Mathematics MCQ Class 12

CBSE Class 12 2023 : Maths Important MCQs for CBSE Board Exam 2023

In this post, CBSE Board 12th Maths Exam 2023 Important MCQs Question Answers have been given.

The purpose of these practice questions and answers is to help the students prepare for their exams and perform well in them.

It is advised to read this resource before the actual paper as it can give students an idea about the type of questions they can expect in the exam.

1 The number of equivalence relations that can be defined in the set A= {1,2,3} which containing the elements (1,2) is

(a) 0

(b) 1

(c) 2

(d) 3

2 The number of one-to-one functions that can be defined from the set {1,2,3,4,5} to {a, b}

a) 5

b) 0

c) 2

d) 3

3 What is the simplified form of cos-1(4x3 − 3x)

(a) 3sin−1 x

(b) 3cos−1 x

(c) π − 3 sin−1 x

(d) π − 3 cos−1 x

4 The number of all possible matrices of order2x3 with entry 1 or 2

1)16

2) 64

3) 6

4) 24

5 If the order of matrix P is 2x3 and the order of matrix Q is 3x4 , find the order of PQ.

1) 2x4

2) 2x2

3) 4x2

4) 3x3

6 Let A is a non – singular matrix of order 3 x 3 then | A ( adj A )| is equal to

a)| A |

b)| A |2

c)| A |3

d)3| A |

7 The function f(x) =[x] is continuous at

a)4

b)-2

c)1

d)1.5

8 The bottom of a rectangular swimming tank is 25 m by 40 m water is pumped into the tank at the rate of 500 cubic meters per minute. Find the rate at which the level of water in the tank is rising?

A) 1⁄4 m/min

B) 2/3 m/min

C) 1/3 m/min

D) 1⁄2 m/min

9 Area of region bound by circle x2+y2=1

a)2 sq units

b) sq units

c)3 sq units

d)4 sq units

10 Integrating factor of the differential equation dy/dx + y tan x – sec x = 0 is:

(A) cosx 

(B) secx 

(C) ecosx

(D) esecx

11 If points A (60 î+ 3 ĵ), (40 î– 8 ĵ) and C ( aî- 52ĵ) are collinear, then ‘a’ is equal to

a) 40

b) -40

c) 20

d) -20

12 Write direction cosines of a line parallel to z-axis.

(a) 1,0,0

(b) 0,0,1

(c) 1,1,0

(d) -1,-1,-1

13 Find the foot of the perpendicular drawn from the point (2,-3,4) on the y-axis.

(a) (2,0,4)

(b) (0.3.0)

(c) (0,-3,0)

(d) (-2,0,-4)

14 If α,β, Υ are the angles that a line makes with the positive direction of x,y,z axis respectively then the direction cosines of the line are

(a) cosα,sinβ, cosΥ

(b) cosα,cosβ, cosΥ

(c) sinα,sinβ, sinΥ

(d) 1, 1, 1

15 The feasible region of the inequality x+y≤1 and x–y≤1 lies in......... quadrants.

(a)Only I and II 

(b)Only I and III

(c)Only II and III

(d)All four

16 Two dice are thrown. If it is known that the sum of numbers on the dice was less than 6, the probability of getting a sum 3 is

a)1/18

b)5/18

c)1/5

d)2

17 The solution set of the inequality 3x + 5y < 4 is

a)an open half-plane not containing the origin.

b)an open half-plane containing the origin.

c)the whole XY-plane not containing the line 3x + 5y = 4.

d)a closed half plane containing the origin.

