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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) $A=\{x: x \in R, 2 x+11=15\}$
(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 $p\}$
Solution:
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Verified Answer
(i) Given $A=\{x: x \in R, 2 x+11=15\}$
Here, $2 x+11=15$
$\Rightarrow \quad 2 x=15-11 \Rightarrow 2 x=4$
$\Rightarrow \quad x=2$
So, $\quad A=\{2\}$
(ii) Given $B=\left\{x \mid x^2=x, x \in R\right\}$
Here, $\quad x^2=x$
$\Rightarrow \quad x^2-x=0 \Rightarrow x(x-1)=0$
$\Rightarrow \quad x=0,1$
So, $\quad B=\{0,1\}$
(iii) Given $C=\{x \mid x$ is a positive factor of prime number $p\}$.
Since, positive factors of a prime number are 1 and the number itself.
So, $\quad C=\{1, p\}$
Here, $2 x+11=15$
$\Rightarrow \quad 2 x=15-11 \Rightarrow 2 x=4$
$\Rightarrow \quad x=2$
So, $\quad A=\{2\}$
(ii) Given $B=\left\{x \mid x^2=x, x \in R\right\}$
Here, $\quad x^2=x$
$\Rightarrow \quad x^2-x=0 \Rightarrow x(x-1)=0$
$\Rightarrow \quad x=0,1$
So, $\quad B=\{0,1\}$
(iii) Given $C=\{x \mid x$ is a positive factor of prime number $p\}$.
Since, positive factors of a prime number are 1 and the number itself.
So, $\quad C=\{1, p\}$
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