1.Algebra Booster

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Matrices and Determinants 7.79 0 1 where B = Ê ˆ Á Ë0 0 ˜ ¯ Since B 2 = O, we get r B = O" r ≥2 2005 1 2005 Thus, A = I + 2005B= Á Ë0 1 ˜ ¯ 51. Given, È1 0 0˘ A = Í 2 1 0 ˙ Í ˙ ÍÎ3 2 1˙˚ fi |A| = 1 π 0 fi A –1 exists Thus, Ê 1 0 0ˆ 1 A - = Á-2 Á 1 0˜ ˜ Ë 1 -2 1¯ Ê1ˆ Ê 1 0 0ˆÊ1ˆ Ê 1ˆ Now, -1 U1 = A Á0˜ = Á- 2 1 0˜Á0˜ = Á-2˜ Á ˜ Á ˜Á ˜ Á ˜ Ë0¯ Ë 1 -2 1¯Ë0¯ Ë 1¯ Ê2ˆ Ê 1 0 0ˆÊ2ˆ Ê 2ˆ -1 U2 = A Á3˜ = Á- 2 1 0˜Á3˜ = Á-1˜ Á ˜ Á ˜Á ˜ Á ˜ Ë0¯ Ë 1 -2 1¯Ë0¯ Ë-4¯ Ê2ˆ Ê 1 0 0ˆÊ2ˆ Ê 2ˆ -1 U3 = A Á3˜ = Á- 2 1 0˜Á3˜ = Á-1˜ Á ˜ Á ˜Á ˜ Á ˜ Ë1¯ Ë 1 -2 1¯Ë1¯ Ë-3¯ Ê 1 2 2ˆ (i) Thus, U = Á-2 -1 -1˜ Á ˜ Ë 1 -4 -3¯ 1 2 2 fi | U | =-2 -1 - 1 = 3 1 -4 -3 Ê 1 2 2ˆ (ii) Given, U = Á-2 -1 -1˜ Á ˜ Ë 1 -4 -3¯ fi Ê-1 -7 9ˆ 1 1 U - = Á-2 -5 6˜ 3Á ˜ Ë 0 -3 3¯ Thus, the sum of the elements of U –1 = 1 ( - 18 + 18) = 0 3 (iii) We have, È3˘ [3 2 0] U Í 2 ˙ Í ˙ ÍÎ0 ˙˚ Ê 1 2 2ˆ Ê3ˆ = (3 2 0) Á-2 -1 -1˜ Á2˜ Á ˜ Á ˜ Ë 1 -4 -3¯ Ë0¯ Ê 7ˆ = (3 2 0) Á-8˜ Á ˜ Ë-5¯ = (21 – 16) = (5) 53. We have, 1 -2 3 D =-1 1 - 2 = 0 1 -3 4 and the determinant 1 3 -1 0 1 k -1 D1 =-1 - 2 k =-1 - 2 k = 3-k 1 4 1 0 3 k + 1 Ê R1Æ R1+ R2ˆ Á ËR Æ R + R ˜ ¯ 3 2 3 If k π 3, D = 0, D π 0, therefore, the system of equations has no solutions. 54. (i) Total number of matrices = 9 + 3 = 12 We have Ê111ˆ Ê011ˆ Ê011ˆ A1= Á1 0 0 ˜, A2= Á1 1 0 ˜, A3= Á1 0 0˜ Á ˜ Á ˜ Á ˜ Ë100¯ Ë100¯ Ë101¯ Ê1 0 1ˆ Ê0 0 1ˆ Ê0 0 1ˆ A4= Á0 0 1 ˜, A5= Á0 1 1 ˜, A6= Á0 0 1˜ Á ˜ Á ˜ Á ˜ Ë1 1 0¯ Ë1 1 0¯ Ë1 1 1¯ Ê1 1 0ˆ Ê0 1 0ˆ Ê0 1 0ˆ A7= Á1 0 1 ˜, A8= Á1 1 1 ˜, A9= Á1 0 1˜ Á ˜ Á ˜ Á ˜ Ë0 1 0¯ Ë0 1 0¯ Ë0 1 1¯ Ê1 1 0ˆ Ê010ˆ Ê010ˆ A10 = Á1 0 1˜, A11 = Á1 1 1 ˜, A12 = Á1 0 1˜ Á ˜ Á ˜ Á ˜ Ë0 1 0¯ Ë010¯ Ë011¯ Alternate method: If two zeros are the entries in the diagonal, then 3 C 2 ¥ 3 C 1 If all the entries in the principle diagonal is 1, then 3 C 1 . Thus, the total number of matrices = 12. (ii) Here, |A 2 | π 0, |A 3 | π 0, |A 4 | π 0 and |A 5 | π 0, |A 7 | π 0, |A 9 | π 0, Thus, there are six matrices A such that Êxˆ Ê1ˆ AÁy˜ = Á0˜ has a unique solution. Á ˜ Á ˜ Ëz¯ Ë0¯
7.80 Algebra <strong>Booster</strong> 55. (iii) Let Êxˆ Ê1ˆ X = Áy˜ and B= Á0˜ Á ˜ Á ˜ Ëz¯ Ë0¯ Thus, A 1 X = B has infinite number of solutions A 6 X = B has no solution A 8 X = B has no solution A 10 X = B has no solution A 12 X = B has infinite number of solutions We know that, det(AdjA) = (detA) n–1 and B is a skew-symmetric matrix, so det (B) = 0 det (adj B) = det B) 3–1 = (det B) 2 = 0 now, 2k - 1 2 k 2 k | A| = 2 k 1 -2k -2 k 2k -1 2k - 1 2 k 2 k = 0 1+ 2 k -(1+ 2 k) ( R2Æ R2 + R3) -2 k 2k -1 2k - 1 2 k 4 k = 0 1+ 2k 0 ( C3Æ C3+ C2) -2 k 2k 2k -1 = (2k – 1)(4k 2 – 1) + 8k(1 + 2k) = (2k + 1)[(2k – 1) 2 + 8k] = (2k + 1)[4k 2 + 4k + 1] = (2k + 1)(2k + 1) 2 = (2k + 1) 3 It is given that, det (adj A) + det (adj B) = 10 6 fi (2k + 1) 6 + 0 = 10 6 fi (2k + 1) 6 = 10 6 fi (2k + 1) = 10 fi k = 4.5 Thus, the value of [k] = [4.5] = 4 56. The given system