Mathematics — Complete Formula Sheet

A relation from A to B is any subset of A × B. A relation in A is a subset of A × A.

Empty: no element is related to any element. Universal: every element is related to every element. Both are called trivial relations.

Every element of A must be related to itself. A single missing (a, a) makes R non-reflexive.

Every pair must have its reverse in R. Pairs of the form (a, a) never break symmetry.

Check every chain a → b → c. If no chain exists at all, the relation is transitive by default.

Typical examples: 'is congruent to', 'is parallel to' (with each line parallel to itself), and 'a − b is divisible by n' on Z.

The equivalence classes are pairwise disjoint and their union is the whole set A. Two classes are either equal or have no element in common.

$n$: Fixed positive integer (the modulus) · $[r]$: Class of integers with remainder r on division by n

An equivalence relation on the integers with exactly n classes. [r] is the set of integers that leave remainder r on division by n.

Distinct inputs give distinct outputs. If two different inputs share an image, f is many-one. A function that is strictly increasing or strictly decreasing on its domain is one-one.

Every element of the co-domain must be hit. Whether f is onto depends on the co-domain, not only on the formula.

Prove the two parts separately. Showing only one of them does not establish bijectivity.

Works only for finite sets mapped to themselves. On N, f(x) = 2x is one-one but not onto.

$m$: Number of elements in A · $n$: Number of elements in B

A × B has mn ordered pairs, and each relation is a subset of it. A relation in A alone (m = n) gives 2^(n²).

$m$: Number of elements in the domain A · $n$: Number of elements in the co-domain B

Each of the m elements of A has n choices of image. For one-one maps the choices fall as n, n − 1, ..., so no image repeats. If m > n there is no one-one function.

A bijection of a finite set onto itself is just a permutation of its elements. For A = {1, 2, 3} there are 3! = 6.

Sine is one-one on [−π/2, π/2], and this interval is the range of sin⁻¹. Negative inputs give negative angles.

The range is [0, π], so cos⁻¹ is never negative. Negative inputs give obtuse angles.

Defined for every real number. The end-points ±π/2 are never reached.

Same interval as cos⁻¹, but open at both ends. cot⁻¹(−1) = 3π/4, not −π/4.

Defined only for |x| ≥ 1. π/2 is left out because sec π/2 is undefined.

Defined only for |x| ≥ 1. 0 is left out because cosec 0 is undefined.

To find a principal value, find the angle in the branch whose sine (or cosine, ...) equals x. The graph of y = sin⁻¹x is the mirror image of the restricted sine graph in the line y = x.

The −1 means inverse function, not reciprocal. The same applies to cos⁻¹, tan⁻¹ and the rest.

Valid wherever the inverse is defined. sin(sin⁻¹ 2) has no meaning, because 2 is outside [−1, 1].

Valid ONLY when x is already in the principal branch. This is the step behind every 'simplest form' question.

Valid only on the branch [0, π] (open for cot). Outside it, shift the angle into the branch first.

Uses sin x = sin(π − x), and π − x lies in [−π/2, π/2]. Example: sin⁻¹(sin 4π/5) = π/5.

Uses cos x = cos(2π − x), and 2π − x lies in [0, π]. Example: cos⁻¹(cos 5π/3) = π/3.

tan has period π, so subtract π to bring x into (−π/2, π/2). Example: tan⁻¹(tan 4π/3) = π/3.

Valid for every x in the domain: −1 ≤ x ≤ 1 for sin⁻¹, all real x for tan⁻¹, |x| ≥ 1 for cosec⁻¹. These branches are symmetric about 0, so a negative input gives the negative of the angle. This follows from the branch table. In a board answer, show the step: 'sin y = −1/2 with y in [−π/2, π/2] gives y = −π/6'.

Valid for every x in the domain: −1 ≤ x ≤ 1 for cos⁻¹, all real x for cot⁻¹, |x| ≥ 1 for sec⁻¹. These branches lie in [0, π], so a negative input gives π minus the angle. NCERT uses the same step for cot⁻¹(−1/√3) = π − π/3 = 2π/3.

For sec⁻¹(g(x)) or cosec⁻¹(g(x)) the condition is |g(x)| ≥ 1. tan⁻¹ and cot⁻¹ accept every real value.

After substituting, simplify with a trig identity and use f⁻¹(f(θ)) = θ. Check that θ (or the angle you end with) lies in the principal branch for the given range of x.

Only defined when A and B have the SAME order. Add corresponding entries.

Multiply every entry by the scalar. Does not affect the order of the matrix.

Defined only when (columns of A) = (rows of B). Result has (rows of A) × (columns of B). NOT commutative.

Rows become columns. Note the REVERSAL of order in (AB)^T = B^T A^T.

