CLASSIFICATION OF REAL NUMBERS - Abhijit Kumar Jha · 2014. 10. 14. · NUMBERS (Theory) CLASSIFICATION OF REAL NUMBERS Real Number are classified into rational and irrational numbers

KVPY NOTES BY ABHIJIT KUMAR JHA


CHAPTER – 1

NUMBERS

(Theory) CLASSIFICATION OF REAL NUMBERS Real Number are classified into rational and irrational numbers. Rational Numbers A number which can be expressed in the form p/q where p and q are integers and q 0 is called a rational number. For example, 4 is a rational number since 4 can be written as 4/1 where 4 and 1 are integers and the denominator 1 0. Similarly, the numbers ¾, - 2/5 etc. are also rational numbers. Recurring decimals are also rational numbers. A recurring decimal is a number in which one or more digits at the end of a number after the decimal point repeats endlessly ( For example, 0.333….., 0.111111…, 0.166666…., etc. are all recurring decimals). Any recurring decimal can be expressed as a fraction of the form p/q and hence it is a rational number. We will study in another section in this chapter the way to convert recurring decimals into fractions. Between any two numbers, there can be infinite number of other rational numbers. Irrational Numbers Numbers which are not rational but which can be represented by points on the number line are called irrational numbers. Examples for irrational numbers are 342, 3, 5, 8 etc. Numbers like , e are also irrational no Between any two numbers, there are infinite numbers of irrational numbers. Another way of looking at rational and irrational numbers is Any terminating or recurring decimal is a rational number. Any non – terminating non – recurring decimal is an irrational number. Integers All integers are rational numbers, integers are classified into negative integers, zero and positive integers Positive integers can be classified as Prime Numbers and Composite Numbers. In problems on Numbers, we very often use the word “number” to mean an “ integer” Prime Numbers A number which does not have any factor apart from one and itself is called a prime number. Examples for prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, etc. There is no general formula that can give prime numbers. Every prime number greater than 3 can be written in the form of (6k + 1) or (6k – 1) where k is an integer. Composite Numbers



Any number other than 1, which is not a prime number is called a composite number, in other words, a composite number is a number which has factors other than one and itself. Examples for composite numbers are 4, 6, 8, 9, 10, 14, 15, etc. Note : The number 1 is neither prime nor composite. The only prime number that is even is 2. There are 15 prime numbers between 1 and 50 and 10 prime numbers between 50 and 100. So, there are a total of 25 prime numbers between 1 and 100. Relative Primes Two numbers are said to be relative prime or co – primes if they do not have any common factor other than 1. For example, the numbers 15 and 16 do not have any common factors and hence they are relative primes. Please note that none of the two numbers may individually be prime and still they can be relative primes. Unity is a relative prime to all numbers. Multiples If one number is divisible exactly by a second number, then the first number is said to be a multiple of the second number, then the first number is said to be a multiple of the second number. For example, 15 is a multiple of 5; 24 is a multiple of 4. Factors If one number divides a second number exactly, then the first number is said to be a factors of the second number. For example, 5 is a factor of 15; 3 is a factor of 18. Factors are also called sub – multiples divisors. Even and odd numbers Numbers divisible by 2 are called even numbers whereas numbers that are not divisible by 2 are called odd numbers. Examples for even number are 2, 4, 6, 8, 10, etc Examples for odd numbers ends in 1, 3, 5, 7, 9, etc NOTE: Every even number ends in 0, 2, 4, 6 or 8 The sum of any number of even numbers is always even. The sum of odd number of odd numbers ( i.e, the sum of 3 odd numbers, the sum of 5 odd numbers, etc) is always odd whereas the sum of even number of odd numbers ( i.e the sum of 2 odd numbers, the sum of 4 odd numbers etc.) is always even. The product of any number of odd numbers is always odd. The product of any number of numbers where there is at least one even number is even. A number is said to be a perfect number if the sum of ALL its factors excluding itself ( but including 1) is equal to the number itself. For example, 6 is a perfect number because the factors of 6, i.e 1, 2 and 3 add up to the number 6 itself. Other examples of perfect numbers are 28, 496, 8128 etc. Rules for divisibility



In a number of situations, we will need to find the factors of a given number. Some of the factors of a given number can, in a number of situations, be found very easily either by observation or by applying simple rules. We will look at some rules for divisibility of numbers. Divisibility by 2 A number divisible by 2 will have an even number as its last digit ( For example 128, 246, 2346 etc) Divisibility by 3 A number is divisible by 3 if the sum of its digits is a multiple of 3. For example, take the number 9123, the sum of the digits is 9 + 1 + 2 + 3 = 15 which is a multiple of 3. Hence, the given number 9123 is divisible by 3. Similarly 342, 789 etc are all divisible by 3. If we take the number 74549. the sum of the digits is 29 which is not a multiple of 3. Hence the number 74549 is not divisible by 3. Divisibility by 4 A number is divisible by 4 if the number formed with its last two digits is divisible by 4. For example, if we take the number 178564, the last two digits form 64 since the number 64 is divisible by 4. the number 178564 is divisible by 4. If we take the number 476854, the last two digits form 54 which is not divisible by 4 and hence the number 476854 is not divisible by 4. Divisibility by 5 A number is divisible by 5 if its last digit is 5 or zero ( eg. 15, 40 etc) Divisibility by 6 A number is divisible by 6 if it is divisible both by 2 and 3 ( 18, 42, 96 etc) Divisibility by 7 If the difference between the number of tens in the number and twice the units digit is divisible by 7, then the given number is divisible by 7. Otherwise, it is not divisible by 7. Take the units digit of the number, double it and subtract this figure from the remaining part of the number. If the result so obtained is divisible by 7, then the original number is divisible by 7. if that result is not divisible by 7, then the number is not divisible by 7. For example, let us take the number 595. The units digit is 5 and when it is doubled, we get 10. The remaining part of the number is 59. If 10( which is the units digit double) is subtracted form 59 we get 49. Since this result 49 is divisible by 7, the original number 595 is also divisible by 7. Similarly, if we take 967,doubing the units digit give 14 which when subtracted from 96 gives a result of 82. Since 82 is not divisible by 7, the number 967 is not divisible by 7. If we take a larger number, the same rule may have to be repeatedly applied till the result comes to a number which we can make out by observation. Whether it is divisible by 7. For example, take 456745. We will write down the figures in various steps as shown below.

Number Twice the units digit Remaining part of the number

Col (3) – Col (2)

456745 45664 4558 439

10 8 16 18

45674 4566 455 43

45664 4558 439 25



Since 25 in the lest step is not divisible by 7, the original number 456745 is not divisible by 7 Divisibility by 8 A number is divisible by 8, if the number formed by the last 3 digits of the number is divisible by 8. For example, the number 3816 is divisible because the last three digits form the number 816. which is divisible by 8. Similarly, the numbers 14328, 18864 etc. are divisible by 8. If we take the number 48764, it is not divisible by 8 because the last three digits number 764 is not divisible by 8. Divisibility by 9 A number is divisible by 9 if the sum of its digits is a multiple of 9. For example, if we take the number 6318, the sum of the digits of this number is 6 + 3 + 1 + 8 which is 18. Since this sum 18 is a multiple of 9. Similarly, the numbers 729, 981, etc are divisible by 9. If we take the number 4763, the sum of the digits of this number is 20 which is not divisible by 9. Hence the number 4763 is not divisible by 9. Divisibility by 10 A number divisible by 10 should be end in zero. Divisibility by 11 A number is divisible by 11 if the sum of the alternate digits is the same or they differ by multiples of 11- that is, the difference between the sum of digits in odd places in the number and the sum of the digits in the even places in the number should be equal to zero or a multiple of 11. For example, if we take the number 132, the sum of the digits in odd places is 1 + 2 = 3 and the sum of the digits in even places is 3. Since these two sums are equal, the given number is divisible by 11. If we take the number 785345, the sum of the digits in odd places is 16 and the sum of the digits in even places is also 16. Since these two sums are equal, the given number is divisible by 11. Hence the number is divisible by 11. If we take the number 89394811, the sum of the digits is odd places is 8 + 3 + 4 + 1, which is equal to 16. The sum of the digits in even places is 9 + 9 + 8 + 1, which is equal to 27. The difference between these two figures is 11 ( 27 – 16), which is a multiple of 11. Hence the given number 89394811 is divisible by 11. The number 74537 is not divisible by 11 because the sum of the digits in odd places is 19 and the sum of the digits in even places is 7 and the difference of these two figures is 12 is not a multiple of 11. Divisibility by numbers like 12, 14, 15 can be checked out by taking factors of the number which are relatively prime and checking the divisibility of the given number by each of the factors. For example, a number is divisible by 12 if it is divisible both by 3 and 4. The next number that is of interest to us from divisibility point of view is 19. Divisibility by 19 If the sum of the number of tens in the number and twice the units digit is divisible by 19, then the given number is divisible by 19. Otherwise it is not. Take the units digit of the number, double it and add this figure to the remaining part of the number. If the result so obtained is divisible by 19, then the original number is divisible by 19. If that result is not divisible by 19, then number is not divisible by 19.



For example let us take the number 665. The units digit is 5 and when it is doubled, we get 10. The remaining part of the number is 66. if 10 ( which is the units digit doubled) is added to 66, we get 76. Since this result 76 is divisible by 19, the number 969 is divisible by 19. If we take 873, double the units digit ( 2 3 6 ) added to the remaining part of the number (87),we get 93 which is not divisible by 19. Hence the original number 873 is not divisible by 19. If we take a larger number, the same rule may have to be repeatedly applied till the result comes to a number which we can make out by observation whether it is divisible by 19. For example, take 456760 . We will write down the figures in various steps as shown below.

Number Twice the digit Remaining part of the number

Col (1) + Col (2)

456760 45676 4579 475

0 12 18 10

45676 4567 457 47

45676 4579 475 57

Since 57 in the last step is divisible by 19, the original number 456760 is divisible by 19. Let us take another example, the number 37895. Let us follow the above process step by step till we reach a manageable number 37895 Double the units digit 5 and add the 10 so obtained to 3789. We get 3799 Double the units digit 9 and add the 18 so obtained to 379. We get 397 Double the units digit 7 and add the 14 so obtained to 39. We get 53. Since 53 is not divisible by 19, 37895 is not divisible by 19. Recurring Decimals A decimal in which a digit or a set of digits is repeated continually is called a recurring decimal. Recurring decimals are written in a shortened form, the digits which are repeated being marked by dots placed over the first and the last of then, thus

8 2.666....... 2.6 2.6;3

or

1 0.142857142857142857..... 0.1428577

In case of 1/7, where the set of digits 142857 is recurring, the dot is placed on top of the first and the last digits of the set or alternatively, a bar is placed over the entire set of the digits that recur. A recurring decimal like 0.3 is called a pure recurring decimal because all the digits after the decimal point are recurring. A recurring decimal like 0.16 ( which is equal to 0.166666….) is called a mixed recurring because some of the digits after the decimal are not recurring ( in this case, only the digit 6 is recurring and the digit 1 is not recurring). A recurring decimal is also called a “ circulator”. The digit , or set of digits, which is repeated is called the “period” of the decimal. In the decimal equivalent to 8/3, the period is 6 and in 1/7 it is 142857. As already discussed, all recurring decimals are rational numbers as they can be expressed in the form p/q, where p and q are integers. The general rule for converting recurring decimals into fractions will be considered later. THE LAST DIGIT OF ANY POWER



The last digits of the powers of any number follow a cyclic pattern – i.e, they repeat after certain number of steps. If we find out after how many steps the last digit of the powers of a number repeat, then we can find out the last digits of any power of any number. Let us look at powers of 2. Last digit of 21 is 2 Last digit of 22 is 4 Last digit of 23 is 8 Last digit of 24 is 6 Last digit of 2 5 is 2 Since last digit of 25 is the same as the last digit of 21 then onwards the last digit will start repeating, i.e digits of 25, 26, 27, 28 will be the same as those of 21, 22, 23, 24. Then the last digit of 29 is again the same as the last digit of 21 an so on. So, we have been able to identify that for powers of 2 the last digits repeat after every 4 steps. In other words whenever the power is a multiple of 4, the last digit of that number will be the same as the last digit of 24. Suppose we want to find out the last digit of 267, we should look at a multiple of 4 which is less than or equal the power 67. Since 64 is a multiple of 4, the last digit of 264 will be the same as the last digit of 24 Then the last digits of 265, 266, 267 will be the same as the least digits of 21, 22, 23 respectively. Hence the last digit of 267 is the same as the last digit of 23 i.e 8. Similarly, we can find out the last digit of 374 by writing down the pattern of the powers of 3. Last digit of 31 is 3 Last digit of 32 is 9 Last digit of 33 is 7 Last digit of 34 is 1 Last digit of 35 is 3 The last digit repeats after 4 steps ( like in the case or powers of 2) To find the last digit of 374, we look for a multiple of 4 which is less than or equal to 74. since 72 is a multiple of 4, the last digit of 372 will be the same as that of 34. Hence the last digit of 374 will be the same as the last digit of 32, i.e 9. LAST DIGIT OF A SUM OR PRODUCT The problem consist of finding the last digit of the sum of two numbers each of which is power of some integer. For example, you may be asked to find out the last digit of the sume 267 + 374 In general, when we want to find out the last digit of the sum to two numbers, we can just take the last digit of the two numbers and add them up. That will be the last digit of the sum. The last digit of 243 + 456 will be the same as the sum of the least digits of the two numbers. i.e., the sum of 3 and 6, which is 9. Similarly, in the case of 267 + 374 also, the last digit will be equal to the sum at the last digits of the two terms 267 and 374. We have already looked at finding out the last digit of powers like 267 and 374. Hence the last digit of 267 + 374 is 8 + 9 i.e 7. Similarly, the last digit of the product 267 374 will be equal to the last digit of the product of the last digit of 267 and last of 374, i.e the last digit of 8 9, i.e., 2 Hence the last digit of 67 742 3 is 2.



