◆ QA · Algebra

Maxima-Minima , formulas + CAT PYQs

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

17CAT PYQs
Algebrachapter
Formulas PYQs (17)↩ 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.
17 CAT questions

Maxima-Minima, CAT PYQs

Maxima-Minima

HardCAT 2001

Let x, y be two positive numbers such that x + y = 1. Then, the minimum value of (x + 1/x)² + (y + 1/y)² is

  • (1) 12
  • (2) 20
  • (3) 12.5
  • (4) 13.3
Show solution
(3) 12.5. By symmetry the minimum is at x = y = 1/2. Each term = (1/2 + 2)² = (5/2)² = 25/4. Sum = 2·25/4 = 25/2 = 12.5.
HardCAT 2002

For three integers x, y and z, x + y + z = 5, and xy + yz + xz = 3. What is the largest value which x can take?

  • (1) 3√13
  • (2) √19
  • (3) 13/3
  • (4) √15
Show solution
(3) 13/3. Substitute z = 5 − x − y into xy + z(x+y) = 3 to get y² + (x−5)y + (x²−5x+3) = 0. For real y, discriminant ≥ 0: (x−5)² − 4(x²−5x+3) ≥ 0 ⇒ 3x² − 10x − 13 ≤ 0 ⇒ (3x−13)(x+1) ≤ 0 ⇒ −1 ≤ x ≤ 13/3. So x_max = 13/3.
HardCAT 2003

If three positive real numbers x, y, z satisfy y − x = z − y and xyz = 4, then what is the minimum possible value of y?

  • (1) 2^(1/3)
  • (2) 2^(2/3)
  • (3) 2^(1/4)
  • (4) 2^(3/4)
Show solution
(2) 2^(2/3). x, y, z are in AP so y is their AM. By AM ≥ GM, y ≥ (xyz)^(1/3) = 4^(1/3) = 2^(2/3). Minimum y = 2^(2/3).
HardCAT 2003

Let a, b, c, d be four integers such that a + b + c + d = 4m + 1 where m is a positive integer. Which one of the following is necessarily true?

  • (1) The minimum possible value of a² + b² + c² + d² is 4m² − 2m + 1
  • (2) The minimum possible value of a² + b² + c² + d² is 4m² + 2m + 1
  • (3) The maximum possible value of a² + b² + c² + d² is 4m² − 2m + 1
  • (4) The maximum possible value of a² + b² + c² + d² is 4m² + 2m + 1
Show solution
(2) The minimum possible value of a² + b² + c² + d² is 4m² + 2m + 1. Sum of squares is minimised when the values are as equal as possible: three equal to m and one equal to m + 1. Then 3·m² + (m + 1)² = 4m² + 2m + 1.
HardCAT 2004

Let f(x) = ax² − b |x|, where a and b are constants. Then at x = 0, f(x) is

  • (1) maximized whenever a > 0, b > 0
  • (2) maximized whenever a > 0, b < 0
  • (3) minimized whenever a > 0, b > 0
  • (4) minimized whenever a > 0, b < 0
Show solution
(4) minimized whenever a > 0, b < 0. For a > 0, b < 0, both branches open upward with no negative dip at 0, so x = 0 is a minimum.
ModerateCAT 2006

Let f(x) = max (2x + 1, 3 − 4x), where x is any real number. Then the minimum possible value of f(x) is:

  • (1) 1/3
  • (2) 1/2
  • (3) 2/3
  • (4) 5/3
Show solution
(4) 5/3. The max of two lines is least where they cross: 2x+1 = 3−4x ⇒ x = 1/3, value = 2(1/3)+1 = 5/3.
ModerateCAT 2007

A quadratic function f(x) attains a maximum of 3 at x = 1. The value of the function at x = 0 is 1. What is the value of f(x) at x = 10?

  • (1) −119
  • (2) −159
  • (3) −110
  • (4) −180
Show solution
(2) −159. Vertex form f(x) = a(x−1)² + 3 with maximum ⇒ a < 0. f(0) = a + 3 = 1 ⇒ a = −2. So f(x) = −2(x−1)² + 3 ⇒ f(10) = −2(81) + 3 = −159.
ModerateCAT 2017TITA

If a, b, c, and d are integers such that a + b + c + d = 30 then the minimum possible value of (a − b)² + (a − c)² + (a − d)² is

Show solution
2. Minimised when values are as equal as possible. 30/4 = 7.5, so take 8, 8, 7, 7. With a = 8 and b,c,d = 8,7,7: 0 + 1 + 1 = 2.
ModerateCAT 2018TITA

Let f(x) = min (2x², 52 − 5x) where x is any positive real number. Then the maximum possible value of f(x) is

Show solution
32. The min-of-two reaches its peak where they meet: 2x² = 52 − 5x ⇒ 2x²+5x−52 = 0 ⇒ x = 4. f(4) = 2·16 = 32.
ModerateCAT 2018TITA

