Answer :

(i): ${2,3,4} \subset{1,2,3,4,5}$

(ii): ${a, b, c} \not \subset{b, c, d}$

(iii): ${x$ : $x$ is a student of class XI of your school $} \subset $ ${x: x$ is student of your school $}$

(iv): ${x: x$ is a circle in the plane $ } \not \subset {x: x$ is a circle in the same plane with radius 1 unit $}$

(v): ${x: x$ is a triangle in a plane $ } \not \subset {x: x $ is a rectangle in the plane $}$

(vi): ${x: x$ is an equilateral triangle in a plane $} \subset {x: x$ in a triangle in the same plane $}$

(vii): ${x: x$ is an even natural number $} \subset {x: x$ is an integer $}$

Answer :

(i): False. Each element of ${a, b}$ is also an element of ${b, c, a}$.

(ii): True. $a, e$ are two vowels of the english alphabet.

(iii): False. $2 \in{1,2,3}$; however, $2 \notin{1,3,5}$

(iv): True. Each element of ${a}$ is also an element of ${a, b, c}$.

(v): False. The elements of ${a, b, c}$ are $a, b, c$. Therefore, ${a} \subset{a, b, c}$

(vi): True

$ {x: x \text{ is an even natural number less than 6 }} = {2,4}$

$ {x:x \text{ is a natural number which divides 36 } } = {1,2,3,4,6,9,12,18,36}$

Answer :

$A={1,2,{3,4}, 5}$

(i): The statement ${3,4} \subset A$ is incorrect because $3 \in{3,4}$; however, $3 \notin A$.

(ii): The statement ${3,4} \in A$ is correct because ${3,4}$ is an element of $A$.

(iii): The statement ${{3,4}} \subset A$ is correct because ${3,4} \in{{3,4}}$ and ${3,4} \in A$.

(iv): The statement $1 \in A$ is correct because $1$ is an element of $A$.

(v): The statement $1\subset A$ is incorrect because an element of a set can never be a subset of itself.

(vi): The statement ${1,2,5} \subset A$ is correct because each element of ${1,2,5}$ is also an element of $A.$

(vii): The statement ${1,2,5} \in A$ is incorrect because ${1,2,5}$ is not an element of $A$.

(viii): The statement ${1,2,3} \subset A$ is incorrect because $3 \in{1,2,3}$; however, $3 \notin A$.

(ix): The statement $\Phi \in A$ is incorrect because $\Phi$ is not an element of $A$.

(x): The statement $\Phi \subset A$ is correct because $\Phi$ is a subset of every set.

(xi): The statement ${\Phi} \subset A$ is incorrect because $\Phi \in{\Phi}$; however, $\Phi \in A$.

Answer :

(i): The subsets of ${a}$ are $\Phi$ and ${a}$.

(ii): The subsets of ${a, b}$ are $\Phi,{a},{b}$, and ${a, b}$.

(iii): The subsets of ${1,2,3}$ are $\Phi,{1},{2},{3},{1,2},{2,3},{1,3}$, and ${1,2,3}$

Answer :

(i): ${x: x \in R,-4<x \leq 6}=(-4,6]$

(ii): ${x: x \in R,-12<x<-10}=(-12,-10)$

(iii): ${x: x \in R, 0 \leq x<7}=[0,7)$

(iv): ${x : x \in$ $R, 3 \leq x \leq 4 }=[3,4]$

Answer :

(i): $(-3,0)={x: x \in R,-3<x<0}$

(ii): $[6,12]={x: x \in R, 6 \leq x \leq 12}$

(iii): $(6,12]={x: x \in R, 6<x \leq 12}$

(iv): $[-23,5)={x: x \in R,-23 \leq x<5}$

Answer :

(i): For the set of right triangles, the universal set can be the set of triangles or the set of polygons.

(ii): For the set of isosceles triangles, the universal set can be the set of triangles or the set of polygons or the set of two-dimensional figures.

Answer :

(i): It can be seen that $A \subset{0,1,2,3,4,5,6}$

$B \subset{0,1,2,3,4,5,6}$

However, C $\not \subset$ ${0,1,2,3,4,5,6}$

Therefore, the set ${0,1,2,3,4,5,6}$ cannot be the universal set for the sets $A, B,$ and $C.$

(ii): $ \ \ A \not \subset \phi, B$ $ \not \subset $ $\Phi, C \not \subset \Phi$

Therefore, $\Phi$ cannot be the universal set for the sets $A, B,$ and $C.$

(iii): $ \ \ \ A\quad \subset{0,1,2,3,4,5,6,7,8,9,10}$

$\qquad \ B$ $\quad \subset{0,1,2,3,4,5,6,7,8,9,10}$

$\qquad \ C$ $\quad\subset{0,1,2,3,4,5,6,7,8,9,10}$

Therefore, the set ${0,1,2,3,4,5,6,7,8,9,10}$ is the universal set for the sets $A, B,$ and $C.$

(iv): $ \ \ A \quad\subset{1,2,3,4,5,6,7,8}$

$ \qquad B \quad\subset{1,2,3,4,5,6,7,8}$

However, $ C$ $\quad\not \subset$ ${1,2,3,4,5,6,7,8}$

Therefore, the set ${1,2,3,4,5,6,7,8}$ cannot be the universal set for the sets $A, B,$ and $C.$