Class 12

Odisha State Board CHSE Odisha Class 12 Logic Solutions Chapter 1 The Theory of Inference Questions and Answers.

CHSE Odisha 12th Class Logic Chapter 1 Question Answer The Theory of Inference

Group – A

Short type Questions with Answers
I. Answer with in Two/Three sentence.

Question 1.
What is called an inference?
Answer:

1. Inference is a mental fact or a mental process or a mental product.
2. It is an indirect way to get the different types of knowledge.
3. Example : By observing the smoke arising out of the hil if we say there is fire in that hill then this is called the process of inference.

Question 2.
What is called in immediate inference?
Answer:
(i) Immediate inference is a kind of deductive inference where the conclusion comes out of the only one premise.

(ii) Immediate inference is classified into 4 types, such as conversion, obversion, inversion and contraposition.

Question 3.
What is called deductive inference?
Answer:

1. Deductive inference is that inference where the conclusion comes out of the premises by the process of all to all or all to some, known to known and observe to observe.,
2. Deductive inference is two type such as immediate and mediate.
3. In deductive inference, the conclusion is less general than the premises.

Question 4.
What is called an inductive inference?
Answer:
(i) When the conclusion is drawn out of the premises by the process of some to all, known to known observe to unobserved that is called an inductive inference.

(ii) Here the conclusion is more general than the premises.

Question 5.
What are the main classifications of immediate inference?
Answer:
(i) Immediate inference is classified into 4 types.
(ii) These are conversion, obversion, inversion, contraposition.

Question 6.
Define conversion. Give an example of it.
Answer:
(i) Conversion is a kind of deductive immediate inference in which there is a legitimate transposition between the subject and predicate of the given proposition.

(ii) Example : A = All is P
∴ I = Some P is S

Question 7.
Write any two rules of conversion.
Answer:
(i) The subject of convertend will be predicate in converse and predicate of convertend will be subject in converse.
(ii) Quality will be same both in convertend and converse.

Question 8.
What is called convertend?
Answer:
(i) The given premise of conversion is called converted.
(b) Example : Converted (A) = All S is P
∴ Coverse (I) = Some P is S

Question 9.
What is called converse?
Answer:
(i) The conclusion of conversion is called converse.
(ii) Example : Converted (E) = No. dogs are cats.
∴ Converse (E) = No cats are dogs.

Question 10.
What is obverstion?
Answer:
(i) Obverstion is a kind of immediate deductive inference where the subject of the given premise will be same in conclusion and the predicate of the given premise will be contradictory form in the conclusion.

(ii) Example : A = All S is P
∴ E = No S is not P

Question 11.
Write any three rules of obversion.
Answer:
(i) The subject of obvertend will be same in obverse.
(ii) The predicate of obvertend will be contradictory form in obverse.
(iii) Quality will be change and quantity will be same.

Question 12.
What is called obvertend?
Answer:
(i) The given premise of obversion is called obvertend.

(ii) Example : Obvertend (A) = All S is P
∴ Obverse (E) = No S is not P

Question 13.
What is called obverse?
Answer:
(i) The conclusion of obversion is called obverse.
(ii) Example : obverted (E) – No S is P.
Obverse (A) = All S is not P

Question 14.
What is called mediate inference?
Answer:
(i) Mediate inference is that inference where the conclusion comes out of the two premises taken jointly but not separately.
(ii) Example : All men are mortal All students are men.
∴ All students are mortal

II. Answer with in Five/Six sentence :

Question 1.
Simple conversion :
Answer:
Simple conversion is that conversion where the quality and quantity of both convertend and converse are same.
For example;
Convertend (E) = No S is P.
Converse (E) = No P is S.
Converted (I) = Some S is P.
Converse (I) = Some P is S.

Question 2.
Partial conversion/conversion per limitation/conversion per accidence.
Answer:
In a conversion, if the quality of both convertend and converse are same but the quantity is different that is called partial conversion.
For example:
Convertend (A) = All S is P.
Conversion (I) = Some P is S.

Question 3.
State the rules of conversion.
Answer:
(i) The subject of convertend becomes the predicate of converse.
(ii) The predicate of convertend becomes the subject of converse.
(iii) Quality will be same both in convertend and in converse.
(iv) The term which is not distributed in convertend that should not be distributed in converse.

Question 4.
State the rules of obversion.
Answer:
(i) The subject of obvertend becomes the subject of obverse.
(ii) The predicate of obvertend becomes the contradictory form in obverse.
(iii) Quality of the obvertend will be change in obverse.
(iv) Quantity will be same both in obvertend and in obverse.
(v) The terms which is not distributed in obvertend that terms should not be distributed in obverse.

