Arihant Maths Ncert Exemplar PDF

 Neha Tyagi Amit Rastogi ARIHANT PRAKASHAN (School Division Series) ARIHANT PRAKASHAN (School Division Series) All Rights Reserved © PUBLISHERS No part of this publication may be re-produced, stored in a retrieval system or distributed in any form or by any means, ele...

 Neha Tyagi Amit Rastogi ARIHANT PRAKASHAN (School Division Series) ARIHANT PRAKASHAN (School Division Series) All Rights Reserved © PUBLISHERS No part of this publication may be re-produced, stored in a retrieval system or distributed in any form or by any means, electronic, mechanical, photocopying, recording, scanning, web or otherwise without the written permission of the publisher. Arihant has obtained all the information in this book from the sources believed to be reliable and true. However, Arihant or its editors or authors or illustrators don’t take any responsibility for the absolute accuracy of any information published and the damages or loss suffered there upon. All disputes subject to Meerut (UP) jurisdiction only. ADMINISTRATIVE & PRODUCTION OFFICES Regd. Office ‘Ramchhaya’ 4577/15, Agarwal Road, Darya Ganj, New Delhi -110002 Tele: 011- 47630600, 43518550; Fax: 011- 23280316 Head Office Kalindi, TP Nagar, Meerut (UP) - 250002 Tele: 0121-2401479, 2512970, 4004199; Fax: 0121-2401648 SALES & SUPPORT OFFICES Agra, Ahmedabad, Bengaluru, Bareilly, Chennai, Delhi, Guwahati, Hyderabad, Jaipur, Jhansi, Kolkata, Lucknow, Meerut, Nagpur & Pune ISBN : 978-93-5176-264-5 Published by Arihant Publications (India) Ltd. For further information about the books published by Arihant log on to www.arihantbooks.com or email to [email protected] /arihantpub /@arihantpub Arihant Publications /arihantpub PREFACE The Department of Education in Science & Mathematics (DESM) & National Council of Educational Research & Training (NCERT) developed Exemplar Problems in Science and Mathematics for Secondary and Senior Secondary Classes with the objective to provide the students a large number of quality problems in various forms and format viz. Multiple Choice Questions, Short Answer Questions, Long Answer Questions etc., with varying levels of difficulty. NCERT Exemplar Problems are very important for both; School & Board Examinations as well as competitive examinations like NTSE, Olympiad etc. The questions given in exemplar book are mainly of higher difficulty order by practicing these problems, you will able to manage with the margin between a good score and a very good or an excellent score. Approx 20% problems asked in any Board Examination or Competitive Examinations are of higher difficulty order, exemplar problems will make you ready to solve these difficult problems. This book NCERT Exemplar Problems-Solutions Mathematics X contains Explanatory & Accurate Solutions to all the questions given in NCERT Exemplar Mathematics book. For the overall benefit of the students’ we have made unique this book in such a way that it presents not only hints and solutions but also detailed and authentic explanations. Through these detailed explanations, students can learn the concepts which will enhance their thinking and learning abilities. We have introduced some additional features with the solutions which are as follows — Thinking Process Along with the solutions to questions we have given thinking process that tell how to approach to solve a problem. Here, we have tried to cover all the loopholes which may lead to confusion. All formulae and hints are discussed in detail. — Note We have provided notes also to solutions in which special points are mentioned which are of great value for the students. For the completion of this book, we would like to thank Priyanshi Garg who helped us at project management level. With the hope that this book will be of great help to the students, we wish great success to our readers. Authors CONTENTS 1. Real Numbers 1-16 2. Polynomials 17-36 3. Pair of Linear Equations in Two Variables 37-75 4. Quadratic Equations 76-97 5. Arithmetic Progressions 98-132 6. Triangles 133-162 7. Coordinate Geometry 163-203 8. Introduction to Trigonometry and Its Applications 204-228 9. Circles 229-253 10. Constructions 254-268 11. Area Related to Circles 269-295 12. Surface Areas and Volumes 296-326 13. Statistics and Probability 327-372  1 Real Numbers Exercise 1.1 Multiple Choice Questions (MCQs) Q. 1 For some integer m, every even integer is of the form (a) m (b) m + 1 (c) 2m (d) 2m + 1 Sol. (c) We know that, even integers are 2, 4, 6,… So, it can be written in the form of 2m. where, m = Integer = Z [since, integer is represented by Z] or m = L, − 1, 0, 1, 2, 3,… ∴ 2 m = L, − 2, 0, 2, 4, 6… Alternate Method Let ‘a’ be a positive integer. On dividing ‘a’ by 2, let m be the quotient and r be the remainder. Then, by Euclid’s division algorithm, we have a = 2 m + r, where a ≤ r < 2 i.e., r = 0 and r = 1. ⇒ a = 2 m or a = 2 m + 1 when, a = 2 m for some integer m, then clearly a is even. Q. 2 For some integer q, every odd integer is of the form (a) q (b) q + 1 (c) 2q (d) 2q + 1 Sol. (d) We know that, odd integers are 1, 3, 5, L So, it can be written in the form of 2q + 1. where, q = integer = Z or q = L, − 1, 0, 1, 2, 3, … ∴ 2q + 1 = L, − 3, − 1, 1, 3, 5, … Alternate Method Let ‘a’ be given positive integer. On dividing ‘a’ by 2, let q be the quotient and r be the remainder. Then, by Euclid’s division algorithm, we have a = 2q + r , where 0 ≤ r < 2 ⇒ a = 2q + r, where r = 0 or r = 1 ⇒ a = 2q or 2q + 1 when a = 2q + 1for some integer q, then clearly a is odd. 2 NCERT Exemplar (Class X) Solutions Q. 3 n2 − 1 is divisible by 8, if n is (a) an integer (b) a natural number (c) an odd integer (d) an even integer Sol. (c) Let a = n2 − 1 Here n can be ever or odd. Case I n = Even i.e., n = 2 k, where k is an integer. ⇒ a = (2 k )2 − 1 ⇒ a = 4k 2 − 1 At k = −1, = 4 (−1) − 1 = 4 − 1 = 3, which is not divisible by 8. 