Mathematics Class 10 ASD PDF

ANIL Super Digest  TM H. M. PUBLICATIONS, AGRA     `        Unit 1 : Number System 1 Real Number EXERCISE 1.1 Multiple...

ANIL Super Digest  TM H. M. PUBLICATIONS, AGRA     `        Unit 1 : Number System 1 Real Number EXERCISE 1.1 Multiple Choice Type Questions We apply euclid’s division algorithm on divisor 224 and the remainder 1. H.C.F of 650 and 1170 is : 32. (a) 130 (b) 140 Dividend = (Divisor × Quotient (c) 80 (d) 160. + Remainder) Sol. We start with the larger no. 1170. 120 = 32 × 7 + 0 By euclid’s division algorithm, we 32 224  7 have 224 Dividend = (Divisor × Quotient 0 + Remainder) HCF (224, 256) = 32. 1170 = 650 × 1 + 520 We apply euclid’s division algorithm 650  1170 1 on divisor 24 and the remainder 8. 650 Dividend = (Divisor × Quotient + Remainder) 520 32 = 24 × 1 + 8 We apply euclid’s division algorithm on divisor 650 and the remainder 24  32 1 24 520. Dividend = (Divisor × Quotient 8 + Remainder) We apply euclid’s division algorithm 650 = 520 × 1 + 130 on divisor 32 and the remainder 24. Dividend = (Divisor × Quotient 520  650 1 + Remainder) 520 24 = 8 × 3 + 0 130 8 24  3 Again, we apply Euclid’s division 24 algorithm on divisor 520 and 0 remainder 130. HCF (120, 224, 256) = 8. Ans. Dividend = (Divisor × Quotient 3. HCF of 455 and 42 is : + Remainder) (a) 5 (b) 11 520 = 130 × 4 + 0 (c) 7 (d) 9. 130  520  4 Sol. We start with larger no. 256. 520 By euclid’s division algorithm, we 0 have, HCF (650, 1170) = 130. Ans. Dividend = (Divisor × Quotient 2. HCF of 120, 224 and 256 is : + Remainder) 455 = 42 × 10 + 35 (a) 6 (b) 8 42 455 10 (c) 14 (d) 16. 420 Sol. We start with larger no. 256. 35 By euclid’s division algorithm, we We apply euclid’s division algorithm have, on divisor 35 and the remainder 7 Dividend = (Divisor × Quotient Dividend = (Divisor × Quotient + Remainder) + Remainder) 256 = 224 × 1 + 32 42 = 35 × 1 + 7 224  256 1 35 42 1 224 35 32 7 6 | Anil Super Digest Mathematics X We apply euclid’s division algorithm Hence, HCF of (120, 244, 256) = 8. on divisor 42 and the remainder 35 Greatest number which divides 121, Dividend = (Divisor × Quotient 226 and 259 and leaves remainder + Remainder) 1, 2 & 3 respectively is = 8. 35 = 7 × 5 + 0 6. Find the greatest possible 7 35  5 number which can divide 76, 35 132 and 160 and leaves 0 remainder same in each case. HCF (455, 42) = 7. Ans. Sol. Let the greatest number which can 4. Find the greatest number which divide 76, 132 and 160 is = p divides 260, 1314, and 1331 and let the remainder = r leaves remainder 5 in each case. The quotient is equal to q1, q2 and q3. Sol. Required number Dividend = Divisor × Quotient = HCF of (260 – 5), (1314 – 5) and + Remainder (1331 – 5) i.e., required number 76 = p × q1 + r...(1) = HCF of 255, 1309, 1326 132 = p × q2 + r...(2) To find HCF of 255, 1309, 1326 first 160 = p × q3 + r...(3) we will find HCF of 255 and 1309 From 1, 2 and 3 we get : and then we will find HCF of (255, p(q2 – q1) = 132 – 76 = 56 1309) and 1326. p(q3 – q2) = 160 – 132 = 28 Step 1. HCF of 255 and 1309 by p(q3 – q1) = 160 – 76 = 84 using euclid’s division algorithm is : Therefore the HCF of 56, 28 & 84 1309 = 255 × 5 + 34 are : 255 = 34 × 7 + 17 56 = 2 × 2 × 2 × 7 34 = 17 × 2 + 0 28 = 2 × 2 × 7 HCF of 1309, 255 is 17. 84 = 2 × 2 × 3 × 7 Step 2. Now, HCF of 17 and 1326 HCF = 2 × 2 × 7 will be : = 28 1326 = 17 × 78 + 0 Therefore the greatest number Hence, HCF (255, 1309, 1326) = 17 which can divide 76, 132 & 160 5. Find the greatest number which leave the same remainder divides 121, 226 and 259 and = 28. leaves remainder 1,2 and 3 7. Using Euclid’s division respectively. algorithm find HCF of 274170 Sol. Required number = HCF of (121 – 1), (226 – 2) and 17017. and (259 – 3) i.e. required number Sol. We have, = HCF of 120, 224, 256. Dividend = 274170 Step 1. To find HCF of 120, 224, Divisor = 17017 256 first we will find HCF of 120 We start with larger number and 224 and then we will find HCF 274170. of (120, 224) and 256. By euclid’s division algorithm, we by using euclid’s division algorithm have is : Dividend = Divisor × Quotient 224 = 120 × 1 + 104 + Remainder 120 = 104 × 1 + 16 274170 = 17017 × 16 + 1898 104 = 16 × 6 + 8 16 = 8 × 2 + 0 17017  274170 16 HCF of 120, 224 is 8. 17017 Step 2. Now HCF of 256 and 8 104000 will be : 102102 256 = 8 × 32 + 0 1898 E:\HM\2019\MATH-10\MATH-01.P65\FINAL PRINT\PARADISE  Real Number | 7 We apply euclid’s division algorithm 1309 = 255 × 5 + 34 on divisor 17017 and the remainder 255 = 34 × 7 + 17 1898. 17 = 17 × 1 + 0 Dividend = Divisor × Quotient Now, we find HCF of 17 and 1326 + Remainder we apply euclid’s division algorithm 17017 = 1898 × 8 + 1833 on 1326 and 17. Dividend = Divisor × Quotient 1898  17017  8 15184 + Remainder 1326 = 17 × 78 + 0 1833 Thus, HCF of 255, 1309, 1326 is Again,we apply euclid’s division = 17. algorithm on divisor 1898 and 9. Show that the square of any remainder 1833. positive integer can not be of Dividend = Divisor × Quotient the form 6q + 2 or 6q + 5 for any + Remainder integer q. 1898 = 1833 × 1 + 65 Sol. Let a be an arbitary positive integer, 1833 1898 1 then by Euclid’s division algorithm, 1833 corresponding to the positive 65 integer a and 6, there exist non- Again,we apply euclids division negative integers q and r such that, algorithm on divisor 1833 and a = 6q + r, where 0 < r < 6 remainder 65. a = 6q + r, where 0  r < 6 Dividend = Divisor × Quotient  a2 = (6q + r)2 + Remainder = 36q2 + 12qr + r2[ (a + b)2 1833 = 65 × 28 + 13 = a2 + b2 + 2ab] 65 1833  28  a = 6 (6q + 2qr) + r2 2 2...