Algebra


  1. For what value(s) of a is   x +
    1
    x + a2   a perfect square ?
    4









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    x +
    1
    x + a2
    4

    = (√x)2 + 2.√x.
    1
    + (a)2
    8

    Clearly a =
    1
    8

    Then, expression =x +
    1
    2
    8

    Correct Option: B

    x +
    1
    x + a2
    4

    = (√x)2 + 2.√x.
    1
    + (a)2
    8

    Clearly a =
    1
    8

    Then, expression =x +
    1
    2
    8


  1. If a ≠ b, then which of the following statements is true ?









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    Arithmetic mean (AM) =
    a + b
    2

    Geometric mean (GM) = √ab
    As AM > GM
    a + b
    > √ab
    2

    Correct Option: C

    Arithmetic mean (AM) =
    a + b
    2

    Geometric mean (GM) = √ab
    As AM > GM
    a + b
    > √ab
    2



  1. If  
    a
    +
    b
    +
    c
    = 1, then the value of
    1
    +
    1
    +
    1
    is
    1 − a1 − b1 − c1 − a1 − b1 − c









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    Tricky Approach

    a
    +
    b
    +
    c
    = 1
    1 − a1 − b1 − c

    ⇒ 
    a
    + 1 +
    b
    + 1 +
    c
    + 1 = 3 + 1 = 4
    1 − a1 − b1 − c

    ⇒ 
    a + 1 − a
    +
    b + 1 − b
    +
    c + 1 − c
    = 4
    1 − a1 − b1 − c

    ⇒ 
    1
    +
    1
    +
    1
    = 4
    1 − a1 − b1 − c

    Correct Option: D

    Tricky Approach

    a
    +
    b
    +
    c
    = 1
    1 − a1 − b1 − c

    ⇒ 
    a
    + 1 +
    b
    + 1 +
    c
    + 1 = 3 + 1 = 4
    1 − a1 − b1 − c

    ⇒ 
    a + 1 − a
    +
    b + 1 − b
    +
    c + 1 − c
    = 4
    1 − a1 − b1 − c

    ⇒ 
    1
    +
    1
    +
    1
    = 4
    1 − a1 − b1 − c


  1. If x, y are two positive real numbers and x1/3 = y1/4 , then which of the following relations is true?









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    x1/3 = y1/4
    ⇒ (x1/3)12 = (y1/4)12 ⇒ x4 = y3
    ⇒ (x4)5 = (y3)5 ⇒ x20 = y15

    Correct Option: D

    x1/3 = y1/4
    ⇒ (x1/3)12 = (y1/4)12 ⇒ x4 = y3
    ⇒ (x4)5 = (y3)5 ⇒ x20 = y15



  1. If x is real, then the minimum value of (x2 – x + 1) is









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    For expression ax2 + bx + c, a > 0, the minimum value is given by

    4ac − b2
    4a

    Here, for x2 – x + 1
    a = 1, b = –1, c = 1
    ∴  Minimum value =
    4 × 1 × 1 − 1
    =
    3
    4 × 14

    Correct Option: A

    For expression ax2 + bx + c, a > 0, the minimum value is given by

    4ac − b2
    4a

    Here, for x2 – x + 1
    a = 1, b = –1, c = 1
    ∴  Minimum value =
    4 × 1 × 1 − 1
    =
    3
    4 × 14