Question

$\begin{array}{cccccc}& \text{Column - I}& & \text{Column - II}& & \text{Column - III}\\ \left(I\right)& y.(y^{\prime }{)}^2-x{y}^{\prime }(1+y)+x^2=& \left(i\right)& \left[y\right]=1,\text{where [.] is greatest}& \left(p\right)& \text{Curve is bounded with area},\pi \\ & 0,y\left(\sqrt{3}\right)=2& & \text{integer function}& & \\ \left(II\right)& y^{\prime }=\frac{y^2-x^2}{2xy},y\left(1\right)=1& \left(ii\right)& \text{Maximum value of y is 3}& \left(Q\right)& \text{Area bounded by curve in first quadrant with}\\ & & & & & \text{co-ordinate axes is}\frac{3\pi }{4}\\ \left(III\right)& y^{\prime }=\frac{-9x}{y},y\left(1\right)=0& \left(iii\right)& \text{Maximum value of y is not defined}& \left(R\right)& \text{Curve is conic with eccentricty},\frac{1}{2}\\ \left(IV\right)& y^{\prime }=\frac{x}{y},y\left(2\right)=0& \left(iv\right)& \text{Maximum value of y is 1}& \left(S\right)& \text{Curve is conic with eccentricity},\sqrt{2}\end{array}$

The correct combination is

• A. (I) (ii) (Q)
• B. (II) (iv) (P)
• C. (III) (i) (R)
• D. (IV) (iv) (Q)

Answer 1

The correct option is B

(II) (iv) (P)

$y^{\prime }=\frac{y^2-x^2}{2xy}$

Let $y=vx$

$\Rightarrow y^{\prime }=v+x\frac{dv}{dx}$

Putting the value of $y^{\prime }$ in original equation, we get,

$v+x\frac{dv}{dx}=\frac{v^2-1}{2v}$

$\Rightarrow x\frac{dv}{dx}=\frac{-v^2-1}{2v}$

$\Rightarrow \int \frac{2v}{v^2+1}dv=-\int \frac{dx}{x}$

Let $v^2+1=t$

$\Rightarrow 2vdv=dt$

Hence, integral becomes $\int \frac{dt}{t}=-\int \frac{dx}{x}$

$\Rightarrow \ln t=-\ln x+C$

$\Rightarrow \ln \left(\frac{x^2+y^2}{x^2}\right)=-\ln x+C$

$\Rightarrow \ln \left(\frac{x^2+y^2}{x}\right)=C$

$\Rightarrow \frac{x^2+y^2}{x}=e^C$

Given, $y\left(1\right)=1$

$\Rightarrow e^C=2$

$\Rightarrow C=\ln 2$

So, the function is $x^2+y^2-2x=0$

It is a circle with centre $\equiv (1,0)$ and radius $=1$

So, area $=\pi$

To find the maximum value, differentiate the function:

$x+y{y}^{\prime }-1=0$

$\Rightarrow y^{\prime }=\frac{1-x}{\sqrt{2x-x^2}}$

For maximum or minimum value, $y^{\prime }=0\Rightarrow x=1$

Differentiating again,

$y{y}^{\prime \prime }+(y^{\prime }{)}^2=-1$

Since, $y^{\prime \prime }$ is negative, so $x=1$ is the maxima.

So, maximum value of $y$ at $x=1$ is 1.

Thus, correct combination is (II) (iv) (P)