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Chapter 2 — Algebra

Class 11 · Mathematics

Overview

This unit on Algebra for Class 11 develops the language and tools used to express and solve mathematical relationships. It begins with algebraic expressions and polynomials, moves through equations and their roots, and covers sequences and series, the binomial theorem, logarithms, inequalities and complex numbers. The unit also introduces matrices and determinants as elementary algebraic structures useful in solving linear systems. Together these topics build skills in manipulation, reasoning, and problem solving that are important for calculus, geometry, probability and many applied fields. Mastery of algebraic techniques helps students simplify expressions, solve equations, model real situations, and understand behavior of functions. The unit emphasizes derivations, worked examples, and problems of increasing difficulty that match ICSE / ISC expectations and prepare students for board examinations and higher studies in mathematics, engineering and sciences.

Learning Objectives

  • Understand and manipulate algebraic expressions and identities with confidence.
  • Apply the Remainder and Factor Theorems to evaluate and factor polynomials.
  • Solve quadratic and higher-degree equations and analyse relationships between roots and coefficients.
  • Recognise and work with arithmetic, geometric and harmonic progressions and compute their sums.
  • Use the binomial theorem for positive integer powers and apply binomial coefficients.
  • Employ logarithms and exponentials to simplify and solve exponential equations.
  • Prove and use basic inequalities such as AM-GM and solve simple inequality problems.
  • Represent and solve linear systems using matrices and determinants.

Topics in this chapter

16 topics · tap a topic title to jump straight to it.

🔣1

Algebraic expressions and operations

Basic ideas. An algebraic expression is a combination of numbers, variables and operations. Variables such as x and y represent numbers that can change. Expressions are built using addition, subtraction, multiplication, division and powers. The goal of algebraic manipulation is to rewrite expressions into standard, simpler or more useful forms so that they can be compared, evaluated or used in equations.

Laws and rules. The fundamental arithmetic laws—commutative, associative and distributive—govern how we combine terms. Commutative law: a + b = b + a and ab = ba. Associative law: (a + b) + c = a + (b + c). Distributive law: a(b + c) = ab + ac. Index rules tell us how to handle powers: x^m·x^n = x^{m+n}, (x^m)^n = x^{mn}, and x^0 = 1 for x ≠ 0. When dividing powers with same base use x^m / x^n = x^{m−n} (provided x ≠ 0).

Like and unlike terms. Terms in an expression are parts separated by + or −. Like terms contain the same variable factors to the same powers and can be combined by adding coefficients. For example 4x^2 and −3x^2 are like terms. Unlike terms, such as 3x^2 and 5x, cannot be directly combined and must be left separate or transformed through factoring.

Factorisation techniques. Factorisation rewrites an expression as a product of simpler expressions. Common techniques include extracting the greatest common factor, factor by grouping, using special identities (difference of squares, perfect square trinomials), and quadratic factoring. Example identities: a^2 − b^2 = (a − b)(a + b), a^2 + 2ab + b^2 = (a + b)^2. For cubic expressions sometimes grouping or standard formulas are used. Recognising patterns speeds up work and reduces algebraic mistakes.

Expansions and simplification. Expanding parentheses requires application of distributive law repeatedly. Careful handling of signs is essential. Collect terms by arranging polynomial terms in descending powers to get standard form. Simplifying rational expressions may require factoring numerator and denominator and canceling common factors, keeping domain restrictions in mind (do not cancel factors that are zero at excluded points).

Algebraic fractions and rational expressions. Work with algebraic fractions as with numerical fractions: find LCM to add/subtract, multiply numerators and denominators, and simplify by canceling common polynomial factors. When solving equations involving rational expressions, multiply both sides by common denominator only after noting values that make denominator zero (these are excluded).

Practical tips. Keep expressions neat, align like terms, check signs after expansion, and verify factorisation by multiplying factors back. Use substitution to test steps for complex manipulations. These foundational skills are used in solving equations, working with functions and later calculus problems.

📌 Examples
  • Simplify 3x2 − 2x + 5x2 + 7x = 8x2 + 5x / 3x2 − 2x + 5x2 + 7x = 8x2 + 5x
  • Factorise x3 − x2 − 6x = x(x2 − x − 6) = x(x−3)(x+2) / Factorise x3 − x2 − 6x = x(x2 − x − 6) = x(x−3)(x+2)
  • Expand (2x + 3)(x − 4) = 2x2 − 5x − 12 / Expand (2x + 3)(x − 4) = 2x2 − 5x − 12
🧮 Formulas
  1. a2 − b2 = (a − b)(a + b)
  2. (a + b)2 = a2 + 2ab + b2
  3. (a − b)2 = a2 − 2ab + b2
📊 Visual ideas
Draw a number line showing positions of 0, 1, −1 and show how adding like terms moves coefficients along the line
Sketch simple bar models to show distributive law: a(b + c) as a×b and a×c grouped
🔢2

Polynomials: definition and degree

What is a polynomial? A polynomial in one variable x is an expression that is a finite sum of terms of the form a_n x^n where n is a whole number (0,1,2,...) and a_n is a constant called the coefficient. Examples include 4x^3 − x + 7, 5 (a constant polynomial), and x^2 + 3x + 2. Polynomials include monomials (single-term), binomials (two-term) and trinomials (three-term).

Degree and leading coefficient. The degree of a polynomial is the highest exponent of x that appears with a non-zero coefficient. For example degree of 7x^4 − 3x^2 + x − 5 is 4. The coefficient of the highest-degree term is called the leading coefficient. Writing polynomial in standard form with descending powers makes degree and leading coefficient obvious.

Behaviour and importance of degree. Degree influences the number of possible roots, the shape of graph and long-term (end) behaviour. A non-zero polynomial of degree n has at most n real roots. The graph of a polynomial of degree n can have at most n−1 turning points. The end behaviour (whether the graph rises or falls at the two ends) depends on the parity of the degree and sign of the leading coefficient: if n is even both ends go same direction; if odd ends go opposite ways.

Operations on polynomials. Addition and subtraction combine like terms; multiplication follows distributive law and increases degree by sum of degrees. Division of polynomials is done by long division or synthetic division when dividing by linear factors. When dividing by (x − a), Remainder Theorem applies. Polynomial factorisation rewrites polynomial as product of lower-degree polynomials whenever possible; irreducible polynomials over reals may be linear or quadratic factors.

Special polynomials and identities. Recognise patterns such as difference of squares a^2 − b^2=(a−b)(a+b), sum and difference of cubes a^3±b^3=(a±b)(a^2∓ab+b^2), and binomial expansions for (x + y)^n which give polynomial coefficients. Factorising using these identities simplifies algebra and solution of equations.

Rational roots and factor testing. Rational Root Theorem gives possible rational roots p/q where p divides constant term and q divides leading coefficient; testing these candidates often finds linear factors and helps factorise polynomials of moderate degree. Synthetic division is an efficient method to divide by (x − c) and obtain quotient and remainder quickly.

Applications and modelling. Polynomials model many real phenomena: quadratics model projectile motion and area problems; cubic and higher-degree polynomials model more complex curves and approximations. Understanding their algebraic properties prepares students to solve equations and to study functions in later chapters.

