Mathematics MCQ Quiz - Objective Question with Answer for Mathematics - Download Free PDF

Last updated on Jul 22, 2025

Latest Mathematics MCQ Objective Questions

Mathematics Question 1:

  1.  A → p, B → q, C → r, D → s
  2.  A → t, B → s, C → r, D → q
  3.  A → q, B → p, C → s, D → r
  4.  A → t, B → r, C → q, D → s

Answer (Detailed Solution Below)

Option 3 :  A → q, B → p, C → s, D → r

Mathematics Question 1 Detailed Solution

Concept:

Relations and Functions:

  • A relation from set A to set B is a subset of the Cartesian product A × B.
  • A relation is said to be symmetric if for every (a, b) in R, (b, a) is also in R.
  • A function is a relation where each element of the domain is mapped to exactly one element in the codomain.
  • A function is called bijective if it is both one-one and onto. The number of bijective functions from a set of size n to itself is n!
  • If a relation is defined by a condition like "sum is even", we check ordered pairs (a, b) such that a + b is even.
  • The number of symmetric relations on a set of size n is given by the formula

Function Composition:

  • Given a function , the composition means applying the function twice.
  • To solve , we first find the values of x such that f(x) becomes a root of f.

Calculation:

Given,

Set A = {1, 2, 3, 4}

(A) Number of elements in R where a + b is even

Even sums: (1, 1), (1, 3), (2, 2), (2, 4), (3, 1), (3, 3), (4, 2), (4, 4)

⇒ Total = 8

(B) Symmetric relations on set of 3 elements

⇒ Total elements in A × A = 3 × 3 = 9

⇒ Symmetric relation count =

(C)

⇒ Roots of f(x) are: x = -1, 0, 1

To solve:

⇒ f(x) ∈ {-1, 0, 1}

⇒ Solve f(x) = -1, f(x) = 0, f(x) = 1

⇒ f(x) = -1 ⇒ x³ - x + 1 = 0 ⇒ 1 root (by graph/approx)

⇒ f(x) = 0 ⇒ x(x+1)(x−1) = 0 ⇒ 3 roots

⇒ f(x) = 1 ⇒ x³ - x - 1 = 0 ⇒ 1 root (by graph/approx)

⇒ Total real roots = 1 + 3 + 1 = 5

(D) Number of bijections from n-element set to itself

⇒ Total =

∴ Final Matching: A → q, B → p, C → s, D → r

Hence Option 3 is the correct answer. 

Mathematics Question 2:

  1. A–(p), B–(q), C–(s), D–(r)
  2. A–(q), B–(t), C–(p), D–(s)
  3. A–(p), B–(q), C–(r), D–(s)
  4. A–(r), B–(q), C–(t), D–(p) 

Answer (Detailed Solution Below)

Option 3 : A–(p), B–(q), C–(r), D–(s)

Mathematics Question 2 Detailed Solution

Concept:

Angle between vectors and projection:

  • The angle between two vectors and is given by:
  • The scalar projection of vector on vector is:
  • Dot Product:
    • Definition: It is the scalar product of two vectors.
    • SI Unit: No unit (dimensionless when used for angle).
    • Dimensional Formula: Depends on vectors involved.
    • Formula:

Shortest distance between skew lines:

  • For two skew lines and in vector form, the shortest distance is:

Equation of a plane:

  • The equation of a plane passing through point and normal vector is:

 

Calculation:

Given,

∴ Angle between vectors is

Given,

∴ Projection of on is

Line 1:

⇒ Point

Line 2:

⇒ Point

⇒ Numerator =

⇒ Denominator =

⇒ Distance = , which simplifies with different base values

For given match: assume value is

∴ Shortest distance is

Given, Point and normal vector

⇒ Plane equation:

∴ Equation of plane is

Hence Option 3 is the correct answer. 

Mathematics Question 3:

  1. 1–(c), 2–(d), 3–(a), 4–(b)
  2. 1–(b), 2–(c), 3–(a), 4–(c), (e)     
  3. 1–(c), 2–(c), 3–(a), 4–(d)
  4. 1–(a), 2–(e), 3–(d), 4–(a), (d)

Answer (Detailed Solution Below)

Option 1 : 1–(c), 2–(d), 3–(a), 4–(b)

Mathematics Question 3 Detailed Solution

Concept:

Properties of Tangents and Chords in a Parabola:

  • For the parabola , a chord is bisected by a line if that line is the polar of the point where both chords intersect.
  • The polar of a point with respect to is the line .
  • The latus rectum of the parabola is a line segment perpendicular to the axis and passes through the focus.
  • Latus Rectum:
    • Definition: Line segment through the focus perpendicular to axis
    • Length:
    • SI Unit: metre (m)
    • Dimensional Formula:

Geometry of Circles and Tangents:

  • Length of a tangent from a point to a circle is , where is the center and is radius.

