Mathematics गणित

Quick summary

Sets and relations are the language of all higher mathematics. This topic revises set operations and then studies relations on a set, their properties (reflexive, symmetric, transitive), equivalence relations, equivalence classes and partitions. It is Unit I of the first algebra paper in most Maharashtra universities.

Key points

• A set is a well-defined collection of objects. A ⊆ B means every element of A is in B; A = B when A ⊆ B and B ⊆ A.
• Operations: union A ∪ B, intersection A ∩ B, difference A − B, complement A′. De Morgan's laws: (A ∪ B)′ = A′ ∩ B′ and (A ∩ B)′ = A′ ∪ B′.
• Power set P(A) is the set of all subsets of A. If A has n elements, P(A) has 2ⁿ elements.
• Cartesian product A × B = {(a, b) : a ∈ A, b ∈ B}. If |A| = m and |B| = n then |A × B| = mn.
• A relation R on A is any subset of A × A. We write a R b when (a, b) ∈ R.
• Reflexive: a R a for every a ∈ A. Symmetric: a R b ⇒ b R a. Transitive: a R b and b R c ⇒ a R c.
• Antisymmetric: a R b and b R a ⇒ a = b. A reflexive, antisymmetric, transitive relation is a partial order (example: ≤ on R, divisibility on N).
• An equivalence relation is reflexive, symmetric and transitive.
• The equivalence class of a is [a] = {x ∈ A : x R a}.
• Any two equivalence classes are either equal or disjoint, and together they cover A. So an equivalence relation partitions A; conversely every partition gives an equivalence relation.
• Congruence mod n (a ≡ b mod n iff n divides a − b) is the most important example; it splits Z into n classes [0], [1], …, [n − 1].

Important terms

• Relation (संबंध) – a subset of A × A.
• Partition – a collection of non-empty, pairwise disjoint subsets whose union is A.
• Quotient set A/R – the set of all equivalence classes.
• Identity relation – {(a, a) : a ∈ A}; it is an equivalence relation.

Formulas

• |A ∪ B| = |A| + |B| − |A ∩ B|
• |P(A)| = 2ⁿ, |A × B| = mn
• Number of relations on a set of n elements = 2^(n²)
• A − (B ∪ C) = (A − B) ∩ (A − C); A − (B ∩ C) = (A − B) ∪ (A − C)

5 important questions

Q1. Define an equivalence relation and give one example.

A relation R on a set A is an equivalence relation if it is reflexive (a R a), symmetric (a R b ⇒ b R a) and transitive (a R b, b R c ⇒ a R c). Example: "is parallel to or equal to" on the set of lines in a plane, or a ≡ b (mod n) on Z.

Q2. Prove that two equivalence classes are either equal or disjoint.

Suppose [a] ∩ [b] is not empty and c lies in both. Then c R a and c R b. By symmetry a R c, and by transitivity a R b. Now if x ∈ [a], then x R a and a R b give x R b, so x ∈ [b]. Hence [a] ⊆ [b]; similarly [b] ⊆ [a]. So [a] = [b].

Q3. On Z define a R b iff 3 divides a − b. Show R is an equivalence relation and find its classes.

1. Reflexive: a − a = 0 = 3 × 0, so a R a.
2. Symmetric: if a − b = 3k, then b − a = 3(−k), so b R a.
3. Transitive: if a − b = 3k and b − c = 3m, then a − c = 3(k + m), so a R c.
4. Classes: [0] = {…, −3, 0, 3, 6, …}, [1] = {…, −2, 1, 4, 7, …}, [2] = {…, −1, 2, 5, 8, …}.
5. These three classes are disjoint and their union is Z, so they partition Z.

Q4. Let A = {1, 2, 3} and R = {(1,1), (2,2), (3,3), (1,2), (2,1)}. Is R an equivalence relation? Find the partition.

1. Reflexive: (1,1), (2,2), (3,3) are all in R ✓.
2. Symmetric: (1,2) ∈ R and (2,1) ∈ R ✓; the other pairs are of the form (a, a).
3. Transitive: (1,2), (2,1) give (1,1) ∈ R ✓; (2,1), (1,2) give (2,2) ∈ R ✓. All other chains involve (a, a) and give back a pair already in R.
4. So R is an equivalence relation. Classes: [1] = [2] = {1, 2} and [3] = {3}.
5. Partition: {{1, 2}, {3}}.

Q5. Prove De Morgan's law (A ∪ B)′ = A′ ∩ B′.

Let x ∈ (A ∪ B)′. Then x ∉ A ∪ B, so x ∉ A and x ∉ B, that is x ∈ A′ and x ∈ B′, so x ∈ A′ ∩ B′. Conversely, if x ∈ A′ ∩ B′ then x ∉ A and x ∉ B, so x ∉ A ∪ B, i.e. x ∈ (A ∪ B)′. Both inclusions hold, so the sets are equal.

Exam tip

In equivalence-relation questions, write the three properties as three separate numbered steps with one line of proof each. Always finish by listing the classes and saying they form a partition.

Quick summary

Elementary number theory studies the integers: divisibility, greatest common divisor, primes and congruences. The division algorithm and Euclid's algorithm give a quick way to find the gcd, and congruences let us do "clock arithmetic" and solve linear congruences. This is part of the first-year algebra paper.

