Question

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

• 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 $\int \frac{dx}{x-3y}$ $=\log (x-3y)+C$
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, $\frac{dy}{dx}=\left(-\frac{y}{x}\right)\cdot \frac{x+y}{x-3y}$.
To prove the integral relation, it is sufficient to show that $\frac{d}{dx}$ RHS. $=\frac{1}{x-3y}$
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. $=\frac{1}{4}\left[\frac{1}{x}-\frac{3}{y}\frac{dy}{dx}\right]$
$=\frac{1}{4}\left[\frac{1}{x}-\frac{3}{y}\left(-\frac{y}{x}\right)\frac{x+y}{x-3y}\right]=\frac{1}{x-3y}$