Algebra


  1. If (x – 1) and (x + 3) are the factors of x2 + k1x + k2 then









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    f (x) = x² +k1x + k1
    (x – 1) is a factor of f(x).
    ∴ f (1) = 0
    ⇒ 1 + k1 + k2 = 0
    ⇒ k1 + k2 = –1 ... (i)
    Again, f (–3) = 0
    ⇒ (– 3)² + k1 (–3) + k2 = 0
    ⇒ 9 – 3k1 + k2 = 0
    ⇒ 3k1 – k2 = 9 ...(ii)
    On adding both equations,
    4k1 = 8
    ⇒ k1 = 2 From equation (i),
    k1 + k2 = –1
    ⇒ 2 + k2 = –1
    ⇒ k2 = –1 – 2 = –3

    Correct Option: B

    f (x) = x² +k1x + k1
    (x – 1) is a factor of f(x).
    ∴ f (1) = 0
    ⇒ 1 + k1 + k2 = 0
    ⇒ k1 + k2 = –1 ... (i)
    Again, f (–3) = 0
    ⇒ (– 3)² + k1 (–3) + k2 = 0
    ⇒ 9 – 3k1 + k2 = 0
    ⇒ 3k1 – k2 = 9 ...(ii)
    On adding both equations,
    4k1 = 8
    ⇒ k1 = 2 From equation (i),
    k1 + k2 = –1
    ⇒ 2 + k2 = –1
    ⇒ k2 = –1 – 2 = –3


  1. The expression x4 – 2x2 + k will be a perfect square if the value of k is









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    Expression = x4 – 2x2 + k
    = (x2)2 – 2.x.2 .1 + (1)2 – (1)2 + k
    For a perfect square,
    –1 + k = 0 ⇒ k = 1

    Correct Option: A

    Expression = x4 – 2x2 + k
    = (x2)2 – 2.x.2 .1 + (1)2 – (1)2 + k
    For a perfect square,
    –1 + k = 0 ⇒ k = 1



  1. Find the value of x for which the expression 2 – 3x – 4x2 has the greatest value.









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    Expression = 2 – 3x – 4x2
    = – (4x2 + 3x – 2)

    = − (2x)2 + 2 × x ×
    3
    +
    3
    2
    3
    2 − 2
    444

    = − 2x +
    3
    2 +
    3
    2 + 2
    44

    The value of expression will be maximum if,
    2x +
    3
    = 0
    4

    ⇒  2x = −
    3
    4

    ⇒  x = −
    3
    8

    Correct Option: C

    Expression = 2 – 3x – 4x2
    = – (4x2 + 3x – 2)

    = − (2x)2 + 2 × x ×
    3
    +
    3
    2
    3
    2 − 2
    444

    = − 2x +
    3
    2 +
    3
    2 + 2
    44

    The value of expression will be maximum if,
    2x +
    3
    = 0
    4

    ⇒  2x = −
    3
    4

    ⇒  x = −
    3
    8


  1. If   x2 +
    1
    x + a2 is a perfect square, then a is
    5









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    x2 +
    1
    x + a2
    5

    = x2 + 2.x.
    1
    +
    1
    2
    1
    2+ a2
    101010

    ∴  a2
    1
    2 = 0 ⇒ a2 =
    1
    2
    1010

    ⇒  a =
    1
    10

    Correct Option: C

    x2 +
    1
    x + a2
    5

    = x2 + 2.x.
    1
    +
    1
    2
    1
    2+ a2
    101010

    ∴  a2
    1
    2 = 0 ⇒ a2 =
    1
    2
    1010

    ⇒  a =
    1
    10



  1. If x = 3t, y =
    1
    (t + 1), then the value of t for which x = 2y is
    2









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    x = 2y

    ⇒  3t = 2 ×
    1
    (t + 1)
    2

    ⇒  3t = t + 1 ⇒ 3t – t = 1
    ⇒  2t = 1 ⇒ t =
    1
    2

    Correct Option: B

    x = 2y

    ⇒  3t = 2 ×
    1
    (t + 1)
    2

    ⇒  3t = t + 1 ⇒ 3t – t = 1
    ⇒  2t = 1 ⇒ t =
    1
    2