Key points

• a divides b (a | b) if b = ak for some integer k. If a | b and a | c then a | (bx + cy) for all integers x, y.
• Division algorithm: for integers a and b > 0 there are unique integers q, r with a = bq + r and 0 ≤ r < b.
• gcd(a, b) is the largest positive integer dividing both. If gcd(a, b) = 1, a and b are relatively prime (co-prime).
• Euclidean algorithm: gcd(a, b) = gcd(b, r) where r is the remainder of a ÷ b; repeat until the remainder is 0. The last non-zero remainder is the gcd.
• Bezout's identity: gcd(a, b) = ax + by for some integers x, y (found by working the Euclidean steps backwards).
• Fundamental theorem of arithmetic: every integer n > 1 is a product of primes in exactly one way (apart from order).
• Congruence: a ≡ b (mod n) if n | (a − b). Congruences can be added, subtracted and multiplied like equations.
• Linear congruence ax ≡ b (mod n) has a solution iff d = gcd(a, n) divides b; then it has exactly d solutions modulo n.
• Fermat's little theorem: if p is prime and p does not divide a, then a^(p−1) ≡ 1 (mod p).
• Euler's theorem: if gcd(a, n) = 1 then a^φ(n) ≡ 1 (mod n), where φ(n) counts the numbers from 1 to n co-prime to n.

Important terms

• Prime (मूळ संख्या) – an integer p > 1 whose only positive divisors are 1 and p.
• Residue class – all integers leaving the same remainder mod n.
• Inverse mod n – a number a′ with aa′ ≡ 1 (mod n); exists iff gcd(a, n) = 1.
• lcm – least common multiple; gcd(a, b) × lcm(a, b) = ab for positive a, b.

Formulas

• a = bq + r, 0 ≤ r < b
• gcd(a, b) = gcd(b, a − bq)
• φ(p) = p − 1 for prime p; φ(pᵏ) = pᵏ − pᵏ⁻¹
• φ(n) = n(1 − 1/p₁)(1 − 1/p₂)… over the distinct primes dividing n

5 important questions

Q1. State the division algorithm.

For any integers a and b with b > 0, there exist unique integers q (quotient) and r (remainder) such that a = bq + r and 0 ≤ r < b.

Q2. State Fermat's little theorem and find the remainder when 2¹⁰⁰ is divided by 7.

If p is prime and p ∤ a, then a^(p−1) ≡ 1 (mod p). Here an even smaller power helps: 2³ = 8 ≡ 1 (mod 7). So 2⁹⁹ = (2³)³³ ≡ 1, and 2¹⁰⁰ = 2⁹⁹ × 2 ≡ 2 (mod 7). Remainder = 2.

Q3. Find gcd(595, 252) and express it as 595x + 252y.

1. 595 = 2 × 252 + 91
2. 252 = 2 × 91 + 70
3. 91 = 1 × 70 + 21
4. 70 = 3 × 21 + 7
5. 21 = 3 × 7 + 0, so gcd = 7.
6. Back-substitute: 7 = 70 − 3 × 21 = 70 − 3(91 − 70) = 4 × 70 − 3 × 91 = 4(252 − 2 × 91) − 3 × 91 = 4 × 252 − 11 × 91 = 4 × 252 − 11(595 − 2 × 252) = 26 × 252 − 11 × 595.
7. So x = −11, y = 26. Check: 26 × 252 = 6552, 11 × 595 = 6545, difference 7 ✓.

Q4. Solve 7x ≡ 3 (mod 15).

1. gcd(7, 15) = 1 and 1 | 3, so there is exactly one solution mod 15.
2. Inverse of 7 mod 15: 7 × 13 = 91 = 6 × 15 + 1, so 7⁻¹ ≡ 13.
3. x ≡ 13 × 3 = 39 ≡ 9 (mod 15).
4. Check: 7 × 9 = 63 = 4 × 15 + 3 ✓.

Q5. Prove that there are infinitely many primes.

Suppose there are only finitely many primes p₁, p₂, …, pₖ. Let N = p₁p₂…pₖ + 1. N > 1, so it has a prime factor p, which must be one of the pᵢ. Then p divides p₁p₂…pₖ and also N, so p divides N − p₁…pₖ = 1, which is impossible. So there are infinitely many primes.

Exam tip

Show every line of the Euclidean algorithm and check the final Bezout combination by multiplying out. In congruence problems, verify the answer by substitution.

Quick summary

A matrix is a rectangular arrangement of numbers in rows and columns. At BSc level we study the rank of a matrix, elementary transformations, solving systems of linear equations, eigenvalues and eigenvectors, and the Cayley–Hamilton theorem.

Key points

• A matrix of order m × n has m rows and n columns. Square, diagonal, scalar, identity, symmetric (Aᵀ = A), skew-symmetric (Aᵀ = −A) are special types.
• Hermitian: A^θ = A (A^θ is the conjugate transpose); skew-Hermitian: A^θ = −A. Orthogonal: AAᵀ = I. Unitary: AA^θ = I.
• Elementary transformations: interchange two rows, multiply a row by a non-zero number, add a multiple of one row to another (same for columns). They do not change the rank.
• Rank (कोटी) r(A): the order of the largest non-zero minor; equals the number of non-zero rows in the echelon form.
• Inverse exists only if |A| ≠ 0: A⁻¹ = adj A/|A|. It can also be found by row operations [A | I] → [I | A⁻¹].
• Consistency of AX = B (n unknowns): if r(A) ≠ r(A|B) → no solution; if r(A) = r(A|B) = n → unique solution; if r(A) = r(A|B) < n → infinitely many solutions.
• Homogeneous system AX = 0 always has the zero solution; it has non-zero solutions if r(A) < n.
• Characteristic equation: |A − λI| = 0. Its roots λ are the eigenvalues; a non-zero X with AX = λX is an eigenvector.
• Sum of eigenvalues = trace of A; product of eigenvalues = |A|.
• Eigenvalues of a real symmetric (or Hermitian) matrix are real; of a triangular matrix are its diagonal entries.
• Cayley–Hamilton theorem: every square matrix satisfies its own characteristic equation. Used to find A⁻¹ and powers of A.

