Vedanta Optional Mathematics Teacher's Guide ~ 10 101 11(a). The sum of the three numbers in G.P. is 38 and their product is 1728. Find the numbers. Solution Let the three numbers b in GP be a r , a, ar According to question a r +a+ar=38 or a+ar+ar2 r =38...............(i) Also a r .a.ar=1728 or, a3 =1728 or, a3 =123 or, a=12 Substituting value of r is eqn(i) or 12(1+r+r2 ) r =38 or, 6+6r+6r2 =19r or, 6r2 –13r+6=0 or, 3r(2r–3)+2(2r–3)=0 or, (2r–3)(3r–2)=0 Either 2r–3=0............(ii) OR, 3r–2=0...........(iii) From (ii) , r= 3 2 From (iii),r= 2 3 when r= 3 2 , numbers are a r = 3 2 9 =8, a=12 and ar=12× 3 2 =18 when r= 2 3 a r = 2 3 12 =18, a=12 and ar=12× 2 3 =18 Hence the required numbers are 8,12 and 18 or 18, 12 and 8. 11(b). Sum of the three consecutive terms in G.P. is 28 and their product is 512. Find the numbers. Solution Let the three consecutive terms in GP be a r , a, ar
102 Vedanta Optional Mathematics Teacher's Guide ~ 10 According to question a r +a+ar=28 ...............(i) Also a r .a.ar=512 or, a3 =512 or, a3 =83 or, a=8 Substituting value of a is eqn(i) 8 r +8+8r=28 or 8+8r+8r2 r =28 or, 2+2r+2r2 =7r or, 2r2 –5r+2=0 or, 2r2 –4r–r+2=0 or, 2r(r–2)–1(r–2)=0 or, (r–2)(2r–1)=0 or, (2r–3)(3r–2)=0 Either r–2=0............(ii) OR, 2r–1=0...........(iii) From (ii) , r=2 From (ii),r= 1 2 when r=2,numbers are a r = 3 2 =4, a=8 and ar=8×2=16 when r= 1 2 a r = 1 2 8 =16, a=8 and ar= 8 2 =4 11(c). The product of three numbers in G.P. is 729 and sum of their squares is 819 ? Find the numbers Solution Let the three numbers is G.P. be a r , a and ar According to question a r .a.ar=729 ...............(i) a r .a.ar=512
Vedanta Optional Mathematics Teacher's Guide ~ 10 103 or, a3 =729 or, a3 =93 or, a=9 Also. Substituting value of a is eqn(i) a r 2 +a2 +(ar)2 =819 9 r 2 +92 +(9r)2 =819 81 r2 =81+81r2 =819 or, 9+9r2 +9r4 =91r2 or, 9r4 – 81r2 –r2 +9=0 or, 9r2 (r2 –9)–1(r2 –9)=0 or, (r2 –9)(9r2 –1)=0 or, (r+9)(r –9)(9r2 –1)=0 or, 2r2 –5r+2=0 Either r+9=0............(ii) or, r–9=0...........(iii) or, 9r2 –1=0.........(iii) From (i) , r = –9 From (ii), r = 9 From (iii), r2 = 1 3 2 , r=± 1 3 where r =–9, numbers are a r = 9 -9 =–1, a=9 and ar=–81 But it doesnot satisfy the second condition. So they arenot the required numbers. Similarly when r = 9, numbers are 1,9 and 81. They also don't satisfy second condition. when r= 1 3 a r = 1 3 9 =27, a=9 and ar=9. 1 3 =3 It satisfies both conditions. So the numbers are 27, 9 and 3. when r=– 1 3 the required numbers are, a r = -1 3 9 =–27, a=9 and ar=9. -1 3 =–3 Hence the required numbers are 27, 9,3 or –27,9,–3
104 Vedanta Optional Mathematics Teacher's Guide ~ 10 Questions for practice 1. Find the 10th term of a geometric sequence 2, 4, 8, 16, 32,..................... 2. If p, p–2,p+1 are in G.P., find the value of p. 3. If 1 3 , p, 3 and q are in G.P., find the values of p and q. 4. If x+6, x, x–3 are three terms of a G.P.,find the value of x. 5. If the first term and third term of a G.P. are 2 and 8–1, find the common ratio of the progression. 6. The 6th and 10th terms of a G.P. are 64 and 1024 respectiely, find the 7th term of the G.P. 7. If 2nd term and 4th term of a G.S. are 1 4 and 1 16, find the 7th term. 8. The product of three numbers in a G.P. is 216 and their sum is 26, find the numbers. 9. The product of three numbers in G.P. is 64 and their sum is 21 , find the numbers. 10. If the 7th term of a G.P. is 1 and its 16 times of 11th term is also 1, find the 20th term. Geometric Means (G.M.'s) Notes: i) G.M. between two numbers a and b, G.M.= ab ii) Let m1 ,m2 ,m3 ,........mn be n G.M.'s between a and b, then common ratio (r)= b a 1 n+1 and geometric means are given by m1 =ar m2 =ar2 . . . mn =arn Some solved problems 1. Find the G.M. between 1 3 and 1 27 . Solution Here, a= 1 3 ,b= 1 27 G.M. between a and b= ab = 1 3 . 1 27 = 1 9 2. Find the value of x,y and z if 3,x,y,z,48 are in G.P.
Vedanta Optional Mathematics Teacher's Guide ~ 10 105 Solution Here x,y,z are in G.M.'s between 3 and 48. Numbers of G.M.'s (n)=3,a=3, b=48 Common ratio(r) = b a 1 n+1 = 48 3 1 3+1 = 1 4 16 =2 Now, first mean (m1 )=x=ar=3.2=6 Second mean (m2 )=y=ar2 =3.22 =12 Third mean (m3 )=z=ar3 =3.23 =24 3.Insert four G.M.'s between 1 and 32. Solution: Here a=1, b=32, Common ratio(r) = b a 1 n+1 = 32 1 1 4+1 = 1 5 25 =2 Let required geometric means be m1 , m2 , m3 and m4 . m1 =ar=1.2=2 m2 =ar2 =1.22 =4 m3 =ar3 =1.23 =8 m4 =ar4 =1.24 =16 4. Third geometric mean between 27 and 1 27 is 1. Find the number of means. Solution Here, number of means(n)=?, m3 =1,a=27, b= 1 27 common ratio (r)= b a 1 n+1 = 27 1 27 1 n+1 = 1 36 1 n+1 1 3 6 n+1
106 Vedanta Optional Mathematics Teacher's Guide ~ 10 third mean (m3 ) =ar3 or, 1=27 1 3 6 n+1 3 or, 1 27 = 1 3 18 n+1 or, 1 3 2 = 1 3 18 n+1 or, 3 = 18 n+1 or, 6 = n+1 ⸫ n=5 5(a). There are three geometric means, between a and b. If the first mean and third mean are 1 4 and 1 64 respectively, find the values of a and b. Solution Here, first term=a, last term =b Let r be the common ratio m1 =ar and m3 = ar3 or, 1 4 =ar...............(i) 1 64 = ar3 ...............(ii) Dividing (ii) by (i), we get ar3 ar = 1 4 1 64 or, r2 = 1 16 ×4 or, r2 = 1 16 or, r= 1 4 put the value of r in equation (i), we get. 1 4 =a. 1 4 a=1 Again, put the value of a and r in
