250+ TOP MCQs on Universal Set & Answers | Class 11 Maths

Mathematics Multiple Choice Questions on “Universal Set”.

1. A set which is superset of all basic sets of that type?
a) Power set
b) Universal set
c) Empty set
d) Singleton set
Answer: b
Clarification: Universal set is the set which is a superset of all basic sets of that type.
Power set is the set of all subsets of given set. Empty set is the set which does not contain any element. Singleton set is the set that contains one element.

2. Which of the following is universal set for integers?
a) Natural numbers
b) Whole numbers
c) Rational numbers
d) Prime numbers
Answer: c
Clarification: Since set of rational numbers is superset for set of integers so it can be a universal set for integers. Rest are subset of integers so they can not act as the universal set for set of integers.

3. Let A={1,2}, B={2,4}, C={4,5,6}. Which of the following may be considered as the universal set for set A, B, C?
a) {1,6,7,8,9}
b) {1,2,3,4}
c) {2,4,5,6}
d) {1,2,3,4,5,6}
Answer: d
Clarification: Universal set is the set which is superset of all basic sets of that type.
{1,2,3,4,5,6} is the set which contains all the elements of set A, B, C.

4. Which of the following is a universal set for the equilateral triangle?
a) Set of isosceles triangles
b) Set of right triangles
c) Set of acute triangles
d) Set of obtuse triangles
Answer: a
Clarification: Set of isosceles triangles can be considered as universal set for set of equilateral triangles because all equilateral triangles are isosceles.

5. Which of the following is considered as universal set for squares?
a) Set of Rhombus
b) Set of Parallelogram
c) Set of Rectangle
d) Set of Trapezium
Answer: c
Clarification: Set of rectangles is considered as universal set for set of squares because all squares are rectangles.

6. Which of the following is universal set for {a, p}?
a) Set of vowels
b) Set of consonants
c) Set of letters of English alphabet
d) Set of numbers
Answer: c
Clarification: Set of letters of English alphabet can be considered as universal set for {a, p} because ‘a’ is a vowel and ‘p’ is a consonant.

7. Which of the following is considered as universal set for set of multiple of 4?
a) Set of multiple of 16
b) Set of multiple of 12
c) Set of multiple of 2
d) Set of multiple of 8
Answer: c
Clarification: Since every multiple of 4 is a multiple of 2 so, set of multiple of 2 is considered a set of multiple of 4.

8. U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Which of the following is not a subset of the universal set?
a) {1,2}
b) {0,1,2,3,4,5,6,7,8,9}
c) {2,3,5,7}
d) {1,2,3,4,5,6,7,8,9,10}
Answer: d
Clarification: In {1,2,3,4,5,6,7,8,9,10}, 10 is not present in set U. So, it is not the subset of universal set. Rest all are subsets of set U as their elements are present in universal set.

9. The set {a, b, e, i, o, u, v, z} is a universal set for a set of vowels.
a) True
b) False
Answer: a
Clarification: Since set {a, b, e, i, o, u, v, z} has all the vowels so it can be considered as the universal set for set of vowels.

10. The set of prime numbers is a universal set for odd numbers.
a) True
b) False
Answer: b
Clarification: Since every prime number is not odd as 2 is even prime number so set of prime numbers can not be considered as universal set for a set of odd numbers.

250+ TOP MCQs on Complex Numbers & Answers | Class 11 Maths

Mathematics Multiple Choice Questions on “Complex Numbers”.

1. The value of x and y if (3y – 2) + i(7 – 2x) = 0
(a) x = 7/2, y = 2/3
(b) x = 2/7, y = 2/3
(c) x = 7/2, y = 3/2
(d) x = 2/7, y = 3/2

Clarification: x = 7/2, y = 2/3
Hint:
Given, (3y – 2) + i(7 – 2x) = 0
Compare real and imaginary part, we get
3y – 2 = 0
⇒ y = 2/3
and 7 – 2x = 0
⇒ x = 7/2
So, the value of x = 7/2 and y = 2/3

2. Is i(iota) a root of 1+x2=0?
a) True
b) False

Answer: a
Clarification: 1+x2 = 0
1 + i2 = 1 – 1 = 0.
So, it is a root of 1 + x2 = 0.

