Write the following sets in the roaster form.
(i) $A=\{x: x \in R, 2 x+11=15\}$
(ii) $B=\left\{x \mid x^2=x, x \in R\right\}$
(iii) $C=\{x \mid x$ is a positive factor of a prime number $p\}$
(i) We have,
$$A=\{x: x \in R, 2 x+11=15\}$$
$$\begin{aligned} \therefore & & 2 x+11 & =15 \\ \Rightarrow & & 2 x & =15-11 \Rightarrow 2 x=4 \\ \Rightarrow & & x & =2 \\ & & A & =\{2\} \end{aligned}$$
(ii) We have,
$$\begin{array}{rlrl} \therefore & x^2 & =x \\ \Rightarrow & x^2-x & =0 \quad \Rightarrow & x(x-1)=0 \\ \Rightarrow & x & =0,1 \\ \therefore & B & =\{0,1\} \end{array}$$
(iii) We have, $C=\{x \mid x$ is a positive factor of prime number $p\}$. Since, positive factors of a prime number are 1 and the number itself.
$$\therefore \quad C=\{1, p\}$$
Write the following sets in the roaster form.
(i) $D=\left\{t \mid t^3=t, t \in R\right\}$
(ii) $E=\left\{w \left\lvert\, \frac{w-2}{w+3}=3\right., w \in R\right\}$
(iii) $F=\left\{x \mid x^4-5 x^2+6=0, x \in R\right\}$
(i) We have, $D=\left\{t \mid t^3=t, t \in R\right\}$
$$\begin{array}{llrl} \therefore & t^3 =t \\ \Rightarrow & t^3-t =0 \Rightarrow t\left(t^2-1\right)=0 \\ \Rightarrow & t(t-1)(t+1) =0 \Rightarrow t=0,1,-1 \\ \therefore & D =\{-1,0,1\} \end{array}$$
(ii) We have,
$$E=\left\{w \left\lvert\, \frac{w-2}{w+3}=3\right., w \in R\right\}$$
$\begin{array}{llll}\therefore & \frac{w-2}{w+3}=3 & \\ \Rightarrow & w-2=3 w+9 & \Rightarrow & w-3 w=9+2 \\ \Rightarrow & -2 w=11 & \Rightarrow & w=\frac{-11}{2}\end{array}$
$\therefore \quad E=\left\{\frac{-11}{2}\right\}$
(iii) We have, $\quad F=\left\{x \mid x^4-5 x^2+6=0, x \in R\right\}$
$\begin{array}{lr}\therefore & x^4-5 x^2+6=0 \\ \Rightarrow & x^4-3 x^2-2 x^2+6=0 \\ \Rightarrow & x^2\left(x^2-3\right)-2\left(x^2-3\right)=0 \\ \Rightarrow & \left(x^2-3\right)\left(x^2-2\right)=0\end{array}$
$\begin{array}{ll}\Rightarrow & x= \pm \sqrt{3}, \pm \sqrt{2} \\ \therefore & F=\{-\sqrt{3},-\sqrt{2}, \sqrt{2}, \sqrt{3}\}\end{array}$
If $Y=\left\{x \mid x\right.$ is a positive factor of the number $2^{p-1}\left(2^p-1\right)$, where $2^p-1$ is a prime number\}. Write $Y$ in the roaster form.
$Y=\left\{x \mid x\right.$ is a positive factor of the number $2^{p-1}\left(2^p-1\right)$, where $2^p-1$ is a prime number $\}$.
So, the factor of $2^{p-1}$ are $1,2,2^2, 2^3, \ldots, 2^{p-1}$.
$$\therefore \quad Y=\left\{1,2,2^2, 2^3, \ldots, 2^{p-1}, 2^p-1\right\}$$
State which of the following statements are true and which are false. Justify your answer.
(i) $35 \in\{x \mid x$ has exactly four positive factors $\}$.
(ii) $128 \in\{y \mid$ the sum of all the positive factors of $y$ is $2 y\}$.
(iii) $3 \notin\left\{x \mid x^4-5 x^3+2 x^2-112 x+6=0\right\}$.
(iv) $496 \notin\{y \mid$ the sum of all the positive factors of $y$ is $2 y\}$.
(i) Since, the factors of 35 are 1, 5, 7 and 35 . So, statement (i) is true.
(ii) Since, the factors of 128 are 1, 2, 4, 8, 16, 32, 64 and 128 .
$$\begin{aligned} \therefore \quad \text { Sum of factors } & =1+2+4+8+16+32+64+128 \\ & =255 \neq 2 \times 128 \end{aligned}$$
So, statement (ii) is false.
(iii) We have,
$$x^4-5 x^3+2 x^2-112 x+6=0$$
$\therefore$ For $x=3$,
$$\begin{aligned} & (3)^4-5(3)^3+2(3)^2-112(3)+6 =0 \\ & \Rightarrow 81-135+18-336+6 =0 \\ & \Rightarrow -346 =0 \end{aligned}$$
which is not true.
Hence, statement (iii) is true.
(iv) $\because \quad 496=2^4 \times 31$
So, the factors of 496 are $1,2,4,8,16,31,62,124,248$ and 496 .
$$\begin{aligned} \therefore \quad \text { Sum of factors } & =1+2+4+8+16+31+62+124+248+496 \\ & =992=2(496) \end{aligned}$$
So, $496 \in\{y \mid$ the sum of all the positive factor of $y$ is $2 y\}$.
Hence, statement (iv) is false.
If $L=\{1,2,3,4\}, M=\{3,4,5,6\}$ and $N=\{1,3,5\}$, then verify that $L-(M \cup N)=(L-M) \cap(L-N)$.
Given, $\quad L=\{1,2,3,4\}, M=\{3,4,5,6\}$ and $N=\{1,3,5\}$
$\therefore \quad M \cup N=\{1,3,4,5,6\}$
$\begin{array}{lrrl} & & L-(M \cup N) & =\{2\} \\ & \text { Now, } & L-M & =\{1,2\}, L-N=\{2,4\} \\ & \therefore & (L-M) \cap(L-N) & =\{2\} \\ & \text { Hence, } & L-(M \cup N) & =(L-M) \cap(L-N) .\end{array}$