A) \[a=b\]
B) \[a+b=1\]
C) \[a-b=1\]
D) \[a+b=0\]
E) \[a+b=2\]
Correct Answer: D
Solution :
\[\frac{a}{x-a}+\frac{b}{x-b}=1\] \[\Rightarrow \]\[a(x-b)+b(x-a)=(x-a)(x-b)\] \[\Rightarrow \]\[ax-ab+ba-ab={{x}^{2}}-ax-bx+ab\] \[\Rightarrow \]\[{{x}^{2}}-2ax-2bx+3ab=0\] \[\Rightarrow \]\[{{x}^{2}}-2(a+b)x+3ab=0\] Let \[\alpha \]and \[-\alpha \] be roots of above equation \[\therefore \] \[\alpha -\alpha =\frac{2(a+b)}{1}\] \[\Rightarrow \] \[a+b=0\]
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