Algebra


  1. If p = 3 +
    1
    , the value of p4 +
    1
    is :
    pp4
    1. 81
    2. 27
    3. 120
    4. 119

  1. View Hint View Answer Discuss in Forum

    p = 3 +
    1
    (Given)
    p

    ∴ p -
    1
    = 3
    p

    On squaring both sides,
    p -
    1
    2 = (3)2 = 9
    p

    ⇒ p2 +
    1
    - 2 = 9
    p2

    ⇒ p2 +
    1
    = 9 + 2 = 11
    p2

    On squaring again,
    p2 +
    1
    2 = (11)2 = 121
    p2

    ⇒ p4 +
    1
    + 2 = 121
    p4

    ⇒ p4 +
    1
    = 121 - 2 = 119
    p4

    Correct Option: D

    p = 3 +
    1
    (Given)
    p

    ∴ p -
    1
    = 3
    p

    On squaring both sides,
    p -
    1
    2 = (3)2 = 9
    p

    ⇒ p2 +
    1
    - 2 = 9
    p2

    ⇒ p2 +
    1
    = 9 + 2 = 11
    p2

    On squaring again,
    p2 +
    1
    2 = (11)2 = 121
    p2

    ⇒ p4 +
    1
    + 2 = 121
    p4

    ⇒ p4 +
    1
    = 121 - 2 = 119
    p4


  1. If x = 1 + √2 + √3, then the value of (2x4 – 8x3 + 26x – 28) is
    1. 2√2
    2. 3√3
    3. 5√5
    4. 6√6

  1. View Hint View Answer Discuss in Forum

    x = 1 + √2 + √3
    ⇒ x - 1 = √3 + √2
    On squaring both sides ,
    x2 – 2x + 1 = 3 + 2 + 2√6
    ⇒ x2 – 2x + 1 - 5 = 2√6
    ⇒ x2 – 2x - 4 = 2√6
    On squaring again,
    (x2 – 2x - 4)2 = (2√6)2
    ⇒ x4 + 4x2 + 16 – 4x3 + 16x - 8x2 = 24
    ⇒ x4 – 4x3 - 4x2 + 16x - 8x2 - 8 = 0
    ⇒ 2x4 – 8x3 - 8x2 + 32x - 16 = 0
    ∴ 2x4 – 8x3 - 5x2 + 32x - 16 + 3x2 - 6x - 12
    = 0 + 3 (x2 – 2x – 4) = 3 × 2√6
    Required answer = 6√6

    Correct Option: D

    x = 1 + √2 + √3
    ⇒ x - 1 = √3 + √2
    On squaring both sides ,
    x2 – 2x + 1 = 3 + 2 + 2√6
    ⇒ x2 – 2x + 1 - 5 = 2√6
    ⇒ x2 – 2x - 4 = 2√6
    On squaring again,
    (x2 – 2x - 4)2 = (2√6)2
    ⇒ x4 + 4x2 + 16 – 4x3 + 16x - 8x2 = 24
    ⇒ x4 – 4x3 - 4x2 + 16x - 8x2 - 8 = 0
    ⇒ 2x4 – 8x3 - 8x2 + 32x - 16 = 0
    ∴ 2x4 – 8x3 - 5x2 + 32x - 16 + 3x2 - 6x - 12
    = 0 + 3 (x2 – 2x – 4) = 3 × 2√6
    Required answer = 6√6



  1. If x +
    1
    = √3 the value of (x18 + x12 + x6 + 1) is
    x
    1. 0
    2. 1
    3. 2
    4. 3

  1. View Hint View Answer Discuss in Forum

    x +
    1
    = √3
    x

    On cubing both sides,
    x +
    1
    3 = (3√3)3
    x

    ⇒ x3 +
    1
    + 3x.
    1
    x +
    1
    = 3√3
    x3xx

    ⇒ x3 +
    1
    + 3 × √3 = 3√3
    x3

    ⇒ x3 +
    1
    = 3√3 - 3√3 = 0
    x3

    ⇒ x6 + 1 = 0
    ∴ x18 + x12 + x6 + 1 = x12( x6 + 1 ) + 1( x6 + 1 )
    x18 + x12 + x6 + 1 = ( x6 + 1 )( x12 + 1 ) = 0

    Correct Option: A

    x +
    1
    = √3
    x

    On cubing both sides,
    x +
    1
    3 = (3√3)3
    x

    ⇒ x3 +
    1
    + 3x.
    1
    x +
    1
    = 3√3
    x3xx

    ⇒ x3 +
    1
    + 3 × √3 = 3√3
    x3

    ⇒ x3 +
    1
    = 3√3 - 3√3 = 0
    x3

    ⇒ x6 + 1 = 0
    ∴ x18 + x12 + x6 + 1 = x12( x6 + 1 ) + 1( x6 + 1 )
    x18 + x12 + x6 + 1 = ( x6 + 1 )( x12 + 1 ) = 0


  1. If a = 299, b = 298, c = 297 then the value of 2a3 + 2b3 + 2c3 – 6abc is
    1. 5154
    2. 5267
    3. 5364
    4. 5456

  1. View Hint View Answer Discuss in Forum

    2a3 + 2b3 + 2c3 – 6abc
    = 2 (a3 + b3 + c3 – 3abc)

    = 2 (a + b + c) ×
    1
    {(a – b)2 + (b –
    c)2 + (c – a)2}
    2

    = (299 + 298 + 297) {(299 – 298)2 + (298 – 297)2 + (297 – 299)2}
    = 894 × (1 + 1 + 4)
    = 894 × 6 = 5364

    Correct Option: C

    2a3 + 2b3 + 2c3 – 6abc
    = 2 (a3 + b3 + c3 – 3abc)

    = 2 (a + b + c) ×
    1
    {(a – b)2 + (b –
    c)2 + (c – a)2}
    2

    = (299 + 298 + 297) {(299 – 298)2 + (298 – 297)2 + (297 – 299)2}
    = 894 × (1 + 1 + 4)
    = 894 × 6 = 5364



  1. If a + b = 1, then a4 + b4 – a3 – b3 – 2a2b2 + ab is equal to
    1. 1
    2. 2
    3. 4
    4. 0

  1. View Hint View Answer Discuss in Forum

    a4 + b4 – a3 – b3 – 2a2b2 + ab
    = a4 + b4 – 2a2b2 – a3 – b3 + ab
    = (a + b)2 (a – b)2 – (a + b) (a2 – ab + b2) + ab
    = (a – b)2 – a2 + ab – b2 + ab
    [∵  a + b = 1]
    = (a – b)2 – (a – b)2 = 0

    Correct Option: D

    a4 + b4 – a3 – b3 – 2a2b2 + ab
    = a4 + b4 – 2a2b2 – a3 – b3 + ab
    = (a + b)2 (a – b)2 – (a + b) (a2 – ab + b2) + ab
    = (a – b)2 – a2 + ab – b2 + ab
    [∵  a + b = 1]
    = (a – b)2 – (a – b)2 = 0