2 At k = 0, a = 4 (0)2 − 1 = 0 − 1 = −1, which is not divisible by 8, which is not Case II n = Odd i.e., n = 2 k + 1, where k is an odd integer. ⇒ a = 2k + 1 ⇒ a = (2 k + 1)2 − 1 ⇒ a = 4K 2 + 4k + 1 − 1 ⇒ a = 4k 2 + 4k ⇒ a = 4k(k + 1) At k = −1, a = 4 (−1)(−1 + 1) = 0 which is divisible by 8. At k = 0, a = 4 (0)(0 + 1) = 4 which is divisible by 8. At k = 1, a = 4 (1)(1 + 1) = 8 which is divisible by 8. Hence, we can conclude from above two cases, if n is odd, then n2 − 1 is divisible by 8. Q. 4 If the HCF of 65 and 117 is expressible in the form 65m − 117, then the value of m is (a) 4 (b) 2 (c) 1 (d) 3 K Thinking Process Apply Euclid’s division algorithm until the remainder is 0. Finally we get divisor, which is the required HCF of 65 and 117. Now, put 65 m − 117 = HCF (65, 117) and get the value of m. Sol. (b) By Euclid’s division algorithm, b = aq + r, 0 ≤ r < a [Qdividend = divisor × quotient + remainder] ⇒ 117 = 65 × 1 + 52 ⇒ 65 = 52 × 1 + 13 ⇒ 52 = 13 × 4 + 0 ∴ H CF (65, 117 ) = 13 …(i) Also, given that, HCF (65, 117 ) = 65 m − 117 …(ii) From Eqs. (i) and (ii), 65 m − 117 = 13 ⇒ 65 m = 130 ⇒ m=2 Real Numbers 3 Q. 5 The largest number which divides 70 and 125, leaving remainders 5 and 8 respectively, is (a) 13 (b) 65 (c) 875 (d) 1750 K Thinking Process First, we subtract the remainders 5 and 8 from corresponding numbers respectively and then get HCF of resulting numbers by using Euclid’s division algorithm, which is the required largest number. Sol. (a) Since, 5 and 8 are the remainders of 70 and 125, respectively. Thus, after subtracting these remainders from the numbers, we have the numbers 65 = ( 70 − 5), 117 = (125 − 8), which is divisible by the required number. Now, required number = HCF of 65, 117 [for the largest number] For this, 117 = 65 × 1 + 52 [Q dividend = divisor × quotient + remainder] ⇒ 65 = 52 × 1 + 13 ⇒ 52 = 13 × 4 + 0 ∴ HCF = 13 Hence, 13 is the largest number which divides 70 and 125, leaving remainders 5 and 8. Q. 6 If two positive integers a and b are written as a = x 3 y 2 and b = xy 3 , where x, y are prime numbers, then HCF (a, b) is (a) xy (b) xy 2 (c) x3y3 (d) x2y 2 Sol. (b) Given that, a = x 3 y2 = x × x × x × y × y and b = xy3 = x × y × y × y ∴ HCF of a and b = HCF (x 3 y2 , xy3 ) = x × y × y = xy2 [since, HCF is the product of the smallest power of each common prime facter involved in the numbers] Q. 7 If two positive integers p and q can be expressed as p = ab 2 and q = a 3b ; where a, b being prime numbers, then LCM (p, q) is equal to (a) ab (b) a2b 2 (c) a3 b 2 (d) a3 b3 Sol. (c) Given that, p = ab 2 = a × b × b and q = a3 b = a × a × a × b ∴ LCM of p and q = LCM (ab 2 , a3 b ) = a × b × b × a × a = a3 b 2 [since, LCM is the product of the greatest power of each prime factor involved in the numbers] Q. 8 The product of a non-zero rational and an irrational number is (a) always irrational (b) always rational (c) rational or irrational (d) one Sol. (a) Product of a non-zero rational and an irrational number is always irrational i.e., 3 3 2 × 2 = (irrational). 4 4 4 NCERT Exemplar (Class X) Solutions Q. 9 The least number that is divisible by all the numbers from 1 to 10 (both inclusive) (a) 10 (b) 100 (c) 504 (d) 2520 Sol. (d) Factors of 1 to 10 numbers 1= 1 2 = 1× 2 3 = 1× 3 4 = 1× 2 × 2 5 = 1× 5 6 = 1× 2 × 3 7 = 1× 7 8 = 1× 2 × 2 × 2 9 = 1× 3 × 3 10 = 1 × 2 × 5 ∴ LCM of number 1 to 10 = LCM (1, 2, 3, 4, 5, 6, 7, 8, 9, 10) = 1 × 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520 14587 Q. 10 The decimal expansion of the rational number will terminate after 1250 (a) one decimal place (b) two decimal places (c) three decimal places (d) four decimal places K Thinking Process In terminating rational number the denominator always have the form 2 m × 5 n. 14587 14587 Sol. (d) Rational number = = 1250 21 × 54 2 1250 14587 (2 )3 5 625 = × 5 125 10 × 5 3 (2 ) 3 14587 × 8 5 25 = 5 5 10 × 1000 116696 1 = = 11.6696 10000 Hence, given rational number will terminate after four decimal places. Exercise 1.2 Very Short Answer Type Questions Q. 1 Write whether every positive integer can be of the form 4q + 2, where q is an integer. Justify your answer. Sol. No, by Euclid’s Lemma, b = aq + r, 0 ≤ r < a [Qdividend = divisor × quotient + remainder] Here, b is any positive integer a = 4, b = 4q + r for 0 ≤ r < 4 i.e., r = 0, 1, 2, 3 So, this must be in the form 4q , 4q + 1, 4q + 2 or 4q + 3. Real Numbers 5 Q. 2 ‘The product of two consecutive positive integers is divisible by 2’. Is this statement true or false? Give reasons. K Thinking Process The product of two consecutive numbers i.e., n (n + 1) will always be even, as one out of n or (n + 1) must be even. Sol. Yes, two consecutive integers can be n, (n + 1.) So, one number of these two must be divisible by 2. Hence, product of numbers is divisible by 2. Q. 3 ‘The product of three consecutive positive integers is divisible by 6’. Is this statement true or false? Justify your answer. Sol. Yes, three consecutive integers can be n, (n + 1) and (n + 2 ). So, one number of these three must be divisible by 2 and another one must be divisible by 3. Hence, product of numbers is divisible by 6. Q. 4 Write whether the square of any positive integer can be of the form 3m + 2, where m is a natural number. Justify your answer. Sol. No, by Euclid’s lemma, b = aq + r,0 ≤ r ≤ a Here, b is any positive integer, a = 3, b = 3q + r for 0 ≤ r ≤ 2 So, any positive integer is of the form 3 k, 3 k + 1 or 3 k + 2. Now, (3 k )2 = 9 k 2 = 3 m [where, m = 3 k 2 ] and (3 k + 1)2 = 9 k 2 + 6 k + 1 = 3 (3 k 2 + 2 k ) + 1 = 3 m + 1 [where, m = 3 k 2 + 2 k] Also , (3 k + 2 ) = 9 k + 12 k + 4 [Q (a + b ) = a + 2 ab + b 2 ] 2 2 2 2 = 9 k 2 + 12 k + 3 + 1 = 3 (3 k 2 + 4 k + 1) + 1 = 3m + 1 [where, m = 3 k 2 + 4 k + 1] which is in the form of 3m and 3m+1. Hence, square of any positive number cannot be of the form 3m+2. Q. 5 A positive integer is of the form 3q + 1, q being a natural number. Can you write its square in any form other than 3m + 1, i.e., 3m or 3m + 2 for some integer m? Justify your answer. Sol. No, by Euclid’s Lemma, b = aq + r, 0 ≤ r < a Here, b is any positive integer a = 3, b = 3 q + r for 0 ≤ r < 3 So, this must be in the form 3 q , 3 q + 1 or 3 q + 2. Now, (3 q )2 = 9 q 2 = 3 m [here, m = 3 q 2 ] and (3 q + 1)2 = 9 q 2 + 6 q + 1 = 3 (3 q 2 + 2 q ) + 1 = 3 m + 1 [where, m = 3 q 2 + 2 q ] Also, (3 q + 2 ) = 9 q + 12 q + 4 2 2 = 9 q 2 + 12 q + 3 + 1 = 3(3 q 2 + 4 q + 1) + 1 = 3m+ 1 [here, m = 3 q 2 + 4 q + 1] Hence, square of a positive integer is of the form 3q + 1 is always in the form 3 m + 1 for some integer m. 6 NCERT Exemplar (Class X) Solutions Q. 6 The numbers 525 and 3000 are both divisible only by 3, 5, 15, 25 and 75. What is HCF (525, 3000)? Justify your answer. Sol. Since, the HCF (525, 3000) = 75 By Euclid’s Lemma, 3000 = 525 × 5 + 375 [Qdividend = divisor × quotient + remainder] 525 = 375 × 1 + 150 375 = 150 × 2 + 75 150 = 75 × 2 + 0 and the numbers 3, 5, 15, 25 and 75 divides the numbers 525 and 3000 that mean these terms are common in both 525 and 3000. So, the highest common factor among these is 75. Q. 7 Explain why 3 × 5 × 7 + 7 is a composite number. K Thinking Process A number which has more than two factors is known as a composite number. Sol. We have, 3 × 5 × 7 + 7 = 105 + 7 = 112 Now, 112 = 2 × 2 × 2 × 2 × 7 = 2 4 × 7 So, it is the product of prime factors 2 and 7. i.e., it has more than two factors. Hence, it is a composite number. Q. 8 Can two numbers have 18 as their HCF and 380 as their LCM? Give reasons. Sol. No, because HCF is always a factor of LCM but here 18 is not a factor of 380. 987 Q. 9 Without actually performing the long division, find ifwill have 10500 terminating or non-terminating (repeating) decimal expansion. Give reasons for your answer. Sol. Yes, after simplification denominator has factor 53 ⋅ 2 2 and which is of the type 2 m ⋅ 5n. So, this is terminating decimal. 