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a^2+b^2=2 then a+b lies in the interval
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there is any options....?
all are closed intervals and the options are -1,1. -2,2. 1,2. -2,2
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i cant get this but similar to this i show one...
If a2 + b2 + c2 = 1, then bc + ca + ab lies in the interval (a) [-1/2,3] (b) [-1,2] (c) [-1/2,1] (d) [-1,1/2]
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Given, a2 + b2 + c2 = 1 We know that, (a + b + c)2 ≥ 0 ⇒ a2 + b2 + c2 + 2(ab + bc + ca) ≥ 0 ⇒ 1 + 2(ab + bc + ca) ≥ 0 ⇒ ab + bc + ca ≥ Again, (b - c)2 + (c - a)2 + (a - b)2 ≥ 0 ⇒ 2(a2 + b2 + c2) - 2(ab + bc + ca) ≥ 0 ⇒ ab + bc + ca ≤ a2 + b2 + c2 ⇒ ab + bc + ca ≤ 1 Hence,
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