A) 2
B) \[2\frac{1}{4}\]
C) 3
D) \[3\frac{1}{4}\]
Correct Answer: B
Solution :
[b] \[{{a}^{2}}+{{b}^{2}}+4{{c}^{2}}=ab+ab-\,4c-\,3\] \[\Rightarrow \,\,\,{{a}^{2}}+{{b}^{2}}+4{{c}^{2}}-\,2a-\,2b+4c+3=0\] \[\Rightarrow \,\,\,{{a}^{2}}-\,2a+1+{{b}^{2}}-\,2b+1+4{{c}^{2}}+4c+1=0\] \[\Rightarrow \,\,\,{{(a-\,1)}^{2}}+{{(b-\,1)}^{2}}+{{(2c+1)}^{2}}=0\] \[\therefore \,\,\,a-\,1=0\Rightarrow \,\,a=1;\] \[b-\,1=0\Rightarrow b=1;\] \[2c+1=0\Rightarrow \,\,c=-\,\frac{1}{2}\] \[\therefore \,\,\,{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1+1+\frac{1}{4}\] \[=2\frac{1}{4}\]
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