Unit I: Sets and Functions

This unit forms the foundation for higher mathematics. It covers set theory, the various relations and functions between sets, and the deep study of trigonometric functions.

Chapters in this Unit:

1. Sets
2. Relations and Functions
3. Trigonometric Functions

Chapter 1: Sets

1.1 Introduction to Sets

A set is a well-defined collection of objects. These objects are called the elements or members of the set. Sets are usually denoted by capital letters: \( A, B, C, X, Y, Z \), etc. The elements of a set are represented by small letters: \( a, b, c, x, y, z \), etc.

If \( a \) is an element of set \( A \), we write \( a \in A \) (read as “\( a \) belongs to \( A \)”). If \( b \) is not an element of \( A \), we write \( b \notin A \).

Representations of Sets

1. Roster or Tabular Form: All elements of the set are listed, separated by commas, and enclosed within braces \( \{ \} \). Example: The set of all vowels in the English alphabet is \( V = \{a, e, i, o, u\} \).
2. Set-Builder Form: All elements possess a single common property which is not possessed by any element outside the set. Example: \( V = \{x : x \text{ is a vowel in English alphabet}\} \).

1.2 Types of Sets

• Empty Set: A set which does not contain any element. Denoted by \( \emptyset \) or \( \{\} \).
• Finite and Infinite Sets: A set which is empty or consists of a definite number of elements is finite, otherwise it is infinite.
• Equal Sets: Two sets \( A \) and \( B \) are equal if they have exactly the same elements. We write \( A = B \).
• Subsets: A set \( A \) is a subset of \( B \) if every element of \( A \) is also an element of \( B \). We write \( A \subseteq B \).
• Universal Set: A broad set that contains all elements under consideration, denoted by \( U \).

Subsets of Real Numbers

Important numerical sets and their standard notations:

• \( \mathbb{N} \): Set of natural numbers \( \{1, 2, 3, \ldots\} \)
• \( \mathbb{Z} \): Set of integers \( \{\ldots, -2, -1, 0, 1, 2, \ldots\} \)
• \( \mathbb{Q} \): Set of rational numbers
• \( \mathbb{R} \): Set of real numbers

Intervals as Subsets of \( \mathbb{R} \)

• Open interval: \( (a, b) = \{x : a < x < b\} \)
• Closed interval: \( [a, b] = \{x : a \le x \le b\} \)
• Semi-open intervals: \( [a, b) = \{x : a \le x < b\} \) and \( (a, b] = \{x : a < x \le b\} \)

1.3 Venn Diagrams

Most relationships between sets can be visualized using diagrams, known as Venn Diagrams. The Universal Set \( U \) is represented by a rectangle, and its subsets are represented by circles.

1.4 Operations on Sets

1. Union of Sets: The union of \( A \) and \( B \) is the set of all elements which are either in \( A \), or in \( B \), or in both. \[ A \cup B = \{x : x \in A \text{ or } x \in B\} \]
2. Intersection of Sets: The intersection is the set of elements common to both \( A \) and \( B \). \[ A \cap B = \{x : x \in A \text{ and } x \in B\} \]
3. Difference of Sets: The difference \( A - B \) is the set of elements which belong to \( A \) but not to \( B \). \[ A - B = \{x : x \in A \text{ and } x \notin B\} \]
4. Complement of a Set: Given a universal set \( U \) and a subset \( A \), the complement of \( A \) is the set of all elements of \( U \) which are not elements of \( A \). \[ A’ = \{x : x \in U \text{ and } x \notin A\} = U - A \]

Properties of Complement

• Complement Laws: \( A \cup A’ = U \) and \( A \cap A’ = \emptyset \)
• De Morgan’s Laws:
  • \( (A \cup B)’ = A’ \cap B’ \)
  • \( (A \cap B)’ = A’ \cup B’ \)

Competency-Based Questions

Question 1:

A marketing firm conducted a survey of 1000 people to find out their preferences for three different brands of coffee: A, B, and C. It was found that 450 preferred brand A, 350 preferred brand B, and 300 preferred brand C. Furthermore, 150 preferred both A and B, 100 preferred both B and C, and 120 preferred both A and C. If 80 people liked all three brands, how many people do not prefer any of the three brands? Identify this via set operations.

Answer 1:

Let \( U \) be the set of surveyed people. Thus, \( n(U) = 1000 \).

Let \( A, B, C \) represent the sets of people preferring brands A, B, and C respectively.

We are given:

\( n(A) = 450 \), \( n(B) = 350 \), \( n(C) = 300 \)

\( n(A \cap B) = 150 \)

\( n(B \cap C) = 100 \)

\( n(A \cap C) = 120 \)

\( n(A \cap B \cap C) = 80 \)

We need to find the number of people who do NOT prefer any brand, which is \( n( (A \cup B \cup C)’ ) \).

First, calculate the union of all three sets:

\[ n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C) \]

\[ n(A \cup B \cup C) = 450 + 350 + 300 - 150 - 100 - 120 + 80 \]

\[ n(A \cup B \cup C) = 1100 - 370 + 80 = 810 \]

The number of people who do not prefer any brand:

\[ n( (A \cup B \cup C)’ ) = n(U) - n(A \cup B \cup C) = 1000 - 810 = 190 \]

Conclusion: 190 people evaluated in the survey did not prefer any of the three brands.

Question 2:

Let \( U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \), \( A = \{2, 4, 6, 8\} \), and \( B = \{2, 3, 5, 7\} \).

Verify De Morgan’s Law: \( (A \cup B)’ = A’ \cap B’ \).

Answer 2:

Step 1: Find \( A \cup B \).

\( A \cup B = \{2, 3, 4, 5, 6, 7, 8\} \)

Step 2: Find \( (A \cup B)’ \).

\( (A \cup B)’ = U - (A \cup B) = \{1, 9\} \)

Step 3: Find \( A’ \) and \( B’ \).

\( A’ = U - A = \{1, 3, 5, 7, 9\} \)

\( B’ = U - B = \{1, 4, 6, 8, 9\} \)

Step 4: Find \( A’ \cap B’ \).

\( A’ \cap B’ = \{1, 9\} \)

Conclusion: Since \( (A \cup B)’ = \{1, 9\} \) and \( A’ \cap B’ = \{1, 9\} \), De Morgan’s Law \( (A \cup B)’ = A’ \cap B’ \) is verified.

Chapter 2: Relations and Functions

2.1 Cartesian Product of Sets

Let \( A \) and \( B \) be two non-empty sets. The Cartesian product \( A \times B \) is the set of all ordered pairs of elements from \( A \) and \( B \). \[ A \times B = \{ (a, b) : a \in A, b \in B \} \] If either \( A \) or \( B \) is the null set, then \( A \times B \) will also be empty set.

• Number of elements: If \( n(A) = p \) and \( n(B) = q \), then \( n(A \times B) = pq \).
• Three-dimensional product: \( A \times A \times A = \{(a, b, c) : a, b, c \in A\} \).

2.2 Relations

A relation \( R \) from a non-empty set \( A \) to a non-empty set \( B \) is a subset of the Cartesian product \( A \times B \). This subset is derived by describing a relationship between the first element and the second element of the ordered pairs in \( A \times B \).

• Domain: The set of all first elements of the ordered pairs in a relation \( R \) from set \( A \) to \( B \) is called the domain.
• Range: The set of all second elements in a relation \( R \).
• Co-domain: The whole set \( B \) is called the co-domain of the relation \( R \). Note that \( \text{Range} \subseteq \text{Co-domain} \).

Pictorial representation: Relations can be visually represented using arrow diagrams.

2.3 Functions

A relation \( f \) from a set \( A \) to a set \( B \) is said to be a function if every element of set \( A \) has one and only one image in set \( B \). If \( (a, b) \in f \), then \( f(a) = b \), where \( b \) is called the image of \( a \) under \( f \), and \( a \) is called the pre-image of \( b \) under \( f \).

Some Functions and their Graphs

• Identity Function: \( f(x) = x \)
• Constant Function: \( f(x) = c \)
• Polynomial Function: \( f(x) = a_n x^n + \dots + a_1 x + a_0 \)
• Rational Function: \( f(x) = \frac{p(x)}{q(x)} \), \( q(x) \neq 0 \)
• Modulus Function: \( f(x) = |x| = \begin{cases} x, & x \ge 0 \ -x, & x < 0 \end{cases} \)
• Signum Function: \( f(x) = \begin{cases} 1, & x > 0 \ 0, & x = 0 \ -1, & x < 0 \end{cases} \)
• Greatest Integer Function: \( f(x) = [x] \), where \( [x] \) assumes the value of the greatest integer less than or equal to \( x \).

