Class IX: Foundations

Number Sets and Radicals

Everything builds on N ⊂ Z ⊂ Q ⊂ R. A number is rational if it can be written as a fraction of integers; irrational numbers (like √2 or π) cannot. When a radical sits in a denominator, rationalize it by multiplying top and bottom by the conjugate. For a single radical: multiply by √a/√a. For a binomial like 1/(√a + √b), multiply by (√a − √b)/(√a − √b), using (x+y)(x−y) = x² − y² to eliminate the radicals.

Worked example: 1/(√5 + √3) → multiply by (√5 − √3)/(√5 − √3) → (√5 − √3)/(5 − 3) = (√5 − √3)/2.

Absolute Value

|x| = x if x ≥ 0, and |x| = −x if x < 0. Geometrically, |x − a| is the distance between x and a on the number line. To solve |x| = k (k > 0): x = k or x = −k. To solve |x| < k: this means −k < x < k. To solve |x| > k: this means x < −k or x > k. Mixing up the AND/OR logic here is one of the most common early errors.

The Linear Function

f(x) = ax + b is a line with slope a and y-intercept b. If a > 0 the function is increasing; if a < 0 it’s decreasing. The root is x = −b/a. To find the equation of a line through two points (x₁,y₁) and (x₂,y₂): compute slope a = (y₂−y₁)/(x₂−x₁), then substitute into y − y₁ = a(x − x₁).

The Quadratic Function

f(x) = ax² + bx + c, with a ≠ 0. The discriminant Δ = b² − 4ac tells you everything: Δ > 0 gives two distinct real roots, Δ = 0 gives one repeated root, Δ < 0 gives no real roots. Roots (when they exist): x₁,₂ = (−b ± √Δ)/2a. The vertex is at (−b/2a, −Δ/4a) — a minimum if a > 0, a maximum if a < 0.

Viète’s relations: for roots x₁ and x₂, sum S = x₁+x₂ = −b/a and product P = x₁·x₂ = c/a. These let you answer questions about the roots (like x₁²+x₂² = S²−2P) without solving for them individually.

Worked example — find m so both roots are positive, for x² − (m+1)x + m = 0: Need Δ ≥ 0 (always true here since it simplifies to (m−1)² ≥ 0), sum > 0 → m+1 > 0 → m > −1, and product > 0 → m > 0. Combined: m > 0. This is the standard pattern for “both roots satisfy X” problems — turn the condition into a system of inequalities on Δ, S, and P.

Basic Trigonometry

In a right triangle: sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent. Know exact values for 30°, 45°, 60°, 90° by heart. Fundamental identity: sin²x + cos²x = 1. Law of Sines: a/sin(A) = b/sin(B) = c/sin(C). Law of Cosines: c² = a² + b² − 2ab·cos(C).

Class X: Extending the Toolkit

Irrational Equations

Isolate the radical, then square both sides. Squaring can introduce extraneous solutions, so always verify your answer in the original equation. Before squaring, note the domain restriction from the radical (expression under the root ≥ 0) and, if the radical is set equal to an expression, that expression must be ≥ 0 too.

Worked example: √(x+3) = x−1. First, x−1 ≥ 0 is required (a root can’t equal a negative), so x ≥ 1. Squaring: x+3 = x²−2x+1 → x²−3x−2 = 0 → x = (3±√17)/2. Only the root satisfying x ≥ 1 survives — check both against that condition before finalizing.

Exponential Functions and Equations

f(x) = aˣ, a > 0, a ≠ 1. Increasing if a > 1, decreasing if 0 < a < 1. Rules: aᵐ·aⁿ = aᵐ⁺ⁿ, aᵐ/aⁿ = aᵐ⁻ⁿ, (aᵐ)ⁿ = aᵐⁿ, a⁻ⁿ = 1/aⁿ. To solve equations, match bases and equate exponents, or substitute t = aˣ (t > 0) to reduce to a polynomial in t.

Worked example: 4ˣ − 3·2ˣ − 4 = 0. Since 4ˣ = (2ˣ)², let t = 2ˣ: t² − 3t − 4 = 0 → (t−4)(t+1) = 0 → t = 4 or t = −1. Reject t = −1 (must be positive). So 2ˣ = 4 → x = 2.

Logarithms and Logarithmic Equations

logₐ(x) = y means aʸ = x, defined only for x > 0, a > 0, a ≠ 1. Properties: logₐ(mn) = logₐm + logₐn, logₐ(m/n) = logₐm − logₐn, logₐ(mⁿ) = n·logₐm, and change of base logₐb = logc(b)/logc(a). Always establish the domain (every log argument > 0) before solving, and check your solution against it afterward — this is the step most often skipped.