3. In z=4+i, what is the real part?
a) 4
b) i
c) 1
d) 4+i

Answer: a
Clarification: In z=a+bi, a is real part and b is imaginary part.
So, in 4+i, real part is 4.

4. In z=4+i, what is imaginary part?
a) 4
b) i
c) 1
d) 4+i

Answer: c
Clarification: In z=a+bi, a is real part and b is imaginary part.
So, in 4+i, imaginary part is 1.

5. (x+3) + i(y-2) = 5+i2, find the values of x and y.
a) x=8 and y=4
b) x=2 and y=4
c) x=2 and y=0
d) x=8 and y=0

Answer: b
Clarification: If two complex numbers are equal, then corresponding parts are equal i.e. real parts of both are equal and imaginary parts of both are equal.
x+3 = 5 and y-2 = 2
x = 5-3 and y = 2+2
x=2 and y=4.

6. If z1 = 2+3i and z2 = 5+2i, then find sum of two complex numbers.
a) 4+8i
b) 3-i
c) 7+5i
d) 7-5i

Answer: c
Clarification: In addition of two complex numbers, corresponding parts of two complex numbers are added i.e. real parts of both are added and imaginary parts of both are added.
So, sum = (2+5) + (3+2) i = 7+5i.

7. 0+0i is ______________________for complex number z.
a) additive inverse
b) additive identity element
c) multiplicative identity element
d) multiplicative inverse

Answer: b
Clarification: On adding zero (0+0i) to a complex number, we get same complex number so 0+0i is additive identity element for complex number z i.e. z+0 = z.

8. 1+0i is _________________ for complex number z.
a) additive inverse
b) additive identity element
c) multiplicative identity element
d) multiplicative inverse

Answer: c
Clarification: On multiplying one (1+0i) to a complex number, we get same complex number so 1+0i is multiplicative identity element for complex number z i.e. z*1=z.

9. -z is _________________ for complex number z.
a) additive inverse
b) additive identity element
c) multiplicative identity element
d) multiplicative inverse

Answer: a
Clarification: On adding negative of complex number (-z) to complex number z, we get additive identity element zero i.e. z+(-z)=0.

10. 1/z is _________________ for complex number z.
a) additive inverse
b) additive identity element
c) multiplicative identity element
d) multiplicative inverse

Answer: d
Clarification: On multiplying reciprocal of complex number (1/z) to complex number z, we get multiplying inverse one i.e. z*1=z.

11. If z1 = 2+3i and z2 = 5+2i, then find z1-z2.
a) -3+1i
b) 3-i
c) 7+5i
d) 7-5i

Answer: a
Clarification: In subtracting one complex number from other, difference of corresponding parts of two complex numbers is calculated. So, z1-z2 = (2-5) + (3-2) i = -3+1i.

12. Value of i(iota) is ____________
a) -1
b) 1
c) (-1)1/2
d) (-1)1/4

Answer: c
Clarification: Iota is used to denote complex number.
The value of i (iota) is (sqrt{-1}) i.e. (-1)1/2.

13. Find real θ such that (3 + 2i × sin θ)/(1 – 2i × sin θ) is imaginary
(a) θ = nπ ± π/2 where n is an integer
(b) θ = nπ ± π/3 where n is an integer
(c) θ = nπ ± π/4 where n is an integer
(d) None of these

14. If {(1 + i)/(1 – i)}n = 1 then the least value of n is
(a) 1
(b) 2
(c) 3
(d) 4

15. If arg (z) < 0, then arg (-z) – arg (z) =
(a) π
(b) -π
(c) -π/2
(d) π/2

16. if x + 1/x = 1 find the value of x2000 + 1/x2000 is
(a) 0
(b) 1
(c) -1
(d) None of these