Algebra of Real Functions

For functions \( f : X \to \mathbb{R} \) and \( g : X \to \mathbb{R} \):

• Addition: \( (f + g)(x) = f(x) + g(x) \)
• Subtraction: \( (f - g)(x) = f(x) - g(x) \)
• Multiplication: \( (fg)(x) = f(x) g(x) \)
• Quotient: \( \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} \), provided \( g(x) \neq 0 \)

Competency-Based Questions

Question 1:

An e-commerce company determines that its daily revenue \( R \) (in thousands of rupees) depends on the unit price \( p \) (in rupees) set for a specific electronic item. The relationship is modeled by the function:

\[ R(p) = -2p^2 + 400p - 15000 \]

(a) Find the domain of the price \( p \) for which the company generates a positive revenue.

(b) What is the maximum possible revenue the company can generate per day?

Answer 1:

(a) Domain for positive revenue:

We need to find the values of \( p \) such that \( R(p) > 0 \).

\[ -2p^2 + 400p - 15000 > 0 \]

Divide by -2 (and reverse the inequality):

\[ p^2 - 200p + 7500 < 0 \]

Factor the quadratic equation:

\[ (p - 50)(p - 150) < 0 \]

The roots are \( p = 50 \) and \( p = 150 \). Since this is an upward-opening parabola opening below the x-axis, the solution is the open interval between the roots.

\[ 50 < p < 150 \]

The domain of \( p \) for positive revenue is \( (50, 150) \).

(b) Maximum revenue:

The function \( R(p) = -2p^2 + 400p - 15000 \) is a downward-facing parabola (since \( a = -2 < 0 \)). The maximum value occurs at the vertex.

The p-coordinate of the vertex is:

\[ p = \frac{-b}{2a} = \frac{-400}{2(-2)} = \frac{-400}{-4} = 100 \]

Substitute \( p = 100 \) into the revenue function to find maximum revenue:

\[ R(100) = -2(100)^2 + 400(100) - 15000 \]

\[ R(100) = -20000 + 40000 - 15000 = 5000 \]

Conclusion: The maximum possible revenue is 5000 thousand rupees (or 5,000,000 rupees), achieved at unit price \( p = 100 \).

Question 2:

Determine the domain and range of the real function \( f(x) = \sqrt{9 - x^2} \).

Answer 2:

Domain:

For purely real values of \( f(x) \), the expression inside the square root must be non-negative.

\[ 9 - x^2 \ge 0 \]

\[ x^2 \le 9 \]

Taking the square root on both sides:

\[ |x| \le 3 \]

\[ -3 \le x \le 3 \]

Domain = \( [-3, 3] \).

Range:

Let \( y = \sqrt{9 - x^2} \).

Since a square root yields a principal (non-negative) value, \( y \ge 0 \).

Square both sides:

\[ y^2 = 9 - x^2 \]

\[ x^2 = 9 - y^2 \]

Since \( x^2 \ge 0 \), we must have \( 9 - y^2 \ge 0 \).

\[ y^2 \le 9 \Rightarrow -3 \le y \le 3 \]

Combining \( y \ge 0 \) and \( -3 \le y \le 3 \), we get \( 0 \le y \le 3 \).

Range = \( [0, 3] \).

Chapter 3: Trigonometric Functions

3.1 Angles

An angle is considered as the figure obtained by rotating a given ray about its endpoint. The original ray is called the initial side and the final position of the ray after rotation is called the terminal side of the angle.

• Positive angle: Anti-clockwise rotation.
• Negative angle: Clockwise rotation.

Measure of an Angle

1. Degree Measure: If a rotation from the initial side to terminal side is \( \frac{1}{360} \) of a revolution, the angle is said to have a measure of one degree (1°).
   • 1 degree = 60 minutes (1° = 60’)
   • 1 minute = 60 seconds (1’ = 60“)
2. Radian Measure: Angle subtended at the center by an arc of length 1 unit in a unit circle is said to have a measure of 1 radian.
   • If an arc of length \( l \) subtends an angle \( \theta \) radians at the center of a circle of radius \( r \), then \( \theta = \frac{l}{r} \).
3. Relation between Degree and Radian: \[ \pi \text{ radians} = 180^\circ \]

3.2 Trigonometric Functions

Let a unit circle with center at origin of the coordinate axes. The angle \( x \) subtends from point \( P(a,b) \) on the unit circle. We define:

• \( \cos x = a \)
• \( \sin x = b \)

Since \( P(a,b) \) lies on the unit circle \( x^2 + y^2 = 1 \), we get: \[ \sin^2 x + \cos^2 x = 1 \]

Domain and Range of Trigonometric Functions

• Sine & Cosine: Domain = \( \mathbb{R} \), Range = \( [-1, 1] \).
• Tangent: Domain = \( \mathbb{R} - { x : x = (2n+1)\frac{\pi}{2}, n \in \mathbb{Z} } \), Range = \( \mathbb{R} \).

Graphs of Trigonometric Functions

The periodic nature of trigonometric functions can be visualized easily:

3.3 Trigonometric Identities

Sum and difference of two angles:

1. \( \sin(x \pm y) = \sin x \cos y \pm \cos x \sin y \)
2. \( \cos(x \pm y) = \cos x \cos y \mp \sin x \sin y \)
3. \( \tan(x \pm y) = \frac{\tan x \pm \tan y}{1 \mp \tan x \tan y} \)

Double and triple angles:

• \( \sin 2x = 2 \sin x \cos x = \frac{2 \tan x}{1 + \tan^2 x} \)
• \( \cos 2x = \cos^2 x - \sin^2 x = 2 \cos^2 x - 1 = 1 - 2 \sin^2 x = \frac{1 - \tan^2 x}{1 + \tan^2 x} \)
• \( \sin 3x = 3 \sin x - 4 \sin^3 x \)

Sums into products:

• \( \sin x + \sin y = 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \)
• \( \cos x + \cos y = 2 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \)

Competency-Based Questions

Question 1:

A Ferris wheel has a radius of 20 meters and its center is 25 meters above the ground. The wheel turns at a constant speed, making one complete rotation every 4 minutes.

If a passenger starts at the lowest point of the wheel, express their height \( h \) above the ground as a function of time \( t \) (in minutes). Construct the equation of height.

Answer 1:

The motion is periodic and circular, so we will use a trigonometric function.

The wheel makes a full revolution in 4 minutes, so its period \( T = 4 \).

The angular speed \( \omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2} \) radians per minute.

The minimum height of the passenger = Center height - Radius = 25 - 20 = 5m.

The maximum height = Center height + Radius = 25 + 20 = 45m.

The amplitude of the function is the radius, \( A = 20 \).

The vertical shift is the center position, \( D = 25 \).

Since the passenger starts at the lowest point at \( t = 0 \), a negative cosine function is the best fit, as \( \cos(0) = 1 \), and we want the minimum value. Structure:

\[ h(t) = -A \cos(\omega t) + D \]

\[ h(t) = -20 \cos\left(\frac{\pi}{2} t\right) + 25 \]

Conclusion: The equation representing the passenger’s height is \( h(t) = 25 - 20 \cos\left(\frac{\pi}{2} t\right) \).

Question 2:

Prove the identity using the properties of trigonometric functions:

\[ \frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x \]

Answer 2:

We use the sum-to-product formulas for sine and cosine:

\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]

\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]

Apply these formulas to the numerator and denominator:

Numerator:

\[ \sin 5x + \sin 3x = 2 \sin\left(\frac{5x+3x}{2}\right) \cos\left(\frac{5x-3x}{2}\right) = 2 \sin 4x \cos x \]

Denominator:

\[ \cos 5x + \cos 3x = 2 \cos\left(\frac{5x+3x}{2}\right) \cos\left(\frac{5x-3x}{2}\right) = 2 \cos 4x \cos x \]

Substitute back into the expression:

\[ \frac{2 \sin 4x \cos x}{2 \cos 4x \cos x} \]

Assuming \( \cos x \neq 0 \), we can cancel \( 2 \cos x \) from both numerator and denominator:

\[ \frac{\sin 4x}{\cos 4x} = \tan 4x \]

Conclusion: Hence proved.

Unit II: Algebra

This unit covers foundational algebraic concepts including complex numbers, inequalities, permutations and combinations, the binomial theorem, and sequences and series.

Chapters in this Unit: 4. Complex Numbers and Quadratic Equations 5. Linear Inequalities 6. Permutations and Combinations 7. Binomial Theorem 8. Sequence and Series

Chapter 4: Complex Numbers and Quadratic Equations

4.1 Introduction

The extension of the real number system to the complex number system arises because there is no real number \( x \) that satisfies the equation \( x^2 + 1 = 0 \). We introduce a symbol \( i \) defined as \( i = \sqrt{-1} \), such that \( i^2 = -1 \). This allows us to define solutions for all polynomial equations.