Product of main diagonal minus product of anti-diagonal.

Expand along ANY row or column — pick the one with most zeros. M_{ij} = minor after deleting row i, column j.

|A| behaves multiplicatively — very useful for showing det ≠ 0 without expanding AB.

Transpose of the cofactor matrix. Row i of C becomes column i of adj(A).

Sanity check on adj(A). If this product is not |A|·I, you made a sign or minor error.

A is invertible ⇔ |A| ≠ 0. If |A| = 0, A is called SINGULAR (no inverse).

Only valid when |A| ≠ 0. If |A| = 0 and (adj A)·B ≠ 0, system is inconsistent.

Three-way test for a square linear system. State the case explicitly in the answer.

Order REVERSES when you take the inverse of a product — same as transpose.

Every square matrix decomposes uniquely as A = ½(A + A^T) + ½(A − A^T) — symmetric plus skew-symmetric.

First-principles definition. Rarely asked directly, but conceptually foundational.

All three must be equal AND finite. If any one differs or is undefined, f is discontinuous at a.

Converse is NOT true. |x| is continuous at 0 but not differentiable there.

Linearity of differentiation. Differentiate term-by-term.

First × derivative of second + second × derivative of first. NOT u'v'.

Low d-high minus high d-low, over low squared. Order matters — sign errors are costly.

Differentiate outer function keeping inner intact, then multiply by derivative of inner. Peel layers one at a time.

Cos gets a MINUS sign. cot, cosec also carry the minus.

cos⁻¹ and cot⁻¹ carry an extra MINUS sign versus their sin⁻¹, tan⁻¹ counterparts.

e^x is its own derivative. General a^x picks up a factor of ln a.

Use when base AND exponent are variable, e.g. x^x, (sin x)^x. Then differentiate implicitly.

When x = f(t), y = g(t), never write dy/dt · dx/dt — it is a QUOTIENT of the two.

Differentiate dy/dx once more with respect to x. Do not divide.

For equations not solved for y (e.g. x² + y² = 25), differentiate both sides, treat y as function of x, then solve for dy/dx.

Evaluate the derivative at the point of tangency. Vertical tangent when dy/dx is infinite.

Normal is perpendicular to tangent, so product of slopes = −1. Horizontal tangent gives vertical normal (undefined slope).

Point-slope form at the point of tangency (x₀, y₀).

Same point, reciprocal-negative slope. If m_t = 0, normal is x = x₀ (vertical).

Related-rates workhorse. Both y and x are functions of time; chain rule links them.

Test on the OPEN interval. Boundary points can be included if f is continuous there.

Candidates for local maxima/minima. Also check ENDPOINTS on a closed interval for absolute extrema.

Sign chart of f' across the critical point. Works even when f'' = 0.

Inconclusive when f''(x_c) = 0 — fall back to first-derivative test.

Compare f at all critical points inside (a, b) PLUS the two endpoints. Same for minimum.

Linear approximation. Used to estimate √25.3 ≈ 5 + (0.3)/(2·5) = 5.03.

Points where f''(x) changes sign are points of inflection.

Raise power by one, divide by new power. Fails at n = −1 (that case gives ln|x|).

The missing case of the power rule. Absolute value is essential — x can be negative.

e^x is invariant. General a^x picks up 1/ln a in the antiderivative.

sin flips sign, cos does not. sec² integrates cleanly to tan.

Recognise the RHS by shape of the integrand. Generalisation: replace 1 with a² and divide by a where appropriate.

Reverses the chain rule. Look for a function whose derivative is also present (up to a constant).

ILATE rule chooses u: Inverse trig, Logarithm, Algebraic, Trig, Exponential (earlier letter → u).

Cover-up method: A = P(a)/(a−b), B = P(b)/(b−a). Works when degree(P) < degree(denominator).

Rule of thumb: sum in denominator ⇒ arctan.

Difference in denominator ⇒ log of ratio.

Square-root difference ⇒ arcsin.

Net signed area under y = f(x) from x = a to x = b.

Substitute t = a − x. Extremely powerful for symmetric integrands like x/(x + (a − x)).

Odd function on symmetric limits gives 0 — no work needed. Check parity first.

Think of the region as many thin strips of height y and width dx. Adding them up (integrating) gives the area.

$A$: Area of the region (square units) · $a,b$: Left and right ordinates (x = a, x = b), with b > a · $f(x)$: Curve y = f(x) bounding the region from above

Region bounded by y = f(x), the x-axis and the ordinates x = a and x = b. Use vertical strips.

$c,d$: Lower and upper lines y = c and y = d, with d > c · $g(y)$: Curve x = g(y) bounding the region on the right

Region bounded by x = g(y), the y-axis and the lines y = c and y = d. Use horizontal strips and write x in terms of y.