of equations can be written in matrix form as AX = B It has either (i) a unique solution or (ii) infinite solutions or(iii) no solution. Thus, there can not exist any matrix A such that Èx˘ È1˘ A Í y ˙ Í 0 ˙ Í ˙ = Í ˙ has two distinct solutions. ÍÎz ˙˚ ÍÎ0 ˙˚ 58. Given MN = NM Now, M 2 N 2 (M T N) –1 (MN –1 ) T = M 2 N 2 N –1 (M T N) –1 (MN –1 ) T = M 2 N ◊ (M T ) –1 (N –1 ) T M T = –M 2 N ◊ (M) –1 (N T ) –1 M T = +M 2 N ◊ (M) –1 N –1 M T = –MNMM –1 N –1 M = –MNN –1 M = –M 2 Êa b cˆ 59. Let M = Ád Á e f˜ ˜ Ëg h i¯ Êa b cˆÊ0ˆ Ê-1ˆ Now, Ád e f˜Á1˜ = Á 2˜ Á ˜Á ˜ Á ˜ Ëg h i¯Ë0¯ Ë 3¯ fi b = –1, e = 2, h = 3 Êa b cˆÊ 1ˆ Ê 1ˆ Also, Ád e f˜Á- 1˜ = Á 1˜ Á ˜Á ˜ Á ˜ Ëg h i¯Ë 0¯ Ë-1¯ fi a = 0, d = 3, g = 2 Êa b cˆÊ1ˆ Ê 0ˆ and, Ád e f˜Á1˜ = Á 0˜ Á ˜Á ˜ Á ˜ Ëg h i¯Ë1¯ Ë12¯ fi g + h + i = 12 fi i = 7 Thus, sum of the main diagonal = a + e + i = 0 + 2 + 7 = 9 60. Given P T = 2P + I fi (P T ) T = (2P + I) T fi P = (2P T + I) fi P = 2(2P + I) + I fi 3(P + I) = O fi (P + I) = O fi PX + X = O fi PX = –X Ê1 4 4ˆ 61. Given adj ( P) = Á2 Á 1 7˜ ˜ Ë1 1 3¯ fi 1 4 4 |adj ( P )| = 2 1 7 = 4 1 1 3 As we know that, |adj(P)| = |P| 3–1 = |P| 2 Thus, |P| 2 = 4 fi |P| = 2, –2
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Algebra Booster with Problems & Sol
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Algebra Booster with Problems & Sol
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Dedicated to light of my life (Rush
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viii Preface I owe a special debt o
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x Contents Descartes Rule of Signs
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xii Contents Chapter 8 Probability
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1.2 Algebra Booster (ii) a 1 - k, a
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1.4 Algebra Booster fi a + (n + 1)d
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1.6 Algebra Booster (i) Maximum Val
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1.8 Algebra Booster 25. In an AP, (
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1.10 Algebra Booster b 86. If b = a
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1.12 Algebra Booster 145. Find the
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1.14 Algebra Booster and n= 0 2n 2n
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1.16 Algebra Booster 51. In a serie
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1.18 Algebra Booster 37. Observing
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1.20 Algebra Booster 7. If a, b and
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1.22 Algebra Booster (B) (C) (D) If
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1.24 Algebra Booster 22. No questio
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1.26 Algebra Booster 17 1 50. Ê ˆ
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1.28 Algebra Booster 8. 3 - 1 2 9.