📌 Examples
  • Identify degree: 7x4 − 3x2 + x − 5 has degree 4 / 7x4 − 3x2 + x − 5 का डिग्री 4 है
  • Multiply: (x + 2)(x2 − x + 3) = x3 + x2 + 5x + 6 / Multiply: (x + 2)(x2 − x + 3) = x3 + x2 + 5x + 6
🧮 Formulas
  1. Degree(f × g) = Degree f + Degree g
  2. If f(x) ≠ 0 has degree n then f(x) has at most n real roots
📊 Visual ideas
Sketch a cubic polynomial with three real roots on x-axis and show end behaviour (one tail up, other down)
Sketch a quadratic opening upwards with vertex labelled and axis of symmetry drawn
🔢3

Remainder and Factor Theorems

Remainder Theorem (statement and idea). The Remainder Theorem states that when a polynomial f(x) is divided by a linear polynomial (x − a), the remainder is simply f(a). This can be seen by writing the polynomial division result f(x) = (x − a)q(x) + r, where r is a constant because the divisor has degree 1. Substituting x = a gives f(a) = r, the remainder. This avoids long division when only the remainder is needed.

Factor Theorem (link to roots). The Factor Theorem follows directly: (x − a) is a factor of f(x) if and only if f(a) = 0. Thus zeros (roots) of the polynomial correspond exactly to linear factors. If f(a) = 0 we can divide f(x) by (x − a) to reduce the polynomial’s degree by one and factor further. Repeated evaluation can reveal multiple linear factors and multiplicities.

Multiplicity of roots. If (x − a)^k divides f(x) but (x − a)^{k+1} does not, then a is a root of multiplicity k. On the graph of the polynomial, odd multiplicity means the graph crosses the x-axis at the root, even multiplicity means it touches and turns around at the root. Knowing multiplicity helps sketching graphs and understanding behaviour near roots.

Rational Root Test and candidate checking. When coefficients are integers, possible rational roots are of the form ±(factors of constant term)/(factors of leading coefficient). This gives a manageable list of candidates to test using the Remainder Theorem instead of arbitrary guesses. Testing each candidate by evaluating f(a) quickly identifies true rational roots.

Techniques for factorisation. Common strategy: list rational root candidates, test them to find a root a with f(a)=0, perform polynomial division (synthetic division is quick for linear divisors) to get a quotient polynomial, and repeat the process on the quotient until polynomial is fully factorised into linear and irreducible quadratic factors over reals. Grouping and known identities also help when rational roots are not small.

Using the theorems in problem solving. Remainder Theorem simplifies evaluating large powers or expressions modulo a linear factor: e.g., remainder of x^{100} divided by x − 1 is 1^{100} = 1. The Factor Theorem converts root-finding problems into factorisation tasks and is used to construct polynomials with given roots. It is also useful to check if a proposed linear factor is correct without full division.

Worked practice tips. Always test small integer candidates first, watch arithmetic errors especially with signs, and when quotient polynomials remain of degree ≥2 try quadratic formula if factorisation is not obvious. For contest-style problems, combine theorems with symmetry or coefficient relations to reach the result with minimal computation.

📌 Examples
  • Find remainder when f(x)=x3−2x2+5x−3 is divided by x−1: f(1)=1−2+5−3=1 => remainder 1 / शेषफल 1 है
  • Show (x−2) is factor of x3−4x2+4x: f(2)=8−16+8=0, so divide to get quotient x2−2x+2 / (x−2) गुणक है
🧮 Formulas
  1. If f(x) divided by (x − a) then remainder = f(a)
  2. (x − a) is factor of f(x) ⇔ f(a) = 0
📊 Visual ideas
Draw polynomial with a simple root at x = a: show x−axis crossing at a
Show multiplicity 2 root: graph touches and turns at the root instead of crossing
🔢4

Relations, Functions and Types

Difference between a relation and a function. A relation between two sets assigns some elements of the first set (domain) to elements of the second set (codomain). A function is a special kind of relation where each input in the domain has exactly one output in the codomain. Not all relations are functions—if one input corresponds to more than one output, it is not a function. Functions are the fundamental idea for describing dependencies between quantities.

Notation and basic terms. A function is written as f: A → B, meaning f maps each element of set A to an element of set B. For an input x in A, the output is f(x). Domain is the set of all allowed inputs; range (image) is the set of outputs actually produced. Codomain is the set that outputs are considered to belong to; range is always a subset of codomain.

Common types of functions. Linear functions have form f(x) = ax + b and give straight-line graphs. Quadratic functions f(x) = ax^2 + bx + c give parabolas. Polynomial functions are built from sums of powers of x. Rational functions are ratios of polynomials and can have vertical and horizontal asymptotes. Exponential functions f(x) = a b^x model rapid growth or decay. Logarithmic functions are inverses of exponentials. Trigonometric functions model periodic behaviour.

Even and odd functions, symmetry. A function is even if f(−x) = f(x); graph is symmetric about the y-axis. It is odd if f(−x) = −f(x); graph is symmetric about origin. Understanding symmetry simplifies graphing and integration tasks later on.

One-to-one, onto and inverse functions. Injective (one-to-one) functions map distinct inputs to distinct outputs—if f(a)=f(b) implies a=b. Surjective (onto) functions have range equal to codomain. Bijective functions are both injective and surjective and therefore have inverses. The inverse function f^{-1} undoes the action of f: f(f^{-1}(y)) = y for y in range. Not all functions have inverses; a necessary condition for an inverse on the whole domain is bijectivity. Restricting domain can make inverse possible.

Composition of functions. Composition combines two functions: (f∘g)(x) = f(g(x)). Composition is not commutative: f∘g usually differs from g∘f. Composition is used to create complicated functions from simpler ones and is helpful in transformations.

Domain and range determination. To find domain, ensure expressions inside roots are non-negative where required and denominators are non-zero. For range, consider solving y = f(x) for x and examine which y values are possible, or use graph behaviour to determine minimum, maximum or asymptotic limits. Domain and range are central to correctly interpreting functions in applied contexts.

Graph sketching and interpretation. Graphs show zeros (where f(x)=0), intercepts, maxima/minima, asymptotes and end behaviour. For rational functions exclude points where denominator zero. For composite models, sketch parts separately and combine. Practice by transforming basic graphs through shifts, stretches and reflections to build complex graphs quickly. Understanding functions prepares students for calculus topics where rates and changes depend on function behaviour.

📌 Examples
  • State domain and range of f(x)=√(x−1): domain [1,∞), range [0,∞) / f(x)=√(x−1) का डोमेन [1,∞), रेंज [0,∞)
  • Is f(x)=x3 one-to-one on R? Yes, because it is strictly increasing / हाँ, यह एक-से-एक है
🧮 Formulas
  1. Composition: (f∘g)(x) = f(g(x))
  2. Inverse: f(f−1(x)) = x for x in domain of f−1
📊 Visual ideas
Sketch y=x (identity), y=x2 (parabola) and show symmetry
Draw y=1/x showing vertical and horizontal asymptotes
🌱5

Quadratic equations and their roots

General form and why quadratics matter. A quadratic equation has form ax^2 + bx + c = 0 where a ≠ 0. Quadratics appear in many physical and geometric situations: motion under constant acceleration, projectile paths, area and optimisation problems. Understanding their algebraic and graphical properties gives tools to analyse maxima/minima and intersections.