4.2 Complex Numbers

A number of the form \( a + ib \), where \( a \) and \( b \) are real numbers, is defined to be a complex number. For the complex number \( z = a + ib \):

• \( a \) is called the real part, denoted by \( \text{Re}(z) \).
• \( b \) is called the imaginary part, denoted by \( \text{Im}(z) \).

Two complex numbers \( z_1 = a + ib \) and \( z_2 = c + id \) are equal if \( a = c \) and \( b = d \).

Algebraic Properties

1. Addition: \( (a + ib) + (c + id) = (a+c) + i(b+d) \)
2. Difference: \( (a + ib) - (c + id) = (a-c) + i(b-d) \)
3. Multiplication: \( (a + ib)(c + id) = (ac - bd) + i(ad + bc) \)
4. Division: \( \frac{z_1}{z_2} = \frac{z_1 \bar{z_2}}{z_2 \bar{z_2}} \)

Modulus and Conjugate

Let \( z = a + ib \).

• The conjugate of \( z \) is denoted by \( \bar{z} = a - ib \).
• The modulus of \( z \) is denoted by \( |z| = \sqrt{a^2 + b^2} \).

Important properties:

• \( z\bar{z} = |z|^2 \)
• \( |z_1 z_2| = |z_1| |z_2| \)
• \( \overline{z_1 \pm z_2} = \bar{z}_1 \pm \bar{z}_2 \)
• \( \overline{z_1 z_2} = \bar{z}_1 \bar{z}_2 \)

4.3 Argand Plane

A complex number \( z = x + iy \) can be represented geometrically by the unique point \( P(x, y) \) in the Cartesian plane, known as the Argand Plane or Complex Plane. The x-axis is called the Real Axis, and the y-axis is called the Imaginary Axis.

4.4 Quadratic Equations

For the quadratic equation \( ax^2 + bx + c = 0 \) with real coefficients, where \( D = b^2 - 4ac < 0 \), the roots are complex conjugate pairs: \[ x = \frac{-b \pm i\sqrt{4ac - b^2}}{2a} \]

Competency-Based Questions

Question 1:

Evaluate the following expression if \( z = 2 - \sqrt{-3} \).

Find the value of \( z^2 - 4z + 13 \).

Answer 1:

First, express \( z \) in the standard form.

\[ z = 2 - \sqrt{3}i \]

We need to find the value of \( z^2 - 4z + 13 \).

Let’s first compute \( z^2 \):

\[ z^2 = (2 - \sqrt{3}i)^2 = 2^2 - 2(2)(\sqrt{3}i) + (\sqrt{3}i)^2 \]

\[ z^2 = 4 - 4\sqrt{3}i + 3i^2 = 4 - 4\sqrt{3}i - 3 = 1 - 4\sqrt{3}i \]

Now calculate \( -4z \):

\[ -4z = -4(2 - \sqrt{3}i) = -8 + 4\sqrt{3}i \]

Substitute everything into the expression:

\[ z^2 - 4z + 13 = (1 - 4\sqrt{3}i) + (-8 + 4\sqrt{3}i) + 13 \]

The imaginary parts cancel out: \( -4\sqrt{3}i + 4\sqrt{3}i = 0 \).

Add the real parts:

\[ 1 - 8 + 13 = 6 \]

Conclusion: The value of the expression is 6.

Question 2:

An electrical circuit has an impedance \( Z_{1} = 4 + 3i \) ohms in series with another impedance \( Z_{2} = 2 - i \) ohms. Find the total impedance, and then find its modulus.

Answer 2:

For series impedances, the total impedance \( Z_{total} \) is the sum:

\[ Z_{total} = Z_{1} + Z_{2} = (4 + 3i) + (2 - i) \]

\[ Z_{total} = (4 + 2) + (3i - i) = 6 + 2i \]

Now compute the modulus:

\[ |Z_{total}| = |6 + 2i| = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \]

Conclusion: The total impedance is \( 6 + 2i \) ohms, and its magnitude is \( 2\sqrt{10} \) ohms.

Chapter 5: Linear Inequalities

5.1 Introduction

Two real numbers or two algebraic expressions related by the symbol ‘( < )’, ‘( > )’, ‘( \le )’ or ‘( \ge )’ form an inequality. Examples of linear inequalities in one variable: \( 3x < 20 \), or \( 4x + 1 \ge 5 \).

5.2 Algebraic Solutions of Linear Inequalities in One Variable

Rules for solving inequalities:

1. Addition/Subtraction: Equal numbers may be added to (or subtracted from) both sides of an inequality without affecting the sign of inequality.
2. Multiplication/Division by Positive Numbers: Both sides of an inequality can be multiplied (or divided) by the same positive number.
3. Multiplication/Division by Negative Numbers: When both sides are multiplied or divided by a negative number, then the sign of inequality is reversed.

Representation on Number Line

The solution to a linear inequality can be represented on a number line.

• A hollow circle ‘\( \circ \)’ is used if the point is excluded (for ‘\( < \)’ or ‘\( > \)’).
• A solid dot ‘\( \bullet \)’ is used if the point is included (for ‘\( \le \)’ or ‘\( \ge \)’).

Example: Represent the solution of \( 3x - 5 < x + 7 \). \[ 3x - x < 7 + 5 \] \[ 2x < 12 \] \[ x < 6 \]

Competency-Based Questions

Question 1:

To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade ‘A’ in the course.

Answer 1:

Let \( x \) be the marks Sunita must obtain in the 5th examination.

Her average marks over five exams must be at least 90.

\[ \frac{87 + 92 + 94 + 95 + x}{5} \ge 90 \]

\[ \frac{368 + x}{5} \ge 90 \]

Multiply both sides by 5:

\[ 368 + x \ge 450 \]

Subtract 368 from both sides:

\[ x \ge 450 - 368 \]

\[ x \ge 82 \]

Conclusion: Sunita must obtain at least 82 marks in her 5th examination to get a Grade ‘A’.

Question 2:

Solve the inequality: \( \frac{5 - 2x}{3} \le \frac{x}{6} - 5 \). Display the solution as an interval.

Answer 2:

Given: \( \frac{5 - 2x}{3} \le \frac{x}{6} - 5 \).

Get a common denominator on the right side:

\[ \frac{5 - 2x}{3} \le \frac{x - 30}{6} \]

Multiply both sides by 6 (since it’s positive, the inequality sign doesn’t change):

\[ 2(5 - 2x) \le x - 30 \]

\[ 10 - 4x \le x - 30 \]

Bring x terms to one side and constants to the other:

\[ -4x - x \le -30 - 10 \]

\[ -5x \le -40 \]

Divide by -5, remembering to reverse the inequality symbol:

\[ x \ge 8 \]

Conclusion: The solution in interval form is \( [8, \infty) \).

Chapter 6: Permutations and Combinations

6.1 Fundamental Principle of Counting

“If an event can occur in \( m \) different ways, following which another event can occur in \( n \) different ways, then the total number of occurrence of the events in the given order is \( m \times n \).” This is known as the Multiplication Principle.

Alternatively, the Addition Principle: “If there are two jobs such that they can be performed independently in \( m \) and \( n \) ways respectively, then either of the two jobs can be performed in \( (m + n) \) ways.”

6.2 Factorial Notation

The notation \( n! \) represents the product of first \( n \) natural numbers. \[ n! = n \times (n-1) \times (n-2) \times \dots \times 3 \times 2 \times 1 \] By convention, \( 0! = 1 \).

6.3 Permutations

A permutation is an arrangement in a definite order of a number of objects taken some or all at a time. The number of permutations of \( n \) different objects taken \( r \) at a time, where \( 0 \le r \le n \), is denoted by \( {}^nP_r \) and is given by: \[ {}^nP_r = \frac{n!}{(n - r)!} \]

Permutations when all objects are not distinct

The number of permutations of \( n \) objects, where \( p_1 \) objects are of one kind, \( p_2 \) are of second kind, …, \( p_k \) are of \( k^{\text{th}} \) kind and the rest, if any, are of different kind is: \[ \frac{n!}{p_1! p_2! \dots p_k!} \]

6.4 Combinations

A combination is a selection of some or all of a number of different objects where the order of selection is ignored. The number of combinations of \( n \) different objects taken \( r \) at a time, where \( 0 \le r \le n \), is denoted by \( {}^nC_r \) and is given by: \[ {}^nC_r = \frac{n!}{r!(n - r)!} \]

Important Properties

• \( {}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r \)
• \( {}^nC_r = {}^nC_{n-r} \) (Choosing \( r \) elements is identical to rejecting \( n-r \) elements)

Competency-Based Questions

Question 1:

A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of exactly 1 man and 2 women?

