Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
How to Use This Page
This page provides chapter-wise exercise solutions for CBSE Class 12 Mathematics Part 1 (NCERT). Each chapter in Part 1 — Relations and Functions, Inverse Trigonometric Functions, Matrices, Determinants, and Continuity and Differentiability — is covered with concept recaps, worked examples, and exam-style questions.
Before diving into the solutions, attempt every question on your own. Use these solutions to verify your method, check your steps, and understand the expected answer format. The board examiner awards marks for correct steps, not just the final answer — so pay close attention to how each solution is structured.
Focus on understanding why a particular method is used. For example, when proving a relation is an equivalence relation, the three properties (reflexive, symmetric, transitive) must each be explicitly verified. When finding the inverse of a matrix using the adjoint method, the determinant must be non-zero — stating this is part of the solution. These small details carry marks in the board exam.
Use the Key Concepts section for quick revision before exams. The MCQ, fill-in-the-blank, and true/false sections test factual recall. The short answer and detailed answer sections mirror the 2-mark and 4/6-mark questions you will face in the exam. Practice writing solutions within the suggested time limits: 1 minute per 1 mark.
Key Concepts (Quick Recap)
Chapter 1: Relations and Functions
- Relation: A subset R of A x A. Key types:
- Reflexive: (a, a) belongs to R for every a in A
- Symmetric: (a, b) in R implies (b, a) in R
- Transitive: (a, b) in R and (b, c) in R implies (a, c) in R
- Equivalence: A relation that is reflexive, symmetric, and transitive
- Equivalence class of element a: [a] = {x in A : (x, a) in R}
- Functions:
- One-one (Injective): f(a) = f(b) implies a = b
- Onto (Surjective): For every b in B, there exists a in A such that f(a) = b
- Bijective: Both one-one and onto
- Composition: (f o g)(x) = f(g(x)); note: f o g is not equal to g o f in general
- Inverse of a bijective function f: f^(-1)(y) = x if and only if f(x) = y
- Binary operation * on A: A x A -> A; must satisfy closure
Chapter 2: Inverse Trigonometric Functions
- Principal value branches (restricted domains for invertibility):
| Function | Domain (Principal Value Branch) | Range | |
|---|---|---|---|
| --- | --- | --- | |
| sin^(-1) x | [-1, 1] | [-pi/2, pi/2] | |
| cos^(-1) x | [-1, 1] | [0, pi] | |
| tan^(-1) x | R | (-pi/2, pi/2) | |
| cosec^(-1) x | R - (-1, 1) | [-pi/2, pi/2] - {0} | |
| sec^(-1) x | R - (-1, 1) | [0, pi] - {pi/2} | |
| cot^(-1) x | R | (0, pi) |
- Key properties:
- sin^(-1) x + cos^(-1) x = pi/2, for x in [-1, 1]
- tan^(-1) x + cot^(-1) x = pi/2, for x in R
- sec^(-1) x + cosec^(-1) x = pi/2, for |x| >= 1
- sin^(-1)(-x) = -sin^(-1) x; cos^(-1)(-x) = pi - cos^(-1) x
- tan^(-1)(-x) = -tan^(-1) x; cot^(-1)(-x) = pi - cot^(-1) x
- tan^(-1) x + tan^(-1) y = tan^(-1)((x + y)/(1 - xy)), if xy < 1