Methods to solve. There are three standard methods: factorisation, completing the square, and the quadratic formula. Factorisation is quickest when integer or simple rational roots exist. Completing the square rewrites ax^2 + bx + c into a(x + b/(2a))^2 plus constant; this reveals vertex and is the origin of the quadratic formula. The quadratic formula provides roots directly: x = [−b ± √(b^2 − 4ac)]/(2a). Always check arithmetic under the square root and reduce fractions.

Discriminant and nature of roots. Discriminant D = b^2 − 4ac determines the type of roots: D > 0 means two distinct real roots; D = 0 means a repeated real root (double root); D < 0 means two complex conjugate roots. Graphically, D > 0 corresponds to parabola crossing x-axis twice, D = 0 touches x-axis at vertex, D < 0 has no real intercepts.

Sum and product relations (Vieta). If roots are α and β then α + β = −b/a and αβ = c/a. These relations help find symmetric combinations of roots without finding the roots themselves—for example α^2 + β^2 = (α + β)^2 − 2αβ. They are useful in problems that request sums of powers or to construct polynomials with given root relations.

Vertex form and graph properties. Writing quadratic as y = a(x − h)^2 + k where h = −b/(2a) and k = f(h) gives vertex (h,k) directly. If a > 0 parabola opens upwards and vertex is minimum; if a < 0 opens downwards with vertex maximum. Axis of symmetry is vertical line x = h. Use vertex and intercepts to sketch accurate graphs quickly.

Applications and word problems. Translate real-world conditions to algebraic form: for projectile motion with initial velocity u and acceleration −g, displacement formula gives quadratic in time. Optimization problems like maximizing area or product under sum constraints lead to quadratics; use vertex formula to find maxima/minima. In word problems be careful to include domain restrictions (e.g., time ≥ 0).

Practice and checking. After solving by formula or factorisation, substitute roots back to original equation to verify. For equations derived from modelling check if roots are physically meaningful. When coefficients are large or fractional, clear denominators early and simplify before applying the quadratic formula to reduce arithmetic errors.

📌 Examples
  • Solve x2 − 5x + 6 = 0 => (x − 2)(x − 3) => x = 2, 3 / हल: x = 2, 3
  • Find vertex of y = 2x2 − 8x + 3: x_v = 8/(4)=2, y_v = 2(4) − 16 + 3 = −5 => (2, −5) / वर्टेक्स (2, −5)
🧮 Formulas
  1. x = [−b ± √(b2 − 4ac)]/(2a)
  2. Sum of roots = −b/a ; Product of roots = c/a
📊 Visual ideas
Sketch a parabola opening upwards with vertex and axis x = −b/(2a) marked
Show two cases: D>0 with two x-intercepts and D<0 with no real x-intercepts
🌱6

Theory of equations and symmetric functions of roots

Polynomials and roots—general view. Consider a polynomial of degree n: a_0 x^n + a_1 x^{n−1} + ... + a_{n−1} x + a_n = 0 with a_0 ≠ 0. Let its roots be r_1, r_2, ..., r_n (counted with multiplicity). The coefficients of the polynomial can be written in terms of symmetric sums of its roots. These relationships are powerful because they let us compute many expressions involving roots without solving for each root explicitly.

Vieta’s formula for symmetric sums. Vieta’s relations express elementary symmetric sums: r_1 + r_2 + ... + r_n = −a_1/a_0, sum of products of roots taken two at a time = a_2/a_0 (with alternating signs), and so on, with the sign alternating for successive symmetric sums, up to r_1 r_2 ... r_n = (−1)^n a_n/a_0. For low degrees these are easy to remember: for quadratic ax^2 + bx + c, sum = −b/a and product = c/a; for cubic ax^3 + bx^2 + cx + d, sum = −b/a, sum pairwise = c/a, product = −d/a.

Using symmetric functions to compute powers. Sometimes problems ask for sums like r_1^2 + r_2^2 + ... or higher powers. These can be computed using identities connecting power sums to elementary symmetric sums. For example, for two roots α and β: α^2 + β^2 = (α + β)^2 − 2αβ. For higher degrees Newton’s identities (or Newton–Girard formulas) provide recursive relations between power sums p_k = ∑ r_i^k and coefficients of polynomial. Learning and applying these identities avoids explicit root solving.

Constructing polynomials from roots. If specific numbers are given as roots (including complex ones), construct the minimal polynomial by forming factors (x − r_i) and multiplying. For conjugate pairs (complex) include both factors to keep coefficients real. For example, if α is root then x − α factor appears; if α+β known or product known, one can build quadratic factors directly using symmetric sums.

Transformations and change of variable. To remove certain terms or to simplify, perform change of variable x = y + k. For example, translating the variable can eliminate the x^{n−1} term (depressed polynomial), which simplifies applying some methods for solving or analyzing roots. Relations between roots of transformed polynomials follow from algebraic substitution.

Applications and problem solving tips. Typical ICSE/ISC problems ask for symmetric expressions like sum of reciprocals of roots, sum of squares or cubes of roots, or to form an equation with given root relations. Use Vieta directly: sum of reciprocals ∑ 1/r_i = (sum of all (n−1)-fold products)/product of all roots, i.e., for quadratic 1/α + 1/β = (α + β)/(αβ) = (−b/a)/(c/a) = −b/c. For sums of powers use identities; for higher degree problems reduce to smaller-degree polynomials when possible.

Caution and verification. Keep track of signs from Vieta, especially the alternating signs. When working with multiplicities, include repeated roots appropriately. Check results by constructing a polynomial with the computed symmetric sums and verifying coefficients match given data.

📌 Examples
  • If roots of x2 − 6x + 8 = 0 are α, β, find α2 + β2 = (α + β)2 − 2αβ = 362 − 2·8 = 28 / परिणाम 28
  • Cubic: x3 − 4x2 + x + 6 = 0 has α + β + γ = 4, αβ + βγ + γα = 1, αβγ = −6 / सम्बन्धित मान
🧮 Formulas
  1. For axn + a1xn−1 + ... + an = 0, sum of roots = −a1/a0, sum of pairwise products = a2/a0 with sign alternating
  2. α2 + β2 = (α + β)2 − 2αβ
📊 Visual ideas
Sketch simple cubic and indicate three real roots and their sum approximately on x-axis
Draw diagram to show pairwise products concept using points labelled α, β, γ
🔢7

Sequences: Arithmetic Progression (AP)

Definition and structure. An arithmetic progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a fixed constant d, called the common difference. If the first term is a_1, the sequence is a_1, a_1 + d, a_1 + 2d, ... . The nth term is given by a_n = a_1 + (n − 1)d. The difference d may be positive, negative or zero, producing increasing, decreasing or constant sequences respectively.

Sum of terms—derivation and formula. The sum S_n of the first n terms can be found by pairing terms from ends: S_n = (a_1 + a_n) + (a_2 + a_{n−1}) + ... . Each pair sums to the same value a_1 + a_n, and there are n/2 such pairs when n is even (if n is odd, middle term accounted separately). This yields S_n = (n/2)(a_1 + a_n). Using a_n formula gives alternative form S_n = n/2 [2a_1 + (n − 1)d]. These forms are foundational for many applied problems.

