1.Algebra Booster

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Binomial Theorem 6.27 n Now, ( P + Q) -( P - Q) = 1 + f – f ¢ Since n is odd, the LHS of the above expression contains an even powers of P. Hence, the terms in LHS and I are integers. Therefore, f – f¢is an integer. But 0 < f < 1 and –1 < f¢ < 0 fi 0 < f < 1 and –1 < –f¢ < 0 fi –1 < f – f¢ < 1 fi f – f¢ = 0 fi f = f¢ Thus, ( I + f) f ¢ = ( n P + Q) ( n P -Q) = {( P + Q)( n P -Q)} = (P – Q 2 ) n = k n 51. Given Q - P< 1 fi 0 < ( Q - P) < 1 n n Let ( Q - P) = f ¢ , where 0 < f¢ < 1. Now, ( P + Q) n + ( n P - Q) = I + f + f ¢ = Even integer But 0 < f < 1 and 0 < f¢ < 1 fi 0 < f + f¢ < 2 fi f + f¢ = 1 fi f¢ = 1 – f Hence, (I + f)(I – f) = (I + f)f¢ = ( P + Q) n ¥ ( n P -Q) = (P – Q 2 ) n = k n 52 Let I and f denote the integral and fractional part of R, respectively. Given f = R – [R] 2n+ 1 Let R = I + f = (5 5 + 11) , 0 < f < 1 2n+ 1 Also, f ¢ = (5 5 -11) , 0 < f¢ < 1 Now, n R – f¢ = (5 5 + 11) - (5 5 -11) = an even integer. But 0 < f < 1 and 0 < f¢ < 1 fi 0 < f < 1 and –1 < –f¢ < 0 fi –1 < f – f¢ < 1 fi f – f¢ = 0 fi f = f¢ Therefore, Rf = Rf¢ n = (5 5 + 11) ◊(5 5 -11) = (125 – 121) 2n+1 = 4 2n+1 n 2 + 1 2n+ 1 2 + 1 2n+ 1 53. Let x = I + f = (8 + 3 7) , where 0 < f < 1, n and f ¢ = (8 - 3 7) , 0 < f ¢ < 1. n Now, 1 + f + f¢ (8 3 7) n n = + + (8 -3 7) = an even integer. But 0 < f < 1 and 0 < f¢ < 1 fi 0 < f + f ¢ < 2 fi f + f ¢ = 1 Therefore, I + f + f ¢ = 2k fi I + 1 = 2k fi I = 2k – 1 fi [x] = 2k – 1 Now, x – x 2 + x[x]=x – x(x – [x]) = x – xf = x(1 – f) = xf¢ n n = (8 + 3 7) ◊(8 -3 7) = (64 – 63) n = 1 n = 1 n 54. Given a + b = (8 + 3 7) , where 0 < b < 1. n Let b¢ = (8 - 3 7) , where 0 < b¢ < 1 . Now, a + b + b ¢ (8 3 7) n n = + + (8 -3 7) = an even integer. Also, 0 < b < 1 and 0 < b ¢ < 1 fi 0 < b + b ¢ < 2 fi b + b ¢ = 1 fi b ¢ = 1 – b Thus, (1 – b)(a + b) (8 3 7) n n = - ◊ (8 + 3 7) = (64 – 63) n = 1 n = 1 5 55. Let I + f = (3 + 5) , where 0 < f < 1, 5 and f ¢ = (3 - 5) , where 0 < f ¢ < 1 . Now, 5 5 I + f + f¢ = (3 + 5) + (3 - 5) 5 5 5 3 5 1 2 = 2( C0 ◊ 3 + C2 ◊3 ◊ 5 + C4 ◊3 ◊5 ) = 2(243 + 10.135 + 5.75) = 2(243 + 1350 + 275) = 3936 5 Thus, the number just grater than the number (3 + 5) is I + f + f ¢ = 3936. 56. The co-efficient of a n in (1 + a) m+n = m+n C n = m+n C m+n–n = m+n C m = the co-efficient of a m in (1 + a) m+n .