Combinatorics

Permutations: Pₙ = n! — arrangements of all n objects. Arrangements: A(n,k) = n!/(n−k)! — choosing and ordering k of n objects. Combinations: C(n,k) = n!/(k!(n−k)!) — choosing k of n objects, order irrelevant. The deciding question: does order matter?

Binomial theorem: (a+b)ⁿ = Σ C(n,k)·aⁿ⁻ᵏ·bᵏ, summed k=0 to n. To find a specific term (the (k+1)-th), use Tₖ₊₁ = C(n,k)·aⁿ⁻ᵏ·bᵏ; to find a term with a given power, set the exponent equal to the target and solve for k.

Probability

P(event) = favorable outcomes / total outcomes, assuming equally likely outcomes. Complement: P(not A) = 1 − P(A). Union: P(A or B) = P(A) + P(B) − P(A and B). Most exam problems reduce to correctly counting outcomes with combinatorics first — the probability formula itself is the easy part.

Analytic Geometry

Distance: d = √[(x₂−x₁)² + (y₂−y₁)²]. Midpoint: ((x₁+x₂)/2, (y₁+y₂)/2). Line through two points: slope m = (y₂−y₁)/(x₂−x₁), then y − y₁ = m(x − x₁). Parallel lines have equal slopes; perpendicular lines have slopes whose product is −1.

Area of triangle A(x₁,y₁), B(x₂,y₂), C(x₃,y₃): ½·|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|.

Class XI: Algebra Deepens, Analysis Begins

Matrices

Add matrices by adding corresponding entries (same size required). Multiply by a scalar by multiplying every entry. Matrix multiplication A·B is defined only when A’s column count equals B’s row count, computed row-by-column via dot products. Matrix multiplication is generally not commutative: A·B ≠ B·A. The transpose Aᵀ flips rows and columns; the identity matrix I satisfies A·I = I·A = A.

Determinants

For [[a,b],[c,d]]: determinant = ad − bc. For 3×3, use Sarrus’ rule — repeat the first two columns to the right, sum the three down-right diagonal products, subtract the three down-left diagonal products. Key properties: swapping two rows flips the sign; a scalar multiple of a row scales the whole determinant; identical or proportional rows give determinant 0; det(A·B) = det(A)·det(B).

Systems of Linear Equations and Cramer’s Rule

Form the coefficient matrix, compute its determinant Δ. If Δ ≠ 0, the system has a unique solution: each unknown xᵢ = Δᵢ/Δ, where Δᵢ replaces the i-th column with the constants column. If Δ = 0, check whether the Δᵢ are also all zero to distinguish “infinitely many solutions” from “no solution.”

Inverse of a Matrix

A is invertible only if det(A) ≠ 0. A⁻¹ = (1/det(A))·adj(A). For a 2×2 matrix [[a,b],[c,d]]: A⁻¹ = (1/(ad−bc))·[[d,−b],[−c,a]] — swap the diagonal entries, negate the off-diagonal ones, divide by the determinant.

Algebraic Structures

A set G with operation * is a group if the operation is associative, there’s an identity element e with a*e = e*a = a, and every element has an inverse. If also commutative, it’s abelian. A ring has two operations where the set is an abelian group under addition, multiplication is associative, and distributes over addition. A field is a ring where every nonzero element also has a multiplicative inverse. Exam problems typically hand you a specific (set, operation) pair and ask you to verify each axiom in sequence — mechanical once you know the checklist.

Polynomials

Remainder theorem: dividing P(x) by (x−a) gives remainder P(a). Factor theorem: (x−a) divides P(x) exactly when P(a) = 0. Generalized Viète’s relations connect a degree-n polynomial’s roots to its coefficients (for degree 3: x₁+x₂+x₃ = −b/a, x₁x₂+x₁x₃+x₂x₃ = c/a, x₁x₂x₃ = −d/a).

Sequences (Progressions)

Arithmetic: constant difference r, aₙ = a₁+(n−1)r, sum Sₙ = n(a₁+aₙ)/2. Geometric: constant ratio q, aₙ = a₁·qⁿ⁻¹, sum (q≠1) Sₙ = a₁(qⁿ−1)/(q−1). Identify the type first by checking whether consecutive differences (arithmetic) or consecutive ratios (geometric) are constant.

Limits of Sequences

Key known limits: lim(1/n) = 0, lim(qⁿ) = 0 if |q|<1, lim(qⁿ) = ∞ if q>1, lim(1+1/n)ⁿ = e. For rational expressions in n, divide numerator and denominator by the highest power of n present to resolve ∞/∞.

Worked example: lim (3n²+n)/(n²+5) as n→∞. Divide every term by n²: (3+1/n)/(1+5/n²) → 3/1 = 3.

Limits of Functions

The remarkable limit sin(x)/x → 1 as x→0 underlies most trigonometric limit problems. For 0/0 forms with polynomials, factor and cancel the shared root. For ∞−∞ forms with radicals, multiply by the conjugate to reach a simplifiable form.