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1.30 Algebra Booster fi a(a 2 - d 2
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1.32 Algebra Booster On subtraction
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1.34 Algebra Booster 1 1 1 1 fi - =
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1.36 Algebra Booster 61. We have (a
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1.38 Algebra Booster fi 3 n > 14001
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1.40 Algebra Booster and 2 sin n n=
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1.42 Algebra Booster 102. We have,
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1.44 Algebra Booster 115. It is giv
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1.46 Algebra Booster Now, Ê a + b
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1.48 Algebra Booster 136. (i) We ha
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1.50 Algebra Booster 152. Let t n 3
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1.52 Algebra Booster fi 161. As we
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1.54 Algebra Booster 177. We have (
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1.56 Algebra Booster 193. We know t
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1.58 Algebra Booster Thus, (1 + x)(
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1.60 Algebra Booster Alternate meth
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1.62 Algebra Booster fi 1007d = a -
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1.64 Algebra Booster 2 2 fi Ê 1 1
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1.66 Algebra Booster and b + br = 9
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1.68 Algebra Booster = 1 ◊ cos(1
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1.70 Algebra Booster 6. We have 1 1
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1.72 Algebra Booster fi fi Dividing
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1.74 Algebra Booster fi 7 4 4 4 (1
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1.76 Algebra Booster n 35. We have
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1.78 Algebra Booster 5. We know tha
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1.80 Algebra Booster Hence, the val
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1.82 Algebra Booster Since the equa
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1.84 Algebra Booster 2 S2 = = 3 1 1
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1.86 Algebra Booster 37. Let a and
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1.88 Algebra Booster fi fi Ê n Ê
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CHAPTER 2 Quadratic Equations and E
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Quadratic Equations and Expressions
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CHAPTER 3 Logarithm 1. INTRODUCTION
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Logarithm 3.3 = log 2 (4) = log 2 (
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Logarithm 3.5 x fi Ê1ˆ Á = 3 Ë
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Logarithm 3.7 fi fi 10 log10x log10
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Logarithm 3.9 16. If a 2 + b 2 = 7a
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Logarithm 3.11 2 log 2 + log 3 y =
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Logarithm 3.13 (Q) (R) (S) The valu
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Logarithm 3.15 1 1 13. { , 2 4} 14.
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Logarithm 3.17 Also, (a - b) = log
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Logarithm 3.19 Again, log a b = 2 f
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Logarithm 3.21 Now, log( a + c) + l
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Logarithm 3.23 fi Ê1 2 ˆ log 3/4
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Logarithm 3.25 fi 2x 2 = 12 fi x 2
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Logarithm 3.27 fi fi 1 2 x 2 = 3 3,
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Logarithm 3.29 = log 10 (64 ¥ 31)
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Logarithm 3.31 fi 2x = 8, 4 = 2 3 ,
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4.2 Algebra Booster EXAMPLE 2: The
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4.4 Algebra Booster (i) If z lies i
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4.6 Algebra Booster (iii) w 3 = 1 (
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4.8 Algebra Booster Note The sum of
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4.10 Algebra Booster In an equilate
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4.12 Algebra Booster 3. Straight Li
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4.14 Algebra Booster (vii) We consi
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4.16 Algebra Booster 55. If |z 1 +
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4.18 Algebra Booster 8. The area of
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4.20 Algebra Booster 53. The number
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4.22 Algebra Booster 49. Find the r
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4.24 Algebra Booster 30. Resolve z
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4.26 Algebra Booster 1. The value o
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4.28 Algebra Booster 12. Show that
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4.30 Algebra Booster X¢ P(-1, 0) Y