Vedanta Optional Mathematics Teacher's Guide ~ 10 107 r= b 1 1 n+1 or, 1 4 = b 1 1 3+1 or, 1 4 = b 1 4 Taking fourth power on both sides, we get. 1 256 = b b= 1 256 a=1, b= 1 256 5(b). There are three G.M.'s between p and q . If the first mean and the third mean are respectively 1 and 125, find the values of p and q. Solution Here, first mean(m1 )=1 Third mean (m3 )=125 Let r be the common ratio. Then r= q p 1 n+1 = q p 1 3+1 = q p 1 4 m1 =ar or, 1=pr..............(i) and m3 = ar3 or 125= pr3 ...............(ii) Dividing (ii) by (i), we get pr3 pr = 125 1 or, r2 =125 r=5 5 Put the value of r in equation(i) 1=p.5 5 p= 1 5 5 Again, r = q 1 5 5 1 4 or, 5 5 = q 1 5 5 1 4 Taking fourth power on both sides, 15625=5 5 q
108 Vedanta Optional Mathematics Teacher's Guide ~ 10 or, q= 15625 5 5 =625 5 p= 1 5 5 and q=625 5 6(a). Some G.M.'s are inserted between 4 and 128. Find the number of means where the ratio of first mean and last mean is 1:8. Solution Let r be common ratio. a=4, b=128. r= b a 1 n+1 = 128 4 1 n+1 = 2 5 n+1 But, m1 mn = 1 8 or, ar arn = 1 8 or, rn–1 = 23 But r= 2 5 n+1 n-1 2 5 n+1 =23 or 5(n-1) n+1 =3 or, 5n–5=3n+3 o, 2n =8 ⸫ n=4 (b). There are n G.M.'s between 1 25 and 25. If the ratio of the first mean to the third mean is 1:25, find the value of n. Solution Here, a= 1 25 , b=25 , common ratio(r)= b a 1 n+1 = 25 25 1 1 n+1 625 4 1 n+1
Vedanta Optional Mathematics Teacher's Guide ~ 10 109 = 5 4 n+1 ∴ r=5 4 n+1 By question, first mean last mean = 1 25 or, ar arn = 1 25 or, rn–1 = 52 But r= 5 4 n+1 4 5n+1 n–1 =52 ⸫ 4(n-1) n+1 =2 or, 4n–4=2n+2 or, 2n =6 ∴ n=3 7(a). Find the two numbers whose A.M. and G.M. are respectively 10 and 8. Solution Let required two numbers be a and b. Then, A.M. = a+b 2 =10 a+b =20...........(i) G.M. = ab =8 or, ab=64.............(ii) we have, a–b= (a+b)2 -4ab = 400-256 = 144 a–b=12............(iii) adding(i) and (iii),we get, 2a =32 a=16 put the value of a in equation(i), we get, b=20–16=4 a=16,b=4
110 Vedanta Optional Mathematics Teacher's Guide ~ 10 (b). Find two numbers whose A.M. and G.M. are respectively 25 and 20. Solution Let a and b two required numbers Then, A.M. = a+b 2 =25 a+b =50...........(i) G.M. = ab =20 or, ab=400.............(ii) we have, a–b= (a+b)2 -4ab = 2500-1600 = 900 a–b=30............(i) adding (i) and (ii),we get, 2a =80 a=40 put the value of a in equation(i), we get, b=20–16=410 a=40, b=10 (c). Find the two numbers whose A.M. and G.M. are respectively 13 and 8. Solution Let a and b two required numbers Then, A.M. = a+b 2 =13 a+b =26...........(i) G.M. = ab =12 or, ab=144.............(ii) we have, a–b= (a+b)2 -4ab = 676-576 = 100 =10 a–b=10............(iii) solving equations (ii) and (iii),we get, a =18, b=8 put the value of a in equation(i), we get, b=20–16=4 a=16,b=4
Vedanta Optional Mathematics Teacher's Guide ~ 10 111 (d). Find two numbers whose A.M. is 34 and G.M. is 16. Solution Let the required two numbers be a and b. Then, A.M. = a+b 2 =34 a+b =68...........(i) Also, G.M. = ab =16 or, ab=256.............(ii) we have, a–b= (a+b)2 -4ab = 4624-1024 = 3600 a–b=60............(iii) adding(i) and (iii),we get, a =64 put the value of a in (i),b=4 b=20–16=4 a=64, b=4 8(a). The ratio of two numbers is 1:16, their geometric mean is 1 4 . Find the numbers. Solution Let the required numbers be a and b. Also, Let a=k, b=16k Then, G.M. = ab = 1 4 or, k.16k = 1 4 or, 4k= 1 4 k= 1 16 Hence a=k= 1 16 b=16k=16. 1 16 =1 a= 1 16 ,b=1 (b). The ratio of two numbers is 1:81, their geometric mean is1 . Find the numbers. Solution Let two numbers be ak and b=81k. Then, G.M. = ab =1 or, k.81k =1
112 Vedanta Optional Mathematics Teacher's Guide ~ 10 or, 9k=1 k= 1 9 Hence the required numbers are a= 1 9 and b=81. 1 9 =9 Questions for practice 1. Find the G.M. between 100 and 400. 2. If p+2, P+8 and 17+p are the geometric sequence, find the value of m. 3. Find the values of x,y,z from the given geometric sequence 1 8 ,x,y,z,2. 4, Insert 3 G.M.'s between 3 and 48. 5. Insert 5 G.M.'s between 35 and 2240. 6. There are some geometric means between 5 and 80. If the second mean is 20, find the number of means. Also find the last mean. 7. Find the number of geometric means inserted between 1 and 64 in which the ratio of first mean to the last means is 1:16. 8. There are 5 geometric means between a and b. If the second mean and last means are 63 and 1701 respectively, find the values of a and b. Sum of Geometric Series Notes i) In a G.P. if the first term =a, common ratio =r, number of terms =n are given, the sum of the first n terms is given by Sn = a(rn-1) r-1 , when r>1 Sn = a(1-rn) 1-r , when r<1 ii) When the last term = l, common ratio =r and the first term =a, then the sum of the first n terms is given by Sn = lr-a r-1 , when r>1 Sn = a-lr 1-r , when r<1 iii) Three consecutive terms of a G.P. are given by a r , a, ar Some solved problems 1. Find the sum of series (a). 1+ 1 2 + 1 4 ............ upto 8 terms.
