📐 CBSE Class 10 Mathematics
Complete Revision Notes with Solved Examples | All 14 Chapters
Based on Latest CBSE Syllabus 2024-25
Real Numbers
📌 The Fundamental Theorem of Arithmetic
Every composite number can be expressed as a product of prime numbers and this factorisation is unique, apart from the order in which the prime factors occur. This is also called the Unique Factorisation Theorem.
Key Statement
Any integer greater than 1 can either be a prime number or can be written as a unique product of prime numbers.
Examples:
- \(2 \times 11 = 22\) is the same as \(11 \times 2 = 22\)
- \(6 = 2 \times 3\) or \(3 \times 2\), where 2 and 3 are prime numbers
- \(15 = 3 \times 5\) or \(5 \times 3\), where 3 and 5 are prime numbers
- \(12 = 2 \times 2 \times 3 = 2^2 \times 3\)
HCF and LCM by Prime Factorisation Method
Key Formulae
Step 1: Find all the prime factors of given numbers.
Step 2: \(\text{HCF} = \text{Product of the smallest power of each common prime factor}\)
Step 3: \(\text{LCM} = \text{Product of the greatest power of each prime factor involved}\)
Relationship:
\[\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b\]
\[\text{HCF}(a,b) = \frac{a \times b}{\text{LCM}(a,b)} \quad \text{and} \quad \text{LCM}(a,b) = \frac{a \times b}{\text{HCF}(a,b)}\]
Number System Hierarchy: N ⊂ W ⊂ Z ⊂ Q ⊂ R
Rational and Irrational Numbers
Rational Numbers
A number in the form \(\dfrac{p}{q}\), where \(p\) and \(q\) are co-prime numbers and \(q \neq 0\), is known as a rational number.
Examples: \(2, -3, \dfrac{3}{7}, \dfrac{2}{5}, -\dfrac{5}{6}\), etc.
Irrational Numbers
A number that cannot be written in the form \(\dfrac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\).
Examples: \(\sqrt{2}, \sqrt{3}, \sqrt{5}\), etc.
📌 Proof of Irrationality
Theorem: \(\sqrt{2}\) is irrational
Proof: Assume \(\sqrt{2} = \dfrac{a}{b}\) in its simplest form.
Squaring both sides: \(2b^2 = a^2\), meaning \(a^2\) is even, so \(a\) is even.
Let \(a = 2c\), then \(2b^2 = 4c^2 \Rightarrow b^2 = 2c^2\), so \(b\) is also even.
This contradicts the assumption that \(\dfrac{a}{b}\) is in simplest form. Hence, \(\sqrt{2}\) is irrational.
Theorem: \(\sqrt{3}\) is irrational
Proof: Assume \(\sqrt{3} = \dfrac{a}{b}\). Squaring: \(3b^2 = a^2\), meaning \(a\) is divisible by 3.
If \(a\) is divisible by 3, then \(b\) is also divisible by 3. This contradicts simplest form. Hence, \(\sqrt{3}\) is irrational.
📌 Solved Examples
Show that \(\dfrac{7}{4}\) is a rational number.
Solution: A rational number is any number that can be expressed as \(\dfrac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\).
Since \(\dfrac{7}{4}\) is in the form \(\dfrac{p}{q}\), with \(p = 7\) and \(q = 4\), and both are integers, \(\dfrac{7}{4}\) is a rational number. \(\blacksquare\)
Prove that \(\sqrt{5}\) is irrational.
Solution: \(\sqrt{5}\) is not a perfect square, meaning it cannot be written as a fraction \(\dfrac{p}{q}\) where \(p\) and \(q\) are integers.
By the method of contradiction (similar to \(\sqrt{2}\)), assume \(\sqrt{5} = \dfrac{a}{b}\) in simplest form.
Squaring: \(5b^2 = a^2\). So 5 divides \(a^2\), which means 5 divides \(a\).
Let \(a = 5c\), then \(5b^2 = 25c^2 \Rightarrow b^2 = 5c^2\), so 5 divides \(b\).
This contradicts that \(\dfrac{a}{b}\) is in simplest form. Hence, \(\sqrt{5}\) is irrational. \(\blacksquare\)
Polynomials
📌 Definition and Types
A polynomial in one variable \(x\) of degree \(n\) is an algebraic expression of the form:
where \(n\) is a whole number and \(a_0, a_1, a_2, \ldots, a_n\) are real numbers.
Based on Number of Terms:
- Monomial: One term — e.g., \(2x, -4x^2, \dfrac{3}{4}x\)
- Binomial: Two terms — e.g., \(2x + 3, 5y^2 - 2y\)
- Trinomial: Three terms — e.g., \(4x^2 - 3x + 4\)
Based on Degree:
- Linear: Degree 1 — e.g., \(2x + 4\)
- Quadratic: Degree 2 — e.g., \(3x^2 - 2x + 4\)
- Cubic: Degree 3 — e.g., \(x^3 - x^2 + x + 1\)
📌 Zeroes of a Polynomial
A real number \(k\) is a zero of polynomial \(p(x)\) if \(p(k) = 0\).
Geometrically, the zeroes are the x-coordinates where the graph of \(y = p(x)\) intersects the x-axis.
Graphs of Linear, Quadratic, and Cubic Polynomials showing zeroes on x-axis
📌 Relationship Between Zeroes and Coefficients
For Quadratic Polynomial \(ax^2 + bx + c\):
\[\text{Sum of zeroes } (\alpha + \beta) = -\frac{b}{a}\]
\[\text{Product of zeroes } (\alpha \beta) = \frac{c}{a}\]
| Type | General Form | Max Zeroes | Relationship |
|---|---|---|---|
| Linear | \(ax + b\) | 1 | \(k = -\dfrac{b}{a}\) |
| Quadratic | \(ax^2 + bx + c\) | 2 | \(\alpha + \beta = -\dfrac{b}{a}\), \(\alpha\beta = \dfrac{c}{a}\) |
Discriminant of a Quadratic Polynomial
For \(f(x) = ax^2 + bx + c\), the discriminant \(D = b^2 - 4ac\) determines the nature of zeroes:
- \(D > 0\): Two distinct real zeroes
- \(D = 0\): Two equal (coincident) zeroes
- \(D < 0\): No real zeroes
📌 Solved Examples
Find the zeros of the polynomial: \(f(x) = 2x^2 - 8x + 6\)
Solution: Set \(f(x) = 0\):
\[2x^2 - 8x + 6 = 0\]
Factorise: \(2(x^2 - 4x + 3) = 2(x - 1)(x - 3) = 0\)
Setting each factor to zero:
\(x - 1 = 0 \Rightarrow x = 1\)
\(x - 3 = 0 \Rightarrow x = 3\)
∴ The zeros are \(x = 1\) and \(x = 3\).
Evaluate \(P(x) = x^3 - 4x^2 + 5x - 2\) for \(x = 2\).
Solution: Substitute \(x = 2\):
\[P(2) = (2)^3 - 4(2)^2 + 5(2) - 2\]
\[= 8 - 16 + 10 - 2 = 0\]
∴ \(P(x) = 0\) when \(x = 2\), confirming \(x = 2\) is a zero.