Example 1. Find the last digit of 3424 6324 Sol: Writing down the powers of 3 and 6 to check the pattern of the last digit, we have Last digit of 31………..3 Last digit of 32……….9 Last digit of 33………..7 Last digit of 34………..1 Last digit of 35………..3 Last digit of 61………..6 Last digit of 62………...6 Last digit of 63………...6 We find that the last digits of powers of 3 repeat after 4 steps; the last digit of any power of 6 is always 6. The last digit of 3424 will be the same as 34 as 424 is multiple of 4. So last digit of 3424 is 1. Last digit of 6 324 will be 6 as seen above. Hence, the last digit of 424 4243 6 Will be equal to the last digit of 1 6 6. F INDING THE REMAINDER IN DIVISIONS INVOLVING POWERS OF NUMBERS There is one particular model of problem explained below with help of an example. Example 1. Find the remainder of the division 342/4 Sol. Let us find the pattern that remainders follow when powers of 3 are divided by 4. Remainder of 31/4 3 Remainder of 32/4 1 Remainder of 33 / 4 3 We find the remainder repeats in the third steps i.e after 2 steps. So, remainder of 342 when divided by 4, is the same as 32 when divided by 4, since 42 is a multiple 2 Hence the remainder is 1. PATTERN METHOD Similar to the last digit of the powers of number repeating in a certain pattern, the remainders of powers of a number also follow a certain pattern. If we identify the pattern in which the remainders repeat. We can find out the remainder of any division given To solve the example given above, let us find the pattern that remainders follow when various powers of 2 are divide by 7. Remainder when 21 is divided by 7 is 2 Remainder when 22 is divided by 7 is 4 Remainder when 23 is divided by 7 is 1 Remainder when 24 is divided by 7 is 2 We find that the remainder repeats in the fourth step, i.e after 3 steps So, - the remainder of 24 when divided by 7 is the same as that when 21 is divided by 7 is the same as that when 21 is divided by 7, i.e ., 2 - the remainder of 25 when divided by 7 is the same as that when 22 is divided by 7, i.e 4 - the remainder of 26 when divided by 7 is the same as that when 23 is divided by 7, i.e., 1 - the remainder of 27 when divided by 7 is the same as that when 21 is divided by 7, i.e., 2 and so on.



If we take 254 since 54 is divisible by3, 254 itself completes a cycle of 3 steps and hence the remainder when 254 is divided by will be the same as that when 23 is divided by 7. Hence the remainder in 1. REMAINDER THEOREM METHOD We can apply Remainder Theorem to find the remainder in problems like the one discussed above. Let us first look at Remainder Theorem and understand it. Remainder Theorem states that when f(x), a polynomial function in x is divided by x –a, the remainder is f(a). A polynomial function in x is a function where x will appear only in the form of xn where x is a positive integer and not in any other form. Let us take an example to under stand remainder Theorem. When the function x2 + 2x – 3 is divided by x – 1, the remainder will be f(1). This is because, as per Remainder Theorem, When the divisor is (x- a), the remainder is f(a). Here the divisor is x – 1 and hence the remainder is f(1). To get f(1), we should substitute x = 1 in the given equation. As we get f(1) = 0, the remainder in this case is 0. ( Note that when f(x) is divided by x –a, if the remainder is 0, then x- 1 is a factor of x2 + 2x – 3). When the function x2 + 2x + 3 is divided by x + 1, the reminder will be f(-1) which is (-1)2 + 2(-1) + 3, i.e, 2 Now let us take the example of finding the remainder when 254 is divided by 7 ( which was solved by the Pattern method above) and solve it by Remainder Theorem Method. In the Division 254/7, the dividend is 254 and the divisor in 7. Since the numerator is in terms of powers of 2, express the denominator also in terms of power of 2.In this case 7 can be written as 8 -1 which is 23 – 1. So, now the denominator is in terms of 23, the numerator, i.e the dividend should be rewritten in terms of 23 Which will be (23)18. Now, the given problem reduces to finding out the remainder when (23)18 is divided by 23 – 1. Here, if we consider 23 as x it is equivalent to finding out the remainder when x18 is divided by x – 1 which, as per Remainder Theorem, is (1), i.e the remainder is obtained by substituting 1 in place of x So, the remainder will be (1)18, i.e ., 1. Examples 1. Find the remainder of the division 269/9. Sol. In the division, since the numerator is in terms of powers of 2, the denominator 9 also should be expressed in terms of powers of 2, i.e, as (23 + 1). Now as the denominator is in terms of 23, the numerator should be rewritten in terms of 23 as (23)23 is divided by {23 – (-1)}. This remainder, as per Remainder Theorem, is (-1)23 = - 1 + 9 = 8. (Divisor is added to get positive remainder) 2. Find the remainder of the division 282/7. Sol. In the is division, since the numerator is in terms of powers of 2, the denominator 7 also should be expressed in terms of 2, as (23 – 1). Now as the denominator is in terms of 23, the numerator should be rewritten as 2. (23)27. The problem now reduces to finding the remainder when 2.(23)27 is divided by (23 – 1). This remainder, as per Remainder Theorem, is 2(1)27 = 2. 3. Find the remainder of the division 294/15. Sol Pattern method.



The remainders of powers of 2 when divided by 15 are as follows, Remainder when 21 is divided by 15 is 2 Remainder when 22 is divided by 15 is 4 Remainder when 23 is divided by 15 is 8 Remainder when 24 is divided by 15 is 1 Remainder when 25 is divided by 15 is 2 The reminder repeats after 4 steps. So, the remainder of 294/15 is the same as that of 22/15 i.e 4 (Since 92 is a multiple of 4 and ( 92 + 2) = 94 Hence the remainder is 4. Remainder Theorem Method In the division 294 / 15, the numerator is in the terms of powers of 2, so denominator is expressed as ( 24 – 1). Now since the denominator is in terms of 24, the numerator is rewritten as 22(24)23. Now the remainder as per Remainder Theorem is 22(1)23 = 22. (1) = 4 4. Find the remainder of the division 2189/5 Sol; Pattern method The remainder of various powers of 2 when divided by 5 are as follows Remainder when 21 is divided by 5 is 2 Remainder when 22 is divided by 5 is 4 Remainder when 23 is divided by 5 is 3 Remainder when 24 is divided by 5 is 1 Remainder when 25 is divided by 5 is 2 The remainder is repeated for every 4 steps. So, the remainder when 2189 is divided by 5 is the same as 21 divided by 5 i.e 2. Remainder Theorem Method: In the division 2189/5, the numerator is in powers of 2, so the denominator should also be in terms of powers of 2 as (22 + 1). Now as the denominator is in terms of 22 the numerator is also rewritten as

2(22)94. So the division is now reduced to

942

2

2. 2

2 1. The remainder of this division as per Remainder

Theorem is 2. (-1)94 = 2 (1) = 2. 5. Find the remainder of the division 393 divided by 10. Sol The remainders of powers of 3 when divided by 10 are as follows. Remainder when 31 is divided by 10 is 3 Remainder when 32 is divided by 10 is 9 Remainder when 33 is divided by 10 is 7 Remainder when 34 is divided by 10 is 1 Remainder when 35 is divided by 10 is 3 Since the remainder is repeating after 4 steps, the remainder of 393/10 is the same in 31/10 = 3 ( Since 93 = 4 23 +1). Remainder Theorem Method In the division, 393/10, the numerator is in terms of powers of 3, so denominator is written as (32 + 1). Since the denominator is now in terms of 32, the numerator is expressed as 3.(32)46. So the remainder of 3.(32)46 divided by (32 + 1) as per the Remainder Theorem, is 3 (-1)46 = 3(1) = 3. As is evident form the above examples, the remainder theorem is more suited to cases where the denominator ( i.e,. the divisor) can be written in the form of one more or one less than some power of



the base in the numerator. For example, in case of 254/7, since the base in the numerator is 2, the denominator 7 has to be written as one more or one less than some power of 2. In this case it can be written as 23 – 1. In cases where it is not possible to write it in this manner, then applying the Pattern Method is the easiest method. SUCCESSIVE DIVISION If the quotient of a division is taken and this is used as the dividend in the next division, such a division is called “ successive division.” A successive division process can continue upto any number of steps – until the quotient in a division become zero for the fist time. i.e the quotient in the fist division is taken and divided in the second division: the quotient in the second division is taken as the dividend in the third division, the quotient in the third division; the quotient in the third division is taken as the dividend in the fourth division and so on. If we say that 2479 is divided successively by 3, 5, 7 and 2, then the quotients and remainders are as follows in the successive division. Dividend Divisor Quotient Remainder 2479 3 826 1 826 5 165 1 165 7 23 4 23 2 11 1 Here we say that when 2479 is successively divided by 3, 5, 7 and 2 the respective remainders are 1, 1, 4 and 2. Examples 1. A number when divided successively by 5 and 2 gives respective remainders of 3 and 1. What will be the remainder when largest such two digit number is divided by 12? Sol We write down the divisors one after the other and their respective remainders below them

5

3 1

2 Divisors

Remainders

Then starting from the last remainder, we go diagonally left upwards to the first row multiplying and then directly below adding the figure already obtained. We continue the process till we reach the figure on the extreme left in the second row. So, we ger 1 5 3 8 So the number is 10k + 8(10 is the product of divisors). So the number is of the form 10k + 8, for k = 0, 1, 2…. The largest 2 – digit number is , 10 (9) + 8 = 98. This when divided by 12 leaves a remainder of 2. 2. A number when successfully divided by 3, 4 and 7 leaves respective reminders of 2, 3 and 1. How many such numbers are there under 1000? Sol Let us write down all the divisors and their respective remainders as follows.



3 4 7

32 1 We start at the bottom right corner 1 and go form 2nd row to 1st row diagonally to the left multiplying. We get 1 4 4 , then we come down to the 2nd row adding, we get, 4 + 3 = 7. Again multiplying diagonally left upwards, we get 7 3 21 . Coming down to 2nd row, adding, we get 21 + 2 = 23. The smallest number that satisfies the given condition is 23. The general form of numbers that satisfy the given condition is got by adding multiples of product of divisors, which is 84, to 23. General form is 84k + 23 For k = 0, 1, 2, ……11, the number is less than 1000. Hence, there are 12 numbers less than 1000 that satisfy this condition. 3. A number when successively divided by 3, 4 and 9 leaves respective remainders of 2, 3 and 7. What will be the remainders if the original number is divided successively by 3, 5 and 7? Sol: Here again, we will first find the smallest number that satisfies the given condition.

3 4 9

32 7

Divisors

remainders

The smallest number is 7 4 3 3 2 95

Now, when 95 is successively divided by 3, 5 and 7, the results are. Divided Divisor Quotient Remainder 95 3 31 2 31 5 6 1 6 7 0 6 The remainders are 2, 1 and 6 respectively 4. A number when successively divided by 4, 5 and 7 leaves respective remainders of 3, 4 and 6. What will be the remainders if the original number is divided successively by 2, 5 and 7? Sol We will fist find the smallest number that satisfies the given condition.

4 5 7

43 6

Divisors

remainders The smallest number is 6 5 4 4 3 139.



