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    Question

    If (a + b – 1)2 + (b + c – 9)2

    + (c + a – 4)2 ≤ 0, then the value √(a + b)c + (c + a)b – 1 is :
    A 7 Correct Answer Incorrect Answer
    B 8 Correct Answer Incorrect Answer
    C 9 Correct Answer Incorrect Answer
    D 6 Correct Answer Incorrect Answer

    Solution

    (a + b – 1) 2  + (b + c – 9) 2  + (c + a – 4) 2 ≤ 0 The value of a, b and c should be 0 or less than 0  for the result to be equal to or less than 0. So, from the given equation: a + b - 1 = 0 a + b = 1      ---- (1) b + c - 9 = 0 b + c = 9      ---- (2) c + a - 4 = 0 c + a = 4       ---- (3) Then, c + a = 4 c + 1 - b = 4       ---- from (1) 9 - b + 1 - b = 4      ---- from (2) - 2b = 4 - 10 b = 3 Substituting the value of b in (1), a + b = 1 a + 3 = 1 a = -2 Substituting the value of b in (2), b + c = 9 3 + c = 9 c = 6 ∴ The value of √ [(a + b)c + (c + a)b - 1] is : => √ [((-2) + 3)6 + (6 + (-2))3   - 1] ⇒ √ [ (1)6 + (4)3 - 1] ⇒ √ [1 + 64 – 1] ⇒ √64 = 8 ∴ The value of √ [(a + b)c + (c + a)b - 1] is 8.

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