Question

Column-I	Column-II
$\left(I\right)$ The curve $C$ which passes through $(1,1)$ and has differential equation as $(2x^{2}y-2y^4)dx+(2x^3+3x{y}^3)dy=0$ is given by $\alpha \ln \left|x\right|+\beta \ln |y|+\frac{y^3}{x^2}=\gamma$, then $(\alpha +\beta +\gamma )$ equals-	(P) $1$
$\left(II\right)$ A curve $y=f\left(x\right)$ passes through $(2,0)$ and slope of tangent at any point $P(x,y)$ on it equals $\frac{(x+1{)}^2+(y-3)}{(x+1)}$, then $f\left(3\right)$ is-	(Q) $3$
$\left(III\right)$ Let $f:R^{+}\to R$ satisfies the functional equation $f\left(xy\right)=e^{xy-y-x}\left(e^{y}f\right(x)+e^{x}f(y\left)\right)$ for all $x,y\in R^{+}$. If $f^{\prime }\left(1\right)=e,$ then $\ln \left(\ln \right(f\left(e\right)\left)\right)$ equals-	(R) $5$
$\left(IV\right)$ A curve $y=f\left(x\right)$ passing through the point $(\frac{1}{2},\frac{1}{4})$ satisfies the differential equation $\frac{x}{y}\frac{dy}{dx}=\frac{3x^2-y}{2y-x^2}$ is a conic whose length of latus rectum is-	(S) $2$
	(T) $4$

Which of the following is only correct combination?

• A. $\left(I\right)\left(P\right)$
• B. $\left(II\right)\left(Q\right)$
• C. $\left(III\right)\left(R\right)$
• D. $\left(IV\right)\left(T\right)$

Answer 1

The correct option is B $\left(II\right)\left(Q\right)$

$\left(I\right)$
$2x^{2}ydx-2y^{4}dx+2x^{3}dy+3x{y}^{3}dy=0$
Dividing both side by $x^{3}y$ we have
$2\frac{dx}{x}-\frac{2y^3}{x^3}dx+2\frac{dy}{y}+\frac{3y^2}{x^2}dy=02d\left(\ln x\right)+2d\left(\ln y\right)+d\left(\frac{y^3}{x^2}\right)=0\Rightarrow 2\ln |x|+2\ln |y|+\frac{y^3}{x^2}=C$
Putting $(1,1),C=1$
So, $\alpha +\beta +\gamma =5$

$\left(II\right)$
$\frac{dy}{dx}=\frac{(x+1{)}^2+(y-3)}{(x+1)}\frac{dy}{dx}+\left(\frac{-1}{1+x}\right)y=\frac{x^2+2x-2}{x+1}I.F=e^{\int \left(\frac{-1}{1+x}\right)dx}=\left(\frac{1}{1+x}\right)\Rightarrow y.\left(\frac{1}{1+x}\right)=\int \frac{(x+1{)}^2-3}{(1+x{)}^2}dx+C\frac{y}{1+x}=x+\frac{3}{x+1}+C$
Putting$(2,0),C=-3$
$\Rightarrow \frac{y}{1+x}=x+\frac{3}{x+1}-3y\left(3\right)=3$

$\left(III\right)$
Differentiating w.r.t $x$ and taking $y$ as constant
$f^{\prime }\left(xy\right)y=e^{xy-y-x}(y-1)\left(e^{y}f\right(x)+e^{x}f(y\left)\right)+e^{xy-y-x}\left(e^y{f}^{\prime }\right(x)+e^{x}f(y\left)\right)$
$\text{Now, at}x=1,y=1\Rightarrow f^{\prime }\left(1\right)=e$
$\text{which gives,}f\left(1\right)=0$
$\text{For,}x=1,\text{we have}$
$f^{\prime }\left(y\right)y=e^{-1}(y-1)\left(ef\right(y\left)\right)+e^{-1}(e{e}^y+ef(y\left)\right)y\to x{f}^{\prime }\left(x\right)-f\left(x\right)=\frac{e^x}{x}\frac{e^x{f}^{\prime }\left(x\right)-e^{x}f\left(x\right)}{e^{2x}}=\frac{1}{x}\Rightarrow \frac{d}{dx}\left(\frac{f\left(x\right)}{e^x}\right)=\frac{1}{x}\Rightarrow f\left(x\right)=e^x\ln x$
$\therefore \ln \left(\ln \right(f\left(e\right)\left)\right)=1$

$\left(IV\right)$
$2xydy-x^{3}dy=3x^{2}ydx-y^{2}dx\Rightarrow 2xydy+y^{2}dx=3x^{2}ydx+x^{3}dy\Rightarrow \int d\left(x{y}^2\right)=\int d\left(x{y}^3\right)+C\Rightarrow x{y}^2=x{y}^3+C\text{At}x=\frac{1}{2},y=\frac{1}{4}\Rightarrow C=0\Rightarrow y=x^2\therefore \text{Length of Latus Rectum}=1$