X CBSE Maths Sem 01.pdf

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(3) Table of contents. Table of Contents CHAPTER 01: REAL NUMBERS.............................................................................. 1 CHAPTER MAP: ........................................................................................................................................... 1 Introduction: ....................................................................................................................................... 1 Euclid’s Division Lemma: .................................................................................................................. 1 Solved Examples 1.1: .............................................................................................................. 2 Unsolved Exercise 1.1: ............................................................................................................ 3 The Fundamental Theorem of Arithmetic:......................................................................................... 3 Solved Examples 1.2: .............................................................................................................. 4 Unsolved Exercise 1.2: ............................................................................................................ 5 Irrational Numbers: ............................................................................................................................ 6 Solved Examples 1.3: .............................................................................................................. 7 Unsolved Exercise 1.3: ............................................................................................................ 8 Rational Numbers and their Decimal Expansions: ............................................................................ 8 Unsolved Exercise 1.4: ............................................................................................................ 8 Miscellaneous Exercise: .......................................................................................................... 9 Multiple Choice Questions: .................................................................................................... 11 Column Matching Questions: ................................................................................................ 13 Answers to Unsolved Exercise: ............................................................................................. 14. CHAPTER 02: POLYNOMIALS .............................................................................. 16 CHAPTER MAP: ......................................................................................................................................... 16 INTRODUCTION: ......................................................................................................................................... 16 Geometrical Meaning of the Zeroes of a Polynomial: ..................................................................... 17 Solved Example 2.1: .............................................................................................................. 18 Unsolved Exercise 2.1: .......................................................................................................... 19 Relationship between Zeroes and Coefficients of a Polynomial: .................................................... 19 Solved Examples 2.2: ............................................................................................................ 20 Unsolved Exercise 2.2: .......................................................................................................... 21 Division Algorithm for Polynomials: ............................................................................................ 22 Solved Examples 2.3: ............................................................................................................ 22 Unsolved Exercise 2.3: .......................................................................................................... 23 Miscellaneous Exercise: ........................................................................................................ 24 Multiple Choice Questions: .................................................................................................... 26 Column Matching Questions: ................................................................................................ 27 Answer to Unsolved Exercise: ............................................................................................... 29. CHAPTER 03: PAIR OF LINEAR EQUATIONS IN TWO VARIABLES .................. 31 CHAPTER MAP: ......................................................................................................................................... 31 BASIC FUNDAMENTALS: ............................................................................................................................ 31 GENERAL FORM OF A LINEAR EQUATION IN TWO VARIABLES: ..................................................................... 31 Volume. Universal Tutorials – X CBSE (2012–13) – Mathematics.

(4) Solved Exercise 3.1: .............................................................................................................. 33 Unsolved Exercise 3.1: .......................................................................................................... 33 Pair of Linear Equations in Two Variables: ..................................................................................... 34 Solution of a Pair of Linear Equation in Two Variables .............................................................. 34 Graphical Method of solving a pair of Linear Equations: ................................................................ 35 Solved Examples 3.2: ............................................................................................................ 36 Unsolved Exercise 3.2: .......................................................................................................... 37 Algebraic Methods of solving a pair of Linear Equations: ............................................................... 39 Solved Exercise 3.3: .............................................................................................................. 39 Unsolved Exercise 3.3: .......................................................................................................... 40 Solved Exercise 3.4: .............................................................................................................. 42 Unsolved Exercise 3.4: .......................................................................................................... 43 Cross Multiplication Method: ...................................................................................................... 43 Solved Examples 3.5: ............................................................................................................ 44 Unsolved Exercise 3.5: .......................................................................................................... 46 Equations Reducible to a Pair of Linear Equations in Two Variables ........................................ 47 Solved Exercise 3.6: .............................................................................................................. 47 Unsolved Exercise 3.6: .......................................................................................................... 48 WORD PROBLEMS:.................................................................................................................................... 49 Solved Examples 3.7: ............................................................................................................ 49 Unsolved Exercise 3.7: .......................................................................................................... 50 Miscellaneous Exercise: ........................................................................................................ 52 Multiple Choice Questions: .................................................................................................... 55 Column Matching Questions: ................................................................................................ 57 Answers to Unsolved Exercises: ........................................................................................... 58. CHAPTER 06: TRIANGLES .................................................................................... 62 CHAPTER MAP: ......................................................................................................................................... 62 Congruent Figures:.......................................................................................................................... 62 Unsolved Exercise 6.1: .......................................................................................................... 63 SIMILARITY OF TRIANGLES: ....................................................................................................................... 63 Congruence of Triangles: ................................................................................................................ 63 Theorem 6.1 Basic Proportionality Theorem (Thales Theorem):............................................... 64 Solved Example 6.2: .............................................................................................................. 65 Unsolved Exercise 6.2: .......................................................................................................... 66 CRITERIA FOR SIMILARITY OF TRIANGLES: ................................................................................................. 68 Tests for Similarity of Triangles: ...................................................................................................... 68 Theorem 6.3: A–A–A Criterion of Similarity ............................................................................... 68 Theorem 6.4: S–S–S Criterion of Similarity ............................................................................... 68 Theorem 6.5: S–A–S Criterion of Similarity ............................................................................... 68 Solved Example 6.3: .............................................................................................................. 69 Unsolved Exercise 6.3: .......................................................................................................... 70 Areas of Similar Triangles: .............................................................................................................. 72 Universal Tutorials – X CBSE (2012–13) – Mathematics. 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(5) Table of contents. Theorem 6.6: .............................................................................................................................. 72 Solved Example 6.4: .............................................................................................................. 73 Unsolved Exercise 6.4: .......................................................................................................... 73 PYTHAGORAS THEOREM: .......................................................................................................................... 75 Similarity in Right angled Triangles: ........................................................................................... 75 Theorem 6.7: .............................................................................................................................. 75 Theorem (6.8) Pythagoras Theorem: ......................................................................................... 75 Theorem 6.9: Converse of Pythagoras Theorem:...................................................................... 75 Solved Example 6.5: .............................................................................................................. 76 Unsolved Exercise 6.5: .......................................................................................................... 77 Proof of Theorems:.......................................................................................................................... 78 Theorem 6.7: Similarity in Right Angled Triangles ..................................................................... 81 Angle Bisector Property: ............................................................................................................ 81 Applications of Pythagoras Theorem: ............................................................................................. 82 Acute Angled Triangle: ............................................................................................................... 82 Obtuse Angled Triangle: ............................................................................................................ 83 Appollonius Principle: ................................................................................................................. 83 Miscellaneous Exercise: ........................................................................................................ 83 Multiple Choice Questions: .................................................................................................... 89 Column Matching Questions: ................................................................................................ 91 Answers to Unsolved Exercise: ............................................................................................. 93. CHAPTER 08: INTRODUCTION TO TRIGONOMETRY ......................................... 94 CHAPTER MAP: ......................................................................................................................................... 94 INTRODUCTION: ......................................................................................................................................... 94 What is Trigonometry? ............................................................................................................... 94 Use of learning trigonometry: ..................................................................................................... 94 Trigonometric Ratios: ...................................................................................................................... 94 Solved Examples 8.1: ............................................................................................................ 96 Unsolved Exercise 8.1: .......................................................................................................... 97 Trigonometric Ratio of some specific angles: ................................................................................. 99 Triangle Method: ...................................................................................................................... 102 Solved Examples 8.2: .......................................................................................................... 102 Unsolved Exercise 8.2: ........................................................................................................ 103 COMPLEMENTARY ANGLES: .................................................................................................................... 105 Trigonometric Inter-relationships:............................................................................................. 105 Solved Examples 8.3: .......................................................................................................... 106 Unsolved Exercise 8.3: ........................................................................................................ 106 TRIGONOMETRIC IDENTITIES: ................................................................................................................... 108 Solved Examples 8.4: .......................................................................................................... 109 Unsolved Exercise 8.4: ........................................................................................................ 111 Miscellaneous Exercise: ...................................................................................................... 112 Multiple Choice Questions: .................................................................................................. 116 Volume. Universal Tutorials – X CBSE (2012–13) – Mathematics.

