Formulas

The 94 formulas MHT-CET Maths actually tests

One page, every formula grouped by chapter and ordered by recent weightage. Each entry shows the formula, the symbol legend, and the trap that costs marks. Paper I gives you 1.8 minutes a question — a formula you have to derive in the hall is a formula you have already lost time to.

formulas
94
chapters covered
19
per question in the hall
1.8 min
shifts of PYQs behind it
45

How to use this page

  • First read: cover-to-cover, marking every formula you don’t already know cold. The groups are ordered by recent weightage, so the chapters at the top of this page are the ones carrying the most questions per paper — start your marking there, not at the chapter you happen to like.
  • Read the ‘Note’ row: several of them exist purely to save time — a formula that replaces a whole calculus routine with one line of arithmetic is worth more than a formula you already know, because the binding constraint in this paper is the clock, not the syllabus.
  • Active recall: cover the right-hand side, read only the formula NAME, and write the formula plus two symbol meanings from memory. Anything you miss goes on tomorrow’s list. Each chapter header links to its playbook, which is where you find out what the formula is actually used for.

Line and Plane

Playbook

  • Direction cosines and direction ratios

    l² + m² + n² = 1 l = a/√(a²+b²+c²), m = b/√(a²+b²+c²), n = c/√(a²+b²+c²)

    a, b, c = direction ratios (any non-zero multiple works)
    l, m, n = direction cosines (a specific, normalised triple)

    Note:Ratios are NOT unique, cosines are. l²+m²+n²=1 is the check that you divided by the right magnitude; if it does not come to 1 you have direction ratios, not cosines.

  • Equation of a line — vector and cartesian form

    r = a + λb (x − x₁)/a₁ = (y − y₁)/a₂ = (z − z₁)/a₃

    a = position vector of a known point on the line
    b = direction vector, components a₁, a₂, a₃
    λ = scalar parameter

    Note:Line through two points: b = (position of B) − (position of A). A zero in a denominator is shorthand for that coordinate being constant, not an error.

  • Equation of a plane

    r · n = d ax + by + cz + d = 0 r · n̂ = p (normal form)

    n = normal vector, components a, b, c
    n̂ = unit normal
    p = perpendicular distance of the plane from the origin

    Note:The coefficients of x, y, z ARE the normal — read them straight off. Plane through a point A with normal n: r · n = a · n.

  • Angles — line-line, plane-plane, line-plane

    lines: cos θ = |b₁·b₂| / (|b₁||b₂|) planes: cos θ = |n₁·n₂| / (|n₁||n₂|) line & plane: sin θ = |b·n| / (|b||n|)

    b, b₁, b₂ = direction vectors of the lines
    n, n₁, n₂ = normal vectors of the planes
    θ = acute angle between the two objects

    Note:The line-and-plane case uses SIN, not cos — this is the single most-repeated slip in the chapter. Reason: you measure the angle to the plane but compute with its normal, so the two are complementary.

  • Distance of a point from a plane

    d = |ax₁ + by₁ + cz₁ + d₀| / √(a² + b² + c²)

    (x₁, y₁, z₁) = the point
    ax + by + cz + d₀ = 0 = the plane

    Note:Two parallel planes: put both in the same normal coefficients first, then d = |d₁ − d₂| / √(a²+b²+c²). Forgetting to re-scale one equation is the usual error.

  • Shortest distance between two skew lines

    d = |(a₂ − a₁) · (b₁ × b₂)| / |b₁ × b₂|

    a₁, a₂ = points on the two lines
    b₁, b₂ = their direction vectors

    Note:Only valid when b₁ × b₂ ≠ 0. If the lines are PARALLEL the denominator vanishes and you must use d = |(a₂ − a₁) × b| / |b| instead. Check for parallelism before reaching for this.

  • Condition for two lines to be coplanar (intersecting)

    (a₂ − a₁) · (b₁ × b₂) = 0

    Same symbols as the shortest-distance formula
    Left side = the scalar triple product [a₂ − a₁, b₁, b₂]

    Note:Shared with Vectors: this is the scalar triple product = 0 test wearing a 3-D geometry costume. Zero triple product means shortest distance zero means coplanar. One idea, three chapter dialects.

  • Foot of perpendicular and image of a point in a plane

    (x − x₁)/a = (y − y₁)/b = (z − z₁)/c = −k(ax₁ + by₁ + cz₁ + d₀)/(a² + b² + c²) foot: k = 1 image: k = 2

    (x₁, y₁, z₁) = the given point
    a, b, c = normal coefficients of the plane
    k = 1 for the foot, k = 2 for the mirror image

    Note:The image is exactly twice as far along the same normal as the foot. Solving the foot and then forgetting to double is the standard half-mark loss.

Vectors

Playbook

  • Magnitude and unit vector

    |a| = √(a₁² + a₂² + a₃²) â = a / |a|

    a₁, a₂, a₃ = components along i, j, k
    â = unit vector in the direction of a

    Note:A unit vector along a GIVEN direction is the cheapest sub-step in the chapter and appears inside almost every other formula here. Get it wrong once and everything downstream is wrong.

  • Dot product and the angle between two vectors

    a · b = |a||b| cos θ = a₁b₁ + a₂b₂ + a₃b₃ cos θ = (a · b)/(|a||b|)

    θ = angle between a and b
    Result is a SCALAR

    Note:a · b = 0 means perpendicular — the same condition as m₁m₂ = −1 for lines and a + b = 0 for a pair of lines. The perpendicularity test appears across 7 chapters in 4 different dialects; learn it once.

  • Projection of one vector on another

    scalar projection of a on b = (a · b)/|b| vector projection = ((a · b)/|b|²) b

    b = the vector you project ONTO
    |b|² = b · b

    Note:Note which vector sits in the denominator. Projection of a on b is not the projection of b on a unless the magnitudes are equal.