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4.32 Algebra Booster 41. (d) 42. (c
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4.34 Algebra Booster HINTS AND SOLU
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4.36 Algebra Booster Hence, the val
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4.38 Algebra Booster p q fi = = l(s
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4.40 Algebra Booster p fi < 2q< p 2
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4.42 Algebra Booster 2 2Êq1- q2ˆ
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4.44 Algebra Booster x◊ 3 p + y
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4.46 Algebra Booster 92. We have x
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4.48 Algebra Booster fi fi fi fi Ê
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4.50 Algebra Booster Putting z 3 =
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4.52 Algebra Booster i e p 122. Let
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4.54 Algebra Booster fi fi 3(2 + co
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4.56 Algebra Booster 19. Let z = r(
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4.58 Algebra Booster 2 1 w w = + +
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4.60 Algebra Booster fi Ê 2p 4p 6p
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4.62 Algebra Booster A B 50. Given
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4.64 Algebra Booster fi 2 2 (3x + y
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4.66 Algebra Booster fi fi |(x - 4)
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4.68 Algebra Booster Comparing the
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4.70 Algebra Booster = |(cos (3q) -
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4.72 Algebra Booster Thus, 18. Let
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4.74 Algebra Booster 2 Ê a ˆ = 2
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4.76 Algebra Booster 1È Ê3pˆ Ê5
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4.78 Algebra Booster 12 Ê2k + 1ˆ
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4.80 Algebra Booster 16. We have, s
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4.82 Algebra Booster 3 fi x =± 2 T
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4.84 Algebra Booster fi fi 4 Ê3z -
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4.86 Algebra Booster 53. fi 2 2 2 (
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4.88 Algebra Booster Now, 1 iq -iq
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CHAPTER 5 Permutations and Combinat
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Permutations and Combinations 5.3 (
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Permutations and Combinations 5.5 =
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Permutations and Combinations 5.7 5
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Permutations and Combinations 5.9 1
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Permutations and Combinations 5.11
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6.2 Algebra Booster 5. Subtracting
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6.4 Algebra Booster Proof 1. ( a +
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6.6 Algebra Booster EXERCISES LEVEL
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6.8 Algebra Booster 80. Find the su
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6.10 Algebra Booster 10. In the exp
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6.12 Algebra Booster 11. Find the c
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6.14 Algebra Booster 64. Find the c
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6.16 Algebra Booster 43. Find the s
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6.18 Algebra Booster 3. Match the f
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6.20 Algebra Booster (a) 0 (c) (-1)
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6.22 Algebra Booster LEVEL IV COMPR
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6.24 Algebra Booster 30 2 30 2 2 30
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6.26 Algebra Booster Now, (33 43 -
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6.28 Algebra Booster 57. We have, 1
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6.34 Algebra Booster Comparing the
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6.36 Algebra Booster 1 1 1 1 + + +
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6.38 Algebra Booster 127. We have,
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6.40 Algebra Booster n n 2 n 3 7. W
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6.42 Algebra Booster n r n r n r n
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6.44 Algebra Booster n n Â Ck k =
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6.46 Algebra Booster A1 Ê 1 1 1 1
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6.48 Algebra Booster Multiplying Eq
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6.50 Algebra Booster Ê12 ˆ Middle
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6.52 Algebra Booster 72. Let t n 1
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6.54 Algebra Booster Putting x = 1,
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6.56 Algebra Booster Â Â Thus, 2I
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6.58 Algebra Booster fi fi 2n 2n Ê
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6.60 Algebra Booster Cn ( ,4) 36. L
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6.62 Algebra Booster Thus, S = t 1