Important terms

• Minor – determinant of a square sub-matrix.
• Echelon form – zero rows at the bottom; each leading non-zero entry lies to the right of the one above.
• Trace – sum of the diagonal elements.
• Singular matrix – |A| = 0.

Formulas

• A⁻¹ = adj A/|A|; (AB)⁻¹ = B⁻¹A⁻¹; (AB)ᵀ = BᵀAᵀ
• For 2 × 2: characteristic equation λ² − (trace)λ + |A| = 0
• If λ is an eigenvalue of A: kλ of kA, λⁿ of Aⁿ, 1/λ of A⁻¹
• Consistency: compare r(A), r(A|B) and n

5 important questions

Q1. Define the rank of a matrix.

The rank of a matrix is r if it has at least one non-zero minor of order r and every minor of order (r + 1) is zero. Equivalently, it is the number of non-zero rows in its echelon form.

Q2. State the Cayley–Hamilton theorem.

Every square matrix satisfies its own characteristic equation. If |A − λI| = λⁿ + a₁λⁿ⁻¹ + … + aₙ = 0, then Aⁿ + a₁Aⁿ⁻¹ + … + aₙI = 0.

Q3. Find the rank of A = [1 2 3; 2 4 6; 1 1 1] (rows separated by semicolons).

1. R₂ → R₂ − 2R₁ gives [0 0 0].
2. R₃ → R₃ − R₁ gives [0 −1 −2].
3. Swap R₂ and R₃: [1 2 3; 0 −1 −2; 0 0 0].
4. This is echelon form with 2 non-zero rows.
5. So r(A) = 2. (Check: |A| = 0, so rank is less than 3.)

Q4. Verify the Cayley–Hamilton theorem for A = [1 2; 3 4] and hence find A⁻¹.

1. Trace = 5, |A| = 4 − 6 = −2. Characteristic equation: λ² − 5λ − 2 = 0.
2. A² = [1·1+2·3, 1·2+2·4; 3·1+4·3, 3·2+4·4] = [7 10; 15 22].
3. A² − 5A − 2I = [7−5−2, 10−10; 15−15, 22−20−2] = [0 0; 0 0] ✓.
4. Multiply A² − 5A − 2I = 0 by A⁻¹: A − 5I − 2A⁻¹ = 0.
5. A⁻¹ = (A − 5I)/2 = ½[−4 2; 3 −1] = [−2 1; 1.5 −0.5].
6. Check: A·A⁻¹ = [1(−2)+2(1.5), 1(1)+2(−0.5); 3(−2)+4(1.5), 3(1)+4(−0.5)] = [1 0; 0 1] ✓.

Q5. Find the eigenvalues and eigenvectors of A = [2 1; 1 2].

1. |A − λI| = (2 − λ)² − 1 = 0 → λ² − 4λ + 3 = 0 → λ = 1, 3.
2. For λ = 3: (A − 3I)X = 0 gives −x + y = 0, so X = k(1, 1).
3. For λ = 1: (A − I)X = 0 gives x + y = 0, so X = k(1, −1).
4. Check: sum 1 + 3 = 4 = trace; product 3 = |A| ✓.

Exam tip

Always check eigenvalues with sum = trace and product = determinant. In consistency problems, write r(A), r(A|B) and n clearly before concluding.

Quick summary

Theory of equations studies polynomial equations f(x) = 0 without always solving them fully. We use relations between roots and coefficients, Descartes' rule of signs, transformation of equations, reciprocal equations and Cardan's method for cubics. Many exam problems give extra information about the roots (A.P., G.P., one root known) and ask us to solve.

Key points

• A polynomial equation of degree n has exactly n roots (real or complex), counted with multiplicity.
• With real coefficients, complex roots occur in conjugate pairs; with rational coefficients, roots a + √b occur in pairs a ± √b.
• Relations for a₀xⁿ + a₁xⁿ⁻¹ + … + aₙ = 0: sum of roots = −a₁/a₀, sum of products two at a time = a₂/a₀, …, product of roots = (−1)ⁿ aₙ/a₀.
• Descartes' rule of signs: the number of positive roots is at most the number of sign changes in f(x), and differs from it by an even number. For negative roots, count sign changes in f(−x).
• Transformations: roots multiplied by k → replace x by x/k; roots increased by h → replace x by x − h; reciprocal roots → replace x by 1/x.
• Removing the second term: put x = y − a₁/(n a₀).
• Horner's process is a synthetic-division method to find f(x + h) and to approximate roots.
• Reciprocal equation: unchanged when x is replaced by 1/x; if α is a root, so is 1/α. Even-degree ones are solved by putting y = x + 1/x.
• Cardan's method solves the reduced cubic y³ + py + q = 0 by putting y = u + v with 3uv = −p.

Important terms

• Multiple root – a repeated root; it is a root of both f(x) = 0 and f′(x) = 0.
• Symmetric function of roots – an expression unchanged when roots are interchanged, e.g. α² + β² + γ².
• Synthetic division – quick division of a polynomial by x − h.

Formulas

• Cubic x³ + px² + qx + r = 0 with roots α, β, γ: α + β + γ = −p, αβ + βγ + γα = q, αβγ = −r
• α² + β² + γ² = (α + β + γ)² − 2(αβ + βγ + γα)
• Roots in A.P.: take a − d, a, a + d. Roots in G.P.: take a/r, a, ar.

5 important questions

Q1. State Descartes' rule of signs.

The number of positive real roots of f(x) = 0 cannot exceed the number of sign changes in the coefficients of f(x), and if less, it is less by an even number. The number of negative roots is found the same way from f(−x).

