1.Algebra Booster

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Binomial Theorem 6.59 EXPONENTIAL SERIES Cn ( ,2) 31. Let tn = ( n + 1)! nn ( - 1) = 2 ( n + 1)! 1 nn ( - 1) = ¥ 2 ( n + 1)! 1 Ê( n + 1)( n - 2) + 2ˆ = ¥ 2 Á Ë ( n + 1)! ˜ ¯ 1 Ê( n - 2) 2 ˆ = ¥ 2 Á + Ë n! ( n + 1)! ˜ ¯ 1 Ê 1 2 2 ˆ = ¥ 2 Á - + Ë( n - 1)! n! ( n + 1)! ˜ ¯ Thus, S = t 1 + t 2 + t 3 + t 4 + … 1Ê1 1 1 ˆ Ê 1 1 1 ˆ = 2Á + + + - + + + Ë1! 2! 3! ˜ ¯ Á Ë2! 4! 6! ˜ ¯ Ê 1 1 1 ˆ + Á + + + Ë3! 4! 5! ˜ ¯ 1 Ê 1ˆ = ( e-1) - ( e- 2) + Áe-1-1- 2 Ë ˜ 2¯ e = -1 2 n n n n n C 32. Let 0+ C1+ C2+ C3+ + Cn tn = n Pn n 2 = n! Thus, S = t 1 + t 2 + t 3 + t 4 + … Ê 2 3 4 2 2 2 2 = Á + + + + Ë1! 2! 3! 4! = |e 2 – 1| 33. We have 3 6 9 x x x a = 1 + + + + 3! 6! 9! and 4 7 10 x x x x b = + + + + 1! 4! 7! 10! 2 5 8 11 x x x x c = + + + + 2! 5! 8! 11! Now, a 3 + b 3 + c 3 – 3abc 2 2 = ( a + b + c)( a + bw + cw )( a + bw + cw) 2 x xw xw = e ◊e ◊w 2 x+ xw+ xw = e 2 x(1 + w+ w ) = e = e x ◊ 0 = 1 ˆ ˜ ¯ 34. We have 1 1.3 1.3.5 + + + 1.2 1.2.3.4 1.2.3.4.5.6 1 1.3 1.3.5 1.3.5.7 = + + + 2! 4! 6! 8! 1.3.5.7 º (2n -1) Let tn = (2 n)! 1.2.3.4.5.6…(2 n - 1)2n = (2 n)!(2.4.6.8…2 n) (2 n)! = (2 n)!(2.4.6.8…2 n) 1 = n 2 (1.2.3… n) n (1/2) = n! Thus, S = t 1 + t 2 + t 3 + t 4 + … 2 3 4 (1/2) (1/2) (1/2) (1/2) = + + + + 1! 2! 3! 4! = e 1/2 –1 = ( e -1) 35. We have n Ê k 2 ˆ an = Â Á k = 0Ë( k)! ¥ ( n - k)! ˜ ¯ 1 2 = + 1! ¥ ( n)! 1! ¥ ( n -1)! 2 3 2 2 + + + 2! ¥ ( n - 2)! 3! ¥ ( n -3)! 2 1 Ê n! 2 ( n!) 2 ( n!) = n! Á + + + Ë1! ¥ ( n)! 1! ¥ ( n - 1)! 2! ¥ ( n - 2)! 1 ( n 2 n 2 1 2 n 2 2 n n = Cn + ◊ Cn- + ◊ Cn- + + ◊ C0 ) n! 1 ( n 2 0 2 n 1 2 n 2 2 n n = C + ◊ C + ◊ C + + ◊ Cn ) n! 1 (1 2) n = + n! n 3 = n! Now, Â a r r = 0 = a 0 + a 1 + a 2 + a 3 + a 4 + … Ê 2 3 3 3 3 = Á1 + + + + Ë 1! 2! 3! = e 3 ˆ ˜ ¯ ˆ ˜ ¯
6.60 Algebra <strong>Booster</strong> Cn ( ,4) 36. Let tn = Pnn ( , ) nn ( -1)( n- 2)( n-3) = 24( n!) nn ( -1)( n- 2)( n-3) = 24[ nn ( -1)( n- 2)( n-3)]( n- 4)! 1 = 24( n - 4)! Thus, S = t 4 + t 5 + t 6 + t 7 + t 8 + … 37. Let 38. Let 1 Ê 1 1 1 1 = 1 24 Á + + + + + Ë 1! 2! 3! 4! e = 24 n tn = (2n + 1)! 1 Ê(2n + 1) -1ˆ = 2 Á Ë (2n + 1)! ˜ ¯ 1Ê 1 1 ˆ = 2 Á - Ë(2 n)! (2n + 1)! ˜ ¯ 1 Ê 1 1 ˆ Thus, S = Â 2 Á - Ë (2 n)! (2n + 1)! ˜ ¯ n= 1 1 ÊÊ 1 1ˆ Ê 1 1ˆ = 2 Á - + - Ë Á Ë 2! 3! ˜ ¯ Á Ë 4! 5! ˜ ¯ Ê 1 1ˆ Ê 1 1ˆ ˆ + Á - + - + Ë6! 7! ˜ ¯ Á Ë8! 9! ˜ ¯ ˜ ¯ 1 ÊÊ 1ˆ Ê 1 1ˆ = 1 2 Á - + - Ë Á Ë 1! ˜ ¯ Á Ë2! 3! ˜ ¯ Ê 1 1ˆ Ê 1 1ˆ ˆ + Á - + - + Ë4! 