987 47 47 2 Q = = × 10500 500 53 ⋅ 2 2 2 94 94 94 = 3 = = = 0.094 5 × 2 3 (10)3 1000 Q. 10 A rational number in its decimal expansion is 327.7081. What can you say about the prime factors of q, when this number is expressed in the p form ? Give reasons. q Sol. 327.7081 is terminating decimal number. So, it represents a rational number and also its denominator must have the form 2 m × 5n. 3277081 p Thus, 327.7081 = = (say) 10000 q ∴ q = 104 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5 = 2 4 × 54 = (2 × 5)4 Hence, the prime factors of q is 2 and 5. Real Numbers 7 Exercise 1.3 Short Answer Type Questions Q. 1 Show that the square of any positive integer is either of the form 4q or 4q + 1 for some integer q. K Thinking Process Use Euclid’s division algorithm, put the value of divisor as 4 and then put the value of remainder from 0 to 3 and get the different form. Now, squaring every different form and get the required form. Sol. Let a be an arbitrary positive integer. Then, by, Euclid’s division algorithm, corresponding to the positive integers a and 4, there exist non-negative integers m and r, such that a = 4m + r, where 0 ≤ r < 4 ⇒ a2 = 16m2 + r 2 + 8mr …(i) where, 0≤ r < 4 [Q(a + b )2 = a2 + 2 ab + b 2 ] Case I When r = 0, then putting r = 0 in Eq. (i), we get a2 = 16 m2 = 4 (4 m2 ) = 4q where, q = 4 m2 is an integer. Case II When r = 1, then putting r = 1in Eq. (i), we get a2 = 16 m2 + 1 + 8m = 4 (4 m2 + 2 m) + 1 = 4q + 1 where, q = (4 m2 + 2 m) is an integer. Case III When r = 2, then putting r = 2 in Eq. (i), we get a2 = 16m2 + 4 + 16 m = 4 (4 m2 + 4 m + 1) = 4q where, q = (4m2 + 4m + 1) is an integer. Case IV When r = 3, then putting r = 3 in Eq. (i), we get a2 = 16m2 + 9 + 24m = 16m2 + 24m + 8 + 1 = 4 (4m2 + 6m + 2 ) + 1 = 4q + 1 where, q = (4m2 + 6m + 2 ) is an integer. Hence, the square of any positive integer is either of the form 4 q or 4 q + 1 for some integer q. Q. 2 Show that cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3, for some integer. K Thinking Process Use Euclid’s division algorithm. Put the value of remainder from 0 to 3, get the different forms of any positive integer. Now cubing to every different form of positive integer and get the required forms. Sol. Let a be an arbitrary positive integer. Then, by Euclid’s division algorithm, corresponding to the positive integers a and 4, there exist non-negative integers q and r such that a = 4q + r, where 0 ≤ r < 4 ⇒ a3 = (4 q + r )3 = 64q 3 + r 3 + 12 qr 2 + 48q 2 r [Q(a + b )3 = a3 + b 3 + 3ab 2 + 3a2 b ] ⇒ a = (64q + 48q r + 12 qr ) + r 3 2 2 2 3 …(i) where, 0≤ r < 4 8 NCERT Exemplar (Class X) Solutions Case I When r = 0, Putting r = 0 in Eq. (i), we get a3 = 64 q 3 = 4(16 q 3 ) ⇒ a3 + 4 m where m = 16q 3 is an integer. Case II When r = 1, then putting r = 1in Eq. (i), we get a3 = 64 q 3 + 48 q 2 + 12 q + 1 = 4 (16 q 3 + 12 q 2 + 3 q ) + 1 = 4m+ 1 where, m = (16 q 2 + 12 q 2 + 3 q ) is an integer. Case III When r = 2, then putting r = 2 in Eq. (i), we get a3 = 64 q 3 + 144 q 2 + 108 q + 27 = 64 q 3 + 144 q 2 + 108 q + 24 + 3 = 4 (16 q 3 + 36 q 2 + 27 q + 6) + 3 = 4m + 3 where, m = (16 q + 36 q 2 + 27 q + 6) is an integer. 3 Hence, the cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m. Q. 3 Show that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q. Sol. Let a be an arbitrary positive integer. Then, by Euclid’s divisions Algorithm, corresponding to the positive integers a and 5, there exist non-negative integers m and r such that a = 5m + r, where 0 ≤ r < 5 ⇒ a2 = (5m + r )2 = 25m2 + r 2 + 10mr [Q(a + b )2 = a2 + 2 ab + b 2 ] ⇒ a2 = 5(5m2 + 2 mr ) + r 2 …(i) where, 0≤ r < 5 Case I When r = 0, then putting r = 0 in Eq. (i), we get a2 = 5(5m2 ) = 5q where, q = 5 m2 is an integer. Case II When r = 1, then putting r = 1is Eq. (i), we get a2 = 5(5m2 + 2 m) + 1 ⇒ q = 5q + 1 where, q = (5m2 + 2 m) is an integer. Case III When r = 2, then putting r = 2 in Eq. (i), we get a2 = 5 (5m2 + 4m) + 4 = 5q + 4 where, q = (5m2 + 4m) is an integer. Case IV When r = 3, then putting r = 3 in Eq. (i), we get a2 = 5(5m2 + 6m) + 9 = 5 (5m2 + 6m) + 5 + 4 = 5(5m2 + 6m + 1) + 4 = 5q + 4 where, q = (5m2 + 6m +1) is an integer. Real Numbers 9 Case V When r = 4, then putting r = 4 in Eq. (i), we get a2 = 5 (5 m2 + 8m) + 16 = 5 (5m2 + 8m) + 15 + 1 ⇒ a2 = 5 (5m2 + 8m + 3) + 1 = 5q + 1 where, q = (5m2 + 8m + 3) is an integer. Hence, the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q. Q. 4 Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m. Sol. Let a be an arbitrary positive integer, then by Euclid’s division algorithm, corresponding to the positive integers a and 6, there exist non-negative integers q and r such that a = 6q + r, where 0 ≤ r < 6 ⇒ a2 = (6q + r )2 = 36q 2 + r 2 + 12 qr [Q(a + b )2 = a2 + 2 ab + b 2 ] ⇒ a2 = 6(6q 2 + 2qr ) + r 2 …(i) where, 0≤ r < 6 Case I When r = 0, then putting r = 0 in Eq. (i), we get a2 = 6 (6q 2 ) = 6m where, m = 6q 2 is an integer. Case II When r = 1, then putting r = 1in Eq. (i), we get a2 = 6 (6q 2 + 2q ) + 1 = 6m + 1 where, m = (6q 2 + 2q ) is an integer. Case III When r = 2, then putting r = 2 in Eq (i), we get a2 = 6(6q 2 + 4q ) + 4 = 6m + 4 where, m = (6q 2 + 4q ) is an integer. Case IV When r = 3, then putting r = 3 in Eq. (i), we get a2 = 6(6q 2 + 6q ) + 9 = 6 (6q 2 + 6 a) + 6 + 3 ⇒ a = 6(6q 2 + 6q + 1) + 3 = 6m + 3 2 where, m = (6q + 6q + 1) is an integer. Case V When r = 4, then putting r = 4 in Eq. (i), we get a2 = 6 (6q 2 + 8q ) + 16 = 6 (6q 2 + 8q ) + 12 + 4 ⇒ a2 = 6(6q 2 + 8q + 2 ) + 4 = 6m + 4 where, m = (6q 2 + 8q + 2 ) is an integer. Case VI When r = 5, then putting r = 5 in Eq. (i), we get a2 = 6 (6q 2 + 10q ) + 25 = 6 (6q 2 + 10q ) + 24 + 1 ⇒ a = 6(6q 2 + 10q + 4) + 1 = 6m + 1 2 where, m = (6q 2 + 10q + 1) is an integer. Hence, the square of any positive integer cannot be of the form 6 m + 2 or 6 m + 5 for any integer m. 10 NCERT Exemplar (Class X) Solutions Q. 5 Show that the square of any odd integer is of the form 4 m + 1, for some integer m. Sol. By Euclid’s division algorithm, we have a = bq + r, where 0 ≤ r < 4 …(i) On putting b = 4 in Eq. (i), we get a = 4q + r, where 0 ≤ r < 4 i.e., r = 0, 1, 2, 3...(ii) If r = 0 ⇒ a = 4q , 4q is divisible by 2 ⇒ 4q is even. If r = 1⇒ a = 4q + 1, (4q + 1) is not divisible by 2. If r = 2 ⇒ a = 4q + 2, 2 (2 q + 1) is divisible by 2 ⇒2 (2 q + 1) is even. If r = 3 ⇒ a = 4q + 3, (4q + 3) is not divisible by 2. So, for any positive integer q, (4q + 1) and (4q + 3) are odd integers. Now, a2 = (4q + 1)2 = 16q 2 + 1 + 8q = 4 (4q 2 + 2 q ) + 1 [Q (a + b )2 = a2 + 2 ab + b 2 ] is a square which is of the form 4m + 1, where m = (4q + 2q ) is an integer. 2 and a2 = (4q + 3)2 = 16q 2 + 9 + 24q = 4(4q 2 + 6q + 2 ) + 1 is a square [Q (a + b )2 = a2 + 2 ab + b 2 ] which is of the form 4m + 1, where m = (4q + 6q + 2 ) is an integer. 