Introduction
In this chapter, we explore the concept of Trigonometric Functions, which are crucial for the study of various aspects of mathematics, particularly in geometry and calculus.
Revision Notes for JEE Main/Advanced & Unit Tests
A set is a well-defined collection of distinct objects. The objects in a set are called its elements or members.
Type | Definition | Example |
---|---|---|
Empty/Null Set | Set with no elements | ∅ or {} |
Singleton Set | Set with one element | {5} |
Finite Set | Set with countable number of elements | {1, 2, 3, 4} |
Infinite Set | Set with unlimited elements | Set of natural numbers |
Equal Sets | Sets with exactly same elements | A = {1, 2}, B = {2, 1} |
Equivalent Sets | Sets with same number of elements | A = {1, 2}, B = {a, b} |
Subset: A is a subset of B (A ⊆ B) if every element of A is in B
Proper Subset: A ⊂ B if A ⊆ B and A ≠ B
Find all subsets of A = {1, 2}
Solution: ∅, {1}, {2}, {1, 2} (Total 4 = 2² subsets)
Operation | Notation | Definition |
---|---|---|
Union | A ∪ B | {x | x ∈ A or x ∈ B} |
Intersection | A ∩ B | {x | x ∈ A and x ∈ B} |
Difference | A - B | {x | x ∈ A and x ∉ B} |
Complement | A' or Ac | {x | x ∉ A in universal set} |
(A ∪ B)' = A' ∩ B'
(A ∩ B)' = A' ∪ B'
Visual representation of sets and their relationships using overlapping circles.
n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(A ∩ C) + n(A ∩ B ∩ C)
n(A - B) = n(A) - n(A ∩ B)
JEE Main 2022: Let A and B be two sets containing 3 and 4 elements respectively. The number of subsets of A × B having at least 5 elements is?
JEE Advanced 2021: For sets A and B, let f: A → B and g: B → A be functions such that f(g(x)) = x for each x ∈ B. Which of the following must be true?
JEE Main/Advanced Preparation Notes
Definition: The Cartesian product A × B of two non-empty sets A and B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B.
A × B = {(a, b) : a ∈ A, b ∈ B}
Number of elements: If n(A) = p and n(B) = q, then n(A × B) = p × q
If A = {1, 2} and B = {3, 4, 5}, find A × B and B × A.
Solution:
A × B = {(1,3), (1,4), (1,5), (2,3), (2,4), (2,5)}
B × A = {(3,1), (3,2), (4,1), (4,2), (5,1), (5,2)}
Definition: A relation R from set A to set B is a subset of A × B.
If (a, b) ∈ R, we say 'a is related to b' and write a R b.
Relation Type | Definition | Example |
---|---|---|
Empty Relation | R = ∅ ⊆ A × A | No elements are related |
Universal Relation | R = A × A | All elements are related |
Identity Relation | IA = {(a, a) : a ∈ A} | Each element related to itself |
Reflexive | (a, a) ∈ R ∀ a ∈ A | Every element related to itself |
Symmetric | (a, b) ∈ R ⇒ (b, a) ∈ R | If a~b then b~a |
Transitive | (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R | If a~b and b~c then a~c |
Equivalence | Reflexive, Symmetric and Transitive | Congruence modulo n |
Definition: A function f: A → B is a relation that associates each element of A to exactly one element of B.
Domain: Set A (input values)
Codomain: Set B (possible output values)
Range: Actual outputs {f(x) : x ∈ A} ⊆ B
Function Type | Definition | Example |
---|---|---|
One-one (Injective) | f(x1) = f(x2) ⇒ x1 = x2 | f(x) = 2x + 3 |
Onto (Surjective) | Range = Codomain | f: R → R, f(x) = x³ |
Bijective | Both one-one and onto | f(x) = x + 1 |
Polynomial | f(x) = anxn + ... + a0 | f(x) = x² - 3x + 2 |
Rational | Ratio of two polynomials | f(x) = (x+1)/(x-2) |
Modulus | f(x) = |x| | f(x) = |x - 3| |
Signum | sgn(x) = -1, 0, or 1 | sgn(x) = x/|x| for x≠0 |
Important Function Graphs:
[Graphs would be inserted here in actual implementation]
1. Linear: f(x) = ax + b
2. Quadratic: f(x) = ax² + bx + c
3. Modulus: f(x) = |x|
4. Greatest Integer: f(x) = [x]
For f: A → B and g: B → C, the composition g∘f: A → C is defined as:
(g∘f)(x) = g(f(x))
Definition: For a bijective function f: A → B, the inverse f-1: B → A is defined by:
f-1(y) = x ⇔ f(x) = y
f∘f-1 = IB and f-1∘f = IA
(g∘f)-1 = f-1∘g-1
Definition: A binary operation * on a set A is a function *: A × A → A.