Now, when 139 is successively divided by 2, 5 and 7 the results are: Dividend Divisor Quotient Remainder 139 2 69 1 69 5 13 4 13 7 1 6 Hence the remainders are 1, 4 and 6. FACTORIAL Factorial is defined for any positive integer, it is denoted by or !. Thus “ Factorial n” is written as n! or n, n! is defined as the product of all the integers from 1 to n. Thus n! = 1.2.3…n ( n – 1)n. 0! is defined to be equal to 1. 0! = 1 and , 1! is also equal to 1. RULES PERTAINING TO aa + ba or aa – ba Sometimes, there will be problems involving numbers that can be written in the form an + bn or an – bn which can be simplified using simple rules. Let us first look at the rules pertaining to both an + bn and an - bn The following rules should be remembered for an – bn 1. It is always divisible by a – b 2. When n is even it is also divisible by a + b 3. When n is odd it is not divisible by a + b . The following rules should be remembered for an + bn 1. It is never divisible by a – b. 2. When n is odd it is divisible by a + b 3. When n is even it is not divisible by a + b. SOME IMPORTANT POINTS TO NOTE Please note the following points also which will be very useful in solving problems on Numbers 1. When any two consecutive integers are taken one of them is odd and the other is even Hence the product of any two consecutive integers is always even i.e divisible by2. Two consecutive integers can be written in the form of n and n – 1 or n and n +1. Hence, any number of the form n (n – 1) or n (n +1) will always be even. 2. Out of any 3 consecutive integers , one of the form is divisible by 3 ( Atleast one of the three is definitely even.) Hence, the product of any 3 consecutive integers is always divisible by 6.



Three consecutive integers can be of the form ( n – 1) n and ( n+1). The product of 3 consecutive integers will be of the form (n-1)n (n + 1) or n ( n2 – 1) or (n3 – n). Hence any number of the form (n -1) n (n +1) or n ( n2 – 1) or (n3 – n) will always be divisible by 6. 3. Out of any n consecutive integers, exactly one number will be divided by n and the product of n consecutive integers will be divisible by n ! 4. Any prime number greater than 3 can be written in the form of 6 k + 1 or 6k –1 . The explanation is Let p be any prime number greater than 3. consider the three consecutive integers ( p -1), p and ( p +1) Since p is a prime number greater than 3, p CANNOT be even. Since p is odd , both ( p-1) and ( p + 1) will be even, i.e both are divisible by2. Also since, out of any three consecutive integers, one number will be divisible by 3, one of the three numbers ( p-1), p or (p+1) will be visible by 3. But, since p is prime number – that too grater than 3 , p can not be divisible by 3. Hence, either ( p -1) or ( p +1 ) , one of them – and only one of them – is definitely divisible by3. If ( p – 1) is divisible by 3, since it is also divisible by 2, it will e divisible by 6, i.e it will be of the form 6k. If ( p – 1) is of the form 6 k, then p will be of the form ( 6k +1).If ( p+1) is divisible by 3, since it is also divisible by 2, it will be divisible by 6, i.e it will be of the form 6K if ( p + 1) is of the form 6 K , then p will be of the form ( 6K – 1) If ( p + 1) is divisible by 3, since it is also divisible by 2, it will be divisible by 6, i.e it will be of the form 6k. if ( p + 1) is of the form 6 k, then p will be of the form ( 6k – 1). Hence any prime number greater than 3 will be of the form (6k + 1) or ( 6k – 1). INDICES – SURDS INDICES If a number ‘a’ is added three times to itself , then we write it as 3a. instead of adding , if we multiply ‘a’ three times with itself, we write it as a3 We say that ‘a’ is expressed as an exponent, Here ‘a’ is called the ‘ base’ and 3 is called the ‘ power’ or or ‘ index’ or ‘exponent’. Similarly ‘a’ can be expressed to any exponent ‘ n’ and accordingly written as an. This is read as ‘a to the power n” or “ a raised to the power n”. nn = .............. aaaa n times For example 23 = 2 22 = 8 and 34 = 3333 = 81 While the example taken is for a positive integer value of n, the powers can also be negative integers or positive or negative fractions. In the sections that follow, we will also see how to deal with numbers where the powers are fractions or negative integers.

342 is evaluated by starting at the topmost level ‘3’. Thus we first calculate 43 as equal to 64. Since 2 is raised to the power 43, we now have 264. Similarly

232 is equal to ‘2 raised to the power 32” or “2 raised to the power 9” or 29 which is equal to 512.



There are certain basic rule / formulae for dealing with numbers having powers. These are called Laws of Indices. Rule / L aw Example 1) nmnm aaa

2) nmn

m

aaa

3) mnnm aa

4) mm

aa 1

5) mm aa /1 6) mmm baab .

7) n

nn

ba

ba

8) )0(1 awherea o 9) aa 1

972 555

49777 2

3

5

632 44

125.081

212 3

3

46464 3/13 444 3.232

169

43

43

2

22

13 o 441

These rues / laws will help you in solving a number of problems. In addition to the above, the student should also remember the following rules. Rule 1 : When the bases of two EQUAL numbers are equal. Then their powers also will be equal. ( if the bases are non - zero and non – unity.) For example : If 2n = 23, then it means n = 3 Rule – 2 When the powers of two equal number are equal ( and not equal to zero), two cases arise (i) If the power is an odd number, then the bases are equal. For example if a3 = 43 then a = 4. (ii) If the power are even numbers, then the bases are numerically, equal but can have different signs. For example, if a4 = 34 then a = +3 or -3. The problems associated with indices are normally of THREE types (a)Simplification Here the problem involves terms with different bases and powers which have to be simplified using the rules/ formulae discussed in the table above. (b)Solving for the value of an unknown



Here, the problem will have an equation where an unknown ( like x or y ) will appear in the base or in the power and using Rule 1 and Rule 2 discussed above values of unknown are the be determind. (c)Comparison of numbers Here two or more quantities will be given – each being a number raised to a certain power. These numbers have to be compared in magnitude- either to find the largest or smallest of the quantities or to arrange the given quantities in ascending or descending order. The following example will make clear the different types of problems that you may be asked. Examples

1. Simplify : 14/13/2

21616

81256

64216

Sol: The individual numbers can be written as follows: 216 = 63; 64 = 43; 256 = 44; 81 = 34 ; 16 = 42 Replacing the numbers as above

2

2

2

2

3/23

3/233/2

64

46

4

664216

3

4

3

481

2564/14

4/144/1

2

31

3

21

46

64

21616

14/13/2

21616

81256

64216

= 72932

332

364

636444

5

5

5

212

32

22

2. Simplify the following

2

2

332

3

24

4

32 ...

mln

lnm

nml

Sol: The given function on opening brackets is,

22

2323

23

2224

34

3332 .2..

m

lnn

mn

ml

= 10366412489666 .... nmlnml

= 103

6

nml

2. In the equation given below. Solve for value of x.



72964

23

3

2x

Sol: The given equation can be written as

6

6

63/2

32

32

23

x

This can be written as 63/2

23

23

x

Now, the bases are the same on both the sides of the equation, hence the powers are equal i.e

2018263

2

xxx

SURDS Any number of the form p/q , where p and q are integers and q 0 is called a rational number. Any real number which is not a rational number is an irrational number. Amongst irrational numbers, of particular interest to us are SURDS Amongst surds, we will specifically be looking at ‘ quadratic surds ‘ – surds of the type a + b and a + b + c , where the terms involve only square roots and not any higer roots. We do not need to go very deep into the area of surds – what is required is basic understanding of some of the operations on surds. If there is surds of the form ba , them a surd of the form ba is called the conjugate of the surd .ba The product of a surd and its conjugate will always be a rational number. RATIONALISATION OF SURD

When there is a surd of the form ba

1 , it is difficult to perform arithmetic operations on it. Hence, the

denominator is converted into a rational number thereby facilitating ease of handing the surd. This process of converting the denominator into a rational number without changing the value of the surd is called rationalization. To convert the denominator of a surd into a rational number. Multiply the denominator and the numerator simultaneously with the conjugate of the surd in the denominator so that denominator gets converted to a rational number without changing the value of the fraction. That is , if there is a surd of the type a + b in the denominator, then both the numerator and the denominator have to be multiplied with a surd of the form a - b or a surd of the type – a + b to convert the denominator into a rational number. If there is a surd of the form cba in the denominator, then the process of multiplying the denominator with its conjugate surd has to be carried out TWICE to rationalize the denominator. SQUARE ROOT OF A SURD If there exist a square root of a surd of the type a + b . Then it will be of the form yx We can

equate the square of yx to a + b and thus solve for x and y. Here on point should be noted- When there is an equation with rational and irrational terms, the rational part on the left hand side is



equal to the rational part of right hand side and the irrational part on the left hand side is equal to the irrational part on the right hand side of the equation.

Here, first the given surd is written in the form of 2 2, .x y or x y Then the square root of the

surd will be x y or x y respectively.

COMPARISON OF SURDS Sometimes we will need to compare two or more surd either in ascending/ descending order. The surd given in such cases will be such that they will be close to each other and hence will not be able to identify the largest one by taking the approximate square root of each of the terms. In such a case, the surds can both be squared and the common rational able to make out the order of the surds. If even at this stage, it is not possible to identify the larger of the two, then the numbers should be squared once more. Important Identities : 2 2 22a b a ab b

2 2 22a b a ab b

2 2 2 2 2 2 2a b c a b c ab bc ca

3 3 3 3 2 2 33 3 3a b a b ab a b a a b ab b

3 3 3 3 2 33 3a b a b ab a b a ab b

2 2a b a b a b 2 2a a b a b b

3 3 2 2a b a b a ab b

3 3 2 2a b a b a ab b

3 3 3 3a b c abc = 2 2 2a b c a b c ab bc ca

3 3 3 3a b c abc , if a + b + c = 0. SOLVED EXAMPLES 1. Express 0. 3 in the form of fraction. Ans. 0. 3 = 0.3333 ……….(1) As the period is one digit, we multiply by 101 i.e 10 10 0.3 3.333 ……….(2) (2) – (1) gives

3 19 0.3 3 0.39 3

2. Express 0.54 in the form of a fraction.



Ans: 0.54 0.545454 ………..(1) As the period is containing 2 digits, we multiply by 102 i.e, 100 100 0.54 54.545454 ………...(2) (2) – (1) gives

54 699 0.54 54 0.5499 11

3. Express the recurring decimal 0.026 in the form of a fraction. Ans: 0.026 0.026026026 ………..(1) As the period in containing 3 digits, we multiply by 103, i.e 1000, therefore 1000 0.026 26.026026 ……….(2) (2)- (1) gives

26999 0.026 26 0.026999

Note: We can now write down the rule for converting a pure recurring decimal into a fraction as follows: A pure recurring decimal is equivalent to a vulgar fraction which has the number formed by the recurring digits ( called the period of the decimal) for its numerator, and for its digits as many nines as there are digits in the period.

Thus 0.37 can be written as equal to 3799

0.225 can be written as equal to 225999

which is the same as 25 .111

63 70.6399 11

A mixed recurring decimal becomes the sum of a whole number and a pure recurring decimal. When it is multiplied by suitable power of 10 which will bring the decimal point to the left of the fist recurring figure. We can then find the equivalent vulgar fraction by the process as explained in case of a pure recurring decimal. 4. Express 0.23 as a fraction. Ans. Let x = 0.23 Then 10x = 2.3 = 2 + 0.3

3 3 1 72 sin 0.3 29 9 3 3

ce

7 7103 30

x x

5. Express 0.136 in the form of a fraction. Ans. Let x = 0.136

36 4 1510 1.36 1 199 11 11

x

15 15 31011 110 22

x x



Note: Now we can write the rule to express a mixed recurring decimal into a ( valgar) fraction as below. In the numerator write the entire given number, formed by the ( recurring and non-recurring parts) and subtract from it the part of the decimal that is not recurring. In the denominator, write as many nines as the period( i.e as many nine s the number of digits recurring) and then place next to it as many zeroes as there are digits without recurring in the given decimal.

156 1 155 31. 0.156990 990 198

i e

73 7 66 110.7390 90 15

6. If 2 2729 81x x

, then find the value of x.

Ans. 2 26 4 6 12 4 83 3 3 3x x x x

Since the bases are equal on both sides we can equate the powers 6 12 4 8 2 20 10x x x x

7. If 3

281 816561

xx

, then find the value of x

Ans. The given equation can be rewritten as

32 2 32 2 4

49 9 9 99

xx x x

Since bases are equal the powers can be equated. 2 3 4 6 2 4 3x x x x x 8. Arrange the following in ascending order. 8 7 1127 , 81 9and Ans. Each of the given bases can be expressed in terms of base3. Hence, the given numbers can be written as

8 7 113 4 23 . 3 3and

24 28 223 ;3 3and As the bases are equal, the values can be compared on the basis of thepowers. Hence 22 24 28 11 8 73 3 3 9 27 81or 9. Arrange the following in descending order. 6 3 449 ; 81 ; 125and



Ans. Each of the given bases can be expressed in exponential form. Hence, the given numbers are:

6 3 42 4 37 ; 3 5and i.e 712 ; 312 and 512 All these numbers have the same power. Hence, 712 > 512 > 312 or (49)6 > (125)4 > (81)3

10. Simplify : 1 12 3 2 3

Ans. 12 3 2 3

= 2 3 2 3 2 3 2 3 2 3

4 3 12 3 2 3

11. Rationalise the denominator of the surd 13 7

Ans Since the denominator of the surd is 3 7 , to rationalise, we multiply both the numerator and the

denominator by 3 7 .