Below the axis the definite integral comes out negative. Take its absolute value, because area is always positive.

Split at every point where the curve crosses the x-axis, then add the absolute values. For a line y = mx + k, the crossing is at x = −k/m. Integrating straight from a to b gives the NET value, not the area.

The parts above and below the axis cancel in the integral but not in the area. The area formula assumes f keeps one sign on [0, a]. Example: y = x³ from −1 to 1 has area 1/2, not 0.

A quick check for regions under y = x², y = x³, y = √x and similar curves, starting at x = 0. Here k > 0 and b > 0.

Each arch of y = sin x or y = cos x encloses 2 square units with the x-axis. So the area between y = sin x and the x-axis from 0 to 2π is 4, even though the integral is 0.

Comes from Chapter 7. Every circle and ellipse area depends on it, so learn it exactly, including the a²/2 factor.

At x = a the first term vanishes and sin⁻¹(1) = π/2. At x = 0 both terms are 0. This is the first-quadrant area of x² + y² = a².

$a$: Radius of the circle (units)

The circle is symmetric about both axes. Take y = +√(a² − x²) in the first quadrant and multiply by 4.

$a$: Semi-axis along x (units) · $b$: Semi-axis along y (units)

Solve for y = (b/a)√(a² − x²) in the first quadrant. A circle is the special case a = b.

$a$: Parabola parameter in y² = 4ax (a > 0) · $h$: Ordinate x = h closing the region (h > 0)

Region bounded by y² = 4ax and the ordinate x = h. The parabola is symmetric about the x-axis, so double the upper half y = 2√(ax). For h = a (the latus rectum) the area is 8a²/3.

Look only at which derivatives appear, not at their powers. Order is always a positive integer.

Degree is the highest power of the HIGHEST-ORDER derivative, once the equation is a polynomial in its derivatives. The 4th power of dy/dx does not count here.

A derivative inside sin, cos, e^( ), log and so on means the equation is not a polynomial in its derivatives, so its degree is not defined.

A particular solution has no arbitrary constants. It is obtained by fixing the constants from given conditions.

Use the condition only after the general solution is complete. Substituting early often loses terms.

$g(x)$: Factor depending on x only · $h(y)$: Factor depending on y only · $C$: Arbitrary constant

Get every y-term with dy and every x-term with dx, then integrate both sides. Requires h(y) ≠ 0.

$\lambda$: Any non-zero constant · $n$: Degree of homogeneity

Replace x by λx and y by λy. If λ factors out as λⁿ, F is homogeneous of degree n. Equivalently, F(x, y) = xⁿ g(y/x).

To test, show F(λx, λy) = λ⁰F(x, y) = F(x, y). Write this test in your answer; it is part of a complete solution.

This turns a homogeneous equation into a separable equation in v and x. Do not forget the 'v +' term.

Integrate, then replace v by y/x to get the general solution.

Use this when the equation is more naturally written as dx/dy, for example when terms like e^(x/y) appear.

$P$: Coefficient of y (function of x or constant) · $Q$: Right-hand side (function of x or constant)

y and dy/dx appear only to the first power and are never multiplied together. Bring the equation to exactly this form first.

Leave out the constant when integrating P. Simplify using e^(log f) = f. Here log means the natural logarithm, as in NCERT.

After multiplying by the I.F., the left side is exactly d/dx[y × I.F.]. Integrate the right side (often by parts), then divide by the I.F. if y is needed explicitly.

Use this when the equation is linear in x but not in y, for example y dx − (x + 2y²) dy = 0. Here P₁ and Q₁ are functions of y only.

Also e^(log sec x) = sec x and e^(−log cos x) = sec x for values where these functions are positive.

Any 3-D vector has three components along the coordinate axes.

Pythagoras in 3-D. Magnitude is always non-negative.

Same direction as a, magnitude 1. Undefined for the zero vector.

Cosines of angles the vector makes with the x, y, z axes. Their squares sum to 1 — good sanity check.

Result is a SCALAR. Zero iff a ⊥ b (or one vector is zero).

Rearranged dot-product formula. Always in [0, π].

Scalar projection — length of shadow of a along direction of b. Sign can be negative.

Result is a VECTOR perpendicular to both a and b. Direction from right-hand rule.

Zero iff a and b are PARALLEL (θ = 0 or π) or one is zero.

Cross product magnitude = area of parallelogram spanned by a and b. Triangle is half of it.

Normalise the cross product. Direction by right-hand rule; other direction is −n̂.

Volume of parallelepiped with edges a, b, c. Zero iff the three vectors are COPLANAR.

Point dividing AB internally in ratio m : n. Same form as coordinates section formula.

Swapping order flips the sign. Dot product is commutative; cross product is not.