Property | Definition | Example |
---|---|---|
Commutative | a*b = b*a ∀ a,b ∈ A | Addition on R |
Associative | (a*b)*c = a*(b*c) | Matrix multiplication |
Identity | ∃ e ∈ A such that a*e = e*a = a | 0 for addition, 1 for multiplication |
Inverse | For each a ∈ A, ∃ b ∈ A such that a*b = b*a = e | -a is additive inverse of a |
In this chapter, we explore the concept of Trigonometric Functions, which are crucial for the study of various aspects of mathematics, particularly in geometry and calculus.
Trigonometric ratios are defined based on a right-angled triangle, where:
Understanding the graphs of trigonometric functions is essential for solving problems related to periodicity, amplitude, and phase shift. Below is an overview of each function:
Mathematical Induction is a method of proof used in mathematics to prove that a statement is true for all natural numbers. It is a powerful technique that helps in proving results in various mathematical domains, including sequences, series, and inequalities.
The process of proving a statement by induction involves two main steps:
When both steps are successfully completed, the statement is proven for all natural numbers starting from the base case.
Let's prove that the sum of the first n natural numbers is given by the formula:
Sum((1 + 2 + 3 + ... + n) = n(n + 1)/2
When n = 1, the left-hand side is just 1, and the right-hand side is 1(1 + 1)/2 = 1. So, the base case holds true.
Assume the formula holds true for n = k. That is, assume:
1 + 2 + 3 + ... + k = k(k + 1)/2
Now, we need to prove that the formula holds true for n = k + 1. The sum of the first k + 1 natural numbers is:
1 + 2 + 3 + ... + k + (k + 1)
Using the inductive hypothesis, we can replace the first part of the sum with k(k + 1)/2. So, the sum becomes:k(k + 1)/2 + (k + 1)
Factor out (k + 1) from the expression:(k + 1)[k/2 + 1]
Simplifying further, we get:(k + 1)(k + 2)/2
This is exactly the formula for the sum of the first k + 1 natural numbers, proving the statement for n = k + 1. Hence, by the principle of mathematical induction, the formula is valid for all natural numbers n.Let's prove that for all integers n ≥ 1, the following inequality holds:
2^n ≥ n + 1
When n = 1, we have:
2^1 = 2 and 1 + 1 = 2.
So, the base case holds true.Assume that the inequality holds true for n = k, i.e., assume:
2^k ≥ k + 1
We need to prove that the inequality holds for n = k + 1. That is, we need to prove:
2^(k + 1) ≥ (k + 1) + 1
Simplifying the right-hand side:2^(k + 1) ≥ k + 2
We know that 2^(k + 1) = 2 * 2^k, so we can write:2 * 2^k ≥ k + 2
By the inductive hypothesis, 2^k ≥ k + 1, so we have:2 * (k + 1) ≥ k + 2
Expanding the left-hand side:2k + 2 ≥ k + 2
Subtracting k + 2 from both sides:k ≥ 0
Since k is a natural number, this is true. Hence, the inequality holds for n = k + 1. By mathematical induction, the inequality is true for all integers n ≥ 1.Sequences and Series are fundamental concepts in mathematics that describe ordered sets of numbers (sequences) and their sums (series). This chapter covers arithmetic and geometric sequences and series, two of the most important types used in mathematical problems, including their general formulas, properties, and applications.
An arithmetic sequence is a sequence in which the difference between any two consecutive terms is constant. This difference is called the common difference (d).
The general form of an arithmetic sequence is:
a, a + d, a + 2d, a + 3d, ...
Where:
The nth term of an arithmetic sequence is given by:
tₙ = a + (n - 1)d
The sum of the first n terms (Sₙ) of an arithmetic sequence is given by:
Sₙ = n/2 × [2a + (n - 1)d]
or alternatively,Sₙ = n/2 × (a + l)
Where:A geometric sequence is a sequence in which each term is obtained by multiplying the previous term by a constant called the common ratio (r).
The general form of a geometric sequence is:
a, ar, ar², ar³, ...
