A) \[\frac{x}{y}\]
B) \[\frac{{{x}^{2}}}{{{y}^{2}}}\]
C) \[\frac{x}{{{y}^{2}}}\]
D) \[x{{y}^{2}}\]
Correct Answer: D
Solution :
Given series is \[y=1+x+{{x}^{2}}+....\infty \] \[\Rightarrow \] \[y=\frac{1}{1-x}\] \[(\because \]infinite\[GP\]series) On differentiating w.r.t.\[x,\] we get \[\frac{dy}{dx}=-\frac{(-1)}{{{(1-x)}^{2}}}=\frac{1}{{{(1-x)}^{2}}}\] \[\therefore \] \[\frac{dy}{dx}-y=\frac{1}{{{(1-x)}^{2}}}-\frac{1}{1-x}=\frac{1-1+x}{{{(1-x)}^{2}}}\] \[=x{{y}^{2}}\] Note: If infinite \[GP\] series is \[y=1+x+{{x}^{2}}+{{x}^{3}}+...\] The sum \[\frac{1}{1-x}\] is possible only when\[|x|\,\,<1\].
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