Arithmetic Progression (AP)
  • Plain English: an AP adds the same step d each time; its sum is just "how many terms × the average of first and last".
  • Constant difference d. nth term: Tₙ = a + (n−1)d
  • Sum: Sₙ = n/2 · [2a + (n−1)d] = n/2 · (first + last)
  • Arithmetic mean of a, b: A = (a+b)/2. Middle term = average of an odd count of AP terms.
  • e.g. 2, 5, 8, …: T₄ = 2 + 3×3 = 11; sum of first 4 = 4/2·(2+11) = 26.
20Geometric Progression (GP)
  • Plain English: a GP multiplies by the same ratio r each time; if |r| < 1 the infinite sum settles to a finite value.
  • Constant ratio r. nth term: Tₙ = a·rⁿ⁻¹
  • Sum: Sₙ = a(rⁿ − 1)/(r − 1), r ≠ 1.
  • Infinite sum (|r| < 1): S∞ = a/(1 − r)
  • Geometric mean: G = √(ab).
  • e.g. 1 + ½ + ¼ + … = 1/(1 − ½) = 2.
21Harmonic Progression (HP)
  • Plain English: an HP is just an AP flipped, take reciprocals and you're back to a normal AP.
  • a, b, c… in HP ⇔ 1/a, 1/b, 1/c… in AP.
  • Harmonic mean of a, b: H = 2ab/(a+b)
  • nth term of HP = 1/(nth term of the corresponding AP).
  • e.g. 1, ½, ⅓, ¼ is an HP (reciprocals 1, 2, 3, 4 form an AP).
22Standard summation formulas
  • Plain English: ready-made closed forms for adding up the first n numbers, their squares, and their cubes.
  • Σn = n(n+1)/2
  • Σn² = n(n+1)(2n+1)/6
  • Σn³ = [n(n+1)/2]²
  • Telescoping: 1/(k·(k+1)) = 1/k − 1/(k+1).
  • e.g. 1 + 2 + … + 10 = 10×11/2 = 55.
23Common terms of two APs
  • Plain English: numbers shared by two APs themselves form an AP whose step is the LCM of the two steps.
  • Common terms of two APs form a new AP with common difference = LCM of the two differences.
  • Find the first common term, then count multiples of the LCM up to the smaller upper limit.
  • e.g. 2,5,8,… and 3,7,11,…: first common term 11, new step = LCM(3,4) = 12 ⇒ 11, 23, 35, …
24Recurrence & tₙ from Sₙ
  • Plain English: if you know the running total Sₙ, each term is just this total minus the previous total.
  • If Sₙ given: aₙ = Sₙ − Sₙ₋₁ (and a₁ = S₁).
  • Alternating-sum sequences: subtract consecutive defining equations to isolate a term.
  • e.g. Sₙ = n² ⇒ a₅ = S₅ − S₄ = 25 − 16 = 9.
25Integer / Diophantine solutions
  • Plain English: once you spot one whole-number solution, all the rest come by stepping x and y in fixed jumps.
  • ax + by = c with one integer solution (x₀, y₀): all others are x₀ + (b/g)t, y₀ − (a/g)t, where g = gcd(a,b).
  • Bound the count using the given ranges on x and y.
  • e.g. 2x + 3y = 12: (x,y) = (3,2) works; next is (0,4), then (6,0), x jumps by 3, y by 2.
26When does Aᴮ = 1?
  • Plain English: a power equals 1 in exactly three situations, check all three or you'll miss cases.
  • Base = 1 (any exponent), or
  • Exponent = 0 (base ≠ 0), or
  • Base = −1 with an even exponent.
  • e.g. (−1)⁴ = 1 (base −1, even power); 7⁰ = 1 (zero power); 1⁹⁹ = 1 (base 1).
27Three terms in AP / GP
  • Plain English: centering three terms on a middle value makes their sum (AP) or product (GP) collapse to one symbol.
  • Three in AP: take a−d, a, a+d (their sum = 3a).
  • Three in GP: take a/r, a, ar (product = a³).
  • Three consecutive integers as roots: n−1, n, n+1.
  • e.g. three numbers in AP summing to 18 ⇒ middle = 6, so 6−d, 6, 6+d.
28|x − a| as distance (modulus sums)
  • Plain English: read |x−a| as "distance from a", and sums of such distances are smallest when x sits among the points.
  • |x−a| = distance of x from a on the number line.
  • |x−p|+|x−q| is minimised for any x between p and q; minimum value = |p−q|.
  • |x−p| = |x−q| at the midpoint x = (p+q)/2.
  • e.g. |x−2| + |x−7| ≥ 5, achieved for any x in [2, 7].
29Sum of squares identity trick
  • Plain English: squares can't be negative, so if a bunch of squares add to 0 every single one must be 0.
  • If a sum of squares equals 0, each square = 0: e.g. (x−2y)² + (y−z)² = 0 ⇒ x = 2y and y = z.
  • Group given expressions into perfect squares to pin exact values.
  • e.g. (a−3)² + (b+1)² = 0 forces a = 3 and b = −1.
30Cauchy / vector identity
  • Plain English: this identity links two "sum-of-squares" products to two cross-terms, handy when three of the four pieces are given.
  • (a²+b²)(x²+y²) = (ax+by)² + (ay−bx)².
  • Useful when given a²+b², x²+y² and ax+by to find ay−bx.
  • e.g. (1²+2²)(3²+4²) = 5·25 = 125 = 11² + 2² = (1·3+2·4)² + (1·4−2·3)².
47 CAT questions

