गणित Mathematics

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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].

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• |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)

५ महत्त्वाचे प्रश्न

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.

परीक्षेसाठी टीप

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.

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• 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

५ महत्त्वाचे प्रश्न

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.

परीक्षेसाठी टीप

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

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• 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

५ महत्त्वाचे प्रश्न

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| ✓.

परीक्षेसाठी टीप

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

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• 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.

५ महत्त्वाचे प्रश्न

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.

परीक्षेसाठी टीप

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.

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• 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)

५ महत्त्वाचे प्रश्न

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.

परीक्षेसाठी टीप

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.

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• 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)

५ महत्त्वाचे प्रश्न

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.

परीक्षेसाठी टीप

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.

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• 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

५ महत्त्वाचे प्रश्न

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.

परीक्षेसाठी टीप

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

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• 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)

५ महत्त्वाचे प्रश्न

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.

परीक्षेसाठी टीप

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.

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• 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).

सूत्रे

• 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! + …

५ महत्त्वाचे प्रश्न

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.).

परीक्षेसाठी टीप

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.

थोडक्यात सारांश

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 ∞/∞.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• 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

५ महत्त्वाचे प्रश्न

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).

परीक्षेसाठी टीप

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.

थोडक्यात सारांश

When a function depends on two or more variables, a partial derivative measures how it changes with one variable while the others are kept constant. This topic covers partial derivatives, Euler's theorem on homogeneous functions, total derivatives and chain rule, Jacobians, and maxima–minima of functions of two variables.

महत्त्वाचे मुद्दे

• For u = f(x, y): ∂u/∂x is found by treating y as a constant; ∂u/∂y by treating x as constant.
• Higher derivatives: ∂²u/∂x², ∂²u/∂y², ∂²u/∂x∂y. Usually ∂²u/∂x∂y = ∂²u/∂y∂x (for smooth functions).
• Homogeneous function of degree n: f(tx, ty) = tⁿ f(x, y).
• Euler's theorem: if u is homogeneous of degree n, x ∂u/∂x + y ∂u/∂y = nu.
• Corollary: x²uₓₓ + 2xy uₓᵧ + y²uᵧᵧ = n(n − 1)u.
• If f(u) is homogeneous of degree n (u itself not), then x uₓ + y uᵧ = n f(u)/f'(u).
• Total derivative and chain rule: if u = f(x, y) with x, y functions of t, du/dt = (∂u/∂x)(dx/dt) + (∂u/∂y)(dy/dt).
• Jacobian J = ∂(u, v)/∂(x, y) measures how areas change under a change of variables.
• Maxima and minima: at a stationary point fₓ = 0, fᵧ = 0. With r = fₓₓ, s = fₓᵧ, t = fᵧᵧ: rt − s² > 0 and r < 0 → maximum; rt − s² > 0 and r > 0 → minimum; rt − s² < 0 → saddle point; rt − s² = 0 → needs more study.
• Lagrange's multipliers handle extremum with a condition (constraint).

महत्त्वाच्या संज्ञा

• Partial derivative (आंशिक अवकलज) – derivative w.r.t. one variable, others fixed.
• Laplace's equation – uₓₓ + uᵧᵧ = 0; its solutions are harmonic functions.
• Saddle point – stationary point that is neither max nor min.

सूत्रे

• Euler: x uₓ + y uᵧ = nu
• Chain rule: du/dt = uₓ (dx/dt) + uᵧ (dy/dt)
• Implicit function f(x, y) = 0: dy/dx = −fₓ/fᵧ
• Jacobian: ∂(u, v)/∂(x, y) = uₓvᵧ − uᵧvₓ; and J × J' = 1
• Polar: x = r cos θ, y = r sin θ → ∂(x, y)/∂(r, θ) = r

५ महत्त्वाचे प्रश्न

Q1. State Euler's theorem on homogeneous functions.

If u = f(x, y) is a homogeneous function of degree n in x and y, then x ∂u/∂x + y ∂u/∂y = nu.

Q2. Find ∂(x, y)/∂(r, θ) for x = r cos θ, y = r sin θ.

xᵣ = cos θ, x_θ = −r sin θ, yᵣ = sin θ, y_θ = r cos θ. J = xᵣy_θ − x_θyᵣ = r cos²θ + r sin²θ = r.

Q3. If u = log(x² + y²), show that uₓₓ + uᵧᵧ = 0.

1. uₓ = 2x/(x² + y²).
2. uₓₓ = [2(x² + y²) − 2x(2x)]/(x² + y²)² = 2(y² − x²)/(x² + y²)².
3. By symmetry, uᵧᵧ = 2(x² − y²)/(x² + y²)².
4. Add: uₓₓ + uᵧᵧ = [2(y² − x²) + 2(x² − y²)]/(x² + y²)² = 0.

Q4. If u = sin⁻¹[(x² + y²)/(x + y)], prove that x uₓ + y uᵧ = tan u.

1. u is not homogeneous, but sin u = (x² + y²)/(x + y) = f is homogeneous of degree 2 − 1 = 1.
2. Apply Euler to f = sin u: x ∂(sin u)/∂x + y ∂(sin u)/∂y = 1 × sin u.
3. ∂(sin u)/∂x = cos u · uₓ; similarly for y.
4. cos u (x uₓ + y uᵧ) = sin u.
5. x uₓ + y uᵧ = tan u. Proved.

Q5. Find the extreme value of f(x, y) = x² + y² − 2x − 4y + 5.