Vedanta Optional Mathematics Teacher's Guide ~ 10 113 Solution Given series is a G.P. First term (a)=1,common ratio (r)= 1 2 <1 number of terms (n)=8 Sum of the 8 terms is given by, Sn = a(1–rn) 1–r S8 = 1 1– 1– 1 2 1 2 8 = 1– 1 256 1 2 = 0 255 128 = 1 127 128 (b). 81+27+..........+ 1 243 Solution Here. First term (a)=81, common ratio (r)= 27 81 = 1 3 <1 last term (l)= 1 243 Now, Sn = a-lr 1-r = 81+ 1 243 . 1 3 1- 1 3 = 59049 729 × 1 3 = 59048 243×2 = 29524 243 3. Find the first n term of a geometric series whose last term is 1792, common ratio 2, sum 3577 Solution Here, in a given series, a =?, r=2, l=1792, sn=3577 By using formula, Now, Sn = lr-a r-1 or, 3577= 1792×2-a 2-1 or, 3577=3584–a or, –7=–a a=7
114 Vedanta Optional Mathematics Teacher's Guide ~ 10 4(a). How many terms of G.P. must be taken in the series 128+64+32+................ so that the sum may be 254. Solution In the given G.S. First term (a)=128,common ratio (r)= 64 128 = 1 2 Sn =254 By using formula, Sn = a(1–rn) 1–r or, 254 = 128 1 – 1– 1 2 1 2 n or, 127 64 = 2 1 2 n 1– or, 127 128 = 1 2 n 1– or, 1 2 n =1– 127 128 or, 1 2 n =1– 1 128 or, 1 2 n = 1 2 7 n=7 (b) How many terms of a G.P. must be taken in the series 64+96+144+216+......... so that the sum may be 2059 ? Solution In the given G.S., a=64, r = 96 64 = 3 2 Sn =254, n= ? By using formula, Sn = a(rn –1) r–1 or,2059 = 64 –1 3 2 – 1 3 2 n or, 2059 64 =2 3 2 n –1
Vedanta Optional Mathematics Teacher's Guide ~ 10 115 or, 2059 128 = 3 2 n –1 or, 2059 128 +1= 3 2 n or, 2187 128 = 3 2 n or, 3 2 7 = 3 2 n n=7 (c). How many terms of G.P. must be taken in the series 3–6+12..... so that the sum may be –63 ? Solution Let Sn = –63 r=– 6 3 =–2 a=3 By using formula, Sn = a(1-rn) 1-r or, –63 = 3(1-(-2)n) 1+2 or,–63 =1–(–2)n or,64 =(–2)n or,(–2)6 =(–2)n n=6 5(a). The first term of a G.P. is 5 and the sum of the first four terms is 780, find the common ratio. Solution Here, a=5, S4 =780 Now, Sn = a(rn -1) r-1 or,780 = 5(r4 -1) r-1 or,156 = (r2 +1)(r-1)(r+1) r-1 or, 156=r3 +r2 +r+1 or, r3 +r2 +r–155=0 or, r3 –5r2 +6r2 –30r+31r–155=0 or, r2 (r–5)+6r(r–5)+31(r–5)=0 or, (r2 +6r+31)(r–5)=0
116 Vedanta Optional Mathematics Teacher's Guide ~ 10 Either r=5 or r2 +6r+31=0 Here, r2 +6r+31 doesnot give real solution r=5 5(b). The first term of a G.P. is 3 and the sum of the first four terms is 120, find the common ratio. Solution Here, in given G.P. , a=3, S4 =120, n=4 Now, Sn = a(rn -1) r-1 or,120 = 3(r4 -1) r-1 or,40 = (r2 +1)(r-1)(r+1) r-1 or, 40=r3 +r+r2 +1 or, r3 +r2 +r–39=0 for r=3, r3 +r2 –39=0 By factor theorem, r–3 is a factor of r3 +r2 +–39=0 Now, r3 –3r2 +4r2 –12r+13r–39=0 or, r2 (r–3)+4r(r–3)+13(r–3)=0 or, (r–3)(r2 +4r+13)=0 Either r–3=0=3 or r2 +6r+31=0 Here, r2 +4r+13=0 does not give real value. r=3 6(a). The sum of the first four terms of a G.P. with common ratio 3 is 118 3 . Find the sum of the first ten terms of the progression. Solution Here, S4 = 118 3 , r=3 Now, Sn = a(rn –1) r–1 , or, Sn = a(34 –1) 3–1 or, 118 3 = a.80 2 or, 118 3 =40a or, a= 118 120 = 59 60 Again, S10 = a(r10 –1) r–1 , or, Sn = a(34 –1) 3–1
Vedanta Optional Mathematics Teacher's Guide ~ 10 117 = 59 120 {310–1} 3–1 = 59 120 {59049–1} 2 = 59.59048 240 = 435479 30 S10 = 435479 30 6(b). The sum of the first three terms of a G.P. with common ratio 2 is 112. Find the sum of the first 8 terms of the progression. Solution Here, S3 =112, r=2 Now, Sn = a(rn –1) r–1 or, S3 = a(23 –1) 2–1 or, 112 = 7a 1 a = 112 7 =16 Again, S8 = 16(28 -1) 2-1 =4080 7(a). In a G.P. the sum of three consecutive terms is 21 and their product is 64. Find them. Solution Let three terms in be a r , a, ar then, a r +a+ar=64 or, a3 =64 a=4 Again, a r .a.ar=21 or, 4 r +4+4r=21 or,4+4r+4r2 =21r or, 4r2 –17r+4=0 or, 4r2 –16r–r+4=0 or, 4r(r–4)–1(r–4)=0
118 Vedanta Optional Mathematics Teacher's Guide ~ 10 or, (r–4)(4r–1)=0 r=4, 1 4 Find r=4, the required numbers are, a r = 4 4 =1, a=4, ar =4.4 =16 when, r= 1 4 , the required numbers are, a r = 4 1 4 =16, a=4, ar=4. 1 4 =16 Hence the required numbers are 1, 4, 16 or 16,4,1 7(b). The sum of three consecutive terms in G.P. is 62 and their product is 1000, find them. Solution Let three terms in G.P. be a r , a, ar then, a r +a+ar=1000 a=10 Again, a r .a.ar=62 or, 10 r +10+10r=62 or,10+10r+10r2 =62r or, 10r2 –52r+10=0 or, 5r2 –26r+5=0 or, 5r2 –25r–r+5=0 or, 5r(r–5)–1(r–5)=0 or, (r–5)(5r–1)=0 r=5, 1 5 Find r=5, the required numbers are, a r =10 5 =2, a=10, ar=10.5=50 when, r= 1 5 , the required numbers are, a r =10 1 5 =50, a=10, ar=10. 1 5 =2 Hence the required numbers are 2, 10 and 50 or 50,10 and 2. 8(a). The sum of the first 2 terms of G.P. is 6 and that of the first four terms is 15 2 . Find the sum of the first six terms. Solution Here, S2 =6 , S4 = 15 2
Vedanta Optional Mathematics Teacher's Guide ~ 10 119 Now, S2 S4 = 15 2 9 or, a(rn –1) r–1 a(rn –1) r–1 = 12 15 or, 1 r2 +1 = 12 15 or, 15=12r2 +12 or, 3=12r2 r2 = 1 4 or, r=± 1 2 Since sum is positive, we take positive value of r= 1 2 Put the value r in s2 =6 By using formula, or, a(r2 -1) r-1 =6 or, a(r2 -1) r-1 =6 or, 1 4 – 1 a 1 2 – 1 =6 or, a. 3 4 .2=6 a=4 Now, Sn = a(1–rn) 1–r or,S6 = 4 1– 1– 1 2 1 4 6 = 1 64 4 1– 1– 1 2 = 63 8 (b).The sum of the first 2 terms of G.P. is 8 and that of the first four terms is 80 . Find the sum of the first 6 terms. Solution Here, S2 =8 and s4 =80 Now, Sn = a(rn –1) r–1 , or, S2 = a(r2 –1) r–1
120 Vedanta Optional Mathematics Teacher's Guide ~ 10 and S4 = a(r4 –1) r–1 or, S2 S4 = a(r4 –1) r–1 a(r2 –1) r–1 or, 12 15 =r2 +1 or, 10=r2 +1 or, r2 =9 r=±3 Since sum are positive, taking positive square roots only. r=3 put the value of r in S2 = a(r2 –1) r–1 or, 8= a(9–1) 3–1 a=2 Again, S6 = a(r6 –1) r–1 , = 2(36 –1) 3–1 =728 9(a). The sum of the first 8 terms of G.P. is 5 times the sum of the first 4 terms. Find the common ratio. Solution Here, S6 =5 s4 or. a(r8 –1) r–1 = 5 a(r4 –1) r–1 or, (r4 +1)(r4 –1)=5(r4 –1) or, r4 +1=5 or, r4 =4 ⸫ r= 2 9(b). The sum of the first 6 terms of G.P. is 9 times the sum of the first 3 terms. Find the common ratio. Solution Here, S6 =9 S3
Vedanta Optional Mathematics Teacher's Guide ~ 10 121 or. a(r6 –1) r–1 = 9. a(r3 –1) r–1 or, r3 +1=9 or, r3 =8 r=2 10(a). The sum of the first 2 terms of G.P. is 9 8 and the sum of the last 2 terms is 1152. There are altogether 12 terms. Find the common ratio and the first term. Solution In given G.P. S2 = 9 8 or. a(r2 –1) r–1 = 9 8 or, a(r+1)= 9 8 ..................(i) Total number of terms =12 and the sum of last two term =1152 t 11+t12=1152 or, ar10+ar11=1152 or, ar10(1+r)=1152............(ii) Dividing (ii) by (i), we get, ar10(1+ r) a(1+a) = 1152 918 or, r10=1024 or, r10=210 r=2 put the value of r in equation (i), we get a(2+1)= 9 8 or,a= 9 8×3 = 3 8 r=2, a= 3 8 10(b). The sum of the first four terms of G.P. is 30 and that of the last four terms is 960.If its first terms is 2 and the last terms is 512,find the common ratio and the number of terms. Solution In a given G.P. First term (a)=3, S4 =30 Let us find common ratio r, Sn = a(rn –1) r–1 or,S4 = 2(r4 –1) r–1 or,30 = 2(r2 +1)(r+1)(r–1) r–1
