Q. For $|x| < 1,\ let\ y = 1 + x + x^{2} + \ldots to\ \infty,\ then\frac{dy}{dx} - y$ is equal to
A) $\frac{x}{y}$
B) $\frac{x^{2}}{y^{2}}$
C) $\frac{x}{y^{2}}$
D) $xy^{2}$
Explanation
(d)
Given, $y = 1 + x + x^{2} + x^{3} + ...\infty$
Since, |$x| < 1$
$\therefore\ \ y = \frac{1}{(1 - x)} = (1 - x)^{- 1}$
$\Rightarrow \ \frac{dy}{dx} = \ - \frac{1}{(1 - x)^{2}}( - 1) = \frac{1}{(1 - x)^{2}}$
$\therefore\ \frac{dy}{dx} - y = \frac{1}{(1 - x)^{2}} - \frac{1}{(1 - x)} = \frac{x}{(1 - x)^{2}}$
$\Rightarrow \ \frac{dy}{dx} - y = \ xy^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.