1 3 7 3 7 3 79 7 23 7 3 7

12 Rationalise the denominator of the surd 12 3 5

Ans. Here, the rationalization process has to be carried out twice (i.e in two steps) before getting a rational denominator. First take 2 3 as one term and 5 as the second term and carry out the

rationalization by multiplying the numerator and denominator by 2 3 5 .

=

2 3 51

2 3 5 2 3 5

=

2 2

2 3 5 2 3 5

4 3 4 3 52 3 5

=

2 3 5 2 3 5

2 4 3 2 1 2 3

Now the denominator has to be rationalized again. We multiply the numerator and denominator by 1 2 3 .

Given fraction is equal to:



2 3 5 1 2 3

2 1 2 3 1 2 3

= 2

4 3 3 5 2 15

2 1 2 3

=

4 3 3 5 2 152 11

= 4 3 3 5 2 1522

13 Find the square root of the surd 12 + 6 3 .

Ans. We will write the given surd as 2a b here, as the irrational part is positive. { if the irrational part is

negative, we would have written it as 2a b }

In the expansion of 2a b we get 2a b term. Since the coefficient is 2, we will keep the

coefficient of the irrational term of the given surd as 2. Take the term 6 3 . Here since the coefficient is 6, we will retain only 2 and take the remaining 3 under the square root. When this goes under the same root, it becomes 9, then the term 6 3 2 27 Thus the given surd is equal to 12 2 27 Now, the figure under the square root ( i.e 27) has to be written as the product of two factors such that the sum of these factors in equal to the rational part ( which is 12 in this case).27 can be written as a product of 9 and 3 such that 9 3 12. Hence the given surd can be written as 12 2 27 9 3 2 9 3

= 2 223 3 2.3. 3 3 3

Hence the square root of 12 +6 3 is 3 3 .

( Please note that although the square root can be with a positive or a negative sign, when written in the

form 12 6 3, positive root is implied.) 14 Compare and find which of the following two surds is greater. 5 19 2 27and

Ans. If we try to take approximate values of both surds, we find that both are more than 6 and we will not be able to judge which is greater. The comparison can be done by squaring the surds and comparing the squares. Now, we get

25 19 5 19 2 95 24 2 95



22 27 2 27 2 54 24 5 2 54

{ The square of the seconds surd is written as the form of 24 + 2 95 . By writing this way we have two terms in the form of 24 certain quantity to compare}.

Since both terms have 24, we only need to compare the remaining parts i.e 2 95 5 2 54and 2 95 4 95 380 We know 380 lies between 19 and 20 ( since 192 = 361 & 202 = 400) 5 2 54 lies between ( 5+27) and ( 5+28) i.e between 19 and 21 ( Since 54 lies between 7 and 8)

Squaring again, we have values which are 380 and 25 2 54 .

25 2 54 . = 25 + 216 +20 54 = 241 + 20 54 241 20 54

Which lies between 241 20 7 and 241 20 8 ; i,e between 381 and 401.

381 > 380; hence 5 2 24 2 95 i.e 2 27 5 19

15. If x = 7 5 and y = 7 5 , evaluate (i) xy (ii) x2 + y2 (iii) (x2 – y2)2 Ans. 7 5, 7 5x y

(i) 7 5 7 5xy

= 2 27 5 2 2a b a b a b

= 7 – 5 = 2.

(ii) 2 22 2 7 5 7 5x y

= 7 5 2 7 5 7 5 2 7 5

= 12 + 12 = 24.

(iii) x2 – y2 = 2 27 5 7 5

= 7 5 2 7 5 7 5 2 7 5

= 12 + 2 35 12 2 35 4 35

222 2 4 35 16 35 560x y

16. If a = 2 + 3 3 3 3 5and b , prove that a2 + b2 – 4a – 6b – 3 = 0. Ans. a = 2 3 5, 3 3 5b

2 3 5, 3 3 5a b



2 22 22 3 3 5 3 5a b

2 22 24 4 6 9 2 3 5a a b b

2 2 4 6 13 2 3 5a b a b

2 2 4 6 13 16a b a b 2 2 4 6 3 0a b a b 17. If 3 2 2 3 2 2,x and y evaluate 4 4 2 26 .x y x y 3 2 2, 3 2 2x y

2 22 2 3 2 2 3 2 2x y

= 2 2

2 3 2 2 2 3 8 22

3 2 2 3 2 2x y

= 2 23 2 2 3 8 5

22 2 5 25x y

Now, 4 4 2 2 4 4 2 2 2 26 2 4x y x y x y x y x y

= 22 2 2 24x y x y = 222 4 25 = 484 + 100 = 585.

18. If 2 3 2 3 3 1,2 3 2 3 3 1

x

find the value of

22 253x

x

2 3 2 3 3 12 3 2 3 3 1

x

=

2 3 2 3 2 3 2 3 3 1 3 1

2 3 2 3 2 3 2 3 3 1 3 1

=

2 2 2

2 2 22 2 2

2 3 2 3 3 1

2 3 2 3 3 1

= 2 2 22 3 2 3 3 1

4 3 4 3 3 1

= 2

2 2 3 1 3 1 2 32 3 2 3 4 3 4 3 4 3 4 32 2

= 4 2 314 14 2 3 16 32

22

1 16 31 1 16 3 16 3 16 3256 3 25316 3 16 3 16 3 16 3x



253 16 3x

Now, 2

2 2 222 253 16 3 16 3 2 16 3xx

= 2 256 3 2 259 518.

19. If 3 1 3 1,3 1 3 1

x and y

find the value of

(i) 2 2x xy y (ii) 2 2xy x y

Ans.

2

2 2

3 1 3 1 3 13 1 3 1 2 33 13 1 3 1 3 1 3 1

x

= 4 2 3 2 32

23 1 3 1 3 13 1 3 1 2 3 2 3

3 1 23 1 3 1 3 1y

(i) 2 22 2 2 3 2 3 2 3 2 3x xy y

= 4 3 4 3 4 3 4 3 4 3

= 7 4 3 1 7 4 3 = 7 + 1 + 7 = 15

(ii) 2 2 2 3 2 3 2 3xy x y xy y x

= 4 3 4 1 4 4. CHAPTER – 2

POLYNOMIALS Theory Students are familiar with algebraic expression like 2x, x2 + 3x, 2x3 – 3x2 + 4x + 5 , - 3x3 + 4x2 – 2x + 1. All these algebraic expressions contain only one variable x and exponents of variable x are whole numbers. The algebraic expressions of this form are called polynomials in one variable i.e, in variable x and are denoted by the symbols f (x), p (x), q (x ), etc. For example, p (x) = - 3 x3 + 4x2 – 2x + 1 is a polynomial in variable x, q (y) = 2 y2 + 3y + 5 is polynomial in variable y and f (u) = 2 u2 – 7 u + 6 is a polynomial in variable u.



In the polynomial f(x) = ax2 + bx + c, the expressions ax2, bx and c are called the terms of the polynomial f (x). Moreover constant c is called the constant term of the polynomial. Each term of a polynomial has coefficient. In polynomial 2x3, - 8x2 + x – 3, we have four terms 2x3 , - 8x2, x and – 3 and their coefficients are 2, - 8, 1 and – 3 respectively. Constant Polynomials

Constants 2, - 2 , 2 , 32

and a can be written as 2 xo, -2xo, 02x , 032

x and ax0 respectively.

Therefore, these constants are expressed as polynomials which contain single term in variable x and the exponent of the variable is 0. Thus, we can define a constant as a constant polynomial. Moreover, a non – zero constant is defined as a polynomial of degree zero. Polynomials of Degree in n ( n is any natural number) We define a polynomial of degree n where n is any whole number. Expression an xn, an-1 xn-1 + …..+ a1x + a0 for any whole number n and an 0 is defined as a polynomial of degree n in one variable x. Here, an xn, an-1xn-1, …a1x and a0 are called the terms of the polynomial. The term an xn ( the term containing highest exponent) called the leading term of the polynomial. The constants an, an – 1 ,……a1, a0are the coefficients of terms containing xn, xn-1, x1

, x0 respectively.

REMARK. The exponent of the variable in the leading term of a polynomial determines the degree of the polynomial Now, we have expressed the algebraic expression f(x) = an xn + an-1 xn-1 + …….+ a1x + a0 as polynomial of degree n for every whole number value of n, i.e., for n = 0, n = 1, n = 2, n = 3, ….Here, we have an 0 and the other coefficients an-1, an-2,…. a0 can be any real constants. In case, we have an = an-1 = …..= a1 = a0 = 0. then the polynomial reduces to 0,i.e a zero polynomial and we know that the degree of zero polynomial is not defined. For n = 0, we have f(x) = a0 and a0 0. ZEROS OF A POLYNOMIAL Let us consider the polynomial p (x) = x3 – 2x2 – 2x – 3. If we replace x by 1 everywhere in the above expression, we get P(1) = (1)3 – 2(1)2 – 2(1) – 3 = 1 – 2 – 2 – 3 = -6 Here, we can say that the value of the polynomial p (x) at x = 1 is – 6. Similarly, we can find that P(0) = (0)3 – 2 (0)2 – 2 (0) – 3 = - 3. P(2) = (2)3 – 2(2)2 – 2(2) – 3 = 8 – 8 – 4 – 3 = - 7, P (3) = (3)3 – 2 (3)2 – 2 (3) – 3 = 27 – 18 – 6 – 3 = 0, P (4) = (4)3 – 2 (4)2 – 2(4) – 3 = 64 – 32 – 8 – 3 = 21 In general, we can write p as the value of the polynomial p (x) at x = where is a real number. A real number is a zero of a polynomial p (x) if the value of the polynomial p (x) is zero at x = i,e., if p ( ) = 0.



Roots of Polynomial Equation A real number is root at a polynomial equation p(x) = 0 if p ( ) = 0 i.e is a zero of the polynomial p (x). Consider the polynomial p (x) = 2x -5, then 2x - 5 = 0 is the corresponding polynomial equation.

Here, p 5 52 5 5 5 02 2

, i.e ., 52

is a zero of the polynomial

We can also say that 52

is a root of the polynomial equation 2x – 5 = 0.

Now, we take another polynomial p (x) = x2 – 3x + 2. It is of degree2. Here, p (1) = (1)2 – 3 (1) + 2 = 1 - 3 + 2 = 0. Therefore, 1 is a zero of the polynomial x2 – 3x + 2 or we can say that 1 is a root of the polynomial equation x2 – 3x + 2 = 0. Similarly, we can verify that, 2 is zero of the polynomial x2 – 3x + 2 and also 2 is root of the polynomial equation x2 – 3x + 2 = 0. Note: 1. is a zero of polynomial p (x) and a root of the corresponding polynomial equation p (x) = 0 if p ( ) = 0 i.e value of the polynomial p (x) is 0 at x = . 2. A non – zero constant polynomial has no zero. 3. Every real number is a zero of the zero polynomial. 4. A linear polynomial ax + b ; a 0 has exactly on zero and it is the root of the linear equation ax +b = 0,

it is x = .ba

5. A quadratic polynomial ax2 + bx + c, a 0 can have at the most two real zeros.In some case, it may not have any real zero. Remainder Theorem Let p (x) be any polynomial of degree 1 and a be any real constant. If the polynomial p (x) is divided by the linear polynomial ( x – a), then the remainder is p (a). Factor Theorem Let p (x) be any polynomial of degree 1 and for any real constant a, ( x – a) is a factor of polynomial p (x) if and only if p (a)= 0. Note: 1. If p (x) be a polynomial of degree 1 and it is divided by linear polynomial

ax + b ( a and b real coefficients), then the remainder obtained is equal to p ba

.

2. If p (x) be a polynomial of degree 1 and ax + b is a factor of p (x), then we have p ba

= 0.

3. If p (x) be a polynomial of degree 1. then Polynomial ( x – a) ( x – b ) is a factor of p (x) if p(a) = 0 and p (b) = 0.



