If a,b,c∈R and a2+b2+c2=1, then ab+bc+ca lies in the nterval
A
[12,2]
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B
[−1,2]
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C
[−12,1]
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D
[−1,12]
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Solution
The correct option is C[−12,1] (a+b+c)2≥0⇒1+2(ab+bc+ca)≥0⇒ab+bc+ca≥−12⋯(1) (a−b)2+(b−c)2+(c−a)2≥0⇒2−2(ab+bc+ca)≥0⇒ab+bc+ca≤1⋯(2) From (1) and (2), ab+bc+ca∈[−12,1]