Q2. Find α² + β² + γ² if α, β, γ are the roots of x³ − 6x² + 11x − 6 = 0.

Σα = 6 and Σαβ = 11. So α² + β² + γ² = 6² − 2 × 11 = 36 − 22 = 14. (Check: the roots are 1, 2, 3 and 1 + 4 + 9 = 14 ✓.)

Q3. Solve x³ − 12x² + 39x − 28 = 0, given that its roots are in A.P.

1. Let the roots be a − d, a, a + d.
2. Sum: 3a = 12, so a = 4.
3. Product: (a − d)a(a + d) = 28, so 4(16 − d²) = 28, 16 − d² = 7, d = ±3.
4. Roots: 1, 4, 7.
5. Check Σαβ: 1×4 + 4×7 + 7×1 = 4 + 28 + 7 = 39 ✓.

Q4. Use Descartes' rule for f(x) = x³ + 2x² − 5x − 6 and verify by solving.

1. Signs of f(x): +, +, −, −. One change, so exactly 1 positive root.
2. f(−x) = −x³ + 2x² + 5x − 6. Signs: −, +, +, −. Two changes, so 2 or 0 negative roots.
3. Try x = 2: 8 + 8 − 10 − 6 = 0, so (x − 2) is a factor.
4. Dividing: f(x) = (x − 2)(x² + 4x + 3) = (x − 2)(x + 1)(x + 3).
5. Roots 2, −1, −3: one positive, two negative ✓.

Q5. Remove the second term of x³ − 6x² + 5x + 8 = 0.

1. Here n = 3 and a₁/a₀ = −6, so put x = y + 6/3 = y + 2.
2. (y + 2)³ − 6(y + 2)² + 5(y + 2) + 8 = y³ + 6y² + 12y + 8 − 6y² − 24y − 24 + 5y + 10 + 8.
3. Collect terms: y³ + (12 − 24 + 5)y + (8 − 24 + 10 + 8) = y³ − 7y + 2 = 0.
4. The y² term is gone; this reduced cubic is now ready for Cardan's method.

Exam tip

When extra information is given (A.P., G.P., sum of two roots zero), choose the roots in that form first. Verify the final roots using all three root–coefficient relations.

Quick summary

Complex numbers z = x + iy can be written in polar form r(cos θ + i sin θ). De Moivre's theorem gives a simple way to find powers and roots of complex numbers. It also helps expand cos nθ and sin nθ, and leads to hyperbolic functions and logarithms of complex numbers.

Key points

• i² = −1. z = x + iy; real part x, imaginary part y; conjugate z̄ = x − iy.
• Modulus |z| = r = √(x² + y²); argument θ = tan⁻¹(y/x) (choose the correct quadrant).
• Polar form z = r(cos θ + i sin θ) = re^(iθ) (Euler's formula: e^(iθ) = cos θ + i sin θ).
• De Moivre's theorem: (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ for every integer n (and gives one value for fractional n).
• Product: moduli multiply, arguments add. Quotient: moduli divide, arguments subtract.
• n-th roots: z^(1/n) has exactly n distinct values, equally spaced on a circle of radius r^(1/n).
• Cube roots of unity: 1, ω, ω², where ω = (−1 + i√3)/2; 1 + ω + ω² = 0 and ω³ = 1.
• Expansions: cos 3θ = 4cos³θ − 3cos θ; sin 3θ = 3sin θ − 4sin³θ.
• Hyperbolic functions: cosh x = (eˣ + e⁻ˣ)/2, sinh x = (eˣ − e⁻ˣ)/2; cosh²x − sinh²x = 1.
• Relation: cos(ix) = cosh x, sin(ix) = i sinh x.
• Logarithm: log z = log r + i(θ + 2nπ); the principal value takes n = 0.

Important terms

• Argand plane – plane where x-axis is real and y-axis is imaginary (संमिश्र प्रतल).
• Principal argument – the value of θ in −π < θ ≤ π.
• Roots of unity – solutions of zⁿ = 1.

Formulas

• zⁿ = rⁿ(cos nθ + i sin nθ)
• n-th roots: z^(1/n) = r^(1/n)[cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)], k = 0, 1, …, n−1
• cos θ = (e^(iθ) + e^(−iθ))/2; sin θ = (e^(iθ) − e^(−iθ))/(2i)
• tanh x = sinh x/cosh x
• Log z = log r + iθ (principal value)

5 important questions

Q1. State De Moivre's theorem.

For any integer n, (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ. For a rational n = p/q, cos(pθ/q) + i sin(pθ/q) is one of the q values of (cos θ + i sin θ)^(p/q).

Q2. Prove that 1 + ω + ω² = 0.

ω is a root of z³ − 1 = 0, i.e. (z − 1)(z² + z + 1) = 0. Since ω ≠ 1, ω² + ω + 1 = 0.

Q3. Evaluate (√3 + i)⁶.

1. r = √(3 + 1) = 2; θ = tan⁻¹(1/√3) = π/6.
2. (√3 + i)⁶ = 2⁶ (cos 6·π/6 + i sin 6·π/6).
3. = 64 (cos π + i sin π) = 64(−1 + 0) = −64.

Q4. Find all the cube roots of 8i.

1. 8i = 8(cos π/2 + i sin π/2); r = 8, θ = π/2.
2. Roots: 2[cos((π/2 + 2kπ)/3) + i sin((π/2 + 2kπ)/3)], k = 0, 1, 2.
3. k = 0: angle π/6 → 2(√3/2 + i/2) = √3 + i.
4. k = 1: angle 5π/6 → 2(−√3/2 + i/2) = −√3 + i.
5. k = 2: angle 3π/2 → 2(0 − i) = −2i.
6. Check: (−2i)³ = −8i³ = −8(−i) = 8i ✓.