1. fₓ = 2x − 2 = 0 → x = 1; fᵧ = 2y − 4 = 0 → y = 2. Stationary point (1, 2).
2. r = fₓₓ = 2, s = fₓᵧ = 0, t = fᵧᵧ = 2.
3. rt − s² = 4 > 0 and r > 0, so (1, 2) is a minimum.
4. f(1, 2) = 1 + 4 − 2 − 8 + 5 = 0. Minimum value = 0.

परीक्षेसाठी टीप

For Euler's theorem questions, first find the degree carefully by replacing x by tx and y by ty. If u is inside sin⁻¹, tan⁻¹ or log, apply Euler to the inner homogeneous function.

थोडक्यात सारांश

A function f(x, y) of two variables can have maximum, minimum and saddle points. We find stationary points using partial derivatives and classify them with the second-derivative test (rt − s²). When the variables must satisfy a condition, Lagrange's method of undetermined multipliers is used. Taylor's theorem for two variables gives the basis for the test.

महत्त्वाचे मुद्दे

• Taylor's theorem (two variables): f(a + h, b + k) = f(a, b) + (h fₓ + k fᵧ) + (1/2!)(h² fₓₓ + 2hk fₓᵧ + k² fᵧᵧ) + …, all derivatives at (a, b).
• Stationary point: where fₓ = 0 and fᵧ = 0 together.
• Notation at a stationary point: r = fₓₓ, s = fₓᵧ, t = fᵧᵧ.
• Test: if rt − s² > 0 and r < 0 → maximum; rt − s² > 0 and r > 0 → minimum; rt − s² < 0 → saddle point (neither); rt − s² = 0 → doubtful, needs further study.
• An extreme value is a local (relative) value, not necessarily the largest or smallest over the whole plane.
• Lagrange's multipliers: to find extreme values of f(x, y, z) subject to φ(x, y, z) = 0, form F = f + λφ and solve ∂F/∂x = 0, ∂F/∂y = 0, ∂F/∂z = 0 together with φ = 0.
• Lagrange's method gives the stationary points; the nature (max or min) is usually decided from the physical meaning of the problem.

महत्त्वाच्या संज्ञा

• Saddle point – a stationary point which is a max along one direction and a min along another.
• Extreme value (कमाल-किमान मूल्य) – a maximum or minimum value.
• Lagrange multiplier λ – an unknown constant introduced to include the condition.

सूत्रे

• Stationary point: fₓ = 0, fᵧ = 0
• r = fₓₓ, s = fₓᵧ, t = fᵧᵧ; decide by the sign of rt − s² and of r
• F = f + λφ; Fₓ = Fᵧ = F_z = 0, φ = 0

५ महत्त्वाचे प्रश्न

Q1. What is a saddle point?

A stationary point (fₓ = fᵧ = 0) at which rt − s² < 0. The function increases in some directions and decreases in others, so it is neither a maximum nor a minimum. Example: (0, 0) for f = x² − y².

Q2. State the conditions for f(x, y) to have a maximum at (a, b).

fₓ = 0 and fᵧ = 0 at (a, b), and at that point rt − s² > 0 with r < 0 (where r = fₓₓ, s = fₓᵧ, t = fᵧᵧ).

Q3. Find the extreme values of f(x, y) = x³ + y³ − 3axy (a > 0).

1. fₓ = 3x² − 3ay = 0 and fᵧ = 3y² − 3ax = 0, so y = x²/a and x = y²/a.
2. Substituting: x = x⁴/a³, so x(x³ − a³) = 0, giving x = 0 or x = a. Points: (0, 0) and (a, a).
3. r = 6x, s = −3a, t = 6y.
4. At (0, 0): rt − s² = 0 − 9a² < 0 → saddle point.
5. At (a, a): r = 6a, t = 6a, rt − s² = 36a² − 9a² = 27a² > 0 and r > 0 → minimum.
6. Minimum value = a³ + a³ − 3a·a·a = −a³.

Q4. Find the minimum of x² + y² + z² subject to x + y + z = 6.

1. F = x² + y² + z² + λ(x + y + z − 6).
2. Fₓ = 2x + λ = 0, Fᵧ = 2y + λ = 0, F_z = 2z + λ = 0, so x = y = z = −λ/2.
3. From the condition: 3x = 6, so x = y = z = 2.
4. Minimum value = 4 + 4 + 4 = 12. (It is a minimum, since x² + y² + z² can be made as large as we like on the plane.)

Q5. Find the maximum of f(x, y) = 2x + 4y − x² − y² + 5.

1. fₓ = 2 − 2x = 0 → x = 1; fᵧ = 4 − 2y = 0 → y = 2.
2. r = −2, s = 0, t = −2; rt − s² = 4 > 0 and r < 0 → maximum.
3. Maximum value = 2 + 8 − 1 − 4 + 5 = 10.

परीक्षेसाठी टीप

Make a small table of each stationary point with r, s, t, rt − s² and the conclusion. In Lagrange problems, first express all variables in terms of λ, then use the condition.

थोडक्यात सारांश

A reduction formula connects an integral containing a power n with a similar integral of lower power (n − 1 or n − 2). Using it again and again, we reduce the power until the integral is easy. The most used results are for sinⁿx, cosⁿx, sinᵐx cosⁿx and tanⁿx, especially between 0 and π/2 (Wallis' formula).

