Question

Let $S_1,S_2,S_3$ and $S_4$ be four sets defined as
$S_1=\{x:x\in Z\text{and}{\log }_2|4-3x|\le 2\}$
$S_2=\{x:x\in Z\text{and}|1-\frac{|x|}{1+|x|}|\ge \frac{1}{3}\}$
$S_3=\left\{x:x^2-3x+2\text{sgn}\left(x\right)=0\right\},$ where $\text{sgn}\left(x\right)$ represents the signum function.
$S_4=\left\{\right(x,y):x,y\in Z,x^2+y^2\le 4\}$.

$\text{List I}$ has four entries and $\text{List II}$ has five entries. Each entry of $\text{List I}$ is to be correctly matched with a unique entry of $\text{List II}.$

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& n\left(S_1\Delta S_2\right)& \text{(P)}& 9\\ \text{(B)}& n((S_1\times S_2)\cap (S_2\times S_1))& \text{(Q)}& 12\\ \text{(C)}& n(S_1\cap S_2\cap S_3^{\prime })& \text{(R)}& 36\\ \text{(D)}& n(S_4\times S_3)& \text{(S)}& 2\\ & & \text{(T)}& 0\end{array}$

Which of the following is the only CORRECT combination?

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

Answer 1

The correct option is C $\text{(D)}\to \text{(R)}$

For $S_1:$
${\log }_2|4-3x|\le 2$
$\Rightarrow |4-3x|\le 2^2\text{and}4-3x\ne 0$
$\Rightarrow -4\le 4-3x\le 4\text{and}3x\ne 4$
$\Rightarrow -8\le -3x\le 0\text{and}3x\ne 4$
$\Rightarrow \frac{8}{3}\ge x\ge 0\text{and}3x\ne 4$
Since $x\in Z,$
we have $S_1=\{0,1,2\}$

For $S_2:$
$|1-\frac{|x|}{1+|x|}|\ge \frac{1}{3}$
$\Rightarrow \left|\frac{1}{1+|x|}\right|\ge \frac{1}{3}$
$\Rightarrow \frac{1}{1+|x|}\ge \frac{1}{3}$

$\Rightarrow 1+|x|\le 3$
$\Rightarrow |x|\le 2\Rightarrow x\in [-2,2]$
Since $x\in Z,$
we have $S_2=\{-2,-1,0,1,2\}$

For $S_3:$
$\left(i\right)$ If $x>0$
$x^2-3x+2\text{sgn}\left(x\right)=0$
$\Rightarrow x^2-3x+2=0$
$\Rightarrow (x-1)(x-2)=0$
$\Rightarrow x=1,2$

$\left(ii\right)$ If $x=0$
$x^2-3x+2\text{sgn}\left(x\right)=0$
$\Rightarrow x^2-3x=0$
$\Rightarrow x=0,3$
$\therefore x=0$

$\left(iii\right)$ If $x<0$
$x^2-3x+2\text{sgn}\left(x\right)=0$
$\Rightarrow x^2-3x-2=0$
$\Rightarrow x=\frac{3\pm \sqrt{17}}{2}$
$\therefore x=\frac{3-\sqrt{17}}{2}$
$\therefore S_3=\{\frac{3-\sqrt{17}}{2},0,1,2\}$

For $S_4:$
$x^2+y^2\le 4,x,y\in Z$
$\Rightarrow S_4=\left\{\right(0,0),(-1,0),(0,1),(0,-1),(1,0),(-2,0),(0,2),(0,-2),(2,0\left)\right\}$

$n(S_1\cap S_2\cap S_3^{\prime })=n(S_1\cap S_2)-n(S_1\cap S_2\cap S_3)$
$=3-3=0$

$n(S_4\times S_3)=n\left(S_4\right)\times n\left(S_3\right)=9\times 4=36$