(i) 130 where, 0 r < 6 533 Case I : When r = 0, then putting 520 13 r = 0 in eq. (i), we get a2 = 6 (6q2) = 6m Now, we apply euclids division Where m = 6q2 is an integer. algorithm on divisor 65 and Case II : When r = 1, then putting remainder 13. r = 1 in eq. (i), we get Dividend = Divisor × Quotient a2 + 6 (6q2 + 2q) + 1 = 6m + 1 + Remainder Where, m = (6q2 + 2q) is an integer. 65 = 13 × 5 + 0 Case III : When r = 2, then putting 13 65  5 r = 2 in eq. (i), we get 65 a2 = 6 (6q2 + 4q) + 4 = 6m + 4 0 Where, m = (6q2 + 4q) is an integer. Thus, HCF of 274170 and 17017 is Case IV : When r = 3, then putting = 13. r = 3 in eq. (i), we get 8. Using euclid’s division algo- a2 = 6 (6q2 + 6q) + 9 rithm find HCF of 255, 1309 and = 6 (6q2 + 6q) + 6 + 3 1326.  a = 6 (6q2 + 6q + 1) + 3 = 6m + 3 2 Sol. We have to find HCF of 255, 1309 Where, m = (6q2 + 6q + 1) is an and 1326 first we find HCF of 255 integer. and 1309 then HCF of (225, 1309 ) Case V : When r = 4, then putting and 1326. r = 4 in eq. (i), we get We apply euclid’s division on 255 a2 = 6 (6q2 + 8q) + 16 and 1309. = 6 (6q2 + 8q) + 12 + 4 Dividend = Divisor × Quotient + Remainder  a = 6 (6q2 + 8q + 2) + 4 = 6m + 4 2 E:\HM\2019\MATH-10\MATH-01.P65\FINAL PRINT\PARADISE 8 | Anil Super Digest Mathematics X Where, m = (6q2 + 8q + 2) is an So, a = 4q or 4q + 1 or 4q + 2 or 4q + 3. integer. (4q)3 = 64q3 = 4 (16q)3 Case VI : When r = 5, then putting = 4q, where q is some integer. r = 5 in eq. (i), we get (4q + 1)3 = 64q3 + 48q2 + 12q + 1 a2 = 6 (6q2 + 10q) + 25 = 4 (16q3 + 12q2 + 3q) + 1 = 6 (6q2 + 10q) + 24 + 1 = 4q + 1, where q is some  a = 6 (6q2 + 10q + 4) + 1 = 6m + 1 2 integer. Where, m = (6q2 + 10q + 4) is an (4q + 3)3 = 64q3 + 144q2 + 108q + 27 integer. = 4 (16q3 + 36q2 + 27q + 6) + 3 Hence,the square of any positive = 4q + 3, where q is some integer cannot be of the form of 6q integer. + 2 or 6q + 5 for any integer q. Hence, the cube of any positive 10. Show that the square of any integer is of the form 4q, 4q + 1, or positive integer can be in the 4q + 3 for some integer q. form of 6q, 6q + 1, 6q + 3, 6q + 4. 12. Show that one and only one Sol. Let a be any positive integer & b = 6 out of n, n + 4, n + 8, n + 12 and then, by euclid’s division algorithm n + 16 is divisible by 5 where n a = 6m + r ; 0  r < 6 is any positive integer. odd positive integer will be of the Sol. Let n, n + 4, n + 8, n + 12, n + 16 form be integers. where n can take the a = 6m + 1, 6m + 3, 6m + 5 form 5q, 5q + 1, 5q + 2, 5q + 3, Case I Where r = 1 & bq = 6m 5q + 4. Case I : Where n = 5q, Then n is a2 = (6m + 1)2 divisible by 5, but neither of 5q + = 36m2 + 12m + 1 1, 5q + 2, 5q + 3 & 5q + 4, is = 6(6m2 + 2m) + 1 divisible by 5. = 6q + 1; Case II  Where n = 5q + 1, Then where q = 6m2 + 2m n is not divisible by 5 Case II Where r = 3 & bq = 6m n + 4 = 5q + 1 + 4 a2 = (6m + 3)2 = 5q + 5 = 36m2 + 36m + 9 = 5 (q + 1) = 6(6m2 + 6m + 1) + 3 which is divisible by 5. = 6q + 3; Case III : Where n = 5q + 2, Then where a = 6m2 + 6m +1 n is not divisible by 5 Case III Where r = 5 & bq = 6m n + 8 = 5q + 2 + 8 a2 = (6m + 5)2 = 5q + 10 = 36m2 + 60m + 25 = 5 (q + 2) = 6(6m2 + 10m + 4) + 1 which is divisible by 5. = 6q + 1; Case IV : Where n = 5q + 3, Then where q = 6m2 + 10m + 4 n is not divisible by 5 Hence, passed that 6q, 6q + 1, n + 12 = 5q + 3 + 12 6q + 3, 6q + 4 can be is the form = 5q + 15 = 5 (q + 3) of squares of positive integer. which is divisible by 5. 11. Show that cube of any positive Case V : Where n = 5q + 4, Then integer is either of the form 4q, n is not divisible by 5 4q + 1 or 4q + 3. n + 16 = 5q + 4 + 16 Sol. Let a be the positive integer and = 5q + 20 b = 4. = 5 (q + 4) Then, by Euclid’s algorithm, a = 4q which is divisible by 5. + r, for some integer q 0 and r = 0, Hence, n, n + 4, n + 8, n + 12 and 1, 2, 3 because 0 r < 4. n + 16 is divisible by 5. E:\HM\2019\MATH-10\MATH-01.P65\FINAL PRINT\PARADISE  Real Number | 9 13. Prove that one and only one a2 = 4(4m2 + 6m + 2) + 1 out of n, n + 2, n + 4 is divisible a2 = 4q + 1 by 3 where n is an positive Where, q = 4m2 + 6m + 2, integer. Hence, square of any positive Sol. We applied euclid’s division integer is in form of 4q or 4q + 1, algorithm on n and 3. where q is any integer. a = bq + r on putting a = n and 15. There are 120 boys and 114 girls b =3 in class X of a school. Principal n = 3q + r, 0 < r < 3 i.e., of the school decided as a policy n = 3q...(1) matter to have maximum n = 3q + 1...(2) number of mixed sections, each n = 3q + 2...(3) section has to accommodate n = 3q is divisible by 3 or n + 2 equal number of boys and equal = 3q + 1 + 2 = number of girls. What is the 3q + 3 also divisible by 3 or n + 4 maximum number of such = 3q + 2 + 4 = 3q + 6 is also divisible by 3. sections ? Hence, n, n + 2, n + 4 are divisible Sol. we have 120 boys and 114 girls. by 3. To find maximum number of each 14. Show that square of any section. We have to find H.C.F of positive integer is of the form 120 & 114. 4q or 4q + 1 for some integer q. First we take H.C.F. of largest Sol. Let positive integer q = 4m + r, by number 120 division algorithm we know here We apply euclid’s division algorithm 0  r 4, So, when r = 0 on 120 & 114. a = 4m Dividend = (Divisor × Quotient Squaring both side, we get + Remainder) a2 = (4m)2 120 = 114 × 1 + 6 a2 = 4(4m)2 114)120(1 a2 = 4q, where q = 4m2. 