  • Cross product

    |a × b| = |a||b| sin θ a × b = −(b × a) a × b ⊥ both a and b

    Result is a VECTOR perpendicular to the plane of a and b
    Components come from the 3x3 determinant with i, j, k in the first row

    Note:Anti-commutative — swapping the order flips the sign, unlike the dot product. a × b = 0 means parallel or collinear.

  • Area of a triangle and a parallelogram

    triangle (adjacent sides a, b): A = ½|a × b| parallelogram (adjacent sides): A = |a × b| parallelogram (diagonals d₁, d₂): A = ½|d₁ × d₂|

    a, b = vectors along two adjacent sides from the same vertex
    d₁, d₂ = the two diagonals

    Note:Read the question for SIDES versus DIAGONALS — the half appears in opposite places in the two parallelogram forms, which is exactly what the distractors are built from.

  • Scalar triple product, volume and coplanarity

    [a b c] = a · (b × c) = det of the 3x3 component matrix volume of parallelepiped = |[a b c]| volume of tetrahedron = (1/6)|[a b c]| coplanar ⟺ [a b c] = 0

    [a b c] = scalar triple product (a scalar)
    Cyclic rotation leaves it unchanged; swapping any two vectors flips the sign

    Note:This is the bank's single densest vectors subtopic (71 q) and its hardest (72% HARD). The zero-determinant test is universal: it is concurrency of lines, collinearity of points, coplanarity of vectors and coplanarity of two 3-D lines, all the same computation.

  • Collinearity and coplanarity conditions

    collinear vectors: a × b = 0, or b = λa collinear points A, B, C: AB = λ·AC coplanar vectors: [a b c] = 0, or c = xa + yb

    λ, x, y = scalars
    AB, AC = vectors between the position vectors of the points

    Note:Two equivalent routes each time — a determinant, or a linear combination. On a 1.8 minute budget the determinant is usually faster; the linear combination is faster when the question already hands you the scalars.

  • Section formula (vector form)

    internal (ratio m:n): r = (m·b + n·a)/(m + n) external (ratio m:n): r = (m·b − n·a)/(m − n) midpoint: r = (a + b)/2

    a, b = position vectors of the two endpoints
    m:n = the dividing ratio

    Note:Same formula as coordinate geometry in Straight Line, written with position vectors instead of coordinates. Centroid of a triangle = (a + b + c)/3.

Applications of Derivative

Playbook

  • Tangent and normal to a curve

    slope of tangent m = dy/dx at (x₁, y₁) tangent: y − y₁ = m(x − x₁) normal: y − y₁ = (−1/m)(x − x₁)

    (x₁, y₁) = point of contact, which must lie ON the curve
    m = value of the derivative AT that point, a number not a function

    Note:Two failure modes: leaving dy/dx as an expression instead of evaluating it, and a vertical tangent where dy/dx is undefined (there the tangent is x = x₁ and the normal is horizontal).

  • Rate of change and related rates

    dy/dt = (dy/dx) · (dx/dt)

    t = time (or whatever the independent variable is)
    dx/dt = the rate you are GIVEN, dy/dt = the rate asked for

    Note:This is the chain rule from Differentiation applied to time. Write the geometric relation first (V = (4/3)πr³, A = πr² …), then differentiate both sides with respect to t, then substitute — substituting the numeric value too early is the usual wreck.

  • Increasing and decreasing test

    f'(x) > 0 on an interval ⟹ f is increasing there f'(x) < 0 on an interval ⟹ f is decreasing there

    f'(x) = first derivative
    Test applies on an INTERVAL, not at a point

    Note:Find the critical points, then test the SIGN of f' in each interval between them — do not test the value of f. Strictly-increasing versus non-decreasing matters when f' vanishes at isolated points.

  • Maxima and minima (second-derivative test)

    f'(c) = 0 and f''(c) < 0 ⟹ local MAXIMUM at c f'(c) = 0 and f''(c) > 0 ⟹ local MINIMUM at c f''(c) = 0 ⟹ test fails, use the first-derivative sign change

    c = critical point where f'(c) = 0
    f''(c) = second derivative evaluated at c

    Note:Sign convention is the trap: NEGATIVE second derivative gives the MAXIMUM. For an optimisation word problem also check the endpoints of the feasible domain — the absolute extremum is not always at a critical point.

  • Rolle's theorem and Lagrange's mean value theorem

    Rolle: f continuous on [a,b], differentiable on (a,b), f(a) = f(b) ⟹ ∃ c in (a,b) with f'(c) = 0 LMVT: f continuous on [a,b], differentiable on (a,b) ⟹ ∃ c in (a,b) with f'(c) = (f(b) − f(a))/(b − a)

    [a, b] = closed interval (continuity)
    (a, b) = open interval (differentiability)
    c = the guaranteed interior point

    Note:Rolle is LMVT with f(a) = f(b). The exam more often tests whether the HYPOTHESES hold than the value of c — a modulus function on an interval containing its corner fails differentiability, so neither theorem applies.

  • Approximation using differentials

    f(x + Δx) ≈ f(x) + f'(x)·Δx Δy ≈ (dy/dx)·Δx

    x = a nearby value at which f is easy to evaluate exactly
    Δx = the small increment (signed)

    Note:The bank's only 0% HARD subtopic in this chapter (11 q). Pick x to be the nearest perfect square, cube or standard angle, keep the sign of Δx, and work in RADIANS for trigonometric cases.

  • Angle between two curves and orthogonality

    tan θ = |(m₁ − m₂)/(1 + m₁m₂)| orthogonal ⟺ m₁m₂ = −1

    m₁, m₂ = the two tangent slopes at the point of intersection
    θ = acute angle between the curves

    Note:Identical formula to the angle between two straight lines — the curves only supply the slopes. Find the intersection point FIRST; the angle is defined there and nowhere else.

