Question

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

• A. $x/y$
• B. $x^2/y^2$
• C. $x/y^2$
• D. $x{y}^2$

Answer 1

The correct option is C

$x/y^2$

Explanation for the correct option:

Step 1. Find the sum of infinite series :

Given $y=1+x+x^2+.....\infty ,$

This is an infinite GP series with $a=1andr=x\left(wherea=firstterm,r=commonratio\right)$

Sum of the series given by

$$\begin{gathered} S_{\infty }=\frac{a}{(1-r)} \\ so,y=\frac{1}{(1-x)} \end{gathered}$$

Step 2. Differentiation with respect to x :

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=\frac{\left[\right(1-x)\times 0-(1\times -1\left)\right]}{(1-x^2)} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{(1-x^2)} \end{gathered}$$

Step 3. : Calculate $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{-}\mathit{y}$ :

$$\begin{gathered} =\left[\frac{1}{{(1-x)}^2}-\frac{1}{(1-x)}\right] \\ =\left[\frac{1-(1-x)}{{(1-x)}^2}\right] \\ =\frac{x}{{(1-x)}^2} \\ =\frac{x}{y^2}\mathbf{}\mathbf{}\left\{\because y=\frac{1}{(1-x)}\right\} \end{gathered}$$

Hence the correct option is C.