114 when, r = 1 a = 4m + 1 6 Squaring both side, we get Now, we apply euclid’s division a2 = (4m + 1)2 algorithm on 114 & 6. a2 = 16m2 + 1 + 8m Dividend = (Divisor × Quotient a2 = 4(4m2 + 2m) + 1 + Remainder) a2 = 4q + 1, 114 = 6 × 19 + 0 where q = 4m2 + 2m when r =2 6 )114(19 a = 4m + 2 6 Squaring both side, we get 54 a2 = (4m + 2)2 54 a2 = 16m2 + 4 + 16m 0 a2 = 4(4m2 + 4m + 1) a2 = 4a Hence, maximum number of each when q = 4m2 + 4m + 1 section is = 6. when r =3 16. Show that square of any odd a = 4m + 3 integer is of the form 4q + 1 for Squaring both side, we get some integer q. a2 = (4m + 3)2 Sol. We know that any positive odd a2 = 16m2 + 9 + 24m integer of the form a2 = 16m2 + 24m + 8 + 1 2m + 1, 2m + 3.......... E:\HM\2019\MATH-10\MATH-01.P65\FINAL PRINT\PARADISE 10 | Anil Super Digest Mathematics X Let a be any odd integer, then  5 = 210 × 5 + 55 × A a = 2m + 1 55 × A = 5 – 1050 Squaring both sides, we get 55 × A = – 1045 a2 = (2m + 1)2 A= 1045 a2 = 4m2 + 4m + 1 55 a2 = 4(m2 + m) + 1 a2 = 4q + 1 A = – 19 [where (m2 + m) = q]  2 – A = 2 – (–19) = 2 + 19 Hence, proved = 21. Ans. 17. HCF of 126 and 35 is H. If H is 19. Find HCF of 81 and 237. Express expressed as the HCF in the form of 237 × p H = 126 × A + 35 × B + 81 × q. Find the value of A ×B (3p + q). then prove that = –2. H Sol. H.C.F. of 81 and 237, by using Sol. HCF of 126 and 35 : Euclid’s division Legorithm, As Prime factor of 35 = 5 × 7 237 = 81 × 2 + 75...(1) Prime factor of 126 = 2 × 3 × 3 × 7 81 = 75 × 1 + 6...(2)  Common factor = 7 75 = 6 × 12 + 3...(3)  HCF = common factor = 7 6=3 × 2 +0...(4) Now, according to question, Hence, H.C.F. of 81 and 237 is 3. HCF = 126A + 35B = 7 From equation (3), we get 18 × 7A + 5 × 7B = 7 3 = 75 – 6 × 12 18A + 5B = 1, here many solutions And, from eq. (2), we get possible because given one equation 6 = 81 – 75 × 1 and two variables So, 3 = 75 – (81 – 75 × 1) × 12 Let A = 2 and B = –7 3 = 75 – (81 × 12 – 75 × 12) then, 18 × 2 – 5 × 7 = 1 3 = 75 × 13 – 81 × 12 So, A = 2 and B = – 7 is a solution From eq. (1), we get of equation 75 = 237 – 81 × 2 So, 3 = (237 – 81 × 2) × 13 – 81 × 12 A×B 3 = (237 × 13 – 81 × 26) – 81 × 12 Now, LHS = H 3 = (237 × 13) (+ 81) × – 38 Put A = 2, B = – 7 and H = 7 [We need an expression of 237 × p A × B 2  7 + 81 + q] Then, = = –2 = RHS Therefore, p = 13, q = – 38 H 7 3p + q = 3 × 13 + (– 38) Hence, proved. = 39 – 38 18. If the HCF of 210 and 55 is 3p + q = 1 Ans. expressible in the form of 210 × 20. A mason has to fit a bathroom 5 + 55 × A, then find the value with square marble tiles of the of [2 – A]. largest possible size. The size Sol. Let us first find the HCF of 210 and of the bathroom is 14·5 ft by 55. Applying euclid division 12·5 ft. What would be the size algorithm on 210 and 55 we get, of tiles in inches ? How many 210 = 55 × 3 + 45 such tiles are required ? 55 = 45 × 1 + 10 Sol. Size of bathroom = 14·5 ft × 12·5 ft 45 = 10 × 4 + 5 1 feet = 12 inches 10 = 5 × 2 + 0 Hence, 14·5 feet = 14·5 × 12 We observe that the remainder at = 174 inches this stage is zero. So, the last 12·5 feet = 12·5 × 12 division i.e., 5 is the H.C.F of 210 = 150 inches and 55. E:\HM\2019\MATH-10\MATH-01.P65\FINAL PRINT\PARADISE  Real Number | 11 To find the side of a square we need saparate bundles and each HCF (174 & 150) we apply euclid’s bundle must contain same division algorithm on 174 & 150 number of books. Find the least Dividend = Divisor × Quotient number of bundles which can + Remainder be made for these 117 books. 174 = 150 × 1 + 24 Sol. We have given 150 = 24 × 6 + 6 maths books = 45 24 = 6 × 4 + 0 physics books = 72 Hence, HCF (174, 150) = 6 To find least number of bundles we The size of tiles in inches have to find HCF of 45 and 72. = 6 inches. To find HCF we apply euclid’s division algorithm on 72 & 45 Number of square tiles Dividend = Divisor × Quotient area of bathroom + Remainder = area of square tiles 72 = 45 × 1 + 27 174  150 45 = 27 × 1 + 18 = 27 = 18 × 1 + 9 66 18 = 9 × 2 + 0 = 725 tiles. Ans.  HCF (72, 45) = 9 21. A bookseller purchased 117 Therefore each bundle contain books out of which 45 books 9 books. are of mathematics and the Bundles can be made remaining 72 books are of physics. Each book has same 117 = size. Mathematics and physics 9 books are to be packed in = 13 Bundles. Ans. EXERCISE 1.2 Multiple Choice Type Questions 2. Prime factor of 176 is : (a) 2 × 2 × 2 × 11 1. Prime factors of 256 is : (b) 2 × 2 × 2 × 2 × 11 (a) 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 (c) 3 × 3 × 9 (b) 2 × 2 × 2 × 2 × 2 (d) 2 × 2 × 2 × 13 (c) 2 × 2 × 2 × 2 × 2 × 2 Sol. We use division method to find (d) 2 × 2 × 2 prime factor Sol. We use division method to find 2 176 prime factor 2 88 2 256 2 44 2 128 2 22 11 11 2 64 1 2 32 Prime factor of 2 16 176 = 2 × 2 × 2 × 2 × 11 Ans. 3. Find the prime factor of 12673. 2 8 Sol. We use division method to find 2 4 prime factor 19 12673 2 2 23 667 1 29 29 Prime factor of 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 1 Ans. Prime factor of 12673 = 19 × 23 × 29. Ans. E:\HM\2019\MATH-10\MATH-01.P65\FINAL PRINT\PARADISE 12 | Anil Super Digest Mathematics X 4. Given that HCF (1261, 1067) 23460 × 2 = 510 × 92 = 97, find L.C.M (1261, 1067). 