17. The value of √(-144) is
(a) 12i
(b) -12i
(c) ±12i
(d) None of these

18. If the cube roots of unity are 1, ω, ω², then the roots of the equation (x – 1)³ + 8 = 0 are
(a) -1, -1 + 2ω, – 1 – 2ω²
(b) – 1, -1, – 1
(c) – 1, 1 – 2ω, 1 – 2ω²
(d) – 1, 1 + 2ω, 1 + 2ω²

19. (1 – w + w²)×(1 – w² + w4)×(1 – w4 + w8) × …………… to 2n factors is equal to
(a) 2n
(b) 22n
(c) 23n
(d) 24n

20. The modulus of 5 + 4i is
(a) 41
(b) -41
(c) √41
(d) -√41

250+ TOP MCQs on Series and Answers | Class 11 Maths

Mathematics Multiple Choice Questions on “Series”.

1. The sequence is same as series.
a) True
b) False

Answer: b
Clarification: No, sequence and series both are not same. When we use addition between the terms of sequence, it is said to be series.

2. A series can also be denoted by symbol ____________
a) π an
b) ∑an
c) Φ an
d) θ an

Answer: b
Clarification: When we use addition between the terms of sequence, it is said to be series.
We know that addition can also be written in the form of sigma so, series can also be denoted by ∑an.

3. 1+2+3+4 or 10 is a series?
a) 1+2+3+4 only
b) 10 only
c) 1+2+3+4 and 10
d) neither 1+2+3+4 nor 10

Answer: a
Clarification: 1+2+3+4 is a finite series of 4 terms.
10 is sum of the terms of this series not a series itself.

4. Find sum of series 2+3+5+7.
a) 5
b) 10
c) 17
d) infinite

Answer: c
Clarification: Sum of the series 2+3+5+7 is finite because given series has finite number of terms. The sum of given 4 terms i.e. 17.

5. Find the sum of first 5 terms of series 2+4+6+………………
a) 14
b) 16
c) 20
d) 30

Answer: d
Clarification: Since 2, 4 and 6 all are even numbers so, given series involve all even number terms.
The next two terms will be 8 and 10 so, sum will be 2+4+6+8+10 = 30.

6. (sum_{i=1}^4) 2n+3 = _____________________
a) 5
b) 12
c) 21
d) 32

Answer: d
Clarification: a1 = 2*1+3 = 5, a2 = 2*2+3 = 7, a3 = 2*3+3 = 9, a4 = 2*4+3 = 11.
Sum = 5+7+9+11 = 32.

7. Which of the following is not a series?
a) Arithmetic series
b) Geometric series
c) Isometric series
d) Harmonic series

Answer: c
Clarification: The isometric series is not a series. Rest all are series i.e. arithmetic series, geometric series and harmonic series.

250+ TOP MCQs on Three Dimensional Geometry – Section Formula & Answers | Class 11 Maths

Mathematics Multiple Choice Questions on “Three Dimensional Geometry – Section Formula”.

1. Find midpoint of (1, 4, 6) and (5, 8, 10).
a) (6, 12, 8)
b) (3, 6, 8)
c) (1, 9, 12)
d) (4, 9, 12)
Answer: b
Clarification: We know, midpoint of (x1, y1, z1) and (x2, y2, z2) is (x1+x2) /2, (y1+y2) /2, (z1+z2)/2).
So, midpoint of (1, 4, 6) and (5, 8, 10) is ((1+5)/ 2, (4+8)/ 2, (6+10)/2) is (3, 6, 8).

2. The coordinates of a point dividing the line segment joining (1, 2, 3) and (4, 5, 6) internally in the ratio 2:1 is ____________________
a) (3, 4, 5)
b) (5, 4, 3)
c) (5, 3, 4)
d) (4, 5, 3)
Answer: a
Clarification: The coordinates of a point dividing the line segment joining (x1, y1, z1) and (x2, y2, z2) internally in the ratio m : n is ((frac{mx_2+nx_1}{m+n},frac{my_2+ny_1}{m+n},frac{mz_2+nz_1}{m+n})).
So, the coordinates of a point dividing the line segment joining (1, 2, 3) and (4, 5, 6) internally in the ratio 2:1 is ((frac{2*4+1*1}{2+1},frac{2*5+1*2}{2+1},frac{2*6+1*3}{2+1})) = (3, 4, 5).