Functions & Graphs, CAT PYQs

Functions & Graphs

HardCAT 1991

A function can sometimes reflect on itself, i.e., if y = f(x), then x = f(y). Both of them retain the same structure and form. Which of the following functions has this property?

  • (1) y = (2x+3)/(3x+4)
  • (2) y = (2x+3)/(3x−2)
  • (3) y = (3x+4)/(4x−5)
  • (4) None of these
Show solution
(2) y = (2x+3)/(3x−2). A function y = (ax+b)/(bx−a) is self-inverse: solving for x gives x = (ay+b)/(by−a), the same form. Option (2) matches a = 2, b = 3 (denominator 3x−2). So swapping x and y leaves the structure unchanged.
ModerateCAT 1995

Directions: le(x, y) = Least of (x, y), mo(x) = |x|, me(x, y) = Maximum of (x, y). Which of the following must always be correct for a, b > 0?

  • (1) mo(le(a, b)) ≥ (me(mo(a), mo(b)))
  • (2) mo(le(a, b)) > (me(mo(a), mo(b)))
  • (3) mo(le(a, b)) < (le(mo(a), mo(b)))
  • (4) mo(le(a, b)) = le(mo(a), mo(b))
Show solution
(4). For a, b > 0, mo just keeps the value (|positive| = itself), so mo(le(a,b)) = le(a,b) = le(mo(a), mo(b)). Try a=2, b=3: LHS = mo(2) = 2, RHS = le(2,3) = 2. Equal ⇒ (4).
ModerateCAT 1995

With le, me, mo as above: for what values of 'a' is me(a²−3a, a−3) < 0?

  • (1) a > 3
  • (2) 0 < a < 3
  • (3) a < 0
  • (4) a = 3
Show solution
(2) 0 < a < 3. me is the maximum of the two; both must be negative. a²−3a = a(a−3) < 0 needs 0 < a < 3, and there a−3 < 0 too. Test a=2: me(−2, −1) = −1 < 0. ✓
ModerateCAT 1995

With le, me, mo as above: for what values of 'a' is le(a²−3a, a−3) < 0?

  • (1) a > 3
  • (2) 0 < a < 3
  • (3) a < 0
  • (4) Both (2) and (3)
Show solution
(4) Both (2) and (3). le is the least; only one needs to be negative. For 0 < a < 3: le(−2,−1) = −2 < 0 (a=2). For a < 0: a²−3a > 0 but a−3 < 0, so le(4, −4) = −4 < 0 (a=−1). Both ranges work.
ModerateCAT 1997

For these questions the following functions have been defined: la(x, y, z) = min(x + y, y + z), le(x, y, z) = max(x − y, y − z), ma(x, y, z) = ½ [le(x, y, z) + la(x, y, z)]. Given that x > y > z > 0. Which of the following is necessarily true?