Answer 1:

Total number of ways to form the committee:

Since we are selecting 3 persons out of 5 (2 men + 3 women) and order does not matter, this is a combination problem.

The total number of ways = \( {}^5C_3 \).

\[ {}^5C_3 = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4 \times 3!}{3! \times 2 \times 1} = \frac{20}{2} = 10 \]

So, there are 10 ways to form the committee.

Committees with exactly 1 man and 2 women:

We need to select 1 man from 2 men AND 2 women from 3 women.

By the Fundamental Principle of Counting, we multiply the combinations:

Number of ways = \( {}^2C_1 \times {}^3C_2 \)

\[ {}^2C_1 = 2 \]

\[ {}^3C_2 = \frac{3!}{2!1!} = 3 \]

Total = \( 2 \times 3 = 6 \)

Conclusion: There are 6 committees that consists of exactly 1 man and 2 women out of total 10 committees.

Question 2:

How many words, with or without meaning, can be formed using all the letters of the word “EQUATION” at a time so that the vowels and consonants occur together?

Answer 2:

The word “EQUATION” has 8 letters.

Vowels (V): E, U, A, I, O (5 vowels)

Consonants (C): Q, T, N (3 consonants)

Since vowels and consonants must occur together, we can block them into two groups: [V] and [C].

These 2 groups can be arranged among themselves in \( 2! \) ways.

Inside group [V], the 5 vowels can be arranged in \( 5! \) ways.

Inside group [C], the 3 consonants can be arranged in \( 3! \) ways.

By the Multiplication Principle, the total number of permutations is:

\[ 2! \times 5! \times 3! = 2 \times 120 \times 6 = 1440 \]

Conclusion: 1440 different words can be formed.

Chapter 7: Binomial Theorem

7.1 Introduction

An algebraic expression consisting of two terms is called a binomial expression. Example: \( x + y \), \( 2a - b \). The expansion of power of a binomial expression is what we study in the Binomial Theorem.

Observations on simple expansions:

• \( (a+b)^0 = 1 \)
• \( (a+b)^1 = a + b \)
• \( (a+b)^2 = a^2 + 2ab + b^2 \)
• \( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \)

7.2 Pascal’s Triangle

The coefficients of the expansions of \( (a+b)^n \) for \( n = 0, 1, 2, \dots \) form a pattern that is easily generated by Pascal’s Triangle.

7.3 Binomial Theorem for Positive Integral Indices

The expansion of \( (a+b)^n \) where \( n \in \mathbb{N} \) is: \[ (a+b)^n = {}^nC_0 a^n + {}^nC_1 a^{n-1}b + {}^nC_2 a^{n-2}b^2 + \dots + {}^nC_{n-1} ab^{n-1} + {}^nC_n b^n \] Or simply: \[ (a+b)^n = \sum_{k=0}^{n} {}^nC_k a^{n-k} b^k \]

Observations:

1. The total number of terms in the expansion is \( n+1 \).
2. The sum of indices of \( a \) and \( b \) in each term is \( n \).
3. The coefficients \( {}^nC_k \) are called the binomial coefficients, and \( {}^nC_r = {}^nC_{n-r} \). Equidistant terms from the beginning and end have equal coefficients.

Competency-Based Questions

Question 1:

Compute \( (98)^5 \) using the Binomial Theorem.

Answer 1:

We express 98 as \( 100 - 2 \) so that we can use the Binomial Theorem easily.

\[ (98)^5 = (100 - 2)^5 \]

Using Binomial Theorem:

\[ (100 - 2)^5 = {}^5C_0 (100)^5 - {}^5C_1 (100)^4 (2) + {}^5C_2 (100)^3 (2)^2 - {}^5C_3 (100)^2 (2)^3 + {}^5C_4 (100)(2)^4 - {}^5C_5 (2)^5 \]

Evaluate the terms:

\[ (100 - 2)^5 = 1 \cdot 10^{10} - 5 \cdot 10^8 \cdot 2 + 10 \cdot 10^6 \cdot 4 - 10 \cdot 10^4 \cdot 8 + 5 \cdot 100 \cdot 16 - 1 \cdot 32 \]

\[ = 10,000,000,000 - 1,000,000,000 + 40,000,000 - 800,000 + 8,000 - 32 \]

Summing the positive terms:

\[ 10,000,000,000 + 40,000,000 + 8,000 = 10,040,008,000 \]

Summing the negative terms:

\[ 1,000,000,000 + 800,000 + 32 = 1,000,800,032 \]

Result:

\[ 10,040,008,000 - 1,000,800,032 = 9,039,207,968 \]

Conclusion: \( 98^5 = 9,039,207,968 \).

Question 2:

Find \( (x + 1)^6 + (x - 1)^6 \). Hence or otherwise evaluate \( (\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6 \).

Answer 2:

Let’s find the expansion of \( (x + 1)^6 + (x - 1)^6 \):

\[ (x + 1)^6 = x^6 + {}^6C_1 x^5 + {}^6C_2 x^4 + {}^6C_3 x^3 + {}^6C_4 x^2 + {}^6C_5 x + 1 \]

\[ (x - 1)^6 = x^6 - {}^6C_1 x^5 + {}^6C_2 x^4 - {}^6C_3 x^3 + {}^6C_4 x^2 - {}^6C_5 x + 1 \]

Upon adding both equations, the odd power terms cancel out to 0:

\[ (x + 1)^6 + (x - 1)^6 = 2 \left( x^6 + {}^6C_2 x^4 + {}^6C_4 x^2 + 1 \right) \]

Substitute combinations values:

\[ (x + 1)^6 + (x - 1)^6 = 2 \left( x^6 + 15x^4 + 15x^2 + 1 \right) \]

Now calculate the value at \( x = \sqrt{2} \):

\[ (\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6 = 2 \left( (\sqrt{2})^6 + 15(\sqrt{2})^4 + 15(\sqrt{2})^2 + 1 \right) \]

\[ = 2 (8 + 15(4) + 15(2) + 1) \]

\[ = 2 (8 + 60 + 30 + 1) = 2 (99) = 198 \]

Conclusion: The value of the expression is 198.

Chapter 8: Sequence and Series

8.1 Introduction

A sequence is an ordered list of numbers obeying a certain rule. Elements of a sequence are called its terms, usually denoted by \( a_1, a_2, a_3, \dots, a_n \). If a sequence has a finite number of terms, it is a finite sequence. Otherwise, it is an infinite sequence.

A series is the sum of elements of a sequence. If \( a_1, a_2, a_3, \dots, a_n \) limit the sequence, then the expression \( a_1 + a_2 + a_3 + \dots + a_n \) is called series.

8.2 Arithmetic Progression (A.P.)

A sequence \( a_1, a_2, \dots \) is an Arithmetic Progression (AP) if \( a_{n+1} - a_n = d \), a constant, for all \( n \ge 1 \). \( d \) is the common difference.

• General Term: \( a_n = a + (n-1)d \)
• Sum of first n terms: \( S_n = \frac{n}{2} [2a + (n-1)d] = \frac{n}{2} (a + l) \), where \( l \) is the last term.
• Arithmetic Mean (A.M.): Given two numbers \( a \) and \( b \), their A.M. is \( A = \frac{a+b}{2} \). We can insert \( n \) numbers between \( a \) and \( b \) to form an A.P.

8.3 Geometric Progression (G.P.)

A sequence \( a_1, a_2, \dots \) with non-zero terms is a Geometric Progression (GP) if \( \frac{a_{k+1}}{a_k} = r \), a constant, for all \( k \ge 1 \). \( r \) is called the common ratio.

• General Term: \( a_n = ar^{n-1} \)
• Sum of first n terms: \[ S_n = \frac{a(r^n - 1)}{r - 1} \text{ if } r \neq 1 \] If \( r = 1 \), then \( S_n = na \).
• Sum of an infinite G.P.: If \( |r| < 1 \), then sum up to infinity is: \[ S_\infty = \frac{a}{1 - r} \]
• Geometric Mean (G.M.): Given two numbers \( a \) and \( b \), their G.M. is \( G = \sqrt{ab} \).

Relationship between A.M. and G.M.

