Algebra


  1. If a, b are rational numbers and ( a − 1) √2 + 3 = b√2 + a , the value of (a + b) is









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    (a − 1)√2 + 3 = b√2 + a
    ⇒  a = 3 ; a – 1 = b
    ⇒  3 – 1 = b ⇒ b = 2
    ∴  a + b = 3 + 2 = 5

    Correct Option: D

    (a − 1)√2 + 3 = b√2 + a
    ⇒  a = 3 ; a – 1 = b
    ⇒  3 – 1 = b ⇒ b = 2
    ∴  a + b = 3 + 2 = 5


  1. If   a =
    5 + 1
      and  b =
    5 − 1
    ,   then the value of
    5 − 15 − 1
    a2 + ab + b2
      is
    a2 − ab + b2









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    a =
    5 + 1
    =
    5 + 1
    ×
    5 + 1
    5 − 15 − 15 + 1

    =
    (√5 + 1)2
    =
    5 + 1 + 2√5
    5 − 14

    =
    3 + √5
    2

    ∴  b =
    5 − 1
    =
    3 − √5
    22

    ∴  a + b =
    3 + √5
    +
    3 − √5
    = 3
    22

    and   ab =
    5 + 1
    ×
    5 − 1
    = 1
    5 − 15 + 1

    ∴  Expression =
    a2 + ab + b2
    =
    (a + b)2 − ab
    a2 − ab + b2(a + b)2 − 3ab

    =
    9 − 1
    =
    8
    =
    4
    9 − 363

    Correct Option: B

    a =
    5 + 1
    =
    5 + 1
    ×
    5 + 1
    5 − 15 − 15 + 1

    =
    (√5 + 1)2
    =
    5 + 1 + 2√5
    5 − 14

    =
    3 + √5
    2

    ∴  b =
    5 − 1
    =
    3 − √5
    22

    ∴  a + b =
    3 + √5
    +
    3 − √5
    = 3
    22

    and   ab =
    5 + 1
    ×
    5 − 1
    = 1
    5 − 15 + 1

    ∴  Expression =
    a2 + ab + b2
    =
    (a + b)2 − ab
    a2 − ab + b2(a + b)2 − 3ab

    =
    9 − 1
    =
    8
    =
    4
    9 − 363



  1. If   64x + 1 =
    64
    ,   then the value of x is
    4x









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    (64)x + 1 =
    64
    4x

    ⇒  (43)x + 1 × 4x = 64
    ⇒  43x + 3 + x = 43
    ⇒  44x + 3 = 43
    ⇒  4x + 3 = 3
    ⇒  x = 0

    Correct Option: B

    (64)x + 1 =
    64
    4x

    ⇒  (43)x + 1 × 4x = 64
    ⇒  43x + 3 + x = 43
    ⇒  44x + 3 = 43
    ⇒  4x + 3 = 3
    ⇒  x = 0


  1. If ax2 + bx + c = a (x – p)2 , then the relation among a, b, c would be









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    ax2 + bx + c = a (x – p)2
    ⇒  ax2 + bx + c = a (x2 – 2px + p2)
    ⇒  ax2 + bx + c = ax2 – 2apx + ap2
    Comparing the corresponding coefficients,
    b = – 2ap and c = ap2

    ⇒  b2 = 4a2p2 and p2 =
    c
    a

    ⇒  p2 =
    b2
    ;
    4a2

    ∴ 
    b2 =
    =
    c
    ⇒  b2 = 4ac
    4a2a

    Correct Option: C

    ax2 + bx + c = a (x – p)2
    ⇒  ax2 + bx + c = a (x2 – 2px + p2)
    ⇒  ax2 + bx + c = ax2 – 2apx + ap2
    Comparing the corresponding coefficients,
    b = – 2ap and c = ap2

    ⇒  b2 = 4a2p2 and p2 =
    c
    a

    ⇒  p2 =
    b2
    ;
    4a2

    ∴ 
    b2 =
    =
    c
    ⇒  b2 = 4ac
    4a2a



  1. If a + b + c + d = 1, then the maximum value of (1 + a) (1 + b) (1 + c) (1 + d) is









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    For maximum value,

    a = b = c = d =
    1
    4

    ∴  (1 + a) (1 + b) (1 + c) (1 + d)
    5
    4
    4

    Correct Option: D

    For maximum value,

    a = b = c = d =
    1
    4

    ∴  (1 + a) (1 + b) (1 + c) (1 + d)
    5
    4
    4