InterviewSolution
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InterviewSolution
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If `a^2+b^2+c^2=1` then `ab+bc+ca` lies in the intervalA. [0,1]B. `[-1//2,1]`C. `[0,1//2]`D. [1,2] |
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Answer» Correct Answer - B We have, `a^(2)+b^(2)+c^(2)-ab-bc-ca=(1)/(2)[(a-b)^(2)+(b-c)^(2)+(c-a)^(2)]ge0` `implies" "a^(2)+b^(2)+c^(2)geab+bc+ca` `implies" "1geab+bc+ca" "[becaise a^(2)+b^(2)+c^(2)=1]` `implies" "ab+bc+cale1" "...(i)` Also, `(a+b+c)^(2)ge0` `implies" "a^(2)+b^(2)+c^(2)+2(ab+bc+ca)ge0` `implies" "1+2(ab+bc+ca)ge0" "[because a^(2)+b^(2)+c^(2)=1]` `implies" "ab+bc+ca ge-(1)/(2)` From (i) and (ii), we obtain `-(1)/(2)geab+bc+cale1impliesab+bc+ca in[-1//2,1].` |
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