2 Hence, for some integer m, the square of any odd integer is of the form 4m + 1. Q. 6 If n is an odd integer, then show that n2 − 1 is divisible by 8. Sol. Let a = n −1 2 …(i) Given that, n is an odd integer. ∴ n = 1, 3, 5,... From Eq. (i), at n = 1, a = (1)2 − 1 = 1 − 1 = 0, which is divisible by 8. From Eq. (i), at n = 3, a = (3)2 − 1 = 9 − 1 = 8, which is divisible by 8. From Eq. (i), at n = 5, a = (5)2 − 1 = 25 − 1 = 24 = 3 × 8, which is divisible by 8. From Eq. (i), at n = 7, a = (7 )2 − 1 = 49 − 1 = 48 = 6 × 8, which is divisible by 8. Hence, (n2 − 1) is divisible by 8, where n is an odd integer. Alternate Method We know that an odd integer n is of the from (4q + 1) or (4q + 3) for some integer q. Case I When n = 4q + 1 In this case, we have (n2 − 1) = (4q + 1)2 − 1 = 16q 2 + 1 + 8q − 1 (a + b ) = a + 2 ab + b ] 2 2 2 = 16q 2 + 8q = 8q (2 q + 1) = 16q 2 + 8q = 8q (2 q + 1) which is clearly, divisible by 8. Real Numbers 11 Case II When n = 4q + 3 In this case, we have (n2 − 1) = (4q + 3)2 − 1 = 16q 2 + 9 + 24q − 1 [Q(a + b )2 = a2 + 2 ab + b 2 ] = 16q 2 + 24q + 8 = 8(2 q 2 + 3q + 1) which is clearly divisible by 8. Hence, (n2 − 1) is divisible by 8. Q. 7 Prove that, if x and y are both odd positive integers, then x 2 + y2 is even but not divisible by 4. K Thinking Process Here we have to take any two consecutive odd positive integers. After that squaring and adding both the number and check it is even but not divisible by 4. Sol. Let x = 2 m + 1and y = 2 m + 3 are odd positive integers, for every positive integer m. Then, x 2 + y2 = (2 m + 1)2 + (2 m + 3)2 = 4m2 + 1 + 4m + 4m2 + 9 + 12 m [Q (a + b )2 = a2 + 2 ab + b 2 ] = 8m2 + 16 m + 10 = even = 2(4m2 + 8m + 5) or 4(2 m2 + 4m + 2 ) + 1 Hence, x + y is even for every positive integer m but not divisible by 4. 2 2 Q. 8 Use Euclid’s division algorithm to find the HCF of 441, 567 and 693. K Thinking Process First we use Euclid’s division algorithm between two larger numbers any of three and get the HCF between these two. After that we take the third number and resulting HCF of two numbers and apply again Euclid’s division algorithm and get the required HCF. Sol. Let a = 693, b = 567 and c = 441 By Euclid’s division algorithms, a = bq + r …(i) [Q dividend = divisor × quotient + remainder] First we take, a = 693 and b = 567 and find their HCF. 693 = 567 × 1 + 126 567 = 126 × 4 + 63 126 = 63 × 2 + 0 ∴ HCF (693, 567 ) = 63 Now, we take c = 441and say d = 63, then find their HCF. Again, using Euclid’s division algorithm, c = dq + r ⇒ 441 = 63 × 7 + 0 ∴ HCF (693, 567, 441) = 63 12 NCERT Exemplar (Class X) Solutions Q. 9 Using Euclid’s division algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3, respectively. Sol. Since, 1, 2 and 3 are the remainders of 1251, 9377 and 15628, respectively. Thus, after subtracting these remainders from the numbers. We have the numbers, 1251 − 1 = 1250, 9377 − 2 = 9375 and 15628 − 3 = 15625 which is divisible by the required number. Now, required number = HCF of 1250, 9375 and 15625 [for the largest number] By Euclid’s division algorithm, a = bq + r...(i) [Qdividend = divisor × quotient + remainder] For largest number, put a = 15625 and b = 9375 15625 = 9375 × 1 + 6250 [ from Eq. (i)] ⇒ 9375 = 6250 × 1 + 3125 ⇒ 6250 = 3125 × 2 + 0 ∴ HCF (15625, 9375) = 3125 Now, we take c = 1250 and d = 3125, then again using Euclid’s division algorithm, d = cq + r [from Eq. (i)] ⇒ 3125 = 1250 × 2 + 625 ⇒ 1250 = 625 × 2 + 0 ∴ HCF (1250, 9375, 15625) = 625 Hence, 625 is the largest number which divides 1251,9377 and 15628 leaving remainder 1, 2 and 3, respectively. Q. 10 Prove that 3 + 5 is irrational. K Thinking Process In this type of question, we use the contradiction method i.e., we assume given number is rational and at last we have to prove our asssumption is wrong, i.e., the number is irrational. Sol. Let us suppose that 3 + 5 is rational. Let 3 + 5 = a, where a is rational. Therefore, 3=a− 5 On squaring both sides, we get ( 3 )2 = (a − 5 )2 ⇒ 3 = a2 + 5 − 2 a 5 [Q (a − b )2 = a2 + b 2 − 2 ab] ⇒ 2a 5 = a + 2 2 a2 + 2 Therefore, 5= which is contradiction. 2a As the right hand side is rational number while 5 is irrational. Since, 3 and 5 are prime numbers. Hence, 3 + 5 is irrational. Q. 11 Show that 12n cannot end with the digit 0 or 5 for any natural number n. Sol. If any number ends with the digit 0 or 5, it is always divisible by 5. If 12 n ends with the digit zero it must be divisible by 5. This is possible only if prime factorisation of 12 n contains the prime number 5. Now, 12 = 2 × 2 × 3 = 2 2 × 3 ⇒ 12 n = (2 2 × 3)n = 2 2 n × 3n [since, there is no term contains 5] Hence, there is no value of n ∈ N for which 12 n ends with digit zero or five. Real Numbers 13 Q. 12 On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. What is the minimum distance each should walk, so that each can cover the same distance in complete steps? Sol. We have to find the LCM of 40 cm, 42 cm and 45 cm to get the required minimum distance. For this, 40 = 2 × 2 × 2 × 5, 42 = 2 × 3 × 7 and 45 = 3 × 3 × 5 ∴ LCM (40, 42, 45) = 2 × 3 × 5 × 2 × 2 × 3 × 7 = 30 × 12 × 7 = 210 × 12 = 2520 Minimum distance each should walk 2520 cm. So that, each can cover the same distance in complete steps. 257 Q. 13 Write the denominator of rational numberin the form 2m × 5n , 5000 where m, n are non-negative integers. Hence, write its decimal expansion, without actual division. 257 Sol. Denominator of the rational number is 5000. 5000 Now, factors of 5000 = 2 × 2 × 2 × 5 × 5 × 5 × 5 = (2 )3 × ( 5)4 , which is of the type 2 m × 5n , where m = 3 and n = 4 are non-negative integers. 257 257 2 ∴ Rational number = = × 5000 2 3 × 54 2 [since, multiplying numerator and denominater by 2] 514 514 = 4 = 2 × 54 (10)4 514 = = 0.0514 10000 257 Hence, which is the required decimal expansion of the rational and it is also a 5000 terminating decimal number. Q. 14 Prove that p + q is irrational, where p and q are primes. Sol. Let us suppose that p + q is rational. Again, let p + q = a, where a is rational. Therefore, q =a− p On squaring both sides, we get q = a2 + p − 2 a p [Q (a − b )2 = a2 + b 2 − 2 ab] a + p−q 2 Therefore, p= , which is a contradiction as the right hand side is rational 2a number while p is irrational, since p and q are prime numbers. Hence, p+ q is irrational. 