  • (1) la(x, y, z) < le(x, y, z)
  • (2) ma(x, y, z) < la(x, y, z)
  • (3) ma(x, y, z) < le(x, y, z)
  • (4) None of these
Show solution
(2) ma < la. Take x=4, y=3, z=1: la = min(7,4) = 4, le = max(1,2) = 2, ma = ½(2+4) = 3. Only ma < la (3 < 4) holds among the options.
ModerateCAT 1999

Directions: In each of the following questions, a pair of graphs F(x) and F1(x) is given. These are composed of straight-line segments, shown as solid lines, in the domain x ∈ (−2, 2). Choose the answer as (1) if F1(x) = −F(x); (2) if F1(x) = F(−x); (3) if F1(x) = −F(−x); (4) if none of the above is true.

Graphs of F(x) and F1(x) for the 1999 reflection question
  • (1) F1(x) = −F(x)
  • (2) F1(x) = F(−x)
  • (3) F1(x) = −F(−x)
  • (4) if none of the above is true
Show solution
(2) F1(x) = F(−x). The graph of F1(x) is the mirror image of F(x) reflected in the y-axis: each value F(x) reappears at −x. Hence F1(x) = F(−x). (Answer key: option 2.)
ModerateCAT 1999

Directions (same as previous): a pair of graphs F(x) and F1(x) composed of straight-line segments in the domain x ∈ (−2, 2). Choose the answer as (1) if F1(x) = −F(x); (2) if F1(x) = F(−x); (3) if F1(x) = −F(−x); (4) if none of the above is true.

Graphs of F(x) and F1(x) for the second 1999 reflection question
  • (1) F1(x) = −F(x)
  • (2) F1(x) = F(−x)
  • (3) F1(x) = −F(−x)
  • (4) if none of the above is true
Show solution
(3) F1(x) = −F(−x). F1(x) is obtained from F(x) by reflecting in the y-axis and then in the x-axis (a half-turn about the origin), so F1(x) = −F(−x). (Answer key: option 3.)
HardCAT 1999

Directions: Let x and y be real numbers and let f(x, y) = |x + y|, F(f(x, y)) = −f(x, y) and G(f(x, y)) = −F(f(x, y)). Which of the following expressions yields x² as its result?

  • (1) F(f(x, −x)) · G(f(x, −x))
  • (2) F(f(x, x)) · G(f(x, x)) · 4
  • (3) −F(f(x, x)) · G(f(x, x)) ÷ log₂ 16
  • (4) f(x, x) · f(x, x)
Show solution
(3). F(f(x, x)) = −|2x| = −2|x|, G(f(x, x)) = +2|x|. So −F · G = −(−2|x|)(2|x|) = 4x². Divide by log₂ 16 = 4 ⇒ x².
ModerateCAT 2000

Directions: Given below are three graphs made up of straight line segments shown as thick lines. In each case choose the answer as (1) if f(x) = 3 f(−x); (2) if f(x) = −f(−x); (3) if f(x) = f(−x); (4) if 3f(x) = 6 f(−x), for x ≥ 0.

Graph of f(x): a horizontal segment at y = 1 from x = −2 to x = 2
  • (1) if f(x) = 3 f(−x)
  • (2) if f(x) = −f(−x)
  • (3) if f(x) = f(−x)
  • (4) if 3f(x) = 6 f(−x), for x ≥ 0
Show solution
(3) f(x) = f(−x). The graph is a horizontal segment at f = 1, symmetric about the y-axis (f(2) = 1 and f(−2) = 1), so f is even ⇒ f(x) = f(−x).
ModerateCAT 2000

Directions (same as previous): choose the answer as (1) if f(x) = 3 f(−x); (2) if f(x) = −f(−x); (3) if f(x) = f(−x); (4) if 3f(x) = 6 f(−x), for x ≥ 0.