Where:
The nth term of a geometric sequence is given by:
tₙ = ar^(n - 1)
The sum of the first n terms (Sₙ) of a geometric sequence is given by:
Sₙ = a × (1 - rⁿ) / (1 - r), for r ≠ 1
If r = 1, the sum is simply:
Sₙ = n × a
An infinite geometric series is the sum of an infinite number of terms of a geometric sequence. It converges (i.e., has a finite sum) only if the common ratio r satisfies |r| < 1.
The sum of an infinite geometric series is given by:
S = a / (1 - r), for |r| < 1
Find the 10th term of the sequence 2, 5, 8, 11, ...
Solution: Here, a = 2, d = 3. Using the formula for the nth term:
tₙ = a + (n - 1)d = 2 + (10 - 1) × 3 = 2 + 27 = 29
Find the sum of the first 6 terms of the sequence 3, 6, 12, 24, ...
Solution: Here, a = 3, r = 2. Using the formula for the sum of the first n terms:
Sₙ = a × (1 - rⁿ) / (1 - r) = 3 × (1 - 2⁶) / (1 - 2) = 3 × (1 - 64) / (-1) = 3 × 63 = 189
Find the sum of the infinite series 4 + 2 + 1 + 0.5 + ...
Solution: Here, a = 4, r = 0.5. Using the formula for the sum of an infinite geometric series:
S = a / (1 - r) = 4 / (1 - 0.5) = 4 / 0.5 = 8
JEE Math Notes - Class 11
Permutations and Combinations are fundamental concepts in combinatorics used to count the possible arrangements and selections of objects. This chapter explores these concepts in-depth and explains how to solve various counting problems. These concepts are important for solving probability problems as well.
n!
.P(n, r) = n! / (n - r)!
C(n, r) = n! / [r!(n - r)!]
m
ways and another event can occur in n
ways, then the number of ways both events can occur is m × n
.A permutation is an arrangement of objects in a specific order. If you are asked to arrange r
objects from a set of n
distinct objects, the number of possible permutations is given by:
P(n, r) = n! / (n - r)!
For example, if you want to arrange 3 objects from a set of 5 distinct objects, you would use the permutation formula:
P(5, 3) = 5! / (5 - 3)! = 5 × 4 × 3 = 60
A combination is a selection of objects where the order does not matter. The number of ways to choose r
objects from a set of n
distinct objects is given by:
C(n, r) = n! / [r!(n - r)!]
For example, if you want to choose 2 objects from a set of 5 distinct objects, the number of possible combinations is:
C(5, 2) = 5! / [2!(5 - 2)!] = 5 × 4 / 2 × 1 = 10
How many ways can 4 people be arranged in a row for a photo shoot?
Solution: This is a permutation problem. The number of ways to arrange 4 people out of 4 is:
P(4, 4) = 4! = 4 × 3 × 2 × 1 = 24
How many ways can 3 students be chosen from a group of 6 students?
Solution: This is a combination problem. The number of ways to choose 3 students from 6 is:
C(6, 3) = 6! / [3!(6 - 3)!] = 6 × 5 × 4 / 3 × 2 × 1 = 20
n! = n × (n - 1) × (n - 2) × ... × 1
P(n, r) = n! / (n - r)!
(If there are identical objects, adjust the formula accordingly.)r
objects from n
distinct objects with repetition is C(n + r - 1, r)
.Permutations and combinations are widely used in problems related to:
JEE Math Notes - Class 11
The Binomial Theorem provides a formula for expanding expressions that are raised to a power. It is a key tool for simplifying expressions in algebra and solving problems involving expansion of binomial expressions. The theorem also plays a significant role in probability theory and combinatorics.
The general form of the binomial expansion for (a + b)^n
is given by:
(a + b)^n = Σ (nCk) * a^(n-k) * b^k, where k ranges from 0 to n
Here, nCk
is the binomial coefficient, which is calculated as:
nCk = n! / (k!(n - k)!)
The binomial expansion gives the terms of the expansion as:
The binomial coefficients, nCk
, are the coefficients that appear in the expansion of the binomial expression. These coefficients are computed using the factorial formula:
nCk = n! / (k!(n - k)!)
Some properties of binomial coefficients include:
nCk = nC(n - k)
n
is 2^n
.Using the Binomial Theorem to expand (x + 2)4
:
(x + 2)4 = x4 + 4x32 + 6x24 + 4x23 + 16
The terms correspond to the binomial coefficients 4C0, 4C1, 4C2, 4C3, 4C4
respectively.