(6) Column Matching Questions: .............................................................................................. 119 Answers to Unsolved Exercises: ......................................................................................... 120. CHAPTER 14: STATISTICS.................................................................................. 123 CHAPTER MAP: ....................................................................................................................................... 123 Calculation of Central tendencies for grouped data ...................................................................... 123 14.1 Mean of grouped data: ..................................................................................................... 123 Solved Examples 14.1: ........................................................................................................ 124 Unsolved Exercise 14.1: ...................................................................................................... 126 To find mean by Assumed mean method: ............................................................................... 127 To determine mean by step deviation method: ........................................................................ 127 Solved Examples 14.2: ........................................................................................................ 128 Unsolved Exercise 14.2: ...................................................................................................... 130 14.2 Mode of grouped data: .......................................................................................................... 132 Solved Examples 14.3: ........................................................................................................ 132 Unsolved Exercise 14.3: ...................................................................................................... 133 14.3 Median of Grouped Data: ...................................................................................................... 134 Solved Examples 14.4: ........................................................................................................ 135 Unsolved Exercise 14.4: ...................................................................................................... 137 Comparative Study: .................................................................................................................. 139 Graphical Representation of Cumulative frequency Distribution: ................................................. 139 Ogive of Less than type: .......................................................................................................... 139 Ogive of more than type: .......................................................................................................... 140 Solved Examples 14.5: ........................................................................................................ 141 Unsolved Exercise 14.5: ...................................................................................................... 142 Miscellaneous: ..................................................................................................................... 143 Multiple Choice Questions: .................................................................................................. 146 Answer to the Unsolved Exercise: ....................................................................................... 147. ANSWER TO THE MCQS: .................................................................................... 149. Universal Tutorials – X CBSE (2012–13) – Mathematics. Volume.

(7) Chapter 01: Real Numbers. 1. Chapter 01: Real Numbers Chapter Map: → Euclid’s Division Lemma → The Fundamental Theorem of Arithmetic → Irrational Numbers → Rational Numbers and their Decimal Expansions. Introduction: ¾ An algorithm is a series of well defined steps which gives a procedure for solving a type of problem. ¾ A lemma is a proven statement used for proving another statement.. Euclid’s Division Lemma: Theorem 1.1 (Euclid’s Division Lemma): z. Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b.. z. Consider the numbers 455 and 42; 455 can be uniquely expressed as 455 = 42 × 10 + 35.. z. If we consider 24 and 6 it can be uniquely expressed as 24 = 6 × 4 + 0.. Note: Euclid’s division lemma is a technique to compute the Highest Common Factor (HCF) of two given positive integers. Recall that the HCF of two positive integers a and b is the largest positive integer d that divides both a and b.. Euclid’s division algorithm: z. Euclid’s Division lemma can be expressed in words as: Dividend = Divisor × Quotient + Remainder. To obtain the HCF of two positive integers, say c and d, with c > d, follow the steps below: Step 1: Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 ≤ r < d. Step 2: If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r. Step 3: Continue the process till the remainder is zero. The divisor at the last stage will be the required HCF. . z z z z. This algorithm works because HCF (c, d) = HCF (d, r) where the symbol HCF (c, d) denotes the HCF of c and d etc. z. Eg: If we consider the numbers 420 and 272; 420 and 272 can be expressed as 420 = 272 × 1 + 148 Again consider 272 = 148 it can be expressed as 272 = 148 × 1 + 124 Again 148 = 124 × 1 + 24. Volume. Universal Tutorials – X CBSE (2012–13) – Mathematics. 1.

(8) 2. Similarly 124 = 24 × 5 + 4 Finally 24 = 4 × 6 + 0 So HCF is 6 i.e. 6 = HCF (24, 4) = HCF (124, 24) = HCF (148, 124) = HCF (420, 272). SOLVED EXAMPLES 1.1: 1) Use Euclid’s algorithm to find the HCF of 4052 and 12576. Sol: Step 1: Since 12576 > 4052, we apply the division lemma to 12576 and 4052, to get 12576 = 4052 × 3 + 420. 2) Sol:. 3) Sol:. Step 2: Since the remainder 420 ≠ 0, we apply the division lemma to 4052 and 420, to get 4052 = 420 × 9 + 272 Step 3: We consider the new divisor 420 and the new remainder 272, and apply the division lemma to get 420 = 272 × 1 + 148 We consider the new divisor 272 and the new remainder 148, and apply the division lemma to get, 272 = 148 × 1 + 124 We consider the new divisor 148 and the new remainder 124, and apply the division lemma to get, 148 = 124 × 1 + 24 We consider the new divisor 124 and the new remainder 24, and apply the division lemma to get, 124 = 24 × 5 + 4 We consider the new divisor 24 and the new remainder 4, and apply the division lemma to get, 24 = 4 × 6 + 0 The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4. Notice that 4 = HCF(24, 4) = HCF(124, 24) = HCF(148, 124) = HCF (272, 148) = HCF (420, 272) = HCF (4052, 420) = HCF (12576, 4052). Euclid’s division algorithm is not only useful for calculating the HCF of very large numbers, but also because it is one of the earliest examples of an algorithm that a computer had been programmed to carry out. Show that every positive even integer is of the form 2q, and that every positive odd integer is of the form 2q + 1, where q is some integer. Let a be any positive integer and b = 2. Then, by Euclid’s algorithm, a = 2q + r, for some integer q ≥ 0, and r = 0 or r = 1, because 0 ≤ r < 2. So, a = 2q or 2q + 1. If a is of the form 2q, then a is an even integer. Also, a positive integer can be either even or odd. Therefore, any positive odd integer is of the form 2q + 1. What is the greatest number which divides each of the numbers 2261 and 2527 exactly? The greatest no. which divides each of the no. 2261 & 2527 exactly is the HCF of 2261 & 2527 By Euclid’s algorithm, 2527 = 2261 × 1 + 266 ⇒ 2261 = 266 × 8 + 133 ⇒ 266 = 133 × 2 + 0 ∴ 133 is the HCF of 2527 and 2261 HCF (2527, 2261) = 133. 2. Universal Tutorials – X CBSE (2012–13) – Mathematics. Volume.

(9) Chapter 01: Real Numbers. 3. UNSOLVED EXERCISE 1.1: CW Exercise: 1) Use Euclid’s division algorithm to find the HCF of: i) 135 and 225 ii) 196 and 38220 iii) 81 and 127 2) Show that any positive integer is of the form 3q, 3q + 1 or 3q + 2 where q is some integer. 3) Show that any odd positive integer p can be expressed in the form i) 6q + 1, or 6q + 3, or 6q + 5 ii) 4q + 1 or 4q + 3 4) Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m. [CBSE–08] 5) What is the greatest number which divides 209 and 1195 leaving remainder 5 in each case? 6) What is the largest number which when divides 63, 77 and 112 leaves 3, 5 and 4 as remainders respectively? 7) Find the HCF of 65 and 117 and express it in the form 65m + 117n. 8) An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? 9) Find the greatest number of 6 digits exactly divisible by 24, 15 and 36. HW Exercise: 1) Use Euclid’s division algorithm to find the HCF of: i) 867 and 255 ii) 3638 and 3587 2) Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8. 3) A sweetseller has 420 kaju barfis and 130 badam barfis. She wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the maximum number of barfis that can be placed in each stack for this purpose? 4) The length, breadth and height of a room are 8m 25 cm, 6m 75 cm and 4 m 50 cm, respectively. Determine the longest tape which can measure the three dimensions of the room exactly. 5) Show that any positive odd integer is of the form 6q + 1 or 6q + 5 or 6q + 3, where q is some integer. 6) Show that the square of any positive integer is of the form 5m, 5m + 1 or 5m + 4 for some integer m. 7) If the HCF of 210 and 55 is expressible in the form 210 × 5 + 55y, find y. 8) If d is the HCF of 56 and 72, find x, y satisfying d = 56x + 72y. Also, show that x and y are not unique. 9) Find the largest number that divides 2053 and 967 and leaves a remainder of 5 and 7 respectively. 10) Find the largest number that will divide 398, 436 and 542 leaving remainder 7, 11 and 15 respectively.. The Fundamental Theorem of Arithmetic: Theorem 1.2 (Fundamental Theorem of Arithmetic): z. Volume. Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur. Universal Tutorials – X CBSE (2012–13) – Mathematics. 3.

