Question

If $y$ is a function of $x$ such that $y(x-y)^2=x$.
Statement - I : $\int \frac{d x}{x-3 y}=\frac{1}{2} \log \left[(x-y)^2-1\right]$ Because
Statement - II : $\int \frac{d x}{x-3 y}=\log (x-3 y)+c$.

• A. Statement-I is True, Statement-II is True ; Statement-II is a correct explantion for Statement-I
• B. Statement-I is True, Statement-II is True ; Statement-II is NOT a correct explantion for Statement-I
• C. Statement-I is True, Statement-II is False.
• D. Statement-I is False, Statement-II is True

Answer 1

Answer: (C)

The statement-II is false since in $\int \frac{ dx }{ x -3 y }$ $=\log (x-3 y)+C$
we are assuming that $y$ is a constant. We will now prove the statement-I.
From the given relation $(x-y)^2=\frac{x}{y}$,
and $2 \log (x-y)=\log x-\log y$ ........(1)
Also, $\frac{d y}{d x}=\left(-\frac{y}{x}\right) \cdot \frac{x+y}{x-3 y}$.
To prove the integral relation, it is sufficient to show that $\frac{ d }{ dx }$ RHS. $=\frac{1}{x-3 y}$
Now, RHS $=\frac{1}{2} \log \left[\frac{x}{y}-1\right]\left(\because(x-y)^2=\frac{x}{y}\right)$
$=\frac{1}{2}[\log (x-y)-\log y]$
$=\frac{1}{2}\left[\frac{\log x-\log y}{2}-\log y\right] $ [From Eq. (1)]
$=\frac{1}{4}[\log x-3 \log y]$
$\Rightarrow \frac{ d }{ dx }$ RHS. $=\frac{1}{4}\left[\frac{1}{x}-\frac{3}{y} \frac{d y}{d x}\right]$
$=\frac{1}{4}\left[\frac{1}{x}-\frac{3}{y}\left(-\frac{y}{x}\right) \frac{x+y}{x-3 y}\right]=\frac{1}{x-3 y}$