Let \( A \) and \( G \) be A.M. and G.M. of two given positive real numbers \( a \) and \( b \), respectively. Then: \[ A = \frac{a+b}{2} \text{ and } G = \sqrt{ab} \] It can be shown that \( A \ge G \), because \[ A - G = \frac{a+b}{2} - \sqrt{ab} = \frac{a+b - 2\sqrt{ab}}{2} = \frac{(\sqrt{a} - \sqrt{b})^2}{2} \ge 0 \]

Competency-Based Questions

Question 1:

The sum of the first \( p, q, r \) terms of an A.P. are \( a, b, c \), respectively. Prove that:

\[ \frac{a}{p}(q-r) + \frac{b}{q}(r-p) + \frac{c}{r}(p-q) = 0 \]

Answer 1:

Let the first term of the A.P. be \( A \) and common difference be \( D \).

Given:

\[ S_p = \frac{p}{2}[2A + (p-1)D] = a \Rightarrow \frac{a}{p} = A + \frac{1}{2}(p-1)D \]

\[ S_q = \frac{q}{2}[2A + (q-1)D] = b \Rightarrow \frac{b}{q} = A + \frac{1}{2}(q-1)D \]

\[ S_r = \frac{r}{2}[2A + (r-1)D] = c \Rightarrow \frac{c}{r} = A + \frac{1}{2}(r-1)D \]

Substitute these values into LHS of the given expression:

\[ LHS = \left[ A + (p-1)\frac{D}{2} \right] (q-r) + \left[ A + (q-1)\frac{D}{2} \right] (r-p) + \left[ A + (r-1)\frac{D}{2} \right] (p-q) \]

Collect A terms:

\[ A(q-r + r-p + p-q) = A(0) = 0 \]

Collect D/2 terms:

\[ \frac{D}{2} [ (p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q) ] \]

Evaluate bracket for D/2 terms:

\[ = pq - pr - q + r + qr - pq - r + p + pr - qr - p + q \]

All terms cancel each other out to yield 0.

\[ LHS = 0 + \frac{D}{2}[0] = 0 \]

Conclusion: Hence proved.

Question 2:

Determine the 10th term of a G.P. whose 3rd term is 24 and whose 6th term is 192.

Answer 2:

Let the first term be \( a \) and common ratio be \( r \).

We are given:

\[ a_3 = ar^2 = 24 \qquad \text{— (1)} \]

\[ a_6 = ar^5 = 192 \qquad \text{— (2)} \]

Divide eq (2) by eq (1):

\[ \frac{ar^5}{ar^2} = \frac{192}{24} \]

\[ r^3 = 8 \Rightarrow r = 2 \]

Substitute \( r=2 \) in eq (1):

\[ a(2^2) = 24 \Rightarrow 4a = 24 \Rightarrow a = 6 \]

Now we must find the 10th term (\( a_{10} \)):

\[ a_{10} = ar^9 = 6(2^9) = 6(512) = 3072 \]

Conclusion: The 10th term of the G.P. is 3072.

Unit III: Coordinate Geometry

This unit covers the algebraic representation of geometric shapes like straight lines, circles, parabolas, ellipses, and hyperbolas, as well as an introduction to 3D coordinate geometry.

Chapters in this Unit: 9. Straight Lines 10. Conic Sections 11. Introduction to Three-dimensional Geometry

Chapter 9: Straight Lines

9.1 Introduction

Coordinate geometry involves the study of geometry using a coordinate system. This bridges the gap between algebra and geometry.

Important formulas from earlier classes:

• Distance Formula: The distance between points \( P(x_1, y_1) \) and \( Q(x_2, y_2) \) is \( PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
• Section Formula: The coordinates of a point dividing the line segment joining \( (x_1, y_1) \) and \( (x_2, y_2) \) internally in the ratio \( m:n \) are \( \left(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}\right) \).
• Area of Triangle: Area = \( \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \).

9.2 Slope of a Line

If \( \theta \) is the angle of inclination of a line \( l \), then \( \tan \theta \) is called the slope or gradient of the line \( l \). Slope is usually denoted by \( m \). \[ m = \tan \theta, ; \text{where } \theta \neq 90^\circ \] The slope of a line passing through points \( (x_1, y_1) \) and \( (x_2, y_2) \) is: \[ m = \frac{y_2 - y_1}{x_2 - x_1}, ; \text{where } x_1 \neq x_2 \]

Angle between Two Lines

If \( m_1 \) and \( m_2 \) are slopes of two non-vertical lines, the acute angle \( \theta \) between them is: \[ \tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \]

• Lines are parallel if \( m_1 = m_2 \).
• Lines are perpendicular if \( m_1 m_2 = -1 \).

9.3 Various Forms of Equations of a Line

1. Horizontal and vertical lines: Equation of x-axis is \( y = 0 \). Equation of y-axis is \( x = 0 \).
2. Point-slope form: Equation of a line passing through \( (x_0, y_0) \) with slope \( m \) is: \[ y - y_0 = m(x - x_0) \]
3. Two-point form: Line passing through \( (x_1, y_1) \) and \( (x_2, y_2) \) is: \[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1) \]
4. Slope-intercept form: Line with slope \( m \) and y-intercept \( c \) is \( y = mx + c \).
5. Intercept form: Line making intercepts \( a \) and \( b \) on x and y axes is: \[ \frac{x}{a} + \frac{y}{b} = 1 \]
6. Normal form: Line uniquely determined by length of perpendicular \( p \) from origin and angle \( \omega \) this perpendicular makes with positive x-axis: \[ x \cos \omega + y \sin \omega = p \]

9.4 General Equation of a Line

Any equation of the form \( Ax + By + C = 0 \), where \( A \) and \( B \) are not simultaneously zero, is called the general linear equation. It always represents a straight line.

• Slope \( m = -\frac{A}{B} \)
• x-intercept \( = -\frac{C}{A} \)
• y-intercept \( = -\frac{C}{B} \)

Distance of a Point From a Line

The perpendicular distance \( d \) from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is: \[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \] Distance between two parallel lines \( Ax + By + C_1 = 0 \) and \( Ax + By + C_2 = 0 \): \[ d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} \]

Competency-Based Questions

Question 1:

A line passes through the point \( P(3,4) \) and cuts off intercepts on the coordinate axes such that their sum is 14. Determine the possible equation(s) of this line.

Answer 1:

Let the intercepts on the x and y axes be \( a \) and \( b \) respectively.

The equation of the line in intercept form is:

\[ \frac{x}{a} + \frac{y}{b} = 1 \]

We are given that \( a + b = 14 \Rightarrow b = 14 - a \).

Also, the line passes through \( (3, 4) \), so we substitute \( x=3 \) and \( y=4 \):

\[ \frac{3}{a} + \frac{4}{14 - a} = 1 \]

Take the LCM:

\[ \frac{3(14 - a) + 4a}{a(14 - a)} = 1 \]

\[ 42 - 3a + 4a = 14a - a^2 \]

\[ 42 + a = 14a - a^2 \]

Rearranging to form a quadratic equation:

\[ a^2 - 13a + 42 = 0 \]

Factorizing the quadratic equation:

\[ (a - 6)(a - 7) = 0 \]

So, \( a = 6 \) or \( a = 7 \).

Case 1: If \( a = 6 \), then \( b = 14 - 6 = 8 \).

The equation is \( \frac{x}{6} + \frac{y}{8} = 1 \Rightarrow 4x + 3y = 24 \).

Case 2: If \( a = 7 \), then \( b = 14 - 7 = 7 \).

The equation is \( \frac{x}{7} + \frac{y}{7} = 1 \Rightarrow x + y = 7 \).

Conclusion: The two possible equations for the line are \( 4x + 3y = 24 \) and \( x + y = 7 \).

Question 2:

Find the angle between the lines \( y - \sqrt{3}x - 5 = 0 \) and \( \sqrt{3}y - x + 6 = 0 \).

Answer 2:

First, rewrite both equations in slope-intercept form (\( y = mx + c \)) to find their slopes.

Line 1: \( y = \sqrt{3}x + 5 \)

Slope \( m_1 = \sqrt{3} \)

Line 2: \( \sqrt{3}y = x - 6 \Rightarrow y = \frac{1}{\sqrt{3}}x - \frac{6}{\sqrt{3}} \)

Slope \( m_2 = \frac{1}{\sqrt{3}} \)

The acute angle \( \theta \) between the lines is given by:

\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

\[ \tan \theta = \left| \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + (\sqrt{3})\left(\frac{1}{\sqrt{3}}\right)} \right| \]

Simplify numerator:

\[ \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3 - 1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \]

Simplify denominator:

\[ 1 + 1 = 2 \]

Thus,

\[ \tan \theta = \left| \frac{ \frac{2}{\sqrt{3}} }{2} \right| = \frac{1}{\sqrt{3}} \]

The angle whose tangent is \( \frac{1}{\sqrt{3}} \) is \( 30^\circ \) or \( \frac{\pi}{6} \) radians.