Geometrical Meaning of the Zeroes of Polynomial For a liner polynomial ax + b, a 0, the graph of y = ax + b is a straight line which intersects the x –

axis at exactly one point, namely, ,0 .ba

Therefore, the linear polynomial ax + b, a 0, has exactly

one zero, namely the x – coordinate of the point where the graph of y = ax + b intersects the x- axis. Consider the quadratic polynomial x2 – 3x – 4. Let us see what the graph* of y = x2 – 3x - 4 looks like. Let us list a few values of y = x2 – 3x – 4 corresponding a few values for x as given in Table 2.1

x - 2 -1 0 1 2 3 4 5 y = x2 – 3x – 4 6 0 -4 -6 -6 -4 0 6

-1

-2

-3

-4

-5

-6

-7

-3 -2 1 2 3 5 6

1

2

3

4

5

6

7

o

(-2, 6)

(4, 0)(-1, 0)

(5, 6)

(3, -4)

(2, -6)

(1, -6)

(0, -4)

X' X

Y

Y'

y=x2 -

3x - 4

In fact, for any quadratic polynomial ax2 + bx + c, a 0, the graph of the corresponding equation y =

ax2+ bx + c has one of the two shapes either open upwards like or open downwards like depending on whether a > 0 or a < 0. ( These curves are called parabolas.) This fact is true for any quadratic polynomial, i.e., the zeroes of a quadratic polynomial ax2 + bx + c, a 0, are precisely the x – coordinates of the points where the parabola representing y = ax2 + bx + c intersects the x – axis From our observation earlier about the shape of the graph of y = ax2 + bx + c , the following three cases can happen. Case (i): Here, the graph cuts x – axis at two distinct points A and A’ The x – coordinates of A and A’ are two zeroes of the quadratic polynomial ax2 + bx + c in this case



X' X

Y

Y'

OA A'

a < 0D > 0

X'

Y

Y'

OAA'

a > 0D > 0

(i) (ii) Case (ii): Here, the graph cuts the x – axis at exactly one point i.e at two coincident points. So, the two points A and A’ of Case (i) coincide here to become one point A ( See Fig 2.4)

X' X

Y

Y'

OA

a < 0D = 0

X'

Y

Y'

O A

a > 0

(i) (ii)

D = 0

The x – coordinate of A is the only zero for the quadratic polynomial ax2 + bx + c in this case. Case (iii): Here, the graph is either completely above the x- axis or completely below the x- axis. So, it does not cut the x- axis at any point ( see Fig 2.5)



X' X

Y

Y'

O

a > 0D < 0

X' X

Y

Y'

O

a < 0

(i) (ii)

D < 0

So, the quadratic polynomial ax2 + bx + c has no zero in this case. Consider the cubic polynomials x3 and x3 – x2. We draw the graphs of y = x3 and y = x3 – x2 in Fig. 2.7 and Fig 2.8 respectively.

-1

-2

-3

-4

-5

-6

-7

-3 -2 1 2 3

1

2

3

4

5

6

7

oX'

Y

Y'

4-3-4 -1

( -1, -1)

( -2, -8)

( 1, -1)

( 2, 8)

-8

8

X

-1

-2

-3

-4

-5

-6

-7

-3 -2 1 2 3

1

2

3

4

5

6

7

oX'

Y

Y'

4-3-4 -1

( -1, -2)

( 1, 0)

( 2, 4)

-8

8

X

Fig. 2.7 Fig. 2.8 Note that 0 is the only zero of the polynomial x3. Also, from Fig 2.7, you can see that 0 is the x- coordinate of the only point where the graph of y = x3 intersects the x – axis. Similarly, since x3 – x2 = (x – 1), 0 and 1 are the only zeroes of the polynomial x3 – x2.these values are the x- coordinates of the only points where the graph of y = x3 – x2 intersects the x – axis. In general, given a polynomial p (x) of degree n, the graph of y = p(x) intersects the x- axis at most n points. Therefore, a polynomial p(x) of degree n has atmost n zeroes.



Example: Look at the graph in Fig 2.9 given below. Each is the graph of y = p(x), where p(x) is a polynomial. For each of the graphs, find the number of zeroes of p(x).

X' X

Y

Y'

X' X

Y

Y'

X

Y

Y'

(i) (ii) (iii)

X' X

Y

Y'

(iv)

X' X

Y

Y'(v)

X

Y

Y'(vi)

O O O

O O O

Solution: (i) The number of zeroes is 1 as the graph intersects the x- axis at one point only. (ii)The number of zeroes is 2 as the graph intersects the x- axis at two points. (iii) The number of zeroes is 3. ( Why?) (iv) The number of zeroes is 1. (Why?) (v) The number of zeroes is 1. (Why?) (vi) The number of zeroes is 4. (Why?) Relationship between zeroes and coefficients of a polynomial Quadratic equation If and are the zeroes of a quadratic polynomial ),0()( 2 acbxaxxP then )(and)( xx are its factors. ),()(2 xxkcbxax where k is a constant

kxkkxxxk

)(])([

2

2

Comparing coefficients of x2, x and constant terms on both sides.

ac

ab

kckbka

),(,



i.e., sum of zeroes = 2xof.coeffx)of.(coeff

product of zeroes = 2xof.coeffterm)(constant

Illustration: Find a quadratic polynomial, the sum and product of whose zeroes are 5 and 6 respectively. Solution: Given,

ac

ab

6

5

If a = 1, then b = -5, c = 6

.65.,.

ispolynomialqadraticthe2

2

xxeicbxax

Cubic polynomial If dcxbxax 23 is a cubic polynomial whose roots are ,,, then

ad

ac

ab

For example consider the cubic polynomial .64)( 23 xxxxP Here )3()1()2()( xxxxP i.e., 2, -1 and 3 are zeroes of P(x). Now here a = 1, b = -4, c = 1, d = 6 Sum of zeroes, 4312

Also 4ab

sum of zeroes = ab

Product, 63)1(2

Also, 6ad

Product ad

Division algorithm for polynomials



Here, we will discuss the method of dividing one polynomial by another with the help of an example. Consider division of .12by2634 223 xxxxx First of all, we arrange the divisor and the dividend in decreasing powers of x. i.e., we change them into their standard form. Step 1: To obtain first term of quotient, divide the highest degree term of dividend by highest

degree term of divisor. i.e., .442

3

xxx

Then, carry out the division process. What remains is

.21011 2 xx Step 2: For second term of quotient, divide the highest degree term of new dividend by highest degree

term of divisor i.e, .11112

2

xx Again carry out the division. What remains is 32x – 9. Now

degree of 32x – 9 is less than the degree of divisor. Therefore, we cannot continue with the division any longer. Here, the divisor is 122 xx , the quotient is 4x – 11, remainder is 32x – 9. We know that dividend = divisor quotient + remainder. Here also, the above relation holds true. If p(x) and q(x) are two polynomials with ,0)( xq then we can find 2 polynomials r(x) and s(x) such that P(x) = r(x) q(x) + s(x). where s(x) will be 0 or degree of s(x) < degree of q(x). This result is known as division algorithm for polynomials. Illustration Divide 36234 24 xbyxxx Solution: First, change given polynomials in standard form,

.3and6324 24 xxxx i.e., Then, division can be done as follows:

122 xx

114 x

2634 23 xxxxxx 484 23

21011 2 xx112211 2 xx932 x

3x

11138124 23 xxx

6324 24 xxx34 124 xx

23 212 xx 23 3612 xx

xx 338 2 xx 11438 2

6111 x333111 x339



Here, quotient = 11138124 23 xxx Remainder = 339. HCF and LCM of polynomials Divisor (Factor) If a polynomial f(x) is a product of two polynomials g(x) and h(x) i.e; f(x) = g(x). h(x) then g(x) and h(x) are called factors of f(x). Ex. If f(x) = x2 – 5x + 6. i.e. )3()2(65)( 2 xxxxxf then (x – 2) and (x – 3) are factors of x2 – 5x + 6. Note: If g(x) is a factor of f(x) then –g(x) is also a factor of f(x). Generally we take g(x) or –g(x) as a factor in which the highest degree term has positive coefficient. Highest common factor (HCF) or Greatest common divisor (GCD) The highest common factor (HCF) of two polynomials f(x) and g(x) is that common factor which has highest degree among all common factors and in which the coefficient of highest degree term is positive. Working Rule: To find the HCF of two or more given polynomials. Step 1. Express each polynomial as a product of powers of irreducible factors (simple factors). Numerical factors, if any, are expressed as product of powers of primes. Step 2. If there is no common factor, then the HCF is 1. If there are common simple factors find the smallest exponents of these simple factors in the factorized from of the polynomials. Step 3. Raise the common simple factors to the smallest exponents found in step 2 and Multiply to get the HCF. Illustration Find the HCF of the polynomials. ).5148()3(84and)3)(16(150 2232 xxxxxx Solution: Let 32 )3()16(150)( xxxxf



And )5148()3(84)( 22 xxxxg Now, f(x) = 150 22 )3()16( xxx 32 )3()13()12(532 xxx )5148()3(84)( 22 xxxxg )54()12()3(732 22 xxx

Common simple factor Least exponent 2 1 3 1 2x + 1 1 x – 3 2

Hence, required HCF 2111 )3.()12(3.2 xx 2)3()12(6 xx LCM of polynomials The least common multiple (LCM) of two or more polynomials is the polynomial of the lowest degree, having smallest numerical coefficient which is exactly divisible by the given polynomials and whose coefficient of highest degree term has the same sign as the sign of the coefficient of highest degree term in their product. Working rule To find out LCM of two or more polynomials, we may use the following three steps Step 1. Express each polynomial as a product of powers of irreducible factors. Express numerical factors, if any, as product of powers of primes. Step 2. Consider all the irreducible factors occurring in the given polynomials each one once only. Find the greatest exponent of each of these simple factors in the factorized form of the given polynomials. Step 3. Raise each irreducible factor to the greatest exponent found in step 2, and multiply to get the LCM. Illustration Find the LCM of the polynomials ).7152()3(140and)12()65(90 2322 xxxxxx Solution: Let 22 )12()65(90)( xxxxf And )7152()3(140)( 23 xxxxg Then 22 )12()3()2(532)( xxxxf And )7()12()3(752)( 32 xxxxg



Irreducible factor Greatest exponent 2 2 3 2 5 1 7 1 x – 2 1 x – 3 3 2x + 1 2 x + 7 1

LCM 12311122 )7.()12.()3(.)2(.7.5.3.2 xxxx LCM )7()12()3()2(1260 23 xxxx Relation between LCM and HCF L.C.M. {of f(x) and g(x)} H.C.F. {of f(x) and g(x)} = f(x) . g(x) = -f(x) . g(x) Thus one should use this result with caution. Four quantities viz, L.C.M, H.C.M, f(x) and g(x) are involved here. If f(x), g(x) and one of L.C.M., H.C.F. are given, the other can be found without ambiguity because L.C.M and H.C.F. are unique, except for a factor of -1. If H.C.F., L.C.M. and one of f(x) and g(x), say f(x), are given, then g(x) can not be determined uniquely. There would be two polynomials by g(x) and -g(x) which f(x), produce the given H.C.F and L.C.M. Let us denote the two polynomials by f(x) and g(x). Let us denote the H.C.F. by h(x) and the L.C.M. by l(x). Then we have )()()()( xgxfxlxh Illustration: The H.C.F. of the polynomials .3is65)(and)2)(3()( 22 xxxxqxxxxp Find the L.C.M. Solution: 2( ) ( 3) ( 2) ( 3)( 1)( 2)p x x x x x x x )2)(3()65()( 2 xxxxxq H.C.F. = (x – 3)



L.C.M. )3(

)2)(3()2)(1)(3(...

)()(

xxxxxx

FCHxqxp

).2)(3)(2)(1( xxxx Illustration: The L.C.M. and H.C.F. of two polynomials, p(x) and q(x) are )1()1(and)1(2 24 xxx respectively. If ,1)( 23 xxxxp final q(x). Solution: )1)(1(1)( 223 xxxxxxp

)22()1(2)1)(1(

)1)(1()1(2)(

......)(

.....)()(

44

2

24

xxxx

xxxxp

FCHMCLxq

FCHMCLxqxp

SOLVED PROBLEMS Problem 1. Divide 1by2673 2234 xxxxx and find the remainder. Solution: the remainder is 9x – 4. Problem 2: Find all the zeroes of 621773 234 xxxx given that 3 and - 3 are two of its zeros. Solution: Given that 3 and - 3 are zeroes, )3()3( xandx are factors of given polynomial i.e., 3)3)(3( 2 xxx is a factor. We divide the given polynomial by x2 – 3.

12 x

632 xx

2673 234 xxxx4x 2x

xxx 663 23 33x x3

296 2 xx26x 6

49 x



)273()3(621773 22234 xxxxxxx )2()13()3( 2 xxx

So its zeroes are given by .2,31,3,3

Problem 3: ,7652)( 23 xxxxf find f(-1). Solution: x = -1 to put in given f(x) for f(-1)

.17652

7)1(6)1(5)1(2)1( 23

f

Problem 4: Find roots of polynomial x2 – 16. Solution: 0162 x 8 x x = 8, -8 are roots. Problem 5: xxx 5 is a polynomial or not? Give reason.

Solution: not, exponents of x in first term is 23 which is not whole number.

Problem 6: Consider cubic polynomial .432 23 xxx What is sum of all roots.