6.28 Algebra <strong>Booster</strong> 57. We have, 10 C r = 10 C r+4 fi r + r + 4 = 10 fi 2r = 10 – 4 = 6 fi r = 3 58. We have, C1 C2 C3 Cn + 2◊ + 3◊ + + n ◊ C C C C - 59. We have 60. We have, 0 1 2 n 1 n Ên -1ˆ Ên - 2ˆ Ê1ˆ = + 2◊ Á + 3 + + n 1 Ë ˜ 2 ¯ Á Ë ˜ Á ˜ 3 ¯ Ë n ¯ = n + (n – 1) + (n – 2) + … + 3 + 2 + 1 = 1 + 2 + 3 + … + n nn+ ( 1) = 2 61. We have, ( )( )( )…( ) Ê ˆÊ ˆ Á Ë ˜Á ¯ ˜ Ë ¯ 62. We have, n! n! = + r! ¥ ( n - r)! ( r - 1)! ¥ ( n - r + 1)! n! Ê1 1 ˆ = + ( r - 1)! ¥ ( n - r)! Á Ër n - r + 1˜ ¯ n! Ên - r + 1+ rˆ = ( r - 1)! ¥ ( n - r)! Ë Á r( n - r + 1) ¯ ˜ ( n + 1) ¥ n! = r ¥ ( r - 1)! ¥ ( n - r + 1) ¥ ( n - r)! ( n + 1)! = r! ¥ ( n- r + 1)! C n + 1 = r 63. We have, Ênˆ Ê n ˆ Ê n ˆ Á + 2 + Ër˜ ¯ Á Ër -1˜ ¯ Á Ër - 2˜ ¯ 15 15 15 15 C1 C2 C3 C15 n n n + 2◊ + 3◊ + + 15◊ = Cr + 2 Cr-1 + Cr-2 15 15 15 15 C0 C1 C2 C14 n n n n = Cr + Cr-1 + Cr-1 + Cr-2 15 ¥ (15 + 1 ¢ ) = n+ 1 n+ 1 2 = ( Cr + Cr-1) 15 ¥ 16 n + 2 = = Cr 2 64. We have, = 120 Ênˆ Ê n ˆ Ê n ˆ Ê n ˆ Á + 3 + 3 + Ër˜ ¯ Á Ër -1˜ ¯ Á Ër - 2˜ ¯ Á Ër -3˜ ¯ Ê C1ˆÊ C2ˆÊ C3 ˆ Ê Cn ˆ Á 1+ 1+ 1 + … 1+ Ë C ˜Á 0¯Ë C ˜Á 1¯Ë C ˜ Á 2¯ Ë C ˜ ÊÊnˆ Ê n ˆˆ ÊÊ n ˆ Ê n ˆˆ n -1¯ = + + 2 + Ë ÁË Ár¯ ˜ Ë Ár -1¯ ˜ ¯ ˜ Ë ÁË Ár -1¯ ˜ Ë Ár - 2¯ ˜ ¯ ˜ Ê nˆÊ n -1ˆÊ n - 2ˆ Ê 1ˆ = Á1+ ˜Á1+ ˜Á1 + ˜… Á1+ Ë ˜ Ê 1¯Ë 2 ¯Ë 3 ¯ Ë n¯ Ê n ˆ Ê n ˆ + Á + Ë Á Ër - 2˜ ¯ Á Ër -3˜ ¯ Ê1+ nˆÊ1+ nˆÊ1+ nˆ Ê1+ nˆ = Á ˜Á ˜Á ˜… Ë Á ˜ 1 ¯Ë 2 ¯Ë 3 ¯ Ë n ¯ Ên + 1ˆ Ên + 1ˆ Ên + 1ˆ = Á + 2 + n ( n + 1) Ë r ˜ ¯ Á Ër -1˜ ¯ Á Ër - 2˜ ¯ = {1.2.3… n} Ên 1ˆ Ên 1ˆ Ên 1ˆ Ên 1ˆ = n Á + + + ( n + 1) Ë r ˜ ¯ Á Ër -1˜ ¯ Á Ër -1˜ ¯ Á Ër - 2˜ ¯ = ( n)! Ên + 2ˆ Ên + 2ˆ = Á + Ë r ˜ ¯ Á Ë r - 1˜ ¯ n n n n n n n n C0+ C1 C1+ C2 C2 + C3 Cn + Cn-1 Ên + 3ˆ = Á n n n n C1 C Ê 2 C ˆ Ê 3 C ˆ Ë r ˜ ¯ n 1+ 1+ 1 + … 1+ n n Á n ˜ Á n ˜ 65. We have, C0 Ë C1¯Ë C2¯ Cn-1 n n n n n ¥ CCC 0 1 2… C + 4 + 6 + 4 + n-1 r r - r - r - r - Ê1+ nˆÊ1+ nˆÊ1+ nˆ Ê1+ nˆ = Á ˜Á ˜Á ˜ Á ˜ ¥ k ÊÊnˆ Ê n ˆˆ ÊÊ n ˆ Ê n ˆˆ Ë 1 ¯Ë 2 ¯Ë 3 ¯ Ë n ¯ = + + 3 + Ë ÁË Ár¯ ˜ Ë Ár -1¯ ˜ ¯ ˜ Ë ÁË Ár -1¯ ˜ Ë Ár - 2¯ ˜ ¯ ˜ n ( n + 1) = ¥ k ( n)! + 3 + + + r r r r n C r + n C r–1 ( ) ( ) Ê + + ˆ Ê + + ˆ Á Ë ˜ ¯ Á Ë ˜ ¯ Ê ˆ Ê ˆ Ê ˆ Ê ˆ Ê ˆ Á Ë ˜ ¯ Á Ë 1˜ ¯ Á Ë 2˜ ¯ Á Ë 3˜ ¯ Á Ë 4˜ ¯ ÊÊ n ˆ Ê n ˆˆ ÊÊ n ˆ Ê n ˆˆ Á Ë Á Ë - 2˜ ¯ Á Ë -3˜ ¯˜ ¯ Á Ë Á Ë -3˜ ¯ Á Ë - 4˜ ¯˜ ¯ Ên + 1ˆ Ên + 1ˆ Ên + 1ˆ Ên + 1ˆ = Á + 3 + 3 + Ë r ˜ ¯ Á Ër -1˜ ¯ Á Ër - 2˜ ¯ Á Ër -3˜ ¯ ˆ ˜ ¯
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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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Page 332 and 333: Permutations and Combinations 5.31
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Page 357 and 358: 6.2 Algebra Booster 5. Subtracting
Page 359 and 360: 6.4 Algebra Booster Proof 1. ( a +
Page 361 and 362: 6.6 Algebra Booster EXERCISES LEVEL
Page 363 and 364: 6.8 Algebra Booster 80. Find the su
Page 365 and 366: 6.10 Algebra Booster 10. In the exp
Page 367 and 368: 6.12 Algebra Booster 11. Find the c
Page 369 and 370: 6.14 Algebra Booster 64. Find the c