Expand (3x - y)3
using the binomial theorem:
The expansion will be:
(3x - y)3 = (3x)3 - 3(3x)2(y) + 3(3x)(y)2 - y3
= 27x3 - 27x2y + 9xy2 - y3
(1 + x)n
when x
is small)JEE Math Notes - Class 11
The chapter on Straight Lines is one of the most essential topics in coordinate geometry. Understanding the properties and equations of straight lines in the Cartesian plane is fundamental to solving various algebraic and geometric problems. In this chapter, we will learn about the slope of a line, different forms of the equation of a straight line, and various applications in geometry.
The equation of a straight line in the Cartesian coordinate system can be expressed in various forms. Some of the most commonly used forms are:
y = mx + c
where m
is the slope and c
is the y-intercept.y - y1 = m(x - x1)
where (x1, y1)
is a point on the line.Ax + By + C = 0
where A
, B
, and C
are constants.(y - y1) = [(y2 - y1)/(x2 - x1)] * (x - x1)
The slope (denoted as m
) of a straight line is a measure of its steepness. It is defined as the change in the y-coordinate divided by the change in the x-coordinate between two distinct points on the line. The formula for the slope is:
m = (y2 - y1) / (x2 - x1)
Some key points about slope:
Two lines are:
m1 = m2
.-1>, i.e., m1 * m2 = -1
.
Find the equation of the line passing through the point (2, 3)
with slope m = 4
.
Solution: Using the point-slope form of the equation:
y - 3 = 4(x - 2)
Expanding this equation:
y - 3 = 4x - 8
y = 4x - 5
This is the equation of the line in slope-intercept form.
Find the slope of the line passing through the points (1, 2)
and (3, 8)
.
Solution: Using the slope formula:
m = (y2 - y1) / (x2 - x1) = (8 - 2) / (3 - 1) = 6 / 2 = 3
The slope of the line is 3
.
JEE Math Notes - Class 11
Conic sections are curves obtained by intersecting a cone with a plane. These curves include the circle, ellipse, parabola, and hyperbola. Understanding these curves is fundamental in various fields, including geometry, physics, and engineering. In this chapter, we will explore the properties and equations of conic sections and their applications.
A circle is the set of all points in a plane that are at a constant distance (radius) from a fixed point (center). The standard equation of a circle is:
(x - h)2 + (y - k)2 = r2
Where:
(h, k)
is the center of the circle.r
is the radius of the circle.The equation of a circle can also be written in general form as:
x2 + y2 + 2gx + 2fy + c = 0
An ellipse is the set of all points for which the sum of the distances to two fixed points (foci) is constant. The standard equation of an ellipse is:
(x - h)2 / a2 + (y - k)2 / b2 = 1
Where:
(h, k)
is the center of the ellipse.a
is the semi-major axis.b
is the semi-minor axis.If a = b
, the ellipse becomes a circle.
A parabola is a curve that is symmetrical and has a single focus. The general equation of a parabola is:
y = ax2 + bx + c
For a parabola with a horizontal axis, the equation is:
x = ay2 + by + c
The vertex of the parabola is the point where the curve changes direction. It can be calculated using the formula:
x = -b / 2a
A hyperbola is the set of all points where the difference of the distances to two fixed points (foci) is constant. The standard equation of a hyperbola is:
(x - h)2 / a2 - (y - k)2 / b2 = 1
Where:
(h, k)
is the center of the hyperbola.a
and b
are constants related to the dimensions of the hyperbola.If a = b
, the hyperbola becomes a rectangular hyperbola.
Find the equation of a circle with center (3, -2)
and radius 5
.
Solution: Using the standard form of the equation of a circle:
(x - 3)2 + (y + 2)2 = 25
This is the equation of the required circle.
Find the equation of an ellipse with center (2, 1)
, semi-major axis 6
, and semi-minor axis 4
.
Solution: Using the standard form of the ellipse equation:
(x - 2)2 / 62 + (y - 1)2 / 42 = 1
This is the equation of the required ellipse.
JEE Math Notes - Class 11
Three-Dimensional Geometry is the branch of mathematics that deals with points, lines, planes, and other geometric objects in three-dimensional space. It is crucial for understanding the properties and relationships of shapes and figures that exist in 3D space, which is vital for fields such as engineering, physics, and computer graphics.
In three-dimensional space, a point is defined by three coordinates: (x, y, z)
. These represent the point's position along the X-axis, Y-axis, and Z-axis respectively. The three axes are perpendicular to each other, and the origin is the point where all three axes intersect: (0, 0, 0)
.
The position of any point in 3D space can be described by its distance from the origin along the three axes.
