Set Theory Test 12

Element/ number three is denoted by $$3$$ and not $$\{3\}$$.
$$\{3\}$$ denotes a subset containing element 3. Hence
$$\{3\}\subseteq \{1,3,5\}$$
And 
$$3\in \{1,3,5\}$$

P = $$\{ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 ..., 60, ...,90, ....,99\}$$
Q = $$\{10, 20, 30, 40, 50, 60, 70, 80, 90 \}$$
Then looking at the multiple choices we get 
$$P \displaystyle \cap Q$$= $$\{30, 60, 90\}$$
$$\displaystyle \Rightarrow $$P $$\displaystyle \cap $$ Q = {x$$\displaystyle \mid $$ x is a multiple of 30 x$$\displaystyle \epsilon $$ N}

Here, $$2^{\left | y \right |}-\left | 2^{y-1}-1 \right |=2^{y-1}+1$$
We know to define modulus, we have three cases as
Case I: y< 0
$$\Rightarrow $$   $$2^{-y}+\left ( 2^{y-1}-1 \right )=2^{y-1}+1$$
$$\Rightarrow $$   $$2^{-y}=2^{1}$$          (as when y< 0 |y|=-y and $$\left | 2^{y-1}-1 \right |=-\left ( 2^{y-1}-1 \right )$$)
Hence, y=-1, which is true when y< 0          (i)
Case II: $$0\leq y< 1$$
$$\Rightarrow $$   $$2^{y}+\left ( 2^{y-1}-1 \right )=2^{y-1}+1$$
$$\Rightarrow $$   $$2^{y}=2$$          (as when $$0\leq y< 1$$ |y|=-y and $$\left | 2^{y-1}-1 \right |=-\left ( 2^{y-1}-1 \right )$$)
$$\Rightarrow $$   $$y=1$$, which shows no solution as,
   $$0\leq y< 1$$          (ii)
Case III: $$y\geq 1$$
$$\Rightarrow $$   $$2^{y}-\left ( 2^{y-1}-1 \right )=2^{y-1}+1$$
$$\Rightarrow $$   $$2^{y}=2^{y-1}+2^{y-1}$$
$$\Rightarrow $$   $$2^{y}=2.2^{y-1}$$          (as when $$y\geq 0$$ |y|=y and $$\left | 2^{y-1}-1 \right |=\left ( 2^{y-1}-1 \right )$$)
$$\Rightarrow $$   $$2^{y}=2^{y}$$, which is an identity therefor, it is true $$\forall y\geq 1$$          (iii)
Hence, from Eqs. (i), (ii), (iii) the solution of set is $$\left \{ y: y\geq 1\cup y=-1 \right \}$$.

We have $$\displaystyle X\cap \left ( Y\cup X \right )'=X\cap \left ( Y'\cap X' \right )$$
= $$\displaystyle X\cap Y'\cap X'=X\cap X'\cap Y'$$
= $$\displaystyle \phi \cap Y'= \phi $$

Since $$\cfrac{1}{y}\ne 0$$, $$\cfrac{1}{y}\ne 2$$, $$\cfrac{1}{y}\ne \cfrac{-2}{3}$$ [$$\because y\in N$$]
$$\therefore \cfrac{1}{y}$$ can be $$1$$

Ans: B

$$x=x$$ for all $$x$$
Hence, there is no element in A. 
A is an empty set .
$$A=\{ \}$$ 

$$\displaystyle 3\quad \subseteq \quad \left\{ 1,3,5 \right\} $$

$$A = \{2, 4, 6, 8, ........\}$$
$$B = \{1, 3, 5, 7, .......\}$$
$$\displaystyle B\cap A=\left \{ 2,4,6,8,..... \right \}\cap \left \{ 1,3,5,7,.... \right \}=\phi $$
$$\displaystyle \xi -A=\left \{ 1,2,3,4,5,6,.... \right \}-\left \{ 2,4,6,8,.... \right \}=\left \{ 1,3,5,7,.... \right \}$$
$$\displaystyle \therefore B\cap A-(\xi -A)=\phi -\left \{ 1,3,5,7,.... \right \}=\phi $$

$$\displaystyle n(A-B)=n(A)-n(A\cap B)$$
$$\displaystyle \Rightarrow 52=120-n(A\cap B)$$
$$\displaystyle \Rightarrow n(A\cap B)=120-52=68$$
Now $$\displaystyle n(A\cup B)=n(A)+n(B)-n(A\cap B)$$
$$\displaystyle =120+250-68$$
$$=302$$

$$x+2<9$$
$$x<7$$
Since $$x$$ is even positive integer then $$x\in [2,7)$$.