(10) 4 z. Eg: Consider the number 32760, it can be expressed as. z. 32760 = 2 ×2 ×2 × 3 × 3 × 5 × 7 × 13 = 23 × 32 × 5 × 7 × 13 This prime factorization is unique apart from the order in which the prime factor occurs.. Note: In general, given a composite number x, we factorise it as x = p1p2 … pn, where p1,p2, …, pn are primes and written in ascending order, i.e. p1 ≤ p2 ≤ … ≤ pn. If we combine the same primes, we will get powers of primes. z z z. HCF of numbers is the Product of the smallest power of each common prime factor in the numbers. LCM of numbers is the Product of the greatest power of each prime factor, involved in the numbers. For any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b. We can use this result to find the LCM of two positive integers, if we have already found the HCF of the two positive integers.. Note: For any positive integers a, b and c. a × b × c ≠ HCF(a, b, c) × LCM (a, b, c) LCM (p, q, r) =. p ⋅ q ⋅ r ⋅ HCF ( p, q, r ) p ⋅ q ⋅ r ⋅ LCM ( p, q, r ) ; HCF (p, q, r) = HCF ( p, q ) ⋅ HCF (q, r ) ⋅ HCF ( p, r ) LCM ( p, q ) ⋅ LCM (q, r ) ⋅ LCM ( p, r ). SOLVED EXAMPLES 1.2: 1) Consider the numbers 4n, where n is a natural number. Check whether there is any value of n for which 4n ends with the digit zero. Sol: If the number 4n, for any n, were to end with the digit zero, then it would be divisible by 5. That is, the prime factorisation of 4n would contain the prime 5. This is not possible because 4n = (2)2n; so the only prime in the factorisation of 4n is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 4n. So, there is no natural number n for which 4n ends with the digit zero. 2) Find the LCM and HCF of 6 and 20 by the prime factorisation method. Sol: We have: 6 = 21 × 31 and 20 = 2 × 2 × 5 = 22 × 51. You can find HCF(6, 20) = 2 and LCM(6, 20) = 2 × 2 × 3 × 5 = 60, as done in your earlier classes Note that HCF(6, 20) = 21 = Product of the smallest power of each common prime factor in the numbers = 2. LCM (6, 20) = 22 × 31 × 51 = Product of the greatest power of each prime factor, involved in the numbers = 60. 3) Find the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM. Sol: The prime factorisation of 96 and 404 gives: 96 = 25 × 3, 404 = 22 × 101 Therefore, the HCF of these two integers is 22 = 4. 96 × 404 96 × 404 = = 9696. Also, LCM (96, 404) = HCF (96, 404) 4 4) Find the HCF and LCM of 6, 72 and 120, using the prime factorization method. Sol: We have: 6 = 2 × 3, 72 = 23 × 32, 120 = 23 × 3 × 5 Here, 21 and 31 are the smallest powers of the common factors 2 and 3 respectively. So, HCF (6, 72, 120) = 21 × 31 = 2 × 3 = 6 23, 32 and 51 are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the three numbers. So, LCM (6, 72, 120) = 23 × 32 × 51 = 360 4. Universal Tutorials – X CBSE (2012–13) – Mathematics. Volume.

(11) Chapter 01: Real Numbers. 5. UNSOLVED EXERCISE 1.2: CW Exercise: 1) Express each number as a product of its prime factors: i) 140 ii) 3825 iii) 372 iv) 9072 v) 462 2) Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. i) 26 and 91 ii) 510 and 92 iii) 336 and 54 iv) 84,144 3) Find the LCM and HCF of the integers given below by applying the prime factorization method. Also verify that for 3 numbers LCM × HCF = product is true or not. i) 12, 15 and 21 ii) 7, 13, and 19 iii) 20, 18, and 75 iv) 52, 75 and 77 4) Check whether 6n can end with the digit 0 for any natural number n. Also count the number of zeros in the number given by 24n × 25n × 26n 5) Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers. 6) Find the smallest number which leaves remainder 8 and 12 when divided by 28 and 32 respectively. 7) Answer the questions given below based on the information given about two numbers x, y: i) HCF(x, y) = 16 and x. y = 3072. Find LCM(x, y). ii) LCM(x, y) = 6, HCF(x, y) = 180 and x = 30 then find the value of y. 8) In a seminar, the number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively. Find the minimum number of rooms required if in each room the same number of participants is to be seated and all of them being in the same subject. 9) On CST–Thane route one cycle of journey for a local train is considered from CST to Thane and back to CST. A fast local takes 90 min. and slow local takes 120 min. to complete a cycle on CST–Thane route. A fast local and a slow local start together from CST at 10:00 am, at what time they will meet again at CST considering there is no time gap between the cycles of journey? 10) Four bells strike at intervals of 6, 8, 9 and 12 minutes. An alarm is set in a mobile phone such that it alarms after every hour. It was noticed that all the bells struck simultaneously when mobile phone alarmed at 10:00 am. At what time all the bells will strike together with the alarm? HW Exercise: 1) Express each number as a product of its prime factors: i) 156 ii) 5005 iii) 7429 iv) 19530 v) 6006 2) Find the LCM and HCF of the 336 and 54 and verify that LCM × HCF = product of the two numbers. 3) Find the LCM and HCF of the following integers by applying the prime factorization method. i) 17, 23 and 29 ii) 8, 9 and 25 iii) 24, 36 and 60 4) Given that HCF (306, 657) = 9, find LCM (306, 657). 5) Explain why 7 × 19 × 23 + 23 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 7 are composite numbers. 6) There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point? 7) Consider the number 4n, where n is a natural number. Check whether there is any value of n for which 4n ends with the digit zero. 8) Three sets of English, Hindi and Mathematics book have to be stacked in such a way that all the books are stored topic wise and the highest of each stack is the same. The number of English book is 96, the number of Hindi book is 240 and the number of Mathematics book is 336. Assuming that the books are of the same thickness, determine the number of stacks of English, Hindi and Mathematics books.. Volume. Universal Tutorials – X CBSE (2012–13) – Mathematics. 5.

(12) 6. Irrational Numbers: ¾ Recall, a number ‘s’ is called irrational if it cannot be written in the form. p , where p and q are q. integers and q ≠ 0. Some examples of irrational numbers, with which you are already familiar, are: 2,. 3,. 15 , π, –. 2 3. , 0, 10110111011110 …, etc.. Theorem 1.3: z z. Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer. Proof : Let the prime factorisation of a be as follows: a = p1p2 . . . pn, where p1, p2, . . ., pn are primes, not necessarily distinct. Therefore, a2 = (p1p2 . . . pn)(p1p2 . . . pn) = p12 … pn2 .. z z z z z z. Now, we are given that p divides a2. Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a2. However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a2 are p1, p2, . . ., pn. So p is one of p1, p2, . . ., pn. Now, since a = p1 p2 . . . pn. Therefore, p divides a.. Theorem 1.4: 2 is irrational: z. Proof : Let us assume, to the contrary, that. 2 is rational.. So, we can find integers r and s (≠ 0) such that. 2 =. r s. Suppose r and s have a common factor other than 1. Then, we divide by the common factor to get,. 2 =. a , where a and b are co prime. b. So, b 2 = a. Squaring on both sides and rearranging, we get 2b2 = a2. Therefore, 2 divides a2. Now, by Theorem 1.3, it follows that 2 divides a. So, we can write a = 2c for some integer c. Substituting for a, we get 2b2 = 4c2, that is, b2 = 2c2. This means that 2 divides b2, and so 2 divides b(again using Theorem 1.3 with p = 2). Therefore, a and b have at least 2 as a common factor. But this contradicts the fact that a and b have no common factors other than 1. This contradiction has arisen because of our incorrect assumption that So, we conclude that. 2 is rational. 2 is irrational.. Note: The sum or difference of a rational and an irrational number is irrational and the product and quotient of a non-zero rational and irrational number is irrational. 6. Universal Tutorials – X CBSE (2012–13) – Mathematics. Volume.