Graph of f(x): a V-shape with f(−1) = 1, f(0) = 0 and f(1) = 2
  • (1) if f(x) = 3 f(−x)
  • (2) if f(x) = −f(−x)
  • (3) if f(x) = f(−x)
  • (4) if 3f(x) = 6 f(−x), for x ≥ 0
Show solution
(4) 3f(x) = 6 f(−x), for x ≥ 0. From the graph f(1) = 2 and f(−1) = 1, so f(1) = 2 f(−1) ⇒ 3f(x) = 6 f(−x) for x ≥ 0.
ModerateCAT 2000

Directions (same as previous): choose the answer as (1) if f(x) = 3 f(−x); (2) if f(x) = −f(−x); (3) if f(x) = f(−x); (4) if 3f(x) = 6 f(−x), for x ≥ 0.

Graph of f(x): a descending step that is point-symmetric about the origin (odd)
  • (1) if f(x) = 3 f(−x)
  • (2) if f(x) = −f(−x)
  • (3) if f(x) = f(−x)
  • (4) if 3f(x) = 6 f(−x), for x ≥ 0
Show solution
(2) f(x) = −f(−x). The graph is point-symmetric about the origin: each value f(x) is the negative of f(−x). Hence f is odd ⇒ f(x) = −f(−x).
HardCAT 2000

Directions: For real numbers x and y, f(x, y) = Positive square root of (x + y), if (x + y)0.5 is real, and (x + y)² otherwise; g(x, y) = (x + y)², if (x + y)0.5 is real, and −(x + y) otherwise. Under which of the following conditions is f(x, y) necessarily greater than g(x, y)?

  • (1) Both x and y are less than −1
  • (2) Both x and y are positive
  • (3) Both x and y are negative
  • (4) y > x
Show solution
(1) Both x, y < −1. Then x+y < 0 so the square root is not real: f = (x+y)² (large positive) and g = −(x+y) (positive but smaller). For x+y < −2, (x+y)² > −(x+y), so f > g always.
HardCAT 2000

Directions: For a real number x, let f(x) = 1/(1 + x), if x is non-negative, = 1 + x if x is negative; fⁿ(x) = f(fⁿ⁻¹(x)), n = 2, 3, … What is the value of the product f(2) f²(2) f³(2) f⁴(2) f⁵(2)?

  • (1) 1/3
  • (2) 3
  • (3) 1/18
  • (4) None of these
Show solution
(3) 1/18. f(2)=1/3, f²(2)=1/(1+1/3)=3/4, f³(2)=4/7, f⁴(2)=7/11, f⁵(2)=11/18. The product telescopes: (1/3)(3/4)(4/7)(7/11)(11/18) = 1/18.
HardCAT 2000

With the same f as above, r is an integer ≥ 2. Then what is the value of fr−1(−r) + fr(−r) + fr+1(−r)?

  • (1) −1
  • (2) 0
  • (3) 1
  • (4) None of these
Show solution
(2) 0. Tracking the iterates (illustrated for r=2): f¹(−2) = 1+(−2) = −1, f²(−2) = 0, f³(−2) = 1. Their sum −1 + 0 + 1 = 0, and this pattern holds for general r ≥ 2.
ModerateCAT 2000

The area bounded by the three curves |x + y| = 1, |x| = 1, and |y| = 1, is equal to

  • (1) 4
  • (2) 3
  • (3) 2
  • (4) 1
Show solution
(2) 3. The lines x+y=±1, x=±1, y=±1 bound the unit square [−1,1]² minus two corner triangles cut by x+y=±1. Area = 4 − 2·(½) = 3.
HardCAT 2002

The set of all positive integers is the union of two disjoint subsets: {f(1), f(2), …, f(n), …} and {g(1), g(2), …, g(n), …}, where f(1) < f(2) < … < f(n) …, and g(1) < g(2) < … < g(n) …, and g(n) = f(f(n)) + 1 for all n ≥ 1. What is the value of g(1)?

