Question

If $x^{13}{y}^7={\left(x+y\right)}^{20}$ then $xy\text{'}-y$ is.

• A. $7$
• B. $13$
• C. $20$
• D. $0$

Answer 1

The correct option is D

$0$

Explanation for the correct option:

Finding the value of $\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathit{y}$:

Given,

$x^{13}{y}^7={\left(x+y\right)}^{20}$

On applying log on both sides

$\begin{array}{rcl}\log \left(x^{13}{y}^7\right)& =& \log {\left(x+y\right)}^{20}\\ & \Rightarrow & 13\log x+7\log y=20\log \left(x+y\right)\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{log}\mathbf{}\mathbf{ab}\mathbf{=}\mathbf{}\mathbf{log}\mathbf{}\mathbf{a}\mathbf{+}\mathbf{}\mathbf{log}\mathbf{}\mathbf{b}\mathbf{}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{log}\mathbf{}{\mathbf{a}}^{\mathbf{n}}\mathbf{=}\mathbf{n}\mathbf{}\mathbf{log}\mathbf{}\mathbf{a}\mathbf{]}\end{array}$

On differentiating on both sides

$$\begin{gathered} \frac{13}{x}+\frac{7}{y}y\text{'}=\left[\frac{20}{\left(x+y\right)}\right]\left[1+y\text{'}\right] \\ \Rightarrow \frac{13}{x}+\frac{7}{y}y\text{'}=\frac{20}{\left(x+y\right)}+\frac{20}{\left(x+y\right)}y\text{'} \\ \Rightarrow \frac{13}{x}-\frac{20}{\left(x+y\right)}=\frac{20}{\left(x+y\right)}y\text{'}-\frac{7}{y}y\text{'} \\ \Rightarrow \frac{13x+13y-20x}{x\left(x+y\right)}=y\text{'}\frac{\left(20y-7x-7y\right)}{\left(x+y\right)y} \\ \Rightarrow \frac{13y-7x}{x}=y\text{'}\frac{13y-7x}{y} \\ \Rightarrow \frac{1}{x}=\frac{y\text{'}}{y} \\ \Rightarrow xy\text{'}-y=0 \end{gathered}$$

Hence, the correct option is D.