(13) Chapter 01: Real Numbers. 7. SOLVED EXAMPLES 1.3: 1) Prove that. 3 is irrational.. [CBSE–08]. 3 is rational.. Sol: Let us assume, to the contrary, that. a . b Suppose a and b have a common factor other than 1, then we can divide by the common factor,. That is, we can find integers a and b (≠ 0) such that. 3 =. 3b=a Squaring on both sides, and rearranging, we get 3b2 = a2. Therefore, a2 is divisible by 3, and by Theorem 1.3, it follows that a is also divisible by 3. So, we can write a = 3c for some integer c. Substituting for a, we get 3b2 = 9c2, that is, b2 = 3c2. This means that b2 is divisible by 3, and so b is also divisible by 3 (using Theorem 1.3 with p = 3) Therefore, a and b have at least 3 as a common factor. But this contradicts the fact that a and b are co prime. This contradiction has arisen because of our incorrect assumption that 3 are rational.. and assume that a and b are co prime. So,. 3 is irrational.. So, we conclude that 2) Show that 5 –. 3 is irrational. 3 is rational.. Sol: Let us assume, to the contrary, that 5 –. That is, we can find co prime a and b (b ≠ 0) such that 5 – Therefore, 5 –. a = b. 3 =. a b. 3. Rearranging this equation, we get. 3 =5–. Since a and b are integers, we get 5 – But this contradicts the fact that. a 5b − a = b b. a is rational, and so b. 3 is rational.. 3 is irrational.. This contradiction has arisen because of our incorrect assumption that 5 – So, we conclude that 5 –. 3 is rational.. 3 is irrational.. 3) Show that 3 2 is irrational. Sol: Let us assume, to the contrary, that 3 2 is rational. That is, we can find co prime a and b (b ≠ 0) such that 3 2 = a . 3b a Since 3, a and b are integers, is rational, and so 3b. Rearranging, we get. a . b. 2 =. But this contradicts the fact that. 2 is rational.. 2 is irrational.. So, we conclude that 3 2 is irrational. Volume. Universal Tutorials – X CBSE (2012–13) – Mathematics. 7.

(14) 8. UNSOLVED EXERCISE 1.3: CW Exercise: 1) Prove that the following are irrationals: 1 ii) 7 5 i) 2 v) 3 + 2 5. vi) 5 3. vii). HW Exercise: 1) Prove that the following are irrationals: 1 i) ii) 7 10 3 v) 5 – 2 3 [CBSE–08] vi) (3 –. iii) 3 +. 5 )2. 2 +. iii) 6 + vii). 2 [CBSE 08] iv). 5 [CBSE 08–09]. 3. 5. 6 2 3. iv) viii). 6 45. Rational Numbers and their Decimal Expansions: Theorem 1.5: z. Let x be a rational number whose decimal expansion terminates then x can be expressed in p the form , where p and q are co prime and the prime factorisation of q is of the form 2n5m, q where n, m are non-negative integers.. Theorem 1.6: z. Let x =. p be a rational number, such that the prime factorization of q is of the form 2n5m, q. where n, m are non-negative integers then x has a decimal expansion which terminates.. Theorem 1.7: z. Let x =. p be a rational number, such that the prime factorization of q is not of the form q. 2n5m, where n, m are non-negative integers then, x has a decimal expansion which is nonterminating repeating (recurring). Note: From theorem 1.5, 1.6 and 1.7, we can conclude that the decimal expansion of every rational number is either terminating or non-terminating repeating.. UNSOLVED EXERCISE 1.4: CW Exercise: 1) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 13 64 15 35 11 23 ii) iii) iv) 3 2 v) vi) i) 3125 455 1600 50 1000 2 5 2) Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.. 8. Universal Tutorials – X CBSE (2012–13) – Mathematics. Volume.

(15) Chapter 01: Real Numbers. 9. 3) The following real numbers have decimal expansions as given below. In each case, decide p whether they are rational or not. If they are rational, and of the form , what can you say about q the prime factors of q? i) 43.123456789 ii) 0.120120012000120000 . . . iii) 43.123456789 HW Exercise: 1) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 17 29 6 77 13 129 i) ii) iii) 2 7 5 iv) v) vi) 8 343 15 210 8000 2 5 7 2) Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions. 3) The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form p/q, what can you say about the prime factors of q? i) 2.12342543. ii) 3.01001000100001 . . .. iii) 25.912345678. MISCELLANEOUS EXERCISE: 1) Use Euclid’s algorithm to find the HCF of: i) 2527 and 1653 ii) 1261 and 442 iii) 576 and 252 2) Use Euclid’s algorithm to find the HCF of: i) 1320 and 935 ii) 1624 and 1276 3) Use Euclid’s algorithm to find the HCF of: i) 963 and 657 ii) 3638 and 3587 iii) 468 and 222 iv) 495 and 657 4) Show that any positive odd integer is of the form 8q + 1, 8q + 3, 8q + 5 or 8q + 7, where q is some integer. 5) Show that the square of any positive odd integer is of the form 8m + 1, for some integer m. [CBSE–09] 6) Find the greater number which will divides 3457 and 9375 leaving 6 and 8 as remainder respectively. 7) Find the greater number which will divide 410, 751 and 1030 so as to leave remainder 7 in each case. 8) Two masses of gold weighing 3318 and 3054 gram respectively are each to be made into medals of the same size. What is the weight of the largest possible medal? 9) Express each number as a product of its prime factor. i) 1560 ii) 3990 10) Find the HCF by prime factorisation method of: i) 81 and 17. ii) 225 and 450. 11) Find the HCF by prime factorisation method of, 106, 159 and 265 12) Find the LCM of 45, 105 and 165 by finding the prime factors. 13) If HCF (12, 15) = 3. Find LCM of (12, 15). 14) Find the HCF of 96 and 404 by prime factorisation method. Hence, find their LCM. 15) Find the HCF and LCM of 6, 72 and 120, using prime factorisation method. 16) Find the HCF of 16 and 40 by prime factorisation method. Hence, find their LCM. 17) Find the smallest number which when divided by 25, 40 and 60 leaves remainder 7 is each case. Volume. Universal Tutorials – X CBSE (2012–13) – Mathematics. 9.

