Question

Match $\text{List I}$ with the $\text{List II}$ and select the correct answer using the code given below the lists :

$\begin{array}{cccc}& \text{List-I}& & \text{List-II}\\ \text{(A)}& \text{Let}P(x,x^2)\text{be a point on the parabola}y=x^2.\text{If}& \text{(P)}& 1\\ & y=\sqrt{(x-3{)}^2+(x^2-2{)}^2}-\sqrt{x^2+(x^2-1{)}^2},\text{then}& & \\ & \text{the maximum value of}{y}^2\text{is}& & \\ \text{(B)}& \text{The area of a right triangle of least area that can be}& \text{(Q)}& 4\\ & \text{drawn so as to circumscribe a rectangle of sides}3& & \\ & \text{and 1,}\text{the right angle of the triangle coinciding with}& & \\ & \text{one of the angles of the rectangle, is equal to}& & \\ \text{(C)}& \text{Let}x+y=3\text{meets}{x}^2+y^2-4x+6y-3=0\text{at}& \text{(R)}& 6\\ & A\text{and}B.\text{A variable line meets the axes at}P\text{and}Q& & \\ & \text{respectively so that}AQ\text{meets}BP\text{at}R\text{at the right}& & \\ & \text{angle. The locus of}R\text{is a circle whose radius is}r,& & \\ & \text{then the value of}{r}^2\text{is}& & \\ \text{(D)}& \text{Let}{x}^2+3y^2=3\text{be the equation of an ellipse in the}& \text{(S)}& 8\\ & xy\text{-plane. If}A\text{and}B\text{are two points whose position}& & \\ & \text{vectors are}-\sqrt{3}\hat{i}\text{and}-\sqrt{3}\hat{i}+2\hat{k},\text{then the}& & \\ & \text{magnitude of the position vector of a point}P\text{on the}& & \\ & \text{ellipse such that}\angle APB=\frac{\pi }{4},\text{is}& & \\ & & \text{(T)}& 10\end{array}$

Which of the following is correct combination?

• A. $\text{(A)}\to \text{(S)}$
• B. $\text{(A)}\to \text{(T)}$
• C. $\text{(B)}\to \text{(P)}$
• D. $\text{(B)}\to \text{(Q)}$

Answer 1

The correct option is B $\text{(A)}\to \text{(T)}$

$\text{(A)}$
$f\left(x\right)=\sqrt{(x^2-2{)}^2+(x-3{)}^2}-\sqrt{x^2+(x^2-1{)}^2}$
Here, $1^{\text{st}}$ radical sign gives the distance between $P(x,x^2)$ and $A(3,2)$ whereas $2^{\text{nd}}$ radical sign describes the distance between $P(x,x^2)$ and $B(0,1)$.
Also $(x,x^2)$ lies on $y=x^2$
But $PA-PB\le AB$ for all possible positions of $P.$
$\therefore f(x{)}_{\text{max}}=$ distance between $AB$
$=\sqrt{9+1}=\sqrt{10}\Rightarrow y_{max}^2=10$


$\text{(A)}\to \text{(T)}$

$\text{(B)}$
$\frac{y}{3}=\frac{1}{x}\Rightarrow xy=3$
$A=(x+3)(y+1)=xy+x+3y+3$
$A\left(x\right)=6+x+\frac{9}{x}=6+{(\frac{3}{\sqrt{x}}-\sqrt{x})}^2+6$
$A$ will be least if $x=3$ and $y=1$
$\therefore \text{Area}=\frac{\left(6\right)\left(2\right)}{2}=6$


$\text{(B)}\to \text{(R)}$