46920 = 46920. Sol. We have, 2 336 2 54 HCF (1267, 1067) = 97 (iii) We know that, 2 168 3 27 Product of LCM & HCF 2 84 3 9 = Product of two number 2 42 3 3 97 × LCM = 1261 × 1067 3 21 1 1261  1067 LCM = 7 7 97 = 13871 1 Hence, LCM (1261, 1067) = 13871. Ans. Prime factors of 336 5. Find the LCM and HCF of the = 2 × 2 × 2 × 2 × 3 × 7. following pairs of integers and Prime factors of 54 verify : = 2 × 3 × 3 × 3 LCM × HCF = Product of two LCM (336, 54) = 2 × 2 × 2 × 2 × 3 × 3 ×3 ×7 numbers. = 3024. (i) 26 and 91 (ii) 510 and 92 HCF (336, 54) = 2 × 3. (iii) 336 and 54 = 6. Sol. To find LCM and HCF we find LCM × HCF = Product of two prime factors. To find prime factors numbers we use division method : 3024 × 6 = 336 × 54 18144 = 18144. 2 26 7 91 6. Find the LCM and HCF of the 13 13 13 13 following pairs of integers by (i) applying the fundamental 1 1 theorem of arithmetic method : Prime factors of 26 = 2 × 13. (i) 455, 78, (ii) 408, 170 (iii) 13, 11. Prime factors of 91 = 7 × 13. Sol. (i) LCM (26, 91) = 2 × 7 × 13 = 182. 455 78 HCF (26, 91) = 13. LCM × HCF = Product of two 5 91 2 39 numbers 182 × 13 = 26 × 91 7 13 3 13 2366 = 2366. Prime factors of 455 = 5 × 7 × 13 2 510 2 92 Prime factors of 78 = 2 × 3 × 13 (ii) 3 255 2 46 LCM (455, 78) = 2 × 3 × 5 × 7 × 13 5 85 23 23 = 2730 HCF (455, 78) = 13 17 17 1 (ii) 1 408 170 Prime factors of 510 2 204 2 85 = 2 × 3 × 5 × 17. Prime factors of 92 = 2 × 2 × 23. LCM (510, 92) 2 102 5 17 = 2 × 2 × 3 × 5 × 17 × 23 = 23460. 2 51 HCF (26, 9) = 2. LCM × HCF = Product of two 3 17 numbers E:\HM\2019\MATH-10\MATH-01.P65\FINAL PRINT\PARADISE  Real Number | 13 Prime factors of 408 = 2 × 2 × 2 × 3 × 17 408 170 =2 × 5 × 17 LCM (408, 170) = 2 × 2 × 2 × 3 × 5 × 17 2 204 = 2040. HCF (408, 170) = 2 × 17 = 34. 2 102 (iii)13 and 11 are prime numbers  Therefore, there are no prime 2 51 factors of 13 and 11. Hence, HCF (13, 11) = 1 LCM (13, 11) = 13 × 11 = 143. 3 17 7. Find the LCM and HCF of the following integers by applying Prime factors of the prime factorisation method : 765 = 3 × 3 × 5 × 17 (i) 275, 225, 175 510 = 2 × 3 × 5 × 17 (ii) 765, 510, 408 (iii) 19, 13, 7 408 = 2 × 2 × 2 × 3 × 17 Sol. (i) LCM (765, 510, 408) = 2 × 2 × 2 × 275 225 3 × 3 × 5 × 17 = 6120. HCF (765, 510, 408) = 3 × 17 5 55 3 75 = 51. (iii) 19, 13, 7 are prime numbers. Therefore there are no prime 5 11 3 25 factors of 19, 13, 7  HCF (19, 13, 7) = 1 5 5 LCM (19, 13, 7) = 19 ×  ×  175 = 1729. Ans. 8. Explain why : 3 × 5 × 7 + 7 × 11 5 35 is a composite number. Sol. A composite number is a number 5 7 that is divisible by another number Prime factors of 275= 5 × 5 × 11 other then by itself and one. The 225 = 3 × 3 × 5 × 5 number 2 is the only even prime 175 = 5 × 5 × 7 number. All other even number Hence, LCM (275, 225, 175) are composite. = 5 × 5 × 3 × 3 × 7 × 11 We have, = 17325.  3 ×  ×  ×  HCF (275, 225, 175)  105 + 77  182 =5 ×5 182 is an even number and = 25. therefore it is a composite number. (ii) other than this 182 is the product 765 510 of 2 × 7 × 13 Hence, it is a prime number. 9. Explain why : 5 × 7 × 11 + 13 × 17 3 255 2 255 is a composite number. Also find smallest divisor. 3 85 3 85 Sol. We have, 5 × 7 × 11 + 13 × 17 5 17 5 17 385 + 221 606. Prime factors of 606 = 2 × 3 × 101 E:\HM\2019\MATH-10\MATH-01.P65\FINAL PRINT\PARADISE 14 | Anil Super Digest Mathematics X So, it is the product of more then It will also be divisible by 2 and 5 two prime numbers 2, 3 and 101 as 2 × 5 = 10 Hence, it is a composite number It can be observed that 5 is not in and 2 is a smallest divisor of 606. the prime factorization of 10. Check whether following 28 = 2 × 2 × 7 or (28)n numbers are prime or Hence, for any value of n, 28n will composite : not be divisible by 5. (i) 2 × 3 + 5 Therefore, 28n cannot end with the (ii) 7 × 11 × 13 + 13 digit 0 for any natural number n. (iii) 3 × 7 + 2 13. Check whether (26)n can end (iv) 13 × 17 × 19 + 23 × 38 with the digit 5 for any n  N. Sol. (i) We have, 2 × 3 + 5 Sol. A number to end with digit 5, the 6 + 5 = 11 prime factorisation of a number 11 has no prime factors. must have 5 as its prime factor. Hence, 11 is a prime numbers. (ii) We have, 7 × 11 × 13 + 13 So, calculating prime factorisation 1001 + 13 = 1014 of 26 we get, So, it is the product of more than 26 = 2 × 13 two prime numbers 2, 3, and 13. Since, 26 has no prime factor 5 in Hence, it is a composite number. its prime factorisation there is no (iii) We have, 3 × 7 + 2 natural number ‘n’ for which 26n 21 + 2 = 23 will end with 5. 23 has no prime factors Short Answer Type Questions Hence, 23 is a prime number. (iv) We have, 13 × 17 × 19 + 23 × 38 14. A rectangular field is 150 m × 4199 + 874 = 5073. 60 m. Two cyclists Karan and So, it is a product of more than two Vijay start together and can prime factors 3,19,89. cycle at speed of 21 m/min and Hence, it is a composite number. 28 m/min, respectively. They 11. Check whether (15)n can end cycle along the rectangular with the digit 0 for any n  N. track, around the field from the Sol. If (15)n ends with digit zero. Then same point and at the same number should be divisible by moment. After how many 2 and 5. minutes will they meet again As, 2 × 5 = 10 at the starting point ?  