The distance between two points P(x₁, y₁, z₁)
and Q(x₂, y₂, z₂)
in 3D space is given by the formula:
d = √[(x₂ - x₁)2 + (y₂ - y₁)2 + (z₂ - z₁)2]
This formula helps us calculate the straight-line distance between any two points in three-dimensional space.
The section formula in three dimensions helps find the coordinates of a point dividing the line segment joining two given points in a given ratio. If a point P(x, y, z)
divides the line joining A(x₁, y₁, z₁)
and B(x₂, y₂, z₂)
in the ratio m:n
, then the coordinates of point P
are:
x = (mx₂ + nx₁) / (m + n)
y = (my₂ + ny₁) / (m + n)
z = (mz₂ + nz₁) / (m + n)
The direction cosines of a line are the cosines of the angles between the line and the coordinate axes. If a line has direction ratios (l, m, n)
, then the direction cosines are (cos α, cos β, cos γ)
, where α, β, and γ are the angles the line makes with the X, Y, and Z axes respectively.
The direction ratios (l, m, n)
are proportional to the direction cosines and are often used to describe the direction of the line in 3D space.
The general equation of a plane in three-dimensional space is:
Ax + By + Cz + D = 0
Where A
, B
, and C
are the direction ratios of a normal to the plane, and D
is a constant.
If the plane passes through a point (x₁, y₁, z₁)
with a normal vector (A, B, C)
, the equation of the plane can be written as:
A(x - x₁) + B(y - y₁) + C(z - z₁) = 0
Find the distance between the points (1, 2, 3)
and (4, 6, 8)
.
Solution: Using the distance formula:
d = √[(4 - 1)2 + (6 - 2)2 + (8 - 3)2]
d = √[32 + 42 + 52] = √[9 + 16 + 25] = √50 ≈ 7.07
The distance between the points is approximately 7.07 units.
JEE Math Notes - Class 11
Limits and derivatives are the foundational concepts of calculus. In this chapter, we will discuss the concept of limits, how they are used to describe the behavior of functions, and how derivatives provide us with the rate of change of a function. These concepts form the basis for understanding continuity, differentiability, and integration in calculus.
The limit of a function describes the behavior of the function as the input approaches a certain value. The limit helps us understand how a function behaves near a particular point, even if the function is not defined at that point.
Mathematically, the limit of a function f(x)
as x
approaches a
is denoted as:
limx→a f(x) = L
This means that as x
approaches a
, the value of f(x)
approaches L
.
limx→a f(x) = L
, then limx→a [f(x) + g(x)] = L + M
.limx→a [cf(x)] = c * limx→a f(x)
, where c
is a constant.limx→a f(x) = L
and limx→a g(x) = M
, then limx→a [f(x) * g(x)] = L * M
.The derivative of a function measures the rate at which the function's value changes as the input changes. It gives us the slope of the tangent line to the function at a particular point.
Mathematically, the derivative of a function f(x)
with respect to x
is denoted as:
f'(x) = limh→0 [f(x + h) - f(x)] / h
This is known as the difference quotient, and it gives the slope of the secant line between two points on the curve.
f(x) = xn
, then f'(x) = n * xn-1
.f(x) = g(x) + h(x)
, then f'(x) = g'(x) + h'(x)
.f(x) = g(x) * h(x)
, then f'(x) = g'(x) * h(x) + g(x) * h'(x)
.f(x) = g(x) / h(x)
, then f'(x) = [g'(x) * h(x) - g(x) * h'(x)] / [h(x)]2
.Find the derivative of the function f(x) = 3x4 - 5x2 + 2x - 7
.
Solution: Using the power rule:
f'(x) = 12x3 - 10x + 2
This is the derivative of the given function.
Find the critical points and determine if they are maxima or minima for the function f(x) = x3 - 6x2 + 9x
.
Solution: First, find the derivative:
f'(x) = 3x2 - 12x + 9
Set f'(x) = 0
to find critical points:
3x2 - 12x + 9 = 0
Solve for x
to get the critical points, then use the second derivative test to classify them.
JEE Math Notes - Class 11
Mathematical induction is a proof technique used to prove statements about natural numbers. It is based on the principle that if a statement holds for a base case, and if assuming it holds for a particular case implies it holds for the next case, then the statement is true for all natural numbers.
