Algebra

 

 

 

 

Introduction


I. Facts :
(a) sum of first ‘n’ natural numbers – n*(n+1)/2
(b) sum of the squares of first ‘n’ natural numbers – n*(n+1)*(2n+1)/6
(c) sum of the cubes of first ‘n’ natural numbers – n^2*(n+1)^2/4
(d) total number of primes between 1 and 100 – 25

 

Basic Formulae of Algebra


  1. (a + b)2 = a2 + b2 + 2ab
  2. (a – b)2 = a2 + b2 – 2ab
  3. (a + b)2 – (a – b)2 = 4ab
  4. (a + b)2 + (a – b)2 = 2 (a2 + b2)
  5. (a2 – b2) = (a + b) (a – b)
  6. (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)
  7. (a3 + b3) = (a +b) (a2 – ab + b2)
  8. (a3 – b3) = (a – b) (a2 + ab + b2)
  9. (a3 + b3 + c3 – 3abc) = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
  10. If a + b + c = 0, then a3 + b3 + c3 = 3abc

 

Linear Equations


Linear Equation is an equation whose graph is a straight line. Each term has a degree of atmost 1. Each term can have degree 0 (constant term) or degree 1. A linear equation in one variable is an equation that involves only one variable x. Geeral form of linear equation can be written as ax + b = 0. There are no higher or lower order terms such as x2, x3 or x1/2.

Example 1: 6x + 5 = 0 is a linear equation. Note that it is in the form ax + b = 0, where a and b are constants. In this case a = 6 and b = 2. Any equation which can be reduced to form ax + b is also called linear eqation.

Example 2: 7x + 3 = 9, this can be written as 7x – 6 = 0

Example 3: 2(x+1) = 6(x-4) is also a linear equation.

Solving Linear Equations

Any linear equation can have only one solution. If you solve a linear equation, you will get one value of x. Solving linear equations is very simple. First, open all brackets and take all terms involving x to Left Hand Side (LHS) and constant terms to Right Hand Side (RHS). Then, using simple additions and subtractions, you will get the value of x.

Example 4:

2(x+1) = 6(x-4)
2x + 2 = 6x – 24
2x – 6x = – 24 – 2
– 4x = – 26
4x = 26
x = 26 / 4
x = 13 / 2 or 6.5

 

Example 5: The ratio of Rita’s age to her mother’s age is 3:8. The difference of their ages is 35 years. The ratio of their ages after 4 years will be:

(i) 7:12

(ii) 5:12

(iii) 38:43

(iv) 42:47

Solution:

Let their ages be 3x and 8x

8x – 3x =35

x =7

Their present ages are 21 and 56 years.

Ratio of their ages after 4 years are 25:60 = 5:12

Example 6 :2x + 5y = 103. Find the number of pairs of positive integers x and y tat satisfy this equation?

  1. 9
  2. 10
  3. 12
  4. 20

 

Rearranging the equation, we get:
2x = 103 – 5y

This says that when you subtract a multiple of 5 from 103, you get an even number. You have to subtract an odd multiple of 5 from 103 in order to get an even number. There are 20 multiples of 5 till 100, ten of which are odd. (Note that you cannot subtract 105, or higher multiples as they result in a negative value for x.)

So, y can have ten integer values. x also has 10 integer values, each corresponding to a particular value of y.
y = 1, gives us a potential value for x, so do y = 3, 5, 7….19. y can take 10 values totally.

 

Hence the answer is “10”

Choice B is the correct answer.

Example 7 : If a, b, c are distinct positive integers, what is the highest value a * b * c can take if a + b + c = 31?

  1. 1080
  2. 1200
  3. 1024
  4. 1056

 

The sum of three numbers is given; the product will be maximum if the numbers are equal.
a+b+c3a+b+c3 > ∛abc
So, if a + b + c is defined, abc will be maximum when all three terms are equal. In this instance, however, with a, b, c being distinct integers, they cannot all be equal.

So, we need to look at a, b, c to be as close to each other as possible.
a = 10, b =10, c = 11 is one possibility, but a, b, c have to be distinct. So, this can be ruled out.
The close options are,
a, b, c: 9, 10, 12; product = 1080
a, b, c: 8, 11, 12; product = 1056
Maximum product = 1080

Choice A is the correct answer.

 

Quadratic Equations


ax2+bx+c=0

Linear Equation can have degree of at most 1 and has only one solution. Quadratic Equation can have degree of at most 2 and has two solutions.

General form of quadratic equation: ax2 + bx + c = 0, where a, b and c are constants. Note that maximum degree of x is 2.

