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