Let f(x) = max(5x, 52 − 2x²), where x is any positive real number. Then the minimum possible value of f(x) is

Show solution
20. The max-of-two is least where they cross: 5x = 52 − 2x² ⇒ 2x²+5x−52 = 0 ⇒ x = 4. f(4) = 5·4 = 20.
HardCAT 2020TITA

If x and y are positive real numbers satisfying x + y = 102, then the minimum possible value of 2601(1 + 1/x)(1 + 1/y) is:

Show solution
2704. To minimise (1+1/x)(1+1/y) maximise x and y, so x = y = 51. Then 2601·(52/51)(52/51) = 2601·(52²/51²) = 52² = 2704.
HardCAT 2020

For real x, the maximum possible value of x/√(1 + x⁴) is:

  • (1) 1/√3
  • (2) 1/√2
  • (3) 1/2
  • (4) 1
Show solution
(2) 1/√2. For x > 0, x/√(1 + x⁴) = 1/√(1/x² + x²). By AM-GM, x² + 1/x² ≥ 2 (equality at x = 1), so the denominator is at least √2 and the expression is at most 1/√2.
ModerateCAT 2021 · Slot 3

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

  • (1) − 15
  • (2) − 20
  • (3) − 16
  • (4) − 12
Show solution
(3) −16. f(g(x)) − 3x = (x+3)² − 7(x+3) − 3x = x² − 4x − 12. Minimum = −D/4a = −(16 + 48)/4 = −16.
ModerateCAT 2022 · Slot 1

Let 0 ≤ a ≤ x ≤ 100 and f(x) = |x − a| + |x − 100| + |x − a − 50|. Then, the maximum value of f(x) becomes 100 when a is equal to:

  • (1) 0
  • (2) 25
  • (3) 100
  • (4) 50
Show solution
(4) 50. Since a ≤ x ≤ 100, |x − a| = x − a and |x − 100| = 100 − x, so f(x) = 100 − a + |x − a − 50|. Its maximum over x ∈ [a, 100] is 100 − a + max(50, |50 − a|). This equals 100 exactly when a = 50 (giving 50 + 50 = 100).
HardCAT 2022 · Slot 3

The minimum possible value of (x² − 6x + 10)/(3 − x), for x < 3, is:

  • (1) −2
  • (2) 2
  • (3) 1/2
  • (4) −1/2
Show solution
(2) 2. Let t = 3 − x > 0. Numerator = (x−3)² + 1 = t² + 1, so expression = (t² + 1)/t = t + 1/t ≥ 2 by AM-GM (equality at t = 1, i.e. x = 2).
HardCAT 2023 · Slot 2TITA

Let k be the largest integer such that the equation (x − 1)² + 2kx + 11 = 0 has no real roots. If y is a positive real number, then the least possible value of k/(4y) + 9y is

Show solution
6. (x − 1)² + 2kx + 11 = x² + (2k − 2)x + 12. No real roots ⇒ D < 0 ⇒ (2k − 2)² − 48 < 0 ⇒ (2k − 2)² < 48, so the largest integer k = 4. Then k/(4y) + 9y = 1/y + 9y ≥ 2√(9) = 6 by AM-GM (equality at y = 1/3).
Genuinely unrecoverable / out-of-scope from this chapter's pages: a cluster of pure arithmetic word problems printed in the Algebra solved-papers but belonging to Arithmetic (cheque-transposition CAT 2000 & 2007; census Chota/Mota Hazri CAT 2001; burger-shake-fries CAT 2001; coin-change CAT 2001; Miso currency CAT 2007; rice-shop CAT 2008; bacteria-generation CAT 1998; salary X & Y CAT 2001; book page-sum CAT 2001; Fibonacci 10th term CAT 2001; telecom operators CAT 2005; cities-and-roads CAT 2000), these are covered better in the Arithmetic chapter and are not algebra-technique items. A few log/indices questions whose option text is too garbled in extraction to transcribe faithfully (CAT 1998 a#b / a∇b symbol operations Q4; CAT 1998 log M/N Q6 & Q7; CAT 1998 series log Q8; CAT 2004 log chain Q14; CAT 2005 smallest-value x=−0.5 Q12; several CAT 2017-2021 log identities printed with broken fraction glyphs) were left out rather than guessed, per the authenticity rule.

CAT 2024 & 2025, recent

ModerateCAT 2025 · Slot 2 TITA

If m and n are integers such that (m + 2n)(2m + n) = 27, then the maximum possible value of 2m − 3n is:

Show solution
17. Let m+2n and 2m+n be integer factors of 27. Their sum (m+2n)+(2m+n) = 3(m+n) must be a multiple of 3, so only the factor pairs (3, 9), (9, 3), (−3, −9), (−9, −3) qualify. Solving each gives (m, n) = (−1, 5), (5, −1), (−5, 1), (1, −5), for which 2m − 3n = −13, 13, −17, 17. The maximum is 17.