122 Vedanta Optional Mathematics Teacher's Guide ~ 10 or, 15=r3 +r2 +r+1 or, r3 +r2 +r–14=0 → For r = 2, 23 + 22 + 2 – 14 = 0 Hence,by factor theorem, (r–2) is a factor of r3 +r2 +r–14 Now, r3 –2r2 +3r2 –6r+7r–14=0 or, r2 (r–2)+3r(r–2)+7(r–2)=0 or, (r–2)(r2 +3r+7) =0 Either r–2=0 ⇒ r=2 or, r2 +3r+7=0 It doesnot give real valueof r. Again, last term, l=arn–1 or, 512=2.2n–1 or, 256=2n–1 or, 28 =2n–1 8=n–1 or, n=9 Then sum of the last four terms is given by ar5 +ar6 +ar7 +ar8 =960(True). ie. 2(25 +26 +27 +28 ) =2(32+64+128+256) =960 Questions for practice 1. Find the sum of the series 27+9+3+.......+ 1 81 . 2. How many terms are there in the series 1+2+4+.....+256. Calculate the sum. 3. A geometric series has 8 terms whose sum of the first three terms 7 and the sum of the sum of the last three terms is 224. Find the first term, common ratio and sum of the series. 4. In a geometric series if the fifth term is 8 times the second term and the sum of the first four terms is 60, then find the positive common ratio and the first term of the series. 5. In a G.P. of the sixth term is 16 times the second term and the sum of the first seven terms is 127 4 then find the positive common ratio and the first term of the series. 6. Find two numbers whose A.M. is 5 and G.M. 4. 7. Find two numbers whose A.M. is 85 and G.M. 40. 8. The sum of three numbers in A.P. 45 and if 17, 11 and 3 are subtracted from them respectively, the resulting numbers will be in G.P. Find the numbers. 9. Find the sum of the first 8 terms of a G.P. whose 5th and 8th terms are respectively 81 and 2187. 10. The ratio of the fifth term and eighth term of a G.P. is 1:8. If its seventh term is 64, find the sum of the first 10 ten terms.
Vedanta Optional Mathematics Teacher's Guide ~ 10 123 Estimated teaching periods : 4 1. Objectives S.N. Level Objectives (i) Knowledge(K) To define inequality To draw graph of linear equation To define linear programming problem. To define feasible region. (ii) Understanding(U) To draw graph of linear inequality To draw graph of set of inequalities. (iii) Application(A) To draw graph of system of linear inequalities to find feasible/convex region. To find the vertices of convex polygon. (iv) Higher Ability(HA) To find maximum and minimum values of an objective function under given conditions graphically. 2. Teaching materials Graph papers, pencils 3. Teaching Learning Strategies: – Review on real number lines. – Draw graphs of the following inequalities (as given in text book). x≥0, x≤0, x≥4, x≤4, y≥4, y≤4 – Discuss on the graph of x>4 and x≥4. – Define linear inequalities with examples – Discuss how to draw graph of 3x+4y ≤ 12, with origin test and shade solution region. – Define system of linear inequalities and their solution, on the basis of graphs of system of linear inequalities–define feasible region and convex polygon, find the vertices of convex polygon. – Define linear programming graphically. – Discuss about mathematical formulation of linear programming problems. (given in text book). Note : (i) The equation of boundary line corresponding to inequalities ax+by ≤ c and ax +by ≥ c. (ii) The inequalities x ≥ 0 and y ≥ 0 represent a feasible region on the first quadrant. (iii) x =0 represents y–axis and y=0 represent x–axis. (iv) Each of the boundary line is drawn on the graph paper. (v) To find the feasible plane region formed by given equatities ax + by +≤ c and ax + by 7 c, we use origin o(0,0) as a testing point, if the boundary line does not pass through the origin. Linear Programming four UNIT
124 Vedanta Optional Mathematics Teacher's Guide ~ 10 (vi) Find the feasible region by shading the area which is satisfied by all the constraints. (vii) Find the coordinates of all the vertices of feasible region / convex polygon. (viii) Find the value of the objective function at each vertex of the feasible region / convex polygon. (ix) Identify the value of x and y of the point for which the objective function has maximum or minimum values. (x) Write the conclusion at which point the objective function has maximum or minimum value. Some solved problems 1. Draw graph of 4y–5x ≥ 20,. Solution Given inequality is 4y–5x ≥ 20 The corresponding boundary line of given inequality is 4y–5x ≥ 20 when x=0, then, y=5 when y=0, then x = –4 Hence the boundary line passes through the points (–4,0) and (5,0). Take 0(0,0) as a testing point for 4y–5x ≥ 20 , we get, 0 ≥ 20 which is false. Hence the solution region is the half plane not containing the origin. Alternatively, The points on the boundary line –5x +4y = 20 can be obtained as x -4 + y 5 =1 which is in the form of x a + y b =1 Hence the boundary line passes through the points (–4,0) and (0,5). 2. Graph the following system of inequalities and find the vertices of convex polygon of exists. (a) x+y+2 ≥ 0 , y ≤ 2x+4, y ≤ 4–4x . Solution The boundary line equations of given system of inequalities are given by x+y+2=0..........(i) y = 2x+4..........(ii) y = 4–4x .........(iii) From boundary line equation (i) x+y+2=0 when x=0, they y=–2 o –5 –5 5 5 y y' x' x 4y–5x≥20
Vedanta Optional Mathematics Teacher's Guide ~ 10 125 when y=0, they x=–2 The boundary line (i) passes through the points (–2,0) and (0,–2). Take 0(0,0) as a testing point for x+y+2 ≥ 0, we get, 2≥0, which is true . Hence the solution of x+y+2 ≥ 0, is the half plane containing the origin. From the boundary line equation (ii), we get, y=2x+4, For x=0, y=4 and for y=0, x=–2 the boundary line (ii) passes through the points (0,4) and (–2,0). Take 0(0,0) as a testing point for y ≤ 2x+4 we get, 0≤4 which is true. Hence the solution of y≤2x+4 is the half plane containing the origin. Again from the boundary line (iii), we get, y = 4–4x For x=0, y=4 and for y=0, x=1. The boundary line (iii) passes through the points (0,4) and (1,0). Take 0(0,0) as a testing point for y≤4–4x. we get, 0≤4 which is true. Hence the solution of y≤4–4x is the half plane containing the origin. All the boundary lines are drawn on the graph the feasible region is ∆ABC whose vertices A(–2,0), B(2,–2) and C(0,4) which is shaded in the graph. (b) x+y≥3 and 2x–y≤4 Solution The boundary line equations of the corresponding inequalities are x+y=3...........(i) 2x–y=4..........(ii) From boundary line equation (i), we get, x 3 + y 3 =1. The boundary line equation (i) passes through the points (3,0) and (0,3). Take 0(0,0) as a testing point for x+y≥3, we get 0 ≥ 3 which is false. Hence solution of x+y≥ 3 is the half plane not containing the origin. Again from boundary equation (ii), we get, x 2 + y -4 =1 The boundary line (ii), passes through the points (2,0) and (0,–4). Take 0(0,0) as a testing o –4 –2 –4 –2 2 4 6 4 2 y y' x' x A C B o –4 –2 –4 3 3 y y' x' x+y=3 x 2x–y=4
126 Vedanta Optional Mathematics Teacher's Guide ~ 10 point for 2x–y≤4 , we get , 0≤4 which is true. Hence the solution of 2x–y≤4 is the half plane containing the origin. The boundary lines are plotted on the graph and common solution region is shaded in the graph which is not closed. The convex polygon does not exist. 3. Maximize z=2x+3y subject to the constraints 2x+y ≤ 14, x+2y ≤ 10, x ≥ 0, y ≥ 0. Solution The given linear constraints are 2x+y ≤ 14 ...........(i) x+2y ≤ 10 ............(ii) x ≥ 0, y ≥ 0 .........(iii) The corresponding linear equations are 2x+y = 14 ...........(i) x+2y =10 ............(ii) x = 0.......(iii) From equation(i), if x=0, then y=14 if y=0, then x=7 Hence the line (i) passes through the points (0,14) and (7,0). From equation (ii), we get, if x=0 then y=5 and if y=0, then x=10 Hence the boundary line (ii) passes through the points (0,5) and (10,0) For determination of the plane region if both of the linear constraints, we get, by using origin as testing point (0,0), put x=0, y=0, 2.0 +0 ≤14 ie. 0≤14 which is true also put the point in x+2y≤10, we get, 0≤10 which is true. Hence the solutions of both 2x+y ≤ 14 and x+2y ≤ 10 contains the origin. Here, x=0 represents equation of y–axis x ≥ 0 has solution of the right half plane of y–axis. All the boundary lines are plotted in the graph. the feasible region is quadrilateral OABCD with vertices O(0,0), A(7,0), B(6,2) and C(0,5). S.N. vertices value of z=2x+3y Remarks 1. O(0,0) 0 Minimum 2. A(7,0) 14 3. B(6,2)18 18 maximum 4. C(0,5) 15 Hence z has the maximum value 18 at B(6,2).