- tan^(-1) x - tan^(-1) y = tan^(-1)((x - y)/(1 + xy)), if xy > -1
- 2 tan^(-1) x = sin^(-1)(2x/(1 + x^2)) = cos^(-1)((1 - x^2)/(1 + x^2)), if |x| <= 1
Chapter 3: Matrices
- Types: Row, Column, Zero, Identity (I), Diagonal, Scalar, Symmetric (A^T = A), Skew-symmetric (A^T = -A)
- Matrix addition/subtraction: Same order; element-wise
- Scalar multiplication: kA = multiply each element by k
- Matrix multiplication: A_{m x n} times B_{n x p} = C_{m x p}; C_{ij} = sum of A_{ik} * B_{kj}
- Transpose: (A^T)^T = A; (kA)^T = kA^T; (A + B)^T = A^T + B^T; (AB)^T = B^T A^T
- Symmetric: A = A^T; Skew-symmetric: A = -A^T
- Every square matrix A can be written as A = (A + A^T)/2 + (A - A^T)/2 (sum of symmetric and skew-symmetric parts)
- Invertible matrix: A^(-1) exists if and only if |A| is not equal to 0; A^(-1) = (1/|A|) adj A
- (AB)^(-1) = B^(-1) A^(-1)
Chapter 4: Determinants
- Determinant of 2x2: |A| = a_{11} a_{22} - a_{12} a_{21}
- Determinant of 3x3 (expansion along row i): |A| = sum of a_{ij} C_{ij} where C_{ij} = (-1)^{i+j} M_{ij}
- Properties:
- |A^T| = |A|
- If any two rows/columns are identical, |A| = 0
- If any row/column is all zeros, |A| = 0
- |kA| = k^n |A| for n x n matrix
- |AB| = |A| |B|
- If A is invertible, |A^(-1)| = 1/|A|
- Area of triangle with vertices (x1,y1), (x2,y2), (x3,y3): Area = (1/2) |det|
- Minor M_{ij}: determinant of submatrix obtained by deleting i-th row and j-th column
- Cofactor C_{ij} = (-1)^{i+j} * M_{ij}
- Adjoint: adj A = transpose of cofactor matrix; A(adj A) = (adj A)A = |A| I
- Inverse: A^(-1) = (adj A)/|A|, provided |A| is not equal to 0
- Cramer's Rule: For AX = B, x_i = |A_i|/|A| where A_i is A with i-th column replaced by B
- Consistency of system AX = B:
- If |A| is not equal to 0: unique solution (consistent)
- If |A| = 0 and (adj A)B = O: infinitely many solutions or no solution
- If |A| = 0 and (adj A)B is not equal to O: no solution (inconsistent)
Chapter 5: Continuity and Differentiability
- Continuity at x = a: lim_{x->a} f(x) exists and equals f(a); i.e., LHL = RHL = f(a)
- Differentiability at x = a: lim_{h->0} [f(a+h) - f(a)]/h exists (left-hand derivative = right-hand derivative)
- If f is differentiable at a, then f is continuous at a (converse is NOT true — e.g., |x| at x = 0)
- Chain rule: d/dx[f(g(x))] = f'(g(x)) * g'(x)
- Implicit differentiation: Differentiate both sides w.r.t. x, then solve for dy/dx
- Logarithmic differentiation: y = u^v, take log both sides: ln y = v ln u, then differentiate
- Derivatives of inverse trig functions:
- d/dx[sin^(-1) x] = 1/sqrt(1 - x^2)
- d/dx[cos^(-1) x] = -1/sqrt(1 - x^2)
- d/dx[tan^(-1) x] = 1/(1 + x^2)
- d/dx[cot^(-1) x] = -1/(1 + x^2)
- d/dx[sec^(-1) x] = 1/(|x| sqrt(x^2 - 1))
- d/dx[cosec^(-1) x] = -1/(|x| sqrt(x^2 - 1))
- Rolle's Theorem: If f is continuous on [a,b], differentiable on (a,b), and f(a) = f(b), then there exists c in (a,b) such that f'(c) = 0
- Mean Value Theorem (Lagrange): If f is continuous on [a,b] and differentiable on (a,b), then there exists c in (a,b) such that f'(c) = [f(b) - f(a)]/(b - a)