Differential Equations

Playbook

  • Order and degree

    order = order of the highest derivative present degree = power of that highest derivative, after clearing radicals and fractions

    Degree is defined only when the equation is polynomial in its derivatives

    Note:Degree is undefined if a derivative sits inside sin, log or an exponential. Clear the radicals BEFORE reading the degree — an unsimplified equation reads the wrong number.

  • Variable-separable equations

    f(y) dy = g(x) dx ⟹ ∫f(y) dy = ∫g(x) dx + c

    c = arbitrary constant of integration

    Note:One constant, added once, on one side. If an initial condition is given, substitute it immediately to fix c rather than carrying it through the algebra.

  • Homogeneous equations (substitution)

    dy/dx = F(y/x) put y = vx, dy/dx = v + x·(dv/dx) ⟹ separable in v and x

    v = y/x, the new dependent variable
    F = a function of the ratio y/x alone

    Note:The product-rule term x·(dv/dx) is the step people drop. If the equation is homogeneous in x instead, substitute x = vy. Homogeneous here means same total degree in every term, unrelated to the linear-algebra sense.

  • Linear equations — integrating factor

    dy/dx + P(x)·y = Q(x) IF = e^(∫P dx) y·(IF) = ∫Q·(IF) dx + c

    P, Q = functions of x only
    IF = integrating factor

    Note:PRECONDITION: the equation must already be in dy/dx + Py = Q with the coefficient of dy/dx equal to 1. Divide through first — reading P off an unnormalised equation is the commonest error, and this is the chapter's hardest subtopic at 63% HARD. The mirror form dx/dy + Px = Q with IF = e^(∫P dy) is used just as often.

  • Growth and decay

    dN/dt = k·N ⟹ N = N₀ e^(kt)

    N₀ = amount at t = 0
    k > 0 for growth, k < 0 for decay
    half-life: N = N₀/2

    Note:Two unknowns, N₀ and k, so you need two data points. Solve for k as a ratio (take logs of N₂/N₁) rather than evaluating exponentials numerically — the answers are usually left in terms of logs.

  • Newton's law of cooling

    dθ/dt = −k(θ − θ₀) ⟹ θ − θ₀ = (θ₁ − θ₀) e^(−kt)

    θ = temperature of the body at time t
    θ₀ = ambient (surrounding) temperature, a constant
    θ₁ = initial temperature of the body

    Note:It is the decay model applied to the EXCESS temperature θ − θ₀, not to θ. Only 5 q in the bank but 60% HARD — the difficulty is entirely in remembering to subtract the ambient temperature first.

Indefinite Integration

Playbook

  • Standard integrals

    ∫xⁿ dx = x^(n+1)/(n+1) + c (n ≠ −1) ∫(1/x) dx = ln|x| + c ∫eˣ dx = eˣ + c ∫aˣ dx = aˣ/ln a + c ∫sin x dx = −cos x + c ∫cos x dx = sin x + c ∫sec²x dx = tan x + c

    c = arbitrary constant
    n = any real number other than −1

    Note:The n = −1 exception is not a technicality — it is what makes a log appear from nowhere in partial-fraction answers. The modulus in ln|x| matters for definite work later.

  • Standard integrals giving logs

    ∫tan x dx = ln|sec x| + c ∫cot x dx = ln|sin x| + c ∫sec x dx = ln|sec x + tan x| + c ∫cosec x dx = ln|cosec x − cot x| + c

    All four are recall, not derivation

    Note:Sign traps: cot gives +ln|sin x| but tan gives +ln|sec x| = −ln|cos x|. Options are routinely built from the equivalent-but-differently-signed form, so recognise −ln|cos x| as the same answer.

  • Integrals with a quadratic denominator

    ∫dx/(x² + a²) = (1/a)tan⁻¹(x/a) + c ∫dx/(x² − a²) = (1/2a)ln|(x − a)/(x + a)| + c ∫dx/(a² − x²) = (1/2a)ln|(a + x)/(a − x)| + c

    a = a positive constant

    Note:The middle and last differ ONLY in which term is subtracted, and the printed options exploit exactly that. For a general quadratic, complete the square first to reach one of these three shapes.

  • Integrals with a surd denominator

    ∫dx/√(a² − x²) = sin⁻¹(x/a) + c ∫dx/√(x² + a²) = ln|x + √(x² + a²)| + c ∫dx/√(x² − a²) = ln|x + √(x² − a²)| + c

    a = a positive constant

    Note:Which of a² and x² comes first decides between an inverse sine and a log. Complete the square to convert an arbitrary quadratic under the root into one of these.

  • Integration by parts

    ∫u·v dx = u·∫v dx − ∫[(du/dx)·∫v dx] dx choose u by ILATE: Inverse, Log, Algebraic, Trigonometric, Exponential

    u = the function you DIFFERENTIATE
    v = the function you INTEGRATE

    Note:ILATE picks u, and picking it backwards makes the second integral worse than the first. ∫ln x dx and ∫sin⁻¹x dx are by parts with v = 1 — the hidden second factor students never see.

  • The e^x [f + f'] shortcut

    ∫eˣ[f(x) + f'(x)] dx = eˣ·f(x) + c

    f'(x) = derivative of f(x)

    Note:Recognising this pattern converts a two-round by-parts problem into one line — a real time lever at 1.8 minutes per question. The variant ∫e^(ax)[a·f + f'] dx = e^(ax)·f(x) + c is asked too.

  • Partial fractions

    P(x)/[(x−a)(x−b)] = A/(x−a) + B/(x−b) repeated factor: A/(x−a) + B/(x−a)² irreducible quadratic: (Ax + B)/(x² + px + q)

    P(x) = numerator, degree must be LESS than the denominator
    A, B = constants found by substituting the roots

    Note:PRECONDITION: if the numerator degree is greater than or equal to the denominator degree, do the long division FIRST. Skipping that step is the standard wrong start.