This means the prime Distance factorisation of (15)n should contain Sol. Speed = Time prime factors 2 and 5. Speed  Karan = 21 m/min  (15)n = (3 × 5)n Vijay = 28 m/min It does not have prime factor 2 but Distance = Perimeter of have 3 and 5. rectangle Since, 2 is not present in the prime factorisation there is no natural = (150 × 2) + (60 × 2) numbers nor which (15)n ends with = 300 + 120 digit zero. Time taken by Karan = 420 m. So, (15)n can not end with digit 420 zero. 21 = Time 12. Check whether (28)n can end with the digit 0 for any n  N. 420 Time = Sol. If any number ends with the digit 0, 21 It should be divisible by 10 or in = 20 min. other words, Time taken by Vijay E:\HM\2019\MATH-10\MATH-01.P65\FINAL PRINT\PARADISE  Real Number | 15 420 Prime factors of 28 = 12 = 2 × 2 × 3 Time 420 9 =3 ×3 Time = LCM (12, 9) = 2 × 2 × 3 × 3 28 = 15 min. = 36. So, they will meet : At the LCM of Therefore, they will meet after 36 (20, 15) minutes. Prime factors of 16. Three sets of english, hindi and 20 = 2 × 2 × 5 mathematics books have to be 15 = 3 × 5 stacked in such a way that all  LCM (20, 15) = 2 × 2 × 3 × 5 the books are stored topic wise = 60. and the height of each stack is Therefore, they will meet again the same. The number of english at the starting point after 60 books is 96, the number of hindi minutes. books is 240 and the number of 15. Radius of a circular track is 63 mathematics books is 336. m. Two cyclists Amit and Ajit Assuming that the books are of start together from the same the same thickness, determine position, at the same time and the number of stacks of english, in the same direction with hindi and mathematics books. speeds 33 m/min and 44 m/min. Depict values. After how many minutes they Sol. There are 3 sets of english, hindi meet again at the starting and maths books which have to be point ? stacked in such a way that the Sol. Given speed : Amit = 33 m/min height of each stack is same. Ajit = 44 m/min There are 96 english books Distance : Perimeter of circle 240 hindi books = 2r 336 maths books 22 First we will find HCF of (96, 240, =2 × × 63 7 336) = 396 m. Prime factorisation Distance 96 = 2 × 2 × 2 × 2 × 2 × 3 Speed = Time 240 = 2 × 2 × 2 × 2 × 3 × 5 Time taken by Amit : 336 = 2 × 2 × 2 × 2 × 3 × 7 396 HCF (96, 240, 336) 33 = = 2 × 2 × 2 × 2 × 3 = 48 Time 396 Now we are going to divide the Time = total number of books of each 33 = 12 min. subject by the HCF which is 48. Time taken by Ajit : English books (number of stacks) = 96 48 396 44 = = 2. Time Hindi books (number of stacks) 396 = 240 48 Time = 44 = 5. = 9 min. Maths books (number of stacks) So, they will meet = At the L.C.M. = 336 48 of (12, 9) = 7. Ans. E:\HM\2019\MATH-10\MATH-01.P65\FINAL PRINT\PARADISE 16 | Anil Super Digest Mathematics X EXERCISE 1.3 Short Answer Type Questions But 11 is a irrational no. 1. Prove that 2 + 6 is an Since irrational rational irrational number. Our assumption is wrong Sol. Let us assume, to the contrary Hence, 3 11 is irrational. 2  6 as rational number Proved p Now, let 2  6 = 3. Prove that ( 3  8) is an q (where p & q are co-primes and q  0) irrational no. p Sol. Let ( 3  8) be any rational 6 = 2 number x. q x = 3 8 p  2q 6 = Squaring both the sides q Since, p and q are integers, x 2 = ( 3  8)2 p  2q x 2 = 3 + 8 – 2 24 therefore, is a rational q x 2 = 11 – 2 24 number. x2 – 11 =  2 24 By fact 6 is irrational number. our assumption is wrong. x2  11 =  24 2 Hence, it is proved that 2  6 is as x is a rational no. So x2 is also irrational no. a rational number, 11 & 2 are 2. Prove that 3 11 is an rational numbers. So  24 must irrational number. also be a rational number as a Sol. We have to prove 3 11 is quotient of two rational numbers is irrational number also a rational number. But  24 Let us assume the opposite. is a irrational number. i.e., 3 11 is a rational number So, we arrive to a contradiction Hence, 3 11 can be written in This shows our assumption is wrong. p form of q So, 3  8 is an irrational number. Where p and q (q 0) are co-prime 4. Prove that 3 10 is an (no common factor other then 1) irrational number. p Sol. Let us assume 3 10 is an rational Hence, 3 11 = no. q So, it can be expressed in the form 1 p 11 =  a 3 q of form where a and b are b p integers. 11 = 3q a So; 3 10 = p b Here, is a rational no. a 3q 10 = 3b E:\HM\2019\MATH-10\MATH-01.P65\FINAL PRINT\PARADISE  Real Number | 17 a 5b2 = 25c2 Since, a and b are integer, so b2 = 5c2 3b is a rational number. Which implies b2 = c2 that 3 10 is also a rational 5 number. Hence, 5 divides b2 By theorem : If p is prime number But it contridicts the facts that 10 and p divides a2, then p divides a, is an irrational number. where a is positive number Therefore, out assumption is wrong. So, 5 divides b also...(2) Hence, it is proved that 3 10 is an By (1) and (2) irrational number. 5 divides both a & b Hence, 5 is a factor of a & b 5. Prove that 5 is an irrational So, a & b have a factor of 5. number. Therefore, a & b are not co-prime. Sol. We have to prove 5 is an Hence, our assumption is wrong. irrational numbers. By contradiction it is proved that Let us assume the opposite i.e. 5 5 is an irrational number. is a rational number. 6. Prove that 7 8 is an Hence, 5 can be written in the irrational number. a Sol. Let us assume that 7  8 is form of where a & b (b 0) be irrational no. b are co-prime. a  7  8   (non common factor other than 1) b a a Hence, 5 = b 7  8 b 5b = a a Squaring both sides and 8 are rational numbers b ( 5 b)2 = a2 But 7 is an irrational number. 