(16) 10. 18) Can two numbers have 14 as their HCF and 204 as their LCM. Give reasons in support of your answer. 19) Three horses run round a circular path, 1760 metres in circumference, at the rate of 440 m, 352 m and 264 m per minute. When will they again be together at the starting point? 20) In a morning walk three persons step off together. Their steps measures 80 cm, 85 cm and 90 cm respectively. What is the minimum distance each should walk so that he can cover the distance in complete steps? 21) Find the smallest length of a rope which can be measured exact number of times by three taps measuring 1 m, 20 cm, 75 cm and 1 m. 22) Telegraph poles occur at equal distances of 220 m along and heaps of stones are put at equal distances of 300 m along the same road. The first heap is at the foot of the first pole. How far from it along the road is the next heap which lies at the foot of the first pole. How far from it along the road is the next heap which lies at the foot of a pole? 23) Determine the number nearest to 110000 which is exactly divisible by each of 8, 15 and 21. 24) Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468. 25) A circular field has a circumference of 360 km. Three cyclists start together and can cycle 48, 60 and 72 km a day, round the field. When will they meet again? 26) The LCM and HCF of two numbers are 180 and 6 respectively. If one of the numbers is 30, Find the other number. 27) Write down the decimal expansions of the following rational numbers by writing their denominators in the form 2m × 5n, where m, n are non–negative integers. 3 8. i). ii). 13 125. iii). 7 80. iv). 14588 625. v). 129 2. 2 × 57. 28) Examine each of the following as rational or irrational i) 5 +. 5. ⎛ 1 ⎞ ⎟⎟ ii) ⎜⎜ 3 + 3⎠ ⎝. 2. ⎛ 1 ⎞ ⎟⎟ iii) ⎜⎜ 3 + 2⎠ ⎝. 2. iv). 15 6 5. 29) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion. i). 7 1250. ii). v). 24 729. vi). 15 24 13 25 5 3. iii). 32 455. iv). vii). 3 8. viii). 18 4000 131 2 5 4 75 3. 71 17 x) 630 1500 30) The following real numbers have decimal expansions as given below. In each case decide, p what can you say abut whether they are rational or not. If they are rational, and of the form q the prime factor of q?. ix). i) 0.0875. ii) 0.130130013000130000. iii) 0. 142857. 31) Prove that the square of any positive integer is of the form 4q or 4q + 1 for some integer q. 32) Prove that the product of three consecutive positive integers is divisible by 6. 33) If the HCF of 408 and 1032 is expressible in the form 1032m – 408 × 5, find m. 34) If the HCF of 657 and 963 is expressible in the form 657x + 963 × – 15, find x. 10. Universal Tutorials – X CBSE (2012–13) – Mathematics. Volume.

(17) Chapter 01: Real Numbers. 11. 35) Find the largest number which divides 280 and 1245 leaving remainders 4 and 3, respectively. 36) What is the largest number that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively. 37) 105 goats, 140 donkeys and 175 cows have to be taken across a river. There is only one boat which will have to make many trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. Can you tell how many animals went in each trip? 38) 15 pastries and 12 biscuit packets have been donated for a school fete. These are to be packed in several smaller identical boxes with the same number of pastries and biscuit packets in each. How many biscuit packets and how many pastries will each box contain? 39) A mason has to fit a bathroom with square marble tiles of the largest possible size. The size of the bathroom is 10 ft. by 8 ft. What would be the size in inches of the tile required that has to be cut and how many such tiles are required? 40) Two brands of chocolates are available in pack of 24 and 15 respectively. If I need to buy an equal number of chocolates of both kinds, what is the least number of boxes of each kind I would need to buy? 41) 144 cartons of Coke Cans and 90 cartons of Pepsi Cans are to be stacked in a Canteen. If each stack is of the same height and is to contain cartons of the same drink, what would be the greatest number of cartons each stack would have? 42) During a sale, colour pencils were being sold in pack of 24 each and crayons in packs of 32 each. If you want full pack of both and the same number of pencils and crayons, how many of each would you need to buy? 43) Find the greatest number of 6 digits exactly divisible by 24, 15 and 36. 44) A rectangular courtyard is 18 m 72 cm long and 13m 20 cm broad. It is to be paved with square tiles of the same size. Find the least possible number of such tiles. 45) What is the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case? 46) Prove that the following are irrational: i). 6. v) 2 +. ii). 3. 1. iii). 5. vi) 3 –. 5. iv). 11. vii) 7 + 3 2. 3 –. 2. [CBSE–09]. MULTIPLE CHOICE QUESTIONS: CW Exercise: 1) 21 .234 78 is a) an integer. b) a rational. c) an irrational. d) none of these.. 8 is a) rational b) irrational c) none of these. 3) Which one of the following is an irrational number? a) x2 = 9 b) y2 = 64 c) z2 = 8 d) none of these. p 4) A rational number is terminating decimals only when prime factors of q are only: q 2). a) 2 or 3. Volume. b) 3 or 5. c) 3 and 4. Universal Tutorials – X CBSE (2012–13) – Mathematics. d) 2 or 5.. 11.

(18) 12. 13 is 8 a) 0.175 b) 0.625 c) 1.625 Which rational number is represented by 11.125 89 98 93 a) b) c) 8 4 8 The LCM of two numbers 26 and 91 is: a) 91 b) 182 c) 1183 The sum of a rational number and irrational number is always a) a rational number b) an irrational number c) an integer 2 If p is an even integer then p is an a) odd integer b) even integer c) multiple of 3. 5) The decimal expansion of 6). 7) 8) 9). d) 1.525. d). 88 9. d) 637. d) none of these. d) none of these.. 10) 4 5 is. a) rational b) irrational c) not real d) none of these. 11) Every composite number can be expressed as a product of a) coprimes b) primes c) none of these. HW Exercise: 1) 0.101001000100001000001 is a) an irrational b) a rational c) an integer d) none of these. 2) The decimal representation of an irrational number is always a) terminating b) terminating, repeating c) non terminating, repeating d) non-terminating, non-repeating. 3) HCF (a, b) × LCM (a, b) = a) a + b b) a – b c) a × b d) none of these. 4) Every terminating decimal is a) an integer b) a rational c) an irrational d) none of these. 5) Which rational number is represented by 3.41 .. 307 338 341 34 b) c) d) 90 99 900 990 The HCF of two numbers 867 and 255 is a) 17 b) 34 c) 51 d) 68. The LCM of three numbers is 28, 44, 132 is a) 528 b) 231 c) 462 d) 924. Circumference of the circle π= is Diameter of the circle a) an irrational b) a rational c) none of these. The decimal expansion of a rational number is always: a) non-terminating b) non-terminating and non repeated c) terminating or non-terminating repeated d) none of these. a). 6) 7) 8) 9). 10) The given number: 3 + 2 2 is an a) rational number b) irrational number c) not real. 11 11) The rational number is a 24 a) terminating decimal b) non-terminating repeating 12. Universal Tutorials – X CBSE (2012–13) – Mathematics. c) none of these. Volume.

(19) Chapter 01: Real Numbers. 13. COLUMN MATCHING QUESTIONS: 1) Listed in column I are some statements. For each statement in column I. Applying Euclid’s division lemma choose all the correct options in column II. Column I. Column II. i). A) 9m or 9m+1 for some integer m. The square of any positive integer is of the form. ii) Any positive odd integer is of the form. B) 6m+1 or 6m+3 or 6m+5 for some m. iii) The cube of any positive integer is of the form. C) 5m, 5m+1, 5m+4 for some integer m D) 3m or 3m+1 for some m E) 4m+1 or 4m+3 for some m. 2) Given in column I are some types of real numbers. For each item in column I. Choose all the correct options in column II. Column I. Column II. i). Irrational Number. A). (5+ 3 ) (5– 3 ). ii). Integer. B). 43 .12345. iii). Rational Number. C) 43.1234567 D) (3– 5 ) (3+ 7 ) E). 18 2. 3) Given in column I are some real numbers. For each item in column I. Choose all correct options in column II. Column I Column II i) ii) iii). π. A) Terminating decimal representation B) Irrational. 135 3. 2 ×5 5. 3. C) Rational number D) Non–terminating and non–repeating decimal representation E) Non–terminating and repeating decimal. Volume. Universal Tutorials – X CBSE (2012–13) – Mathematics. 13.

