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) (iii) (R)
• B. (II) (ii) (Q)
• C. (III) (ii) (Q)
• D. (IV) (i) (P)

Answer 1

The correct option is C

(III) (ii) (Q)

$y^{\prime }=-\frac{9x}{y}$

$\Rightarrow y{y}^{\prime }=-9x$

$\Rightarrow \int ydy=-9\int xdx$

$\Rightarrow \frac{y^2}{2}=-\frac{9x^2}{2}+C$

Given, $y\left(1\right)=0$, so, $C=\frac{9}{2}$

Hence, the function becomes $\frac{x^2}{1}+\frac{y^2}{9}=1$

It is an ellipse, with $a=1$ and $b=3$ [Comparison with standard equation of ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$]

So, area of the ellipse is $\pi ab=3\pi$

Area of the ellipse in the first quadrant $=\frac{3\pi }{4}$

For maximum or minimum value of the funtion, $y^{\prime }=0$

$\Rightarrow x=0$

Differentiating again,

$y{y}^{\prime \prime }+(y^{\prime }{)}^2+9=0$

$\Rightarrow y^{\prime \prime }=\frac{-9-(y^{\prime }{)}^2}{y}$

So, $y^{\prime \prime }$ is negative, hence, $x=0$ is the point of maxima.

Thus, the maximum value is $y=3$

The correct combination is (III) (ii) (Q)