Conclusion: The angle between the two lines is \( 30^\circ \).

Chapter 10: Conic Sections

10.1 Introduction

The curves we obtain by slicing a double-napped right circular cone with a plane are called conic sections. Depending on the angle of the intersecting plane relative to the axis of the cone, we get circles, ellipses, parabolas, and hyperbolas.

10.2 Sections of a Cone

Let \( \alpha \) be the angle made by the generator with the axis of the cone, and \( \beta \) be the angle made by the intersecting plane with the axis.

• Circle: When \( \beta = 90^\circ \), the section is a circle.
• Ellipse: When \( \alpha < \beta < 90^\circ \), the section is an ellipse.
• Parabola: When \( \beta = \alpha \), the section is a parabola.
• Hyperbola: When \( 0 \le \beta < \alpha \), the plane intersects both nappes, forming a hyperbola.

10.3 Circle

A circle is the set of all points in a plane that are equidistant from a fixed point in the plane. The fixed point is the center and the fixed distance is the radius.

Equation of a circle with center \( (h, k) \) and radius \( r \): \[ (x - h)^2 + (y - k)^2 = r^2 \]

10.4 Parabola

A parabola is the set of all points in a plane that are equidistant from a fixed line (the directrix) and a fixed point (the focus).

• Vertex: The point of intersection of the parabola with its axis of symmetry.
• Standard Equations:
  1. \( y^2 = 4ax \) (Opens right)
  2. \( y^2 = -4ax \) (Opens left)
  3. \( x^2 = 4ay \) (Opens up)
  4. \( x^2 = -4ay \) (Opens down)
• Latus Rectum: The focal chord perpendicular to the axis of the parabola. Its length is \( 4a \).

10.5 Ellipse

An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points (the foci) in the plane is a constant.

• Standard Equation: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)
• Vertices: \( (\pm a, 0) \)
• Foci: \( (\pm c, 0) \) where \( c^2 = a^2 - b^2 \)
• Eccentricity: \( e = \frac{c}{a} < 1 \)
• Length of Latus Rectum: \( \frac{2b^2}{a} \)

10.6 Hyperbola

A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points (the foci) in the plane is a constant.

• Standard Equation: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
• Vertices: \( (\pm a, 0) \)
• Foci: \( (\pm c, 0) \) where \( c^2 = a^2 + b^2 \)
• Eccentricity: \( e = \frac{c}{a} > 1 \)

Competency-Based Questions

Question 1:

Find the equation of a circle passing through the points \( (4,1) \) and \( (6,5) \) and whose center is on the line \( 4x + y = 16 \).

Answer 1:

Let the equation of the circle be \( (x-h)^2 + (y-k)^2 = r^2 \).

Since the center \( (h, k) \) lies on the line \( 4x + y = 16 \), we have:

\[ 4h + k = 16 \Rightarrow k = 16 - 4h \qquad \text{— (1)} \]

The circle passes through \( (4,1) \):

\[ (4-h)^2 + (1-k)^2 = r^2 \qquad \text{— (2)} \]

The circle passes through \( (6,5) \):

\[ (6-h)^2 + (5-k)^2 = r^2 \qquad \text{— (3)} \]

Equating (2) and (3) since both equal \( r^2 \):

\[ (4-h)^2 + (1-k)^2 = (6-h)^2 + (5-k)^2 \]

\[ 16 - 8h + h^2 + 1 - 2k + k^2 = 36 - 12h + h^2 + 25 - 10k + k^2 \]

Cancel \( h^2 \) and \( k^2 \):

\[ 17 - 8h - 2k = 61 - 12h - 10k \]

Bring all terms to one side:

\[ 4h + 8k = 44 \Rightarrow h + 2k = 11 \qquad \text{— (4)} \]

Substitute value of \( k \) from (1) into (4):

\[ h + 2(16 - 4h) = 11 \]

\[ h + 32 - 8h = 11 \]

\[ -7h = -21 \Rightarrow h = 3 \]

Now find \( k \):

\[ k = 16 - 4(3) = 16 - 12 = 4 \]

The center is \( (3, 4) \).

Find radius squared \( r^2 \) using eq (2):

\[ r^2 = (4-3)^2 + (1-4)^2 = 1^2 + (-3)^2 = 1 + 9 = 10 \]

The equation of the circle is \( (x-3)^2 + (y-4)^2 = 10 \), which can be expanded as:

\[ x^2 - 6x + 9 + y^2 - 8y + 16 = 10 \]

\[ x^2 + y^2 - 6x - 8y + 15 = 0 \]

Conclusion: The equation of the circle is \( x^2 + y^2 - 6x - 8y + 15 = 0 \).

Question 2:

Determine the coordinates of the focus, axis, the equation of directrix and latus rectum of the parabola \( y^2 = -8x \).

Answer 2:

The equation involves \( y^2 \), so its axis is the x-axis. The coefficient of \( x \) is negative, meaning the parabola opens towards the left.

Compare \( y^2 = -8x \) with the standard form \( y^2 = -4ax \).

\[ -4a = -8 \Rightarrow a = 2 \]

• Focus: Since it opens to the left, focus \( F(-a, 0) \rightarrow (-2, 0) \).

• Axis: The axis of symmetry is the x-axis, whose equation is \( y = 0 \).

• Directrix: The equation is \( x = a \rightarrow x = 2 \).

• Latus Rectum: Length \( = 4a = 4(2) = 8 \).

Conclusion: Focus: \( (-2, 0) \), Axis: \( y=0 \), Directrix: \( x=2 \), Latus Rectum: 8.

Chapter 11: Introduction to Three-Dimensional Geometry

11.1 Introduction

In two-dimensional geometry, we use two intersecting coordinate axes (X-axis and Y-axis) to locate a point in a plane. To locate a point in space, we require a system of three intersecting, mutually perpendicular lines, forming the three-dimensional coordinate system.

11.2 Coordinate Axes and Coordinate Planes

Let X’OX, Y’OY, and Z’OZ be three mutually perpendicular lines passing through the origin O. These three lines form the coordinate axes (X-axis, Y-axis, Z-axis). The planes determined by pairs of axes are called Coordinate Planes:

1. XY-plane: The plane determined by the X-axis and Y-axis. Every point on this plane has a z-coordinate of zero.
2. YZ-plane: Determined by Y and Z axes. The x-coordinate is zero.
3. ZX-plane: Determined by Z and X axes. The y-coordinate is zero.

These three planes divide space into eight parts called octants.

Coordinates of a Point in Space

The coordinates of a point \( P \) in 3D space are always written in the form of an ordered triplet \( (x, y, z) \). Here, \( x, y, z \) represent the perpendicular distances from the YZ, ZX, and XY planes respectively.

11.3 Distance between Two Points

The distance \( PQ \) between two points \( P(x_1, y_1, z_1) \) and \( Q(x_2, y_2, z_2) \) in space is given by the 3D distance formula: \[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]

This is a direct derivation from the 2D distance formula by applying the Pythagorean Theorem twice in space.

11.4 Section Formula

The coordinates of a point \( R \) that divides the line segment joining points \( P(x_1, y_1, z_1) \) and \( Q(x_2, y_2, z_2) \) internally in the ratio \( m:n \) are given by: \[ \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right) \]

If point \( R \) divides the segment externally in the ratio \( m:n \), then the coordinates are: \[ \left( \frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n}, \frac{mz_2 - nz_1}{m-n} \right) \]

Mid-point Formula: If \( R \) is the midpoint of segment \( PQ \), the ratio is \( 1:1 \), and the coordinates are: \[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \]

Competency-Based Questions

Question 1:

Show that the points \( P(-2, 3, 5) \), \( Q(1, 2, 3) \), and \( R(7, 0, -1) \) are collinear.

Answer 1:

Points are collinear if they lie on the same straight line. We can check this by calculating the distances between the points. If the sum of any two distances is equal to the third distance, the points are collinear.

Calculate \( PQ \):

\[ PQ = \sqrt{(1 - (-2))^2 + (2 - 3)^2 + (3 - 5)^2} \]

\[ PQ = \sqrt{3^2 + (-1)^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \]

Calculate \( QR \):

\[ QR = \sqrt{(7 - 1)^2 + (0 - 2)^2 + (-1 - 3)^2} \]

\[ QR = \sqrt{6^2 + (-2)^2 + (-4)^2} = \sqrt{36 + 4 + 16} = \sqrt{56} = 2\sqrt{14} \]

Calculate \( PR \):

\[ PR = \sqrt{(7 - (-2))^2 + (0 - 3)^2 + (-1 - 5)^2} \]

\[ PR = \sqrt{9^2 + (-3)^2 + (-6)^2} = \sqrt{81 + 9 + 36} = \sqrt{126} = 3\sqrt{14} \]

Notice that \( PQ + QR = \sqrt{14} + 2\sqrt{14} = 3\sqrt{14} \).