Page 371 and 372: 6.16 Algebra Booster 43. Find the s
Page 373 and 374: 6.18 Algebra Booster 3. Match the f
Page 375 and 376: 6.20 Algebra Booster (a) 0 (c) (-1)
Page 377 and 378: 6.22 Algebra Booster LEVEL IV COMPR
Page 379 and 380: 6.24 Algebra Booster 30 2 30 2 2 30
Page 381: 6.26 Algebra Booster Now, (33 43 -
Page 385 and 386: 6.30 Algebra Booster 78. We have, 1
Page 389 and 390: 6.34 Algebra Booster Comparing the
Page 391 and 392: 6.36 Algebra Booster 1 1 1 1 + + +
Page 393 and 394: 6.38 Algebra Booster 127. We have,
Page 395 and 396: 6.40 Algebra Booster n n 2 n 3 7. W
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Page 399 and 400: 6.44 Algebra Booster n n Â Ck k =
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Page 403 and 404: 6.48 Algebra Booster Multiplying Eq
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Page 407 and 408: 6.52 Algebra Booster 72. Let t n 1
Page 409 and 410: 6.54 Algebra Booster Putting x = 1,
Page 411 and 412: 6.56 Algebra Booster Â Â Thus, 2I
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Page 415 and 416: 6.60 Algebra Booster Cn ( ,4) 36. L
Page 417 and 418: 6.62 Algebra Booster Thus, S = t 1
Page 419 and 420: 6.64 Algebra Booster 7. We have 10
Page 421 and 422: 6.66 Algebra Booster = ( 49 C 4 + 4
Page 423 and 424: 6.68 Algebra Booster Differentiatin
Page 427 and 428: 6.72 Algebra Booster 51. Let the th
Page 429 and 430: 7.2 Algebra Booster (vii) Identity
Page 431 and 432: 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 (
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7.28 Algebra Booster 17. The determ
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7.30 Algebra Booster 43. If a 0 A
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7.32 Algebra Booster 65. Let M be a
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7.34 Algebra Booster 12. Then AB =
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7.36 Algebra Booster Êa bˆÊ1 2ˆ
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7.38 Algebra Booster 1 a a 47 The g
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7.40 Algebra Booster 2 2 2 1 0 = (1
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7.42 Algebra Booster Thus, D1 ( d -
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7.44 Algebra Booster 77. We have, 2
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7.46 Algebra Booster 90. We know th
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7.52 Algebra Booster 4. We have, 2
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7.54 Algebra Booster = ([x] + [y] +
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7.56 Algebra Booster It is given th
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7.58 Algebra Booster 27. The given
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7.60 Algebra Booster and 12 -3 5 D1
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7.62 Algebra Booster fi Ê a ˆ Ê
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7.64 Algebra Booster a b c b c a =