महत्त्वाचे मुद्दे

• Reduction formulae are obtained by integration by parts, writing sinⁿx = sinⁿ⁻¹x · sin x and so on.
• ∫ sinⁿx dx = −(sinⁿ⁻¹x cos x)/n + ((n − 1)/n) ∫ sinⁿ⁻²x dx.
• ∫ cosⁿx dx = (cosⁿ⁻¹x sin x)/n + ((n − 1)/n) ∫ cosⁿ⁻²x dx.
• For definite integrals from 0 to π/2 the first term vanishes, so Iₙ = ((n − 1)/n) Iₙ₋₂, with I₀ = π/2 and I₁ = 1.
• ∫₀^(π/2) sinⁿx dx = ∫₀^(π/2) cosⁿx dx (by the property ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a − x) dx).
• ∫ tanⁿx dx = tanⁿ⁻¹x/(n − 1) − ∫ tanⁿ⁻²x dx. For 0 to π/4: Iₙ + Iₙ₋₂ = 1/(n − 1).
• Wallis' formula for sinᵐx cosⁿx (0 to π/2): numerator = (m − 1)(m − 3)… × (n − 1)(n − 3)…, denominator = (m + n)(m + n − 2)…; multiply by π/2 only when m and n are both even.
• Each product of factors stops at 1 or 2.

महत्त्वाच्या संज्ञा

• Reduction formula (लघुकरण सूत्र) – a formula expressing Iₙ through Iₙ₋₁ or Iₙ₋₂.
• Wallis' formula – direct value of ∫₀^(π/2) sinⁿx dx or sinᵐx cosⁿx dx.

सूत्रे

• n even: ∫₀^(π/2) sinⁿx dx = [(n − 1)(n − 3)…3·1]/[n(n − 2)…4·2] × π/2
• n odd: ∫₀^(π/2) sinⁿx dx = [(n − 1)(n − 3)…4·2]/[n(n − 2)…5·3]
• ∫₀^(π/4) tanⁿx dx: Iₙ = 1/(n − 1) − Iₙ₋₂, I₀ = π/4, I₁ = ½ log 2

५ महत्त्वाचे प्रश्न

Q1. Evaluate ∫₀^(π/2) sin⁶x dx.

n = 6 (even): (5·3·1)/(6·4·2) × π/2 = (15/48) × π/2 = 5π/32.

Q2. Evaluate ∫₀^(π/2) cos⁵x dx.

n = 5 (odd): (4·2)/(5·3) = 8/15.

Q3. Obtain the reduction formula for Iₙ = ∫₀^(π/2) sinⁿx dx.

1. Write Iₙ = ∫ sinⁿ⁻¹x · sin x dx and integrate by parts with u = sinⁿ⁻¹x, dv = sin x dx.
2. Iₙ = [−sinⁿ⁻¹x cos x]₀^(π/2) + (n − 1)∫ sinⁿ⁻²x cos²x dx.
3. The bracket is 0 at both limits. Put cos²x = 1 − sin²x: Iₙ = (n − 1)(Iₙ₋₂ − Iₙ).
4. So nIₙ = (n − 1)Iₙ₋₂, i.e. Iₙ = ((n − 1)/n) Iₙ₋₂.

Q4. Evaluate ∫₀^(π/2) sin⁴x cos²x dx.

1. m = 4, n = 2, both even, so multiply by π/2.
2. Numerator: (3·1)(1) = 3. Denominator: (6)(4)(2) = 48.
3. Value = (3/48) × π/2 = π/32.

Q5. Evaluate ∫₀^(π/4) tan⁴x dx.

1. Use Iₙ = 1/(n − 1) − Iₙ₋₂ with I₀ = ∫₀^(π/4) dx = π/4.
2. I₂ = 1/1 − I₀ = 1 − π/4.
3. I₄ = 1/3 − I₂ = 1/3 − 1 + π/4 = π/4 − 2/3.
4. Check: π/4 − 2/3 ≈ 0.785 − 0.667 = 0.118, positive and small, as expected since 0 ≤ tan⁴x ≤ 1 there.

परीक्षेसाठी टीप

For Wallis' formula, write the numerator and denominator factors in two clear lines and decide "× π/2 or not" before simplifying. Derivations are asked often — learn the integration-by-parts step for sinⁿx well.

थोडक्यात सारांश

Beta and Gamma functions are special functions defined by improper integrals. Many hard definite integrals can be written in terms of them and evaluated quickly. The key facts are Γ(n + 1) = nΓ(n), Γ(½) = √π and the relation B(m, n) = Γ(m)Γ(n)/Γ(m + n).

महत्त्वाचे मुद्दे

• Gamma function: Γ(n) = ∫₀^∞ e^(−x) x^(n−1) dx, n > 0.
• Γ(1) = 1 and Γ(n + 1) = nΓ(n). For a positive integer n, Γ(n + 1) = n!.
• Γ(½) = √π. So Γ(3/2) = ½√π, Γ(5/2) = (3/4)√π.
• Beta function: B(m, n) = ∫₀¹ x^(m−1)(1 − x)^(n−1) dx, m, n > 0. It is symmetric: B(m, n) = B(n, m).
• Relation: B(m, n) = Γ(m)Γ(n)/Γ(m + n).
• Trigonometric form: B(m, n) = 2∫₀^(π/2) sin^(2m−1)θ cos^(2n−1)θ dθ. Hence ∫₀^(π/2) sinᵖθ cos^qθ dθ = ½ B((p + 1)/2, (q + 1)/2).
• Other form: B(m, n) = ∫₀^∞ x^(m−1)/(1 + x)^(m+n) dx.
• Duplication formula: Γ(m)Γ(m + ½) = √π Γ(2m)/2^(2m−1).
• Γ(n)Γ(1 − n) = π/sin nπ for 0 < n < 1.

