Question

"""'"''50 MARKS""""''''

answer it

Answer 1

(16)

Given a^2 + b^2 + c^2 = 14.

⇒ (a + b + c)^2 ≥ 0.

We know that (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)

⇒ 14 + 2(ab + bc + ca) ≥ 0

⇒ 2(ab + bc + ca) ≥ -14

⇒ ab + bc + ca ≥ -7     ------ (1)

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⇒ (a - b)^2 + (b - c)^2 + (c - a)^2 ≥ 0

⇒ a^2 + b^2 - 2ab + b^2 + c^2 - 2bc + c^2 + a^2 - 2ac ≥ 0

⇒ 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca ≥ 0

⇒ 2[a^2 + b^2 + c^2 - ab - bc - ca] ≥ 0

⇒ 2[14 - (ab + bc + ca)] ≥ 0

⇒ -(ab + bc + ca) ≥ - 14

⇒ ab + bc + ca ≤ 14.

Therefore, the range is [-7,14]

Hope this helps!

Answer 2

answer in the attachment..

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