18 If A is a square matrix of order 3 and |A| = 5, then |adjA| =

(a) 5

(b) 25

(c) 125

(d) ⅕

19 The area of a triangle with vertices (2, −6), (5,4) and (k, 4) is 35 square units then , k is

A.12

B. −2

C. −12, −2

D. 12, −2

20 The vector having initial and terminal points as (2,5,0) and (-3,7,4) respectively is

A. 5î+ 2ĵ− 4k̂

B. −î+ 12ĵ+ 4k̂

C. −5î+ 2ĵ+ 4k̂

D.−5î+ 12ĵ+ 4k̂

21. Let A = { 1 , 2, 3 } and consider the relation R = {(1 , 1), (2 , 2), ( 3 , 3), (1 , 2), (2 , 3), (1,3)} Then , R is

(a) Reflexive but not symmetric

(b) Reflexive but not transitive

(c) Symmetric and transitive

(d)Neither symmetric nor transitive

22. If A is a skew – symmetric matrix , then A² is

(a) Symmetric

(b) Skew – symmetric

(c) A² = A

(d) A² ≠ A

23. The number of arbitrary constants in the general solution of a differential equation of fourth order are :

(a) 0

(b) 2

(c) 3

(d) 4

24. If y = log x/ (1+x) , then dy/ dx is equal to

(a) x/ (1+x)²

(b) 1/ (1+x) ²

(c) 2x/ (1+x)²

(d) 1/ x(1+x)

25. Corner points of the region for an LPP are (0, 2), (3, 0), (6, 0),(6, 8), and (0, 5) . Let F = 4x + 6y be the objective function. The minimum value of F occurs at

(a) (0,2) only

(b)(3,0) only

(c) The midpoint of the line segment joining the points (0,2) and ( 3, 0) only

(d) Any point on the line segment joining the points (0,2) and (3, 0) only.

26. f(x) = xˣ has a stationary point at

(a) x = e

(b) x = 1/ e

(c) x = 1

(d) x = √e

27. If A and B are events such that P ( A/ B) = P (B/ A) , then ………

(a) P(A) = P(B)

(b) P(A) ≠ P(B)

(c) P(A) + P(B) = 1

(d) None of these

28. If R = {(x , y ): x + 2y = 8 } is a relation N , then the range of R is :

(a){3, 2 , 1}

(b) { 3 , 2}

(c) {2 , 8 , 1}

(d) {3}

29. Let f ∶ R → R be defined as f(x) = 3x. Choose the correct answer.

(a) f is one – one onto

(b) f is many-one onto

(c) f is one – one but not onto

(d) f is neither one-one nor onto

30. The principal value of [tan-1 √3 − cot-1, (−√3)] is :

(a)π

(b) − π/2

(c) 0

(d) 2√3

31. If cos (sin-1 2/ 5 + cos-1 x) = 0 , then x is equal to

(a) 1/ 5

(b)2/ 5

(c) 0

(d) 1

32. The value of ∫ (cos 2x)/ (sinx +cosx)² dx is

(a) log | cos x + sin x | + C

(b) log | cos x − sin x | + C

(c) log | cos x + sin x |² + C

(d) log | cos x + sin x |-2 + C

33. The derivative of 3ˣ  w.r.t x is :

(a) log 3

(b) x. 3ˣ-1

(c) 0

(d) 3ˣ log 3

34. A is a square matrix of order 2, then adj(adj A) is :

(a) O

(b) I

(c) A-1

(d) A

35. The number of arbitrary constants in the general solution of a differential equation of fourth order is :

(a) 1

(b) 2

(c) 3

(d) 4

 

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1. Chemical Reactions and Equations

Chemical Equations 

When a magnesium ribbon is burnt in oxygen, it gets converted to magnesium oxide. This description of a chemical reaction in a sentence form is quite long. It can be written in a shorter form. The simplest way to do this is to write it in the form of a word-equation.

The word-equation for the above reaction would be –

Magnesium + Oxygen → 

(reactants)

Magnesium Oxide 

(product)

The substances that undergo chemical change in the reaction, magnesium and oxygen, are the reactants. The new substance is magnesium oxide, formed during the reaction, as a product.

A word-equation shows change of reactants to products through an arrow placed between them. The reactants are written on the left-hand side (LHS) with a plus sign (+) between them. Similarly, products are written on the right-hand side (RHS) with a plus sign (+) between them. The arrowhead points towards the products, and shows the direction of the reaction.

