A) \[a=b\ne c\]
B) \[a\ne b=c\]
C) \[a=b=c\]
D) \[a\ne b\ne c\]
Correct Answer: C
Solution :
| [c] \[3\,({{a}^{2}}+{{b}^{2}}+{{c}^{2}})={{(a+b+c)}^{2}}\] \[\Rightarrow \]\[3{{a}^{2}}+2{{b}^{2}}+3{{c}^{2}}={{a}^{2}}+{{b}^{2}}+{{b}^{2}}+{{c}^{2}}2ab+2bc+2ca\] \[\Rightarrow \] \[2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2ab-2bc-2ca=0\] \[\Rightarrow \]\[{{a}^{2}}-2ab+{{b}^{2}}+{{b}^{2}}=2bc+{{c}^{2}}+{{c}^{2}}-2ca+{{a}^{2}}=0\] \[\Rightarrow \] \[{{(a-b)}^{2}}ab+{{(b-c)}^{2}}+{{(c-a)}^{2}}=0\] \[\therefore \] \[a=b=c\] |
You need to login to perform this action.
You will be redirected in
3 sec