Since \( PQ + QR = PR \), the points \( P, Q, \) and \( R \) are collinear.

Conclusion: The given points are collinear.

Question 2:

Find the coordinates of the point which divides the line segment joining the points \( (1, -2, 3) \) and \( (3, 4, -5) \) in the ratio \( 2:3 \) internally.

Answer 2:

Let the given points be \( A(1, -2, 3) \) and \( B(3, 4, -5) \).

The ratio is \( m:n = 2:3 \).

Using the internal section formula, the coordinates \( (x,y,z) \) of the point \( P \) are:

\[ x = \frac{mx_2 + nx_1}{m+n} = \frac{2(3) + 3(1)}{2+3} = \frac{6 + 3}{5} = \frac{9}{5} \]

\[ y = \frac{my_2 + ny_1}{m+n} = \frac{2(4) + 3(-2)}{2+3} = \frac{8 - 6}{5} = \frac{2}{5} \]

\[ z = \frac{mz_2 + nz_1}{m+n} = \frac{2(-5) + 3(3)}{2+3} = \frac{-10 + 9}{5} = \frac{-1}{5} \]

Conclusion: The coordinates of the point are \( \left(\frac{9}{5}, \frac{2}{5}, \frac{-1}{5}\right) \).

Unit IV: Calculus

Calculus is the mathematical study of continuous change. This introductory unit for Class 11 focuses on the foundational concepts of limits and the calculation of derivatives.

Chapters in this Unit: 12. Limits and Derivatives

Chapter 12: Limits and Derivatives

12.1 Introduction to Limits

Calculus is that branch of mathematics which mainly deals with the study of change in the value of a function as the points in the domain change. The concept of a limit is fundamental to calculus. It allows us to determine the value that a function approaches as its input approaches some value.

Let \( y = f(x) \). If \( f(x) \) gets closer and closer to a real number \( L \) as \( x \) gets closer and closer to \( a \) (from both sides), we say that the limit of \( f(x) \) as \( x \) approaches \( a \) is \( L \), and we write: \[ \lim_{x \to a} f(x) = L \]

Algebra of Limits

Let \( \lim_{x \to a} f(x) = l \) and \( \lim_{x \to a} g(x) = m \).

1. Limit of sum: \( \lim_{x \to a} [f(x) + g(x)] = l + m \)
2. Limit of difference: \( \lim_{x \to a} [f(x) - g(x)] = l - m \)
3. Limit of product: \( \lim_{x \to a} [f(x) \cdot g(x)] = l \cdot m \)
4. Limit of quotient: \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{l}{m} \), provided \( m \neq 0 \).

Standard Limits

1. \[ \lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1} \]
2. \[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \] (where x is in radians)
3. \[ \lim_{x \to 0} \frac{1 - \cos x}{x} = 0 \]

12.2 Derivatives

Derivative of a function \( f(x) \) at a point \( a \) represents the rate of change of \( f(x) \) with respect to \( x \) at \( x = a \). It is geometrically the slope of the tangent to the curve \( y = f(x) \) at that point.

The derivative of a function \( f \) at \( a \) is defined as: \[ f’(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \] provided this limit exists.

If we view derivative as a function, then the derivative of \( f(x) \) with respect to \( x \) is: \[ f’(x) = \frac{d}{dx} f(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] This process of finding the derivative is called differentiation from first principles.

Algebra of Derivatives

1. Sum/Difference Rule: \( \frac{d}{dx}[u \pm v] = \frac{du}{dx} \pm \frac{dv}{dx} \)
2. Product Rule (Leibniz Rule): \( \frac{d}{dx}[uv] = u \frac{dv}{dx} + v \frac{du}{dx} \)
3. Quotient Rule: \( \frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \), where \( v \neq 0 \).

Derivatives of Standard Functions

• \( \frac{d}{dx}(x^n) = n x^{n-1} \)
• \( \frac{d}{dx}(\sin x) = \cos x \)
• \( \frac{d}{dx}(\cos x) = -\sin x \)
• \( \frac{d}{dx}(\tan x) = \sec^2 x \)
• \( \frac{d}{dx}(\text{constant}) = 0 \)

Competency-Based Questions

Question 1:

Evaluate the limit: \( \lim_{x \to 2} \frac{x^3 - 2x^2}{x^2 - 5x + 6} \).

Answer 1:

Let \( f(x) = \frac{x^3 - 2x^2}{x^2 - 5x + 6} \).

If we substitute \( x = 2 \) directly, we get:

Numerator: \( 2^3 - 2(2^2) = 8 - 8 = 0 \)

Denominator: \( 2^2 - 5(2) + 6 = 4 - 10 + 6 = 0 \)

This gives a \( \frac{0}{0} \) indeterminate form, which means \( (x-2) \) is a common factor in both numerator and denominator. Let’s factorize.

Numerator: \( x^3 - 2x^2 = x^2(x - 2) \)

Denominator: \( x^2 - 5x + 6 = x^2 - 3x - 2x + 6 = x(x-3) - 2(x-3) = (x-2)(x-3) \)

So, the function can be written as:

\[ \lim_{x \to 2} \frac{x^2(x - 2)}{(x-2)(x-3)} \]

Cancel the common factor \( (x-2) \) for \( x \neq 2 \):

\[ \lim_{x \to 2} \frac{x^2}{x-3} \]

Now substitute \( x = 2 \):

\[ = \frac{2^2}{2 - 3} = \frac{4}{-1} = -4 \]

Conclusion: The value of the limit is -4.

Question 2:

Find the derivative of the function \( f(x) = \frac{x + \cos x}{\tan x} \) with respect to \( x \).

Answer 2:

Here we use the Quotient Rule: \( \frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v \cdot u’ - u \cdot v’}{v^2} \).

Let \( u = x + \cos x \) and \( v = \tan x \).

Differentiate \( u \):

\[ u’ = \frac{d}{dx}(x + \cos x) = 1 - \sin x \]

Differentiate \( v \):

\[ v’ = \frac{d}{dx}(\tan x) = \sec^2 x \]

Apply the rule:

\[ f’(x) = \frac{(\tan x)(1 - \sin x) - (x + \cos x)(\sec^2 x)}{(\tan x)^2} \]

Expand the numerator:

\[ = \frac{\tan x - \tan x \sin x - x \sec^2 x - \cos x \sec^2 x}{\tan^2 x} \]

We know \( \cos x \sec^2 x = \cos x \cdot \frac{1}{\cos^2 x} = \frac{1}{\cos x} = \sec x \). And \( \tan x \sin x = \frac{\sin^2 x}{\cos x} \).

\[ f’(x) = \frac{\tan x - \frac{\sin^2 x}{\cos x} - x \sec^2 x - \sec x}{\tan^2 x} \]

Conclusion: The final derivative is \( f’(x) = \frac{\tan x - \frac{\sin^2 x}{\cos x} - x \sec^2 x - \sec x}{\tan^2 x} \).

Unit V: Statistics and Probability

This final unit discusses the measures of dispersion in data and the theoretical probability of different events.

Chapters in this Unit: 13. Statistics 14. Probability

Chapter 13: Statistics

13.1 Measures of Dispersion

In earlier classes, we have studied measures of central tendency: Mean, Median, and Mode. These measures give us a single value around which the data is clustered. However, they do not tell us how the data is scattered around the central value. The scattering of data is called dispersion.

Some important measures of dispersion are:

1. Range
2. Mean Deviation
3. Variance
4. Standard Deviation

13.2 Range

The range is the simplest measure of dispersion. It is the difference between the maximum and minimum values of a series. \[ \text{Range} = x_{\text{max}} - x_{\text{min}} \]

13.3 Mean Deviation

Mean deviation is the arithmetic mean of the absolute deviations of the observations from an appropriate measure of central tendency (usually mean or median).

For Ungrouped Data

Let \( x_1, x_2, \dots, x_n \) be \( n \) observations. Let \( a \) be the measure of central tendency (Mean \( \bar{x} \) or Median \( M \)). Then, Mean Deviation about \( a \) is: \[ M.D.(a) = \frac{1}{n} \sum_{i=1}^{n} |x_i - a| \]

For Grouped Data

Let the observations \( x_1, x_2, \dots, x_n \) have frequencies \( f_1, f_2, \dots, f_n \) respectively. Let \( N = \sum f_i \). Then: \[ M.D.(a) = \frac{1}{N} \sum_{i=1}^{n} f_i |x_i - a| \]

13.4 Variance and Standard Deviation

The square of the standard deviation is called variance. It overcomes the drawback of taking absolute values in Mean Deviation by squaring the deviations before averaging them.

