Algebra


  1. If a + b + c = 3, a2 + b2 + c2 = 6 and
    1
    +
    1
    +
    1
    = 1, where a, b, c are all non-zero, then ‘abc’ is equal to
    abc









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    a + b + c = 3; a2 + b2 + c2 = 6
    ∴  (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
    ⇒  32 = 6 + 2 (ab + bc + ca)
    ⇒  9 – 6 = 2(ab + bc + ca)

    ⇒ ab + bc + ca =
    3
    2

    ∴ 
    1
    +
    1
    +
    1
    = 1
    abc

    ⇒ 
    bc + ac + ab
    = 1
    abc

    ⇒  abc = ab + bc + ca =
    3
    2

    Correct Option: B

    a + b + c = 3; a2 + b2 + c2 = 6
    ∴  (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
    ⇒  32 = 6 + 2 (ab + bc + ca)
    ⇒  9 – 6 = 2(ab + bc + ca)

    ⇒ ab + bc + ca =
    3
    2

    ∴ 
    1
    +
    1
    +
    1
    = 1
    abc

    ⇒ 
    bc + ac + ab
    = 1
    abc

    ⇒  abc = ab + bc + ca =
    3
    2


  1. If a2 + b2 + c2 = 2a - 2b - 2,then the value of 3a - 2b + c is









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    a2 + b2 + c2 = 2a – 2b – 2
    ⇒  a2 + b2 + c2 – 2a + 2b + 2 = 0
    ⇒  a2 – 2a + 1 + b2 + 2b + 1 + c2 = 0
    ⇒  (a – 1)2 + (b + 1)2 + c2 = 0
    ⇒  a – 1 = 0 ⇒ a = 1;
    ⇒  b + 1 = 0 ⇒ b = –1;
    and c = 0
    ∴  3a – 2b + c = 3 × 1 – 2 (–1) + 0
    = 3 + 2 = 5

    Correct Option: C

    a2 + b2 + c2 = 2a – 2b – 2
    ⇒  a2 + b2 + c2 – 2a + 2b + 2 = 0
    ⇒  a2 – 2a + 1 + b2 + 2b + 1 + c2 = 0
    ⇒  (a – 1)2 + (b + 1)2 + c2 = 0
    ⇒  a – 1 = 0 ⇒ a = 1;
    ⇒  b + 1 = 0 ⇒ b = –1;
    and c = 0
    ∴  3a – 2b + c = 3 × 1 – 2 (–1) + 0
    = 3 + 2 = 5



  1. If a = √6 + √5 , b = √6 − √5 then 2a2 – 5 ab + 2b2 = ?









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    a = √6 + √5 , b = √6 − √5
    a − b = √6 + √5 − √6 + √5 = 2√5
    ab = (√6 + √5)(√6 − √5)
    = 6 – 5 = 1
    ∴  2a2 – 5ab + 2b2

    = 2 a2
    5
    ab + b2
    2

    = 2 a2 − 2ab + b2
    1
    ab
    2

    = 2 (a2 – 2ab + b2) – ab
    = 2 (a – b)2 – ab
    = 2 × (2√5)2 − 1
    = 2 × 4 × 5 – 1 = 40 – 1 = 39

    Correct Option: B

    a = √6 + √5 , b = √6 − √5
    a − b = √6 + √5 − √6 + √5 = 2√5
    ab = (√6 + √5)(√6 − √5)
    = 6 – 5 = 1
    ∴  2a2 – 5ab + 2b2

    = 2 a2
    5
    ab + b2
    2

    = 2 a2 − 2ab + b2
    1
    ab
    2

    = 2 (a2 – 2ab + b2) – ab
    = 2 (a – b)2 – ab
    = 2 × (2√5)2 − 1
    = 2 × 4 × 5 – 1 = 40 – 1 = 39


  1. If a1/3 + b1/3 + c1/3 = 0, then a relation among a, b , c is









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    If a + b + c = 0
    a3 + b3 + c3 = 3abc
    ∴  If   a1/3 + b1/3 + c1/3 = 0
    ⇒  (a1/3)3 + (b1/3)3 + (c1/3)3
    = 3.a1/3.b1/3.c1/3
    ⇒  a + b + c = 3.a1/3.b1/3.c1/3
    ⇒  (a + b + c)3
    = 33.(a1/3.b1/3.c1/3)3 = 27abc

    Correct Option: B

    If a + b + c = 0
    a3 + b3 + c3 = 3abc
    ∴  If   a1/3 + b1/3 + c1/3 = 0
    ⇒  (a1/3)3 + (b1/3)3 + (c1/3)3
    = 3.a1/3.b1/3.c1/3
    ⇒  a + b + c = 3.a1/3.b1/3.c1/3
    ⇒  (a + b + c)3
    = 33.(a1/3.b1/3.c1/3)3 = 27abc



  1. If a + b + c + d = 4, then find the value of
    1
    +
    1
    +
    1
    +
    1
    .
    (1 − a)(1 − b)(1 − c)(1 − b)(1 − c)(1 − d)(1 − c)(1 − d)(1 − a)(1 − d)(1 − a)(1 − b)









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    a + b + c + d = 4 (Given)

    Expression =
    1
    +
    1
    +
    1
    +
    1
    .
    (1 − a)(1 − b)(1 − c)(1 − b)(1 − c)(1 − d)(1 − c)(1 − d)(1 − a)(1 − d)(1 − a)(1 − b)

    =
    1 − d + 1 − a + 1 − b + 1 − c
    (1 − a)(1 − b)(1 − c)(1 − d)

    =
    4 − (a + b + c + d)
    (1 − a)(1 − b)(1 − c)(1 − d)

    =
    4 − 4
    = 0
    (1 − a)(1 − b)(1 − c)(1 − d)

    Correct Option: A

    a + b + c + d = 4 (Given)

    Expression =
    1
    +
    1
    +
    1
    +
    1
    .
    (1 − a)(1 − b)(1 − c)(1 − b)(1 − c)(1 − d)(1 − c)(1 − d)(1 − a)(1 − d)(1 − a)(1 − b)

    =
    1 − d + 1 − a + 1 − b + 1 − c
    (1 − a)(1 − b)(1 − c)(1 − d)

    =
    4 − (a + b + c + d)
    (1 − a)(1 − b)(1 − c)(1 − d)

    =
    4 − 4
    = 0
    (1 − a)(1 − b)(1 − c)(1 − d)