Q5. Using De Moivre's theorem, express cos 3θ and sin 3θ in powers of cos θ and sin θ.

1. cos 3θ + i sin 3θ = (cos θ + i sin θ)³. Let c = cos θ, s = sin θ.
2. Expand: c³ + 3c²(is) + 3c(is)² + (is)³ = c³ + 3ic²s − 3cs² − is³.
3. Real parts: cos 3θ = c³ − 3cs² = c³ − 3c(1 − c²) = 4c³ − 3c.
4. Imaginary parts: sin 3θ = 3c²s − s³ = 3(1 − s²)s − s³ = 3s − 4s³.

Exam tip

Always find the argument in the correct quadrant before applying De Moivre. For n-th roots, list every k from 0 to n−1 — missing a root loses marks.

Quick summary

Using Euler's formula e^(iθ) = cos θ + i sin θ, we extend trigonometric, hyperbolic and logarithmic functions to complex numbers. This topic covers hyperbolic functions and their identities, the link between circular and hyperbolic functions, separating a complex function into real and imaginary parts, the logarithm of a complex number and inverse hyperbolic functions.

Key points

• Hyperbolic functions: sinh x = (eˣ − e⁻ˣ)/2, cosh x = (eˣ + e⁻ˣ)/2, tanh x = sinh x/cosh x. Also sech, cosech, coth as reciprocals.
• sinh is odd, cosh is even; sinh 0 = 0, cosh 0 = 1.
• Basic identity: cosh²x − sinh²x = 1 (compare cos²x + sin²x = 1). Also sech²x + tanh²x = 1.
• Link with circular functions: sin(ix) = i sinh x, cos(ix) = cosh x, sinh(ix) = i sin x, cosh(ix) = cos x.
• Derivatives: d/dx (sinh x) = cosh x, d/dx (cosh x) = sinh x (no minus sign).
• Separation: sin(x + iy) = sin x cosh y + i cos x sinh y; cos(x + iy) = cos x cosh y − i sin x sinh y.
• Logarithm: if z = re^(iθ), then Log z = log r + i(θ + 2nπ), n ∈ Z. It is many-valued. The principal value (n = 0, −π < θ ≤ π) is log z = log r + iθ.
• Complex powers: a^z = e^(z Log a).
• Inverse hyperbolic functions: sinh⁻¹x = log(x + √(x² + 1)), cosh⁻¹x = log(x + √(x² − 1)) for x ≥ 1, tanh⁻¹x = ½ log((1 + x)/(1 − x)) for |x| < 1.

Important terms

• Principal argument – the value of arg z in (−π, π].
• Many-valued function – gives more than one value for one input (like Log z).
• Hyperbolic function (अतिपरवलयिक फल) – functions defined from eˣ and e⁻ˣ.

Formulas

• sinh 2x = 2 sinh x cosh x; cosh 2x = cosh²x + sinh²x = 2cosh²x − 1
• cosh(x ± y) = cosh x cosh y ± sinh x sinh y
• Log(x + iy) = ½ log(x² + y²) + i(tan⁻¹(y/x) + 2nπ) (for x > 0)

5 important questions

Q1. Prove that cosh²x − sinh²x = 1.

cosh²x − sinh²x = (cosh x + sinh x)(cosh x − sinh x) = (eˣ)(e⁻ˣ) = 1, since cosh x + sinh x = eˣ and cosh x − sinh x = e⁻ˣ.

Q2. Find the principal value of log(−1) and of log(1 + i).

−1 = 1·e^(iπ), so log(−1) = log 1 + iπ = iπ (general value (2n + 1)πi).
1 + i = √2 e^(iπ/4), so log(1 + i) = log √2 + iπ/4 = ½ log 2 + iπ/4.

Q3. Show that iⁱ is real and find its principal value.

1. i = e^(iπ/2), so Log i = i(π/2 + 2nπ).
2. iⁱ = e^(i Log i) = e^(i · i(π/2 + 2nπ)) = e^(−(π/2 + 2nπ)).
3. Every value is real. Principal value (n = 0): e^(−π/2) ≈ 0.2079.

Q4. Separate cos(π/4 + i log 2) into real and imaginary parts.

1. Use cos(x + iy) = cos x cosh y − i sin x sinh y with x = π/4, y = log 2.
2. e^(log 2) = 2, e^(−log 2) = ½. So cosh(log 2) = (2 + ½)/2 = 5/4 and sinh(log 2) = (2 − ½)/2 = 3/4.
3. cos(π/4) = sin(π/4) = 1/√2.
4. Result: 5/(4√2) − i·3/(4√2). Real part 5/(4√2), imaginary part −3/(4√2).

Q5. If sin(A + iB) = x + iy, prove that x²/cosh²B + y²/sinh²B = 1.

1. sin(A + iB) = sin A cosh B + i cos A sinh B.
2. Comparing: x = sin A cosh B, y = cos A sinh B.
3. So x/cosh B = sin A and y/sinh B = cos A.
4. Squaring and adding: x²/cosh²B + y²/sinh²B = sin²A + cos²A = 1. Hence proved.

Exam tip

Remember the signs: cos(x + iy) has a minus before the imaginary part, and d/dx cosh x = + sinh x. For logarithms, always write the general value first, then put n = 0 for the principal value.

Quick summary

Successive differentiation means differentiating a function again and again: y₁ = dy/dx, y₂ = d²y/dx², … yₙ. This topic gives standard n-th derivatives, the method of partial fractions, and Leibniz's theorem for the n-th derivative of a product.