Vedanta Optional Mathematics Teacher's Guide ~ 10 127 o –5 5 10 –5 5 y y' x' x 10 15 x+2y=10 2x+y=14 C(0,14) x=0 C(0,5) B(6,2) A(7,0) y=0 4. Minimize F=2x+y subject to the constraints 2x+y ≤20, 2x+3y≤24. x ≥ 0, y ≥ 0. Solution The given linear constraints are 2x+y ≤20, 2x+3y≤24. x ≥ 0, y ≥ 0. The corresponding boundary line equations of above constraints are 2x+y =20...................(i) 2x+3y=24..................(ii) x= 0....................(iii) y = 0....................(iv) From equation (i), if x=0, then y=20 if y=0, then x=10 The boundary line equation(i) passes through the points (0,20) and (10,0). Taking O(0,0) for 2x+y ≤20, we get, 0≤ 20 which is true. Hence the solution of 2x+y ≤ 0 is the half plane containing the origin. From equation (ii), if x=0, they y=8, if y=0, then x=10. Hence the boundary line (ii) passes through the paints.(0,8) and (10,0) Taking O(0,0) as a testing paint for 2x+3y≤24, we get 0≤ 24, which is true. x=0 and y=0 represents y–axis and x–axis respectively. As x ≥ 0 and y ≥ 0 a feasible region lies in the first quadrant. Hence the solution of 2x+3y ≤ 24 is the half plane containing the origin. All the boundary lines are plotted in the graph. The feasible region of given system of constraints is quadrilateral OABC with vertices O(0,0), A(10,0), B(9,2) and C(0,8).
128 Vedanta Optional Mathematics Teacher's Guide ~ 10 S.N. vertices value of Objectives function F=2x+y Remarks 1. O(0,0) 0 Minimum 2. A(10,0) 20 Maximum 3. B(9,2) 20 maximum 4. C(0,8) 8 Hence the given objective function F has minimum value o at origin O(0,0). –5 o 5 10 –5 5 y y' x' x 10 15 20 2x+y=20 2x+3y=24 C(0,8) (0,20) x=0 B(9,2) A(10,0) y=0 5. Find the maximum and minimum values (extreme values) of F=16x–2y+40 subject to the constraints 3x+5y ≤ 24, 0≤ x ≤ 7, 0 ≤ y ≤ 4. Solution Given objective function is F=16x–2y+40 Given linear constraints are 3x+5y ≤ 24 0≤ x ≤ 7 ie, x ≥ 0, x≤7 0 ≤ y ≤ 4 ie. y ≥ 0, y≤4 The corresponding boundary line equations of above consttraints are
Vedanta Optional Mathematics Teacher's Guide ~ 10 129 3x+5y = 24,...................(i) x=7..................(ii) y=4....................(iii) x=0, y = 0....................(iv) From equation (i), if y=0, then x=8 if x=–2, then y=6 Hence the boundary line equation(i) passes through the points (8,0) and (–2,6). Taking O(0,0) for 3x+5y ≤24, we get, 0≤ 24, which is true. So, solution of 3x+5y ≤ 24 is the half plane containing the origin. x=7, is a line parallel to y–axis and x ≤ 7 has the solution left if x=7. y=4 is the boundary line parallel to x–axis and y≤4 has the solution lower half plane of y=4. As x ≥ 0 and y ≥ 0, a feasible region lies in the first quadrant. All the boundary lines are plotted in the graph and the feasible region is pentagon OABC with vertices O(0,0), A(7,0), B(7, 3 5 ) and C( 4 3 ,4) and D(0,4) S.N. vertices value of Objectives function F=2x+3y Remarks 1. O(0,0) 40 2. A(7,0) 152 Maximum 3. B(7, 3 5 ) 150.8 4. C( 4 3 ,4) 53.3 5. D(0,4) 32 Minimum From above table, we conclude that the maximum value is 152 at A(7,0) and minimum value 32 at (0,4). –2 o 5 10 –5 5 x' x 8 D(0,4) 3x+5y=24 B(7, 3 5 ) C( 3 5 , 4) A(7,0) y=4 x=7 6(a). In the given graph the coordinates of the points O, A, B, C, D and E are respectively
130 Vedanta Optional Mathematics Teacher's Guide ~ 10 (0,0), (2,0), (0,3). (0,5), ( 20 19 , 45 19 ), (5,0) respectively. OADB is feasible region. Find all the inequalities and minimize p=6x+10y+20 Solution In the graph feasible region is OADB. We have to find the equations of linear constraints. For line AC, it cuts x–axis at A(2,0) and y–axis at c(0,5). So equation of boundary line AC is x 2 + y 5 =1. or, 5x+2y ≤ 10 Let 5x+2y ≤ 10, taking O(0,0) as the testing point we get, 0 ≤ 10 which is true. So 5x+2y ≤ 10 is the inequality associated with line AC. For line BE, it cuts X–axis at E (5,0) and Y–axis at B (0,3).So equation of boundary line is given by x 5 + y 3 =1 or, 5x+3y ≤ 15 Let 5x+3y ≤ 15, taking O(0,0) as the testing point we get, 0 ≤ 15 which is true. So 5x+3y ≤ 15 is the inequality associated with boundary line BE. Feasible region is quadrilateral OADB which is the first quatrant. Hence x ≥0 and y ≥ 0 . Now, we find the minimum value of p =6x+10y+20. S.N. vertices value of Objectives function F=2x+3y Remarks 1. O(0,0) 20 2. A(2,0) 32 Maximum 3. B( 20 19 , 45 19 ) 50 4. B(0,4) 50 From above table, we conclude that P has minimum value is 20 at (0,0) . o 5 E 10 10 y y' x' x 5 C B A D
Vedanta Optional Mathematics Teacher's Guide ~ 10 131 6(b) text book 8b (Page 95) Solution In the graph ∆ABC is the feasible region is AB is parallel to the y–axis at distance of 5 units right of the y–axis. Hence its equation is x=5. . As feasible region is left of AB,we write x ≤ 5. F. For line AC, we have A(5,0) and C(0,5), equation of AC is given by x 5 + y 5 =1. or, x+y = 5 Let x+y = 5, taking testing point O(0,0), we get, 0 ≤ 5 which is true. Hence x+y = 5 is the linear constraint associated with line AC. Again for line OB it passes through the origin, we have O(0,0) and B (5,10). The equation of OB is given by y=mx It passes through the point (5,10), So we get, 10=m.5 m=2 Hence the equation of OB is y=2x. let 2x–y ≤0, take the testing point (1,0), we get, 2 ≤0, which is false . Hence required linear constraints is 2x–y ≥ 0 Questions for practice 1. Find the maximum value of the objective function Z=5x+8y subject to constraints 2x +5y ≤ 15, x ≥ 2,y ≥ 1. (Ans 33 at (5,1)) 2. Maximum the objective function F=x+2y subject to constraints 2x +y ≤ 14, x +2y ≤ 10 and x ≥ 0, y ≥ 0. (Ans 10 at (6,2)and (0,5)) 3. Minimize F=100x+600y subject to constraints x +y ≤ 10, x +3y ≤ 16 , x ≤ 3, y ≥ 2. (Ans 1500 at (3,2)) 4. Minimize the function F(x,y)=5x+2y subject to constraints 2x +y ≤ 4, x –2y ≤ 2 , x ≥ 0, y ≥ 0 (Ans 0 at (3,2)) 5. Find the extreme values of the function Z=5x+7y subject to constraints x + 2y ≤ 20, x +y ≤ 16 , x ≥ 0, y ≥ 0.