Question 5.
What is material obversion?
Answer:
Material obversion is a fallacious form of obversion in which the meaning of subject and predicate of conclusion are opposite of the subject and predicate of the premise and the quality remains same. This fallacy is given by the logician Bain.
For example;
Knowledge is good.
∴ Ignorance is bad.

Question 6.
Why ‘O’ proposition cannot be converted?
Answer:
If we convert the ‘O’ proposition then the conclusion will be ‘o’ proposition, in which the predicate ‘S’ term will be distribute. But as this term is not distributed in the premise, so it leads the fallacy, which violates the rules of conversion. Hence ‘o’ proposition cannot be converted.

Question 7.
Distinguish between Immediate and mediate inference.
Answer:
(i) Immediate inference is a kind of deductive inference in which the conclusion is drawn out of the only one premise.

(ii) Immediate inference is divided into 4 types, such as conversion, obversion, inversion and contraposition.

(iii) For example;
All men are mortal.
∴ Some mortal beings are men.

(iv) Mediate inference is a kind of deductive inference in which the conclusion is drawn from two premises taken jointly but not separately. It is otherwise called as syllogism.
For example;
All men are mortal.
All students are men.
All students are mortal

Question 8.
What are the different kinds of inferences?
Answer:
Inference is mainly divided into two types, such as deductive inference and inductive inference. Again deductive inference is divided into 2 types, such as, immediate inference and mediate inference. Mediate inference which is called as syllogism can be pure and mixed.

Question 9.
Distinguish between convertend and converse.
Answer:
The given premise of conversion is called convertend. But the conclusion of converse is called converse. For example;
Convertend (E) = No Dogs are Cats.
Converse (E) = No Cats are Dogs.

Question 10.
Define Obversion.
Answer:
Obversion is a kind of immediate deductive inference where the subject of the given premise will be same in the conclusion but the predicate of the given premise will be the contradictory form in the conclusion.
For example:
Obvertend (A) = All S is P.
∴ Obverse (E) = No S is not P.

Group – B

Long Type Questions With Answers

Question 1.
What is meant by inference ? Is immediate inference an inference at all ? Discuss.
Answer:
Inference is a valid source of knowledge. In most cases we depend upon inferential knowledge. For example, where there is smoke, there is fire. By observing the smoke arising out of the hill, we can infer that there is fire in the hill. So it is a process from something known to unknown. The word ‘inference’ has a double meaning. It is used either as a mental process or a mental product.

(1) As a mental process inference means the process of thought by which we pass from something known to something unknown. The known truths are called the premises and the unknown truth which is inferred from the known truths is called conclusion. In other words, inference is the process of thought by which we derive the conclusion on the basis if one or more premises. So it is a form of reasoning.

(2) As a mental product inference means the product or the result of the mental process. The conclusion alone is the product or result of our thinking. The conclusion which is Justified by the premises is valid. An inference requires more than one propositions and when it is expressed in Language that is called an argument. So an argument consists of more than one propositions. The given proposition is called premise and the proposition which we derive is called conclusion. So it is said that Logic is directly concerned with argument but indirectly concerned with inference.

Classification of inference : Inference is mainly divided into two types; such as :
(i) Deductive inference.
(ii) Inductive inference.

(i) Deductive inference : In deductive inference the conclusion is not more general than the premises. It is implied by the premise or premises. So the conclusion adds nothing new to our knowledge.
Example:
All men are mortal.
Mohan is a man.
∴ Mohan is a man.

(ii) Inductive inference : In inductive inference the conclusion is more general than the premises. It asserts more than what is implied in the premises. It adds something new to our knowledge.
Example:
Ram is mortal.
Hari is mortal.
∴ All men are mortal.

Deductive inference again has been subdivided into two classes viz; (1) Immediate and (2) Mediate. An immediate inference is a kind of deductive inference in which the conclusion is drawn from one premise only.
Example :
All crows are black.
∴ Some crows are black.

A Mediate inference is a kind of deductive inference in which the conclusion is drawn from more than one premise.
Example :
All men are honest.
Madhu is a man.
∴ Madhu is a man.

Some Logicians like Bain and Mill are of the opinion that immediate inference is not an inference all. There is only re-arrangement of terms in the conclusion. The conclusion does not tell anything new, Mill says, it is “inference improperly so-called”. Bain says that the conclusion never goes beyond what is asserted in the premise. But such type of objection are not Justified. One cannot simply ignore the usefulness of immediate inference by criticising that they do not state anything new. Welton says, “In immediate inference the conclusion helps in making the meaning explicit what was implicitly contained in the premises’. Hence it is said that immediate inference is a true form of inference.