3. In which ratio (3, 4, 5) divides the line segment joining (1, 2, 3) and (4, 5, 6) internally?
a) 1:2
b) 2:1
c) 3:4
d) 4:3
Answer: b
Clarification: The coordinates of a point dividing the line segment joining (x1, y1, z1) and (x2, y2, z2) internally in the ratio m: n is ((frac{mx_2+nx_1}{m+n},frac{my_2+ny_1}{m+n},frac{mz_2+nz_1}{m+n})).
Let the ratio be k : 1.So, the coordinates of a point dividing the line segment joining (1, 2, 3) and (4, 5, 6) internally in the ratio k: 1 is ((frac{k*4+1*1}{k+1},frac{k*5+1*2}{k+1},frac{k*6+1*3}{k+1}))
=> ((frac{k*4+1*1}{k+1},frac{k*5+1*2}{k+1},frac{k*6+1*3}{k+1})) is same as (3, 4, 5).
=> (4k+1)/(k+1) = 3
=> 4k+1 = 3k+3
=> k = 2
So, ratio is 2:1.

4. The coordinates of a point dividing the line segment joining (1, 2, 3) and (4, 5, 6) externally in the ratio 2:1 is ____________________
a) (4, 5, 6)
b) (6, 8, 9)
c) (7, 8, 9)
d) (8, 6, 4)
Answer: c
Clarification: The coordinates of a point dividing the line segment joining (x1, y1, z1) and (x2, y2, z2) externally in the ratio m : n is ((frac{mx_2-nx_1}{m-n},frac{my_2-ny_1}{m-n},frac{mz_2-nz_1}{m-n})).
So, the coordinates of a point dividing the line segment joining (1, 2, 3) and (4, 5, 6) externally in the ratio 2:1 is ((frac{2*4-1*1}{2-1},frac{2*5-1*2}{2-1},frac{2*6-1*3}{2-1})) = (7, 8, 9).

5. If coordinates of vertices of a triangle are (7, 6, 4), (5, 4, 6), (9, 5, 8), find the coordinates of centroid of the triangle.
a) (7, 5, 3)
b) (7, 3, 5)
c) (5, 3, 7)
d) (3, 5, 7)
Answer: a
Clarification: If coordinates of vertices of a triangle are (x1, y1, z1), (x2, y2, z2), (x3, y3, z3) the coordinates of centroid of the triangle are ((x1+x2+x3)/3, (y1+y2+y3)/3, (z1+z2+z3)/3)
So, coordinates of centroid of the given triangle are ((7+5+9)/3, (6+4+5)/3, (4+6+8)/3) = (7, 5, 3).

6. The ratio in which line joining (1, 2, 3) and (4, 5, 6) divide X-Y plane is ________
a) 2
b) -2
c) 1/2
d) -1/2
Answer: d
Clarification: The coordinates of a point dividing the line segment joining (x1, y1, z1) and (x2, y2, z2) internally in the ratio m : n is ((frac{mx_2+nx_1}{m+n},frac{my_2+ny_1}{m+n},frac{mz_2+nz_1}{m+n})).
Let ratio be k : 1.
So, z-coordinate of the point will be (k*6+1*3)/(k+1).
We know, for X-Y plane, z coordinate is zero.
(6k+1*3)/(k+1) = 0 => k=-1/2

7. Find the points which trisects the line joining (4, 9, 8) and (13, 27, -4).
a) (7, 4, 15)
b) (7, 15, 4)
c) (4, 15, 7)
d) (4, 7, 15)
Answer: b
Clarification: Points which trisect the line divides it into 2:1 and 1:2.
The coordinates of a point dividing the line segment joining (x1, y1, z1) and (x2, y2, z2) internally in the ratio m : n is ((frac{mx_2+nx_1}{m+n},frac{my_2+ny_1}{m+n},frac{mz_2+nz_1}{m+n})).
For 1:2, coordinates of point are ((frac{1*13+2*4}{1+2},frac{1*27+2*9}{1+2},frac{-4+2*8}{1+2})) = (7, 15, 4)
For 2:1, coordinates of point are ((frac{2*13+1*4}{1+2},frac{2*27+1*9}{1+2},frac{-8+1*8}{1+2})) = (10, 21, 0)

