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(i) Universal set for a set of right angle triangles can be set of polygons or set of all triangles. (ii)  Universal set for a set of isosceles angle triangles can be set of polygons.
The given intervals can be written in set builder form as : (i) (– 3, 0)   (ii) [6 , 12]   (iii) (6, 12]  (iv) [–23, 5)
The following can be written in interval as :  (i) {x : x R, – 4  x  6}  (ii) {x : x  R, – 12 x –10}   (iii) {x : x  R, 0  x  7}  (iv) (iv) {x : x  R, 3   x   4}
Let the elements in set A be m,then n m then, number of elements in power set of A         n  Here,  A =    so    n 0 n  1 Hence,we conclude P(A) has 1 element.
Subset  of   is   only. The subset of a null set is null set itself
{ 1, 2, { 3, 4 }, 5 }. but  is not an element  of { 1, 2, { 3, 4 }, 5 }. Hence,the above statement is incorrect.
Subsets of
Subsets of  . Thus the given set has 4 subsets
Subsets of .
{3, 4} is element of  A. So, {3, 4} A. Hence,the statement is correct.
is subset of all sets. Hence,the above statement is correct.
is not an element of { 1, 2, { 3, 4 }, 5 }. So,. Hence,the above statement is incorrect.
Here, 3 {1,2,3} but  3   { 1, 2, { 3, 4 }, 5 }. So, {1,2,3}  A Hence,the given statement is incorrect.
Here,{1,2,5} is not an element of  { 1, 2, { 3, 4 }, 5 }. So,{1,2,5} A . Hence, statement is incorrect.
All elements of {1,2,5} are present in { 1, 2, { 3, 4 }, 5 }. So, {1,2,5} { 1, 2, { 3, 4 }, 5 }. Hence,the statement given is correct.
Here, 1 is element of set A = { 1, 2, { 3, 4 }, 5 }.So,elements of set A cannot be subset of set A. 1  { 1, 2, { 3, 4 }, 5 }. Hence,the statement given is incorrect.
Given, 1 is element of  { 1, 2, { 3, 4 }, 5 }. So,1  A. Hence,statement is correct.
Here, { 3, 4 } { 1, 2, { 3, 4 }, 5 }  and { 3, 4 }   {{3, 4}} So, {{3, 4}}  A. Hence,the statement is correct.
3  {3,4} but 3 {1,2,{3,4},5}. SO, {3, 4}  A Hence,the statement is incorrect.