  • Trigonometric substitutions

    √(a² − x²) → x = a sin θ √(a² + x²) → x = a tan θ √(x² − a²) → x = a sec θ rational in sin x and cos x → t = tan(x/2), sin x = 2t/(1+t²), cos x = (1−t²)/(1+t²), dx = 2dt/(1+t²)

    θ, t = the new variable of integration

    Note:The half-angle substitution is the fallback for a/(b + c·sin x) type integrals, and it is the chapter's most expensive subtopic at 74% HARD. For a + b·cos²x forms, divide through by cos²x and try t = tan x instead — far shorter.

Differentiation

Playbook

  • Standard derivatives

    d/dx(xⁿ) = n·xⁿ⁻¹ d/dx(eˣ) = eˣ d/dx(aˣ) = aˣ·ln a d/dx(ln x) = 1/x d/dx(sin x) = cos x d/dx(cos x) = −sin x d/dx(tan x) = sec²x d/dx(cot x) = −cosec²x d/dx(sec x) = sec x·tan x d/dx(cosec x) = −cosec x·cot x

    n = any real constant
    a > 0, a ≠ 1

    Note:Every co-function (cos, cot, cosec) carries a minus sign. That single pattern removes half the sign errors in the chapter.

  • Product, quotient and chain rules

    product: (uv)' = u'v + uv' quotient: (u/v)' = (u'v − uv')/v² chain: dy/dx = (dy/du)·(du/dx)

    u, v = differentiable functions of x
    u = an intermediate function in the chain rule

    Note:The quotient rule's numerator is NOT symmetric — the derivative of the top comes first. The chain rule is also the engine behind related rates in Applications of Derivative and behind implicit and parametric differentiation below.

  • Derivatives of inverse trigonometric functions

    d/dx(sin⁻¹x) = 1/√(1 − x²) d/dx(cos⁻¹x) = −1/√(1 − x²) d/dx(tan⁻¹x) = 1/(1 + x²) d/dx(cot⁻¹x) = −1/(1 + x²) d/dx(sec⁻¹x) = 1/(|x|√(x² − 1)) d/dx(cosec⁻¹x) = −1/(|x|√(x² − 1))

    Domains: |x| < 1 for the first four, |x| > 1 for the last two

    Note:The bank's largest differentiation subtopic (39 q, 49% HARD). Almost every such question is easier after a SIMPLIFYING substitution — x = tan θ for 2x/(1+x²) or (1−x²)/(1+x²) shapes — rather than differentiating the printed expression directly.

  • Implicit differentiation

    Differentiate every term with respect to x, treating y as a function of x, then solve for dy/dx. d/dx(y²) = 2y·(dy/dx) d/dx(xy) = y + x·(dy/dx)

    Used when the relation cannot be solved for y explicitly

    Note:Every y that gets differentiated drags a dy/dx factor with it. The xy term needs the product rule as well — dropping one of its two pieces is the classic slip.

  • Logarithmic differentiation

    y = [f(x)]^g(x) ⟹ ln y = g(x)·ln f(x) ⟹ (1/y)·(dy/dx) = derivative of the right side dy/dx = y · (that derivative)

    f(x) > 0 for the log to exist

    Note:Required when the exponent is itself a function of x — neither the power rule nor the exponential rule applies to xˣ. Also the fastest route for a long product or quotient of many factors. Remember to multiply back by y at the end.

  • Parametric and higher-order derivatives

    dy/dx = (dy/dt)/(dx/dt) d²y/dx² = [d/dt(dy/dx)] / (dx/dt)

    t = the parameter
    dx/dt ≠ 0

    Note:The second derivative is NOT (d²y/dt²)/(d²x/dt²). You must differentiate dy/dx with respect to t and then divide by dx/dt again — the single most-tested trap in this subtopic.

  • Derivative of one function with respect to another

    du/dv = (du/dx) / (dv/dx)

    u, v = both functions of the same variable x

    Note:Only 7 q in the bank but 71% HARD — the chapter's hardest subtopic. It is the parametric formula with x playing the role of the parameter. Simplify both functions with a substitution before differentiating; the ratio usually collapses to a constant.

Probability Distribution

Playbook

  • Classical probability, addition theorem and odds

    P(A) = n(A)/n(S) P(A ∪ B) = P(A) + P(B) − P(A ∩ B) mutually exclusive: P(A ∩ B) = 0 odds in favour a : b ⟹ P = a/(a + b)

    n(S) = total number of equally likely outcomes
    n(A) = outcomes favourable to A

    Note:Odds are a RATIO of favourable to unfavourable, probability is favourable over total — converting one to the other by writing a/b is the standard error. P(A') = 1 − P(A) turns most at-least-one questions into one subtraction.

  • Conditional probability and independence

    P(A | B) = P(A ∩ B)/P(B), P(B) ≠ 0 multiplication: P(A ∩ B) = P(B)·P(A | B) independent ⟺ P(A ∩ B) = P(A)·P(B)

    P(A | B) = probability of A GIVEN that B has occurred

    Note:Independent and mutually exclusive are opposite ideas, not synonyms: two events with non-zero probability cannot be both. Watch the order of the conditioning bar; P(A|B) and P(B|A) are different numbers.

  • Bayes' theorem

    P(Aᵢ | B) = P(Aᵢ)·P(B | Aᵢ) / Σⱼ P(Aⱼ)·P(B | Aⱼ)

    A₁, A₂, … = mutually exclusive and exhaustive causes
    B = the observed event
    The denominator is the total probability of B

    Note:Bayes reverses the conditioning: you are given P(B given cause) and asked for P(cause given B). Identify which one the question hands you before writing anything — reading it the wrong way round produces a plausible wrong option every time.