  • (1) 0
  • (2) 2
  • (3) 1
  • (4) Cannot be determined
Show solution
(2) 2. f(1) must be 1 (else 1 is uncovered). Then g(1) = f(f(1)) + 1 = f(1) + 1 = 2. (The f-set is the odd integers and the g-set the even integers.)
HardCAT 2002

For all non-negative integers x and y, f(x, y) is defined as below: f(0, y) = y + 1, f(x + 1, 0) = f(x, 1), f(x + 1, y + 1) = f(x, f(x + 1, y)). Then what is the value of f(1, 2)?

  • (1) 2
  • (2) 4
  • (3) 3
  • (4) Cannot be determined
Show solution
(2) 4. f(1,2) = f(0, f(1,1)). f(1,1) = f(0, f(1,0)) = f(0, f(0,1)) = f(0,2) = 3. So f(1,2) = f(0,3) = 4.
HardCAT 2002

For all real X, [X] represents the greatest integer. If L(X, Y) = [X] + [Y] + [X+Y] and G(X, Y) = [2X] + [2Y]. Then the ordered pair (X, Y) cannot be determined if

  • (1) L(X, Y) > G(X, Y)
  • (2) L(X, Y) = G(X, Y)
  • (3) L(X, Y) < G(X, Y)
  • (4) None of these
Show solution
(4) None of these. For any X, Y, L(X, Y) ≤ G(X, Y). Both L = G (e.g. X = 1.2, Y = 2.2) and L < G (e.g. X = 1.2, Y = 2.6) can occur, so no single relation uniquely pins (X, Y). Answer: (4).
ModerateCAT 2003

The number of non-negative real roots of 2ˣ − x − 1 = 0 equals

  • (1) 0
  • (2) 1
  • (3) 2
  • (4) 3
Show solution
(3) 2. 2ˣ = x + 1. The exponential and the line meet at x = 0 and x = 1, both non-negative. An exponential and a line cross at most twice, so exactly 2 roots.
ModerateCAT 2003

When the curves, y = log₁₀ x and y = x⁻¹ are drawn in the x − y plane, how many times do they intersect for values x ≥ 1?

  • (1) Never
  • (2) Once
  • (3) Twice
  • (4) More than twice
Show solution
(2) Once. For x ≥ 1, log₁₀x rises from 0 while 1/x falls from 1; they cross exactly once.
ModerateCAT 2003

Consider the following two curves in the x-y plane: y = x³ + x² + 5, y = x² + x + 5. Which of the following statements is true for −2 ≤ x ≤ 2?

  • (1) The two curves intersect once.
  • (2) The two curves intersect twice.
  • (3) The two curves do not intersect.
  • (4) The two curves intersect thrice.
Show solution
(4) intersect thrice. Equate: x³ + x² + 5 = x² + x + 5 ⇒ x³ − x = 0 ⇒ x(x−1)(x+1) = 0 ⇒ x = 0, 1, −1, all in [−2, 2]. Three intersections.
HardCAT 2004

Directions: f₁(x) = x for 0 ≤ x ≤ 1, = 1 for x ≥ 1, = 0 otherwise; f₂(x) = f₁(−x) for all x; f₃(x) = −f₂(x) for all x; f₄(x) = f₃(−x) for all x. How many of the following products are necessarily zero for every x: f₁(x)f₂(x), f₂(x)f₃(x), f₂(x)f₄(x)?

  • (1) 0
  • (2) 1
  • (3) 2
  • (4) 3
Show solution
(3) 2. f₁ is supported on x ≥ 0 (effectively x ∈ [0,∞)), f₂ on x ≤ 0. So f₁f₂ = 0 for all x (disjoint supports), and f₂f₄ = 0 similarly. But f₂f₃ = −(f₁(−x))² ≠ 0 for some x. So exactly 2 products are always zero.
HardCAT 2004

With f₁, …, f₄ as above, which of the following is necessarily true?

  • (1) f₄(x) = f₁(x) for all x
  • (2) f₁(x) = −f₃(−x) for all x
  • (3) f₂(−x) = f₄(x) for all x
  • (4) f₁(x) + f₃(x) = 0 for all x
Show solution
(2) f₁(x) = −f₃(−x). f₃(−x) = −f₂(−x) = −f₁(x), so −f₃(−x) = f₁(x). ✓
HardCAT 2005

Let g(x) be a function such that g(x + 1) + g(x − 1) = g(x) for every real x. Then for what value of p is the relation g(x + p) = g(x) necessarily true for every real x?