Solution: Sum ab

212

Problem 7: If x = 2 is root of ,)( 3 kxxxf then find k. Solution: x = 2 is root of f(x)

273 2 xx

32 x 621773 234 xxxx43x 29x

xxx 2127 23 37x x21

22x22x

66

0



6028

02)2(0)2(

3

kk

kf

Problem 8: Find the quadratic polynomial if sum and product of zeroes are 25

23 and

respectively.

Solution: Given ab

23

5,3,2

25.

cbaac

Quadratic polynomial is cbxax 2 i.e., .532 2 xx Problem 9: Check whether first polynomial is factor of second polynomial or not x = 3, x2 – 9. Solution: Second polynomial can be written as )3)(3(92 xxx which contains x – 3, hence x -3 is factor of x2 – 9. Problem 10. Find HCF of the polynomials .)3()16(64 3223 xxxandxxx Solution: Let 64)( 23 xxxxf

2 3

3

( 1) ( 2) ( 3)( ) (6 1)( 3)

(2 1)(3 1) ( 3)

x x xand g x x x x

x x x

(Common simple factor least exponent x – 3 1 Hence HCF required = x – 3. Problem 11. The zeroes of the polynomial 71712 2 xx are

(a) 47,

31 (b)

47,

31

(c) 47,

31 (d) 1 7,

3 4

Solution:

.

47

31

)74()13(71712 2

andarezeroesthe

xxxx



Problem 12. If ,27)( 3 xxxp then sum of its zeroes is (a) -7 (b) 2 (c) 0 (d) none of these. Solution: If dcxbxaxxp 23)(

Sum of roots ab

Here b = 0 Sum of roots = 0. Problem 13. Which of the following polynomials divides 4159 2 xx completely (a) 3x – 1 (b) 3x + 1 (c) 3x – 7 (d) 3x + 7 Solution: 4159)( 2 xxxp On dividing P(x) by 3x + 1, the remainder obtained is zero. Hence (3x + 1) divides P(x) completely. Problem 14. If and are the roots of p(x) = ,13 2 x then is

(a) 31 (b)

31

(c) zero (d) none of these Solution: 1.03)( 2 xxxp

Sum of roots 30 sum = 0.

Problem 15. The number of zeroes of a cubic polynomial is (a) atmost 3 (b) exactly 3 (c) exactly 2 (d) none of these Solution: The number of zeroes of a polynomial of degree n is atmost n. Problem 16. The number of zeros in given polynomial is /are (a) 1 (b) 2 (c) 3 (d) 4 Solution: (D). As polynomial cuts x-axis at four different points. Problem 17. The quadratic polynomial whose sum of zeros is -1 and product of zeros is -10 (a) 102 xx (b) 102 xx



(c) 102 xx (d) none of these Solution: (B). Sum of zeroes s = -1 Product of zero p = -10

2

2

2

( ) 8( 1) 10

10.

p x x x px xx x

FILL IN THE BLANKS Problem 18. The number of zeros of 456 2 tt are …… and they are ……., ……….

Solution: 34,

21,2

)43()12(0456 2 tttt

zeroes are .34,

21

TRUE / FALSE Problem 19. LCM of polynomials )3()5()1()5()3()5()1( 22 xxxisxxandxx Solution: (True). 2)5)(1()( xxxf g(x) = (x + 3) (x + 5) irreducible factor greatest exponent x – 1 1 x + 5 2 x + 3 1 LCM )3()5()1( 2 xxx Problem 20. Consider graphs

y

xx'

y'

(0, 0)

y

xx'

y'



(a) (b) Graph (b) has more number of zeros than (a) Solution: (False). Graph (a) is y = x3 has 3 coincident zeroes at x = 0 while (b) has only two zeroes. CHAPTER – 3 LINEAR EQUATIONS & INEQUATIONS INTRODUCTION In this chapter, we shall study about linear equations in two variables and their solutions. We shall also learn various methods for finding common solutions of a pair of linear equations, known as simultaneous equations. Then we shall tackle several types of practical problems which may be reduced to simultaneous equations for solving them. Linear equation in two variables The general form of a linear equation in two variables x and y is ax + by + c = 0, baba ,,0,0 and c being real numbers. For example, 253,24932,0843 yxyxyxyx etc, are linear equations in two variables x and y. A solution of such an equation is a pair of values, one for x and other for y, which makes the two sides of the equation equal. An equation in two variables cannot determine absolutely the value of either variable or unknown, but would give the value of one in terms of other.

For example, the equation 2x + 3y = 6 can be put in the form .326 xy

In this form the value of y is

obtained in terms of x. Giving various values to x = 0, 3, 6, 9, …., we get y = 2, 0, -2, -4, …. Thus for any value given to x, there is a corresponding value of y. Similarly, we can put the equation in the form

,236 yx

so that for any value of y, we can get corresponding value of x.

Thus for linear equation involving two variables x and y, there is an infinite number of pairs of x and y which will satisfy the equation and hence a linear equation in two variables has infinite number of solutions. Graphical solution of a linear equation in two variables



Let the equation be ax + by + c = 0, ,0,0 ba This can be put in the form .

b

caxy

For any two values of x, we calculate the corresponding values of y form the above expression to obtain two solutions, say ).,(and),( 2211 Now we plot these two points ),(and),( 2211 BA on a graph paper and join them to get a line AB.

We observe that every solution x = r, y = s of the given equation determines a point (r, s), that lies on this line. Also every point (m, n) lying on the line AB, determines a solution x = m, y = n of the given equation. This line AB is known as the graph of the equation ax + by + c = 0. Since there are infinitely many points on this line, the equation has infinite solutions.

Simultaneous Equations When two (or more) equations are satisfied by the same values of the unknown quantities, they are called simultaneous equation. They are said to be of the first degree, if the unknowns occur in the first degree only. Example: Let us consider a linear equation in two variables, x + 3y = 7. We see that x = 1, y = 2 is a solution of this equation. Now we take another equation 2x + y = 4. The pair of values x = 1, y = 2 satisfy this equation also. That means x = 1, y = 2 is a solution of both the equations, or we can say that this pair of values of x and y satisfy both the equations simultaneously. The equations of this type are known as simultaneous equations. Note: The solve simultaneous equations involving two unknown quantities, we must have at least two equations. Graphical solution of simultaneous linear equations in two variables We known that the graph of a liner equation in x and y is a straight line and also we know that two straight lines in one plane can intersect only in one point. Thus the coordinates of the point of intersection satisfy both these equations and hence they give the solution of the equations. So, to solve two equations graphically, we draw the graph of these two lines using the same pair of axes. Now three different cases may arise: Case – 1: When both the line intersect at one point When the graph of a system of linear equations are a pair of intersecting lines, the point of intersection gives the unique solution of the given system. Such a system of equations is known as consistent system having a unique solution. For example the graph of the system of equations x – y = 0 and x + y = 2 is a pair of straight lines, which intersect at point (1, 1). Therefore, it is a consistent system having a unique solution (1,1).

y

x

y'

x'

1

1

(0, 2)

x - y = 0

(1, 1)

(2, 0)O

y

xx'

y'

(0, 2)

O (2, 0)

x + y = 2

3x + 3y = 6



Case – II: When the two lines are coincident When the graph consists of two coincident lines, every point on the lines is common to both the lines. Therefore, every point gives a solution. Such system is called a consistent system having infinitely many solutions. For example: the graph of the system of equations x + y = 2 and 3x + 3y = 6 is a pair of coincident lines. Therefore it is a consistent system having infinitely many solutions. Note: We observe that the above equations are not independent, for the second can be obtained from the first by multiplying throughout by 3. So any pair of values which will satisfy one equation will satisfy the other. Case – III: When the two lines are parallel When the graph of a system is a pair of parallel lines, there is non common point of intersection and so the system has no solution. This is called an inconsistent system of equations. For example: the graph of the system of equations x + y = 1 and x + y = 4 is a pair of parallel lines. Therefore, it is inconsistent system of equations having no solution. Illustration : Find three different solutions of 5x + 3y = 4. Solution: Taking x = -1, the given equation reduces to

33993435 yyy

3,1 yx is a solution of the given equation. Taking x = 2, the given equation reduces to 2634310 yyy 2,2 yx is a solution of the given equation. Again taking x = 5, the given equation reduces to 72134325 yyy 7,5 yx is a solution of the given equation. Thus we get three solutions of the given equation. Illustration: Graphically show that the system of equations

01332

,01064

yxyx

Has no solution.

y

xx'

y'

(0, 4)

O

x + y = 4

(0, 1)x + y = 1

(4, 0)(1, 0)



Solution: The given equations are 4x+ 6y – 10 = 0 and 2x + 3y + 13 = 0 Table of values of x and y for the equation

4x + 6y = 10 or isyyx235

4610

And table of values of x and y for the equation 2x + 3y + 13 = 0 or

2

313 yx is

Plotting the ordered pairs (1, 1) and (4, -1) and joining the points, we get the graph of the equation 4x + 6y – 10 = 0 Again, plotting the ordered pairs (-5, -1) and (-2, -3) and joining points obtained. We get the graph, we see that graph of two equations are parallel lines, i.e., they have no finite point of intersection. Hence the system has no solution. Illustration Solve graphically the following system of equations 2x – 6y + 10 = 0, 3x – 9y + 15 = 0. Solution: The given system of equations are

)2........(01593)1.......(01062

yxyx

We have the following table of values for (x, y) for the equations (1) and (2).

x 1 4 y 1 -1

x -5 -2 y -1 -3



Table for 2x – 6y + 10 = 0 Table for 3x – 9y + 15 = 0 Now, we plot the points A(1, 2), B(-5, 0) and join them to get the line AB as graph of 2x – 6y + 10 = 0. Again, we plot the points C(4, 3) and D(-2, 1). We find that points lie on the line BA. Thus the two lines coincide with each other.

Hence the given system of equations are consistent and have an infinite number of solutions. Every solution of 2x – 6y + 10 = 0 is the solution of the system. Illustration: Solve the following system of linear equations graphically 2x + 4y = 10 and 3x + 6y = 12. Solution: Plotting the two lines since the lines are parallel and hence there is no common point therefore there exists no real solution for the system of equations.

x

y

1 2 3 4 5

123

2x + 4y = 10

3x + 6y = 12

Algebraic methods of solution of system of simultaneous linear equations The graphical method of solution of simultaneous equations has certain limitations. It can not be applied in all the cases, because sometimes it becomes difficult to find integer solutions in order to draw the

x 1 -5 y 2 0

x 4 -2 y 3 -1



required lines. Also, there is chance of making mistakes while reading the coordinates of point of

intersection like

6421,

13511 etc.

To avoid such disadvantages of graphical method and to find accurate solution, we use algebraic methods. The most commonly used algebraic methods of solving simultaneous linear equations in two variables are: (i) Elimination method (by substitution) (ii) Elimination method (by equating the coefficients) (iii) Elimination method (by comparison) (iv) Cross multiplication method. We shall discuss these method one by one Method of elimination by substitution In this method, we find the value of one variable in terms of the other variable from any one of the given equations and substitute this value in the other equation so that we get a liner equation in one variable only. Generally, that equation is preferred, which provides simpler form of substitution. Then we solve this equation and find the value of that variable. We substitute the value of variable so obtained, in any one of the equations and solve the resulting equation for the second variable. This method of solving a system of linear equations is known as the method of elimination by substitution. Illustration: Solve the following system of linear equations using elimination by substitution:

.1232

12

yxyx

Solution: The given equations are x + 2y = -1 …(1) 2x – 3y = 12 …(2) Here, we shall eliminate x by substituting its value from one equation into the other. From (1), we get x = -1 – 2y …(3) Now, we substitute this value of x from (3) in equation (2). We get 2(-1 -2y) – 3y = 12 Or, -2 -4y -3y =12 Or, -7y = 14 y = -2 Now, substituting this value of y in (3), we get x = -1 -2 (-2) = 3 Thus, the solution of the system is x =3 and y = -2. Verification: Substituting x = 3 and y = -2, we find that both the equations (1) and (2) are satisfied . Hence, the solution is correct. Illustration



Solve the following system of linear equations using elimination by substitution:

)2....(53)1...(732

yxyx

Solution: Here, it is more convenient to eliminate y from (2) and substitute its value in (1) From (2), we get y = 3x -5 …(3) Substituting the value of y from (3) in (1), we get 2x + 3(3x – 5) = 7