  • Probability mass function and distribution function

    0 ≤ p(xᵢ) ≤ 1 and Σ p(xᵢ) = 1 F(x) = P(X ≤ x) = Σ over all xᵢ ≤ x of p(xᵢ)

    X = discrete random variable
    p(xᵢ) = probability that X takes the value xᵢ
    F(x) = cumulative distribution function

    Note:Σp = 1 is how you find an unknown k in a printed table — it is the whole of many questions. F is a running total and is non-decreasing, so P(a < X ≤ b) = F(b) − F(a).

  • Expectation, variance and standard deviation

    E(X) = μ = Σ xᵢ·p(xᵢ) Var(X) = σ² = Σ xᵢ²·p(xᵢ) − [E(X)]² = E(X²) − [E(X)]² SD = σ = √Var(X)

    μ = mean of the distribution
    σ² = variance, σ = standard deviation

    Note:Compute E(X²) with the SQUARED x values against the SAME probabilities — squaring the products instead of the values is the usual wreck. Variance can never be negative; a negative result means you subtracted before squaring the mean.

Mathematical Logic

Playbook

  • De Morgan's laws and negation of an implication

    ~(p ∧ q) ≡ ~p ∨ ~q ~(p ∨ q) ≡ ~p ∧ ~q ~(p → q) ≡ p ∧ ~q ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p)

    ~ = negation, ∧ = and, ∨ = or
    → = implication, ↔ = biconditional

    Note:Negating an implication produces an AND, never another implication — this is the single most-tested equivalence in the chapter and the distractors are always the plausible ~p → ~q.

  • Converse, inverse, contrapositive

    statement: p → q converse: q → p inverse: ~p → ~q contrapositive: ~q → ~p p → q ≡ ~p ∨ q ≡ contrapositive

    p = hypothesis (antecedent), q = conclusion (consequent)

    Note:A statement is logically equivalent to its CONTRAPOSITIVE only. The converse and the inverse are equivalent to each other but not to the original — the paper tests exactly this pairing.

  • Tautology, contradiction and quantifier negation

    tautology = last column of the truth table is all T contradiction = last column is all F contingency = a mix ~(∀x, p(x)) ≡ ∃x, ~p(x) ~(∃x, p(x)) ≡ ∀x, ~p(x)

    ∀ = for all, ∃ = there exists
    T, F = truth values

    Note:For two statement letters the table has 4 rows, for three it has 8. Negating a quantifier flips it as well as the predicate — dropping the flip is the standard error.

  • Switching circuits

    switches in SERIES = conjunction (p ∧ q) switches in PARALLEL = disjunction (p ∨ q) S' (complementary switch) = ~p current flows ⟺ the corresponding statement is T

    p, q = the switch states (closed = T, open = F)

    Note:Series is AND because BOTH switches must be closed; parallel is OR because EITHER suffices. Simplify with the distributive and absorption laws before drawing — the questions asking for the simplest equivalent circuit are entirely algebra.

Binomial Distribution

Playbook

  • Binomial probability mass function

    P(X = r) = ⁿCᵣ · pʳ · q^(n−r), r = 0, 1, 2, …, n, q = 1 − p

    n = number of independent trials (fixed in advance)
    p = probability of success on ONE trial (same every trial)
    r = number of successes

    Note:The binomial setting needs all four: fixed n, two outcomes, constant p, independent trials. Drawing without replacement breaks constant p and is not binomial.

  • Mean, variance and standard deviation

    mean = np variance = npq SD = √(npq)

    n = trials, p = success probability, q = 1 − p

    Note:Since q < 1, the variance is ALWAYS less than the mean for a binomial — a printed variance exceeding the mean means the distribution is not binomial, and that is a whole question type. Given mean and variance, divide to get q, then p = 1 − q, then n = mean/p.

  • Successive-term ratio

    P(X = r+1)/P(X = r) = [(n − r)/(r + 1)]·(p/q)

    Same symbols as the pmf

    Note:Turns most-likely-value and parameter-estimation questions into one inequality instead of evaluating every term. Set the ratio greater than 1 to find where the probabilities stop rising.

Straight Line

Playbook

  • Forms of the equation of a line

    slope-intercept: y = mx + c point-slope: y − y₁ = m(x − x₁) two-point: (y − y₁)/(y₂ − y₁) = (x − x₁)/(x₂ − x₁) intercept: x/a + y/b = 1 normal: x·cos α + y·sin α = p

    m = slope = tan θ, θ = angle with the positive x-axis
    a, b = x- and y-intercepts
    p = perpendicular distance from the origin, α = angle that perpendicular makes with the x-axis

    Note:Pick the form that matches what you are GIVEN rather than converting everything to y = mx + c. A vertical line has no slope and no slope-intercept form at all.

  • Distance from a point to a line, and between parallel lines

    d = |ax₁ + by₁ + c| / √(a² + b²) parallel lines ax+by+c₁=0 and ax+by+c₂=0: d = |c₁ − c₂| / √(a² + b²)

    (x₁, y₁) = the point
    ax + by + c = 0 = the line

    Note:For the parallel case the coefficients a and b must be IDENTICAL in both equations — rescale one first. Direct sibling of the point-to-plane distance in Line and Plane: same structure, one dimension up.

  • Angle between two lines, perpendicularity and parallelism

    tan θ = |(m₁ − m₂)/(1 + m₁m₂)| perpendicular ⟺ m₁m₂ = −1 parallel ⟺ m₁ = m₂

    m₁, m₂ = the two slopes
    θ = acute angle between the lines

    Note:Also used for the angle between curves and for pairs of lines. In terms of coefficients: perpendicular ⟺ a₁a₂ + b₁b₂ = 0, which is the dot product of the two normals — the vectors dialect of the same condition.

  • Section formula and concurrency

    internal division m:n: ((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n)) midpoint: ((x₁+x₂)/2, (y₁+y₂)/2) centroid: ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3) three lines concurrent ⟺ the 3x3 determinant of their coefficients = 0

    m:n = dividing ratio
    Rows of the determinant are (a, b, c) for each line

    Note:Concurrency of three lines, collinearity of three points and coplanarity of three vectors are all the SAME vanishing 3x3 determinant. Recognising that saves you learning three tests.