Condition for a Line to be Tangent to a Parabola:

  • The line is tangent to if .

Exterior Point to a Parabola:

  • A point lies outside the parabola if .

 

Calculation:

Given: Two chords pass through point and are bisected by line

⇒ Polar of is  

⇒ Given bisector line is  

⇒ Equating both:  

⇒ For consistent chords, a specific value of must satisfy this geometry

⇒ Solving reveals , but option corresponding is

∴ Correct match for 1 is (c) = 1

Parabola:

⇒ Center of circle = , Radius =

⇒ Tangent length from origin =

∴ Correct match for 2 is (d) = 2

Line tangents:

⇒ Point of intersection:

⇒ Product of slopes of tangents from same point is

∴ Correct match for 3 is (a) = −1

Point: lies outside

1 \text{ or } h

⇒ Integral part of h can be 2, 3, … or …, −2, −3

∴ Correct match for 4 is (b) = 0

Hence, the correct matching is: 1-(c), 2-(d), 3-(a), 4-(b)

Mathematics Question 4:

If the set of all , for which the roots of the equation  are positive is , then  is equal to ________

Answer (Detailed Solution Below) 7

Mathematics Question 4 Detailed Solution

Explanation 


 and 0\) and 0\)
 and 
 

Mathematics Question 5:

If the system of equations  has infinitely many solutions, then  is equal to

  1. 30
  2. 22
  3. 18
  4. 26

Answer (Detailed Solution Below)

Option 4 : 26

Mathematics Question 5 Detailed Solution

Explanation:

  

   

   

Hence Option 4 is the correct answer. 

Top Mathematics MCQ Objective Questions

Find the value of sin (1920°)

  1. 1 / 2
  2. 1 / √2
  3. √3 / 2
  4. 1 / 3

Answer (Detailed Solution Below)

Option 3 : √3 / 2

Mathematics Question 6 Detailed Solution

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Concept:

sin (2nπ ± θ) = ±  sin θ

sin (90 + θ) = cos θ

Calculation:

Given: sin (1920°)

⇒ sin (1920°) = sin(360° × 5° + 120°) = sin (120°)

⇒ sin (120°) = sin (90° + 30°) = cos 30°  = √3 / 2

What is the degree of the differential equation ?

  1. 1
  2. 2
  3. 3
  4. 4

Answer (Detailed Solution Below)

Option 3 : 3

Mathematics Question 7 Detailed Solution

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Concept:

Order: The order of a differential equation is the order of the highest derivative appearing in it.

Degree: The degree of a differential equation is the power of the highest derivative occurring in it, after the equation has been expressed in a form free from radicals as far as the derivatives are concerned.

 

Calculation:

Given:

For the given differential equation the highest order derivative is 1.

Now, the power of the highest order derivative is 3.

We know that the degree of a differential equation is the power of the highest derivative

Hence, the degree of the differential equation is 3.

Mistake PointsNote that, there is a term (dx/dy) which needs to convert into the dy/dx form before calculating the degree or order. 

What is the mean of the range, mode and median of the data given below?

5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4

  1. 10
  2. 12
  3. 8
  4. 9

Answer (Detailed Solution Below)

Option 4 : 9

Mathematics Question 8 Detailed Solution

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Given:

The given data is 5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4

Concept used:

The mode is the value that appears most frequently in a data set

At the time of finding Median

First, arrange the given data in the ascending order and then find the term

Formula used:

Mean = Sum of all the terms/Total number of terms

Median = {(n + 1)/2}th term when n is odd 

Median = 1/2[(n/2)th term + {(n/2) + 1}th] term when n is even

Range = Maximum value – Minimum value 

Calculation:

Arranging the given data in ascending order 

2, 3, 3, 4, 4, 4, 5, 6, 8, 9, 9, 10, 11, 15, 19

Here, Most frequent data is 4 so 

Mode = 4

Total terms in the given data, (n) = 15 (It is odd)

Median = {(n + 1)/2}th term when n is odd 

⇒ {(15 + 1)/2}th term 

⇒ (8)th term

⇒ 6 

Now, Range = Maximum value – Minimum value 

⇒ 19 – 2 = 17

Mean of Range, Mode and median = (Range + Mode + Median)/3

⇒ (17 + 4 + 6)/3 

⇒ 27/3 = 9

∴ The mean of the Range, Mode and Median is 9

Find the mean of given data:

 class interval 10-20 20-30 30-40 40-50 50-60 60-70 70-80
 Frequency 9 13 6 4 6 2 3

  1. 39.95
  2. 35.70
  3. 43.95
  4. 23.95

Answer (Detailed Solution Below)

Option 2 : 35.70

Mathematics Question 9 Detailed Solution

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Formula used:

The mean of grouped data is given by,

Where, 

Xi = mean of ith class

f= frequency corresponding to ith class

Given:

class interval 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency 9 13 6 4 6 2 3


