Algebra


  1. If x is a rational number and
    (x + 1)3 - (x - 1)3
    = 2 , then the sum of numerator and denominator of x is
    (x + 1)2 - (x - 1)2










  1. View Hint View Answer Discuss in Forum

    ( x + 1 )3 - ( x - 1 )3
    = 2
    ( x + 1 )2 - ( x - 1 )2

    ( x3 + 3x2 + 3x + 1 ) - ( x3 - 3x2 + 3x - 1 )
    = 2
    ( x2 + 2x + 1 ) - ( x2 - 2x + 1 )

    x3 + 3x2 + 3x + 1 - x3 + 3x2 - 3x + 1
    = 2
    x2 + 2x + 1 - x2 + 2x - 1

    6x2 + 2
    = 2
    4x

    3x2 + 1
    = 1 ⇒ 3x2 + 1 = 4x
    4x

    ⇒ 3x2 - 4x + 1 = 0
    ⇒ 3x2 - 3x - x + 1 = 0
    ⇒ 3x(x – 1) – 1(x – 1) = 0
    ⇒ (3x – 1)(x – 1) = 0
    ⇒ 3x – 1 = 0, or, x – 1 = 0
    ⇒ x =
    1
    or 1
    3

    Hence, sum of the numerator and denominator = 1 + 3 = 4 or, 1 + 1 = 2

    Correct Option: B

    ( x + 1 )3 - ( x - 1 )3
    = 2
    ( x + 1 )2 - ( x - 1 )2

    ( x3 + 3x2 + 3x + 1 ) - ( x3 - 3x2 + 3x - 1 )
    = 2
    ( x2 + 2x + 1 ) - ( x2 - 2x + 1 )

    x3 + 3x2 + 3x + 1 - x3 + 3x2 - 3x + 1
    = 2
    x2 + 2x + 1 - x2 + 2x - 1

    6x2 + 2
    = 2
    4x

    3x2 + 1
    = 1 ⇒ 3x2 + 1 = 4x
    4x

    ⇒ 3x2 - 4x + 1 = 0
    ⇒ 3x2 - 3x - x + 1 = 0
    ⇒ 3x(x – 1) – 1(x – 1) = 0
    ⇒ (3x – 1)(x – 1) = 0
    ⇒ 3x – 1 = 0, or, x – 1 = 0
    ⇒ x =
    1
    or 1
    3

    Hence, sum of the numerator and denominator = 1 + 3 = 4 or, 1 + 1 = 2


  1. If
    p
    +
    q
    +
    r
    = 1 and
    a
    +
    b
    +
    c
    = 0
    abcpqr

    where p, q, r and a, b, c are non-zero, then the value of
    p2
    +
    q2
    +
    r2
    is
    a2b2c2










  1. View Hint View Answer Discuss in Forum

    Let
    p
    = x , 
    q
    = y , 
    r
    = z
    abc

    ∴ x + y + z = 1
    and
    1
    +
    1
    +
    1
    = 0
    xyz

    yz + xz + xy
    = 0
    xyz

    ⇒ yz + xz + xy = 0
    ∴ x + y + z = 1
    On squaring both sides ,
    x2 + y2 + z2 + 2xy + 2yz + 2zx = 1
    ⇒ x2 + y2 + z2 + 0 = 1
    ⇒ x2 + y2 + z2 = 1

    Correct Option: C

    Let
    p
    = x , 
    q
    = y , 
    r
    = z
    abc

    ∴ x + y + z = 1
    and
    1
    +
    1
    +
    1
    = 0
    xyz

    yz + xz + xy
    = 0
    xyz

    ⇒ yz + xz + xy = 0
    ∴ x + y + z = 1
    On squaring both sides ,
    x2 + y2 + z2 + 2xy + 2yz + 2zx = 1
    ⇒ x2 + y2 + z2 + 0 = 1
    ⇒ x2 + y2 + z2 = 1