Question 2.
Define conversion. Explain its rules and determine the converse of A, E, I and C propositions.
Answer:
Definition : Conversion is a kind of immediate deductive inference in which there is a legitimate transposition of the subject and the predicated of the given proposition.
Conversion is an inference, because here we draw a conclusion from a given premise. It is an immediate inference because here the conclusion is drawn from one premise. It is a deductive inference, because here the conclusion is never more general than the premise.
(i) The name of the proposition which is given in conversion is called convertend.
(ii) The conclusion of conversion is called converse.

Rules of Conversion :

1. The subject of the convertend will be the predicate in the converse.
2. The predicate of the convertend will be the subject in the converse.
3. Quality will be remains same.
4. If a term is not distributed in the convertend, it should not be distributed in the converse.

Explanation of the rules :

1. Rule first + second state the defining characteristics of conversion.
2. Rule third states that the premise and the conclusion have exactly the same terms. Only their positions are interchanged. So if they be positively related in the premise, they cannot be negatively related in the conclusion. Therefore, the quality of the premise cannot be changed in the conclusion.
3. As the conversion is deductive in character so the conclusion cannot be wider than the premise.

Converse of Propositions :
Premise = A = All S is P.
The conclusion must be affirmative (A or I). If we make it ‘A’ (All P is S), then the term ‘P’ will be distributed in the conclusion, without being distributed in the premise. So it must be T (Some P is S). Therefore the converse of A is I. Converse of A is I.

Convertend = ‘A’ = All mangoes are fruits.
Converse = T = Some fruits are Mangoes.
Premise = L = No S is P.
Conclusion = E = No P is S.
Here no rule is violated.
Convertend = ‘E’ = No cats are dogs.
Converse = ‘E’ = No dogs are cats.
Premise T = Some S is P.
Conclusion T = Some P is S.
Here no rule is Violated.
Convertend T = Some fruits are sweet.
∴ Converse T =Some sweet things are fruits.
Premise ‘O’ = Some S is not P.
Conclusion = Nothing.

Convertend T = Some students are not intelligent.
Converse = Nothing.
Here out of the premise (O), if we draw any conclusion then it will be ‘O’ proposition in which ‘S’ term will be distributed, which will never be distributed in the convertend or premise. Therefore ‘O’ Proposition cannot be converted.

Out of the above analysis it is concluded that A given I, I gives I, E gives E, ‘O’ does not give any proposition. I = I, E = E are the example of simple conversion because here the quality and quantity if both convertend and converse are equal. But A = I is the example of partial conversion because here the quantity of both convertend and Converse are different from each other but the quality is same.

Question 3.
State the rules of obversion. Apply them A, E, I and O propositions.
Answer:
Obversion is a kind of immediate inference where the predicate of the conclusion is the contradictory of the predicate of the premise, the subject remaining same. The premise of the obversion is called obvertend and the conclusion is called obverse. There are some rules which are to be followed while obverting a proposition. Those rule are discussed below.

• Rule-1: The subject of the obvertend becomes the subject of the obverse. The predicate of the obverse is the contradictory of the predicate of obvertend.
• Rule-2 : The quality change. If the premise is affirmative, the conclusion is negative and if the premise is negative the conclusion is affirmative.
• Rule-3 : The quantity of the obverse remains same as the quantity of the obvertend.
• Rule-4 : The term which is not distributed in the obvertend can not be distributed in the obverse.

These rules of obversion can be applied to different propositions and obversion can be done in the following way.

Obversion of ‘A’ proposition :
A-All swans are white (obvertened)
∴ E – No swans are not-white (obverse)
By the application of the rules of obversion ‘A’ proposition can be validly obverted to ‘E’ proposition. By the rule-I the subject ‘Swans’ remained same in the obverse. But the predicate became contradictory from ‘white’ to not – white. By the rule-2, the quality changed. The premise is affirmative whereas the conclusion is negative. By the 3rd rule the quantity of both is universal. Again no term is distributed in the obverse without being distributed in the obvertend.

Obversion of ‘E’ proposition :
E-No swans are white (obvertend)
∴ A-All swans are not-white (obverse)
As we see here, in obversion ‘E’ becomes ‘A’. Here all the rules of obversion are followed. The subject ‘swan’ is distribute in both the places.

Obversion of ‘I’ proposition :
I-some swans are white (obvertend).
∴ O-some swans are not not-white (obvertend).
Here again all the rules are followed. By the application of the rules we get ‘O’ proposition from T proposition.

Obversion of ‘O’ proposition :
O-some swans are not white, (obvertend).
I-some swans are not – white, (obverse).
By the application of all rules, in obversion ‘O’ proposition gives ‘T’ propositions.