Variance (\( \sigma^2 \))

For ungrouped data: \[ \sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 \] A commonly used alternative formula is: \[ \sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2 \]

Standard Deviation (\( \sigma \))

The standard deviation is the positive square root of the variance. \[ \sigma = \sqrt{ \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 } \]

Coefficient of Variation (C.V.)

To compare the variability or dispersion of two or more series, we use the Coefficient of Variation. \[ C.V. = \frac{\sigma}{\bar{x}} \times 100 \] A series with lesser C.V. is more consistent (less dispersed).

Competency-Based Questions

Question 1:

The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation if the wrong item is omitted.

Answer 1:

Given: Number of observations \( n = 20 \)

Incorrect Mean \( \bar{x} = 10 \)

Incorrect Standard Deviation \( \sigma = 2 \)

Incorrect observation \( = 8 \)

First, let’s find the correct sum of observations.

\[ \text{Incorrect sum of observations } \sum x = n \bar{x} = 20 \times 10 = 200 \]

Since observation 8 is omitted, the new number of observations \( n’ = 19 \).

\[ \text{Correct sum of observations } \sum x’ = 200 - 8 = 192 \]

\[ \text{Correct Mean } \bar{x}’ = \frac{192}{19} \approx 10.105 \]

Now calculate the correct variance. We know:

\[ \sigma^2 = \frac{1}{n} \sum x^2 - (\bar{x})^2 \]

\[ 2^2 = \frac{1}{20} \sum x^2 - (10)^2 \]

\[ 4 = \frac{1}{20} \sum x^2 - 100 \]

\[ 104 = \frac{1}{20} \sum x^2 \Rightarrow \text{Incorrect } \sum x^2 = 2080 \]

When the item 8 is removed, the correct sum of squares:

\[ \text{Correct } \sum x’^2 = 2080 - 8^2 = 2080 - 64 = 2016 \]

Now find the new variance:

\[ \text{New Variance } \sigma’^2 = \frac{\text{Correct } \sum x’^2}{n’} - (\bar{x}’)^2 \]

\[ \sigma’^2 = \frac{2016}{19} - \left(\frac{192}{19}\right)^2 = \frac{2016}{19} - \frac{36864}{361} \]

\[ \sigma’^2 = \frac{2016 \times 19 - 36864}{361} = \frac{38304 - 36864}{361} = \frac{1440}{361} \approx 3.988 \]

The new standard deviation is:

\[ \sigma’ = \sqrt{ \frac{1440}{361} } = \frac{\sqrt{1440}}{19} = \frac{12\sqrt{10}}{19} \approx \frac{37.947}{19} \approx 1.997 \]

Conclusion: The correct mean is 10.105 and the correct standard deviation is 1.997.

Question 2:

Determine the range and mean deviation about the mean for the simple data set: 6, 7, 10, 12, 13, 4, 8, 12.

Answer 2:

Range:

Maximum value = 13, Minimum value = 4.

Range = 13 - 4 = 9.

Mean Deviation about mean:

Data points: 4, 6, 7, 8, 10, 12, 12, 13.

Number of observations \( n = 8 \).

Mean \( \bar{x} = \frac{4 + 6 + 7 + 8 + 10 + 12 + 12 + 13}{8} = \frac{72}{8} = 9 \).

Deviations \( |x_i - \bar{x}| \):

\( |4 - 9| = 5 \)

\( |6 - 9| = 3 \)

\( |7 - 9| = 2 \)

\( |8 - 9| = 1 \)

\( |10 - 9| = 1 \)

\( |12 - 9| = 3 \)

\( |12 - 9| = 3 \)

\( |13 - 9| = 4 \)

Sum of absolute deviations = 5 + 3 + 2 + 1 + 1 + 3 + 3 + 4 = 22.

Mean Deviation = \( \frac{22}{8} = 2.75 \).

Conclusion: Range = 9, Mean Deviation about Mean = 2.75.

Chapter 14: Probability

14.1 Introduction

Probability is the measure of uncertainty of various phenomena. In mathematical terms, probability defines the likelihood of occurrence of an event.

• Random Experiment: An experiment whose outcomes cannot be predicted with certainty.
• Sample Space (S): The set of all possible outcomes of a random experiment.
• Event (E): A subset of the sample space.

14.2 Types of Events

• Impossible Event: The empty set \( \emptyset \). It indicates an event that cannot occur.
• Sure Event: The entire sample space \( S \).
• Simple Event: An event containing only one sample point.
• Compound Event: An event containing more than one sample point.

Algebra of Events

• Complementary Event (Not E): The event ‘\( \text{not } E \)’ occurs when \( E \) does not occur. Denoted by \( E’ \). \( E’ = S - E \).
• Event ‘A or B’: Occurs if either \( A \) occurs, or \( B \) occurs, or both occur. Denoted by \( A \cup B \).
• Event ‘A and B’: Occurs if both \( A \) and \( B \) occur simultaneously. Denoted by \( A \cap B \).
• Mutually Exclusive Events: Two events \( A \) and \( B \) are mutually exclusive if they cannot occur simultaneously. \( A \cap B = \emptyset \).
• Exhaustive Events: Events \( E_1, E_2, \dots, E_n \) are exhaustive if their union equals the sample space. \( E_1 \cup E_2 \cup \dots \cup E_n = S \).

14.3 Axiomatic Approach to Probability

Let \( S \) be the sample space of a random experiment. A probability \( P \) is a real valued function whose domain is the power set of \( S \) and range is the interval \( [0, 1] \) satisfying the following axioms:

1. For any event \( E \), \( P(E) \ge 0 \).
2. \( P(S) = 1 \).
3. If \( E \) and \( F \) are mutually exclusive events, then \( P(E \cup F) = P(E) + P(F) \).

Important Theorems

• \( P(E’) = 1 - P(E) \)
• \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
• If \( A \) and \( B \) are mutually exclusive, \( P(A \cap B) = 0 \), so \( P(A \cup B) = P(A) + P(B) \).

Competency-Based Questions

Question 1:

Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that:

(a) Both अनिल and Ashima will not qualify the examination.

(b) At least one of them will not qualify the examination.

Answer 1:

Let \( A \) be the event “Anil qualifies” and \( B \) be the event “Ashima qualifies”.

Given:

\( P(A) = 0.05 \)

\( P(B) = 0.10 \)

\( P(A \cap B) = 0.02 \)

(a) Both will not qualify:

This is the event \( A’ \cap B’ \). By De Morgan’s Laws, \( A’ \cap B’ = (A \cup B)’ \).

First, find \( P(A \cup B) \):

\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.05 + 0.10 - 0.02 = 0.13 \]

Now, \( P(A’ \cap B’) = P((A \cup B)’) = 1 - P(A \cup B) = 1 - 0.13 = 0.87 \]

.

(b) At least one of them will not qualify:

This is the event “not both qualify”, which is \( (A \cap B)’ \).

\[ P((A \cap B)’) = 1 - P(A \cap B) = 1 - 0.02 = 0.98 \]

.

Conclusion: The probability that neither qualifies is 0.87, and the probability that at least one does not qualify is 0.98.

Question 2:

A bag contains 9 discs of which 4 are red, 3 are blue and 2 are yellow. The discs are similar in shape and size. A disc is drawn at random from the bag. Calculate the probability that it will be:

(i) red

(ii) yellow

(iii) blue

(iv) not blue

(v) either red or blue

Answer 2:

Total number of discs = \( 4 + 3 + 2 = 9 \).

(i) Probability of red:

\( P(R) = \frac{\text{Number of Red Discs}}{\text{Total Discs}} = \frac{4}{9} \)

(ii) Probability of yellow:

\( P(Y) = \frac{2}{9} \)

(iii) Probability of blue:

\( P(B) = \frac{3}{9} = \frac{1}{3} \)

(iv) Probability of not blue:

\( P(B’) = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3} \)

(v) Probability of either red or blue:

Since a single disc cannot be both red and blue, events R and B are mutually exclusive.

\( P(R \cup B) = P(R) + P(B) = \frac{4}{9} + \frac{3}{9} = \frac{7}{9} \)

Conclusion: The answers are (i) 4/9, (ii) 2/9, (iii) 1/3, (iv) 2/3, and (v) 7/9.