(20) 14. 4) Given in col. I are values of missing entries named x & y.Choose all correct options in column II Column I Column II i) x = 7, y = 2 A) 156 x. 78 2. 39 y. ii). x = 2, y = 3. B). 13. 240 2. 120 2. 60 2. 30 y. 15 x. iii). x = 3, y = 2. C). 5. 210 x. 105 y. 35 5. D). 7. 462 3. 154 y. 77 x. E). 11. 372 x. 186 2. 93 y. 31. ANSWERS TO UNSOLVED EXERCISE: CW Exercise 1.1: 1) (i) 45 (ii) 196 (iii) 1 5) 34 6) 12 7) HCF=13, m=2, n –1 8) 8 columns HW Exercise 1.1: 1) (i) 51 ii) 17 3) 10 4) 75 7) y = –19 8) x = –68, y = 53 9) 64 10) 17 CW Exercise 1.2: 1) (i) 22 × 5 × 7 (ii) 32 × 52 × 17 (iii) 22 × 3 × 31 (iv) 24 × 34 × 7 (v) 3 × 2 × 7 × 11 14. Universal Tutorials – X CBSE (2012–13) – Mathematics. Volume.

(21) Chapter 01: Real Numbers. 15. 2) (i) 182, 13 (ii) 23460, 2 (iii) 3024, 6 (iv) 1008, 2 3) (i) 420, 3 (ii) 1729, 1 (iii) 900, 1 (iv) 300300, 1 4) 2n 6) 204 7) (i)192 (ii) 36 8) 21. 9) 4:00 pm. 10) 4:00 pm. HW Exercise 1.2: 1) (i) 22 × 3 × 13 (ii) 5 × 7 × 11 × 13 (iii) 17 × 19 × 23 (iv) 2 × 32 × 5 × 7 × 31 (v) 2 × 3 × 7 × 11 × 13 2) 3024, 6 3) (i) 11339, 1 (ii) 1800, 1 (iii) 360, 12 (iv) 4) 22338 6) 36 minutes 8) 2, 5, 7 CW Exercise 1.4: 1) (i, iii, iv, v, vi) Terminating (ii) Non–terminating 2) (i) 0.00416 (iii) 0.009375 (iv) 0.115 (v) 0.7 3) (i) Rational (ii) Non–rational (iii) Rational HW Exercise 1.4: 1) (i, iv, vi) Terminating (ii, iii, v) Non–terminating 2) (i) 2.125 (iv) 0.4 3) (i) Rational (ii) Non–rational (iii) Rational Miscellaneous: 1) (i) 19 (ii) 13 (iii) 36 2) (i) 55 (ii) 116 3) (i) 9 (ii) 17 (iii) 6 (iv) 45 6) 493 7) 31 8) 6 gm 17) 607 18) (i) Two numbers cannot have 14 as HCF and 204 as LCM (ii) Can never end with 0 19) After 20 minutes 20) After covering a distance of 12240 cm. from the staring point 21) 6 metres 22) 3300 m from first telegraph pole 23) 110040 24) 4663 25) 60 days 26) 36 28) i, iii, iv) Irrational (ii) Rational 29) i, ii, iv, vi, vii) Terminating iii, v, viii, ix, x) Non-terminating repeating 30) i) Rational, prime factors of q will be either 2 or 5 or both only (ii) Not rational (iii) Rational prime factors of q will also have a factor other than 2 or 5. 33) 2 34) 22 35) 138 36) 625 37) 35 38) 4 Biscuit packets, 5 Pastries 39) 24 inches, 20 tiles 40) 8 of second kind, 5 of 1st kind 41) 18 42) 4 packets of Pencils, 3 packets of Crayons 43) 999720 44) 4290 45) 3647 Column Matching Question: 1) i–CD; ii–BE; iii–A 3) i–BD; ii–AC; iii–BD. Volume. 2) i–D; ii–AE; iii–ABEC 4) i–D; ii–ACE; iii–B. Universal Tutorials – X CBSE (2012–13) – Mathematics. 15.

(22) 16. Chapter 02: Polynomials Chapter Map: → Introduction → Geometrical meaning of Zeroes of a Polynomial → Relationship between zeroes and coefficients of a Polynomial → Division Algorithm for Polynomials. Introduction: Degree of a Polynomial: z. If p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of the polynomial p(x).. Linear polynomial: z. A polynomial of degree 1 is called a linear polynomial.. z. For example, 2x – 3,. z. General form of linear polynomial is ax + b, where a, b are real numbers and a ≠ 0.. 3 x + 5, y +. 2,x–. 2 2 , 3z + 4, u + 1 etc. are linear polynomials. 11 3. Quadratic polynomial: z z z. z. A polynomial of degree 2 is called a quadratic polynomial. The name ‘quadratic’ has been derived from the word ‘quadrate’, which means ‘square’. 2 2 u 2 1 , y – 2, 2 – x2 + 3 x, − 2u2 + 5, 5 v2 – v, 4z2 + are some examples 5 3 3 7 of quadratic polynomials (whose coefficients are real numbers). Generally, any quadratic polynomial in x is of the form ax2 + bx + c, where a, b, c are real numbers and a ≠ 0.. 2x2 + 3x –. Cubic polynomial: z. A polynomial of degree 3 is called a cubic polynomial.. z. Some examples of a cubic polynomial are 2 – x3, x3,. z. In fact, the most general form of a cubic polynomial is ax3 + bx2 + cx + d, where, a, b, c, d are real numbers and a ≠ 0.. 3. 2. 3. 3. 2. 2 x , 3 – x + x , 3x – 2x + x – 1.. Value of a polynomial: z z z. 16. If p(x) is a polynomial in x, and if k is any real number, then the value obtained by replacing x by k in p(x), is called the value of p(x) at x = k, and is denoted by p(k). Consider the polynomial P(x) = x2 – 3x – 4 value of the polynomial at x = 3 is denoted by P(3) and P(3) = 32 – 3(3) – 4 = –4. Value of the above polynomial at x = –1 is given by P(–1) = (–1)2 – 3(–1) – 4 = 0.. Universal Tutorials – X CBSE (2012–13) – Mathematics. Volume.

(23) Chapter 02: Polynomials. 17. Zero of a polynomial: z z z z. z. A real number k is said to be a zero of a polynomial p(x), if p(k) = 0. If x – a is a factor of the polynomial p(x), then p(a) = 0 If x + a is a factor of the polynomial p(x), then p(–a) = 0 Consider the polynomial P(x) = x2 – 5x + 6 P(2) = 22 – 5(2) + 6 = 0 and P(3) = 32 – 5(3) + 6 = 0 As P(2) = 0 and P(3) = 0, 2 and 3 are called the zeros of the polynomial x2 – 5x + 6.. Geometrical Meaning of the Zeroes of a Polynomial: ¾ ¾ ¾ ¾. Now, let us look for the geometrical meaning of a zero of a quadratic polynomial. 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 to a few values for x as given in Table. –2 –1 0 1 2 3 4 5 x 2 6 0 –4 –6 –6 –4 0 6 y = x – 3x – 4 (–2,6). (5, 6). 6 5 4 3 2. (–1,0). Note from the figure that the curve intersect x–axis at the points –1 and 4. Thus the zeros of the Polynomial x2 – 3x – 4 are –1 and 4.. 1. –3 –2 –1 0 –1. (4,0) 1. 2. 3. 4. 5. –2 –3 –4. (0,–4). (3,–4). –5 –6 (1,–6). (2,–6). ¾ From our observation earlier about the shape of the graph of y = ax2 + bx + c, the following three cases can happen: ¾ Case (i) : Here, the graph cuts x-axis at two distinct points A and A′. ¾ The x-coordinates of A and A′ are the two zeroes of the quadratic polynomial ax2 + bx + c in this case (see given below). Y. A X′. 0. Volume. A′. A′ X. Y′ (i). Y. X′. A 0. X. Y′ (ii). Universal Tutorials – X CBSE (2012–13) – Mathematics. 17.

