Category: Class 11 Mathematics Objective Questions

Mathematics Multiple Choice Questions on “Three Dimensional Geometry – Distance between Two Points”.

1. Do we have distance formula in 3-D geometry also?
a) True
b) False
Answer: a
Clarification: Yes, we have distance formula in 3-D geometry also. Distance between two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is (sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}).

2. Find the distance between two points (5, 6, 7) and (2, 6, 3).
a) 3 units
b) 0 units
c) 4 units
d) 5 units
Answer: d
Clarification: We know, distance between two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is (sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}).
So, distance between two points (5, 6, 7) and (2, 6, 3) will be (sqrt{(5-2)^2+(6-6)^2+(7-3)^2}) = (sqrt{(3)^2+(4)^2}) = 5 units.

3. The points A (3, 2, 1), B (5, 3, -2) and C (-1, 0, 7) are collinear.
a) True
b) False
Answer: a
Clarification: We know, distance between two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is (sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}).
Distance AB = (sqrt{(3-5)^2+(2-3)^2+(1+2)^2} = sqrt{14})
Distance BC = (sqrt{(5+1)^2+(3-0)^2+(-2-7)^2} = 3sqrt{14})
Distance AC = (sqrt{(3+1)^2+(2-0)^2+(1-7)^2} = 2sqrt{14})
Since AB+AC = BC so, the three points are collinear.

4. The three points A (1, 2, 3), B (3, 1, 2), C (2, 3, 1) form ________________
a) equilateral triangle
b) right angled triangle
c) isosceles triangle
d) right angled isosceles triangle
Answer: a
Clarification: We know, distance between two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is (sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}).
Distance AB = (sqrt{(1-3)^2+(2-1)^2+(3-2)^2} = sqrt{6})
Distance BC = (sqrt{(3-2)^2+(1-3)^2+(2-1)^2} = sqrt{6})
Distance CA = (sqrt{(2-1)^2+(3-2)^2+(1-3)^2} = sqrt{6})
Since AB=BC=CA so, it forms equilateral triangle.

5. The three points A (3, 0, 3), B (5, 3, 2), C (6, 5, 5) form ________________
a) equilateral triangle
b) right angled triangle
c) isosceles triangle
d) right angled isosceles triangle
Answer: c
Clarification: We know, distance between two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is (sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}).
Distance AB = (sqrt{(3-5)^2+(0-3)^2+(3-2)^2} = sqrt{14})
Distance BC =(sqrt{(5-6)^2+(3-5)^2+(2-5)^2} = sqrt{14})
Distance AC =(sqrt{(3-6)^2+(0-5)^2+(3-5)^2} = sqrt{38})
Since AB =BC so, it forms isosceles triangle.

6. The three points A (7, 0, 10), B (6, -1, 6), C (9, -4, 6) form ________________
a) equilateral triangle
b) right angled triangle
c) isosceles triangle
d) right angled isosceles triangle
Answer: d
Clarification: We know, distance between two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is (sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}).
Distance AB = (sqrt{(7-6)^2+(0+1)^2+(10-6)^2} = sqrt{18})
Distance BC = (sqrt{(6-9)^2+(-1+4)^2+(6-6)^2} = sqrt{18})
Distance AC = (sqrt{(7-9)^2+(0+4)^2+(10-6)^2} = sqrt{36})
Since AB = BC and AB²+BC² = AC² so, it forms right angled isosceles triangle.

7. The points A (1, 2, -1), B (5, -2, 1), C (8, -7, 4), D (4, -3, 2) form_____________
a) trapezium
b) rhombus
c) square
d) parallelogram
Answer: d
Clarification: We know, distance between two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is (sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}).
Distance AB = (sqrt{(1-5)^2+(2+2)^2+(-1-1)^2}) = 6
Distance BC = (sqrt{(5-8)^2+(-2+7)^2+(1-4)^2} = sqrt{43})
Distance CD = (sqrt{(8-4)^2+(-7+3)^2+(4-2)^2}) = 6
Distance AD = (sqrt{(1-4)^2+(2+3)^2+(-1-2)^2} = sqrt{43})
Since AB=CD and BC=AD i.e. opposite two sides are equal so, it is a parallelogram.

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.

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 a_n
b) ∑a_n
c) Φ a_n
d) θ a_n

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 ∑a_n.

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.

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}).