14 NCERT Exemplar (Class X) Solutions Exercise 1.4 Long Answer Type Questions Q. 1 Show that the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r. Sol. Let a be an arbitrary positive integer. Then, by Euclid’s division algorithm, corresponding to the positive integers ‘a’ and 6, there exist non-negative integers q and r such that a = 6 q + r, where, 0 ≤ r < 6 ⇒ a3 = (6 q + r )3 = 216 q 3 + r 3 + 3 ⋅ 6 q ⋅ r (6 q + r ) [Q (a + b )3 = a3 + b 3 + 3 ab (a + b )] ⇒ a3 = (216 q 3 + 108 q 2 r + 18qr 2 ) + r 3 …(i) where, 0 ≤ r < 6 Case I When r = 0, then putting r = 0 in Eq. (i), we get a3 = 216q 3 = 6 (36q 3 ) = 6m where, m = 36q 3 is an integer. Case II When r = 1, then putting r = 1 in Eq. (i), we get a3 = (216q 3 + 108q 3 + 18q ) + 1 = 6 (36q 3 + 18q 3 + 3q ) + 1 ⇒ a3 = 6m + 1, where m = (36q 3 + 18q 3 + 3q ) is an integer. Case III When r = 2, then putting r = 2 in Eq. (i), we get a3 = (216q 3 + 216q 2 + 72 q ) + 8 a3 = (216 q 3 + 216q 2 + 72 q + 6) + 2 ⇒ a3 = 6(36q 3 + 36q 2 + 12 q + 1) + 2 = 6m + 2 where, m = (36q 2 + 36q 2 + 12 q + 1) is an integer. Case IV When r = 3, then putting r = 3 in Eq. (i), we get a3 = (216q 3 + 324q 2 + 162 q ) + 27= (216q 3 + 324q 2 + 162 q + 24) + 3 = 6 (36q 3 + 54q 2 + 27 q + 4) + 3 = 6m + 3 where, m = (36 q 2 + 54 q 2 + 27q + 4) is an integer. Case V When r = 4, then putting r = 4 in Eq. (i), we get a3 = (216q 2 + 432 q 2 + 288q ) + 64 = 6 (36q 3 + 72 q 2 + 48q ) + 60 + 4 = a3 6(36q 3 + 72 q 2 + 48q + 10) + 4 = 6m + 4 where, m = (36q 3 + 72q 2 + 48 q + 10) is an integer. Case VI When r = 5, then putting r = 5 in Eq. (i), we get a3 = (216q 3 + 540q 2 + 450 q ) + 125 ⇒ a3 = (216q 3 + 540q 2 + 450q ) + 120 + 5 ⇒ a3 = 6 (36q 3 + 90q 2 + 75q + 20) + 5 ⇒ a3 = 6m + 5 where, m = (36q 3 + 90q 2 + 75q + 20) is an integer. Hence, the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the forms 6m, 6 m + 1, 6 m + 2, 6 m + 3,6 m + 4 and 6 m + 5 i.e., 6 m + r. Real Numbers 15 Q. 2 Prove that one and only one out of n, (n + 2) and (n + 4) is divisible by 3, where n is any positive integer. K Thinking Process Since, n is positive integer put n = 1, 2, 3,… in the given numbers and makes the order triplets, then we see that any one digit in a triplet is divisible by 3. Sol. Let a = n, b = n + 2 and c = n + 4 ∴ Order triplet is (a, b, c) = (n, n + 2, n + 4)...(i) Where, n is any positive integer i.e., n = 1, 2, 3,... At n = 1; (a, b, c ) = (1, 1 + 2, 1 + 4) = (1, 3, 5) At n = 2; (a, b, c ) = (2, 2 + 2, 2 + 4) = (2, 4, 6) At n = 3; (a, b, c ) = (3, 3 + 2, 3 + 4) = (3, 5, 7 ) At n = 4; (a, b, c ) = (4, 4 + 2, 4 + 4) = (4, 6, 8) At n = 5; (a, b, c ) = (5, 5 + 2, 5 + 4) = (5, 7, 9) At n = 6; (a, b, c ) = (6, 6 + 2, 6 + 4) = (6, 8, 10) At n = 7; (a, b, c ) = (7, 7 + 2, 7 + 4) = (7, 9, 11) At n = 8; (a, b, c ) = (8, 8 + 2, 8 + 4) = (8, 10, 12 ) We observe that each triplet consist of one and only one number which is multiple of 3 i.e., divisible by 3. Hence, one and only one out of n,(n + 2 ) and (n + 4) is divisible by 3, where, n is any positive integer. Alternate Method On dividing ‘n’ by 3, let q be the quotient and r be the remainder. Then, n = 3q + r, where, 0 ≤ r < 3 ⇒ n = 3q + r, where, r = 0, 1, 2 ⇒ n = 3q or n = 3q + 1or n = 3q + 2 Case I If n = 3 q , then n is only divisible by 3. but n + 2 and n + 4 are not divisible by 3. Case II If n = 3q + 1, then (n + 2 ) = 3q + 3 = 3(q + 1,) which is only divisible by 3, but n and n + 4 are not divisible by 3. So, in this case, (n + 2 ) is divisible by 3. Case III When n = 3q + 2, then (n + 4) = 3q + 6 = 3(q + 2 ), which is only divisible by 3, but n and (n + 2 ) are not divisible by 3. So, in this case, (n + 4) is divisible by 3. Hence, one and only one out of n, (n + 2 ) and (n + 4) is divisible by 3. Q. 3 Prove that one of any three consecutive positive integers must be divisible by 3. Sol. Any three consecutive positive integers must be of the form n, (n + 1) and (n + 2 ), where n is any natural number. i.e., n = 1, 2, 3,... Let, a = n, b = n + 1and c = n + 2 ∴ Order triplet is (a, b, c ) = (n, n + 1, n + 2 ), where n = 1, 2, 3,......(i) At n = 1; (a, b, c ) = (1, 1 + 1, 1 + 2 ) = (1, 2, 3) At n = 2; (a, b, c) = (1, 2 + 1, 2 + 2 ) = (2, 3, 4) At n = 3; (a, b, c) = (3, 3 + 1, 3 + 2 ) = (3, 4, 5) At n = 4; (a, b, c) = (4, 4 + 1, 4 + 2 ) = (4, 5, 6) At n = 5; (a, b, c) = (5, 5 + 1, 5 + 2 ) = (5, 6, 7 ) 16 NCERT Exemplar (Class X) Solutions At n = 6; (a, b, c) = (6, 6 + 1, 6 + 2 ) = (6, 7, 8) At n = 7; (a, b, c) = (7, 7 + 1, 7 + 2 ) = (7, 8, 9) At n = 8; (a, b, c) = (8, 8 + 1, 8 + 2 ) = (8, 9, 10) We observe that each triplet consist of one and only one number which is multiple of 3 i.e., divisible by 3. Hence, one of any three consecutive positive integers must be divisible by 3. Q. 4 For any positive integer n, prove that n3 − n is divisible by 6. Sol. Let a = n3 − n ⇒ a = n ⋅(n2 − 1) ⇒ a = n ⋅ (n − 1) ⋅ (n + 1) [Q (a2 − b 2 ) = (a − b ) (a + b )] ⇒ a = (n − 1) ⋅ n ⋅ (n + 1)...(i) We know that, I. If a number is completely divisible by 2 and 3, then it is also divisible by 6. II. If the sum of digits of any number is divisible by 3, then it is also divisible by 3. III. If one of the factor of any number is an even number, then it is also divisible by 2. ∴ a = (n − 1) ⋅ n ⋅ (n + 1) [from Eq. (i)] Now, sum of the digits = n − 1 + n + n + 1= 3 ⋅ n = multiple of 3, where n is any positive integer. and (n − 1) ⋅ n ⋅ (n + 1) will always be even, as one out of (n − 1) or n or (n + 1) must of even. Since, conditions II and III is completely satisfy the Eq. (i). Hence, by condition I the number n3 − n is always divisible by 6, where n is any positive integer. Hence proved. Q. 5 Show that one and only one out of n, n + 4 , n + 8, n + 12 and n + 16 is divisible by 5, where is any positive integer. Sol. Given numbers are n, (n + 4), (n + 8), (n + 12 ) and (n + 16), where n is any positive integer. Then, let n = 5q , 5q + 1, 5q + 2, 5q + 3, 5q + 4 for q ∈ N [by Euclid’s algorithm] Then, in each case if we put the different values of n in the given numbers. We definitely get one and only one of given numbers is divisible by 5. Hence, one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5. Alternate Method On dividing on n by 5, let q be the quotient and r be the remainder. Then n = 5 q + r, where 0 ≤ r < 5. ⇒ n = 5 q + r, where r = 0, 1, 2, 3, 4. ⇒ n = 5 q or 5 q + 1 or 5 q + 2 or 5 q + 3 or 5q + 4. Case I If n = 5 q , then n is only divisible by 5. Case II If n = 5 q + 1, then n + 4 = 5 q + 1 + 4 = 5 q + 5 = 5 (q + 1), which is only divisible by 5. So, in this case, (n + 4) is divisible by 5. Case III If n = 5 q + 3, then n + 2 = 5 q + 3 + 12 = 5 q + 15 = 5 (q + 3), which is divisible by 5. So, in this case (n + 12 ) is only divisible by 5. Case IV If n = 5 q + 4, then n + 16 = 5 q + 4 + 16 = 5 q + 20 = 5 (q + 4), which is divisible by 5. So, in this case, (n + 16) is only divisible by 5. Hence, one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.  