1.Algebra Booster

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Binomial Theorem 6.71 44. Given, t 5 + t 6 = 0 fi n C 4 a n – 4 (–b) 4 + n C 5 a n – 5 (–b) 5 = 0 fi n C 4 a n – 4 b 4 – n C 5 a n – 5 b 5 = 0 fi n C 4 a n – 4 b 4 = n C 5 a n – 5 b 5 fi n C 4 a = n C 5 b fi n a C5 ( n - 4) = = n b C 5 4 45. We have Â m Ê10ˆÊ 20 ˆ Á r = 0Ë i ˜Á ¯Ëm - i˜ ¯ = the number of ways of selecting m persons out of 10 men and 20 women = the number of ways of selecting m out of 30 persons = 30 C m . But 30 C m is maximum when m = 15. 46. (1 + t 2 ) 12 (1 + t 12 )(1 + t 24 ) = (1 + t 2 ) 12 (1 + t 12 + t 24 + t 36 ) = [1 + 12 C 1 t 2 + 12 C 2 (t 2 ) 2 + … + 12 C 6 (t 2 ) 6 + … + 12 C 12 (t 2 ) 12 + …] ¥ (1 + t 12 + t 24 + t 36 ) Thus the co-efficients of t 24 = 12 C 12 + 1 + 12 C 6 = 12 C 6 + 2 47. For r ≥ 0, let r k- rÊnˆÊn- rˆ tr = (-1) 2 Á Ër˜Á ¯Ëk - r˜ ¯ r k- r Ê n! ˆ Ê ( n- r)! ˆ = (- 1) 2 ¥ Á ¥ Ër! ¥ ( n-r)! ˜ ¯ Á Ë( k- r)! ¥ ( n-k)! ˜ ¯ r k- r Ê n! ˆ Ê k! ˆ = (- 1) 2 ¥ Á ¥ Ëk! ¥ ( n-k)! ˜ ¯ Á Ë( k - r)! ¥ r! ˜ ¯ r k- r Ênˆ Êkˆ = (- 1) 2 ¥ Á ¥ Ëk˜ ¯ Á Ër˜ ¯ Thus, k r k- r Ênˆ Êkˆ S = Â (- 1) 2 ¥ Á ¥ Ëk˜ ¯ Á Ër˜ ¯ r = 0 k Ênˆ r k- r Êkˆ = Á (- 1) 2 ¥ k˜Â Ë ¯ Á Ër˜ ¯ r = 0 Ênˆ k Ênˆ = Á ¥ (2 - 1) = Ëk˜ ¯ Á Ëk˜ ¯ 48. We have, n C r + 1 = (k 2 – 3) n – 1 C r n C 2 r + 1 n fi ( k - 3) = = n-1 Cr r + 1 Here, 1 £ r £ n + 1 fi 0 < n r 1 n+ 1 £ n+ 1 £ Thus, 0 < k 2 – 3 £ 1 fi 3 < k 2 £ 4 fi 3< k £ 2 49. We have, (1 + x) 30 = 30 C 0 + 30 C 1 x + 30 C 2 x 2 + 30 C 3 x 3 + … + 30 C 20 x 30 + … and (1 – x) 30 = 30 C 0 x 30 + … + 30 C 10 x 20 – 30 C 11 x 19 + … + 30 C 30 Now, (1 + x) 30 (1 – x) 30 30 30 30 30 = ( C ◊ C - C ◊ C 0 10 1 11 30 30 30 30 C2 C12 C10 C30 + ◊ - + ◊ 19 30 ) x + (…) x + + (...) x + Comparing the co-efficients of x 20 from both the sides, we get 30 30 30 30 C0◊ C10- C1◊ C11 30 30 30 30 30 + C ◊ C - + C ◊ C = C 50. We have, A r = 10 C r , B r = 10 C r , C r = 30 C r Now, 2 12 10 30 10 10 Â AB r r = AB 1 1 + A2 B2 + + A10B10 r = 1 10 20 10 20 10 20 = C1 C1+ C2 C2+ C3 C3 10 20 + ... + C10 C10 Co-efficient of x 20 in the expansion of (1 + x) 10 (1 + x) 20 – 1 = 30 C 20 – 1 = 30 C 10 – 1 = C 10 – 1 Also, 10 Â r = 1 ( A ) = r 2 10 10 2 Â( Cr ) r = 1 10 2 10 2 10 2 10 2 ( C1) ( C2) ( C3) ( C10) = + + + + Co-efficient of x 10 in the expansion of (1 + x) 10 (1 + x) 10 – 1 = 20 B 10 – 1 = B 10 – 1 Therefore, 10 Â r = 1 A ( B B - C A ) r 10 r 10 r 10 2 Â B10( ArBr) C10( Ar) r = 1 = - = B 10 (C 10 – 1) – C 10 (B 10 – 1) = (C 10 – B 10 ) Ú 20
6.72 Algebra <strong>Booster</strong> 51. Let the three consecutive terms are t r , t r–1 , t r–2 respectively. Given, fi n + 5 n + 5 n + 5 Cr- 1 Cr Cr+ 1 = n+ 5 n+ 5 C C r r + 1 and n+ 5 n+ 5 Cr-1 Cr : : 5 :10 :14 10 14 = = 5 10 fi ( n+ 5) - r + 1 ( n+ 5) -( r + 1) + 1 7 = 2and = r r + 1 5 fi n+ 6 ( n+ 6) 12 = 3and = r r + 1 5 fi 12 3 r = ( r + 1) 5 fi r = 4 and n + 6 = 12 Thus, r = 4 and n = 6. 52. We have, (1 + x 2 ) 4 (1 + x 3 ) 7 (1 + x 4 ) 12 = (1 + 4 C 1 x 2 + 4 C 2 x 4 + 4 C 3 x 6 + 4 C 4 x 8 ) ¥ (1 + 7 C 1 x 3 + 7 C 2 x 6 + 7 C 3 x 9 + 7 C 4 x 12 + …) ¥ (1 + 12 C 1 x 4 + 12 C 2 x 8 + …) = (1 + 4 C 1 x 2 + 4 C 2 x 4 + 4 C 3 x 6 + 4 C 4 x 8 ) ¥ [1 + 12 C 1 x 4 + 12 C 2 x 8 + 7 C 1 x 3 + ( 7 C 1 ¥ 12 C 1 )x 7 + ( 7 C 1 ¥ 12 C 2 )x 11 + 7 C 3 x 9 + …] Thus, the co-efficient of x 11 =( 7 C 1 ¥ 12 C 2 ) + ( 4 C 1 ¥ 7 C 3 ) + ( 7 C 1 ¥ 12 C 1 ¥ 4 C 2 ) + ( 4 C 4 ¥ 7 C 1 ) = 7 ¥ 66 + 4 ¥ 35 + 7 ¥ 12 ¥ 6 + 7 = 462 + 140 + 504 + 7 = 1113
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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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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
Page 413 and 414: 6.58 Algebra Booster fi fi 2n 2n Ê
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 425: 6.70 Algebra Booster 37. We have, 3
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
Page 451 and 452: 7.24 Algebra Booster 2 2 2 2 2 2 a
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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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 473 and 474: 7.46 Algebra Booster 90. We know th
Page 475 and 476: 7.48 Algebra Booster 108. We have,
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7.50 Algebra Booster 118. We have,
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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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