I. Choose the Best Answer (MCQs)
1. If R is a relation on the set A = {1, 2, 3} given by R = {(1, 1), (2, 2), (1, 2), (2, 1)}, then R is:
(a) Reflexive and symmetric only (b) Reflexive and transitive only (c) Symmetric and transitive only (d) Reflexive, symmetric, and transitive
2. The principal value of sin^(-1)(-1/2) is:
(a) pi/6 (b) -pi/6 (c) 5pi/6 (d) -5pi/6
| 3. If A is a 2 x 2 matrix such that | A | = 5, then | 3A | equals: |
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(a) 5 (b) 15 (c) 45 (d) 25
| Answer: (c) — For a 2 x 2 matrix, | kA | = k^2 | A | . So | 3A | = 9 * 5 = 45. |
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4. If A is a square matrix such that A^2 = A, then (I + A)^3 - 7A equals:
(a) I (b) A (c) I - A (d) I + A
5. The value of tan^(-1)(1) + tan^(-1)(2) + tan^(-1)(3) is:
(a) pi (b) pi/2 (c) 0 (d) pi/4
| 6. If f(x) = | x - 2 | , then f is: |
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(a) Continuous and differentiable everywhere (b) Continuous everywhere but not differentiable at x = 2 (c) Not continuous at x = 2 (d) Differentiable everywhere but not continuous at x = 2
| Answer: (b) — | x - 2 | is continuous for all x, but at x = 2 the left-hand derivative is -1 and the right-hand derivative is +1, so it is not differentiable at x = 2. |
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| 7. The derivative of sin^(-1)(2x / (1 + x^2)) with respect to x, where | x | < 1, is: |
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(a) 2/(1 + x^2) (b) 2/(1 - x^2) (c) 1/(1 + x^2) (d) -2/(1 + x^2)
| 8. If A is a 3 x 3 matrix and | A | = 4, then | adj A | equals: |
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(a) 4 (b) 16 (c) 64 (d) 1/4
| Answer: (b) — | adj A | = | A | ^{n-1} = | A | ^2 = 16 (for n = 3). |
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9. If the function f(x) = {(x^2 - 4)/(x - 2), x is not equal to 2; k, x = 2} is continuous at x = 2, then k equals:
(a) 0 (b) 2 (c) 4 (d) 8
10. Which of the following is NOT a property of determinants?
| (a) | A^T | = | A | |||
|---|---|---|---|---|---|---|
| (b) | AB | = | A | B | ||
| (c) | A + B | = | A | + | B | |
| (d) If two rows are identical, | A | = 0 |
| Answer: (c) — | A + B | is NOT equal to | A | + | B | in general. This is a common misconception. |
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II. Fill in the Blanks
1. A relation R on a set A is called an equivalence relation if it is reflexive, symmetric, and transitive.
2. The principal value of cos^(-1)(1/2) is pi/3.
3. If A and B are invertible matrices of same order, then (AB)^(-1) = B^(-1) A^(-1).
| 4. The value of the determinant | {1, 2, 3}, {0, 4, 5}, {0, 0, 6} | is 24 — upper triangular determinant equals product of diagonal elements: 1 4 6 = 24. |
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5. If f(x) is continuous on [a, b] and differentiable on (a, b) with f(a) = f(b), then by Rolle's Theorem there exists c in (a, b) such that f'(c) = 0.