5 b2 = a2 It is a contradiction a2 Therefore, our assumption 7  8 b2 = is an irrational number. 5 Hence, 5 divides a2 5 By theorem : If p is a prime no 7. Prove that is an irrational 2 and p divides a where a is positive number. numbers So, 5 shall divide ‘a’ also...(1) Sol. To prove 5 is irrational Hence, we can say that 2 a Let us assume that = c, where c is same 2 as rational 5 integer 1 p = (where p and q So, a = 5c 2 q Now, we know that are co-prime) 5b2 = a2 q = 2p Putting a = 5c 5b2 = (5c)2 Squaring both the sides E:\HM\2019\MATH-10\MATH-01.P65\FINAL PRINT\PARADISE 18 | Anil Super Digest Mathematics X q2 = 2p2...(1) q By theorem 13 = a is divisible by 2 p  a = 2c Squaring both the sides (where c is an integer) Putting the value of 2 in equation1 q2 13 = 2p2 = q2 = 2c2 = 4c2 p2 4 c2 13p2 = q2...(1) p2 = 2 q is a multiple of 13 2 q is also a multiple of 13 = 2c2 Let q2 = 13 x, where x is an p2 integer = c2 Put in (1) 2 By theorem p is also divisible by 2 13p2 = (13 x)2 But p are co-prime. 13p2 = 169 x2 This is contradiction which has arisen 169 2 due the our assumption wrong. p2 = x 13 5 p2 = 13x2  is an irrational number. p2is a multiple of 13 2 p is also a multiple of 13 8. Prove that 13 + 17 is an But this contradicts our assumption irrational number. Hence, our assumption is wrong. Sol. We can prove this by method of contradiction. So, 13 is an irrational number. Let us assume that 13 is a rational Hence, it is proved that 13  17 number. is an irrational number. EXERCISE 1.4 Multiple Choice Type Questions 131 Sol. = 1·091. 141 120 1. The decimal expansion of So, it will terminate after 3 digits. 120 will terminate after how many Ans. places of decimals ? 3. The decimal expansion of the (a) 3 (b) 4 11 rational number will (c) 1 (d) 2. 2 3 ·5 2 terminate after : 141 Sol. first it can be expressed in (a) One decimal place 120 (b) Two decimal places 47 (c) Three decimal places simplest form. 40 (d) More than three decimal places. 47 11 11 Now, = 1·175. 40 Sol. 3 2 = 2 5 2  2  2 5  5 So, it will terminate after 3 digit of decimals. Ans. 11 = 131 200 2. The decimal expansion of = 0·055. 120 will terminate after how many 11 places of decimals. So, will terminate after 3 23 ·52 (a) 3 (b) 4 (c) 1 (d) 2. decimal places. E:\HM\2019\MATH-10\MATH-01.P65\FINAL PRINT\PARADISE  Real Number | 19 4. The decimal expansion of the 52 43 (x) rational number 4 will 2  54 3 2 × 53 terminate after : 54 Sol. (i) = Since, the factor of (a) 3 places (b) 4 places 343 (c) 5 places (d) 1 place. denominator 343 is not in the form 43 43 54 Sol. 4 3 = 2n × 5m. Therefore, is a non 2 5 2222555 343 43 terminating repeating decimal. = 35 2000 (ii) = Since, the factor of = 0·0215 800 So, it will terminate after 4 places. denominator 800 are 2 5 × 5 2. Ans. 35 5. The decimal expansion of the Therefore, is a terminating. 23 800 rational numbers will 127 2 2 ·5 (iii) = Since, the factor of terminate after : 910 (a) one decimal place denominator 910 are 51 × 21 × 71 (b) two decimal places 127 (c) three decimal places × 131. Therefore, is a non (d) more than three decimal places. 910 terminating repeating decimal. 23 23 Sol. = 23 2 2 5 2  25 (iv) = Since, the factor of 2048 23 denominator 2048 are 211 × 50. = 20 23 = 1·15. Therefore, is terminating 2048 So, it will terminate after 2 decimal decimal. places. 19 Short Answer Type Questions (v) = Since, the factor of 250 6. Without actually performing denominator 250 are 2 1 × 5 3. the long division, state whether 19 the following rational numbers Therefore, is a terminating 250 will have a terminating decimal decimal. expansion or a non-terminating 922 repeating decimal expansion. (vi) 1 = Since, the 54 35 2  5  73  114 2 (i) (ii) factor of denominator are 343 800 21 × 52 ×73 × 114. 127 23 922 (iii) (iv) Therefore, 1 are 910 2048 2  5  73  114 2 19 non terminating repeating decimal. (v) 1001 250 922 (vii) 0 = Since, the (vi) 2  53  115 21  52  73  114 factor of denominator are 20 × 53 1001 (vii) 1001 20  5 3  115 ×115. Therefore, are 2  53  115 0 328 (viii) non terminating repeating decimal. 22  53  74 328 1111 (viii) = Since, the factor (ix) 4 2 2  53  74 2 7  13 of denominator are 22 × 53 × 74. E:\HM\2019\MATH-10\MATH-01.P65\FINAL PRINT\PARADISE 20 | Anil Super Digest Mathematics X 328 23243456789 Therefore, 2 3 4 are non 23·243456789 = 2 5 7 1000000000 terminating repeating decimal. q = 1000000000 1111 = 29 × 59. (ix) 4 = Since, the factor of 8. A rational number in its 7  132 decimal expansion is 327·7081. denominator are 74 × 132 What can you say about the 1111 Therefore, are non prime factor of q. When this 7  132 4 number is expressed in the terminating repeating decimal. form p/q ? Give reasons. 52 Sol. We know that terminating decimal (x) 3 = Since, the factor of 2  54 must be the one which has a finite denominator are 23 × 54. Therefore decimal digits which ends exactly at 52 a point and we can not find any is terminating factor. repetitions of the decimal values 2  54 3 7. Which of the following decimal thereafter. The terminating decimal expansions are rational. If they can be represented in the form of p a fraction. Since, the value of the are rational q , then write fraction terminates at some point. Since, the rational number 327·7081 down the prime factors of q. is a terminating decimal, it has to (i) 0·240240024000240000 be in the form of a/b where, b is (ii) 11· 21478 the denominator must be of (iii) 12.123456789 structure 2x × 5y. 9. Without actually performing (iv) 125·085 (v) 23·243456789 987 the long division, find if Sol. (i) 0·240240024000240000 is non 10500 terminating non repeating decimal. will have terminating or non Therefore, it is irrational number. terminating (repeating) deci- (ii) 11· 21478 is repeating decimal. mal expansion. Give reasons for So it represents as rational number. your answers. 