Day-02
Date-04/12/21
Q1.Find the no of elements that satisfy the given set, where x, y & z are the distinct elements.
K   x, y, z,  x, y ,  x, y, z
A.3
B.5
C.8
D.4
Solution: B
Explanation:
The collection of subset of a set is also count as an element. So, x, y and x, y, z are encounter as 2
elements.
∴Hence, total no of elements is 5.
Q2.Find the no of elements tha satisfy the given set:
W   
A.1
B.3
C.2
D.0
Solution: A
Explanation:

 It is a singleton set means a set contain only one element.
∴Hence, the total no element is 1.
Q3.Which of the following option is correct list of all the elements for the given set:
P   x : x is a consonant in the word AXE
A.A, X, E
B.A, E
C.X
D.
Solution: C
Explanation:
Consonants are the English alphabet which exclude vowels.
Q4.Find the no of elements that satisfy given set
M   x : x is a english alphabet which precceds 'a'
A.1
B.25
C.26
D.0
Solution: D
Explanation:
No any English alphabet comes before ‘a’ alphabet.
∴Hence, the given set is an empty set.
Q5.Choose the correct set of elements for the given question:
The set of vertices present in a triangle and a square.
A.1, 2,3, 4
B.4,3
C.2,3
D.None of the above
Solution: B
Explanation:
No of vertices present in a triangle and a square is 3, 4 respectively.
Q6.Find the no of elements that satisfy the given set:
M   x : x is a vowel in the word STUDENT
A.3
B.2
C.1
D.0
Solution: B
Explanation:
Vowels are a, e, i, o & u.
Q7.Find the no of elements that satisfy the given set:
N   y : y 2 =4 and y  Z 
A.4
B.2
C.3
D.None of the above
Solution: D
Explanation:
Given:
y 2 =4 and y  Z
 y  2
Z means all positive integers exclude 0.
Hence, there is one element present in the given set.
Q8.Choose the correct no of elements according to this question:
The no. of months in a year having 31 days.
A.7
B.4
C.5
D.6
Solution: A
Explanation:
We know the months of a year having 31 days are January, March, May, July, August, October & December.
Q9.Find the no of elements that satisfy the given set:
S   x : x  R and 3x 2  2 x  6
A.1
B.2
C.0
D.None of the above
Solution: B
Explanation:
Given:
x  R and 3x 2  2 x  6
3x 2  2 x  6
 3x 2  2 x  6  0
Here a  3, b  2 & c  6.
For checking no of roots, we have to find the value of D(discriminant).
D  b 2  4ac
put the value of a, b and c in this above equation.
D  22  4* 3*(6)
 D  4  72
 D  76  0, which means 2 roots.
Hence, the no of elements is 2.
Q10. Choose the correct option which satisfy the given question:
X   x : x is a even no and 0 < x  2
A.0,1, 2
B.

C.0, 2
D.2
Solution: B
Explanation:
Here, the range is from 0 to 2 and there is no even no in that range. So, this is a null set.