Properties and derived formulas. Useful derived relations include a_n + a_{m} = a_{k} + a_{l} when indices sum appropriately; arithmetic means: the term midway between two terms is their average. If three numbers are in AP, the middle equals average of the other two. For solving problems giving two arbitrary terms, use a_i = a_1 + (i−1)d and a_j = a_1 + (j−1)d to set up linear equations to find a_1 and d.

Applications and modelling. AP models linear growth such as saving a fixed amount each month, equally spaced measurement values, arithmetic sequences in pattern problems, or distributing items equally with a fixed increment. Word problems often give sum or particular term and ask for count or difference—translate conditions into formulas and solve resulting linear or quadratic equations for n, a_1 or d.

Problems with unknown n. When n is unknown and appears in sum formula, the resulting equation may be quadratic in n. For example if S_n is given, substitute S_n = n/2(2a_1 + (n−1)d) and solve for n; choose positive integer solutions since n counts terms.

Arithmetic means and relation to other sequences. The arithmetic mean between two numbers x and y is (x+y)/2 and is the middle term of the AP x, (x+y)/2, y. AP relates to GP and HP in mixed problems and conversions using reciprocals. For three-term sequences explore relationships via substitution and algebra.

Practical tips and pitfalls. Always identify whether sequence begins at n=1 or n=0 and write term formula appropriately. Watch sign of d when computing later terms. When manipulating sums algebraically, keep terms in standard order to avoid sign mistakes. Practise extracting a_1 and d from two given terms to build confidence.

📌 Examples
  • Find 10th term of AP 3, 7, 11,... => a1=3, d=4 => a10=3+(9)4=39 / a10=39
  • Sum of first 20 terms of AP with a1=5, d=3 => S20=20/2[2·5+(19)3]=10[10+57]=670 / S20=670
🧮 Formulas
  1. an = a1 + (n − 1)d
  2. Sn = n/2[2a1 + (n − 1)d] = n/2(an + a1)
📊 Visual ideas
Plot terms of AP on index vs value graph: points lie on a straight line
Illustrate pairing method for Sn using first and last terms added to equal sums
🔢8

Sequences: Geometric Progression (GP)

Definition and basic form. A geometric progression (GP) is a sequence where each term after the first is obtained by multiplying the previous term by a fixed constant r, the common ratio. A GP with first term a_1 has terms a_1, a_1 r, a_1 r^2, ..., and the nth term is a_n = a_1 r^{n−1}. The ratio r may be positive, negative or fractional; if |r|<1 terms approach zero in absolute value as n grows.

Sum of finite GP and derivation. Sum of first n terms S_n = a_1 + a_1 r + a_1 r^2 + ... + a_1 r^{n−1}. Multiply both sides by r and subtract from original sum to obtain S_n(1 − r) = a_1(1 − r^n) when r ≠ 1. Thus S_n = a_1(1 − r^n)/(1 − r). This formula is often used in finance (compound interest), repeated percentage change problems, and geometric models.

Infinite GP and convergence. For |r| < 1 the infinite sum converges to S_∞ = a_1/(1 − r). For |r| ≥ 1 the infinite series diverges. The infinite GP formula is used to evaluate steady-state totals, present value of perpetuities in finance and certain repeating decimal conversions.

Properties and relations. The geometric mean of two positive numbers a and c is √(ac); if three consecutive terms are in GP, the middle term b satisfies b^2 = ac. For negative ratios the terms alternate sign. Ratio tests and comparisons often determine whether large-n behavior is dominated by base or exponent in applied contexts.

Solving problems involving n. If r^n appears, solving for n requires logarithms. For integer n constraints (e.g., number of terms), ensure solutions are natural numbers. For example find smallest n such that partial sum exceeds a threshold: substitute and solve inequality involving r^n.

Applications and conversion to other forms. Use GP to model exponential growth or decay: population growth, radioactive decay, compound interest (A = P(1 + r/n)^{nt}), and scaling problems. Recognise geometric behaviour in word problems by checking multiplicative rather than additive changes between successive terms.

Practical advice. Keep track of whether sequence indexing starts at n=1; check sign of r; for alternating sign sequences use absolute values when needed. When combining GP with AP in problems, separate parts or use generating functions or calculus techniques in advanced cases. Practise using the derivative trick for sums involving nk r^{n} when necessary.

📌 Examples
  • Find 5th term of GP 3, 6, 12,... => a1=3, r=2 => a5=3·2^4=48 / a5=48
  • Sum of first 4 terms: 3+6+12+24 = 45 => S4 = 3(1−2^4)/(1−2)=3(1−16)/(−1)=45 / S4=45
🧮 Formulas
  1. an = a1 r^(n−1)
  2. Sn = a1(1 − r^n)/(1 − r) for r ≠ 1 ; S∞ = a1/(1 − r) for |r| < 1
📊 Visual ideas
Plot a GP with r>1 showing exponential rise; x-axis is n, y-axis an
Plot a decaying GP with 0<r<1 showing terms approaching zero
🔢9

Harmonic Progression and relations between AP, GP, HP

Definition and basic transformation. A sequence is said to be in harmonic progression (HP) if the reciprocals of its terms form an arithmetic progression (AP). If the reciprocals follow a_1', a_1' + d', a_1' + 2d', ... then the HP terms are 1/a_1', 1/(a_1' + d'), 1/(a_1' + 2d'), ... . HP often appears in problems involving rates, speeds and resistances where additive relationships between reciprocals are natural.

Relation to AP and GP. HP is linked to AP by reciprocation. If three numbers a, b, c are in HP then 1/a, 1/b, 1/c are in AP, hence 2/b = 1/a + 1/c. If numbers are in GP their reciprocals are also in GP with reciprocal ratio. Many problems use these connections: convert HP to AP by taking reciprocals and use AP formulas to solve, or compare means—harmonic mean (HM), geometric mean (GM) and arithmetic mean (AM) satisfy HM ≤ GM ≤ AM for positive numbers.

Harmonic mean and averaging. For two positive numbers x and y, harmonic mean is 2/(1/x + 1/y) = 2xy/(x + y). HM is appropriate when averaging rates (e.g., average speed over equal distances). In mixed-problem contexts selecting correct mean is crucial: for equal time intervals use arithmetic mean; for equal distance intervals use harmonic mean in speed problems.

Problems and techniques. Typical HP problems give first and third terms and ask for middle term: if first is a and third is c, then middle b satisfies 1/b = (1/a + 1/c)/2 → b = 2ac/(a + c). For HP of n terms with given reciprocals forming AP use AP formulas on reciprocals then invert terms. When denominators include variables, ensure none are zero.

Applications in rates and resistances. If two taps fill a tank in times t1 and t2 separately, then combined rate is 1/t1 + 1/t2 and time taken t satisfies 1/t = 1/t1 + 1/t2. Here times are in HP when rates are in AP. In electrical circuits, resistances in parallel combine like harmonic means; converting to reciprocals simplifies algebra.

Cautions and algebraic care. When reciprocating expressions watch signs and domain: reciprocals undefined at zero. HP terms may not be in simple closed form if reciprocals form AP with terms that can be zero; avoid dividing by zero. For negative terms harmonic relationships still hold algebraically but mean inequalities require positivity.