Key points

• Notation: yₙ = dⁿy/dxⁿ = Dⁿy.
• Standard results are found by differentiating a few times and spotting the pattern; they are proved by induction.
• For rational functions, first break into partial fractions, then use the result for 1/(ax + b).
• For sin, cos with eᵃˣ, write a = r cos φ, b = r sin φ.
• Leibniz's theorem gives (uv)ₙ in binomial form; it is useful when one factor is a polynomial (its derivatives end quickly).
• A classic exam pattern: prove a relation like (1 − x²)y₂ − xy₁ = 0, then differentiate it n times using Leibniz.

Important terms

• n-th derivative (n वा अवकलज) – the result of differentiating n times.
• Mathematical induction – method of proving the general n-th derivative formula.
• Partial fractions – splitting a rational function into simpler fractions.

Formulas

• y = (ax + b)ᵐ: yₙ = m(m−1)…(m−n+1) aⁿ (ax + b)^(m−n)
• y = 1/(ax + b): yₙ = (−1)ⁿ n! aⁿ/(ax + b)^(n+1)
• y = log(ax + b): yₙ = (−1)^(n−1) (n−1)! aⁿ/(ax + b)ⁿ
• y = eᵃˣ: yₙ = aⁿeᵃˣ
• y = sin(ax + b): yₙ = aⁿ sin(ax + b + nπ/2); y = cos(ax + b): yₙ = aⁿ cos(ax + b + nπ/2)
• y = eᵃˣ sin(bx + c): yₙ = rⁿ eᵃˣ sin(bx + c + nφ), r = √(a² + b²), φ = tan⁻¹(b/a) (same for cos)
• Leibniz: (uv)ₙ = uₙv + ⁿC₁uₙ₋₁v₁ + ⁿC₂uₙ₋₂v₂ + … + uvₙ

5 important questions

Q1. Find the n-th derivative of e^(2x) sin 3x.

Here a = 2, b = 3; r = √(4 + 9) = √13; φ = tan⁻¹(3/2). So yₙ = (√13)ⁿ e^(2x) sin(3x + n tan⁻¹(3/2)) = 13^(n/2) e^(2x) sin(3x + nφ).

Q2. State Leibniz's theorem.

If u and v are functions of x having n-th derivatives, then (uv)ₙ = Σ (r = 0 to n) ⁿCᵣ uₙ₋ᵣ vᵣ = uₙv + ⁿC₁uₙ₋₁v₁ + … + uvₙ.

Q3. Find the n-th derivative of y = 1/(x² − 1).

1. Partial fractions: 1/(x² − 1) = ½[1/(x − 1) − 1/(x + 1)].
2. Use 1/(ax + b) result with a = 1.
3. yₙ = ½[(−1)ⁿ n!/(x − 1)^(n+1) − (−1)ⁿ n!/(x + 1)^(n+1)].
4. yₙ = ((−1)ⁿ n!/2)[1/(x − 1)^(n+1) − 1/(x + 1)^(n+1)].

Q4. Find the n-th derivative of y = x² eˣ.

1. Take u = eˣ, v = x². Then uₙ = eˣ for all n; v₁ = 2x, v₂ = 2, v₃ = 0.
2. Leibniz: yₙ = uₙv + n uₙ₋₁v₁ + (n(n−1)/2) uₙ₋₂v₂.
3. yₙ = eˣx² + n eˣ(2x) + (n(n−1)/2) eˣ(2).
4. yₙ = eˣ[x² + 2nx + n(n − 1)].
5. Check n = 1: eˣ(x² + 2x), which is correct.

Q5. If y = sin⁻¹x, prove that (1 − x²)yₙ₊₂ − (2n + 1)x yₙ₊₁ − n²yₙ = 0.

1. y₁ = 1/√(1 − x²), so (1 − x²)y₁² = 1.
2. Differentiate: (1 − x²)2y₁y₂ − 2xy₁² = 0; divide by 2y₁: (1 − x²)y₂ − xy₁ = 0.
3. Differentiate n times by Leibniz: (1 − x²)yₙ₊₂ + n(−2x)yₙ₊₁ + (n(n−1)/2)(−2)yₙ.
4. And for xy₁: x yₙ₊₁ + n yₙ.
5. Combine: (1 − x²)yₙ₊₂ − (2n + 1)x yₙ₊₁ − (n² − n + n)yₙ = 0.
6. So (1 − x²)yₙ₊₂ − (2n + 1)x yₙ₊₁ − n²yₙ = 0. Proved.

Exam tip

In Leibniz problems choose v as the polynomial so the series stops early. Squaring y₁ before differentiating (as in Q5) avoids messy roots.

Quick summary

Mean value theorems connect the values of a function at the end points of an interval with its derivative at some point inside. Rolle's theorem, Lagrange's mean value theorem and Cauchy's mean value theorem are the three main results. They are used to prove inequalities and form the base of Taylor's theorem.

Key points

• Rolle's theorem: if f is continuous on [a, b], differentiable on (a, b) and f(a) = f(b), then there is at least one c in (a, b) with f'(c) = 0.
• Geometric meaning of Rolle's: there is a point where the tangent is parallel to the x-axis.
• Lagrange's mean value theorem (LMVT): if f is continuous on [a, b] and differentiable on (a, b), then f'(c) = (f(b) − f(a))/(b − a) for some c in (a, b).
• Geometric meaning of LMVT: there is a point where the tangent is parallel to the chord joining (a, f(a)) and (b, f(b)).
• Rolle's theorem is a special case of LMVT (when f(a) = f(b)).
• Cauchy's mean value theorem: if f and g are continuous on [a, b], differentiable on (a, b) and g'(x) ≠ 0, then (f(b) − f(a))/(g(b) − g(a)) = f'(c)/g'(c) for some c in (a, b).
• LMVT is the special case g(x) = x of Cauchy's theorem.
• Consequences of LMVT: if f'(x) = 0 on an interval, f is constant; if f'(x) > 0, f is increasing.
• All conditions must hold. Example: f(x) = |x| on [−1, 1] has f(−1) = f(1) but is not differentiable at 0, so Rolle's theorem does not apply.