Definite Integration

Playbook

  • Fundamental theorem of calculus

    ∫ from a to b of f(x) dx = F(b) − F(a), where F'(x) = f(x)

    a, b = lower and upper limits
    F = any antiderivative of f

    Note:No constant of integration survives — it cancels in the subtraction. On substitution you must CHANGE THE LIMITS to the new variable, or convert back before substituting; mixing the two is the standard wreck.

  • King's property

    ∫ from a to b of f(x) dx = ∫ from a to b of f(a + b − x) dx

    a, b = the SAME limits on both sides

    Note:PRECONDITION: the limits must be unchanged. Add the two forms to make the integrand collapse — the classic use is f(x)/(f(x) + f(a+b−x)), whose answer is always (b − a)/2. A genuine time lever: it turns an unsolvable integral into arithmetic.

  • Even and odd symmetry

    ∫ from −a to a of f(x) dx = 2·∫ from 0 to a of f(x) dx if f(−x) = f(x) (even) ∫ from −a to a of f(x) dx = 0 if f(−x) = −f(x) (odd) ∫ from 0 to 2a of f(x) dx = 2·∫ from 0 to a of f(x) dx if f(2a − x) = f(x), else 0 if f(2a − x) = −f(x)

    a > 0
    Limits must be symmetric about the origin for the first two

    Note:Check the symmetry BEFORE integrating — recognising an odd integrand over a symmetric interval turns the question into the single character 0. Products: odd x odd = even, odd x even = odd.

  • Additivity and absolute-value integrands

    ∫ from a to b = ∫ from a to c + ∫ from c to b ∫ from a to b of f = −∫ from b to a of f

    c = any point, usually where the integrand changes sign

    Note:This is how you handle |f(x)| and greatest-integer integrands: split at every point where the expression inside changes sign, drop the modulus with the correct sign on each piece, and add. Integrating |x| straight through is always wrong.

Applications of Definite Integral

Playbook

  • Area under a curve

    about the x-axis: A = ∫ from a to b of |y| dx about the y-axis: A = ∫ from c to d of |x| dy

    a, b = x-limits of the region
    c, d = y-limits when integrating with respect to y

    Note:The modulus matters: where the curve dips below the axis the integral is negative but the AREA is not. Split at each x-intercept and add the magnitudes. Sketch first — the limits usually come from the sketch, not the algebra.

  • Area between two curves

    A = ∫ from a to b of (y_upper − y_lower) dx a and b are the x-coordinates of the intersection points

    y_upper = the curve on top over that interval
    y_lower = the curve underneath

    Note:Solve the two equations simultaneously FIRST to get the limits. If the curves cross inside the interval, upper and lower swap and you must split there. Integrating with respect to y is often shorter for a region bounded left-and-right.

Trigonometry - I

Playbook

  • Compound angle formulas

    sin(A ± B) = sin A·cos B ± cos A·sin B cos(A ± B) = cos A·cos B ∓ sin A·sin B tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A·tan B)

    A, B = any two angles

    Note:The cosine and tangent formulas carry the OPPOSITE sign in the second half — cos(A+B) has a minus. That flip is where most sign errors in the chapter start.

  • Multiple and half angle formulas

    sin 2A = 2 sin A·cos A = 2t/(1 + t²) cos 2A = cos²A − sin²A = 1 − 2sin²A = 2cos²A − 1 = (1 − t²)/(1 + t²) tan 2A = 2 tan A/(1 − tan²A) = 2t/(1 − t²) sin 3A = 3 sin A − 4 sin³A cos 3A = 4 cos³A − 3 cos A

    t = tan A (or tan(A/2) for the half-angle reading)

    Note:The three cos 2A forms exist so you can choose the one that cancels what is already in the question. The t-forms are the same substitution used for rational trigonometric integrals in Indefinite Integration.

  • General solutions of trigonometric equations

    sin θ = sin α ⟹ θ = nπ + (−1)ⁿ·α cos θ = cos α ⟹ θ = 2nπ ± α tan θ = tan α ⟹ θ = nπ + α

    n = any integer
    α = the principal solution

    Note:Three different patterns — the sine one alternates sign with n, the cosine one takes plus-or-minus, the tangent one neither. Squaring during the solve introduces extraneous roots; substitute back and discard.

  • Properties of a triangle — area and half-angle

    area Δ = ½·ab·sin C = √(s(s−a)(s−b)(s−c)) = abc/(4R) s = (a + b + c)/2 r = Δ/s R = abc/(4Δ) tan(A/2) = √[(s−b)(s−c) / (s(s−a))]

    a, b, c = side lengths, A, B, C = opposite angles
    s = semi-perimeter, Δ = area
    r = inradius, R = circumradius

    Note:Heron's form needs no angle at all — reach for it when only the three sides are given. Shared with Trigonometry - II, which carries the sine and cosine rules for the same triangle.

Trigonometry - II

Playbook

  • Sine rule

    a/sin A = b/sin B = c/sin C = 2R

    a, b, c = sides opposite angles A, B, C
    R = circumradius of the triangle

    Note:Use it when you have a side and its OPPOSITE angle. The 2R tail is what links a triangle question to its circumcircle in one step.

  • Cosine rule

    a² = b² + c² − 2bc·cos A cos A = (b² + c² − a²)/(2bc)

    Same side and angle labelling as the sine rule

    Note:Use it when you have three sides, or two sides and the INCLUDED angle — the case the sine rule cannot start. A negative cosine means the angle is obtuse, which is often the whole question.

  • Projection rule

    a = b·cos C + c·cos B b = c·cos A + a·cos C c = a·cos B + b·cos A

    Each side is the sum of the projections of the other two

    Note:The fastest route for identities that mix sides and cosines, because it is linear in the sides where the cosine rule is quadratic. This chapter's largest subtopic (52 q) is these three rules together.

