A) \[\left[ \frac{1}{2},2 \right]\]
B) \[[-1,2]\]
C) \[\left[ -\frac{1}{2},1 \right]\]
D) \[\left[ -1,\frac{1}{2} \right]\]
Correct Answer: C
Solution :
Given, \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1\] ...(i) \[{{(a+b+c)}^{2}}\ge 0\] \[\Rightarrow \] \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2(ab+bc+ca)\ge 0\] \[\Rightarrow \] \[1+2(ab+bc+ca)\ge 0\] \[\Rightarrow \] \[ab+bc+ca\ge -\frac{1}{2}\] ?..(ii) and \[{{(a-b)}^{2}}\ge 0\] \[\Rightarrow \] \[{{a}^{2}}+{{b}^{2}}-2ab\ge 0\] \[\Rightarrow \] \[{{a}^{2}}+{{b}^{2}}\ge 2ab\] Similarly, \[{{b}^{2}}+{{c}^{2}}\ge 2bc,{{c}^{2}}+{{a}^{2}}\ge 2ca\] \[\therefore \] \[2({{a}^{2}}+{{b}^{2}}+{{c}^{2}})\ge 2ab+2bc+2ca\] \[\Rightarrow \] \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}\ge ab+bc+ca\] \[\Rightarrow \] \[1\ge ab+bc+ca\] ...(iii) Hence, from Eqs. (ii) and (iii), \[-\frac{1}{2}\le ab+bc+ca\le 1\] \[\therefore \] Interval\[=\left[ -\frac{1}{2},1 \right]\]
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