Solving Quadratic Equations

Unlike linear equations, any quadratic equation always has two solutions called roots of quadratic equation. After solving quadratic equation, you will get two values of x. To solve quadratic equation, you can use directly quadratic formula.

x = [-b ± √(b2-4ac)]/2a

Discriminant

In the above quadratic formula, the expression underneath the square root sign is called the discriminant of the quadratic equation. Discriminant is used to find the nature of roots.

Δ = b2 – 4ac

Case 1: Δ > 0

Real and distinct roots

Case 2: Δ = 0

Real and one distinct root (two same roots)

Case 3: Δ < 0

Roots are imaginary and occur as complex conjugates of each other

Sum and Product of Roots

Let α and β be the roots of quadratic equation x2 + px + q = 0

x2 + px + q = (x-α)(x-β)

x2 + px + q = x2 – (α+β)x + αβ

p= -(α+β)

q = αβ

Sum of the roots} = -p

Product of roots} = q

Example : Sum of the roots of a quadratic equation is 5 less than the product of the roots. If one root is 1 more than the other root, find the product of the roots?

  1. 6 or 3
  2. 12 or 2
  3. 8 or 4
  4. 12 or 4

Solution :

Sum of the roots = (product of the roots) – 5
Let r1 and r2 be the two roots
r1 = r2 + 1

Therefore,
r1 + 1 + r2 = (r1r2) * r2 – 5
r2 + 1 + r2 = (r2 + 1) * r2 – 5
2r2 + 6 = (r2)2 + r2
(r2)2 – r2 – 6 = 0
Solving, we get r2 = 3 or -2
Therefore, r2 = 4 or -1

Hence product of the two roots is either 12 or 2.

Choice B is the correct answer.

 

Inequalities


comparison relationship between two algebraic expressions or quantities is known as an Inequalities.

There are two main types of inequalities:

  1. Greater than or greater than equal to: > or ≥
  2.  Less than or less than equal to: < or ≤

Example: 3x + 1 > x − 3 or x2 − 3x ≤ 0

As is the case with equations, inequalities are ordered by degree and by the number of unknowns. In the above two examples, the first is a linear inequality with one unknown and the second is a quadratic inequality with one unknown.

Solving Inequations

On the basis of the laws of inequality, there are following working rules.

  1. Rule of Addition and Subtraction

Adding or subtracting a fixed number to each side of an inequality produces an equivalent inequality.

Example: Adding 2 to each side of the inequality x – 2 ≤ 1 is equivalent to x ≤ 3.

  1. Rule of multiplication or division by a positive number

All terms on both sides of an inequality can be multiplied or divided by a positive number.

  1. Rule of multiplication or division by a negative number

If all terms on both sides of an inequality are multiplied or divided by a negative number, the sign of the inequality will be reversed.

Example: If 6x – 4 < – 8x + 6 then dividing by -2, it becomes –3x + 2 > 4x – 3

Solution Sets

The solution of a linear inequation in one variable consists of many real numbers but may not all real numbers. As such the set of solutions is always a subset of real numbers. Solving an inequality means describing a set not just finding a number. This set is the solution set of the problem.

The solution of any inequation is always expressed in a solution set.

  1. Ranges where the ends are excluded

If the value of x is denoted as (1, 2) it means 1 < x < 2 i.e. x is greater than 1 but smaller than 2.

  1. Ranges where the ends are included

[2, 5] means 2 ≤ x ≤ 5

  1. Mixed Ranges

(3, 21] means 3 < x ≤ 21

  1. If solution set = {1, 2} means x = 1 and 2 only.

Example 1: Solve for x, –2x > 4

–2x > 4

Divide both sides by –2,

⇒ x < –2

 

Indices and Surds


Laws of Indices

am x an = a(m+n)

am / an = a(m-n)

(am)n = a(mn)

a(-m) = 1/(am)

a0 = 1

a1 = a

Logarithms


x = by

log x = log by

log x = y log b

(log x)/(log b) = y

logb x = y

Therefore, if x = by, then y = logb x

Logarithmic Identities

  1. logb (xy) = logb (x) + logb (y)
  2. logb (xd) = d logb (x)
  3. x(logb y) = y(logb x)
  4. c logb x + d logb y = logb (xc x yd)
  5. logb 1 = 0
  6. logb b = 1
  7. b(logb x) = x
  8. logb (bx) = x
  9. loga b = (logc b)/(logc a)

 

Surds


Law of Radicals

 

As we have seen that surds can be expressed with fractional exponents (indices), the laws of indices are therefore applicable to surds also. The laws of radicals are very useful to simplify a given radical or to reduce two given radicals to the same form.

 

Ist Law: For any positive integer ‘n’ and a positive rational number ‘a’ ( n a )n = n an = a.