6. The derivative of e^(sin x) with respect to x is *e^(sin x) cos x**.
7. A square matrix A is said to be skew-symmetric if A^T = -A.
8. If y = log(cos x), then dy/dx = -tan x.
9. The area of the triangle with vertices (1, 0), (6, 0), and (4, 3) is 15/2 or 7.5 square units.
| 10. For the system of equations AX = B to have a unique solution, the determinant | A | must be non-zero (not equal to 0). |
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III. True or False
| 1. False — Every differentiable function is continuous, but every continuous function is NOT necessarily differentiable. (Correction: | x | is continuous at x = 0 but not differentiable there.) |
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| 5. False — If | A | = 0, it does not necessarily mean A = O (zero matrix). (Correction: A singular matrix ( | A | = 0) can be a non-zero matrix, e.g., A = [{1, 2}, {2, 4}] has | A | = 0 but A is not zero.) |
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| 6. True — If f(x) = | x | , then f is continuous at x = 0 but not differentiable at x = 0. |
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IV. Match the Following
| Column A | Column B | ||||
|---|---|---|---|---|---|
| (1) sin^(-1) x + cos^(-1) x | (e) pi/2 | ||||
| (2) | adj A | for a 3 x 3 matrix with \ | A\ | = k | (f) k^2 |
| (3) d/dx[log x] | (a) 1/x | ||||
| (4) A matrix with A^T = -A | (d) Skew-symmetric | ||||
| (5) tan^(-1)(1) + tan^(-1)(sqrt(3)) | (b) 5pi/12 | ||||
| (6) \ | kA\ | for n x n matrix A | (c) k^n \ | A\ |
Matches: 1-e, 2-f, 3-a, 4-d, 5-b, 6-c
Explanation for (5): tan^(-1)(1) = pi/4, tan^(-1)(sqrt(3)) = pi/3. Since sqrt(3) * 1 = sqrt(3) < 1 is false (sqrt(3) > 1), but we compute directly: pi/4 + pi/3 = 3pi/12 + 4pi/12 = 7pi/12. Wait — let me recalculate. tan^(-1)(1) = pi/4, tan^(-1)(sqrt(3)) = pi/3. Sum = pi/4 + pi/3 = (3pi + 4pi)/12 = 7pi/12. The match value should be 7pi/12. Revised match: (5) - (g) 7pi/12.
V. Answer Briefly
| Q1. Show that the relation R on the set A = {1, 2, 3} given by R = {(a, b) : | a - b | is divisible by 3} is an equivalence relation. |
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R = {(a, b) : |a - b| is divisible by 3} Reflexive: For any a in A, |a - a| = 0 which is divisible by 3 (0 = 3 * 0). So (a, a) belongs to R. R is reflexive. Symmetric: If (a, b) belongs to R, then |a - b| is divisible by 3. Since |a - b| = |b - a|, we get |b - a| is divisible by 3. So (b, a) belongs to R. R is symmetric. Transitive: If (a, b) and (b, c) belong to R, then |a - b| and |b - c| are divisible by 3. Now |a - c| = |a - b + b - c| which is divisible by 3 (since both |a - b| and |b - c| are divisible by 3, and the sum of two multiples of 3 is a multiple of 3). So (a, c) belongs to R. R is transitive. Since R is reflexive, symmetric, and transitive, R is an equivalence relation.
Q2. Find the principal value of tan^(-1)(-1) + cot^(-1)(-1).
We know tan^(-1) x + cot^(-1) x = pi/2 for all x in R. So tan^(-1)(-1) + cot^(-1)(-1) = pi/2. Verification: tan^(-1)(-1) = -pi/4 (since tan(-pi/4) = -1 and -pi/4 is in (-pi/2, pi/2)). cot^(-1)(-1) = 3pi/4 (since cot(3pi/4) = -1 and 3pi/4 is in (0, pi)). Sum = -pi/4 + 3pi/4 = 2pi/4 = pi/2. Confirmed.
Q3. If A = [{1, 2}, {3, 4}], find A^2 - 5A + 2I.
A^2 = A * A = [{1+6, 2+8}, {3+12, 6+16}] = [{7, 10}, {15, 22}] 5A = [{5, 10}, {15, 20}] 2I = [{2, 0}, {0, 2}] A^2 - 5A + 2I = [{7-5+2, 10-10+0}, {15-15+0, 22-20+2}] = [{4, 0}, {0, 4}] = 4I This verifies the Cayley-Hamilton result: The characteristic equation is lambda^2 - 5lambda + 2 = 0 is NOT the right one — actually we compute: trace = 1 + 4 = 5, det = 4 - 6 = -2. Characteristic equation: lambda^2 - 5lambda - 2 = 0 would give A^2 - 5A - 2I = O. But A^2 - 5A = [{2, 0}, {0, 2}] = 2I, so A^2 - 5A - 2I = O. So A^2 - 5A + 2I = 4I.
Q4. Using determinants, find the area of the triangle with vertices (2, 3), (-1, 0), and (2, -4).
