RRB Non Technical Model Papers 8 with solutions

Question : 75 of 100

Marks: +1, -0

Given that a + b + c = 2 and ab + bc + ca = 1, then the value of

(a + b)^2 + (b + c)^2 + (c + a)^2

is:

Solution:

(2)

(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab+ bc + ca) ∴ 22 = a^2 + b^2 + c^2 + 2×1 ⇒ a^2 + b^2 + c^2 = 4 – 2 = 2 Now,(a + b)^2 + (b + c)^2 + (c + a)^2 = 2(a^2 +b^2 + c^2) + 2(ab + bc + ca) = 2 × 2 +2 × 1 = 6

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