Important terms

• Continuous on [a, b] – no breaks in the graph on the closed interval (सातत्य).
• Differentiable on (a, b) – derivative exists at every interior point (no corners).
• Chord – straight line joining two points of a curve.

Formulas

• Rolle: f'(c) = 0
• LMVT: f(b) − f(a) = (b − a) f'(c), a < c < b
• Another form: f(a + h) = f(a) + h f'(a + θh), 0 < θ < 1
• Cauchy: [f(b) − f(a)]/[g(b) − g(a)] = f'(c)/g'(c)

5 important questions

Q1. State Rolle's theorem and give its geometrical meaning.

If f is continuous on [a, b], differentiable on (a, b) and f(a) = f(b), there exists c in (a, b) with f'(c) = 0. Geometrically, the curve has at least one point between a and b where the tangent is horizontal.

Q2. Can Rolle's theorem be applied to f(x) = |x| on [−1, 1]? Why?

No. Though f(−1) = f(1) = 1 and f is continuous, it is not differentiable at x = 0 (sharp corner). One condition fails, so the theorem cannot be applied.

Q3. Verify Rolle's theorem for f(x) = x² − 4x + 3 on [1, 3].

1. f is a polynomial, so continuous on [1, 3] and differentiable on (1, 3).
2. f(1) = 1 − 4 + 3 = 0; f(3) = 9 − 12 + 3 = 0. So f(1) = f(3).
3. f'(x) = 2x − 4. Put f'(c) = 0: c = 2.
4. c = 2 lies in (1, 3). Rolle's theorem is verified.

Q4. Verify Cauchy's mean value theorem for f(x) = x², g(x) = x³ on [1, 2].

1. Both are polynomials: continuous and differentiable; g'(x) = 3x² ≠ 0 in (1, 2).
2. (f(2) − f(1))/(g(2) − g(1)) = (4 − 1)/(8 − 1) = 3/7.
3. f'(c)/g'(c) = 2c/(3c²) = 2/(3c).
4. 2/(3c) = 3/7 → 9c = 14 → c = 14/9 ≈ 1.56.
5. c lies in (1, 2). Verified.

Q5. Using LMVT, prove that (b − a)/(1 + b²) < tan⁻¹b − tan⁻¹a < (b − a)/(1 + a²) for 0 < a < b.

1. Take f(x) = tan⁻¹x on [a, b]; it is continuous and differentiable; f'(x) = 1/(1 + x²).
2. By LMVT: (tan⁻¹b − tan⁻¹a)/(b − a) = 1/(1 + c²), with a < c < b.
3. Since a < c < b, 1 + a² < 1 + c² < 1 + b².
4. So 1/(1 + b²) < 1/(1 + c²) < 1/(1 + a²).
5. Multiply by (b − a) > 0 to get the result.

Exam tip

Before finding c, write all the conditions and check each one — examiners give marks for this. Always show that the c you get lies inside the open interval.

Quick summary

Taylor's theorem expresses a function as a power series about a point using its derivatives at that point. Maclaurin's series is the Taylor series about x = 0. These series are used to expand standard functions, to approximate values and to evaluate limits.

Key points

• Taylor's theorem (with remainder): f(a + h) = f(a) + h f'(a) + (h²/2!) f''(a) + … + (hⁿ⁻¹/(n−1)!) f⁽ⁿ⁻¹⁾(a) + Rₙ.
• Lagrange's form of remainder: Rₙ = (hⁿ/n!) f⁽ⁿ⁾(a + θh), 0 < θ < 1.
• Cauchy's form: Rₙ = (hⁿ(1 − θ)ⁿ⁻¹/(n − 1)!) f⁽ⁿ⁾(a + θh).
• If Rₙ → 0 as n → ∞, we get the infinite Taylor series.
• In powers of (x − a): f(x) = f(a) + (x − a) f'(a) + ((x − a)²/2!) f''(a) + …
• Maclaurin's series (a = 0): f(x) = f(0) + x f'(0) + (x²/2!) f''(0) + …
• Method: find f(0), f'(0), f''(0), … (or use known series and combine them).
• Expansions help in indeterminate forms (0/0, ∞/∞) together with L'Hospital's rule.
• Series give quick approximations, e.g. √(1.02) or sin 0.1.

Important terms

• Power series (घात श्रेणी) – an infinite series in powers of x (or x − a).
• Remainder term – the error after taking n terms.
• Radius of convergence – range of x for which the series converges (e.g. log(1 + x) needs −1 < x ≤ 1).

Formulas

• eˣ = 1 + x + x²/2! + x³/3! + … (all x)
• sin x = x − x³/3! + x⁵/5! − … ; cos x = 1 − x²/2! + x⁴/4! − …
• log(1 + x) = x − x²/2 + x³/3 − … (−1 < x ≤ 1)
• (1 + x)ⁿ = 1 + nx + n(n − 1)x²/2! + … (|x| < 1)
• tan x = x + x³/3 + 2x⁵/15 + …
• sinh x = x + x³/3! + …; cosh x = 1 + x²/2! + …

5 important questions

Q1. State Maclaurin's theorem.

If f and its derivatives up to order n exist near 0, then f(x) = f(0) + x f'(0) + (x²/2!) f''(0) + … + (xⁿ⁻¹/(n−1)!) f⁽ⁿ⁻¹⁾(0) + (xⁿ/n!) f⁽ⁿ⁾(θx), 0 < θ < 1.