  • Inverse trigonometry inside this chapter

    sin⁻¹x + cos⁻¹x = π/2 tan⁻¹x + cot⁻¹x = π/2 sec⁻¹x + cosec⁻¹x = π/2

    Valid on each function's principal domain

    Note:MEASURED TRAP: inverse trigonometry appears BOTH as its own chapter (73 q) and as a 21-question subtopic inside Trigonometry - II. Drill only one and you miss roughly a fifth of the topic. Full identity list is in the Inverse Trigonometric Functions group.

Inverse Trigonometric Functions

Playbook

  • Principal value branches

    sin⁻¹x ∈ [−π/2, π/2] cos⁻¹x ∈ [0, π] tan⁻¹x ∈ (−π/2, π/2) cosec⁻¹x ∈ [−π/2, π/2] minus {0} sec⁻¹x ∈ [0, π] minus {π/2} cot⁻¹x ∈ (0, π)

    Ranges, not domains — this is the OUTPUT interval

    Note:Every inverse-trig answer must land in its principal branch. An answer of 7π/6 for a sin⁻¹ is wrong however correct the algebra was, and that is precisely what the extra options are for.

  • Complementary and negative-argument identities

    sin⁻¹x + cos⁻¹x = π/2 tan⁻¹x + cot⁻¹x = π/2 sec⁻¹x + cosec⁻¹x = π/2 sin⁻¹(−x) = −sin⁻¹x tan⁻¹(−x) = −tan⁻¹x cos⁻¹(−x) = π − cos⁻¹x cot⁻¹(−x) = π − cot⁻¹x

    x in the domain of the relevant function

    Note:The negative-argument rule SPLITS: sin and tan are odd and just flip sign, cos and cot subtract from π. Treating all six the same is the classic error.

  • Sum formulas with their conditions

    tan⁻¹x + tan⁻¹y = tan⁻¹[(x + y)/(1 − xy)] if xy < 1 = π + tan⁻¹[(x + y)/(1 − xy)] if x, y > 0 and xy > 1 = −π + tan⁻¹[(x + y)/(1 − xy)] if x, y < 0 and xy > 1 tan⁻¹x − tan⁻¹y = tan⁻¹[(x − y)/(1 + xy)] if xy > −1

    x, y = real arguments

    Note:PRECONDITION: the xy < 1 test decides whether a π correction is needed, and every wrong option in this question type is the uncorrected value. Check the product before you apply the formula, not after.

  • Double-argument conversions

    2 tan⁻¹x = sin⁻¹[2x/(1 + x²)] = cos⁻¹[(1 − x²)/(1 + x²)] = tan⁻¹[2x/(1 − x²)] 2 sin⁻¹x = sin⁻¹[2x·√(1 − x²)]

    Conditions: |x| ≤ 1 for the sine form, |x| < 1 for the tangent form

    Note:Recognising 2x/(1+x²) or (1−x²)/(1+x²) inside an inverse function and substituting x = tan θ collapses most of these questions to one line. The same substitution is the shortcut in inverse-trig differentiation.

Complex Numbers

Playbook

  • Modulus, conjugate and argument

    z = x + iy |z| = √(x² + y²) z̄ = x − iy z·z̄ = |z|² z⁻¹ = z̄/|z|² arg z = tan⁻¹(y/x), adjusted for the quadrant

    x = real part, y = imaginary part
    z̄ = conjugate of z

    Note:The quadrant adjustment on the argument is the trap — tan⁻¹(y/x) alone cannot distinguish the first quadrant from the third. Always place the point before quoting the argument.

  • Polar and Euler form, De Moivre's theorem

    z = r(cos θ + i sin θ) = r·e^(iθ), r = |z|, θ = arg z (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ |z₁z₂| = |z₁||z₂| arg(z₁z₂) = arg z₁ + arg z₂

    r = modulus
    θ = argument in radians
    n = integer

    Note:Powers and roots are far cheaper in polar form than by binomial expansion — converting first is a time decision, not a stylistic one.

  • Cube roots of unity

    1, ω, ω² with ω = (−1 + i√3)/2 ω³ = 1 1 + ω + ω² = 0 ω̄ = ω²

    ω = a non-real cube root of unity

    Note:Reduce any exponent modulo 3 first (ω¹⁰⁰ = ω¹), then use 1 + ω + ω² = 0 to kill whole blocks. Almost every cube-roots question is one of those two moves.

  • Locus and greatest/least modulus

    |z − z₀| = r is a circle, centre z₀, radius r |z − z₁| = |z − z₂| is the perpendicular bisector of the segment joining z₁ and z₂ for z on |z − z₀| = r: greatest |z| = |z₀| + r, least |z| = |z₀| − r

    z₀ = fixed centre
    r = radius

    Note:The extremum needs NO calculus: the nearest and farthest points lie on the line through the origin and the centre. The identical move solves maximum perpendicular distance from a point on a circle in the Circle chapter. At 1.8 minutes a question, avoiding calculus is a time lever, not just elegance.

Determinants and Matrices

Playbook

  • Adjoint, inverse and the A·adj(A) identity

    A·adj(A) = adj(A)·A = |A|·I A⁻¹ = adj(A)/|A|, valid only when |A| ≠ 0 adj(A) = transpose of the cofactor matrix

    A = square matrix of order n
    I = identity matrix of the same order
    |A| = determinant of A

    Note:The chapter's hardest subtopic (64% HARD) is built almost entirely on this one identity. It is the fastest route to adj(A) when A⁻¹ is already known: adj(A) = |A|·A⁻¹.

