Question

For $ |x|<1, $ let $ y=1+x+{{x}^{2}}+..... $ to $ \infty , $ then $ \frac{dy}{dx}-y $ is equal to:

• A. $ \frac{x}{y} $
• B. $ \frac{{{x}^{2}}}{{{y}^{2}}} $
• C. $ \frac{x}{{{y}^{2}}} $
• D. $ x{{y}^{2}} $
• E. $ \frac{x-1}{{{y}^{2}}} $

Answer 1

Answer: (C)

$ \because $ $ y=1+x+{{x}^{2}}+....\infty $ $ \therefore $ $ y=\frac{1}{1-x}={{(1-x)}^{-1}} $ $ \Rightarrow $ $ \frac{dy}{dx}=-\frac{1}{{{(1-x)}^{2}}}(-1)=\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}}} $ $ =\frac{x}{{{(1-x)}^{2}}} $ $ \Rightarrow $ $ \frac{dy}{dx}-y=\frac{x}{{{y}^{2}}} $