(24) 18 z. z. 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. given below). The x-coordinate of A is the only zero for the quadratic polynomial ax2 + bx + c in this case. Y. Y. A 0. X′. Y′. z. X. Y′ (ii). (i) z. A. 0. X′. X. Case (iii) : Here, the graph is either completely above the x-axis or completely below the xaxis. So, it does not cut the x-axis at any point (see Fig. given below). So, the quadratic polynomial ax2 + bx + c have no zero in this case. Y. Y. 0. X′. 0. X′. X. Y′. X. Y′ (ii). (i). Number of Zeros of a Polynomial: z z. 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 a degree n has at most n zeros. That is, a quadratic polynomial can have at most two zeros and so on.. SOLVED EXAMPLE 2.1: 1) Look at the graphs in Fig. 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). Y. X′. 0. X. Y′ (i). 18. Y. X′. 0. Y. X. X′. Y′ (ii). Universal Tutorials – X CBSE (2012–13) – Mathematics. 0. X. Y′ (ii). Volume.

(25) Chapter 02: Polynomials. 19 Y. Y. 0. X′. 0. X′. X. Y. Y′ (iv). 0. X′. X. Y′ (v). X. Y′ (vi). Sol: 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. iv) The number of zeroes is 1. v) The number of zeroes is 1. vi) The number of zeroes is 4.. UNSOLVED EXERCISE 2.1: 1) The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case. Y. X′. 0. Y. X. X′. Y′. 0. Y. 0. Y′ (iv). X. X′. Y′ (ii). (i). X′. Y. 0. Y′ (ii). Y. X. X′. 0. X. Y. X. X′. Y′ (v). 0. X. Y′ (vi). Relationship between Zeroes and Coefficients of a Polynomial: ¾ If α and β are the zeroes of the quadratic polynomial p(x) = ax2 + bx + c, a ≠ 0, then you know that x – α and x – β are the factors of p(x). ¾ Therefore, ax2 + bx + c = k(x – α) (x – β), where k is a constant. = k[x2 – (α + β)x + αβ] = kx2 – k(α + β)x + kαβ = k[x2 – (sum of zero)x + (Product of zero)] Volume. Universal Tutorials – X CBSE (2012–13) – Mathematics. 19.

(26) 20. ¾ Comparing the coefficients of x2, x and constant terms on both the sides, we get a = k, b = – k(α + β) and c = kαβ. b c − ¾ This gives α + β = a , αβ = a ¾ i.e., sum of zeroes = α + β = − ¾ Product of zeroes = αβ =. b −(Coefficien t of x) = , a Coefficien t of x 2. c Consta nt term = a Coefficien t of x 2. In general, it can be proved that if α, β, γ are the zeroes of the cubic polynomial b c d ax3 + bx2 + cx + d, then α + β + γ = − , αβ + βγ + γα = , α β γ = − a a a. SOLVED EXAMPLES 2.2: 1) Find the zeroes of the quadratic polynomial x2 + 7x + 10, and verify the relationship between the zeroes and the coefficients.. Sol: We have, x2 + 7x + 10 = (x + 2)(x + 5) So, the value of x2 + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0 i.e., when x = – 2. or. x = –5. 2. ∴ the zeroes of x + 7x + 10 are – 2 and – 5. −(7) −(Coefficien t of x) = 1 Coefficien t of x 2. Now, sum of zeroes = –2 + (–5) = – (7) = Product of zeroes = (–2) × (–5) = 10 =. 10 Consta nt term = 1 Coefficien t of x 2. 2) Find the zeroes of the polynomial x2 – 3 and verify the relationship between the zeroes and the coefficients.. Sol: Recall the identity a2 – b2 = (a – b)(a + b) Using it, we can write: x2 – 3 = (x –. 3 )(x +. So, the value of x2 – 3 is zero when x = Therefore, the zeroes of x2 – 3 are Now, sum of zeroes =. 3 –. 3). 3 or x = – 3. 3 and − 3. 3 =0=. −(Coefficien t of x) Coefficien t of x 2. Product of zeroes = ( 3 )(– 3 ) = –3 = −. 3 Consta nt term = 1 Coefficien t of x 2. 3) Find a quadratic polynomial, the sum and product of whose zeroes are –3 and 2, respectively.. Sol: Let the quadratic polynomial be ax2 + bx + c, and its zeroes be α and β. We have, α + β = – 3 = −. b a. and. αβ = 2 =. c . a. If a = 1, then b = 3 and c = 2. So, one quadratic polynomial which fits the given conditions is x2 + 3x + 2. 20. Universal Tutorials – X CBSE (2012–13) – Mathematics. Volume.

(27) Chapter 02: Polynomials. 21 2. 4) If α and β are zero of the Quadratic Polynomial f(x) = ax + bx + c, then evaluate (i) α + β2 α β (ii) + . β α. Sol: α and β are zero of ax2 + bx + c. ∴α+β=– 2. ⎛ b⎞ ⎛c ⎞ i) α2 + β2 = (α + β)2 – 2αβ = ⎜ − ⎟ – 2 ⎜ ⎟ a ⎝ ⎠ ⎝a⎠. ∴ α2 + β2 =. 2. b c , αβ = a a. b 2 − 2ac a2. b 2 − 2ac 2. ii). α α +β β + = β α αβ. 2. a2 c a. =. =. b 2 − 2ac ac. UNSOLVED EXERCISE 2.2: CW Exercise 1) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. i) x2 – 2x – 8 ii) 4s2 – 4s + 1 iii) 3x2 – x – 4 iv) 7z2 – 343 2 v) 5p + 20p 2) Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. 1 1 1 ii) – , iii) 4, 1 i) 2 , 3 4 4. 3) Find the zeros of the polynomial f(x) = 4 3 x 2 + 5 x − 2 3 , and verify the relationship between the zeros and its coefficients. 4) Find the zeros of the quadratic polynomial f(x) = abx2 + (b2 – ac) x – bc, and verify the relationship between the zeros and its coefficients. 5) If α and β are the zeros of the quadratic polynomial f(x) = x2 – px + q, then find the values of 1 1 ii) + i) α2 + β2 α β 6) If α and β are the zeros of the quadratic polynomial f(x) = ax2 + bx + c, then evaluate: i) α2 + β2. ii). β α + α β. iii). α2 β2 + β α. 7) If α and β are the zeros of the quadratic polynomial f(x) = ax2 + bx + c, then evaluate: i) α4 + β4. ii). α2 β. 2. +. β2 α2. 8) If α and β are the zeros of the quadratic polynomial f(x) = kx2 + 4x + 4 such that α2 + β2 = 24, find the value of k. 9) If α and β are the zero of the quadratic polynomial f(x) = x2 – x – 2, find a polynomial whose zero are 2α + 1 and 2β + 1. HW Exercise: 1) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. i) 6x2 – 3 – 7x [CBSE 08] ii) 4u2 + 8u iii) t2 – 15 Volume. Universal Tutorials – X CBSE (2012–13) – Mathematics. 21.