महत्त्वाच्या संज्ञा

• Improper integral – an integral with an infinite limit or an unbounded integrand.
• Gamma function (गॅमा फल) – a generalisation of factorial to non-integer values.

सूत्रे

• Γ(n + 1) = nΓ(n); Γ(n + 1) = n!
• Γ(½) = √π
• B(m, n) = Γ(m)Γ(n)/Γ(m + n)
• ∫₀^∞ e^(−ax) x^(n−1) dx = Γ(n)/aⁿ

५ महत्त्वाचे प्रश्न

Q1. Find Γ(7/2).

Γ(7/2) = (5/2)(3/2)(1/2)Γ(½) = (15/8)√π.

Q2. Evaluate ∫₀^∞ x⁴ e^(−x) dx.

This is Γ(5) = 4! = 24.

Q3. Evaluate ∫₀¹ x³(1 − x)⁴ dx.

1. Compare with B(m, n): m − 1 = 3, n − 1 = 4, so m = 4, n = 5.
2. B(4, 5) = Γ(4)Γ(5)/Γ(9) = 3! × 4!/8!.
3. = 6 × 24/40320 = 144/40320 = 1/280.

Q4. Show that ∫₀^∞ e^(−x²) dx = √π/2.

1. Put x² = t, so x = √t and dx = dt/(2√t).
2. Integral = ∫₀^∞ e^(−t) · 1/(2√t) dt = ½ ∫₀^∞ e^(−t) t^(−1/2) dt.
3. This is ½ Γ(½) = ½ √π = √π/2.

Q5. Evaluate ∫₀^(π/2) sin⁴θ cos²θ dθ using Beta and Gamma functions.

1. Here p = 4, q = 2, so the integral = ½ B(5/2, 3/2).
2. = ½ × Γ(5/2)Γ(3/2)/Γ(4).
3. Γ(5/2) = (3/4)√π, Γ(3/2) = (1/2)√π, Γ(4) = 3! = 6.
4. = ½ × (3/4)(1/2)π/6 = ½ × (3π/8)/6 = 3π/96 = π/32.
5. This matches Wallis' formula: (3·1·1)/(6·4·2) × π/2 = π/32 ✓.

परीक्षेसाठी टीप

Always write which standard form you are matching (m − 1 = …, n − 1 = …) before substituting. Learn Γ(½) = √π by heart; most answers contain π or √π.

थोडक्यात सारांश

A double integral adds up a function over a region of the plane; a triple integral does the same over a solid. They give areas, volumes, mass and centres of gravity. We learn to evaluate them as repeated integrals, change the order of integration, and change variables (polar, cylindrical, spherical) using the Jacobian.

महत्त्वाचे मुद्दे

• ∬_R f(x, y) dx dy is evaluated as a repeated integral: integrate with respect to one variable (treating the other as constant), then the other.
• When the inner limits depend on the outer variable, the inner integral must be done first.
• Area of R = ∬_R dx dy; volume under z = f(x, y) = ∬_R f dx dy; volume of a solid V = ∭ dx dy dz.
• Change of order: draw the region, then describe it again with the other variable outside. This often turns an impossible inner integral into an easy one.
• Change of variables: dx dy = |J| du dv where J = ∂(x, y)/∂(u, v).
• Polar: x = r cos θ, y = r sin θ, dx dy = r dr dθ. Useful for circles and for x² + y².
• Cylindrical: dx dy dz = r dr dθ dz. Spherical: dx dy dz = r² sin θ dr dθ dφ.
• Order of limits in a triple integral: innermost first, from variable limits to constant limits.

महत्त्वाच्या संज्ञा

• Region of integration – the area (or solid) over which we integrate.
• Jacobian (याकोबियन) – the scale factor between areas (or volumes) in old and new variables.
• Repeated (iterated) integral – a double integral written as two single integrals.

सूत्रे

• Area = ∬ dx dy = ∬ r dr dθ
• Mass = ∬ ρ dx dy; x̄ = (∬ xρ dx dy)/mass
• ∫₀^∞ e^(−x²) dx = √π/2 (proved using polar coordinates)

५ महत्त्वाचे प्रश्न

Q1. Evaluate ∫₀¹ ∫₀ˣ (x + y) dy dx.

Inner: ∫₀ˣ (x + y) dy = xy + y²/2 from 0 to x = x² + x²/2 = 3x²/2. Outer: ∫₀¹ 3x²/2 dx = ½. Answer: ½.

Q2. Find the area of a circle of radius a using a double integral.

In polar coordinates: Area = ∫₀^(2π) ∫₀ᵃ r dr dθ = ∫₀^(2π) (a²/2) dθ = (a²/2)(2π) = πa².

Q3. Change the order and evaluate ∫₀¹ ∫ₓ¹ e^(y²) dy dx.

1. The inner integral ∫ e^(y²) dy has no elementary form, so change the order.
2. Region: 0 ≤ x ≤ 1, x ≤ y ≤ 1, i.e. the triangle with vertices (0,0), (0,1), (1,1).
3. Described the other way: 0 ≤ y ≤ 1, 0 ≤ x ≤ y.
4. Integral = ∫₀¹ ∫₀^y e^(y²) dx dy = ∫₀¹ y e^(y²) dy.
5. Put y² = t: = ½ ∫₀¹ eᵗ dt = (e − 1)/2.