Area = (1/2) |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| = (1/2) |2(0 - (-4)) + (-1)((-4) - 3) + 2(3 - 0)| = (1/2) |2(4) + (-1)(-7) + 2(3)| = (1/2) |8 + 7 + 6| = (1/2)(21) = 21/2 = 10.5 square units
| Q5. Show that f(x) = | x - 1 | is continuous but not differentiable at x = 1. |
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Continuity: f(1) = |1 - 1| = 0. LHL = lim_{x->1^-} |x - 1| = lim_{x->1^-} (1 - x) = 0. RHL = lim_{x->1^+} |x - 1| = lim_{x->1^+} (x - 1) = 0. Since LHL = RHL = f(1) = 0, f is continuous at x = 1. Differentiability: Lf'(1) = lim_{h->0^-} (f(1+h) - f(1))/h = lim_{h->0^-} (|h| - 0)/h = lim_{h->0^-} (-h)/h = -1. Rf'(1) = lim_{h->0^+} (|h| - 0)/h = lim_{h->0^+} h/h = 1. Since Lf'(1) = -1 is not equal to Rf'(1) = 1, f is not differentiable at x = 1.
Q6. Find dy/dx if x = a(cos theta + theta sin theta) and y = a(sin theta - theta cos theta).
dx/dtheta = a(-sin theta + sin theta + theta cos theta) = a theta cos theta dy/dtheta = a(cos theta - cos theta + theta sin theta) = a theta sin theta dy/dx = (dy/dtheta)/(dx/dtheta) = (a theta sin theta)/(a theta cos theta) = tan theta
Q7. If y = (sin^(-1) x)^2, find d^2y/dx^2.
y = (sin^(-1) x)^2 dy/dx = 2 sin^(-1) x (1/sqrt(1 - x^2)) d^2y/dx^2 = 2 [(1/sqrt(1 - x^2)) (1/sqrt(1 - x^2)) + sin^(-1) x d/dx(1/sqrt(1 - x^2))] = 2 [1/(1 - x^2) + sin^(-1) x (2x)/((1 - x^2)^{3/2})] = 2/(1 - x^2) + (2x sin^(-1) x)/((1 - x^2)^{3/2}) = *2(1 - x^2 + x sin^(-1) x sqrt(1 - x^2))/((1 - x^2)^{3/2})** — more commonly written as: d^2y/dx^2 = 2/(1 - x^2) + (2x sin^(-1) x)/((1 - x^2)^{3/2})
VI. Answer in Detail
Q1. Prove that the relation R on the set of integers Z defined by R = {(a, b) : a - b is divisible by 3} is an equivalence relation. Also find the equivalence class of 0, 1, and 2.
| Q2. Solve: tan^(-1)(2x/(1 - x^2)) + tan^(-1)(2y/(1 - y^2)) = pi/4, given that | x | < 1 and | y | < 1. |
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Let x = tan(alpha) and y = tan(beta), where |alpha| < pi/4 and |beta| < pi/4 (since |x| < 1). Then 2x/(1 - x^2) = 2tan(alpha)/(1 - tan^2(alpha)) = tan(2alpha) (double angle formula). Similarly, 2y/(1 - y^2) = tan(2beta). So the equation becomes: tan^(-1)(tan(2alpha)) + tan^(-1)(tan(2beta)) = pi/4 Since |x| < 1, we have |alpha| < pi/4, so |2alpha| < pi/2, meaning 2alpha is in the principal value range. Hence tan^(-1)(tan(2alpha)) = 2alpha. Similarly, tan^(-1)(tan(2beta)) = 2beta. Therefore: 2alpha + 2beta = pi/4 tan^(-1) x + tan^(-1) y = pi/8 Since xy < 1 (because |x| < 1 and |y| < 1), we can write: tan^(-1) x + tan^(-1) y = tan^(-1)((x + y)/(1 - xy)) = pi/8 Therefore: (x + y)/(1 - xy) = tan(pi/8) = sqrt(2) - 1 So x + y = (sqrt(2) - 1)(1 - xy) is the required relation between x and y.
Q3. If A = [{3, -4}, {1, -1}], verify that A^2 - 2A + I = O. Hence find A^(-1).