.2

221171592

xx

xx

Now, substituting this value of x in (3), we get 1523 y Thus, the solution of the system is x = 2, y = 1 Method of elimination by equation coefficients This method of elimination is sometimes more convenient than the above method. In this method, we multiply the equation by such numbers that will make the coefficients of one variable numerically equal. Than we add the equations or subtract one from the other according as the like terms having same coefficients are opposite in sign or of the same sign respectively, so that we get a linear equation in single variable, which can be solved. The value of the variable so obtained is substituted in any one of the given equations and then solved for the second variable. Illustration: Solve the system of equations 3x + 7y = 27 …(1) 5x + 2y = 16 …(2) Using elimination y equating coefficients. Solution: To eliminate x, we multiply equation (1) by 5 and equation (2) by 3, so as to make the coefficients of x in both equation equal. This gives. 15x + 35y = 135 …(3) 15x + 6y = 48 …(4) Now, subtracting (4) from (3), we get 29y = 87 y = 3. To find x, we substitute this value of y in either of the given equations Thus from (1), 3x + 21 = 27 263 xx Hence x = 2 and y = 3 is the required solution. Illustration Solve the following system of equations using the elimination method by equating coefficients



7x + 2y = 47 …(1) 5x – 4y = 1 …(2) Solution: Here, it will be more convenient to eliminate y Multiplying (1) by 2, we get 14x + 4y = 94 And from (2), 5x – 4y = 1 Adding these equations, we get 19x = 95 x = 5. Now, substituting this value of x in (1), we get 35 + 2y = 47 .6122 yy Hence x = 5, y = 6 is the required solution. Remark: We could have eliminated x by multiplying (1) by 5 and (2) by 7, and subtracting one from the other. But the calculation would have been a bit difficult. So we should try simpler calculation while eliminating the variable. Solving system of equations reducible to system of simultaneous linear equations If the given system of equations contains fractional equations of the reciprocal type (having variables in the denominator) we use the reciprocals as unknown quantities (or variables)and eliminate them by any method. Illustration Solve the system of equations

0,0,7610

,198

yxyx

yx

Solution: The given equations are

)4....(7610and)3...(198so,

1,1putwe

)2...(7610and

)1...(198

vuvu

vy

ux

yx

yx

Thus, the equations become simultaneous linear equations in u and v



Now, multiplying equation (3) by 2 and equation (4) by 3 and adding them, we get 46u = 23

21

u

Substituting the value of u in (3), we get 4 – 9v = 1

3139 vv

Hence 31and21

vy

ux

3,2 yx is the required solution of the given equation. Solutions of the equations of the form ax + by = c and bx + ay = d, where a b In these type of equations coefficient of x and y in one are interchanged in the other. They can be solved by the following method. Let the equations be ax + by =c and bx + ay = d Adding the two equations, we get (a + b)x + (a + b)y = c + d

)....(Abadcyx

Again, subtracting the second equation from the first, we get (a – b)x + (b – a) y = c – d

)......(Bbadcyx

Now, we add (A) and (B) to get x and subtract (B) from (A) to get y. Illustration Solve the equations: 127x + 59y = 1928 …(1) and 59x + 127y = 1792 …(2) Solution: Adding equations (1) and (2), we get 186x + 186y = 3720 20 yx …(3) Subtracting (2) from (1), we get 68x – 68y = 136 2 yx …(4) Thus the given system of equations reduces to the system of equations given by (3) and (4). Adding (3) and (4), we get 2x = 22 11 x Subtracting (4) from (3), we get 2y = 18 .9 y Hence x = 11, y = 9 is the required solution of the given set of equations. Method of elimination by comparison



In this method, we find the value of one variable in terms of the other from both equations and equate them to get the value of that variable. Substitution of this value in any one equation will give the value of the other variable. Illustration Solve: 2x + 3y = 13 4x – y = 5 Solution: Given system of equations are 2x + 3y = 13 …(1) And 4x – y = 5 …(2) From equation (1), we have 3y = 13 – 2x

3

213 xy …(3)

From equation (2), we get y = 4x – 5 …(4) equation these two values of y from (3) and (4),

.22814

1512213

543

213

xx

xx

xx

Substituting this values in equation (1), we get

.393

1334

yy

y

Hence the required solution is x = 2, y = 3. Method of cross multiplication The algebraic methods given above are effective only when the system has a unique solution. So we shall describe another algebraic method known as method of cress multiplication. It also tells whether the system has unique solution, no solution or infinitely many solution: Let the two equations be

)2...(0,0,0

)1...(0,0,0

22222

11111

bacybxabacybxa

Multiplying equation (1) by b2 and equation (1) by b1, we get

)4...(0)3...(0

212112

122121

cbybbxbacbybbxba

Subtracting (4) from (3), we get

)5....()(

0)()(

12211221

21121221

cbcbxbabacbcbxbaba

Similarly, eliminating x, we get

)6...(0)()(

0)()(

12211221

21121221

acacybabaacacybaba



Now, different cases may arise:

Case – I: when 2

1

2

11221 ..0

bb

aaeibaba then we can divide both sides of (5) and (6) by this non-

zero number

1221

1221

1221

1221

1221 getwethatso,

babaacac

yandbabacbcb

x

baba

These relations can be rewritten as

122121121221

1babacaca

ycbcb

x

The above result can be represented diagrammatically as follows:

1 1 1 1 1 1

1x yb c c a a b

b2 c2 a2 b2c2 a2 The arrows between two numbers indicate that the numbers are to be multiplied. The numbers with downward arrow are multiplied first and from their product, the product of numbers with upward arrows is subtracted. The values of x and y so obtained a give the solution of the given equations. The in this case, the system has a unique solution.

Case – II: when 2

1

2

11221 ..,0

bb

aaeibaba

In this case, we cannot divide equations (5) and (6) by 1221 baba to get the values of x and y

Now, let kbb

aa

2

1

2

1

2121 kbbandkaa Now, there are two sub cases. Subcase (1)

If ,then,2

1

2

1

2

121 k

cc

bb

aakcc and the equation (1) reduces to

),(sin0 212121222 kccandkbbkaacekcykbxka Or 0)( 222 cbxak Or, )0(0222 kascbxa Which is the same as equation (2) Thus every solution of one equation is a solution of the other also. Hence, the system has infinitely many solutions. Subcase (2)

If ,..2

1

2

1

2

121 c

cbb

aaeikcc

Then the equation (1) becomes



2 2 1 1 2 1 20 (since , )ka x kb y c a ka b kb 0 or, 0)( 122 cybxak or. 0)( 12 cck (from equation (2) ) or, ,21 kcc which is not true. Therefore no solution exists. Thus from the above discussion, we have the following conclusion:

(i). if ,2

1

2

1

bb

aa

the system has exactly one i.e. a unique solution and the system is consistent.

(ii) if ,2

1

2

1

2

1

cc

bb

aa

the system is consistent and has infinite solutions.

(iii) if ,2

1

2

1

2

1

cc

bb

aa

the system is inconsistent i.e. it has no solution.

Illustration: Find a value of a so that the equation 3ax + 2ay = 21 may have x = 1, y = 2 as solution. Solution: Since x = 1, y = 2 is s solution of the given equation 3ax + 2ay = 21 and hence it satisfy the equation.

.32172143

21)2(2)1(3

aaaa

aa

Illustration: Solve the following system of equations by cross multiplication method: 3x + 4y = 10 2x – 3y = 1 Solution: Here 3,2,4,3 2211 baba 017891221 baba Hence, the given system of equations has a unique solution. So, here we apply cross multiplication method to solve the equations. We have 3x + 4y – 10 = 0 2x – 3y – 1 = 0 By cross multiplication method, we get

14 10 10 3 3 4

x y

-3 -1 -1 2 2 -3

891

)3()20()30()4(

yx



11717

21734

171

1734

yand

x

yx

The solution is x =2, y =1. Illustration: Determine the value of k for which the given system of equations 5x + 2y = k 10x + 4y = 3 has infinitely many solutions. Solution: The given system can be written as 5x + 2y – k = 0, 10x + 4y – 3 =0 Here .3,4,10and,2,5 222111 cbakcba For infinitely many solutions we must have

23,

321,

342

105..

2

1

2

1

2

1

korkorkeicc

bb

aa

Hence the given system of equations will have infinitely many

solutions if .23

k

LINEAR INEQUATIONS A statement of inequality between two expressions involving a single variable x with highest power 1, is called a linear inequation.

e.g., (i) 2x + 5 < 3 (ii) 3x + 2 8 (iii) 52

x (iv) 4x – 3 > 7

The general forms of linear inequations are: (i) ax + b > c (ii) ax + b < c (iii) ax + b c (iv) ax + b c, Where a, b, care real numbers and .0a Replacement Set or Domain of the Variable The set from which the values of the variable x are replaced in an inequation, is called the replacement set or the domain of the variable This replacement set is always given to us. Solution Set The set of all those values of x from the replacement set which satisfy the given inequation, is called the solution set of the inequation. Solution set is always a subset of the replacement set. Example: Write down the solution set of x < 5, when the replacement set is (i) N (ii) W (iii) I. Solution: (i) Solution set = }.4,3,2,1{}5:{ xNx



(ii) Solution set = }.4,3,2,1,0{}5:{ xWx (iii) Solution set = }.4,3,2,1,0,1,2{...,}5:{ xIx Properties of Inequations (i) Adding the same number or expression to each side of an inequation does not change the inequality. (ii) Subtracting the same number or expression from each side of an inequation does not change the inequality. (iii) Multiplying (or dividing) each side of an inequation by the same positive number does not change the inequality. (iv) Multiplying (or dividing) each side of an inequation by the same negative number reverses the inequality. e.g., 22 xx [Multiplying both sides by -1] and, 482 xx [Dividing both sides by -2] Remarks: (i) .abba (ii) .abba Thus, .55and33 xxxx MEHTOD OF SOLVING A LINEAR INEQUATION IN ONE VARIABLE (i) Simplify both sides by removing group symbols and collecting like terms. (ii) Remove fractions (or decimals) by multiplying both sides by an appropriate factor (L.C.M. of fractions or a power of 10 in case of decimals). (iii) Collect all variable terms on one side and all constants on the other side of the inequality sign. Collect like terms when possible. (iv) Make the coefficient of the variable 1. (v) Choose the solution set from the replacement set. Some Special Sets of Numbers Shown on Number Line We would like to have a glimpse on how we represent sets of numbers on number line. Some sets of number and their graphs are given below: Graphs of Subsets of N, W and I (i) Graph of }3,2,1{},41:{ Nxxx -2 -1 0 1 2 3 4 The darkened circles indicate the natural numbers contained in the set. (ii) Graph of }.3,2,1,0{},41:{ Wxxx -2 -1 0 1 2 3 The darkened circles indicates the whole numbers contained in the set.



(iii) Graph of ,.....}.6,5,4,3{},2:{ Nxxx -1 0 1 2 3 4 5 6 The darkened circles indicate the natural numbers contained in the set and three dark dots above the right part of the line show that natural numbers are continued indefinitely. (iv) Graph of .....}.,5,4,3,2{},1:{ Ixxx -5 -4 -3 -2 -1 0 1 The darkened circles show the integers contained in the set and three dark dots above the left part of the line show the indefinite continuity of negative integers. Graphs of Set of All Real Numbers Between Two Given Numbers We show the end points of the set by two circles, namely a hollow circle for the number not contained in the set and the darkened circle for the number contained in the set. And, the line segment between these circles is darkened. (v) Graph of }.,31:{ Rxxx -2 -1 0 1 2 3 4 5 Note that the end points -1 and 3 are both contained in the set. (vi) Graph of }.,32:{ Rxxx

-3 -2 -1 0 1 2 3 4

Note that -2 is there in the given set while 3 is not contained in it. (vii) Graph of }.,51:{ Rxxx

-1 0 1 2 3 4 5 6

Here 1 as well as 5 is not contained in the given set. (viii) Graph of }.,40:{ Rxxx



-2 1- 0 1 2 3 4 5 6

Here 0 is not contained in the set while 4 is contained in it. Graphs of : ,x x a x R and : ,x x a x R Such a set has one end point. The ray to the right of this point is darkened. If the end point is contained in the set, then it is shown by a darkened circle, otherwise by a hallow circle. (ix) Graph of : 2,x x x R .

-1 0 1 2 3 4

(x) Graph of : 3,x x x R

-1 0 1 2 3 4 5

Here the end point 3 is not contained in the set. Graphs of }.,:{and},:{ RxaxxRxaxx Such a set has one end point. The ray to the right of this point is darkened. If the end point is contained in the set, then it is shown by a darkened circle, otherwise by a hollow circle. (xi) Graph of }.,3:{ Rxxx -2 -1 0 1 2 3 4 5 Here the end point 3 is contained in the set. (xii) Graph of }.,1:{ Rxxx

-1-2-3 0 1 2 3

Here the end point 1 is not contained in the set.