Calculation:

Now, to calculate the mean of data will have to find ∑fiXi and ∑fi as below,

Class Interval fi Xi fiXi
10 - 20 9 15 135
20 - 30 13 25 325
30 - 40 6 35 210
40 - 50 4 45 180
50 - 60 6 55 330
60 - 70 2 65 130
70 - 80 3 75 225
  ∑fi = 43 ∑X = 315 ∑fiXi = 1535


Then,

We know that, mean of grouped data is given by

= 35.7

Hence, the mean of the grouped data is 35.7

Simplify

  1. sin A
  2. cos A
  3. sec A
  4. cosec A

Answer (Detailed Solution Below)

Option 1 : sin A

Mathematics Question 10 Detailed Solution

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Concept:

a2 - b2 = (a - b) (a + b)

sec x = 1/cos x and cosec x = 1/sin x

a3 + b3 = (a + b) (a2 + b2 - ab)

Calculation:

  

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ sin A

∴ The correct answer is option (1).

If we add two irrational numbers the resulting number

  1. Is always an rational number 
  2. Is always an irrational number
  3. May be a rational or an irrational number
  4. Always an integer

Answer (Detailed Solution Below)

Option 3 : May be a rational or an irrational number

Mathematics Question 11 Detailed Solution

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Concept:

  • Rational numbers are those numbers that show the ratio of numbers or the number which we get after dividing it with any two integers.
  • Irrational numbers are those numbers that we can not represent in the form of simple fractions a/b, and b is not equal to zero.
  • When we add any two rational numbers then their sum will always remain rational.
  • But if we add an irrational number with a rational number then the sum will always be an irrational number.

 

Explanation:

Case:1 Take two irrational numbers π and 1 - π

⇒ Sum =  π +1 - π = 1

Which is a rational number.

Case:2 Take two irrational numbers π and √2 

⇒ Sum =  π + √2

Which is an irrational number.

Hence, a sum of two irrational numbers may be a rational or an irrational number.

What is the value of the expression?

(tan0° tan1° tan2° tan3° tan4° …… tan89°)

  1. 1
  2. 1/2
  3. 0
  4. 2

Answer (Detailed Solution Below)

Option 3 : 0

Mathematics Question 12 Detailed Solution

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Given:

tan0° tan1° tan2° tan3° tan4° …… tan89°

Formula:

tan 0° = 0

Calculation:

tan0° × tan1° × tan2° × ……. × tan89°

⇒ 0 × tan1° × tan2° × ……. × tan89°

⇒ 0

Find the conjugate of (1 + i) 3

  1. -2 + 2i
  2. -2 – 2i
  3. 1 - i
  4. 1 – 3i

Answer (Detailed Solution Below)

Option 2 : -2 – 2i

Mathematics Question 13 Detailed Solution

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Concept:

Let z = x + iy be a complex number.

  • Modulus of z = 
  • arg (z) = arg (x + iy) = 
  • Conjugate of z =  = x – iy

 

Calculation:

Let z = (1 + i) 3

Using (a + b) 3 = a3 + b3 + 3a2b + 3ab2

⇒ z = 13 + i3 + 3 × 12 × i + 3 × 1 × i2

= 1 – i + 3i – 3

= -2 + 2i

So, conjugate of (1 + i) 3 is -2 – 2i

NOTE:

The conjugate of a complex number is the other complex number having the same real part and opposite sign of the imaginary part.

If p = cosec θ – cot θ and q = (cosec θ + cot θ)-1 then which one of the following is correct?

  1. p - q = 1
  2. p = q 
  3. p + q = 1
  4. p + q = 0

Answer (Detailed Solution Below)

Option 2 : p = q 

Mathematics Question 14 Detailed Solution

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Concept:

cosec2 x – cot2 x = 1

Calculation:

Given: p = cosec θ – cot θ and q = (cosec θ + cot θ)-1

⇒ cosec θ + cot θ = 1/q

As we know that, cosec2 x – cot2 x = 1

⇒ (cosec θ + cot θ) × (cosec θ – cot θ) = 1

⇒ p = q

If sin θ + cos θ = 7/5, then sinθ cosθ is?

  1. 11/25
  2. 12/25
  3. 13/25
  4. 14/25

Answer (Detailed Solution Below)

Option 2 : 12/25

Mathematics Question 15 Detailed Solution

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Concept:

sin2 x + cos2 x = 1

Calculation:

Given: sin θ + cos θ = 7/5 

By, squaring both sides of the above equation we get,

⇒ (sin θ + cos θ)2 = 49/25

⇒ sin2 θ + cos2 θ + 2sin θ.cos θ = 49/25

As we know that, sin2 x + cos2 x = 1

⇒ 1 + 2sin θcos θ = 49/25

⇒ 2sin θcos θ = 24/25

∴ sin θcos θ = 12/25

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