IInd Law: If n is a positive integer and a, b are rational numbers, then

 

IIIrd Law: If n is a positive integer and a, b are rational numbers, then

IVth Law: If m, n are positive integers and a is a positive rational number, then

Vth Law: If m, n are positive integers and a is a positive rational number, then

n m (ap )m    = n ap  = mn apm

Progressions


A Progression is a sequence of numbers which have some kinf of relation. This relation determines what kind of a progression is. Generally, there are two types of progressions:

  1. Arithmetic Progression (AP)
  2. Geometric Progression (GP)

Any progression (AP or GP) can be generally expressed as

a1 + a2 + a3 + … + a(n-1) + an

Total Terms: n

First Term: a1

Last Term: an

Arithmetic Progression

In AP, the relation amoung sequence of numbers is that the difference between any two successive numbers is same.

Example: 3, 5, 7, 9, 11, 13, … is an AP with difference 2. This difference is called common difference.

an = a1 + (n – 1)d

Sn = n/2(a1 + an) = n/2[2a1 + (n-1)d]

Geometric Progression

In GP, each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio.

Example: 2, 6, 18, 54, …

a + ar + ar2 + ar3 + ar4 + …

an = ar(n-1)

 

Examples :

 

1.How many even integers n, where 100 n 200, are divisible neither by seven nor by nine ?

 

(1) 40 (2) 37 (3) 39 (4) 38

Sol.There are 101 integers such that 100 n 200. Out of which 51 are even

n(div 7 or div 9) = n(div 7) + n(div9) – n(div7 and 9)

= 7 + 6 – 1 = 12 Total number of numbers neither divisible by 7 nor by 9 = 51 – 12 = 39. Ans.(3)

 

2.Consider the sets Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1, 2, 3,…., 96. How many of these sets contain 6 or any integral multiple thereof (i.e., any one of the numbers 6, 12, 18, …)?

(1) 80 (2) 81 (3) 82 (4) 83

Sol.T1 = {1, 2, 3, 4, 5}, T 2 = {2, 3, 4, 5, 6}, T 3 = {3, 4, 5, 6, 7},

T4 = {4, 5, 6, 7, 8}, T 5 = {5, 6, 7, 8, 9}, T 6 = {6, 7, 8, 9, 10}

From T1 to T6 out of 6 sets 5 contain multiple of 6

out of 96 sets 80 will contain multiple of 6. Ans.(1)

 

  1. Each of the series S1= 2 + 4 + 6 + 9 +… and S2= 3 + 6 + 9 + 12 + 15 … are continued to 100 terms. Find how many terms are identical in both the series.

(1) 32 (2) 33 (3) 34 (4) 31

Sol.s1= 2 + 4 + 6 + 8 + 10 + 12…………..

s2= 3 + 6 + 9 + 12 + 15 + 18……….

Every third term in s 1 is same as every second term in s 2. Therefore identical term in will

100/2= 50 or100/3 = 33 out of 50 & 33, lower value is 33. Ans.(2)

 

4.One hundred boxes each contain an equal number of coins. Someone takes a certain number of coins out of the first box, twice this number out of the second box, three times the number out of the third box, and so on up to one hundred times the number out of the hundredth box, there is one coin left in the last box and 14950 coins left altogether in the boxes. How many coins were taken out of the first box?

(1) 4 (2) 5 (3) 3 (4) None of these

Sol.Let the number of coins in each box be y

and the number of coins taken out from the first box be x

Total number of coins in 100 boxes = 100y

and total number of coins taken out

= x + 2x + 3x + ……+ 100x = 5050x

and also y – 100x = 1 ….(1)

According to problem we have,

100y – 5050x = 14950 ….(2)

Solving (1) and (2) we have,

x = 3. Ans.(3)

 

  1. Suresh is asked to find the arithmetic mean of 20 consecutive two-digit numbers. He made a mistake while counting two numbers of different digits – he interchanged the digits of the number. He got an arithmetic mean which is 2 more than the actual one. What is the difference between the original numbers that he wrongly interpreted?

(1) 1 (2) 14 (3) 16 (4) Not possible

Sol.Let the numbers be from x to x + 19 then the arithmetic means will be x + 9.5.

But Suresh found the arithmetic mean to be x + 11.5 This means the sum is increased by 40 as the average increased by 2. But when a number is interchanged, the difference between the two is always a multiple of 9. Hence, if digits of two 2-digit numbers are interchange then their sum will increase or decrease by 18.