  1. If x4 +
    1
    = 119, and x > 1, then find the positive value of x3 -
    1
    .
    x4x3










  1. View Hint View Answer Discuss in Forum

    x4 +
    1
    = 119
    x4

    x2 +
    1
    2 - 2 = 119
    x2

    x2 +
    1
    2 = 119 + 2 = 121
    x2

    x2 +
    1
    2 = (11)2
    x2

    ⇒ x2 +
    1
    = 11
    x2

    x -
    1
    2 + 2 = 11
    x

    x -
    1
    2 = 11 - 2 = 9 = 32
    x

    ⇒ x -
    1
    = 3
    x

    On cubing both sides,
    x -
    1
    3 = 27
    x

    ⇒ x3 -
    1
    - 3x.
    1
    x -
    1
    = 27
    x3xx

    ⇒ x3 -
    1
    - 3 × 3 = 27
    x3

    ⇒ x3 -
    1
    = 27 + 9 = 36
    x3

    Correct Option: C

    x4 +
    1
    = 119
    x4

    x2 +
    1
    2 - 2 = 119
    x2

    x2 +
    1
    2 = 119 + 2 = 121
    x2

    x2 +
    1
    2 = (11)2
    x2

    ⇒ x2 +
    1
    = 11
    x2

    x -
    1
    2 + 2 = 11
    x

    x -
    1
    2 = 11 - 2 = 9 = 32
    x

    ⇒ x -
    1
    = 3
    x

    On cubing both sides,
    x -
    1
    3 = 27
    x

    ⇒ x3 -
    1
    - 3x.
    1
    x -
    1
    = 27
    x3xx

    ⇒ x3 -
    1
    - 3 × 3 = 27
    x3

    ⇒ x3 -
    1
    = 27 + 9 = 36
    x3


  1. If x2 – 3x + 1 = 0, then the value of
    x6 + x4 + x2 + 1
    will be
    x3










  1. View Hint View Answer Discuss in Forum

    x2 – 3 x + 1 = 0
    ⇒ x2 + 1 = 3 x

    x2 + 1
    = 3
    x

    ⇒ x +
    1
    = 3 ........... (i)
    x

    ∴ Expression =
    x6 + x4 + x2 + 1
    x3

    Expression =
    x6
    +
    x4
    +
    x2
    +
    1
    x3x3x3x3

    Expression = x3 + x +
    1
    +
    1
    xx3

    Expression = x3 +
    1
    + x +
    1
    = 2 × 2 = 4
    x3x

    Expression = x +
    1
    3 - 3 . x .
    1
    x +
    1
    + x +
    1
    xxxx

    Expression = 33 – 3 × 3 + 3 = 27 – 9 + 3 = 21

    Correct Option: C

    x2 – 3 x + 1 = 0
    ⇒ x2 + 1 = 3 x

    x2 + 1
    = 3
    x

    ⇒ x +
    1
    = 3 ........... (i)
    x

    ∴ Expression =
    x6 + x4 + x2 + 1
    x3

    Expression =
    x6
    +
    x4
    +
    x2
    +
    1
    x3x3x3x3

    Expression = x3 + x +
    1
    +
    1
    xx3

    Expression = x3 +
    1
    + x +
    1
    = 2 × 2 = 4
    x3x

    Expression = x +
    1
    3 - 3 . x .
    1
    x +
    1
    + x +
    1
    xxxx

    Expression = 33 – 3 × 3 + 3 = 27 – 9 + 3 = 21



  1. If x +
    1
    = 5, then x6 +
    1
    is
    xx6










  1. View Hint View Answer Discuss in Forum

    x +
    1
    = 5
    x

    On cubing both sides,
    x +
    1
    3 = (5)3
    x

    ⇒ x3 +
    1
    + 3x.
    1
    x +
    1
    = 125
    x3xx

    ⇒ x3 +
    1
    + 3 × 5 = 125
    x3

    ⇒ x3 +
    1
    = 125 - 15 = 110
    x3

    On squaring both sides,
    x6 +
    1
    + 2.x3.
    1
    = 12100
    x6x3

    ⇒ x6 +
    1
    = 12100 – 2 = 12098
    x6

    Correct Option: A

    x +
    1
    = 5
    x

    On cubing both sides,
    x +
    1
    3 = (5)3
    x

    ⇒ x3 +
    1
    + 3x.
    1
    x +
    1
    = 125
    x3xx

    ⇒ x3 +
    1
    + 3 × 5 = 125
    x3

    ⇒ x3 +
    1
    = 125 - 15 = 110
    x3

    On squaring both sides,
    x6 +
    1
    + 2.x3.
    1
    = 12100
    x6x3

    ⇒ x6 +
    1
    = 12100 – 2 = 12098
    x6