$\alpha ,\beta ,\gamma$: Angles with the positive x, y and z-axes · $l,m,n$: Direction cosines (pure numbers)

α, β, γ are the angles a directed line makes with the positive x, y and z-axes. Reversing the direction changes the sign of all three.

True for every line. Use it to find a missing direction cosine, or to check your answer.

$a,b,c$: Direction ratios of the line

Direction ratios a, b, c are any numbers proportional to l, m, n. Take the same sign (all + or all −) in all three.

For the line through P(x₁, y₁, z₁) and Q(x₂, y₂, z₂). Subtracting the other way round, x₁ − x₂ and so on, is also valid.

$PQ$: Distance between the two points (units)

Divide each direction ratio by the distance PQ.

Proportional direction ratios make AB parallel to BC, and B is common to both, so the three points lie on one line.

$\vec{a}$: Position vector of a known point on the line · $\vec{b}$: Vector parallel to the line (its components are direction ratios) · $\lambda$: Real parameter

The line through the point with position vector a, parallel to b. Each real value of λ gives one point of the line.

The line through (x₁, y₁, z₁) with direction ratios a, b, c. The coefficient of x, y and z must be +1 in each numerator.

Parametric form. Use it when you need a point on the line that satisfies an extra condition.

This is the point–direction form with the direction taken as the vector from the first point to the second.

For the lines r = a₁ + λb₁ and r = a₂ + μb₂. The modulus gives the acute angle.

The Cartesian form of the same result. Read the direction ratios from the denominators in standard form.

No denominator is needed, because l² + m² + n² = 1 for each line.

Perpendicular: the dot product of the direction vectors is zero. Parallel: the direction ratios are proportional.

${\vec{a}}_1,{\vec{a}}_2$: Position vectors of known points on the two lines · ${\vec{b}}_1,{\vec{b}}_2$: Direction vectors of the two lines

Skew lines are neither parallel nor intersecting. The shortest distance is the projection of a₂ − a₁ on the common perpendicular b₁ × b₂.

Same result as the vector form. The numerator is the scalar triple product and the denominator is |b₁ × b₂|.

For r = a₁ + λb and r = a₂ + μb with the same direction b. If the second direction is a multiple of the first (for example 2b), use b.

$Z$: Objective function (profit, cost, ...) · $x,y$: Decision variables (non-negative) · $a,b$: Constants (for example profit per unit)

The linear function to be maximised or minimised. a and b are constants, and x and y are the decision variables.

Linear inequalities that the variables must satisfy. The conditions x ≥ 0 and y ≥ 0 are the non-negativity constraints.

The common region of all the half-planes. Each point in it, including points on its boundary, is a feasible solution. Points outside it are infeasible.

Join the two intercepts to draw the boundary line, for a, b, c ≠ 0. For a line through the origin, such as x = y, plot one more point.

If the inequality is true at the origin, shade the side containing the origin. If it is false, shade the other side. If the line passes through the origin, test another point, such as (1, 0).

Solution of a₁x + b₁y = c₁ and a₂x + b₂y = c₂. Elimination works just as well. Then check that the point satisfies all the other constraints.

This is why only the corner points need to be tested, not the infinitely many points inside the region.

$M$: Largest value of Z among the corner points · $m$: Smallest value of Z among the corner points

M and m are the largest and smallest values of Z at the corner points. A bounded region can be enclosed in a circle, and both optimal values exist.

Draw the dotted line ax + by = M. If any part of R lies on the side where ax + by > M, then Z has NO maximum.

Draw the dotted line ax + by = m. If any part of R lies on the side where ax + by < m, then Z has NO minimum.

Happens when the objective line is parallel to an edge of the region. Report both corners and the whole segment.

If the shaded half-planes have no common part, the problem has no solution. State this rather than forcing a corner table.

Follow these steps in order and write a one-line conclusion: 'Maximum Z = ... at (x, y)'.

Probability of A given B has happened. Restrict the sample space to B.

Chain of two events. Useful when one probability is easy to compute and the conditional is given.

Equivalent forms: P(A|B) = P(A), P(B|A) = P(B). Very different from mutually exclusive.

For a partition A₁, A₂, ..., Aₙ of the sample space, sum weighted conditional probabilities.

Reverses the direction of conditioning. Numerator: prior × likelihood; denominator: total probability of B.

'At least one' problems ⇒ complement first. Almost always shorter than direct counting.

Weighted average of possible values, weights = probabilities.

Standard shortcut: compute E(X²) and E(X) separately. Always non-negative.

Fixed n trials, independent, each with success probability p. X counts successes.

Intuitive: expected number of successes in n independent trials.

Maximum when p = ½ (highest uncertainty per trial).

A single yes/no experiment. Binomial with n = 1. Basis for the binomial distribution.