Q4. Evaluate ∫₀^∞ ∫₀^∞ e^(−(x² + y²)) dx dy and deduce ∫₀^∞ e^(−x²) dx.

1. The region is the first quadrant: 0 ≤ r < ∞, 0 ≤ θ ≤ π/2.
2. Integral = ∫₀^(π/2) ∫₀^∞ e^(−r²) r dr dθ.
3. ∫₀^∞ e^(−r²) r dr = ½ (put r² = t). So the integral = (π/2)(½) = π/4.
4. But the integral also equals (∫₀^∞ e^(−x²) dx)² = I². So I² = π/4 and I = √π/2.

Q5. Find the volume of the tetrahedron bounded by x = 0, y = 0, z = 0 and x + y + z = 1.

1. V = ∫₀¹ ∫₀^(1−x) ∫₀^(1−x−y) dz dy dx.
2. Inner: ∫ dz = 1 − x − y.
3. Middle: ∫₀^(1−x) (1 − x − y) dy = (1 − x)² − (1 − x)²/2 = (1 − x)²/2.
4. Outer: ∫₀¹ (1 − x)²/2 dx = ½ × ⅓ = 1/6.
5. Volume = 1/6 cubic units.

परीक्षेसाठी टीप

Always sketch the region before changing the order of integration — most mistakes come from wrong limits. In polar problems, never forget the extra factor r.

थोडक्यात सारांश

A differential equation relates a function with its derivatives. This topic covers first-order, first-degree equations and the standard methods to solve them: variables separable, homogeneous, linear, Bernoulli's and exact equations. Simple applications such as orthogonal trajectories and growth–decay problems are also included.

महत्त्वाचे मुद्दे

• Order = highest derivative present; degree = power of the highest derivative (after removing radicals and fractions).
• The general solution has as many arbitrary constants as the order; a particular solution fixes them using given conditions.
• Variables separable: f(x) dx = g(y) dy → integrate both sides.
• Homogeneous equation: dy/dx = f(y/x). Put y = vx, dy/dx = v + x dv/dx, then separate.
• Linear equation: dy/dx + Py = Q (P, Q functions of x). Integrating factor IF = e^(∫P dx). Solution: y × IF = ∫ Q × IF dx + c.
• Bernoulli's equation: dy/dx + Py = Qyⁿ. Divide by yⁿ and put z = y^(1−n); it becomes linear in z.
• Exact equation: M dx + N dy = 0 is exact if ∂M/∂y = ∂N/∂x.
• Solution of exact equation: ∫M dx (y constant) + ∫(terms of N not containing x) dy = c.
• If not exact, an integrating factor may make it exact.
• Orthogonal trajectories: curves cutting a given family at right angles; replace dy/dx by −dx/dy in the family's equation.

महत्त्वाच्या संज्ञा

• Differential equation (अवकल समीकरण) – equation containing derivatives.
• Integrating factor (IF) – a multiplier that makes an equation directly integrable.
• Exact differential – M dx + N dy that is the total differential du of some u(x, y).

सूत्रे

• Linear: IF = e^(∫P dx); y·IF = ∫Q·IF dx + c
• Linear in x: dx/dy + P₁x = Q₁ → x·e^(∫P₁dy) = ∫Q₁ e^(∫P₁dy) dy + c
• Bernoulli: z = y^(1−n) → dz/dx + (1 − n)Pz = (1 − n)Q
• Exactness test: ∂M/∂y = ∂N/∂x
• Growth/decay: dN/dt = kN → N = N₀e^(kt)

५ महत्त्वाचे प्रश्न

Q1. Find the order and degree of (d²y/dx²)³ + (dy/dx)⁴ + y = 0.

The highest derivative is d²y/dx², so the order is 2. Its power is 3, so the degree is 3.

Q2. Solve dy/dx = xy.

Separate: dy/y = x dx. Integrate: log y = x²/2 + c. So y = A e^(x²/2), where A is a constant.

Q3. Solve dy/dx + y = eˣ.

1. Linear with P = 1, Q = eˣ.
2. IF = e^(∫1 dx) = eˣ.
3. y eˣ = ∫eˣ · eˣ dx + c = e^(2x)/2 + c.
4. y = eˣ/2 + c e^(−x).
5. Check: y' = eˣ/2 − c e^(−x); y' + y = eˣ ✓.

Q4. Solve (2xy + 3) dx + (x² + 4y) dy = 0.

1. M = 2xy + 3, N = x² + 4y.
2. ∂M/∂y = 2x, ∂N/∂x = 2x. Equal, so the equation is exact.
3. ∫M dx (y constant) = x²y + 3x.
4. Terms of N without x: 4y; ∫4y dy = 2y².
5. Solution: x²y + 3x + 2y² = c.

Q5. Solve the homogeneous equation dy/dx = (x + y)/x.

1. Put y = vx, dy/dx = v + x dv/dx.
2. v + x dv/dx = (x + vx)/x = 1 + v.
3. x dv/dx = 1 → dv = dx/x.
4. v = log x + c → y/x = log x + c.
5. Solution: y = x log x + cx.

परीक्षेसाठी टीप

First identify the type (separable, homogeneous, linear, Bernoulli, exact) — write it as the first line of your answer. For linear equations, always write P, Q and IF clearly.

थोडक्यात सारांश

A first-order differential equation of higher degree contains powers of p = dy/dx, such as p² or p³. Such equations are solved by factorising in p (solvable for p), or by making them solvable for y or for x. Clairaut's equation y = px + f(p) is the most important special case: it has a general solution (a family of straight lines) and a singular solution (their envelope).