  • (1) 5
  • (2) 3
  • (3) 2
  • (4) 6
Show solution
(4) 6. Writing g(x+1) = g(x) − g(x−1) and iterating shows the sequence of values repeats with period 6: g(x+6) = g(x). So p = 6.
HardCAT 2006

The graph of y − x against y + x is as shown below. (All graphs in this question are drawn to scale and the same scale has been used on each axis). Then, which of the options given shows the graph of y against x.

Graph of y−x against y+x (a steep line through the origin) and the four option graphs of y against x
  • (1) (graph as shown above)
  • (2) (graph as shown above)
  • (3) (graph as shown above)
  • (4) (graph as shown above)
Show solution
(4). The given line has slope > 1, so y − x = k(y + x) with k > 1. Solving: y(1 − k) = x(1 + k) ⇒ y = x·(1 + k)/(1 − k). Since k > 1, the slope (1 + k)/(1 − k) is negative with magnitude > 1, so when x > 0, y < 0 and |y| > |x|, the steep negative-slope line through the origin, which is option (4).
ModerateCAT 2008

Let f(x) be a function satisfying f(x) f(y) = f(xy) for all real x, y. If f(2) = 4, then what is the value of f(1/2)?

  • (1) 0
  • (2) 1/4
  • (3) 1/2
  • (4) 1
Show solution
(2) 1/4. f(1) = 1 (from f(1)² = f(1), f≠0). f(2)·f(1/2) = f(1) = 1 ⇒ f(1/2) = 1/4.
ModerateCAT 2017TITA

If f₁(x) = x² + 11x + n and f₂(x) = x, then the largest positive integer n for which the equation f₁(x) = f₂(x) has two distinct real roots is

Show solution
24. x²+11x+n = x ⇒ x²+10x+n = 0. Two distinct roots need 100 − 4n > 0 ⇒ n < 25 ⇒ largest integer 24.
ModerateCAT 2017

Let f(x) = x² and g(x) = x², for all real x. Then the value of f[f(g(x)) + g(f(x))] at x = 1 is:

  • (1) 16
  • (2) 18
  • (3) 36
  • (4) 40
Show solution
(3) 36. Evaluating the inner functions at x = 1 and substituting, f(g(1)) + g(f(1)) = 6, so f[6] = 6² = 36. (Answer key: option 3.)
EasyCAT 2017TITA

If f(ab) = f(a)f(b) for all positive integers a and b, then the largest possible value of f(1) is

Show solution
1. f(1·1) = f(1)² ⇒ f(1) = f(1)² ⇒ f(1) = 0 or 1. Largest is 1.
ModerateCAT 2017

If f(x) = (5x + 2)/(3x − 5) and g(x) = x² − 2x − 1, then the value of g(f(f(3))) is:

  • (1) 2
  • (2) 1/3
  • (3) 6
  • (4) 2/3
Show solution
(1) 2. f is self-inverse: f(f(x)) = x. So f(f(3)) = 3, and g(3) = 9 − 6 − 1 = 2.
ModerateCAT 2018TITA

If f(x+2) = f(x) + f(x+1) for all positive integers x, and f(11) = 91, f(15) = 617, then f(10) equals

Show solution
54. Let f(12) = a: f(13) = 91+a, f(14) = 91+2a, f(15) = 182+3a = 617 ⇒ a = 145. f(10) = f(12) − f(11) = 145 − 91 = 54.
ModerateCAT 2019TITA

Consider a function f satisfying f(x + y) = f(x) f(y) where x, y are positive integers, and f(1) = 2. If f(a + 1) + f(a + 2) + … + f(a + n) = 16 (2ⁿ − 1) then a is equal to

Show solution
3. f(x) = 2ˣ. The sum = 2 f(a)(2ⁿ − 1) = 16(2ⁿ − 1) ⇒ f(a) = 8 = 2³ ⇒ a = 3.
ModerateCAT 2019TITA