SOLVED EXAMPLES Problem 1. Find the positive integral solution of 13x + 15y = 189 for which the value of y is largest. Solution: We have 13x + 15y = 189 15y = 189 - 13x …(i) When y is largest, x must have least positive integral value. Since 15y is divisible by 15. R.H.S. of the equation also should be divisible by 15. The least value of x for which R.H.S. is divisible by 15 is x = 3. .103 yx Hence the required solution is x = 3 and y = 10. Problem 2. The sum of the digits of a two – digit number is 9. If 27 is subtracted from the number, the digits are reversed. Find the number. Solution: Let the digit in the units place be x and the digit in the ten’s place y. The number = 10y + x. When 27 is subtracted from the number, the digits are reversed. 10y + x – 27 = 10x + y 2799 xy

(given))...(9

)...(3iiyxixy

Adding (i) and (ii), we get y = 6 Put y = 6 in (ii), we get x = 3. Hence the required number is 63. Problem 3. Find the value of a so that x = -1, y = 3 is a solution of equation 5(x – a) + a(y + 6) = 7. Solution: Since x = -1, y = 3 is a solution of the given equation, these values will satisfy the equation. Putting x = -1 and y =3 in the equation, we get 5(-1 –a) + a(3 + 6) = 7 76355 aaa .312574 aa Problem 4. Find four solutions of the equation 3(x – 2) – 4 (2 – 3y) = 10. Solution: The given equation is 3(x – 2) – 4 (2 – 3y) – 10

84

241231012863

yxyx

yx

Taking x = 0, we get 4y = 8 y = 2



x = 0, y = 2 is a solution. Taking x = 4, we get 4 + 4y = 8

.solutionais1,4

144

yx

yy

Taking x = -4, we get -4 + 4y = 8 3124 yy x = -4, y = 3 is also a solution. Taking x = 8, we get 8 + 4y = 8 004 yy x = 8, y = 0 is a solution. Thus we get four solutions of the given equation.

Problem 5. Solve the system of equations .2194and932xyyxxyyx

Solution: We have xyyx932

…(1)

.2194xyyx

…(2)

Multiplying both sides of (1) and (2) by xy, we get 2y + 3x = 9 …(3) 4y + 9x = 21 …(4)

From (3) 239 xy

Substituting the value of y in (4) we get

133

2192394

xx

xx

Put x = 1 in (3) we get value of y i.e. y =3 Hence x = 1 and y = 3 is the required solution. Problem 6. The area of a rectangle gets reduced 80 sq. units if its length is reduced by 5 units and the breath is increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, the area is increased by 50square units. Find the length and breadth of the rectangle. Solution: Let the length and breadth of the rectangle be x units and y units respectively. Then area of the rectangle = xy sq. units Case I: When the length is reduced by 5 units and the breadth is increased by 2 units. In this case, new length = (x – 5) units, And new breadth (y + 2) units new area = (x – 5) (y + 2) sq. units. Decrease in area = [xy – (x – 5) (y + 2) = 80 According to the question xy – (x – 5) (y + 2) = 80 5y – 2x + 10 = 80



5y – 2x = 70 …(i) Case II: When the length is increased by 10 units and the breadth is decreased by 5 units. In this case, new length = (x + 10) units & new breadth = (y – 5) units new area = (x + 10) (y – 5) sq. units So increase in area = [(x + 10) (y – 5) – xy] sq. units. (x + 10) (y – 5) 9 xy = 50 (given) 10y – 5x = 100 2y – x = 20 …(ii) Multiplying (ii) by 2 and subtracting from (i) we get y = 30. Substituting y = 30 in (ii) we get. 4020302 xx Length of the rectangle = 40 units and the breadth of the rectangle = 30 units. Problem 7. If 1 is added to each of the two given numbers, their ratio becomes 1:2 and if 5 is subtracted form each, the ratio becomes 5:11. Find the numbers. Solution: Let the given numbers be x and y. Then, according to the condition of the problem,

21

11

yx ..(i)

And 115

55

yx ..(ii)

From (i), 2x + 2 = y + 1 2x – y + 1 = 0 ..(iii) From (ii), 11x – 55 = 5y – 25 11x – 5y – 3 = 0 ..(iv) By cross-multiplication, from (iii) and (iv), we get

11 1 1 2 2 1x y

-5 -30-30 11 11 -5

1110

16011530

yx

3513511

7135 xyx

x = 71 1 = 71 hence, the required numbers are 35 and 71. Problem 8. Determine graphically the vertices of the triangle, the equation of whose sides are given below: 2y – x – 8 = 0, 5y – x - 14 = 0 and y – 2x = 1. Solution: The given system of equations are 2y – x – 8 = 0 …(1) 5y – x – 14 = 0 …(2) y – 2x = 1 …(3) Since all the three equations are linear, their graphs must be straight lines.



From (1), 2y = x + 8 2

8

xy

When x = 0, we have 42

80

y


82

y

Thus we have the following table: We plot the points (0, 4) and (2, 5) and join then to get a straight line l1

Again, from (2), 5y – x = 14 514

xy


y

When x = 6, we have .45146

y

Thus, we have the following table: We plot the points (1, 3) and (6, 4) on the same graph paper and join them to get a straight line 12’. Laslty , we consider equation (iii), y – 2x = 1 y = 2x + 1 When x = 0, we have y = 2(0) + 1 =1 When x = 1, we have y = 2(1) + 1 = 3 Thus, we have the following table: We plot the points (0, 1), (1, 3) on the same graph and join them to obtain the straight line l3

From the graph we observe that the lines l1, l2 and l3 interest I points

x 0 2 y 4 5

x 1 6 y 3 4

x 0 1 y 1 3



A(2, 5), B(-4, 2) and C (1, 3) respectively. Hence the vertices of triangle ABC are (2, 5), (-4, 2) and (1, 3). Hence the vertices of the triangle ABC, are (2, 5), (-4, 2) and (1, 3). Problem 9. Solve the following system of equations by the method of substitution: x – y = 0.9

.1)(2

11

yx

Solution: The given equations are x – y = 0.9 …(i)

.1)(2

11

yx …(ii)

From (i), we get x = y + 0.9 Substituting this value of x in equation (ii), we get

3.2118.14

)9.02(211

1)9.0(2

11

yyyyy

Substituting this value of y in equation (i), we get x – 2.3 = 0.9 x = 2.3 + 0.9 = 3.2 Hence x = 3.2 and y 2.3 is the required solution. Problem 10. Solve the following system of equations:

)2...()()(2)()(

)1...()(11

3322

22

yxmllmlmyx

lmml

ymlx

lm

Solution: From equation (1), we get )()( 2222 lmlmylxm …(3) From equation (2), we get

)4...()(2

)(2)()( 3322

lmyxlmmllmyx

Multiplying (4), by m2 and subtracting from (3), we get

2 2 2 2 2

2 2 2 2

( ) ( ) ( 2 )( ) ( ) ( )

l m y m l m l ml m y m l l m

y l m

Putting the value of y in (4), we get mlx Hence x = l + m, y = l + m is the required solution. Problem 11. Draw the graph of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Calculate the area bounded by these lines and x-axis.



Solution: The given equations are x – y + 1 = 0 …(1) and 3x + 2y – 12 = 0 …(2) table for x – y + 1 = 0 Table fo 3x + 2y – 12 = 0 Now we plot the points A(-1, 0), B(2, 3), C(-2, 9) and D(4, 0). The points A(-1, 0) and B(2, 3) lie on the graph of (1) and the line AB is the graph of x – y + 1 = 0. Also, the points (-2, 9) and D(4, 0) lie on the graph of (2) and so the line joining C and D is the graph of (2) We extend the lines AB and CD to both sides. The line AB cuts the x-axis at the point A(-1, 0) and the line CD cut it at the point D(4, 0).Also the lines AB and CD cut each other at B(2, 3). Hence A(-1, 0), B(2, 3) and D(4, 0) are the vertices of the triangle, so formed.

From B(2, 3), we draw perpendicular on the x-axis meeting the x-axis at P. Now, clearly PB is the altitude of BAD and AD being the base of this triangle. Length of AD = 4 – (-1) = 5 units And altitude PB = 3 units

.5.72

153521

21 unitssquarealtitutebaseBADofArea

Problem 12. For k = 6 the system of equation kx + 3y = k – 3 and 12x + ky = k have (a) a unique solution (b) more than one solution (c) no solution (d) infinitely many solutions

x -1 2 y 0 3

x -2 4 y 9 0



Solution: (D). for k = 6, Equations can be written as 6x + 3y = 3 and 12x + 6y = 6

63

63

126

infinitely many solution. Problem 13. The cost of 5 oranges and 3 apples is Rs 35 and the cost of 2 oranges and 4 apples is Rs 28. The cost of an orange is (a) Rs 4 (b) Rs 5 (c) Rs 3 (d) Rs 3.50 Solution: (A). Let the cost of an orange = Rs x Let the cost of an apple = Rs y

2842

3535

yxyx

It givens x = 4 and y = 5 Hence cost of an orange = Rs 4. Problem 14. If the pair of equation 4x + py + 8 = 0 and 2x + 2y + 2 = 0 has unique solution. Then the value of p is (a) p = 4 (b) p = 2 (c) for all values of p except 2. (d) for all values of except 4 Solution: (D). For unique solution:

4

224

p

p

Problem 15. Meena can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. The speed of the current is

(a) sec/9

10 m (b) sec/1825 m

(c) sec/1815 m (d) sec/

1736 m



Solution: (A). Speed of current = x km/hr speed of rowing = y km/hr

./4

22)(4102)(20

hrkmxxyxy

yxyx

Speed of current = sec/9

10 m .

Problem. 16 If the lines are parallel, then the pair of equation has no solution. In their case, the pair of equation is ___________________. Solution: inconsistent

Problem. 17 A fraction becomes 31 when 1 is subtracted from the numerator and it

becomes 41 when 8 is denominator. The fraction is __________.

Solution: Let fraction be yx

84

41

8

33311

yxy

x

yxy

x

On soling x = 5, y = 12

Fraction is .125

Problem 18. The pair of equation 3

3611,17215

baand

ba is given, then the value

a + b = ____________________. Solution:

536

17215

1,15

3611

17215

vu

vu

vb

ua

Put

ba

ba

on solving we get 751

andvu



71,5 ba

7

36715 ba

Example 19. Find the solution set of 3(x - 2)< 1, where }.6,5,4,3,2,1{x Solution: We have: 1631)2(3 xx

.312

3]bysidesboth[Dividing37

sides]bothon6[Adding73

x

x

x

Solution set = }6,5,4,3,2,1{,,312:

AwhereAxxx

}.2,1{

Example 20. Solve the inequation .,3565112 Rxxx Represent the solution set on a number line.

Solution: We have:

xxxx 356111235

65112

]7[6]18[427]72[4218116]bysidesbothng[Multiplyi18301172

bysidesbothDividingxsidesthonxAddingxsidesbothonAddingxx

xx

o

Solution set = }.,6:{ Rxxx This set can be represented on the number line, as shown below:

-1 0 1 2 3 4 5 6 7



Example 21. Solve the inequation .,9233 Rxx Represent the solution set on a number line. Solution: .923and2339233 xxx [Adding -3 to both sides] Now, xx 26233

]2[)...(3

62

bysidesbothDividingix

x

Again, sides]bothto3[Adding62923 xx

)...(3

2]bysidesboth[Dividing3iix

x

From (i) and (ii), we get .33 x }.,33:{setSolution Rxxx This set can be represented on the number line, as shown below:

-4 -3 -2 -1 0 1 2 3 4

Example 22. Solve and graph the solution set of 3x - 4 > 11 and 5 – 2x .7 Solution: We have: 1531143 xx [Adding 4 to both sides] .5 x [Dividing both sides by 3] Let }.,5:{ RxxxA Again, 22725 xx [Adding -5 on both sides] 1 x [Dividing both side by -2] Let B = }.,1:{ Rxxx Required solution set = BA = }.1:{}5:{ xRxxRx The graph of this set is given below:

-1 0 1 2 3 4 5

Example 23. Write down the range of real values of x for which the inequations x > 3 and 52 x are both true.



Solution: Clearly, we have to find the common values of the solution sets of the given inequations. The graph of these inequations are given below:

-2 -1 0 1 2 3 4

-2 -1 0 1 2 3 4 5

The common portion of the two graphs is the set }.,53:{ Rxxx Hence, the required set is }.,53:{ Rxxx Example 24. The following diagram represents two inequations A and B on number lines. A = -1 0 1 2 3 4 5 6 B = -1 0 1 2 3 4 5 6 7 8 (i) Write down A, B and 'B in set builder form. (ii) Represent 'BAandBA on two different number lines. Solution: (i) Clearly, we have:

}7:{}2:{

}.72:{},51:{

'

xRxxRxBxRxB

xRxA

(ii) From the graphs, it is clear that



}52:{ xRxBA And, )21:(' xRxBA The graphs of 'BAandBA may be drawn as shown below: BA = -1 0 1 2 3 4 5 6

'BA -1 0 1 2 3 4 5 6 7 8

Documents

CLASSIFICATION OF REAL NUMBERS - Abhijit Kumar Jha · 2014. 10. 14. · NUMBERS (Theory) CLASSIFICATION OF REAL NUMBERS Real Number are classified into rational and irrational numbers