132 Vedanta Optional Mathematics Teacher's Guide ~ 10 Estimated teaching periods : 3 1. Objectives S.N. Level Objectives (i) Knowledge(K) To define a quadratic function To define a Cubic function To define a line of symmetry of a parabola. To write vartex of parabola with equation y=ax2 +bx+c, a ≠ 0. To define the point of intersection of a curve and a straight line. (ii) Understanding(U) To draw graph of a straight line (iii) Application(A) To draw graph of given equation of parabola. To draw graph of given Cubic function. To find solution of a parabola and a straight line graphically. To find the equation of parabola passing through given points. (iv) Higher Ability(H.A) To find equation of given curve of parabola. 2. Teaching materials i) Graph papers, functions ii) Chart papers with graph of parabola and cubic function. 3. Teaching learning strategies. – Review definitions of linear function, quadratic function and cubic function with graphs. – Discuss graph of a straight line like 4x+3y =12. – Discuss graph of a parabola ln the form of y=ax2 , put a=±1, ±2, ±3. – Discuss and conclude the conclusions about nature of curves. eg. Draw graph of y=x2 +4, y=4x2 –8x+3, y=(x+1)3 – Review the solution of a straight line and a quadratic equation by substitution method. – Discuss how to solve a linear equations and quadratic equation by using graph, with an example (y=x2 +2x–8, y=–5) – Let the student give to do some problems in each class as class work and the teacher supervise them and give necessary feedback. Notes : 1. General equation of parabola is y=ax2 +bx+c, a≠0 2. Equations of parabola vertex i) y=ax2 , a≠0 O(0,0) ii) y=a(x–h)2 +k, a≠0, (h,k) Quadratic Functions and Graphs five UNIT
Vedanta Optional Mathematics Teacher's Guide ~ 10 133 iii)y=ax2 +c a≠0, (0,c) iv) y=ax2 + bx+c, a≠0, – b 2a , 4ac–b2 4a v)y=ax2 +bx, a≠0, – b 2a ,–b2 4a 3. General Equation of cubic function is y=ax3 +bx2 +cx+d, a≠0 Some solved problems 1. Observe the given curve and answer the following questions. (a) Name the curve (b) Find the vertex of the curve (c) Write the coordinates of points M and N. (d) Name the line PQ which is parallel to y-axis. Solutions (a) The name of curve is parabola. (b) The vertex or turning point of the parabola is A(1,0) (c) The coordinates of M and N are (2,1) and (3,4) respectively. (d) The line PQ which divides the curve into two equal parts is called the line of symmetry. It is parallel to y–axis. Its equation in this figure is x=1. 2(a). Draw graph of the given linear equations 4x+3y=12 Solution Here, 4x+3y=12 or, 3y=12–4x or, y= 1 3 (12–4x) take x 0 3 6 –3 y 4 0 –4 8 The points (0,4), (3,0),(6,–4) and (–3,8) are plotted in a graph paper and joined them, we get a straight line. o –4 4 5 10 y P Q M N y' x' x o y y' x' x (6,–4 –5,8 0,4 3,0
134 Vedanta Optional Mathematics Teacher's Guide ~ 10 (b)y = 3x Solution, x 0 1 2 –1 –2 y 0 3 6 –3 –6 3. Find the graphs for the quadratic equations (a) y=x2 Solution Here, y=x2 represents a parabola take x 0 ±1 ±2 ±3 y 0 1 4 9 The points from above table are plotted in a graph paper and joined them, we get a curve called parabola. (b) y=–x2 Solution Here, y=–x2 it represents a parabola with vertex at the origin. x 0 ±1 ±2 ±3 y 0 –1 –4 –9 The points from above table are plotted in a graph paper and joined them, we get a curved as shown in the graph along side. o y y' x' x (–2,6) (–1,3) (2,6) (1,3) y=3x o y (3, 9) (1, 1) y=x2 (–3, 9) (–2, 4) y' x' x O y' (1, –1) (0,0) y=–x2 (–2,–4) y x' x
Vedanta Optional Mathematics Teacher's Guide ~ 10 135 (c) y=x2 +4 Solution Here, y=x2 +4, comparing it with y=ax2 +bx=c,we get, a=1, b=0, c=4 Vertex= - b 2a , 4ac-b2 4a = - 0 2.1 , 4.1.4-02 4.1 =(0,4) x 0 ±1 ±2 ±3 y 4 5 8 13 The points from above table are plotted in a graph paper and joined them, we get a curved graph along side. (d) y = x2–4 Solution Here, y=x2 –4 x ±1 0 ±2 ±3 ±4 y –3 –4 0 5 12 o y y' x' x (–1,5) (–2,8) (–3,13) (0,4) (1,5) (2,8) (3,13) o y y' x' x (–1,5) (–2,0) (0,–4) (–2,8) (2,0) y=x2 –4
136 Vedanta Optional Mathematics Teacher's Guide ~ 10 (e) y=(x+1)2 Solution Here, y=(x+1)2 . It is in the form of y=a(x–h)2 +k vertex =(h,k)=(–1,0), where , a=1, h=–1, k=0 Table x 0 1 2 –1 –2 –3 –4 y 1 4 9 0 1 4 9 The points from above table are plotted in a graph paper and joined them, we get a curved as shown along side. (f) y= 1 2 x2 Solution Here, y= 1 2 x2 . It is in the form of y=ax2 , where , a= y= 1 2 , vertex =(0,0) y= 1 2 x2 Table x 0 ±1 ±2 ±4 y 0 1 2 2 8 The points from above table are plotted in a graph paper and joined them, we get a curved as shown along side. 4. Draw the graphs for the following equations (a) y=x3 Solution Here, y=x3 . This is a equation of cubic function Table x –2 –1 0 1 2 y –8 –1 0 1 –8 The points from the above table are plotted on a graph and joined them, we get a curve as shown alongside. O y (–3,4) (1,4) (–1,0) y' x' x o y (4,8) (2,2) y= 1 2 x2 (–4,8) (–2,2) y' x' x O y y' x' x (–2,–8) (–1,–1) (2,8) (1,1)
Vedanta Optional Mathematics Teacher's Guide ~ 10 137 (b) y=–x3 Solution Here, y=–x3 ............(i). It is an equation of cubic function Table x –2 –1 0 1 2 y 8 1 0 –1 –8 The points from above table are plotted in a graph and joined them, we get a curve as shown the graph alongside. (c) y=(x+1)3 Solution Here, y=(x+1)3 ...............(i) Table x –3 –2 –1 0 2 y –8 –1 0 1 8 The points from the above table are plotted ia graph and joined them, we get a curve as shown in the graph alongside. O y y' x' x y=x3 O y y' x' x (–3,–8) (–2,–1) (–1,0) (1,8) (0,1)
138 Vedanta Optional Mathematics Teacher's Guide ~ 10 (d) y=–(x+1)3 Solution Here, y=–(x+1)3 .=..............(i) Table x –3 –2 –1 0 1 y 8 1 0 –1 –8 The points from the above table are plotted ia graph and joined them, we get a curve as shown in the graph alongside. 5.Draw the graphs for the following quadratic functions (a) y=4x2 – 8x+3 Solution This equation is in the form of y=bx2 –8b+c, where, a=4, b=8 and c=3. Vertex= - b 2a , 4ac-b2 4a = - -8 2.4 , 4.1.4-(-8)2 4.1 =(1,–1) Table x –3 –2 –1 0 1 2 3 y 69 35 15 3 –1 3 15 Plotting the above points in the graph paper and joining them, we get a curved curve as shown alongside. 0 2 4 6 8 –2 2 –2 –4 –6 –8 –4 4 6 y y' x' x (1,–8) (0,–1) (–2,1) (–3,8) (–1,0) O y y' x' x (1,–1) (0,3) (2,3) y=4x2 –8x+3 (–1,15) (3,15)