Practice tip. Convert HP problems to AP by reciprocals first, solve using AP formulas, then invert back. Use harmonic mean formula for two-term problems and remember relation HM ≤ GM ≤ AM for positive numbers to bound values where needed.

📌 Examples
  • If 1, 1/3, 1/5 form AP then corresponding HP is 1, 3, 5 / AP 1, 1/3, 1/5 से HP 1,3,5 बनेगा
  • If numbers are in HP with first term 1 and common difference in reciprocals 0.5, third term = 1/(1+2·0.5)=1/2 / तीसरा पद 1/2
🧮 Formulas
  1. If reciprocals form AP with a' and d', then nth term of HP = 1/(a' + (n − 1)d')
  2. If a, b, c in HP then 2/b = 1/a + 1/c
📊 Visual ideas
Show three numbers in HP on number line and their reciprocals forming AP
Sketch behaviour of terms of HP with increasing n: reciprocals linear, original values decay
🔢10

Sums and series: arithmetic and geometric combined problems

Mixed series and general approach. Some series combine arithmetic and geometric behaviour or more complex forms like terms involving n·r^{n} or alternating signs. The general strategy is to write the general term explicitly, identify parts that form known sums (AP, GP), and split the series into simpler sums that can be evaluated. For sums involving n multiplied by r^{n} use differentiation of the GP sum with respect to r to derive closed forms.

Differentiation trick for power-weighted GPs. Consider S(r) = ∑_{k=0}^{n−1} r^k = (1 − r^n)/(1 − r). Differentiating with respect to r (treating n fixed) gives ∑_{k=0}^{n−1} k r^{k−1} related formula; multiplying by r yields ∑ k r^{k}. These derived formulas help compute series like ∑ k r^{k−1} or ∑ k r^{k} used in many algebra problems.

Telescoping sums. A telescoping series is one where terms cancel when written in expanded form, leaving only a few terms from the beginning and end. A typical pattern is ∑ (a_k − a_{k+1}) = a_1 − a_{n+1}. Recognising telescoping structure simplifies evaluation drastically. Decompose complicated rational expressions into partial fractions to reveal telescoping behaviour.

Alternating series and convergence. Alternating series have terms with alternating signs, often modelled as GP with negative ratio. For infinite series, tests for convergence (ratio test, root test) are advanced topics; for ICSE/ISC often we rely on |r|<1 for GP convergence. For alternating series with decreasing absolute terms approaching zero, partial sums oscillate and converge in some cases.

Sums involving powers of integers. Some problems require sums like ∑_{k=1}^{n} k, ∑ k^2, ∑ k^3. Standard formulas are: ∑ k = n(n+1)/2, ∑ k^2 = n(n+1)(2n+1)/6, ∑ k^3 = [n(n+1)/2]^2. These are used often in combination with GP-based sums or to simplify polynomial expressions in series.

Combined AP-GP terms. When a term is product of arithmetic and geometric parts such as (a + bk) r^{k}, separate into a r^{k} part and b k r^{k} part and use GP sum and differentiated GP formula respectively. For alternating weighted sums ensure sign pattern is handled by using r negative.

Applications and problem-solving tips. Use these techniques in problems on finance (installments, present value), probability (expected values), and sequences appearing in contest questions. Always check conditions for convergence in infinite series and prefer exact algebraic simplification to decimal approximations in board answers. Break the problem into parts, apply the right formula to each, and combine results.

📌 Examples
  • Evaluate S = ∑_{k=1}^{n} k·r^{k−1} for |r|≠1: S = [1 − (n+1)r^n + nr^{n+1}]/(1 − r)^2 (derivable by differentiating GP sum) / योग सूत्र
  • Telescoping: ∑_{k=1}^{n} [1/k − 1/(k+1)] = 1 − 1/(n+1) / परिणाम
🧮 Formulas
  1. \[∑_{k=1}^{n} k r^{k−1} = [1 − (n+1)r^n + nr^{n+1}]/(1 − r)^2 for r ≠ 1\]
  2. \[Telescoping series reduce by cancellation: ∑(a_k − a_{k+1}) = a_1 − a_{n+1}\]
📊 Visual ideas
Plot partial sums of a convergent GP to show they approach S∞
Diagram showing cancellation in telescoping sum with arrows between terms
🔋11

Binomial theorem for positive integer powers

Statement and intuition. The binomial theorem gives a formula to expand (a + b)^n when n is a non-negative integer. It says (a + b)^n = ∑_{k=0}^{n} C(n,k) a^{n−k} b^k where C(n,k) = n!/(k!(n−k)!) are binomial coefficients. Each term corresponds to choosing k factors of b from n identical places; combinatorics explains the coefficients as number of ways to choose positions.

Pascal’s triangle and properties. Binomial coefficients form Pascal’s triangle where each entry is sum of two above entries. Properties include symmetry C(n,k)=C(n,n−k) and the sum of coefficients ∑_{k=0}^{n} C(n,k) = 2^n (substitute a=b=1). Alternate sum ∑ (−1)^k C(n,k) = 0 for n ≥ 1. These properties help quick computations and proofs of identities.

General term and middle term. The general (k+1)-th term is C(n,k) a^{n−k} b^k. Middle term depends on parity of n: if n is even, there is a unique middle term at k = n/2; if n is odd, there are two middle terms. Coefficient of specific power of variable can be read directly using binomial coefficients, useful in coefficient extraction problems.

Applications and problem solving. Binomial expansion is used to expand expressions, evaluate limits approximately for small b/a, and compute coefficients in products of binomials. It underlies discrete probability distributions (binomial distribution) and is extended to multinomial theorem for more than two terms. Problems often ask for specific coefficients or constant term when variable terms cancel.

Special cases and identities. For small n, memorize expansions: (a + b)^2 = a^2 + 2ab + b^2, (a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3. Use substitution a = 1 or b = 1 to compute sums of coefficients quickly. Use binomial theorem to prove combinatorial identities by equating coefficients after expanding expressions in two ways.

Technique for exams. For coefficient problems, identify which term yields desired power by solving n−k = power of a and k = power of b. If terms involve more factors, consider multinomial or repeated binomial multiplication. When numbers are big, simplify factorial expressions by canceling common terms and prefer combinatorial reasoning when possible.

📌 Examples
  • Expand (x + 2)^4 = x4 + 8x3 + 24x2 + 32x + 16 / विस्तार
  • Coefficient of x2 in (1 + x)^5(1 + 2x)^3 can be found by multiplying expansions and collecting x2 terms / गुणांकों का निर्धारण
🧮 Formulas
  1. \[(a + b)^n = ∑_{k=0}^{n} C(n\]
    \[k) a^{n−k} b^k\]
  2. C(n, k) = n!/(k!(n − k)!) ; C(n, k) = C(n, n − k)
📊 Visual ideas
Draw first six rows of Pascal's triangle and label coefficients for n = 0..5
Sketch binomial coefficient symmetry by mirroring row about centre
🟰12

Logarithms and exponential equations

Definitions and basic understanding. For a positive base a ≠ 1, the logarithm log_a x is defined as the exponent to which a must be raised to obtain x: a^{log_a x} = x. Logarithms turn multiplication into addition and powers into products, which is extremely helpful in solving equations where the unknown appears in exponents. Natural logarithm ln uses base e (e ≈ 2.718...), and common logarithm uses base 10.

