Question

Assertion :

If $y$ is function of $x$ such that
$y(x-y{)}^2=x$.

$\int \frac{dx}{x-3y}=\frac{1}{2}\log \left[\right(x-y{)}^2-1]$ Reason: $\int \frac{dx}{x-3y}=\log (x-3y)+c$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

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 $(x-y{)}^2=\frac{x}{y}\Rightarrow 2\log (x-y)=\log x-\log y....\left(1\right)$
Also, $\frac{dy}{dx}=(-\frac{y}{x})\frac{x+y}{x-3y}$
To prove the integral relation, it is sufficient to show that $\frac{d}{dx}$(RHS) $=\frac{1}{x-3y}$
Now, RHS $=\frac{1}{2}\log \left[\frac{x}{y}-1\right](\because (x-y{)}^2=\frac{x}{y})$
$=\frac{1}{2}\left[\log \right(x-y)-\log y]$
$=\frac{1}{2}\left[\frac{\log x-\log y}{2}-\log y\right]$ [From Eq. $\left(1\right)$]
$=\frac{1}{4}[\log x-3\log y]$
$\Rightarrow \frac{d}{dx}$(RHS) $=\frac{1}{4}\left[\frac{1}{x}-\frac{3}{y}\frac{dy}{dx}\right]$
$=\frac{1}{4}\left[\frac{1}{x}-\frac{3}{y}(-\frac{y}{x})\frac{x+y}{x-3y}\right]=\frac{1}{x-3y}$
Thus , statement - I is true.
Ans: $C$