8. Find the points which trisects the line joining (4, 9, 8) and (13, 27, -4).
a) (0, 21, 10)
b) (0, 21, 4)
c) (10, 21, 0)
d) (4, 4, 0)
Answer: c
Clarification: Points which trisect the line divides it into 2:1 and 1:2.
The coordinates of a point dividing the line segment joining (x1, y1, z1) and (x2, y2, z2) internally in the ratio m : n is ((frac{mx_2+nx_1}{m+n},frac{my_2+ny_1}{m+n},frac{mz_2+nz_1}{m+n})).
For 1:2, coordinates of point are ((frac{1*13+2*4}{1+2},frac{1*27+2*9}{1+2},frac{-4+2*8}{1+2})) = (7, 15, 4)
For 2:1, coordinates of point are ((frac{2*13+1*4}{1+2},frac{2*27+1*9}{1+2},frac{-8+1*8}{1+2})) = (10, 21, 0)

9. If P (2, 3, 9), Q (2, 5, 5) and R (8, 5, 3) are vertices of a triangle then find the length of median through P.
a) (sqrt{24})
b) (sqrt{38})
c) (sqrt{11})
d) (sqrt{53})
Answer: b
Clarification: We know, midpoint of (x1, y1, z1) and (x2, y2, z2) is ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2).
Midpoint of line QR is (5, 5, 4).
Length of median through P is distance between midpoint of QR and P i.e. (sqrt{(5-2)^2+(5-3)^2+(4-9)^2} = sqrt{(3)^2+(2)^2+(-5)^2} = sqrt{38})

10. If P (2, 3, 9), Q (2, 5, 5) and R (8, 5, 3) are vertices of a triangle then find the length of median through Q.
a) (sqrt{24})
b) (sqrt{38})
c) (sqrt{11})
d) (sqrt{53})
Answer: c
Clarification: We know, midpoint of (x1, y1, z1) and (x2, y2, z2) is ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2).
Midpoint of line PR is (5, 4, 6).
Length of median through Q is distance between midpoint of PR and Q i.e. (sqrt{(5-2)^2+(4-5)^2+(6-5)^2} = sqrt{(3)^2+(-1)^2+(1)^2} = sqrt{11}).

250+ TOP MCQs on Operation on Sets-1 & Answers | Class 11 Maths

Mathematics Multiple Choice Questions on “Operation on Sets-1”.
1. If A = {1,2,3} and B = {3,4,5,6}. Find A∪B.
a) {1,2,3}
b) {3}
c) {1,2,3,4,5,6}
d) { }
View Answer
Answer: c
Clarification: Union of set A and B is a set that contains all the elements of set A and set B.
A = {1,2,3}
B = {3,4,5,6}
A∪B = {1,2,3,4,5,6}.
2. Let A be the set of odd numbers and B be the set of even numbers then find A∩B.
a) Set of prime numbers
b) Set of real numbers
c) Empty set
d) Set of natural numbers
View Answer
Answer: c
Clarification: Intersection of set A and B is a set that contains elements which is common to both set A and set B. Set of odd numbers and even numbers does not have any common element so, A∩B is an empty set.
3. If A={a, e, i, o, u} and B={a, e, u} then A∪B = __________
a) A
b) B
c) Φ
d) A∩B
View Answer
Answer: a
Clarification: Union of set A and B is a set that contains all the elements of set A and set B.
A={a, e, i, o, u}
B={a, e, u}
A∪B={a, e, i, o, u} = A.
4. If A = {a, e, i, o, u} and B = {a, e, u} then A∩B=__________
a) A
b) B
c) Φ
d) A∪B
View Answer
Answer: b
Clarification: Intersection of set A and B is a set that contains elements which is common to both set A and set B.
A = {a, e, i, o, u}
B = {a, e, u}
A∩B = {a, e, u} = B.
5. If A = {1,2,3} and B = {3,4,5,6}. Find A∩B.
a) {1,2,3}
b) { }
c) {1,2,3,4,5,6}
d) {3}
View Answer
Answer: d
Clarification: Intersection of set A and B is a set that contains elements which is common to both set A and set B.
A = {1,2,3}
B = {3,4,5,6}
Here, 3 is common to both sets so A∩B = {3}.
6. Is A∪B = B∪A?
a) True
b) False
View Answer
Answer: a
Clarification: Let A = {1,2} and B = {2,3}. Here A∪B = {1,2,3} = B∪A.
A∪B or B∪A is same set that contains all the elements of set A and set B.
7. Is A∩B = B∩A?
a) True
b) False
View Answer
Answer: a
Clarification: Let A = {1,2} and B = {2,3}. Here A∩B = {2} = B∩A.
A∩B or B∩A is same set that contains elements which are common to both set A and B.