A = [{3, -4}, {1, -1}] A^2 = [{33 + (-4)1, 3(-4) + (-4)(-1)}, {13 + (-1)1, 1(-4) + (-1)(-1)}] = [{9 - 4, -12 + 4}, {3 - 1, -4 + 1}] = [{5, -8}, {2, -3}] 2A = [{6, -8}, {2, -2}] I = [{1, 0}, {0, 1}] A^2 - 2A + I = [{5-6+1, -8-(-8)+0}, {2-2+0, -3-(-2)+1}] = [{0, 0}, {0, 0}] = O. Verified. Since A^2 - 2A + I = O, we get A^2 = 2A - I. Multiplying both sides by A^(-1): A = 2I - A^(-1) So A^(-1) = 2I - A = [{2-3, 0-(-4)}, {0-1, 2-(-1)}] = [{-1, 4}, {-1, 3}] A^(-1) = [{-1, 4}, {-1, 3}] Verification: A A^(-1) = [{3(-1) + (-4)(-1), 34 + (-4)3}, {1(-1) + (-1)(-1), 14 + (-1)*3}] = [{-3+4, 12-12}, {-1+1, 4-3}] = [{1, 0}, {0, 1}] = I. Confirmed.
Q4. Using Cramer's rule, solve the system of equations: x + y + z = 6, x - y + z = 2, x + 2y + 3z = 14.
The system can be written as AX = B where: A = [{1, 1, 1}, {1, -1, 1}, {1, 2, 3}], X = [{x}, {y}, {z}], B = [{6}, {2}, {14}] Step 1: Find |A| |A| = 1(-3 - 2) - 1(3 - 1) + 1(2 - (-1)) = -5 - 2 + 3 = -4 Since |A| = -4 is not equal to 0, the system has a unique solution. Step 2: Find |A1| (replace column 1 with B) |A1| = |{6, 1, 1}, {2, -1, 1}, {14, 2, 3}| = 6(-3 - 2) - 1(6 - 14) + 1(4 - (-14)) = 6(-5) - 1(-8) + 1(18) = -30 + 8 + 18 = -4 Step 3: Find |A2| (replace column 2 with B) |A2| = |{1, 6, 1}, {1, 2, 1}, {1, 14, 3}| = 1(6 - 14) - 6(3 - 1) + 1(14 - 2) = -8 - 12 + 12 = -8 Step 4: Find |A3| (replace column 3 with B) |A3| = |{1, 1, 6}, {1, -1, 2}, {1, 2, 14}| = 1(-14 - 4) - 1(14 - 2) + 6(2 - (-1)) = -18 - 12 + 18 = -12 Step 5: Apply Cramer's rule x = |A1|/|A| = -4/(-4) = 1 y = |A2|/|A| = -8/(-4) = 2 z = |A3|/|A| = -12/(-4) = 3 Solution: x = 1, y = 2, z = 3 Verification: 1 + 2 + 3 = 6 (satisfied), 1 - 2 + 3 = 2 (satisfied), 1 + 4 + 9 = 14 (satisfied).
| Q5. Prove that the function f(x) = { | x | , x is not equal to 0; 0, x = 0} is continuous at x = 0 but not differentiable at x = 0. Also, discuss the differentiability of f(x) = x * | x | at x = 0. |
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Part (i): f(x) = |x| Continuity at x = 0: f(0) = 0 LHL = lim_{x->0^-} |x| = lim_{x->0^-} (-x) = 0 RHL = lim_{x->0^+} |x| = lim_{x->0^+} x = 0 Since LHL = RHL = f(0) = 0, f is continuous at x = 0. Differentiability at x = 0: Lf'(0) = lim_{h->0} (f(0 - h) - f(0))/(-h) = lim_{h->0} (|(-h)| - 0)/(-h) = lim_{h->0} h/(-h) = -1 Rf'(0) = lim_{h->0} (f(0 + h) - f(0))/h = lim_{h->0} (|h| - 0)/h = lim_{h->0} h/h = 1 Since Lf'(0) = -1 is not equal to Rf'(0) = 1, f is not differentiable at x = 0. *Part (ii): g(x) = x |x| g(x) = x|x| = {x^2, if x >= 0; -x^2, if x < 0} Continuity at x = 0: g(0) = 0 LHL = lim_{x->0^-} -x^2 = 0 RHL = lim_{x->0^+} x^2 = 0 So g is continuous at x = 0. Differentiability at x = 0: Lg'(0) = lim_{h->0} (g(0-h) - g(0))/(-h) = lim_{h->0} (-h^2 - 0)/(-h) = lim_{h->0} h = 0 Rg'(0) = lim_{h->0} (g(0+h) - g(0))/h = lim_{h->0} (h^2 - 0)/h = lim_{h->0} h = 0 Since Lg'(0) = Rg'(0) = 0, g is differentiable** at x = 0, and g'(0) = 0.