1121478 987 47 11· 21478 = Sol. = 99000 10500 500 q = 99000 47 2 = 3  = 23 × 32 × 53 × 11. 5 2 2 20 (iii) 12.123456789 It is repeating 94 = 3 decimal. Therefore, it is a rational no. 5  23 12123456789 94 12.123456789 = = 999999999 103 q = 999999999 94 = 32 × 111111111. = 1000 (iv) 125·085 is terminating decimal. = 0·094 So, it represents a rational number. A rational no. in the form of p/q 125085 with p & q co-primes q 0 can be 125·085 = 1000 expressed in the form of terminating q = 1000 decimal expansion if q has either = 23 × 53. only 25 or only 55 or both. (v) 23·243456789 is terminating Hence, this is a terminating decimal decimal. So, it represents a rational factor. number. E:\HM\2019\MATH-10\MATH-01.P65\FINAL PRINT\PARADISE  Unit 2 : Algebra 2 Polynomials EXERCISE 2.1 1. The graph of y = p(x) are given Sol. It is a linear polynomial. below. For which of these p(x) The number of zeroes is 1 as the is linear or quadratic ? Also graph intersect the x-axis at one point only. find the number of zeroes of 4. p(x) in each case. Y X X O Sol. This polynomial is neither quadratic non linear. Y The number of zeroes is 1 as the graph intersect the x-axis at one Sol. It is quadratic polynomial. point only. The number of zeroes is 2 as the 5. graph intersects the x-axis at 2 points. 2. Sol. This polynomial is a quadratic polynomial. The number of zeroes is 0 as the graph does not intersect the x-axis. 6. Sol. It is a quadratic polynomial. The number of zeroes is 0, as the graph does not intersects at x-axis. 3. Sol. This polynomial is a quadratic polynomial. The number of zeroes is 2, as the graph intersect the x-axis at two points. 22 | Anil Super Digest Mathematics X 7. Sol. This polynomial is linear polynomial. The number of zeroes in this polynomial is 1 as it touches x-axis at one point only. 9. Sol. This polynomial is quadratic polynomial. The number of zeroes is 0 as it does not touches the x-axis. 8. Sol. This polynomial is quadratic polynomial. The number of zeroes in this polynomial is 1 as it touches x-axis at one point only. EXERCISE 2.2 Multiple Choice Type Questions a = 1, b = –3 1. The number of polynomials and c = – 10 having zeroes -2 and 5 is : p(x) = ax2 + bx + c (a) 1 (b) 2 = 1·x2 – 3x – 10 (c) 3 (d) More than 3. = x2 – 3x – 10 Sol. Let p(x) = ax2 + bx + c be the But we know that, if we multiply required polynomial whose zeroes or divide any polynomial by any are –2 and 3. arbitary constant. Then, the zeroes of polynomial never change. b  p(x) = kx 2 – 3kx – 10k  Sum of zeroes = a (where, k is real no.) b  2  5  x2 3 10  = –  p(x) =  x – a 1  k k k (where k is non zero real no.) 2–5 Hence, the required number of = polynomial are infinite i.e., more 1 than 3. 3 2. If 1 is zero of the polynomial =–...(1) 1 p(x)= ax2 – 3 (a – 1) x – 1, then c the value of ‘a’ is : and product of zeroes = (a) 1 (b) – 1 a c (c) 2 (d) – 2. = –2 × 5 Sol. p(x) = ax2 – 3 (a – 1) x – 1 a a(1)2 – 3 (a – 1) 1 – 1 = 0 10 a – 3a + 3 – 1 = 0 =...(2) – 2a + 2 = 0 1 From equations (i) and (ii) 2 = 2a a = 1. E:\HM\2019\COMPUTER\COM-02.P65\IIIRD PRINT\PARADISE  Polynomials | 23 3. If ,  are zeroes of x2 – 6x + k. (a) 3z2 – 2z – 1 (b) 8y2 – 3y What is the value of k if 3 + 2 (c) 9x2 – 5 (d) x2 – 7x – 8. = 20. Sol. (a) 3z2 – 2z – 1 (a) – 16 (b) 8 we have, 3z2 – 2z – 1, (c) – 2 (d) – 8. where a = 3, b = – 2, c = – 1 Sol. Polynomial p(x) = x2 – 6x + k 3z2 – 2z – 1 = 0 Given  &  are the roots. 3z(z – 1) + 1 (z – 1) = 0 To find k, if 3 + 2 = 20...(1) (3z + 1) (z – 1) = 0 From the quadratic expression 3z + 1 = 0 a+ =6...(2) z– 1 =0 and a = k...(3) 1 Multiply equation (2) by 2 and z = 3 subtracting by equ. (1) :  = 20 – 12 z = 1. = 8. Verification : Substitute the value of  in (2) to (1) Sum of zeroes get : coefficient of z = –  =6 – 8 coefficient of z2 = – 2. 1 (2) Substitute these value of  and  1 = – 3 3 in equ. (3) we get : k = a 2 2 =. = – 16. Ans. 3 3 4. If one zero of 2x2 – 3x + k is (2) Product of zeroes. reciprocal to the other, then the constant term value of k is : = coefficient of z2 2 1 1 (a) 2 (b) 1 = 3 3 3 3 (c) (d) – 3. 1 1 2 =. 2 Sol. Given—2x – 3x + k...(1) 3 3 1 Sol. (b) 8y2 – 3y Let the 2 zeroes are a & we have, 8y2 – 3y; where a = 8, b a = – 3, c = 0 Quadratic form : 8y2 – 3y = 0 ax2 + bx + c...(2) y(8y – 3) = 0 On comparing equation (1) & (2) we get : 0 8y – 3 = y a = 2, b = –3, c = k 8y – 3 = 0 1 Product of zeroes = a  = 1 8y = 3 a 3 c y = Product of zeroes = 8 a y =0 k Verification : 1 = (1) Sum of zeroes 2 k =2 coefficient of y =–  The value of k = 2. Ans. coefficient of y2 5. Find the zeroes of the following 3 (– 3) quadratic polynomials and 0 = – 8 8 verify the relationship between the zeroes and the coefficient. 3 3 =. 