महत्त्वाचे मुद्दे

• Notation: p = dy/dx. The equation has the form F(x, y, p) = 0 of degree n in p.
• Solvable for p: factorise into (p − f₁)(p − f₂)…(p − fₙ) = 0, solve each first-order, first-degree equation, and write the general solution as the product of all the solutions = 0, using one constant c.
• Solvable for y: write y = f(x, p), differentiate with respect to x to get an equation in x and p, solve it, then eliminate p (or keep p as a parameter).
• Solvable for x: write x = f(y, p), differentiate with respect to y using dx/dy = 1/p.
• Clairaut's equation: y = px + f(p). The general solution is obtained by simply replacing p by c: y = cx + f(c).
• Singular solution of Clairaut's equation: eliminate p between y = px + f(p) and x + f′(p) = 0. It contains no arbitrary constant and is not a special case of the general solution.
• Geometrically, the singular solution is the envelope of the family of lines y = cx + f(c).
• Some equations reduce to Clairaut's form by a substitution such as X = x², Y = y².

महत्त्वाच्या संज्ञा

• Degree of a DE – the power of the highest derivative after making the equation free of radicals and fractions.
• Singular solution (विचित्र उकल) – a solution with no arbitrary constant that cannot be got from the general solution.
• Envelope – a curve touching every member of a family of curves.

सूत्रे

• Solvable for y: y = f(x, p) ⇒ p = ∂f/∂x + (∂f/∂p)(dp/dx)
• Clairaut: y = px + f(p) ⇒ general y = cx + f(c); singular: x + f′(p) = 0 with the given equation

५ महत्त्वाचे प्रश्न

Q1. Solve p² − 5p + 6 = 0.

(p − 2)(p − 3) = 0, so dy/dx = 2 or 3, giving y = 2x + c or y = 3x + c. General solution: (y − 2x − c)(y − 3x − c) = 0.

Q2. What is Clairaut's equation? Write its general solution.

An equation of the form y = px + f(p), where p = dy/dx. Its general solution is y = cx + f(c), found by replacing p with the constant c.

Q3. Solve y = px + a/p and find its singular solution.

1. This is Clairaut's form with f(p) = a/p. General solution: y = cx + a/c.
2. For the singular solution, differentiate the right side with respect to p: x − a/p² = 0, so p² = a/x and p = √(a/x).
3. Substitute: y = x√(a/x) + a√(x/a) = √(ax) + √(ax) = 2√(ax).
4. Squaring: y² = 4ax, a parabola. It is the envelope of the lines y = cx + a/c.

Q4. Solve y = px + p² and find the singular solution.

1. General solution: y = cx + c².
2. Singular: x + 2p = 0 gives p = −x/2.
3. Then y = (−x/2)x + x²/4 = −x²/4, i.e. x² + 4y = 0.

Q5. Solve p² + py − x² − xy = 0.

1. Factorise: (p − x)(p + x + y) = 0. (Check: p² + px + py − px − x² − xy ✓.)
2. p = x gives y = x²/2 + c, i.e. 2y − x² − c = 0.
3. p + y = −x is linear: dy/dx + y = −x, IF = eˣ. So y eˣ = −∫x eˣ dx = −(x − 1)eˣ + c, giving y = 1 − x + c e^(−x).
4. General solution: (2y − x² − c)(y + x − 1 − c e^(−x)) = 0.

परीक्षेसाठी टीप

When you see y = px + f(p), write the general solution in one line and spend your effort on the singular solution. In "solvable for p" problems, check your factorisation by expanding before solving.

थोडक्यात सारांश

When three vectors are multiplied, we get either a scalar (scalar triple product) or a vector (vector triple product). The scalar triple product gives the volume of a parallelepiped and tests whether vectors are coplanar. The vector triple product has a useful expansion formula. Products of four vectors and reciprocal vectors are built from these.

महत्त्वाचे मुद्दे

• Scalar triple product: [a b c] = a · (b × c). For a = a₁i + a₂j + a₃k etc., it equals the 3 × 3 determinant with rows (a₁, a₂, a₃), (b₁, b₂, b₃), (c₁, c₂, c₃).
• Cyclic order does not change it: [a b c] = [b c a] = [c a b]; swapping two vectors changes the sign. Dot and cross can be interchanged: a · (b × c) = (a × b) · c.
• |[a b c]| = volume of the parallelepiped with edges a, b, c. Volume of the tetrahedron = (1/6)|[a b c]|.
• Three vectors are coplanar iff [a b c] = 0. If two of the three vectors are equal, the product is 0.
• Vector triple product: a × (b × c) = (a · c)b − (a · b)c. It lies in the plane of b and c.
• In general a × (b × c) ≠ (a × b) × c, so the cross product is not associative.
• Four vectors: (a × b) · (c × d) = (a · c)(b · d) − (a · d)(b · c) (Lagrange's identity); (a × b) × (c × d) = [a b d]c − [a b c]d.
• Reciprocal vectors: a′ = (b × c)/[a b c], b′ = (c × a)/[a b c], c′ = (a × b)/[a b c]; then a · a′ = 1 and a · b′ = 0.

महत्त्वाच्या संज्ञा

• Box product – another name for [a b c].
• Coplanar vectors (समतलीय सदिश) – vectors lying in the same plane.
• Parallelepiped – a solid with six parallelogram faces.