Vedanta Optional Mathematics Teacher's Guide ~ 10 139 (b) y=x2 +2x–8 Solution Here, y=x2 +2x–8..............(i). It is in the form of y=bx2 +8b+c , where, a=1, b=2 and c=–8. Vertex= – b 2a , 4ac-b2 4a = – 2 2.1 , 4.1.(–8)–22 4.1 =(–1,–9) x 1 2 3 0 –1 4 –2 –3 –4 –5 y –5 0 7 –8 –9 16 –8 –5 0 7 Plotting the points from above table and joining them, we get a curved curve as shown alongside. (c) y=–x2 –2x+5 Solution The equation is, y=–x2 –2x+5..............(i). It is in the form of y=bx2 +bx+c , where, a=–1, b=–2 and c=5. Vertex= - b 2a , 4ac-b2 4a = -2 2.1 , 4(-1)(5)-(-2)2 4×(-1) =(–1,6) Table x –5 –4 –3 –2 –1 0 1 2 3 y –10 –3 2 5 6 5 2 –3 –10 Plotting the points from above table and joining them, we get a curved curve as shown alongside. O y y' x' x (–4,0) (2,0) (–3,–5) (1,–5) (–2,–8) (0,–8) (–1,–9) (–5,7) (3,7) y=x2 +2x–8 O y y' x' x (1,2) (–4,–3) (2,–3) (–5,–10) (3,–10) (–1,6) (–2,5) (–3,2) (0,5) (3,7) y=x2 +2x–8
140 Vedanta Optional Mathematics Teacher's Guide ~ 10 (d) y=4x2 +8x+5 Solution The eqn. is y=4x2 +8x+5..............(i). It is in the form of y=ax2 –bx+c , where, a=4, b=8 and c=5. Vertex= - b 2a , 4ac-b2 4a = - -8 2.4 , 4×4×5-(8)2 4×(-1) =(–1,1) Table x –3 –2 –1 0 1 y 17 5 1 5 17 Plotting the points from above table and joining them, we get a curve as shown alongside. (e) y=–(x+3)2 Solution The eqn is y=–(x+3)2 ..............(i). or, y=–[x2 +6x+9] or, y= –x2 –6x–9..............(ii) To find vertex of parabola, Vertex= – b 2a , 4ac–b2 4a = 6 –2 , 4×9–(–6)2 –4 =(–3,0) eqn. (i) in table; x –6 –5 –4 –3 –2 –1 0 y –9 –4 –1 0 –1 –4 –9 O y y' x' x (–1,1) (–2,5) (0,5) 5 10 15 y=4x2 +8x+5 (–3,17) (1,17) O y' (0,–9) y=(x+3)2 (–6,–9) (–5,–4) (–1,–4) (–4,–1) (–3,0) (0,–1) y x' x
Vedanta Optional Mathematics Teacher's Guide ~ 10 141 6. Find the equation of parabola under the given conditions (a). Vertex at O(0,0) and passing through (2,3). Solution The eqn. of the parabola whose vertex is at origin is given by:– (i) y=ax2 The line/parabola passes through (2,3) 3=a.(2)2 or, 3 4=a a= 3 4 The eqn. is, y=ax2 put the value of a in the eqn. y= 3 4.x2 y= 3x2 4 is the required equation. 6.(b) Vertex =(2,–1), passing through (1,0). Solution Let, the eqn. of the parabola be, y=a(x–h)2 +k............(i) vertex=(h,k)=(2,–1) the eqn. is y=a(x–2)2 –1............(ii) The parabola passes through (1,0) Put value of x and y such that x=1.y=0, 0=a(1–2)2 –1 or. 0=a(–1)2 –1 or, 1=a Put the value of a in (ii) y=1(x–2)2 –1 or, y=x2 –4x+4–1 or, y=x2 –4x+3 y=x2 –4x+3, is the required eqn. 6.(c) Vertex =(–1,–4), passing through (2,5). Solution let, vertex =(–1,–4)=(h,k). Let, the eqn. of the parabola be,
142 Vedanta Optional Mathematics Teacher's Guide ~ 10 y=a(x–h)2 +k............(i) Put value of hand k in equation (i), we get y=a(x+1)2 +(–4) or, y=a(x+1)2 –4............(ii) (2,5) lies in the parabola. when x=2, y=5 in (ii) Put value of x and y such that x=1.y=0, 5=a(x+1)2 –4 or. 9=a(2+1)2 or, 9=a×9 a=1 Put the value of a in (ii) y=1(x+1)2 –4 or, y=x2 2x+1–4 or, y=x2 +2x33 y=x2 +2x–3, is the required eqn. 6.(d) passes through the points (2,4),(–1,–2) and (3,10). Solution let, the eqn. of parabola be : y=ax2 +bx+c............(i) (2,4) lies in eqn. (i) when x=2, y=4 in (i); 4=a.(2)2 +b(2)+c or, 4=a×4 +2b+c 4=4a +2b+c............(ii) (–1,–2) lies in eqn. (i) when x=–1, y=–2 in (i); –2=(2)2 a +b(–1)+c –2=a – b+c......(iii) (3,10) lies in eqn. (i) ie, when x=3, y=10 in (i); 10=a(3)2 +b(3)+c 10=9a +3b+c......(iv) Subtract (ii) from (iii) -2 = a - b + c 4 = 4a + 2b +c - - - - -6 = -4a+a-2b-b
Vedanta Optional Mathematics Teacher's Guide ~ 10 143 or, –6=–3a–3b or, a+b=2.............(v) Subtract (iv) from (iii) –2 = a – b + c 10 = 9a + 3b +c – – – – –12 =–9a+a – 3b–b or, –12=–8a–4b or, 3=2a+b 2a+b=3 ............. (vi) Subtract (vi) from (v); a + b = 2 2a + b =3 – – – –a = –1 a=1 Put the value of a in (iv); 3=2(1)+b b=1 Put the value of a and b in (iii); –2 = 1–1+c c=–2 Put the value of a, b and c in (i) y=1×x2 +1×x–2 or, y=x2 +x–2 which is the required equation. 6(e). passes through the points (1,4),(2,1) and (4,1). Solution let, the parabola represented by (a) (i) y=ax2 +bx+c............(a) (1,4) lies in eqn. (i) when x=1, y=4 in (i); 4=a.(1)2 +b(1)+c or, 4=1.a +1.b+c 4=a +b+c............(i) (2,1) lies in eqn. (a) when x=2, y=1 or, 1=a.(2)2 +2.b+c or, 1=4a +2b+c......(ii) Subtract (ii) from (i)
144 Vedanta Optional Mathematics Teacher's Guide ~ 10 4 = a + b + c 1 = 4a + 2b +c – – – – 3 = -3a -b 3a+b=–3.....................(iv) (4,1)0) lies in eqn. (i) ie, when x=4, y=1; 1=a.(4)2 +4.b+c 1=16a +4b+c......(iv) Subtract (iv) from (iii) 1 = 4a + 2b + c 1 = 16a + 4b +c – – – – 0 = -12a - 2b or, 6a + b =0 or, 3=2a+b.............(v) Subtract (iii) from (v); 6a + b = 0 3a + b =–3 – – + 3a = 3 a=1 Put the value of a in (v); 6 × 1 +b=0 b=–6 Put the value of a , b and c in (i); a + b+ c =4 or, 1–6+c=4 or, –5+c =4 c= 9 Put the value of a, b and c in (a) y=1×x2 – 6×x +9 or, y=x2 – 6x + 2 y=x2 – 6x + 2 which is the required equation. 7. (a) From the given graphs of parabola, find their equations. Solution From the graph, it can be cleared that (0,0) is the vertex of the parabola (1,1),(2,4) and (3,9) lies in the parabola. Since the vertex is origin, the eqn. is given by,
Vedanta Optional Mathematics Teacher's Guide ~ 10 145 y=ax2 ............(i) (1,1) lies in the parabola, when y=1, x=1, 1=a×1 a=1 The value of a also satisfies the condition For (2,4) and (3,9) according to the equation. The eqn. of the parabola y=1×x2 i.e., y=x2 (b) Solution here, vertex =(–1,) let, (–1,–4) = (h,k) Also, (–2,–3),(0,–3), (–3,0), (1,0) and (–4,5) lies in the parabola. let, the eqn. of the parabola be y=a(x–h)2 + k...............(i) where, (h,k)=vertex Then, When h=–1, k=–4 in (i), y=a(x+1)2 + k...............(ii) (–2,–3) lies in the parabola, when x=–2, y=–3 in (ii); –3=a(–2+1)2 –4 or, 1 = a×1 a=1 put the value of a , b and k in (i); y=1(x+1)2 –4 or, y=x2 +2x+1 – 4 or, y=x2 +2x – 3 which is the required eqn.
