A) \[{{a}^{2}}-{{b}^{2}}\]
B) \[2{{a}^{2}}+{{b}^{2}}\]
C) \[{{a}^{2}}+{{b}^{2}}\]
D) \[{{a}^{2}}+2{{b}^{2}}\]
Correct Answer: C
Solution :
\[\because \]\[p(x)\times q(x)=LCM\times HCF\] \[x\times y=b\times a\] \[\therefore xy=ab\] and, \[x+y=a+b\] (Given) Squaring both sides, \[\therefore \] \[{{(x+y)}^{2}}={{(a+b)}^{2}}\] or,\[{{x}^{2}}+{{y}^{2}}+2xy={{a}^{2}}+{{b}^{2}}+2ab\] or,\[{{x}^{2}}+{{y}^{2}}+2ab={{a}^{2}}+{{b}^{2}}+2ab\] \[(\because \,\,xy=ab)\] \[\therefore \]\[{{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}\]
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