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6.64 Algebra Booster 7. We have 10
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6.66 Algebra Booster = ( 49 C 4 + 4
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6.68 Algebra Booster Differentiatin
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6.72 Algebra Booster 51. Let the th
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7.2 Algebra Booster (vii) Identity
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7.4 Algebra Booster (xii) If A is s
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7.6 Algebra Booster where Proof: fi
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7.8 Algebra Booster Replace A by ad
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7.10 Algebra Booster 10. Unitary ma
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7.12 Algebra Booster 46. Expand the
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7.14 Algebra Booster 88. If A be a
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7.16 Algebra Booster 7. For non-zer
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7.18 Algebra Booster 2x + 4p p + 6a
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7.20 Algebra Booster 4. Prove that
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7.22 Algebra Booster 4. If S r = a
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7.24 Algebra Booster 2 2 2 2 2 2 a
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7.26 Algebra Booster The value of (
Page 455 and 456: 7.28 Algebra Booster 17. The determ
Page 457 and 458: 7.30 Algebra Booster 43. If a 0 A
Page 459 and 460: 7.32 Algebra Booster 65. Let M be a
Page 461 and 462: 7.34 Algebra Booster 12. Then AB =
Page 463 and 464: 7.36 Algebra Booster Êa bˆÊ1 2ˆ
Page 465 and 466: 7.38 Algebra Booster 1 a a 47 The g
Page 467 and 468: 7.40 Algebra Booster 2 2 2 1 0 = (1
Page 469 and 470: 7.42 Algebra Booster Thus, D1 ( d -
Page 471 and 472: 7.44 Algebra Booster 77. We have, 2
Page 473 and 474: 7.46 Algebra Booster 90. We know th
Page 475 and 476: 7.48 Algebra Booster 108. We have,
Page 477 and 478: 7.50 Algebra Booster 118. We have,
Page 479 and 480: 7.52 Algebra Booster 4. We have, 2
Page 481 and 482: 7.54 Algebra Booster = ([x] + [y] +
Page 483 and 484: 7.56 Algebra Booster It is given th
Page 485 and 486: 7.58 Algebra Booster 27. The given
Page 487 and 488: 7.60 Algebra Booster and 12 -3 5 D1
Page 489 and 490: 7.62 Algebra Booster fi Ê a ˆ Ê
Page 491 and 492: 7.64 Algebra Booster a b c b c a =
Page 493 and 494: 7.66 Algebra Booster 2bc -2c -2b 2
Page 495 and 496: 7.68 Algebra Booster 1 1 1 1 = 2 n
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Page 499 and 500: 7.72 Algebra Booster fi (49 - 42)si
Page 501 and 502: 7.74 Algebra Booster fi p + q + r =
Page 503 and 504: 7.76 Algebra Booster 34. We have fi
Page 505: 7.78 Algebra Booster where a 2 + b
Page 509 and 510: 7.82 Algebra Booster fi 2 2 (1 + a)
Page 511 and 512: 8.2 Algebra Booster For example, th
Page 513 and 514: 8.4 Algebra Booster (ii) P(AB) £ P
Page 515 and 516: 8.6 Algebra Booster (ii) a black ca
Page 517 and 518: 8.8 Algebra Booster 64. If two dice
Page 519 and 520: 8.10 Algebra Booster ferred from th
Page 521 and 522: 8.12 Algebra Booster 168. The proba
Page 523 and 524: 8.14 Algebra Booster 28. A fair coi
Page 525 and 526: 8.16 Algebra Booster product is 0.7
Page 527 and 528: 8.18 Algebra Booster event that thr
Page 529 and 530: 8.20 Algebra Booster Questions aske
Page 531 and 532: 8.22 Algebra Booster The probabilit
Page 533 and 534: 8.24 Algebra Booster 72. A is targe
Page 535 and 536: 8.26 Algebra Booster 5, 6, 7. A car
Page 537 and 538: 8.28 Algebra Booster 137. Ê 9 m ˆ
Page 539 and 540: 8.30 Algebra Booster 24. 25. 3 10 1
Page 541 and 542: 8.32 Algebra Booster So, the probab
Page 543 and 544: 8.34 Algebra Booster (iv) Let D be
Page 545 and 546: 8.36 Algebra Booster Hence, the req
Page 547 and 548: 8.38 Algebra Booster 58. Here, S =
Page 549 and 550: 8.40 Algebra Booster 81. Here, S =
Page 551 and 552: 8.42 Algebra Booster The probabilit
Page 553 and 554: 8.44 Algebra Booster Thus, 1 36 =
Page 555 and 556: 8.46 Algebra Booster 128. Hence, th
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8.48 Algebra Booster tively and let
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8.50 Algebra Booster Hence, the pro
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8.52 Algebra Booster Ê 5 1 ˆ = 1-
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8.54 Algebra Booster We are interes
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8.56 Algebra Booster 193. Clearly,
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8.58 Algebra Booster 15. Let E be t
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8.60 Algebra Booster = P( A or CA o
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8.62 Algebra Booster 20. Out of 2 c
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8.64 Algebra Booster Clearly, a = 5
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8.66 Algebra Booster = (0.17) ¥ (0
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8.68 Algebra Booster 1 Ê13ˆÊ1ˆ
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8.70 Algebra Booster 46. We have, P
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8.72 Algebra Booster 3 2 3 3 2 1 1
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8.74 Algebra Booster and the number
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8.76 Algebra Booster 88. Let G = Or
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