146 Vedanta Optional Mathematics Teacher's Guide ~ 10 Solution of linear and Quadratic Equations Some solved problems 1. Solve the following equations by substitution method. (a) y=x2 – 3x and y=x Solution Given equation are y=x2 – 3x ............(i) and y=x..............(ii) put the value of y in equation (i), from (ii), we get, x=x2 – 3x or, x2 – 4x =0 or, x(x–4)=0 Either x=0 or, x=4 x=0,4 when x=0, then y=0 when x=4, then y=4 The required solutions are (0,0) and (4,4). (b) y=x2 +7x+12. x–y+7 =0 Solution Given equation are y=x2 +7x+12 ............(i) or, . x–y+7 =0..............(ii) From equation (ii) y=x+7 put the value of y in equation (i), from (ii), we get, x +7=x2 +7x+12 or, x2 +6x +5 =0 or, x2 +5x +x +5 =0 or, x(x+5) +1(x+5)=0 or, (x+5)(x+1)=0 Either x + 5 = 0, x=–5 or, x + 1=0 x=–1 x=0,4 when x=–5, then y=–5 + 7 =2 when x=–1, then y=–1 +7 =6 Hence the required solution (–5,2) and (–1,6).
Vedanta Optional Mathematics Teacher's Guide ~ 10 147 2. Solve the folowing equations graphically (a) x2 + 6x + 8 =0 Solution Let y=x2 +6x+8=0, i.e. y=0 y=x2 +6x+8=0 represents an equation of parabola, comparing at with y=ax2 +bx+c, we get, a=1, b=6 and c=8 First let us find the vertex of the parabola. x–coordinate of vertex (h)=– b 2a =– 6 2.1 =–3 y–coordinate of vertex (k) = 4ac -b2 4a = 4×1×8 -62 4.1 = 32 -36 4 =–1 Vertex (h,k)=(–3,–1) Table to draw curve of parabola x –5 –4 –3 –2 –1 0 1 y 0 –1 0 3 8 15 From the above curve we observed that the curve cuts the x–axis at points (–2,0) and (–4,0). x = –2,–4 Alternate Method Here, x2 + 6x +8 =0 or, x2 =–6x=8 Let y=x2 = –6x–8 Then, y=x2 ................(i) and y=– 6x–8.............(ii) The equation(i) represents a parabola and (ii) represents a straight line. From equation (i) x ±4 ±3 ±2 ±1 0 y 16 9 4 1 0 plotting the above points on graph paper and joining them, we get a curve of parabola. From equation (ii) y=–6x–8 x –1 –2 –3 y –2 4 10 plotting there points on the same graph, we get a straight line. The straight line cuts O y y' x' –5 x –3 5 5 10 15 y=x2 +6x+8 O y y' x' –6 –4 –2 x 10 8 6 4 2 2 4 6 (–4,16) 16 (–2,4) 14 12
148 Vedanta Optional Mathematics Teacher's Guide ~ 10 the curve of parabola at points (–2,4) and (–4,16). Hence the required solution of given equation area =–2,–4 (b) x2 – x – 2 =0 Solution Let x2 – x – 2 =0, i.e. y= x2 – x – 2, y=0 It represents an equation of parabola, in the term of y=ax2 +bx+c, where, a=1, b=–1,c=–2 Let us find the vertex of the parabola. Vertex (h,k)= - b 2a , 4ac-b2 4a = - (-1) 2.1 , 4.1.(-2)-(-1)2 4.1.(-2) = 1 2 ,- 9 4 To draw a curve of parabola x –2 –1 0 1 2 3 y 4 0 –2 –2 0 4 The points from the above table are plotted in a graph and joined them, we get a curve of parabola. o –5 –4 –3 –2 –1 1 2 3 4 5 –1 –2 –3 3 2 1 x' x 4 5 6 y=x2 –x+2 1 2 ,– 9 4 The curve cuts x–axis at points (–1,0) and (2,0). Hence the required solutions are x=–1,2.
Vedanta Optional Mathematics Teacher's Guide ~ 10 149 3. Solve the following equations graphically. (a) y= x2 – 5 and y=2x+3 Solution Given equations are y= x2 – 5 ......................(i) and y=2x+3 ................(ii) Equation (i) represents a parabola and (ii) represents a straight line. The point of the intersection of the curve of parabola and straight line is the solution of given equations. From equation (i), Vertex (h,k)= - b 2a , 4ac-b2 4a = - 0 2.1 , 4.1.(-5)-0 4.1 =(0,–5) To draw the curve of parabola x –3 –2 –1 0 1 2 3 y 4 –1 –4 –5 –4 –1 4 Plotting the points on graph and joining them we get a curve of parabola. From equation (i) y=2x + 3 x –3 0 1 y 3 3 5 o –5 5 –1,–1 –5 5 x' x 10 4,11 y=x2 –5
150 Vedanta Optional Mathematics Teacher's Guide ~ 10 From the graph, we observed that the point of intersection of the parabola and the straight lines are (–2,–1) and (4,11). Hence the required solutions are (–2,–1) and (4,11). (g) y=4x2 +8x +5 ,x + y=3 Solution Given equations are 4x2 +8x +5 ......................(i) and x + y=3................(ii) In above equation, equation (i) represents a parabola and equation (ii), represents a straight line. From equation (i), y=4x2 +8x +5 x –3 –2 –1 0 1 y 17 5 1 5 17 Vertex of parabola (h,k) = – b 2a , 4ac–b2 4a = – 8 2.4 , 4.4.5–82 4.4 =(–1,1) Then plotting above points from table and joining them, we get a curve of parabola. From equation(ii), y =3–x x 0 –1 3 y 3 4 0 plotting these points in same graph, we get a straight line. From the graph, the curve and the straight lines intersects at points (–2,5) and – 1 4 , 13 4 . Not e. I f t h e p o i n t o f intersection of a curve and a straight line is not clear to read their coordinates, we can solve them by substitution method in rough. o –5 5 –5 (–2,5) 5 y y' x' x 10 15 20 – 1 4 , 13 4