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7.66 Algebra Booster 2bc -2c -2b 2
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7.68 Algebra Booster 1 1 1 1 = 2 n
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7.70 Algebra Booster 3 m m m 3 m =
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7.72 Algebra Booster fi (49 - 42)si
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7.74 Algebra Booster fi p + q + r =
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7.76 Algebra Booster 34. We have fi
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7.78 Algebra Booster where a 2 + b
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7.80 Algebra Booster 55. (iii) Let
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7.82 Algebra Booster fi 2 2 (1 + a)
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8.2 Algebra Booster For example, th
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8.4 Algebra Booster (ii) P(AB) £ P
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8.6 Algebra Booster (ii) a black ca
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8.8 Algebra Booster 64. If two dice
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8.10 Algebra Booster ferred from th
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8.12 Algebra Booster 168. The proba
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8.14 Algebra Booster 28. A fair coi
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8.16 Algebra Booster product is 0.7
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8.18 Algebra Booster event that thr
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8.20 Algebra Booster Questions aske
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8.22 Algebra Booster The probabilit
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8.24 Algebra Booster 72. A is targe
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8.26 Algebra Booster 5, 6, 7. A car
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8.28 Algebra Booster 137. Ê 9 m ˆ
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8.30 Algebra Booster 24. 25. 3 10 1
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8.32 Algebra Booster So, the probab
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8.34 Algebra Booster (iv) Let D be
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8.36 Algebra Booster Hence, the req
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8.38 Algebra Booster 58. Here, S =
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8.40 Algebra Booster 81. Here, S =
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8.42 Algebra Booster The probabilit
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8.44 Algebra Booster Thus, 1 36 =
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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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1.Algebra Booster

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