Chapter 1

REAL NUMBERS

Euclid’s Division Lemma

Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b.Euclid’s division algorithm

To obtain the HCF of two positive integers, say c and d, with c > d,Step 1 : Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 ≤ r < d. Step 2 : If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r. Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF. The Fundamental Theorem of Arithmetic

Every composite number can be expressed ( factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.Theorem : Let p be a prime number. If p divides a 2 , then p divides a, where a is a positive integer

Proof:

Let the prime factorization of a be

a = p1.p2.p3………pn where p1, p2, p3………pn are prime, not necessarily distinct.

Therefore,

a2= (p1.p2.p3………pn) (p1.p2.p3………pn) = p12.p2

2.p32………pn

2

Given that p divides a2

Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a2

Using the uniqueness part of the Fundamental Theorem of Arithmetic, the only prime factors of a2 are p1, p2, p3………pn.

So p is one of p1, p2, p3………pn

Now, since a = p1.p2.p3………pn

p divides a.

Rational Numbers and Their Decimal Expansions

Chapter 2Polynomials

Polynomial –Poly means many and nomial means terms.

Polynomial is an expression of more than two algebraic terms.A polynomial can have: constants   (like   3 ,   -20 , or   ½ ), variables   (like   x   and   y ), exponents   (like the 2 in y 2 ), but only   0, 1, 2, 3, ...   etc are allowed ,that can be combined using   addition, subtraction, multiplication and division except division by a variable.

The degree of a polynomial with only one variable is the largest exponent of that variable.

The Standard Form for writing a polynomial is to put the terms with the highest degree first.General form of a linear polynomial is ax + b , Quadratic polynomial is ax2 + bx + c, cubic polynomial is  ax3 + bx2 + cx + d Zero of a polynomial

A real number k is said to be a zero of a polynomial p(x), if p(k) = 0.

Zero of the linear polynomial ax + b is −ba =−Constant termCoefficeint of x

If α and β are the zeroes of the quadratic polynomial ax2 + bx + c, then

Sum of zeroes α+β=−ba,∧product of zeroesαβ= c

a

If α, β and γ are the zeroes of the cubic polynomial

ax3 + bx2 + cx + d then α+β+γ=−ba , αβ+βγ+γα= c

a , αβγ=−da

CHAPTER 3

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

An equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is called a linear equation in two variable.

• The graph of a linear equation in two variables plotted on a Cartesian plane is a straight line

The general form for a pair of linear equations in two variables x and y is

a 1x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 where a1, b1, c1,a2,b2,c2 are all real numbers and a1

2 + b12≠0 , a2

2 + b22≠0

The pair of linear equations represented by these lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

1. If a1a2 ≠b1b2

  then the pair of linear equations has exactly one solution.

2. If a1a2=b1b2

=c1c2

 then the pair of linear equations has infinitely many

solutions.

3. If a1a2=b1b2≠c1c2

  then the pair of linear equations has no solution.

Terms using in word problems

Chapter 6

TRIANGLESTwo figures are said to be congruent, if they have the same shape and the same size.

Two figures having the same shape (and not necessarily the same size) are called similar figures.

Two polygons of the same number of sides are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion).

Criteria for Similarity of Triangles

AAA SIMILARITY

If the corresponding angles of two triangles are equal, then their corresponding sides are proportional and the triangles are similar.

∆ABC ∆DEF when

A = D, B = E and

C = F

SAS SIMILARITY

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the triangles are similar.  ∆ABC ∆DEF when ABDE= AC

DF and A = D

SSS SIMILARITY

If sides of one triangle are proportional to the sides of the other triangles, then their corresponding angles are equal and the triangles are similar.

∆ABC ∆DEF when ABDE= BCEF= ACDF

Basic Proportionality Theorem (Thales Theorem) :-

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Given:-

Given a triangle ABC , BC intersects other two sides AB and AC at D and E respectively and BC// DE

To Prove :-

ADDB= AEEC

Construction:-

Join BE and CD ,Draw DM ⊥ AC and EN ⊥ AB

.

Proof

Δ BDE and DEC are on the same base DE and between the same parallels BC and DE.