Key properties and identities. Fundamental logarithm rules include: log_a(xy) = log_a x + log_a y, log_a(x/y) = log_a x − log_a y, log_a(x^k) = k log_a x. Also log_a 1 = 0 and log_a a = 1. Change of base formula allows conversion between bases: log_b x = log_a x / log_a b, which is useful for computation using natural or common logs on calculators.

Solving exponential equations. When the variable occurs in the exponent, take logarithms to bring it down. For example solve a^{f(x)} = b by f(x) = log_a b. If both sides can be written with same base, equate exponents. When constants are present, take natural logarithm (or log base 10) to solve for x: e^{2x} = 5 ⇒ 2x = ln 5 ⇒ x = (1/2) ln 5. Be mindful of domain: logarithm arguments must be positive, so any solution must satisfy that restriction where applicable.

Graphical view and inverses. Exponential functions y = a^x (a > 1) are increasing and pass through (0,1). Their inverses are logarithmic functions y = log_a x which pass through (1,0) and are defined for x > 0. Logs grow slowly while exponentials grow rapidly; these qualitative features help estimate and check solutions.

Applications. Exponentials model population growth, compound interest, radioactive decay and cooling laws. Logarithms linearise exponential relationships, enabling straight-line fits in data analysis (e.g., taking ln of both sides). In finance, the formula A = P(1 + r/n)^{nt} uses exponentials and solving for t or r requires logs.

Inequalities and monotonicity. When taking logarithms on both sides of an inequality, remember monotonicity: for base a > 1, log_a is increasing, so inequality direction remains same; for 0 < a < 1, log_a is decreasing and inequality reverses direction. This is important in solving exponential inequalities correctly.

Practical problem-solving tips. Check units and domain, convert bases when helpful, avoid taking logs of negative numbers, and when working symbolically use ln for convenience. For exact answers leave results in log form if decimal approximations are not required in board answers. Use change of base to compute values without special base calculators.

📌 Examples
  • Solve 2^{x+1} = 16 => 2^{x+1} = 2^4 => x+1 = 4 => x = 3 / x = 3
  • Solve e^{2x} = 5 => 2x = ln 5 => x = (1/2) ln 5 / x = (1/2) ln 5
🧮 Formulas
  1. log_a(xy)=log_a x + log_a y ; log_a(x/y)=log_a x − log_a y
  2. log_a(x^k)=k log_a x ; log_b x = log_a x / log_a b
📊 Visual ideas
Sketch y=a^x for a>1 (increasing, passes through (0,1)) and its inverse y=log_a x (increasing, passes through (1,0))
Show ln x curve with asymptote at x=0
🔢13

Inequalities: linear and quadratic, AM-GM inequality

Linear inequalities. Linear inequalities in one variable are similar to linear equations but require attention to the direction of inequality when multiplying or dividing by negative numbers. For ax + b < 0 solve by isolating x: ax < −b then x < −b/a if a > 0, or x > −b/a if a < 0. Represent solution sets on a number line and use open or closed dots depending on strictness of inequality.

Quadratic inequalities. To solve inequalities like ax^2 + bx + c > 0, first find roots of the corresponding quadratic equation ax^2 + bx + c = 0. Use factorisation if possible, or quadratic formula to find real roots. Then use sign analysis: because the parabola is continuous and its sign changes only at roots, test intervals determined by roots to decide where expression is positive or negative. Alternatively, examine leading coefficient: if a>0 the parabola opens upward, so expression ≥0 outside roots, ≤0 between roots (depending on strictness).

Sign chart method. Create a sign chart by marking roots on number line and testing a point in each interval. For multiple factors, use multiplicity: if a root has even multiplicity the sign does not change across it; if odd multiplicity sign changes. Write solution in interval notation or inequality form as required.

AM-GM inequality and applications. The arithmetic mean—geometric mean inequality (AM-GM) states that for non-negative numbers x_1, x_2, ..., x_n, (x_1 + x_2 + ... + x_n)/n ≥ (x_1 x_2 ... x_n)^{1/n}, with equality when all x_i are equal. For two positive numbers a and b, (a + b)/2 ≥ √(ab). AM-GM is a very effective tool to find minimum or maximum values of expressions under given constraints. For example to minimize x + k/x for x>0 use AM-GM to show x + k/x ≥ 2√k with equality at x = √k.

Strategy for optimization problems. When optimizing expressions with positive variables, try to rewrite the expression to apply AM-GM, or use calculus for continuous differentiable cases at higher levels. In contest-style algebra, AM-GM often gives elegant bounds and equality conditions lead to specific values achieving optimum.

Other inequality tools and caution. Additional inequalities like Cauchy-Schwarz give stronger bounds in some cases but are beyond basic syllabus; however ICSE problems may occasionally require simple applications. Always ensure domain conditions (non-negativity for AM-GM) are satisfied before applying inequalities. When manipulating inequalities, reverse the inequality sign whenever multiplying or dividing by negative numbers, and be cautious when squaring both sides—ensure correct sign implications.

Graphical interpretation. For quadratic inequalities sketch parabola and mark intervals of interest. For AM-GM based problems, consider symmetry: optimum occurs when variables are equal, giving a visual cue for equality condition. Practice solving a variety of inequalities, represent solutions clearly and check boundary points when inequality is non-strict.

📌 Examples
  • Solve x2 − 5x + 6 ≥ 0. Roots 2 and 3. Parabola opens up => expression ≥0 for x ≤ 2 or x ≥ 3 / समाधान
  • Use AM-GM to find minimum of x + 4/x for x>0: by AM-GM, x + 4/x ≥ 2√(x·4/x)=4 => minimum 4 at x=2 / न्यूनतम 4
🧮 Formulas
  1. AM-GM for two positives: (a + b)/2 ≥ √(ab) ; equality when a = b
  2. For quadratic ax2 + bx + c, sign determined by roots and leading coefficient a
📊 Visual ideas
Draw parabola y = x2 − 5x + 6 and shade regions where y ≥ 0
Sketch y = x + 4/x for x>0 showing minimum point at x=2
🔢14

Complex numbers: algebraic form and operations

Definition and motivation. Complex numbers extend the real numbers to include solutions of equations like x^2 + 1 = 0. A complex number z is written as z = x + i y where x and y are real numbers and i is defined by i^2 = −1. The real part is Re(z) = x and the imaginary part is Im(z) = y. Complex numbers allow us to work algebraically with roots that are not real and are essential in solving polynomials completely.

Algebraic operations. Addition and subtraction are performed component-wise: (a + ib) + (c + id) = (a + c) + i(b + d). Multiplication uses distributive law with i^2 = −1: (a + ib)(c + id) = (ac − bd) + i(ad + bc). Division is done by multiplying numerator and denominator by conjugate of denominator: (a + ib)/(c + id) = [(a + ib)(c − id)]/(c^2 + d^2). Conjugate of z is z̄ = x − i y and satisfies z z̄ = x^2 + y^2 = |z|^2.