सूत्रे

• [a b c] = det[a₁ a₂ a₃; b₁ b₂ b₃; c₁ c₂ c₃]
• a × (b × c) = (a · c)b − (a · b)c
• [a + b, b + c, c + a] = 2[a b c]

५ महत्त्वाचे प्रश्न

Q1. Show that [a b c] = 0 if a, b, c are coplanar.

b × c is perpendicular to the plane of b and c. If a lies in the same plane, a is perpendicular to b × c, so a · (b × c) = 0.

Q2. State the formula for the vector triple product.

a × (b × c) = (a · c)b − (a · b)c. Memory aid: "outer dot far, times near, minus outer dot near, times far".

Q3. Find the volume of the parallelepiped with edges a = 2i − 3j + 4k, b = i + 2j − k, c = 3i − j + 2k.

1. [a b c] = det[2 −3 4; 1 2 −1; 3 −1 2].
2. = 2(2·2 − (−1)(−1)) − (−3)(1·2 − (−1)·3) + 4(1·(−1) − 2·3).
3. = 2(4 − 1) + 3(2 + 3) + 4(−1 − 6) = 6 + 15 − 28 = −7.
4. Volume = |−7| = 7 cubic units.

Q4. Find λ so that a = 2i − j + k, b = i + 2j − 3k, c = 3i + λj + 5k are coplanar.

1. Need det[2 −1 1; 1 2 −3; 3 λ 5] = 0.
2. = 2(10 + 3λ) + 1(5 + 9) + 1(λ − 6) = 20 + 6λ + 14 + λ − 6 = 7λ + 28.
3. 7λ + 28 = 0 gives λ = −4.

Q5. For a = i − 2j + k, b = 2i + j + k, c = i + 2j − k, find a × (b × c) and verify.

1. a · c = 1 − 4 − 1 = −4; a · b = 2 − 2 + 1 = 1.
2. Formula: (−4)b − (1)c = (−8i − 4j − 4k) − (i + 2j − k) = −9i − 6j − 3k.
3. Direct: b × c = i(1·(−1) − 1·2) − j(2·(−1) − 1·1) + k(2·2 − 1·1) = −3i + 3j + 3k.
4. a × (−3i + 3j + 3k) = i(−2·3 − 1·3) − j(1·3 − 1·(−3)) + k(1·3 − (−2)(−3)) = −9i − 6j − 3k ✓.

परीक्षेसाठी टीप

Write the determinant neatly and expand along the first row with signs + − +. For coplanarity questions, set the determinant to zero; for volume, take the absolute value.

थोडक्यात सारांश

Vector differentiation deals with vectors that change with a variable (such as time) and with fields that vary in space. The del operator ∇ gives three key ideas: gradient of a scalar field, and divergence and curl of a vector field. These are basic tools in physics (fluid flow, electricity, magnetism).

महत्त्वाचे मुद्दे

• If r(t) = x(t)i + y(t)j + z(t)k, then dr/dt = (dx/dt)i + (dy/dt)j + (dz/dt)k. For a moving particle, velocity v = dr/dt, acceleration a = d²r/dt².
• A vector of constant magnitude is perpendicular to its derivative.
• Del operator: ∇ = i ∂/∂x + j ∂/∂y + k ∂/∂z.
• Gradient (प्रवणता) of scalar φ: ∇φ, a vector normal to the level surface φ = c, pointing in the direction of fastest increase.
• Directional derivative of φ along unit vector â is ∇φ · â. Its maximum value is |∇φ|.
• Divergence of F: ∇·F, a scalar; measures the net outflow per unit volume. If ∇·F = 0, F is solenoidal.
• Curl of F: ∇ × F, a vector; measures rotation. If ∇ × F = 0, F is irrotational (and F = ∇φ for some scalar potential φ).
• Important identities: curl(grad φ) = 0; div(curl F) = 0.
• ∇·r = 3 and ∇ × r = 0 for the position vector r = xi + yj + zk; ∇r = r/r (r = |r|).
• Laplacian: ∇²φ = φₓₓ + φᵧᵧ + φ_zz.

महत्त्वाच्या संज्ञा

• Scalar field – a number at each point (temperature).
• Vector field – a vector at each point (velocity of a fluid).
• Level surface – surface on which φ has a constant value.
• Unit normal – n̂ = ∇φ/|∇φ|.

सूत्रे

• ∇φ = φₓ i + φᵧ j + φ_z k
• ∇·F = ∂F₁/∂x + ∂F₂/∂y + ∂F₃/∂z
• ∇ × F = i(∂F₃/∂y − ∂F₂/∂z) − j(∂F₃/∂x − ∂F₁/∂z) + k(∂F₂/∂x − ∂F₁/∂y)
• Directional derivative = ∇φ · â
• ∇·(φF) = φ(∇·F) + F·∇φ; ∇ × (φF) = φ(∇ × F) + ∇φ × F

५ महत्त्वाचे प्रश्न

Q1. Define solenoidal and irrotational vector fields.

F is solenoidal if div F = 0 everywhere (no sources or sinks). F is irrotational if curl F = 0 everywhere; then F can be written as the gradient of a scalar potential.

Q2. Find the unit normal to the surface x² + y² + z² = 9 at (2, 1, 2).

φ = x² + y² + z²; ∇φ = 2xi + 2yj + 2zk = 4i + 2j + 4k at the point. |∇φ| = √(16 + 4 + 16) = 6. Unit normal = (2i + j + 2k)/3.

Q3. Find the directional derivative of φ = x²y + yz² at (1, 1, 1) in the direction of i + 2j + 2k.