So, ar(BDE) = ar(DEC) (3)

Therefore, from (1), (2) and (3), we have

Converse of BPT:-

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Areas of Similar Triangles:-Theorem : The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Given:-

Two triangles ABC and PQR such that Δ ABC ~ Δ PQR

To Prove:-

ar (∆ ABC )ar (∆PQR )

=( ABPQ )2

=( BCQR )2

=(CARP )2

Construction:-

Draw altitudes AM and PN of the triangles

Proof:-

Theorem : If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other

Pythagoras TheoremIn a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given:-

A right triangle ABC right angled at B

To Prove :-

AC2=AB2+BC 2

Construction:-

Draw BD AC

Proof :-

(1)

(2)Adding (1) and (2)

Converse of Pythagoras theorem

In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle

Given

Given a triangle ABC in which AC2=AB2+BC 2

To prove ∠ B = 90°

Construction

Construct a Δ PQR right angled at Q such that PQ = AB and QR = BC

Proof

Chapter 14

StatisticsMean

1. Direct Method

If x1 , x2. ., xn are observations with respective frequencies

f 1 , f2 . . ., fn , then mean observation is

x = f i x iN , N = f I

2. Assumed Mean Method x = a + f id i

N

N = f I

Where a= assumed meand i= x i−a

3. Step deviation Method x = a + ( f iuiN )h

N = f I

Where a= assumed meanui= xi−a

h

h= width of the class

Mode of Grouped Data

Median

Median is a measure of central tendency which gives the value of the middle-most observation in the data.

Cumulative frequency

The cumulative frequency of a class is the frequency obtained by adding the frequencies of all the classes preceding the given class.

Median of grouped data

1. Locate the class whose cumulative frequency is greater than (and nearest to). This is called the median class.

2. After finding the median class, we use the formula for calculating the median.

Empirical Formulae

There is a empirical relationship between the three measures of central tendency

Mode = 3 median – 2 mean

Graphical Representation of Cumulative Frequency Distribution

For cumulative frequency curve, or an ogive (of the less than type)

• Write the upper limits of the respective class intervals.

• Find the corresponding cumulative frequencies.

To represent the data in the table graphically,

• Mark the upper limits of the class intervals on the horizontal axis (x-axis)

• Corresponding cumulative frequencies on the vertical axis ( y-axis), choosing a convenient scale.

For cumulative frequency curve, or an ogive (of the more than type)

• Write the lower limits of the respective class intervals.

• Find the corresponding cumulative frequencies.

To represent the data in the table graphically,

• Mark the lower limits of the class intervals on the horizontal axis (x-axis)

• Corresponding cumulative frequencies on the vertical axis ( y-axis), choosing a convenient scale.

Method 2

Draw both ogives (i.e., of the less than type and of the more than type) on the same axis. The two ogives will intersect each other at a point. From this point, if we draw a perpendicular on the x-axis, the point at which it cuts the x-axis gives us the median

Introduction to trigonometry

Sinθ = Opp . sideHypotenuse cosec θ=

HypotenuseOpp. side

Cos θ= adj . sideHypotenuse Sec θ=Hypotenuse

adj . side

Tan θ = Opp. sideadj . side Cot θ = adj . sideOpp. side

Sin 30 = 12 Sin 60 = √32 Sin 45 = 1√2

Cos 30 = √32 Cos 60 = 12 Cos 45 = 1√2

Tan 30 = 1√3 Tan 60 = √3 tan 45 = 1

Sin 0 = 0 Sin 90 = 1

Cos 0 = 1 Cos 90 = 0

Tan 0 = 0 Tan 90 =

Tan θ = sin θcosθ Sin θ= 1Cosec θ Cosec θ= 1

sin θ

Cot θ = cosθsin θ Cos θ = 1Sec θ Sec θ = 1

cosθ

Tan θ = 1cot θ Cot θ = 1

tan θ

Sin θ Cosec θ = 1 Sin ( 90 – θ ) = Cos θ Cosec ( 90 – θ ) = Sec θ

Cos θ Sec θ = 1 Cos ( 90 – θ ) = Sin θ Sec ( 90 – θ ) = Cosec θ

Tan θ Cot θ = 1 tan ( 90 – θ ) = Cot θ Cot ( 90 – θ ) = tan θ

sin 2θ + cos2θ = 1 sec 2θ - tan2θ = 1

sin 2θ=1−cos 2θ sec 2θ=1+ tan 2θ

cos 2θ=1−sin 2θ tan 2θ=sec 2θ−1

Cosec 2θ−cot 2θ=1

Cosec 2θ=1+cot 2θ

cot 2θ=Cose 2θ−1