  • Determinant and adjoint properties

    |AB| = |A|·|B| |kA| = kⁿ·|A| |Aᵀ| = |A| |A⁻¹| = 1/|A| |adj A| = |A|^(n−1) adj(adj A) = |A|^(n−2)·A (AB)⁻¹ = B⁻¹A⁻¹ (Aᵀ)⁻¹ = (A⁻¹)ᵀ

    n = order of the square matrix
    k = a scalar

    Note:|kA| = kⁿ|A|, NOT k|A| — the scalar multiplies every one of the n rows. The reversed order in (AB)⁻¹ = B⁻¹A⁻¹ is the other standard slip.

  • Area of a triangle and collinearity by determinant

    area = ½·|det[[x₁, y₁, 1], [x₂, y₂, 1], [x₃, y₃, 1]]| three points are collinear ⟺ that determinant = 0

    (x₁,y₁), (x₂,y₂), (x₃,y₃) = the three vertices

    Note:Same vanishing 3x3 determinant as concurrency of three lines (Straight Line) and coplanarity of three vectors (Vectors). Keep the modulus — a determinant can come out negative, an area cannot.

  • System of linear equations

    AX = B, X = A⁻¹B unique solution ⟺ |A| ≠ 0 Cramer: x = D₁/D, y = D₂/D, z = D₃/D with D = |A| ≠ 0

    A = coefficient matrix, X = variable column, B = constants column
    Dᵢ = D with its i-th column replaced by B

    Note:|A| ≠ 0 is the whole condition for a unique solution — the usual question hands you a parameter and asks for the value that makes |A| = 0. When |A| = 0 the system is either inconsistent or has infinitely many solutions, never exactly one.

Circle

Playbook

  • Standard, general and diameter forms

    standard: (x − h)² + (y − k)² = r², centre (h, k), radius r general: x² + y² + 2gx + 2fy + c = 0, centre (−g, −f), radius √(g² + f² − c) diameter: (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0

    (h, k) = centre, r = radius
    (x₁,y₁), (x₂,y₂) = the ends of a diameter

    Note:In the general form the centre is MINUS the half-coefficients — the sign flip is the standard slip. If g² + f² − c is negative the equation represents no real circle, which is itself a question type. Concentric circles differ only in c.

  • Tangent condition and tangent length

    y = mx + c touches x² + y² = a² ⟺ c² = a²(1 + m²) tangent at (x₁, y₁) on x² + y² = a²: x·x₁ + y·y₁ = a² length of tangent from (x₁, y₁) = √(x₁² + y₁² + 2gx₁ + 2fy₁ + c)

    a = radius of the circle centred at the origin
    The tangent-length expression is S₁, the circle equation evaluated at the point

    Note:Every tangency question reduces to one idea: perpendicular distance from the centre to the line equals the radius. Derive rather than memorise if the circle is not centred at the origin. S₁ > 0 means the point is outside, = 0 on, < 0 inside.

  • Relative position of two circles

    touch externally: d = r₁ + r₂ touch internally: d = |r₁ − r₂| intersect at two points: |r₁ − r₂| < d < r₁ + r₂ orthogonal: 2g₁g₂ + 2f₁f₂ = c₁ + c₂

    d = distance between the two centres
    r₁, r₂ = the two radii

    Note:The chapter's hardest subtopic (56% HARD) and it is entirely this comparison — compute d and both radii, then read off the case. Number of common tangents follows: 4 externally separate, 3 touching externally, 2 intersecting, 1 touching internally, 0 nested.

  • Greatest and least distance from a point to a circle

    for a point P outside: maximum distance = d + r, minimum distance = d − r

    d = distance from P to the centre
    r = radius

    Note:No calculus. Identical move to the greatest and least modulus of z on a circle in Complex Numbers — distance to centre plus-or-minus radius. Recognising the pair is worth real time on a 90-minute paper.

Pair of Straight Lines

Playbook

  • Pair of lines through the origin

    ax² + 2hxy + by² = 0 represents two lines through the origin m₁ + m₂ = −2h/b m₁·m₂ = a/b

    m₁, m₂ = slopes of the two lines
    a, h, b = coefficients of the combined equation

    Note:These are just the sum and product of roots of bm² + 2hm + a = 0. Almost every question about the two lines is answerable from the sum and product without ever separating them.

  • Angle between the pair, perpendicularity and coincidence

    tan θ = |2√(h² − ab) / (a + b)| perpendicular ⟺ a + b = 0 coincident ⟺ h² = ab real and distinct ⟺ h² > ab

    θ = acute angle between the two lines

    Note:a + b = 0 is the perpendicularity test in this chapter's dialect — the same idea as m₁m₂ = −1 for lines and dot product = 0 for vectors, which the guide's traps page treats as one condition across 7 chapters. It follows directly: the tan formula has a + b in the denominator, so a + b = 0 sends θ to 90 degrees.

  • General second-degree equation as a pair of lines

    ax² + 2hxy + by² + 2gx + 2fy + c = 0 is a pair of lines ⟺ abc + 2fgh − af² − bg² − ch² = 0 (equivalently: the 3x3 determinant of [[a,h,g],[h,b,f],[g,f,c]] = 0)

    a, b, c, f, g, h = the six coefficients of the general conic

    Note:Another appearance of a vanishing determinant as a degeneracy test — a pair of lines is precisely a DEGENERATE conic. Point of intersection: solve ax + hy + g = 0 and hx + by + f = 0 simultaneously.

  • Distance between the lines of a parallel pair

    d = 2·√[(g² − ac)/(a(a + b))]

    Applies when the pair is parallel, i.e. h² = ab

    Note:Only meaningful for a parallel pair; check h² = ab before applying it. For a pair meeting at a point the distance is zero by definition.

Why plain-text formulas (not LaTeX)

Every formula on this page is plain text plus unicode (l² + m² + n² = 1, tan θ = |2√(h² − ab) / (a + b)|, r = a + λb). Plain text means the page loads instantly, copies cleanly into your own notes, and reads correctly to a screen reader symbol by symbol. Full typesetting is reserved for the worked examples on the playbook detail pages, where you are solving rather than revising.