A) \[a\]
B) \[2a\]
C) \[a+b\]
D) \[a-b\]
Correct Answer: C
Solution :
Put \[\frac{1}{x}=A\]and \[\frac{1}{y}=B\] \[\Rightarrow \] \[aA-bB=0\] ?(i) and \[a{{b}^{2}}A+{{a}^{2}}bB={{a}^{2}}+{{b}^{2}}\] ?(ii) Multiplying Eq. (i) by a2 and adding in Eq. (ii), we get \[{{a}^{3}}A+a{{b}^{2}}A={{a}^{2}}+{{b}^{2}}\] \[aA\,\,({{a}^{2}}+{{b}^{2}})=({{a}^{2}}+{{b}^{2}})\] \[\Rightarrow \] \[A=\frac{1}{a}\] Put the value of A in Eq. (i), we get \[1-bB=0\] \[\Rightarrow \] \[B=\frac{1}{b}\] \[\Rightarrow \] \[x=a\]and \[y=b\] \[\Rightarrow \] \[x+y=a+b\]
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