Q2. Expand eˣ in powers of (x − 1).

Take a = 1. All derivatives of eˣ are eˣ, so each equals e at x = 1. Hence eˣ = e[1 + (x − 1) + (x − 1)²/2! + (x − 1)³/3! + …].

Q3. Derive the Maclaurin series of log(1 + x).

1. f(x) = log(1 + x), f(0) = 0.
2. f'(x) = 1/(1 + x), f'(0) = 1.
3. f''(x) = −1/(1 + x)², f''(0) = −1.
4. f'''(x) = 2/(1 + x)³, f'''(0) = 2; f⁗(0) = −6.
5. log(1 + x) = 0 + x − x²/2! + 2x³/3! − 6x⁴/4! + … = x − x²/2 + x³/3 − x⁴/4 + …

Q4. Expand e^(sin x) up to the term in x³.

1. sin x = x − x³/6 + … Let u = sin x.
2. eᵘ = 1 + u + u²/2 + u³/6 + …
3. u = x − x³/6; u² = x² + (terms of x⁴ and higher); u³ = x³ + …
4. e^(sin x) = 1 + (x − x³/6) + x²/2 + x³/6 + …
5. The x³ terms cancel: e^(sin x) = 1 + x + x²/2 + (0)x³ + …

Q5. Using the binomial series, find √1.02 correct to 4 decimal places.

1. √1.02 = (1 + 0.02)^(1/2), with x = 0.02, n = ½.
2. = 1 + ½(0.02) + (½)(−½)/2 × (0.02)² + …
3. = 1 + 0.01 − (1/8)(0.0004) = 1 + 0.01 − 0.00005.
4. √1.02 ≈ 1.00995 ≈ 1.0100 (4 d.p.).

Exam tip

Learn the five basic series (eˣ, sin x, cos x, log(1 + x), (1 + x)ⁿ) by heart — most expansions are built by combining them instead of finding many derivatives.

Quick summary

Some limits give meaningless forms like 0/0 or ∞/∞ when we substitute directly. These are called indeterminate forms. L'Hospital's rule finds such limits by differentiating the numerator and denominator separately. Other forms (0 × ∞, ∞ − ∞, 1^∞, 0⁰, ∞⁰) are first converted to 0/0 or ∞/∞.

Key points

• The seven indeterminate forms: 0/0, ∞/∞, 0 × ∞, ∞ − ∞, 0⁰, 1^∞, ∞⁰.
• L'Hospital's rule: if f(a) = g(a) = 0 (or both → ∞) and f, g are differentiable near a with g′(x) ≠ 0, then lim f(x)/g(x) = lim f′(x)/g′(x), provided the right side exists.
• Differentiate numerator and denominator separately — never use the quotient rule here.
• The rule may be applied again if the new limit is still 0/0 or ∞/∞; stop as soon as the form is determinate.
• 0 × ∞: write f·g as f/(1/g) or g/(1/f).
• ∞ − ∞: combine into a single fraction (take LCM) to get 0/0.
• 1^∞, 0⁰, ∞⁰: let L = lim f^g; then log L = lim g·log f, which is of type 0 × ∞. Find it, then L = e^(that value).
• Standard series (sin x = x − x³/6 + …, eˣ = 1 + x + x²/2 + …) can also give such limits quickly and are useful to check answers.

Important terms

• Indeterminate form (अनिश्चित रूप) – an expression whose value cannot be found by direct substitution.
• Determinate form – a form with a definite value, e.g. 1/0 → ∞ or 0/5 = 0.

Formulas

• lim(x→a) f/g = lim(x→a) f′/g′ (for 0/0 or ∞/∞)
• lim(x→0) sin x/x = 1; lim(x→0) (1 + x)^(1/x) = e
• If lim g log f = k, then lim f^g = e^k

5 important questions

Q1. State L'Hospital's rule.

If f(x) → 0 and g(x) → 0 (or both → ∞) as x → a, f and g are differentiable near a and g′(x) ≠ 0, then lim(x→a) f(x)/g(x) = lim(x→a) f′(x)/g′(x), provided the latter limit exists.

Q2. Evaluate lim(x→0⁺) x log x.

This is 0 × (−∞). Write it as log x/(1/x), which is ∞/∞. By the rule: (1/x)/(−1/x²) = −x → 0. So the limit is 0.

Q3. Evaluate lim(x→0) (x − sin x)/x³.

1. Form 0/0. Differentiate: (1 − cos x)/(3x²), still 0/0.
2. Again: sin x/(6x), still 0/0.
3. Again: cos x/6 → 1/6.
4. Answer: 1/6. (Check by series: x − sin x = x³/6 − …, so the ratio → 1/6 ✓.)

Q4. Evaluate lim(x→0) (1/x − 1/sin x).

1. Form ∞ − ∞. Combine: (sin x − x)/(x sin x), form 0/0.
2. Differentiate: (cos x − 1)/(sin x + x cos x), still 0/0.
3. Again: (−sin x)/(cos x + cos x − x sin x) = (−sin x)/(2 cos x − x sin x).
4. At x = 0: 0/2 = 0. Answer: 0.

Q5. Evaluate lim(x→0) (tan x/x)^(1/x²).

1. Form 1^∞. Let L be the limit; log L = lim log(tan x/x)/x².
2. Use tan x = x + x³/3 + …, so tan x/x = 1 + x²/3 + …
3. log(1 + x²/3 + …) ≈ x²/3 for small x.
4. So log L = lim (x²/3)/x² = 1/3.
5. L = e^(1/3).

Exam tip

Before each application of the rule, write the form you got (0/0 or ∞/∞) — examiners give marks for it. For power forms, always take log first and remember to convert back with e at the end.