Writing a Chemical Equation

Chemical equations can be made more concise and useful if we use chemical formulae instead of words. A chemical equation represents a chemical reaction. If you recall formulae of magnesium, oxygen and magnesium oxide, the above word-equation can be written as –

Mg + O2 → MgO

Count and compare the number of atoms of each element on the LHS and RHS of the arrow. Is the number of atoms of each element the same on both the sides? If yes, then the equation is balanced. If not, then the equation is unbalanced because the mass is not the same on both sides of the equation. Such a chemical equation is a skeletal chemical equation for a reaction. Equation is a skeletal chemical equation for the burning of magnesium in air.

Balanced Chemical Equations

Recall the law of conservation of mass that you studied in Class IX; mass can neither be created nor destroyed in a chemical reaction. That is, the total mass of the elements present in the products of a chemical reaction has to be equal to the total mass of the elements present in the reactants.

In other words, the number of atoms of each element remains the same, before and after a chemical reaction. Hence, we need to balance a skeletal chemical equation. Is the chemical Eq balanced? Let us learn about balancing a chemical equation step by step.

The word-equation for Activity may be represented as –

Zinc + Sulphuric acid → Zinc sulphate + Hydrogen

The above word-equation may be represented by the following chemical equation –

Zn + H2SO4 → ZnSO4 + H2 

Let us examine the number of atoms of different elements on both sides of the arrow.

As the number of atoms of each element is the same on both sides of the arrow, Eq.is a balanced chemical equation.

Let us try to balance the following chemical equation –

Fe + H2O → Fe3O4 + H2 

Step I: To balance a chemical equation, first draw boxes around each formula. Do not change anything inside the boxes while balancing the equation.

Step II: List the number of atoms of different elements present in the unbalanced equation.

Step III: It is often convenient to start balancing with the compound that contains the maximum number of atoms. It may be a reactant or a product. In that compound, select the element which has the maximum number of atoms. Using these criteria, we select Fe3O4 and the element oxygen in it. There are four oxygen atoms on the RHS and only one on the LHS.

To balance the oxygen atoms –

To equalise the number of atoms, it must be remembered that we cannot alter the formulae of the compounds or elements involved in the reactions. For example, to balance oxygen atoms we can put coefficient ‘4’ as 4 H2O and not H2O4 or (H2O)4. Now the partly balanced equation becomes–

Step IV: Fe and H atoms are still not balanced. Pick any of these elements to proceed further. Let us balance hydrogen atoms in the partly balanced equation.

To equalise the number of H atoms, make the number of molecules of hydrogen as four on the RHS.

The equation would be –

Step V: Examine the above equation and pick up the third element which is not balanced. You find that only one element is left to be balanced, that is, iron.

To equalise Fe, we take three atoms of Fe on the LHS.

Step VI: Finally, to check the correctness of the balanced equation, we count atoms of each element on both sides of the equation.

The numbers of atoms of elements on both sides of Eq. (1.9) are equal. This equation is now balanced. This method of balancing chemical equations is called hit-and-trial method as we make trials to balance the equation by using the smallest whole number coefficient.

Step VII: Writing Symbols of Physical States Carefully examine the above balanced Eq. (1.9). Does this equation tell us anything about the physical state of each reactant and product? No information has been given in this equation about their physical states.

To make a chemical equation more informative, the physical states of the reactants and products are mentioned along with their chemical formulae. The gaseous, liquid, aqueous and solid states of reactants and products are represented by the notations (g), (l), (aq) and (s), respectively. The word aqueous (aq) is written if the reactant or product is present as a solution in water.

The balanced Eq. (1.9) becomes

3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g) (1.10)

Note that the symbol (g) is used with H2O to indicate that in this reaction water is used in the form of steam.

Usually physical states are not included in a chemical equation unless it is necessary to specify them.

Sometimes the reaction conditions, such as temperature, pressure, catalyst, etc., for the reaction are indicated above and/or below the arrow in the equation. For example 

Using these steps, can you balance Eq. (1.2) given in the text earlier?