5! ˜ ¯ Á Ë6! 7! ˜ ¯ ˜ ¯ 1 -1 1 = ¥ e = 2 2e 2n - 1 tn = 2 n! 2n 1 = - 2 n! 2 n! 1 1 = - (2n - 1)! 2 n! Thus, Ê 1 1 ˆ S = Â Á - Ë(2n - 1)! 2 n! ˜ ¯ n= 1 ÊÊ1 1ˆ Ê 1 1ˆ = ÁË - + - Ë Á 1! 2! ˜ ¯ Á Ë 3! 4! ˜ ¯ Ê 1 1ˆ Ê 1 1ˆ ˆ + Á - + - + Ë5! 6! ˜ ¯ Á Ë7! 8! ˜ ¯ ˜ ¯ ˆ ˜ ¯ Ê 1 1 1 1 1 1 1 1 = -Á- + - + - + - + + Ë 1! 2! 3! 4! 5! 6! 7! 8! Ê 1ˆ = Á1 - Ë ˜ e¯ 39. We have, n-1 3 Cn ( ,2) ¥ ( n)! n -1 nn ( - 1) 3 = ¥ 2 ( n)! n -1 1 3 = ¥ 2 ( n - 2)! n-1 3 Thus S = Â C( n,2) ¥ ( n)! n= 0 1 2 3 4 5 1 Ê3 3 3 3 3 ˆ = ¥ 2 Á + + + + + Ë1! 1! 2! 3! 4! ˜ ¯ 2 3 4 3 Ê 1 3 3 3 3 ˆ = ¥ 2 Á + + + + + Ë1! 1! 2! 3! 4! ˜ ¯ 3 3e = 2 2 2 2 3 5 7 40. Consider + + + 1! 3! 5! 2 (2n + 1) Let tn = (2n - 1)! (2n + 1)(2n + 1) = (2n - 1)! 2 4n + 4n + 1 = (2n - 1)! 2 (4n - 1) + (4n + 2) = (2n - 1)! (2n - 1)(2n + 1) + 2(2n + 1) = (2n - 1)! (2n + 1) 2(2n - 1) + 4 = + (2n - 2)! (2n -1)! (2n - 2) + 3 2 4 = + + (2n - 2)! (2n - 2)! (2n -1)! 1 5 4 = + + (2n -3)! (2n - 2)! (2n -1)! Thus, S = t 1 + t 2 + t 3 + t 4 + t 5 + … Ê 1 1 1 1 ˆ = Á + + + + Ë1! 3! 5! 7! ˜ ¯ Ê 1 1 1 1 ˆ + 5Á + + + + Ë 1! 2! 4! 6! ˜ ¯ Ê 1 1 1 1 ˆ + 4 Á + + + + Ë 1! 3! 5! 7! ˜ ¯ 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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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
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
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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 and 382: 6.26 Algebra Booster Now, (33 43 -
Page 383 and 384: 6.28 Algebra Booster 57. 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 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
Page 433 and 434: 7.6 Algebra Booster where Proof: fi
Page 435 and 436: 7.8 Algebra Booster Replace A by ad
Page 437 and 438: 7.10 Algebra Booster 10. Unitary ma
Page 439 and 440: 7.12 Algebra Booster 46. Expand the
Page 441 and 442: 7.14 Algebra Booster 88. If A be a
Page 443 and 444: 7.16 Algebra Booster 7. For non-zer
Page 445 and 446: 7.18 Algebra Booster 2x + 4p p + 6a
Page 447 and 448: 7.20 Algebra Booster 4. Prove that
Page 449 and 450: 7.22 Algebra Booster 4. If S r = a
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Page 453 and 454: 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
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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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