Modulus and argument; polar form. Modulus |z| = √(x^2 + y^2) gives the distance from origin in complex plane. Argument arg(z) = θ is the angle that line from origin to point (x,y) makes with positive real axis. In polar form z = r(cos θ + i sin θ) = r cis θ where r = |z|. Euler’s formula e^{iθ} = cos θ + i sin θ connects trigonometry and complex exponentials, and is used to convert between forms.

De Moivre’s theorem and roots. De Moivre’s theorem states that (r cis θ)^n = r^n cis(nθ) for integer n; this is useful for computing powers and for extracting roots. The nth roots of a complex number are found by taking r^{1/n} and dividing the argument by n, giving n distinct roots equally spaced on a circle of radius r^{1/n} in the complex plane. Roots of unity (solutions of z^n = 1) are particularly important and lie on the unit circle at angles 2πk/n.

Complex conjugates and polynomials. If coefficients of a polynomial are real and z is a complex root, then its conjugate z̄ is also a root. This ensures complex roots occur in conjugate pairs and helps factor polynomials into real quadratic factors when possible. Use conjugates to rationalise denominators and simplify expressions.

Geometric interpretation and applications. Complex numbers represent points or vectors in plane and multiplication by complex numbers corresponds to scaling and rotation. This is handy in geometry, signal processing and solving polynomial equations. Complex methods simplify trigonometric sums and are used in advanced topics like Fourier analysis. For ICSE level focus on algebraic operations, modulus/argument, De Moivre and basic root extraction.

📌 Examples
  • Compute (1+2i)(3−i)=(3−i+6i−2i2)=(3+5i+2) =5+5i / परिणाम 5+5i
  • Find modulus of 3 − 4i: |z| = √(9 + 16) = 5 / मान 5
🧮 Formulas
  1. z = x + iy ; z̄ = x − iy ; |z| = √(x2 + y2)
  2. De Moivre: [r cis θ]^n = r^n cis(nθ)
📊 Visual ideas
Plot z = x + iy on complex plane with real axis and imaginary axis; mark modulus and argument
Show nth roots of unity on unit circle equally spaced
⚖️15

Matrices: basic operations

What is a matrix? A matrix is a rectangular array of numbers arranged in rows and columns and is written with brackets. An m×n matrix has m rows and n columns. Matrices represent data compactly and are particularly useful to encode linear systems, transformations and relations between vector quantities. Elements are denoted a_{ij} for entry in i-th row and j-th column.

Basic operations: addition and scalar multiple. Two matrices of the same size can be added or subtracted by adding or subtracting corresponding entries: (A + B)_{ij} = A_{ij} + B_{ij}. Scalar multiplication multiplies every entry by a real number k: (kA)_{ij} = k·A_{ij}. These operations follow vector-space like rules such as distributivity and associativity.

Matrix multiplication. Product AB of an m×n matrix A and an n×p matrix B is defined and gives an m×p matrix whose (i,j) entry is the dot product of i-th row of A with j-th column of B: (AB)_{ij} = ∑_{k=1}^{n} A_{ik} B_{kj}. Matrix multiplication is associative but not commutative in general: AB ≠ BA usually. Multiplying by identity matrices I acts like multiplying by 1: I_n A = A and A I_p = A provided sizes match.

Transpose and special matrices. Transpose A^T flips rows and columns: (A^T)_{ij} = A_{ji}. A square matrix that equals its transpose is symmetric (A^T = A); one that equals negative transpose is skew-symmetric (A^T = −A), which forces diagonal entries to be zero. Identity matrix I_n has ones on diagonal and zeros elsewhere. Zero matrix O has all zeros.

Determinant and inverse ideas (2×2 focus). For 2×2 matrix [[a,b],[c,d]] determinant det = ad − bc. If determinant ≠ 0 the matrix is invertible and inverse is (1/det)[[d, −b], [−c, a]]. For larger matrices inverse exists when determinant non-zero but computing via adjoint or row reduction is used. Inverse satisfies A A^{-1} = I and when invertible, linear systems A x = b have unique solution x = A^{-1} b.

Applications and interpretation. Matrices represent linear transformations: a 2×2 matrix maps unit square to a parallelogram whose area is absolute value of determinant. Matrices encode systems of linear equations and are central to computer graphics, engineering and data science. At this stage focus on performing operations accurately and understanding multiplication dimension constraints.

Computational tips. Check dimensions before multiplying, use row-by-column rule carefully, and simplify calculations by choosing rows or columns with zeros when expanding determinants or using cofactor methods. For exam work show steps and verify by multiplying back when appropriate.

📌 Examples
  • Add A=[[1,2],[3,4]] and B=[[2,0],[1,−1]] => A+B=[[3,2],[4,3]] / जोड़
  • Multiply A=[[1,2],[0,1]] and B=[[3,1],[2,2]] => AB=[[1·3+2·2,1·1+2·2],[0·3+1·2,0·1+1·2]] = [[7,5],[2,2]] / गुणा
🧮 Formulas
  1. det([[a,b],[c,d]]) = ad − bc
  2. \[Inverse for 2×2: A^{-1} = (1/det)[[d,−b],[−c,a]] if det ≠ 0\]
📊 Visual ideas
Represent a 2×2 matrix as linear map sending unit square to a parallelogram in plane
Show effect of multiplying by scalar as stretching or compressing that parallelogram
🔢16

Determinants and Cramer's rule

What is a determinant? The determinant is a scalar value associated with a square matrix that captures key properties of the linear transformation the matrix defines. For a 2×2 matrix [[a,b],[c,d]] the determinant is ad − bc. For 3×3 and higher, determinants are computed by expansion using minors and cofactors or specific rules like Sarrus’ for 3×3. Determinant zero indicates the matrix is singular (non-invertible) and the linear transformation collapses volume to zero.

Properties of determinants. Determinants are multilinear in rows (or columns), change sign when two rows are swapped, and are multiplied by a scalar when a row is multiplied by that scalar. Also det(AB) = det(A) det(B) and det(A^T) = det(A). Row operations have predictable effects: adding a multiple of one row to another leaves determinant unchanged; swapping rows changes sign; multiplying a row by k multiplies determinant by k.

Cramer’s rule for solving linear systems. For a system of n linear equations in n unknowns with coefficient matrix A and constant column vector b, if det(A) ≠ 0 then the unique solution x = (x_1,...,x_n) has coordinates given by x_i = det(A_i)/det(A). Here A_i is obtained from A by replacing its i-th column with b. Cramer’s rule gives explicit formulas and is especially convenient for 2×2 and 3×3 systems in exam problems.

Using determinants to check solvability. Determinant non-zero implies unique solution; determinant zero requires further analysis: either infinitely many solutions or none depending on consistency. Use augmented matrix and row reduction to determine solvability when det(A)=0. Cramer’s rule is not applicable in singular cases.

Computational techniques. For 3×3 determinants use Sarrus’ rule (write first two columns again and sum products of diagonals) or cofactor expansion along a row or column—choose one with zeros to simplify. For larger matrices use row reduction to triangular form and multiply diagonal entries to get determinant, accounting for row operation effects.

Geometric interpretation and applications. Determinant magnitude gives scale factor for area (2×2) or volume (3×3) under linear transformation. Sign indicates orientation change (reflection). Cramer’s rule and determinants are used in physics and engineering for solving small systems symbolically and in theoretical proofs relating linear independence and invertibility.