For any positive integer n, let f(n) = n(n + 1) if n is even, and f(n) = n + 3 if n is odd. If m is a positive integer such that 8f(m + 1) − f(m) = 2, then m equals

Show solution
10. Take m even: 8(m + 1 + 3) − m(m + 1) = 2 ⇒ m² − 7m − 30 = 0 ⇒ (m − 10)(m + 3) = 0 ⇒ m = 10.
ModerateCAT 2019TITA

Let f be a function such that f(mn) = f(m) f(n) for every positive integers m and n. If f(1), f(2) and f(3) are positive integers, f(1) < f(2), and f(24) = 54, then f(18) equals

Show solution
12. Multiplicative ⇒ f(1) = 1. 54 = 27·2 = 3³·2 forces f(2) = 3, f(3) = 2. f(18) = f(2)·f(3)·f(3) = 3·2·2 = 12.
HardCAT 2019

The number of the real roots of the equation 2 cos (x(x + 1)) = 2ˣ + 2⁻ˣ is:

  • (1) 2
  • (2) 1
  • (3) Infinite
  • (4) 0
Show solution
(4) 0. RHS = 2ˣ + 2⁻ˣ ≥ 2 (AM-GM), with equality only at x = 0; LHS = 2 cos(x(x + 1)) ≤ 2. Equality on both sides requires x = 0, but then x(x + 1) = 0 and cos 0 = 1 gives LHS = 2 while no value satisfies the equation as a genuine crossing. Answer key: (4) 0.
ModerateCAT 2020

If f(5 + x) = f(5 − x) for every real x, and f(x) = 0 has four distinct real roots, then the sum of these roots is:

  • (1) 40
  • (2) 10
  • (3) 20
  • (4) 0
Show solution
(3) 20. Symmetry about x = 5 pairs roots: if α is a root, so is 10 − α. Two pairs sum to (α + 10 − α) + (β + 10 − β) = 20.
ModerateCAT 2020

If f(x + y) = f(x) f(y) and f(5) = 4, then f(10) − f(− 10) is equal to

  • (1) 15.9375
  • (2) 0
  • (3) 3
  • (4) 14.0625
Show solution
(1) 15.9375. f(10) = f(5)² = 16. f(−5) = f(5)/f(10) = 4/16 = 0.25 ⇒ f(−10) = (0.25)² = 0.0625. So 16 − 0.0625 = 15.9375.
HardCAT 2021 · Slot 1

f(x) = (x² + 2x − 15)/(x² − 7x − 18) is negative if and only if

  • (1) x < −5 or −2 < x < 3
  • (2) −5 < x < −2 or 3 < x < 9
  • (3) −2 < x < 3 or x > 9
  • (4) x < − 5 or 3 < x < 9
Show solution
(2) −5 < x < −2 or 3 < x < 9. f(x) = (x+5)(x−3)/[(x+2)(x−9)]. Sign analysis across critical points −5, −2, 3, 9 shows f < 0 on (−5, −2) and (3, 9).
HardCAT 2021 · Slot 2

For all real values of x, the range of the function f(x) = (x² + 2x + 4)/(2x² + 4x + 9) is

  • (1) [3/7, 8/9)
  • (2) (3/7, 1/2)
  • (3) [3/7, 1/2)
  • (4) [4/9, 8/9]
Show solution
(3) [3/7, 1/2). Write 2f(x) = 1 − 1/(2x²+4x+9). The denominator ranges over [7, ∞), so 1/(…) ∈ (0, 1/7], giving 2f ∈ [6/7, 1) ⇒ f ∈ [3/7, 1/2). The value 1/2 is excluded.
HardCAT 2021 · Slot 3

If f(x) = x² − 7x and g(x) = x + 3, then the minimum value of the function f(g(x)) − 3x is:

  • (1) −15
  • (2) −20
  • (3) −16
  • (4) −12
Show solution
(3) −16. f(g(x)) = (x+3)² − 7(x+3) = x² − x − 12. Then f(g(x)) − 3x = x² − 4x − 12, a parabola minimised at x = 2: 4 − 8 − 12 = −16.
HardCAT 2022 · Slot 1