8 8 E:\HM\2019\COMPUTER\COM-02.P65\IIIRD PRINT\PARADISE 24 | Anil Super Digest Mathematics X (2) Product of zeroes Verification : constant term (1) Sum of zeroes = coefficient of x coefficient of y2 =– 3 0 coefficient of x2 0 = 8 8 (7) 0 = 0. – 1 +8 =– 1 Sol. (c) 9x2 – 5 7 = 7. we have, 9x2 – 5, where a = 9, (2) Product of zeroes b = – 5, c = 0 9x2 – 5 = 0 constant term = 9x2 = 5 coefficient of x2 5 8 x2 = – 1 ×8 = 9 1 – 8 = – 8. 5 6. Find the polynomials whose x =  9 zeroes are given as under :  = 2,  = 3 5 x = Sol. Let the polynomial be ax2 + bx + c 3 and its zeroes be  and  5 Here  +  = 2 + 3 = 5 x = –. &  = 2 × 3 = 6 3 Thus, the polynomial formed = Verification : x2 – (sum of zeroes) x + product of (1) Sum of zeroes zeroes coefficient of x x2 – 5x + 6. =– 7.  = – 4,  = 5 coefficient of x2 Sol. Let the polynomial be ax2 + bx + c 5 5 0 and its zeroes be  and   =– 3 3 9 Here  +  = – 4 + 5 = 1 0 = 0.  = – 4 × 5 = – 20 (2) Product of zeroes Thus, the polynomial formed = x2 – (sum of zeroes) x + product constant term = of zeroes coefficient of x2 x2 – x – 20. 5  5 5 8.  = – 1,  = – 2  = Sol. Let the polynomial be ax2 + bx + c 3 3 9 and the zeroes be  and  5 5 Here  +  = – 1 + (– 2) = – 3 =.  = – 1 × – 2 = 2 9 9 2 Sol. (d) x – 7x – 8 Thus, the polynomial formed we have, x2 – 7x – 8, where a = 1, = x2 – (sum of zeroes) x + product b = – 7, c = – 8 of zeroes x2 – 7x – 8 = 0 x2 – (– 3x) + 2. x2 – 8x + x – 8 = 0 x2 + 3x + 2. x(x – 8) + 1 (x – 8) = 0 9.  = 0,  = 5 (x + 1) (x – 8) = 0 Sol. Let the polynomial be ax2 + bx + c x + 1 =0 and the zeroes be  and  x– 8 =0 Here  +  = 0 + 5 = 5 x =– 1  = 0 × 5 = 0 x =8 Thus, the polynomial formed = E:\HM\2019\COMPUTER\COM-02.P65\IIIRD PRINT\PARADISE  Polynomials | 25 = x2 – (sum of zeroes) x2 – (sum of zeroes) x + product of x + product of zeroes zeroes = x2 – 5x + 0 1 = x2 – 5x. Ans. x2 – 1x + 4 10.  = 4,  = – 4 Sol. Let the polynomial be ax2 + bx + 1 x2 – x +. Ans. c and the zeroes be  and  4 Here,  +  = 4 + (– 4) = 0 5 5  = 4 × – 4 = – 16 12.  = ,  = 3 3 Thus, the polynomial formed Sol. Let the polynomial be ax2 + bx + c = x2 – (sum of zeroes) x + product and the zeroes be  and  of zeroes x2 – 0x + (– 16) 5  5  10 Here,  +  =   = – x2 – 16. Ans. 3  3  3 1 1 5 5 25 11.  =  =  =  = 2 2 3 3 9 Sol. Let the polynomial be ax2 + bx + c Thus, the polynomial formed and the zeroes be  and  = x2 – (sum of zeroes) 1 1 x + product of zeroes Here,  +  =  = 1 2 2  10  25 = x2 –   x  1 1 1 3 9  =  = 10 25 2 2 4 = x2 + x. Ans. Thus, the polynomial formed 3 9 EXERCISE 2.3 Multiple Choice Type Questions Sol. 1. If p(x) = x2 + 6x + 9 and q(x) = x2  x  1 x + 3 then remainder will be x  1 x3  1 ( when p(x) is divided by q(x) : (a) – 1 (b) 0 x3  x 2 (c) 11 (d) 2.   Sol. Dividend = x2 + 6x + 9, Divisor – x2  1 = x + 3 Here, dividend and divisor both in  x2  x standard form so, we have   x  1 x3 x  1 2 x  3 x  6 x  9  – – x2  3 x   0 3x  9 3x  9   So, the remainder is 0. 0 3. Dividing x3 + 3x + 3 by (x + 2), the remainder will be : So, the remainder will be 0. (a) – 2 2. Dividing (x3 + 1) by (x + 1) the (b) – 11 remainder will be : (c) 0 (a) – 1 (b) 11 (d) 1. (c) 0 (d) – 2. E:\HM\2019\COMPUTER\COM-02.P65\IIIRD PRINT\PARADISE 26 | Anil Super Digest Mathematics X Short Answer Type Questions 2 x  2x  7 6. In each of the following, divide Sol. x  2 x3  3x  3  the polynomial p by g and find the quotient and the remainder. Find x3  2x2 in which case g is the factor of p.   (i) p(x) = (x4 + 1) and g(x) = (x + 1) 2x2  3x  3 2x2  4 x x3 – x 2  x – 1   Sol. x  1 x 4  1  7x  3 4 x  x 3 7x  14 – –    11 – x3  1 – x3 – x2 So, the remainder is –11.   4. Dividing x3 – 7x + 6 by (x – 1), x2  1 the remainder will be : (a) 6 (b) – 6 x2  x (c) 1 (d) 0. – – –x 1 – x–1 x2  x  6   Sol. x – 1 x3  7 x  6 2 x3  x 2   The quotient is x3 – x2 + x – 1 and the remainder is 2. x2 – 7 x  6 (ii) p(x) = 10x2 – 5x + 2 x2  x and g(x) = 5x.    6x  6 2x  1  6x  6 Sol. 5 x) 10 x2  5 x  2 (  – 10 x 2 0  – 5x  2 So, the remainder will be 0. – 5x 5. Dividing x3 – 3x2 – x + 3 by x2  – 4x + 3 the remainder will be : 2 (a) – 3 (b) 3 (c) 1 (d) 0.  The quotient is 2x – 1 and the remainder is 2. (iii) p(x) = 7x2 – 3x + 9 x1 and g(x) = x – 2. Sol. x 2  4 x  3 x  3 x 2  x  3 3 7 x  11 x3  4 x2  3 x Sol. x – 2) 7 x 2  3 x  9 (    7 x 2 – 14 x x2  4 x  3   11 x  9 x2 – 4 x  3 11 x – 22 –  – –  0 31 The quotient is 7x + 11 and the So, the remainder will be 0. remainder is 31. E:\HM\2019\COMPUTER\COM-02.P65\IIIRD PRINT\PARADISE  Polynomials | 27 Long Answer Type Questions The quotient is y2 + 5y – 2 and 7. In each of the following divide the the remainder is 0. polynomial p by g and find the Hence, the remainder is 0 so, quotient and the remainder. Find g(y) is the factor of p(y). in which cases g is a factor of p. (iv) p(t) = t6 + 3t2 + 10 (i) p(t)= t3 – 3t2 + 4t + 2 and g(t) = t3 + 1. and g(t) = t – 1. t3 – 1 t 2 – 2t  2 Sol. t  1 t – 3t  4t  2  3 2  Sol. t3  1 t6  3t2  10  t6  t 3 t3 – t2 – – –  – 2t 2  4 t  2  t3  3t2  10 – 2 t 2  2t  t3 1  –   2t  2 3t 2  11 2t – 2  The quotient is t3 – 1 and the –  remainder is 3t2 + 11. 4 8. (i) p(x) = x5 + x4 + x3 + x2 + 2x + 2 The quotient is t2 – 2t + 2 and and g(x) = x3 + 1. remainder is 4. (ii) p(u) = u3 – 14u2 + 37u – 60 and x2  x  1 g(u) = u – 2  5 4 3 2 Sol. x3  1 x  x  x  x  2 x  2  u2  12u  13 x 5 x 2 Sol. u – 2 u3  14u2  37u  60 – – 4 3 u3 – 2u2 x  x  2x  2 –  x4 x  12u2  37u  60 – – 3  12u2  24u x  x 2  – x3 1 13u  60 – – 13u – 26 x1 –  2 The quotient