2 Polynomials Exercise 2.1 Multiple Choice Questions (MCQs) Q. 1 If one of the zeroes of the quadratic polynomial (k − 1)x 2 + kx + 1 is −3, then the value of k is 4 −4 2 −2 (a) (b) (c) (d) 3 3 3 3 K Thinking Process If α is the one of the zeroes of the quadratic polynomial f(x) = ax2 + bx + c. Then, f(α ) must be equal to 0. Sol. (a) Given that, one of the zeroes of the quadratic polynomial say p(x ) = (k − 1) x 2 + kx + 1 is −3, then p (−3) = 0 ⇒ (k − 1)(−3)2 + k (−3) + 1 = 0 ⇒ 9(k − 1) − 3 k + 1 = 0 ⇒ 9k − 9 − 3k + 1 = 0 ⇒ 6k − 8 = 0 ∴ k = 4/ 3 Q. 2 A quadratic polynomial, whose zeroes are −3 and 4, is x2 x (a) x2 − x + 12 (b) x2 + x + 12 (c) − −6 (d) 2x2 + 2x − 24 2 2 Sol. (c) Let ax 2 + bx + c be a required polynomial whose zeroes are −3 and 4. Then, sum of zeroes = − 3 + 4 = 1 Q sum of zeroes = − b   a  −b 1 −b (−1) ⇒ = ⇒ =− …(i) a 1 a 1 and product of zeroes = − 3 × 4 = − 12 Q product of zeroes = c   a  c −12 ⇒ = …(ii) a 1 From Eqs. (i) and (ii), a = 1, b = − 1 and c = − 12 = ax 2 + bx + c 18 NCERT Exemplar (Class X) Solutions ∴ Required polynomial = 1⋅ x 2 − 1⋅ x − 12 = x 2 − x − 12 x2 x = − −6 2 2 We know that, if we multiply/divide any polynomial by any constant, then the zeroes of polynomial do not change. Alternate Method Let the zeroes of a quadratic polynomial are α = − 3 and β = 4. Then, sum of zeroes = α + β = − 3 + 4 =1 and product of zeroes = αβ = (−3) (4) = − 12 ∴ Required polynomial = x 2 − (sum of zeroes)x + (product of zeroes) = x 2 − (1) x + (−12 ) = x 2 − x − 12 x2 x = − −6 2 2 Q. 3 If the zeroes of the quadratic polynomial x 2 + (a + 1)x + b are 2 and −3, then (a) a = − 7, b = − 1 (b) a = 5, b = − 1 (c) a = 2, b = − 6 (d) a = 0 , b = − 6 Sol. (d) Let p(x) = x 2 + (a + 1)x + b Given that, 2 and −3 are the zeroes of the quadratic polynomial p(x). ∴ p (2 ) = 0 and p (−3) = 0 ⇒ 2 2 + (a + 1)(2 ) + b = 0 ⇒ 4 + 2a + 2 + b = 0 ⇒ 2a + b = − 6 …(i) and (−3)2 + (a + 1)(−3) + b = 0 ⇒ 9 − 3a − 3 + b = 0 ⇒ 3a − b = 6 …(ii) On adding Eqs. (i) and (ii), we get 5a = 0 ⇒ a = 0 Put the value of a in Eq. (i), we get 2 × 0+ b= −6 ⇒ b= −6 So, the required values are a = 0 and b = − 6. Q. 4 The number of polynomials having zeroes as −2 and 5 is (a) 1 (b) 2 (c) 3 (d) more than 3 Sol. (d) Let p (x ) = ax 2 + bx + c be the required polynomial whose zeroes are −2 and 5. −b ∴ Sum of zeroes = a −b 3 −(−3) ⇒ = −2 + 5 = = …(i) a 1 1 c and product of zeroes = a c −10 ⇒ = −2 × 5 = (ii) a 1 Polynomials 19 From Eqs. (i) and (ii), a = 1, b = − 3 and c = − 10 ∴ p(x ) = ax 2 + bx + c = 1⋅ x 2 − 3x − 10 = x 2 − 3x − 10 But we know that, if we multiply/divide any polynomial by any arbitrary constant. Then, the zeroes of polynomial never change. ∴ p(x ) = kx 2 − 3 kx − 10 k [where, k is a real number] x2 3 10 ⇒ p(x ) = − x − , [where, k is a non-zero real number] k k k Hence, the required number of polynomials are infinite i.e., more than 3. Q. 5 If one of the zeroes of the cubic polynomial ax 3 + bx 2 + cx + d is zero, the product of then other two zeroes is −c c −b (a) (b) (c) 0 (d) a a a K Thinking Process Firstly, we find the sum of product of two zeroes at a time and put the value of one of the zeroes i. e. , zero, we get the required product of the other two zeroes. Sol. (b) Let p(x) = ax 3 + bx 2 + cx + d Given that, one of the zeroes of the cubic polynomial p(x) is zero. Let α, β and γ are the zeroes of cubic polynomial p(x ), where a = 0. We know that, c Sum of product of two zeroes at a time = a c ⇒ αβ + βγ + γα = a c ⇒ 0 × β + βγ + γ × 0 = [Q α = 0, given] a c ⇒ 0 + βγ + 0 = a c ⇒ βγ = a c Hence, product of other two zeroes = a Q. 6 If one of the zeroes of the cubic polynomial x 3 + ax 2 + bx + c is −1, then the product of the other two zeroes is (a) b − a + 1 (b) b − a − 1 (c) a − b + 1 (d) a − b − 1 K Thinking Process Firstly, we find the value of constant term ‘c’, by using p(− 1) = 0. After that we find the product of all zeroes and put the value of one of the zeroes. Finally, we get the required result. Sol. (a) Let p(x) = x 3 + ax 2 + bx + c Let α, β and γ be the zeroes of the given cubic polynomial p(x). ∴ α = −1 [given] and p(−1) = 0 20 NCERT Exemplar (Class X) Solutions ⇒ (−1)3 + a(−1)2 + b (−1) + c = 0 ⇒ −1 + a − b + c = 0 ⇒ c = 1− a + b …(i) We know that, Constant term c Product of all zeroes = (−1)3. =− Coefficient of x 3 1 αβγ = − c ⇒ (−1) βγ = − c [Q α = − 1] ⇒ βγ = c ⇒ βγ = 1 − a + b [from Eq. (i)] Hence, product of the other two roots is 1 − a + b. Alternate Method Since, − 1 is one of the zeroes of the cubic polynomial f(x ) = x 2 + ax 2 + bx + c i.e., (x + 1) is a factor of f(x ). Now, using division algorithm, x 2 + (a − 1)x + (b − a + 1) x + 1 x 3 + ax 2 + bx + c x3 + x2 (a − 1)x 2 + bx (a − 1)x 2 + (a − 1) x (b − a + 1)x + c (b − a + 1) x (b − a + 1) (c − b + a − 1) ∴ x 3 + ax 2 + bx + c = (x + 1) × {x 2 + (a − 1)x + (b − a + 1)} + (c − b + a − 1) ⇒ x 3 + ax 2 + bx + (b − a + 1) = (x + 1) {x 2 + (a − 1)x + (b − a + 1)} Let α and β be the other two zeroes of the given polynomial, then − Constant term Product of zeroes = (− 1) α ⋅ β = Coefficient of x 3 − (b − a + 1) ⇒ − α ⋅β = 1 ⇒ αβ = − a + b + 1 Hence, the required product of other two roots is (− a + b + 1.) Q. 7 The zeroes of the quadratic polynomial x 2 + 99x + 127 are (a) both positive (b) both negative (c) one positive and one negative (d) both equal Sol. (b) Let given quadratic polynomial be p(x) = x 2 + 99x + 127. On comparing p(x) with ax 2 + bx + c, we get a = 1, b = 99 and c = 127 Polynomials 21 −b ± b 2 − 4ac We know that, x= [by quadratic formula] 2a −99 ± (99)2 − 4 × 1 × 127 = 2 ×1 −99 ± 9801 − 508 = 2 −99 ± 9293 −99 ± 96.4 = = 2 2 −99 + 96.4 −99 − 96.4 = , 2 2 −2.6 −195.4 = , 2 2 = − 13. , −97.7 Hence, both zeroes of the given quadratic polynomial p(x) are negative. Alternate Method We know that, a> 0 b > 0, c > 0 In quadratic polynomial, if or , then both zeroes are negative. a< 0 b < 0, c < 0 In given polynomial, we see that a = 1>0, b = 99>0 and c = 127>0 which satisfy the above condition. So, both zeroes of the given quadratic polynomial are negative. Q. 8 The zeroes of the quadratic polynomial x 2 + k x + k where k ≠ 0, (a) cannot both be positive (b) cannot both be negative (c) are always unequal (d) are always equal Sol. (a) Let p(x) = x 2 + kx + k, k ≠ 0 On comparing p(x) with ax 2 + bx + c, we get a = 1, b = k and c = k −b ± b 2 − 4ac Now, x= [by quadratic formula] 2a −k ± k 2 − 4k = 2 ×1 + – + −k ± k (k − 4) –∞ 0 4 +∞ = ,k≠0 2 Here, we see that k (k − 4) > 0 ⇒ k ∈ (−∞, 0) ∪ (4, ∞ ) Now, we know that In quadratic polynomial ax 2 + bx + c If a > 0, b > 0, c > 0 or a < 0, b < 0, c < 0, then the polynomial has always all negative zeroes. and if a > 0, c < 0 or a < 0, c > 0, then the polynomial has always zeroes of opposite sign. 22 NCERT Exemplar (Class X) Solutions Case I If k ∈ (−∞, 0) i.e., k < 0 ⇒ a = 1 > 0, b, c = k < 0 So, both zeroes are of opposite sign. Case II If k ∈ (4, ∞ ) i.e., k ≥ 4 ⇒ a = 1 > 0, b,c ≥ 4 So, both zeroes are negative. Hence, in any case zeroes of the given quadratic polynomial cannot both be positive. Q. 9 If the zeroes of the quadratic polynomial ax 2 + bx + c, where c ≠ 0, are equal, then (a) c and a have opposite signs (b) c and b have opposite signs (c) c and a have same signs (d) c and b have the same signs Sol. (c) The zeroes of the given quadratic polynomial ax 2 + bx + c, c ≠ 0 are equal. If coefficient of x 2 and constant term have the same sign i.e., c and a have the same sign. While b i.e., coefficient of x can be positive/negative but not zero. e.g., (i) x 2 + 4x + 4 = 0 (ii) x 2 − 4x + 4 = 0 ⇒ (x + 2 )2 = 0 ⇒ (x − 2 )2 = 0 ⇒ x = − 2, − 2 ⇒ x = 2, 2 Alternate Method