Continuity

f is continuous at a if f(a) is defined, the limit as x→a exists, and that limit equals f(a). For piecewise functions, compute the left-hand and right-hand limits at the boundary and set both equal to each other and to the function’s value there — exactly how “find m such that f is continuous” problems are solved.

Class XII: Calculus and Synthesis

Derivatives

f'(a) = limit of [f(x)−f(a)]/(x−a) as x→a — the slope of the tangent at that point. Rules: (f+g)’ = f’+g’; (fg)’ = f’g+fg’ (product rule); (f/g)’ = (f’g−fg’)/g² (quotient rule); (f(g(x)))’ = f'(g(x))·g'(x) (chain rule). Know cold: (xⁿ)’ = nxⁿ⁻¹, (eˣ)’ = eˣ, (ln x)’ = 1/x, (sin x)’ = cos x, (cos x)’ = −sin x. Tangent line at (a, f(a)): y = f(a) + f'(a)(x−a).

Monotony and Extrema

Where f'(x) > 0, f is increasing; where f'(x) < 0, decreasing. Critical points (f’=0 or undefined) are candidates for local extrema — sign change positive-to-negative gives a local max, negative-to-positive a local min. The second derivative refines this: f”(a) > 0 at a critical point signals a local minimum, f”(a) < 0 a local maximum. Where f” changes sign, you have an inflection point.

Worked example — study f(x) = x·eˣ on ℝ: f'(x) = eˣ(x+1). Since eˣ > 0 always, sign of f’ matches sign of (x+1): decreasing for x<−1, increasing for x>−1 → local minimum at x=−1, value −1/e. f”(x) = eˣ(x+2), changing sign at x=−2 → inflection point at x=−2, value −2/e². Note the full chain here: compute f’, build the sign table, state the conclusion, then repeat for f” — every step written out, not skipped.

Full Function Study

Always follow this order: (1) domain, (2) limits at boundaries and infinity to find asymptotes, (3) f’ and a sign table for monotony/extrema, (4) f” for concavity/inflection points, (5) the full variation table combining everything, (6) the graph sketch built from that table. This exact sequence is what “studiul funcției” problems are asking for, and skipping a step (even an “obvious” one) is where points get lost.

Antiderivatives (Primitives)

F'(x) = f(x) defines an antiderivative. Core formulas: ∫xⁿdx = xⁿ⁺¹/(n+1)+C (n≠−1), ∫(1/x)dx = ln|x|+C, ∫eˣdx = eˣ+C, ∫sin(x)dx = −cos(x)+C, ∫cos(x)dx = sin(x)+C.

Substitution: when an inner function and its derivative both appear in the integral, substitute u = (inner function) and rewrite entirely in terms of u before integrating.

Integration by parts: ∫u dv = uv − ∫v du. Choose u as the part that simplifies under differentiation (often a polynomial), dv as the part that’s easy to integrate (often exponential or trigonometric).

Definite Integrals

Fundamental theorem of calculus: ∫ₐᵇ f(x)dx = F(b) − F(a). All antiderivative techniques (substitution, by parts) carry over — just remember to either change your bounds when substituting, or substitute back before plugging in the original bounds.

Worked example: ∫₀² (2x+3)dx. F(x) = x²+3x. F(2)−F(0) = (4+6)−0 = 10.

Applications of Integrals

Area between a curve and the x-axis over [a,b]: ∫ₐᵇ |f(x)| dx — take the absolute value seriously, since a portion of the curve below the axis must count as positive area, not be subtracted out. Area between two curves f and g: ∫ₐᵇ |f(x)−g(x)| dx. Volume of a solid formed by rotating f(x) around the x-axis over [a,b]: V = π·∫ₐᵇ [f(x)]² dx.

Threads That Run Through Everything

Mathematical Induction

To prove P(n) for all n ≥ n₀: (1) verify the base case P(n₀) directly, (2) assume P(k) is true (induction hypothesis), (3) prove that assumption forces P(k+1) to be true. Both steps matter — skipping the base case, or not actually using the hypothesis in step 3, are the two most common ways points are lost here.

Domain Discipline

Across irrational, logarithmic, and rational expressions: determine the domain before solving, and check your final answer against it before submitting. An algebraically correct solution that falls outside the valid domain is not a correct solution — this single habit prevents a large share of avoidable point losses across the entire exam.

Showing Justification

Subjects II and III specifically reward visible reasoning — writing out the sign table, stating which theorem justifies a step, showing the substitution explicitly — rather than jumping straight to a final numeric answer. A correct method with a small arithmetic slip typically scores far better than a correct-looking final answer with no visible work, because partial credit is built around seeing the intermediate steps.