Q6. Differentiate (log x)^x + x^(log x) with respect to x using logarithmic differentiation.
Let u = (log x)^x and v = x^(log x). We need to find d/dx(u + v) = du/dx + dv/dx. For u = (log x)^x: Take log: ln u = x ln(log x) Differentiate: (1/u) du/dx = ln(log x) + x (1/log x) (1/x) = ln(log x) + 1/log x du/dx = (log x)^x [ln(log x) + 1/log x] For v = x^(log x): Take log: ln v = (log x) (ln x) — note: log x means natural log here, or if log means log base 10, adjust accordingly. In NCERT convention, log x = log_e x = ln x. ln v = (ln x)(ln x) = (ln x)^2 Differentiate: (1/v) dv/dx = 2(ln x) (1/x) dv/dx = x^(ln x) 2(ln x)/x Final answer: d/dx[(log x)^x + x^(log x)] = (log x)^x [ln(log x) + 1/log x] + x^(log x) 2(log x)/x*
Common Mistakes to Avoid
1. Forgetting to check all three properties for equivalence relations — You must verify reflexive, symmetric, AND transitive separately. Skipping any one costs marks. Also, check the property for ALL elements of the set, not just a few.
2. Using wrong principal value branch for inverse trigonometric functions — Always remember sin^(-1) has range [-pi/2, pi/2], cos^(-1) has range [0, pi], tan^(-1) has range (-pi/2, pi/2). Writing sin^(-1)(-1/2) = 5pi/6 is wrong; the correct principal value is -pi/6.
3. Matrix multiplication is NOT commutative — AB is not equal to BA in general. Always multiply in the correct order. Also, AB = O does not imply A = O or B = O (zero divisors exist in matrices).
4. Confusing minor and cofactor — Cofactor C_{ij} = (-1)^{i+j} * M_{ij}. The sign alternates. When computing adj A, remember it is the TRANSPOSE of the cofactor matrix, not the cofactor matrix itself.
| 5. **Assuming | A + B | = | A | + | B | ** — This is FALSE. Determinant is multiplicative, not additive. The only valid properties are | AB | = | A | B | and | kA | = k^n | A | (note the power n, not 1). |
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6. Forgetting to check continuity before differentiability — If a question asks you to check differentiability, first verify continuity. A function that is not continuous at a point cannot be differentiable there. Always compute LHD and RHD separately to check differentiability.
Exam Checklist
- [ ] Can I verify all three properties (reflexive, symmetric, transitive) for an equivalence relation with complete proofs?
- [ ] Do I know the principal value branches and domains for all six inverse trigonometric functions?
- [ ] Can I correctly apply the formulas for tan^(-1) x + tan^(-1) y, including the three cases (xy < 1, xy > 1, xy = 1)?
- [ ] Can I multiply 2 x 2 and 3 x 3 matrices correctly and find the transpose, adjoint, and inverse?
- [ ] Do I remember that |kA| = k^n|A| (not k|A|) and that A(adj A) = |A|I?
- [ ] Can I expand a determinant along any row or column and correctly apply all properties (especially row/column operations)?
- [ ] Can I use Cramer's rule and check consistency of a system of linear equations?
- [ ] Can I apply the chain rule, implicit differentiation, and logarithmic differentiation correctly, and verify continuity/differentiability at a point using LHL/RHL and LHD/RHD?