A) \[\frac{xy}{x+y-1}\]
B) \[\frac{xy}{x-y-1}\]
C) \[\frac{xy}{x-y+1}\]
D) \[\frac{xy}{x+y+1}\]
Correct Answer: A
Solution :
Key Idea: \[{{(1-x)}^{-1}}=1+x+{{x}^{2}}+.....\infty \] Given, \[x=1+a+{{a}^{2}}+......\infty \] \[y=1+b+{{b}^{2}}+.....\infty \] \[\therefore \] \[x=\frac{1}{1-a},y=\frac{1}{1-b}\] \[\Rightarrow \] \[x-ax=1\]and \[y-yb=1\] \[\Rightarrow \] \[a=\frac{x-1}{x}\]and \[b=\frac{y-1}{y}\] ...(i) Also, \[1+ab+{{a}^{2}}{{\text{b}}^{2}}+...\text{ }\infty \] \[=\frac{1}{1-ab}=\frac{1}{1-\left( \frac{x-1}{x} \right)\left( \frac{y-1}{y} \right)}\] [from(i)] \[=\frac{xy}{xy-xy+y+x-1}=\frac{xy}{x+y-1}\] \[\therefore \] \[1+ab+{{a}^{2}}{{b}^{2}}+\infty =\frac{xy}{x+y-1}\]
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