And since 40 is not a multiple of 9. It is not possible to have such combination of numbers. Ans.(4)

 

  1. Six numbers are in AP such that their sum is 3. The first number is four times the third number. The fifth number of the series is

(1) –15 (2) –3 (3) 9 (4) –4

Sol. Let the six numbers be, a – 5d , a – 3d , a– d , a + d , a + 3d , a + 5d

Now, a – 5d + a – 3d + a – d + a+d + a+3d + a+5d = 3 or, 6a = 3 a = .5

According to the problem we have, a – 5d = 4( a – d) on solving we have,

3a = d or d = –1.5

Fifth number of the series = a + 3d = .5 + (3 × {–1.5}) = –4. Ans. (4)

 

8.On a straight road XY, 100 metres in length, 5 stones are kept begining from the end X. The distance between two adjacent stones is 2 metres. A man is asked to collect the stones one at a time and put at the end Y. What is the distance covered by him?

(1) 460 metres (2) 540 metres (3) 860 metres (4) 920 metres

Sol.

s1 s2 s3 s4 s5

X 2m 2m 2m 2m 92 m Y

To put s1 at Y and coming back to s2 total distance covered will be 100 + 98 = 198

Similarly adding distance for all the stone we get

198 + 194 + 190 + 186 + 92 = 860m . Ans.(3)

 

 

Functions


Introduction

The phrase “y is a function of x” means that the value of y depends upon the value of x, so y can be written in terms of x (e.g. y = 3x ).
If f(x) = 3x, and y is a function of x (i.e. y = f(x) ), then the value of y when x is 4 is f(4), which is found by replacing x”s by 4″s .

Example

If f(x) = 3x + 4, find f(5) and f(x + 1).

f(5) = 3(5) + 4 = 19
f(x + 1) = 3(x + 1) + 4 = 3x + 7

Domain and Range

The domain of a function is the set of values which you are allowed to put into the function (so all of the values that x can take). The range of the function is the set of all values that the function can take, in other words all of the possible values of y when y = f(x). So if y = x2, we can choose the domain to be all of the real numbers. The range is all of the real numbers greater than (or equal to) zero, since if y = x2, y cannot be negative.

One-to-One

We say that a function is one-to-one if, for every point y in the range of the function, there is only one value of x such that y = f(x). f(x) = x2 is not one to one because, for example, there are two values of x such that f(x) = 4 (namely –2 and 2). On a graph, a function is one to one if any horizontal line cuts the graph only once.

Composing Functions

fg means carry out function g, then function f. Sometimes, fg is written as fog

Example

If f(x) = x2 and g(x) = x – 1 then
gf(x) = g(x2) = x2 – 1
fg(x) = f(x – 1) = (x – 1)2

As you can see, fg does not necessarily equal gf

The Inverse of a Function

The inverse of a function is the function which reverses the effect of the original function. For example the inverse of y = 2x is y = ½ x .
To find the inverse of a function, swap the x”s and y”s and make y the subject of the formula.

Example

Find the inverse of f(x) = 2x + 1
Let y = f(x), therefore y = 2x + 1
swap the x”s and y”s:
x = 2y + 1
Make y the subject of the formula:
2y = x – 1, so y = ½(x – 1)
Therefore f -1(x) = ½(x – 1)

f-1(x) is the standard notation for the inverse of f(x). The inverse is said to exist if and only there is a function f-1 with ff-1(x) = f-1f(x) = x

Note that the graph of f-1 will be the reflection of f in the line y = x.

Graphs


Functions can be graphed. A function is continuous if its graph has no breaks in it. An example of a discontinuous graph is y = 1/x, since the graph cannot be drawn without taking your pencil off the paper:

A function is periodic if its graph repeats itself at regular intervals, this interval being known as the period.
A function is even if it is unchanged when x is replaced by -x . The graph of such a function will be symmetrical in the y-axis. Even functions which are polynomials have even degrees (e.g. y = x²).
A function is odd if the sign of the function is changed when x is replaced by -x . The graph of the function will have rotational symmetry about the origin (e.g. y = x³).

The Modulus Function

The modulus of a number is the magnitude of that number. For example, the modulus of -1 ( |-1| ) is 1. The modulus of x, |x|, is x for values of x which are positive and -x for values of x which are negative. So the graph of y = |x| is y = x for all positive values of x and y = -x for all negative values of x:

Transforming Graphs

If y = f(x), the graph of y = f(x) + c (where c is a constant) will be the graph of y = f(x) shifted c units upwards (in the direction of the y-axis).
If y = f(x), the graph of y = f(x + c) will be the graph of y = f(x) shifted c units to the left.
If y = f(x), the graph of y = f(x – c) will be the graph of y = f(x) shifted c units to the right.
If y = f(x), the graph of y = af(x) is a stretch of the graph of y = f(x), scale factor (1/a), parallel to the x-axis. [Scale factor 1/a means that the “stretch” actually causes the graph to be squashed if a is a number greater than 1]

Example

The graph of y = |x – 1| would be the same as the above graph, but shifted one unit to the right (so the point of the V will hit the x-axis at 1 rather than 0).

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