1. φₓ = 2xy = 2; φᵧ = x² + z² = 2; φ_z = 2yz = 2 at (1, 1, 1).
2. ∇φ = 2i + 2j + 2k.
3. Unit vector â = (i + 2j + 2k)/√(1 + 4 + 4) = (i + 2j + 2k)/3.
4. Directional derivative = ∇φ · â = (2 + 4 + 4)/3 = 10/3.
5. Maximum directional derivative = |∇φ| = 2√3.

Q4. Show that F = yz i + zx j + xy k is both solenoidal and irrotational.

1. div F = ∂(yz)/∂x + ∂(zx)/∂y + ∂(xy)/∂z = 0 + 0 + 0 = 0 → solenoidal.
2. curl F, i-component: ∂(xy)/∂y − ∂(zx)/∂z = x − x = 0.
3. j-component: −[∂(xy)/∂x − ∂(yz)/∂z] = −(y − y) = 0.
4. k-component: ∂(zx)/∂x − ∂(yz)/∂y = z − z = 0.
5. curl F = 0 → irrotational. (Its potential is φ = xyz.)

Q5. Prove that curl(grad φ) = 0.

1. grad φ = φₓ i + φᵧ j + φ_z k.
2. i-component of curl: ∂(φ_z)/∂y − ∂(φᵧ)/∂z = φ_zy − φᵧ_z = 0.
3. Similarly the j and k components are φ_zx − φₓ_z = 0 and φᵧₓ − φₓᵧ = 0.
4. (Mixed partial derivatives are equal for smooth φ.)
5. Hence curl(grad φ) = 0.

परीक्षेसाठी टीप

Write curl as a 3 × 3 determinant with i, j, k in the first row and ∂/∂x, ∂/∂y, ∂/∂z in the second — it avoids sign mistakes. Always reduce the direction to a unit vector first.

थोडक्यात सारांश

Vector integration extends integration to curves, surfaces and volumes. Line integrals give work done by a force; surface integrals give flux through a surface. Three great theorems — Green's, Stokes' and Gauss's divergence theorem — connect these integrals to each other.

महत्त्वाचे मुद्दे

• Line integral ∫_C F·dr = ∫(F₁ dx + F₂ dy + F₃ dz) along curve C. If F is a force, it gives the work done.
• If F is conservative (F = ∇φ, curl F = 0), the line integral depends only on the end points, and around a closed curve it is zero.
• Surface integral ∬_S F·n̂ dS gives the flux of F across surface S.
• Volume integral ∭_V φ dV.
• Green's theorem (plane): relates a line integral round a closed curve C to a double integral over the region R it encloses.
• Stokes' theorem: the line integral of F round the boundary of an open surface equals the surface integral of curl F over that surface.
• Gauss's divergence theorem: the outward flux of F through a closed surface equals the volume integral of div F over the volume inside.
• Green's theorem is the plane (2-D) case of Stokes' theorem.
• Curve C is taken anticlockwise (positive direction) in Green's theorem.

महत्त्वाच्या संज्ञा

• Closed curve – starts and ends at the same point (∮).
• Flux (अभिवाह) – amount of field passing through a surface.
• Outward normal – unit normal pointing out of a closed surface.
• Simply connected region – region without holes.

सूत्रे

• Work W = ∫_C F·dr
• Green: ∮_C (M dx + N dy) = ∬_R (∂N/∂x − ∂M/∂y) dx dy
• Area by Green: A = ½∮(x dy − y dx)
• Stokes: ∮_C F·dr = ∬_S (∇ × F)·n̂ dS
• Gauss: ∯_S F·n̂ dS = ∭_V (∇·F) dV

५ महत्त्वाचे प्रश्न

Q1. State Gauss's divergence theorem.

If V is a volume bounded by a closed surface S and F has continuous first derivatives, then ∯_S F·n̂ dS = ∭_V div F dV, where n̂ is the outward unit normal.

Q2. When is a line integral independent of path?

When F is conservative, i.e. F = ∇φ for some scalar φ (equivalently curl F = 0 in a simply connected region). Then ∫ from A to B of F·dr = φ(B) − φ(A).

Q3. Evaluate ∫_C F·dr for F = x² i + xy j along the line y = x from (0, 0) to (1, 1).

1. F·dr = x² dx + xy dy.
2. On y = x: dy = dx, xy = x².
3. Integral = ∫ from 0 to 1 of (x² + x²) dx = ∫₀¹ 2x² dx.
4. = 2[x³/3]₀¹ = 2/3.

Q4. Using Green's theorem, evaluate ∮_C (−y dx + x dy) where C is the circle x² + y² = a².

1. M = −y, N = x.
2. ∂N/∂x − ∂M/∂y = 1 − (−1) = 2.
3. ∮ = ∬_R 2 dx dy = 2 × (area of circle).
4. = 2πa².
5. (This also shows area = ½∮(x dy − y dx) = πa².)

Q5. Using the divergence theorem, find the flux of F = x i + y j + z k over the sphere x² + y² + z² = a².

1. div F = 1 + 1 + 1 = 3.
2. Flux = ∭_V 3 dV = 3 × volume of sphere.
3. = 3 × (4/3)πa³ = 4πa³.
4. (Direct check: on the sphere F·n̂ = a, so flux = a × 4πa² = 4πa³ ✓.)

परीक्षेसाठी टीप

For "verify" questions, do both sides separately and show they are equal. For "evaluate using the theorem", convert to the easier side — usually the double or volume integral.