Practical exam tips. For Cramer’s rule compute determinants carefully, show intermediate steps, and check det(A) ≠ 0 before applying. For 2×2 systems use Cramer quickly; for larger systems if computation looks heavy, prefer elimination unless Cramer is explicitly asked. Verify solutions by substitution to catch arithmetic mistakes.

📌 Examples
  • Solve by Cramer: x + 2y = 5 ; 3x − y = 4. det(A)=1·(−1)−2·3=−1−6=−7; det(Ax)=5·(−1)−2·4=−5−8=−13 => x=det(Ax)/det(A)= (−13)/(−7)=13/7 / हल
  • det([[1,2,3],[0,1,4],[5,6,0]]) compute via expansion gives value −1 / मान
🧮 Formulas
  1. xi = det(Ai)/det(A) provided det(A) ≠ 0
  2. det(AB) = det(A)det(B) ; det(A^T) = det(A)
📊 Visual ideas
Show area scaling by determinant for 2×2 matrix by mapping unit square to parallelogram
Diagram of row operations and their effect on determinant sign or value

Key Concepts

Polynomial
An expression that is a finite sum of terms ax^n with non-negative integer exponents.
Degree
The highest exponent of the variable in a polynomial with non-zero coefficient.
Remainder Theorem
If polynomial f(x) is divided by (x−a) the remainder equals f(a).
Factor Theorem
(x−a) is a factor of f(x) if and only if f(a)=0.
Function
A relation assigning each element of a domain exactly one element of the codomain.
Discriminant
For ax^2+bx+c, discriminant D=b^2−4ac determines nature of roots.
Arithmetic Progression
A sequence where successive terms differ by a constant d.
Geometric Progression
A sequence where successive terms have a constant ratio r.
Binomial Coefficient
C(n,k)=n!/(k!(n−k)!) gives coefficients in expansion of (a+b)^n.
Logarithm
log_a x is the exponent to which base a must be raised to obtain x.
AM-GM inequality
Arithmetic mean of non-negative numbers is at least their geometric mean.
Complex number
A number of the form x + iy where i^2 = −1 and x,y are real.
Matrix
A rectangular array of numbers arranged in rows and columns.
Determinant
A scalar associated with a square matrix that indicates invertibility and scale change.
Cramer's rule
A method to solve linear systems using determinants when coefficient determinant is non-zero.

Practice Questions

  1. Factorise x^3 − 3x^2 − 4x + 12 / x^3 − 3x^2 − 4x + 12 को फैक्टर करें
    Show answer

    First check possible rational roots ±1, ±2, ±3, ±4, ±6, ±12. f(2)=8−12−8+12=0 so (x−2) is factor. Divide polynomial by (x−2) to get x^2−x−6 = (x−3)(x+2). Hence factorisation is (x−2)(x−3)(x+2) / पहले संभावित निराश्रित मूल परखें। f(2)=0 से (x−2) गुणक है। भाग देने पर शेष x^2−x−6 = (x−3)(x+2) मिलता है। अतः (x−2)(x−3)(x+2)।

  2. Find sum of first 50 terms of AP 5, 8, 11,... / AP 5, 8, 11,... के पहले 50 पदों का योग ज्ञात कीजिए
    Show answer

    Here a1=5, d=3, n=50. Last term a50 = 5 + (50−1)·3 = 5 + 147 = 152. Sum S50 = n/2 (a1 + a50) = 50/2 (5 + 152) = 25·157 = 3925 / a1=5, d=3, a50=152। S50=25×157=3925।

  3. If α and β are roots of x^2 − px + q = 0, express α^2 + β^2 in terms of p and q / यदि α और β x^2 − px + q = 0 के मूल हैं, तो α^2 + β^2 को p और q में व्यक्त करें
    Show answer

    We have α + β = p and αβ = q. Then α^2 + β^2 = (α + β)^2 − 2αβ = p^2 − 2q / α + β = p, αβ = q से α^2 + β^2 = p^2 − 2q।

  4. Solve inequality x^2 − 4x + 3 < 0 / असमामिक x^2 − 4x + 3 < 0 को हल करें
    Show answer

    Factorise: x^2 − 4x + 3 = (x − 1)(x − 3). Roots at x=1 and x=3. Parabola opens upwards so expression < 0 between roots. Solution: 1 < x < 3 / कारक रूप (x−1)(x−3)। पराबोला ऊपर खुलती है, अतः 1<x<3।

  5. Expand (1 + x)^5 and find coefficient of x^3 / (1 + x)^5 का विस्तार करें और x^3 का गुणांक बताइए
    Show answer

    Coefficients are C(5,k). Expansion: 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5. Coefficient of x^3 is 10 / C(5,3)=10। अतः विस्तार 1+5x+10x^2+10x^3+5x^4+x^5 और x^3 का गुणांक 10 है।

  6. Solve 2^{x+2} = 8^{x−1} / 2^{x+2} = 8^{x−1} को हल कीजिए
    Show answer

    Write 8 as 2^3: right side = (2^3)^{x−1} = 2^{3(x−1)}. Equate exponents: x + 2 = 3(x − 1) => x + 2 = 3x − 3 => 2 + 3 = 3x − x => 5 = 2x => x = 5/2. / 8=2^3 से 2^{x+2}=2^{3x−3} ⇒ x+2=3x−3 ⇒ 2x=5 ⇒ x=5/2।

  7. Compute determinant of [[2,3],[5,7]] and find inverse if it exists / [[2,3],[5,7]] का डिटर्मिनेंट निकालिए और यदि इनवर्स हो तो बताइए
    Show answer

    det = 2·7 − 3·5 = 14 − 15 = −1 which is non-zero so inverse exists. A^{-1} = (1/det)[[7, −3], [−5, 2]] = −1·[[7, −3], [−5, 2]] = [[−7, 3], [5, −2]] / det=−1 है, अतः इनवर्स मौजूद: [[−7,3],[5,−2]]।

  8. Find complex roots of x^2 + 4 = 0 / x^2 + 4 = 0 के जटिल मूल ज्ञात कीजिए
    Show answer

    x^2 = −4 ⇒ x = ±√(−4) = ±2i. So roots are 2i and −2i / x = ±2i।

  9. Use Remainder Theorem to find remainder when f(x)=x^{100}−1 is divided by x−1 / Remainder Theorem का प्रयोग कर f(x)=x^{100}−1 को x−1 से भाग देने पर शेषफल ज्ञात कीजिए
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    Remainder equals f(1) = 1^{100} − 1 = 0. So remainder is 0, hence x−1 is a factor. / f(1)=0 से शेष 0 है और (x−1) गुणक है।

  10. If three consecutive terms are in GP and the first is 2 and third is 18, find the middle term / यदि तीन लगातार पद GP में हैं और पहला 2 तथा तीसरा 18 है, तो मध्य पद ज्ञात कीजिए
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    Let terms be 2, 2r, 2r^2=18. Then r^2 = 9 ⇒ r = 3 or r = −3. Middle term = 2r = 6 or −6. So possible middle terms are 6 and −6. / r^2=9 से r=±3 और मध्य पद 2r=6 या −6।

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