Given that, the zeroes of the quadratic polynomial ax 2 + bx + c, where c ≠ 0, are equal i.e., discriminant (D) = 0 ⇒ b 2 − 4ac = 0 ⇒ b 2 = 4ac b2 ⇒ ac = 4 ⇒ ac > 0 which is only possible when a and c have the same signs. Q. 10 If one of the zeroes of a quadratic polynomial of the form x 2 + ax + b is the negative of the other, then it (a) has no linear term and the constant term is negative (b) has no linear term and the constant term is positive (c) can have a linear term but the constant term is negative (d) can have a linear term but the constant term is positive Sol. (a) Let p(x) = x 2 + ax + b. Constant term Now, product of zeroes = Coefficient of x 2 Let α and β be the zeroes of p(x). b ∴ Product of zeroes (α ⋅ β ) = 1 ⇒ αβ = b …(i) Given that, one of the zeroes of a quadratic polynomial p(x) is negative of the other. ∴ αβ < 0 So, b< 0 [from Eq. (i)] Hence, b should be negative Polynomials 23 Put a = 0, then, p(x) = x 2 + b = 0 ⇒ x2 = − b ⇒ x = ± −b [Qb < 0] Hence, if one of the zeroes of quadratic polynomial p(x) is the negative of the other, then it has no linear term i.e., a = 0 and the constant term is negative i.e., b < 0. Alternate Method Let f(x ) = x 2 + ax + b and by given condition the zeroes are α and − α. ∴ Sum of the zeroes = α − α = a ⇒ a=0 ∴ f(x ) = x 2 + b, which cannot be linear. and product of zeroes = α ⋅ (− α ) = b ⇒ − α2 = b which is possible when, b < 0. Hence, it has no linear term and the constant term is negative. Q. 11 Which of the following is not the graph of a quadratic polynomial? (a) (b) (c) (d) Sol. (d) For any quadratic polynomial ax 2 + bx + c, a ≠ 0, the graph of the Corresponding equation y = ax 2 + 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. So, option (d) cannot be possible. Also, the curve of a quadratic polynomial crosses the X-axis on at most two points but in option (d) the curve crosses the X-axis on the three points, so it does not represent the quadratic polynomial. 24 NCERT Exemplar (Class X) Solutions Exercise 2.2 Very Short Answer Type Questions Q. 1 Answer the following and justify. (i) Can x 2 − 1 be the quotient on division of x 6 + 2x 3 + x − 1 by a polynomial in x of degree 5? (ii) What will the quotient and remainder be on division of ax 2 + bx + c by px 3 + qx 2 + rx + s, p ≠ 0 ? (iii) If on division of a polynomial (x) by a polynomial (x), the quotient is zero, what is the relation between the degree of (x) and (x)? (iv) If on division of a non-zero polynomial (x) by a polynomial (x), the remainder is zero, what is the relation between the degrees of (x) and (x) ? (v) Can the quadratic polynomial x 2 + kx + k have equal zeroes for some odd integer k > 1? Sol. (i) No, because whenever we divide a polynomial x 6 + 2 x 3 + x − 1 by a polynomial in x of degree 5, then we get quotient always as in linear form i.e., polynomial in x of degree 1. Let divisor = a polynomial in x of degree 5 = ax 5 + bx 4 + cx 3 + dx 2 + ex + f quotient = x 2 − 1 and dividend = x 6 + 2 x 3 + x − 1 By division algorithm for polynomials, Dividend = Divisor × Quotient + Remainder = (ax 5 + bx 4 + cx 3 + dx 2 + ex + f ) × (x 2 − 1) + Remainder = (a polynomial of degree 7) + Remainder [in division algorithm, degree of divisor > degree of remainder] = ( a polynomial of degree 7) But dividend = a polynomial of degree 6 So, division algorithm is not satisfied. Hence, x 2 − 1 is not a required quotient. (ii) Given that, Divisor px 3 + qx 2 + rx + s, p ≠ 0 and dividend = ax 2 + bx + c We see that, Degree of divisor > Degree of dividend So, by division algorithm, quotient = 0 and remainder = ax 2 + bx + c If degree of dividend < degree of divisor, then quotient will be zero and remainder as same as dividend. (iii) If division of a polynomial p(x) by a polynomial g(x), the quotient is zero, then relation between the degrees of p(x) and g(x) is degree of p(x) < degree of g(x). (iv) If division of a non-zero polynomial p(x ) by a polynomial g (x ), the remainder is zero, then g(x) is a factor of p(x) and has degree less than or equal to the degree of p(x). i.e., degree of g (x ) ≤ degree of p(x ). Polynomials 25 (v) No, let p(x) = x 2 + kx + k If p(x ) has equal zeroes, then its discriminant should be zero. ∴ D = B2 − 4 AC = 0 …(i) On comparing p(x) with Ax + Bx + C, we get 2 A = 1, B = k and C = k ∴ (k )2 − 4(1)(k ) = 0 [from Eq. (i)] ⇒ k(k − 4) = 0 ⇒ k = 0, 4 So, the quadratic polynomial p(x) have equal zeroes only at k = 0, 4. Q. 2 Are the following statements ‘True’ or ‘False’? Justify your answer. (i) If the zeroes of a quadratic polynomial ax 2 + bx + c are both positive, then a, b and c all have the same sign. (ii) If the graph of a polynomial intersects the -axis at only one point, it cannot be a quadratic polynomial. (iii) If the graph of a polynomial intersects the X-axis at exactly two points, it need not be a quadratic polynomial. (iv) If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms. (v) If all the zeroes of a cubic polynomial are negative, then all the coefficients and the constant term of the polynomial have the same sign. (vi) If all three zeroes of a cubic polynomial x 3 + ax 2 − bx + c are positive, then atleast one of a, b and c is non-negative. (vii) The only value of k for which the quadratic polynomial kx 2 + x + k has 1 equal zeroes is. 2 Sol. (i) False, if the zeroes of a quadratic polynomial ax 2 + bx + c are both positive, then b c α+β=− and α ⋅ β = a a where α and β are the zeroes of quadratic polynomial. ∴ c < 0, a 0 or c > 0, a > 0 and b < 0 (ii) True, if the graph of a polynomial intersects the X-axis at only one point, then it cannot be a quadratic polynomial because a quadratic polynomial may touch the X-axis at exactly one point or intersects X-axis at exactly two points or do not touch the X-axis. (iii) True, if the graph of a polynomial intersects the X-axis at exactly two points, then it may or may not be a quadratic polynomial. As, a polynomial of degree more than z is possible which intersects the X-axis at exactly two points when it has two real roots and other imaginary roots. 