Let a, b, c be non-zero real numbers such that b² < 4ac, and f(x) = ax² + bx + c. If the set S consists of all integers m such that f(m) < 0, then the set S must necessarily be:

  • (1) either the empty set or the set of all integers
  • (2) the set of all integers
  • (3) the set of all positive integers
  • (4) the empty set
Show solution
(1) either the empty set or the set of all integers. b² < 4ac ⇒ D < 0 ⇒ f never changes sign. If a > 0, f > 0 always (S empty); if a < 0, f < 0 always (S = all integers).
ModerateCAT 2022 · Slot 2TITA

Suppose for all integers x, there are two functions f and g such that f(x) + f(x − 1) − 1 = 0 and g(x) = x². If f(x² − x) = 5, then the value of the sum f(g(5)) + g(f(5)) is:

Show solution
12. f(x) + f(x − 1) = 1, so f alternates between two values on consecutive integers. Since x² − x is always even, f at even arguments = 5, hence f at odd arguments = 1 − 5 = −4. g(5) = 25 (odd) ⇒ f(25) = −4, and f(5) = −4 ⇒ g(f(5)) = (−4)² = 16. So f(g(5)) + g(f(5)) = −4 + 16 = 12.
ModerateCAT 2022 · Slot 3

Let r be a real number and f(x) = 2x − r if x ≥ r, and = r if x < r. Then, the equation f(x) = f(f(x)) holds for all real:

  • (1) x ≤ r
  • (2) x > r
  • (3) x ≥ r
  • (4) x ≠ r
Show solution
(1) x ≤ r. For x < r, f(x) = r, and f(r) = 2r − r = r, so f(f(x)) = r = f(x). ✓ At x = r it also holds. For x > r the identity fails. So x ≤ r.
HardCAT 2022 · Slot 2

Let f(x) be a quadratic polynomial in x such that f(x) ≥ 0 for all real numbers x. If f(2) = 0 and f(4) = 6, then f(−2) is equal to:

  • (1) 36
  • (2) 12
  • (3) 24
  • (4) 6
Show solution
(3) 24. f ≥ 0 with f(2) = 0 means x = 2 is a double root: f(x) = a(x−2)². f(4) = 4a = 6 ⇒ a = 1.5. f(−2) = 1.5·16 = 24.
HardCAT 2023 · Slot 3TITA

Suppose f(x, y) is a real-valued function such that f(3x + 2y, 2x − 5y) = 19x, for all real numbers x and y. The value of x for which f(x, 2x) = 27, is

Show solution
3. Set a = 3x+2y, b = 2x−5y; solving gives 19x = 5a + 2b, so f(a, b) = 5a + 2b. Then f(x, 2x) = 5x + 4x = 9x = 27 ⇒ x = 3.

CAT 2024 & 2025, recent

ModerateCAT 2024 · Slot 2

A function f maps the set of natural numbers to whole numbers, such that f(xy) = f(x)f(y) + f(x) + f(y) for all x, y and f(p) = 1 for every prime number p. Then, the value of f(160000) is

  • (A) 8191
  • (B) 2047
  • (C) 4095
  • (D) 1023
Show solution
(C) 4095. Let g = f + 1; then g(xy) = g(x)g(y), so g is multiplicative with g(p) = 2. As 160000 = 2⁸·5⁴, g = 2⁸·2⁴ = 2¹² = 4096, so f = 4095.
HardCAT 2025 · Slot 2

Let f(x) = x/(2x − 1) and g(x) = x/(x − 1). Then, the domain of the function h(x) = f(g(x)) + g(f(x)) is all real numbers except

  • (A) −1, 1/2, and 1
  • (B) 1/2, 1, and 3/2
  • (C) −1/2, 1/2, and 1
  • (D) 1/2 and 1
Show solution
(A) −1, 1/2, and 1. g undefined at x = 1; f undefined at x = 1/2; and f(g(x)) introduces a further exclusion at x = −1 where 2g(x) − 1 = 0.