(28) 22. 2) Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. 1 ii) 0, 5 iii) 1, 1 i) , –1 4 1 1 3) If α and β are the zeros of the quadratic polynomial f(x) = x2 – x – 4, find the value of + − αβ . α β 4) If α & β are the zeros of the quadratic polynomial P(x) = 4x2 –5x – 1, find the value of α2β + αβ2. 5) If α & β are the zeros of the polynomial f(x) = x2 – 5x + k such that α – β = 1, find the value of k. 21 6) If α, β are the zeros of the polynomial f(x) = 2x2 + 5x +k satisfying the relation α2 + β2 + αβ = , 4 then find the value of k for this to be possible. 7) If sum of the square of the zeros of the quadratic polynomial f(x) = x2 – 8x + k is 40, find the value of k. 8) If α and β are the zeros of the quadratic polynomial f(x) = 2x2 – 5x + 7, find a polynomial whose zeros are 2α + 3β and 3α + 2β.. Division Algorithm for Polynomials: If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that p(x) = g(x) × q(x) + r(x), where r(x) = 0 or degree of r(x) < degree of g(x). This result is known as the Division Algorithm for polynomials. i.e. Dividend = Divisor × Quotient + Reminder. SOLVED EXAMPLES 2.3: 3. 1) Divide 3x + x + 2x + 5 by 1 + 2x + x2. Sol: We first arrange the terms of the dividend and the divisor in the decreasing order of their degrees. Recall that arranging the terms in this order is called writing the polynomials in standard form. In this example, the dividend is already in standard form, and the divisor, in standard form, is x2 + 2x + 1. Step 1: To obtain the first term of the quotient, divide the highest degree term of the dividend (i.e. 3x3) by the highest degree term of the divisor (i.e. x2). This is 3x. Then carry out the division process. What remains is – 5x2 – x + 5. Step 2: Now, to obtain the second term of the quotient, divide the highest degree term of the new dividend (i.e. –5x2) by the highest degree term of the divisor (i.e. x2). This gives –5. Again carry out the division process with –5x2 – x + 5. Step 3: What remains is 9x + 10. Now, the degree of 9x + 10 is less than the degree of the divisor x2 + 2x + 1. So, we cannot continue the division any further. So, the quotient is 3x – 5 and the remainder is 9x + 10. Also, (x2 + 2x + 1) × (3x – 5) + (9x + 10) = 3x3 + 6x2 + 3x – 5x2 – 10x – 5 + 9x + 10 = 3x3 + x2 + 2x + 5 Here again, we see that, Dividend = Divisor × Quotient + Remainder 2) Divide 3x2 – x3 – 3x + 5 by x – 1 – x2, and verify the division algorithm. Sol: Note that the given polynomials are not in standard form. To carry out division, we first write both the dividend and divisor in decreasing orders of their degrees. So, dividend = –x3 + 3x2 – 3x + 5 and divisor = –x2 + x – 1. Division process is shown on the right side. We stop here since degree (3) = 0 < 2 = degree (–x2 + x – 1). So, quotient = x – 2, remainder = 3. 22. 2. Universal Tutorials – X CBSE (2012–13) – Mathematics. Volume.

(29) Chapter 02: Polynomials. 23 2. Now, Divisor × Quotient + Remainder = (–x + x – 1) (x – 2) + 3 = –x3 + x2 – x + 2x2 – 2x + 2 + 3 = –x3 + 3x2 – 3x + 5 = Dividend In this way, the division algorithm is verified. 3) Find all the zeroes of 2x4 – 3x3 – 3x2 + 6x – 2, if you know that two of its zeroes are. 2. [CBSE–08]. and – 2 . 2. Sol: Since two zeroes are 2 and − 2 , (x – 2 )(x + 2 ) = x – 2 is a factor of the given polynomial. Now, we divide the given polynomial by x2 – 2. First term of quotient is. 2x 4 = 2x2 x2. Second term of quotient is. − 3x 3 = –3x x2. x2 =1 x2 So, 2x4 – 3x3 – 3x2 + 6x – 2 = (x2 – 2)(2x2 – 3x + 1). Now, by splitting –3x, we factorise 2x2 – 3x + 1 as (2x – 1)(x – 1).. Third term of quotient is. So, its zeroes are given by x =. 1 2. and. x = 1.. Therefore, the zeroes of the given polynomial are. 2,– 2,. 1 and 1. 2. UNSOLVED EXERCISE 2.3: CW Exercise: 1) Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following: ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2 2) Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial: ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2 i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12 4 3 3) The graph of a polynomial f(x) = 3x + 6x – 2x2 – 10x – 5, intersects x axis at four different ⎛ 5 ⎞ ⎛ 5 ⎞⎟ , 0 ⎟ and ⎜ − ,0 points P, Q, R and S. If the co-ordinates of points P and Q are ⎜ ⎜ 3 ⎟ ⎜ 3 ⎟⎠ ⎝ ⎠ ⎝ respectively then find the co-ordinates of R and S. 4) Polynomial f(x) = x3 – 2x – 4 on dividing by another polynomial g(x) gives equal quotient and remainder. Find out the value of g(x) if the remainder of the division is x-2. 5) Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and i) deg p(x) = deg q(x) ii) deg q(x) = deg r(x) iii) deg r(x) = 0 6) Apply the division algorithm to find the quotient remainder on dividing f(x) by g(x) as given below. ii) f(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x i) f(x) = x3 – 6x2 + 11x – 6, g(x) = x + 2 7) By applying division algorithm prove that the polynomial g(x) = x2 + 3x + 1 is a factor of the polynomial f(x) = 3x4 + 5x3 – 7x2 + 2x + 2. 8) What must be subtracted from 8x4 + 14x3 – 2x2 + 7x – 8 so that the resulting polynomial is exactly divisible by 4x2 + 3x – 2. 9) Find the values of a and b so that x4 + x3 + 8x2 + ax + b is divisible by x2 + 1. Volume. Universal Tutorials – X CBSE (2012–13) – Mathematics. 23.

(30) 24. 10) A polynomial f(x) = x4 – 3x3 + 6x – 4 is factorized into three polynomials such that f(x) = p(x).q(x).r(x). If p(x) = x2 – 3x + 2 and q(x) = x – 2 , then find r(x). 11) For which value of a, (x + a) is a factor of 2x2 + 2ax + 5x + 10? HW Exercise: 1) Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in p(x) = x4 – 5x + 6, g(x) = 2 – x2. 2) Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1. [CBSE–08] 3) Obtain all other zeroes of x4 + x3 – 34x2 – 4x + 120, if two of its zeroes 2 and – 2. 4) On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. Find g(x). 5) If (x – 2) is a factor of polynomial x3 + ax2 + bx + 16 and a – b = 6 then find the value of a and b. 6) Apply the division algorithm to find the quotient remainder on dividing f(x) by g(x) as given below. ii) f(x) = x4 – 5x + 6, g(x) = 2 – x2 i) f(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2 4 3 2 7) What must be added to f(x) = 4x + 2x – 2x + x – 1 so that the resulting polynomial is divisible by g(x) = x2 + 2x – 3. 8) If the polynomial f(x) = x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a.. MISCELLANEOUS EXERCISE: 1) Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case: 1 ii) x3 – 4x2 + 5x – 2; 2, 1, 1 i) 2x3 + x2 – 5x + 2; , 1, – 2 2 2) Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively. 3) If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b, find a and b. 4) If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ± 3 find other zeroes. 5) If the polynomial 6x4 + 8x3 + 17x2 + 21x + 7 is divided by another polynomial 3x2 + 4x + 1, the remainder comes out to be (ax + b), find a and b. [CBSE 09] 6) Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients: i) f(x) = x2 – 2x – 8 ii) g(s) = 4s2 – 4s + 1 iii) h(t) = t2 – 15. iv) p(x) = x2 + 2 2 x – 6. v) q(x) = 3 x2 + 10x + 7 3 vi) f(x) = x2 – ( 3 + 1)x + 3 vii) g(x) = a(x2 + 1) – x(a2 + 1) 7) If α and β are the zeros of the quadratic polynomials f(x) = ax2 + bx + c, then evaluate: i) α – β. ii). 1 1 − α β. iii). 1 1 + − 2αβ α β. iv) α2β + αβ2. 8) If α and β are the zeros of the quadratic polynomial f(x) = 6x2 + x – 2, find the value of 9) If α and β are the zeros of the quadratic polynomial f(x) = x2 + x – 2, find the value of. α β + . β α. 1 1 − . α β. 10) If α and β are the zeros of the quadratic polynomial f(x) = x2 – 5x + 4, find the value of 1 1 + – 2αβ. α β 24. Universal Tutorials – X CBSE (2012–13) – Mathematics. Volume.

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