26 NCERT Exemplar (Class X) Solutions (iv) True, let α, β and γ be the zeroes of the cubic polynomial and given that two of the zeroes have value 0. Let β=γ=0 and f(x ) = (x − α )(x − β )(x − γ ) = (x − α )(x − 0)(x − 0) = x 3 − ax 2 which does not have linear and constant terms. (v) True, if f(x ) = ax 3 + bx 2 + cx + d. Then, for all negative roots, a, b, c and d must have same sign. (vi) False, let α, β and γ be the three zeroes of cubic polynomial x 3 + ax 2 − bx + c. Constant term Then, product of zeroes = (−1)3 Coefficient of x 3 (+ c ) ⇒ αβγ = − 1 ⇒ αβγ = − c …(i) Given that, all three zeroes are positive. So, the product of all three zeroes is also positive i.e., αβγ > 0 ⇒ −c > 0 [from Eq. (i)] ⇒ c< 0 Coefficient of x 2 Now, sum of the zeroes = α + β + γ = (−1) Coefficient of x 3 a ⇒ α+β+ γ=− =−a 1 But α, β and γ are all positive. Thus, its sum is also positive. So, α+β+ γ>0 ⇒ − a> 0 ⇒ a< 0 Coefficient of x −b and sum of the product of two zeroes at a time = (− 1)2 ⋅ = Coefficient of x 3 1 ⇒ αβ + βγ + γα = − b Q αβ + βγ + αγ > 0 ⇒ − b> 0 ⇒ b< 0 So, the cubic polynomial x 3 + ax 2 − bx + c has all three zeroes which are positive only when all constants a, b and c are negative. (vii) False, let f(x ) = kx 2 + x + k For equal roots. Its discriminant should be zero i.e., D = b 2 − 4ac = 0 ⇒ 1 − 4k ⋅ k = 0 1 ⇒ k=± 2 So, for two values of k, given quadratic polynomial has equal zeroes. Polynomials 27 Exercise 2.3 Short Answer Type Questions Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials (i) 4 x 2 − 3x − 1. K Thinking Process Firstly, we use the factorisation method i.e., splitting the middle term of quadratic polynomial and find its zeroes. After that use the formula of sum of zeroes and product of zeroes for verification. Sol. Let f(x ) = 4x 2 − 3x − 1 = 4x 2 − 4x + x − 1 [by splitting the middle term] = 4x(x − 1) + 1(x − 1) = (x − 1) (4x + 1) 1 So, the value of 4x 2 − 3x − 1 is zero when x − 1 = 0 or 4x + 1 = 0 i.e., when x = 1or x = −. 4 1 So, the zeroes of 4x 2 − 3x − 1 are 1 and −. 4 1 3 −(−3) ∴ Sum of zeroes = 1 − = = 4 4 4  Coefficient of x  = (−1)    Coefficient of x 2  product of zeroes = (1)  −  = − 1 1 and  4 4  Constant term  = (−1)2.    Coefficient of x 2  Hence, verified the relations between the zeroes and the coefficients of the polynomial. (ii) 3x 2 + 4 x − 4. Sol. Let f(x ) = 3x 2 + 4x − 4 = 3x 2 + 6x − 2 x − 4 [by splitting the middle term] = 3x(x + 2 ) − 2(x + 2 ) = (x + 2 ) (3x − 2 ) So, the value of 3x 2 + 4x − 4 is zero when x + 2 = 0 or 3x − 2 = 0, i.e., when x = − 2 or 2 2 x =. So, the zeroes of 3x 2 + 4x − 4 are −2 and. 3 3 2 4 ∴ Sum of zeroes = − 2 + = − 3 3  = (−1).  Coefficient of x    Coefficient of x 2  −4 product of zeroes = (−2 )   = 2 and  3 3 = (−1)2.  Constant term    Coefficient of x 2  Hence, verified the relations between the zeroes and the coefficients of the polynomial. 28 NCERT Exemplar (Class X) Solutions (iii) 5 t 2 + 12 t + 7. Sol. Let f(t ) = 5t 2 + 12 t + 7 = 5t 2 + 7t + 5t + 7 [by splitting the middle term] = t (5t + 7 ) + 1(5 t + 7 ) = (5 t + 7 ) (t + 1) So, the value of 5t 2 + 12 t + 7 is zero when 5t + 7 = 0 or t + 1 = 0, −7 i.e., when t = or t = − 1. 5 So, the zeroes of 5t 2 + 12 t + 7 are −7 / 5 and −1. 7 −12 ∴ Sum of zeroes = − − 1 = 5 5 = (−1).  Coefficient of t    Coefficient of t 2  product of zeroes =  −  (−1) = 7 7 and  5 5 2  Constant term  = (−1).    Coefficient of t 2  Hence, verified the relations between the zeroes and the coefficients of the polynomial. (iv) t 3 − 2 t 2 − 15 t. Sol. Let f(t ) = t 3 − 2t 2 − 15 t = t ( t 2 − 2t − 15) = t (t 2 − 5t + 3t − 15) [by splitting the middle term] = t [t ( t − 5) + 3(t − 5)] = t (t − 5) (t + 3) So, the value of t 3 − 2 t 2 − 15 t is zero when t = 0 or t − 5 = 0 or t + 3 = 0 i.e., when t = 0 or t = 5 or t = − 3. So, the zeroes of t 3 − 2 t 2 − 15 t are −3, 0 and 5. −(−2 ) ∴ Sum of zeroes = − 3 + 0 + 5 = 2 = 1  Coefficient of t 2  = (−1).    Coefficient of t 3  Sum of product of two zeroes at a time = (−3) (0) + (0) (5) + (5) (−3) = 0 + 0 − 15 = − 15 = (−1)2.  Coefficient of t    Coefficient of t 3  and product of zeroes = (−3) (0) (5) = 0 = (−1)3  Constant term    Coefficient of t 3  Hence, verified the relations between the zeroes and the coefficients of the polynomial. Polynomials 29 7 3 (v) 2x 2 + x+. 2 4 Sol. 7 3 Let f( x ) = 2 x 2 +x + = 8x 2 + 14x + 3 2 4 = 8x 2 + 12 x + 2 x + 3 [by splitting the middle term] = 4x (2 x + 3) + 1 (2 x + 3) = (2 x + 3) (4x + 1) So, the value of 8x 2 + 14x + 3 is zero when 2 x + 3 = 0 or 4x + 1 = 0, 3 1 i.e., when x = − or x = −. 2 4 3 1 So, the zeroes of 8x + 14x + 3 are − and −. 2 2 4 3 1 7 −7 ∴ Sum of zeroes = − − = − = 2 4 4 2 ×2 (Coefficient of x ) =− (Coefficient of x 2 ) roduct of zeroes =  −   −  = = 3 1 3 3 And  2   4 8 2 × 4 Constant term = Coefficient of x 2 Hence, verified the relations between the zeroes and the coefficients of the polynomial. (vi) 4 x 2 + 5 2x − 3. Sol. Let f(x ) = 4x 2 + 5 2 x − 3 = 4x 2 + 6 2 x − 2 x − 3 [by splitting the middle term] = 2 2 x ( 2 x + 3) − 1( 2 x + 3) = ( 2 x + 3) (2 2 ⋅ x − 1) So, the value of 4x 2 + 5 2 x − 3 is zero when 2 x + 3 = 0 or 2 2 ⋅ x − 1 = 0, 3 1 i.e., when x = − or x =. 2 2 2 3 1 So, the zeroes of 4x 2 + 5 2 x − 3 are − and. 2 2 2 3 1 ∴ Sum of zeroes = − + 2 2 2 5 −5 2 =− = 2 2 4 (Coefficient of x ) =− (Coefficient of x 2 ) 3 1 3 and product of zeroes = − ⋅ =− 2 2 2 4 Constant term = Coefficient of x 2 Hence, verified the relations between the zeroes and the coefficients of the polynomial. 30 NCERT Exemplar (Class X) Solutions (vii) 2 s2 − (1 + 2 2)s + 2. Sol. Let f(s ) = 2 s 2 − (1 + 2 2 ) s + 2 = 2s − s − 2 2 s + 2 2 = s (2 s − 1) − 2 (2 s − 1) = (2 s − 1) (s − 2 ) So, the value of 2 s 2 − (1 + 2 2 ) s + 2 is zero when 2 s − 1 = 0 or s − 2 = 0, 1 i.e., when s = or s = 2. 2 1 So, the zeroes of 2 s 2 − (1 + 2 2 ) s + 2 are and 2. 2 1 1 + 2 2 − [− (1 + 2 2 )] (Coefficient of s ) ∴ Sum of zeroes = + 2 = = = 2 2 2 (Coefficient of s 2 ) 1 1 Constant term and product of zeroes = ⋅ 2 = = 2 2 Coefficient of s 2 Hence, verified the relations between the zeroes and the coefficients of the polynomial. (viii) v 2 + 4 3v − 15. Sol. Let f(v ) = v 2 + 4 3v − 15 = v 2 + (5 3 − 3 ) v − 15 [by splitting the middle term] = v 2 + 5 3v − 3v − 15 = v(v + 5 3 ) − 3(v + 5 3 ) = (v + 5 3 )(v − 3 ) So, the value of v 2 + 4 3v − 15 is zero when v + 5 3 = 0 or v − 3 = 0, i.e., when v = − 5 3 or v = 3. So, the zeroes of v 2 + 4 3v − 15 are −5 3 and 3. ∴ Sum of zeroes = − 5 3 + 3 = − 4 3 = (−1).  Coefficient of v    Coefficient of v 2  and product of zeroes = (−5 3 ) ( 3 ) = − 5 × 3 = − 15 = (−1)2.  Constant term    Coefficient of v 2  Hence, verified the relations between the zeroes and the coefficients of the polynomial. 3 (ix) y 2 + 5y − 5. 2 Sol. 3 Let f( y) = y2 + 5y − 5 2 = 2 y2 + 3 5 y − 10 = 2 y2 + 4 5 y − 5 y − 10 [by splitting the middle term] = 2 y ( y + 2 5) − 5 ( y + 2 5) = ( y + 2 5 ) (2 y − 5 ) 3 So, the value of y2 + 5 y − 5 is zero when ( y + 2 5 ) = 0 or (2 y − 5 ) = 0, 2 5 i.e., when y = − 2 5 or y =. 2 Polynomials 31 5 So, the zeroes of 2 y2 + 3 5 y − 10 are − 2 5 and. 2 5 −3 5 (Coefficient of y) ∴ Sum of zeroes = − 2 5 + = =− 2 2 (Coefficient of y2 ) 5 Constant term And product of zeroes = − 2 5 × =−5= 2 Coefficient of y2 Hence, verified the relations between the zeroes and the coefficients of the polynomial. 11 2 (x) 7y 2 − y−. 3 3 11 2 Sol. Let f( y) = 7 y2 − y− 3 3 = 21y2 − 11y − 2 = 21y2 − 14 y + 3 y − 2 [by splitting the middle term] = 7 y(3 y − 2 ) + 1(3 y − 2 ) = (3 y − 2 ) (7 y + 1) 11 2 So, the value of 7 y − 2 y − is zero when 3 y − 2 = 0 or 7 y + 1 = 0, 3 3 2 1 i.e., when y = or y = −. 3 7 11 2 2 1 So, the zeroes of 7 y2 − y − are and −. 3 3 3 7 2 1 14 − 3 11 −11  ∴ Sum of zeroes = − = = = −   3 7 21 21  3×7