The principle of mathematical induction consists of two steps:
n
, usually n = 1
.n = k
, and then prove that the statement holds for n = k + 1
.If both of these steps are true, the statement is true for all natural numbers starting from the base case.
n = 1
).n = k
.n = k + 1
using the assumption made in Step 2.Prove that the sum of the first n
natural numbers is given by:
1 + 2 + 3 + ... + n = n(n + 1) / 2
Base Case: For n = 1
, we have:
1 = 1(1 + 1) / 2 = 1
The base case holds.
Inductive Step: Assume the formula is true for n = k
, i.e.,
1 + 2 + 3 + ... + k = k(k + 1) / 2
Now, prove the formula holds for n = k + 1
:
1 + 2 + 3 + ... + k + (k + 1) = (k(k + 1) / 2) + (k + 1)
After simplification, we get:
(k + 1)(k + 2) / 2
Thus, the formula holds for n = k + 1
, completing the proof by induction.
Prove that the sum of the squares of the first n
natural numbers is given by:
12 + 22 + 32 + ... + n2 = n(n + 1)(2n + 1) / 6
Base Case: For n = 1
, we have:
12 = 1(1 + 1)(2(1) + 1) / 6 = 1
The base case holds.
Inductive Step: Assume the formula holds for n = k
, i.e.,
12 + 22 + 32 + ... + k2 = k(k + 1)(2k + 1) / 6
Now, prove the formula holds for n = k + 1
:
12 + 22 + ... + k2 + (k + 1)2 = k(k + 1)(2k + 1) / 6 + (k + 1)2
After simplification, we get:
(k + 1)(k + 2)(2k + 3) / 6
This proves the formula by induction.
JEE Math Notes - Class 11
Probability is the branch of mathematics that deals with the likelihood or chance of different outcomes in an experiment. It is a measure of how likely an event is to occur. This chapter explores the concepts of random experiments, outcomes, events, and calculating probabilities in different situations.
The probability of an event is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes in the sample space.
Mathematically, the probability of an event E
is given by:
P(E) = (Number of favorable outcomes) / (Total number of possible outcomes)
Note: The probability of any event is always a number between 0 and 1, inclusive.
P(E) = n(E) / n(S)
, where n(E)
is the number of favorable outcomes, and n(S)
is the total number of possible outcomes in the sample space.P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
, where A
and B
are two events.P(A ∩ B) = P(A) * P(B)
, where A
and B
are independent events.When tossing a fair coin, the sample space is S = {H, T}
, where H
stands for heads and T
stands for tails.
The probability of getting heads P(H)> is:
P(H) = 1/2
The probability of getting tails P(T)> is:
P(T) = 1/2
When rolling a fair 6-sided dice, the sample space is S = {1, 2, 3, 4, 5, 6}
.
The probability of getting an even number is:
P(Even) = 3/6 = 1/2
In a deck of 52 playing cards, what is the probability of drawing a heart?
The number of favorable outcomes (hearts) is 13, and the total number of possible outcomes is 52.
P(Heart) = 13/52 = 1/4
JEE Math Notes - Class 11
Statistics is the branch of mathematics that deals with collecting, analyzing, interpreting, presenting, and organizing data. This chapter focuses on measures of central tendency, dispersion, and the basics of frequency distributions.
Measures of central tendency help in summarizing a large set of data with a single value that represents the center or typical value of the data. The three main measures are:
Mean = (Σx) / n
, where Σx
is the sum of all observations and n
is the number of observations.Consider the data: {5, 7, 10, 15, 20}
The mean is calculated as:
Mean = (5 + 7 + 10 + 15 + 20) / 5 = 57 / 5 = 11.4
Consider the data: {5, 7, 10, 15, 20}
Arranged in ascending order: {5, 7, 10, 15, 20}
The middle value is 10
, so the median is 10.
Dispersion refers to the spread or variability in a dataset. The key measures of dispersion are:
Range = Maximum Value - Minimum Value
.Variance = Σ(x - Mean)2 / n
.Standard Deviation = √Variance
.Consider the data: {5, 7, 10, 15, 20}
The mean is 11.4 (as calculated earlier).
Now, calculate the variance:
Variance = [(5 - 11.4)2 + (7 - 11.4)2 + (10 - 11.4)2 + (15 - 11.4)2 + (20 - 11.4)2] / 5 = 34.16
Therefore, the standard deviation is:
Standard Deviation = √34.16 = 5.84
A frequency distribution is a way to represent data by organizing it into classes or intervals. It helps in understanding the distribution of data in a dataset.
The table below shows an example of a frequency distribution:
Class Interval | Frequency |
---|---|
0 - 10 | 5 |
10 - 20 | 8 |
20 - 30 | 3 |