250+ TOP MCQs on Complex Numbers-2 & Answers | Class 11 Maths

Mathematics Question Bank on “Complex Numbers-2”.

1. z1=1+2i and z2=2+3i. Find z1z2.
a) 2+6i
b) -4+0i
c) -4+7i
d) 8+7i
Answer: c
Clarification: z1z2 = (1+2i) (2+3i)
= 2 + 3i + 4i -6
= -4 + 7i.

2. i2 =______________________
a) 1
b) -1
c) i
d) -i
Answer: b
Clarification: We know, i = (sqrt{-1})
=> i2 = -1.

3. i7 =______________
a) 1
b) -1
c) i
d) -i
Answer: d
Clarification: We know, i = (sqrt{-1})
=> i2 = -1 => i4 = 1.
So, i7 = i4.i3 = 1*i2*i = (-1)*i = -i.

4. i241 =________________
a) 1
b) -1
c) i
d) -i
Answer: c
Clarification: We know, i = (sqrt{-1})
=> i2 = -1 => i4 = 1.
So, i241 = (i4)60 * i = 1 * i = i.

5. Square roots of -7 are____________
a) 7i and -7i
b) (sqrt{7}) i
c) –(sqrt{7}) i
d) (sqrt{7}) i and –(sqrt{7}) i
Answer: d
Clarification: We know, i2 = -1.
-7 = 7(i2)
Square root of i2 is ±i so, square root of -7 are (sqrt{7})i and –(sqrt{7})i.

6. (-i) (8+5i) =________________
a) 8+5i
b) -8-5i
c) -5-8i
d) 5-8i
Answer: d
Clarification: (-i) (8+5i) = -8i – 5 i2
= -8i -5(-1) = 5-8i.

7. (2-i)3 =________________
a) 2-3i
b) 8-i
c) 2-11i
d) 2+11i
Answer: c
Clarification: We know, (a-b)3 = a3-b3-3ab(a-b)
So, (2-i)3 = 23-(i)3-3(2)(i) (2-i)
= 8-(-i)-6i(2-i)
= 8+i-12i-6
= 2-11i.

8. Is z*(bar{z}) = |z|2?
a) True
b) False
Answer: a
Clarification: Let z=a+ bi
=>(bar{z}) = a-bi
So, z*(bar{z}) = (a+bi) (a-bi) = a2-(bi)2 = a2-(b2) (-1) = a2+b2
|z|=(sqrt{a^2+b^2}) => |z|2 = a2+b2
Hence, z*(bar{z}) = |z|2.

9. Find multiplicative inverse of 3+5i.
a) 87+145i
b) 87-145i
c) 145-87i
d) 145+87i
Answer: b
Clarification: We know, z*(bar{z}) = |z|2.
(1/z) = (bar{z})|z|2
z-1=(3-5i) (32+52) = (3-5i) (29) = 87-145i.

10. i-35 =___________________
a) 1
b) -1
c) i
d) -i
Answer: c
Clarification: We know, i-35= 1/i35 = i/i36
= i/(i4)9 = i/1 = i.

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