Question 4.
State and explain the rules of contraposition.
Answer:
Contraposition is a logical rule that involves transforming a given proposition to an equivalent form. It is particularly useful in formal logic and is employed in various deductive reasoning processes. The rules of contraposition are applied to categorical propositions, which are statements that assert or deny the inclusion or exclusion of a particular subject within a specified class. These propositions are usually expressed in the form “All S is P,” “No S is P,” “Some S is P,” or “Some S is not P.” The contraposition rule primarily applies to universal affirmative and universal negative propositions. Let’s explore the rules of contraposition in detail.

Universal Affirmative Proposition (A-type):
The contraposition of a universal affirmative proposition “All S is P” is derived by transforming it into its logically equivalent form. The contrapositive statement is “All non-P is non-S.”

For example, if we start with the proposition “All birds are animals,” the contrapositive would be “All non-animals are non-birds.” This transformation maintains the logical equivalence between the original statement and its contrapositive.The reasoning behind this lies in recognizing that if everything belonging to class S is also in class P, then everything outside of class P is also outside of class S.

Universal Negative Proposition (E-type):
The contraposition of a universal negative proposition “No S is P” involves transforming it into the logically equivalent form “No non-P is non-S.”

Consider the proposition “No humans are immortal.” The contrapositive would be “No non-immortals are non-humAnswer:” In this case, the contraposition maintains the logical relationship between the absence of inclusion in class P and the absence of inclusion in class S.The contraposition of a universal negative proposition reflects the idea that if no members of S are in P, then no members outside of P are in S.

Particular Affirmative Proposition (I-type) :
Contraposition is not directly applicable to particular affirmative propositions (“Some S is P”). However, it is essential to note that the contrapositive of a particular affirmative proposition is not necessarily logically equivalent to the original statement. The contrapositive of “Some S is P” would be “Some non-P is non-S,” but this does not necessarily preserve the logical relationship between the classes. Due to this limitation, contraposition is most commonly and effectively applied to universal propositions.

Particular Negative Proposition (O-type) :
Similarly to particular affirmative propositions, contraposition is not directly applicable to particular negative propositions (“Some S is not P”). The contrapositive of “Some S is not P” would be “Some non-P is not non-S,” but this does not maintain a clear logical equivalence.

In practice, contraposition is most confidently applied when dealing with universal propositions, where the transformation retains the logical relationship between the classes involved.

In conclusion, contraposition is a valuable rule in logic, particularly when working with universal propositions. It allows for the transformation of statements while preserving logical equivalence. Universal affirmative propositions are contraposed by stating that everything outside of the predicate class is also outside of the subject class. Similarly, universal negative propositions are contraposed by asserting that nothing outside of the predicate class is inside the subject class. It is important to recognize the limitations of contraposition when dealing with particular propositions, as the contrapositives may not maintain a clear logical relationship.

Question 5.
Distinguish between :
(a) Mediate and immediate inference
(b) Simple conversion and conversion by limitation.
Answer:
(a) Mediate and immediate inference.
In logic, mediate inference and immediate inference are two types of logical reasoning processes that involve drawing conclusions from given propositions. These forms of inference play crucial roles in decretive reasoning and are integral to understanding and constructing logical arguments.

Immediate Inference :
Immediate inference involves drawing conclusions directly from a single proposition without the need for an additional premise. It is an inference where the conclusion follows immediately from the given statement. Immediate inferences are typically based on the conversion, obversion, or contraposition of a given proposition. These processes allow us to manipulate the original proposition to derive an immediate inference.

1. Conversion : Conversion is an immediate inference that involves switching the subject and predicate terms of a proposition while maintaining its quality. There are two types of conversion: simple conversion and conversion by limitation.

1. Simple Conversion: In simple conversion, the subject and predicate terms are switched without any change in quantity or quality. For example, from the proposition “All men are mortal,” we can immediately infer “All mortals are men.”
2. Conversion by Limitation: In conversion by limitation, the original proposition is converted, and a limiting term is added. For instance, from “No birds are mammals,” we can infer “No mammals are birds of any kind.”

2. Obversion: Obversion is another immediate inference that involves negating the predicate term of ^ proposition while maintaining the same subject and quality. Additionally, a new term, called the “term of obversion,” is introduced by negating the original predicate. For example, from the proposition “Some cats are black,” we can immediately infer “Some cats are not non-black.”

3. Contraposition : Contraposition is an immediate inference primarily applied to universal affirmative and universal negative propositions. It involves switching the subject and predicate terms and negating both. For instance, from the proposition “All humans are mortal,” we can infer “All non-mortals are non-humAnswer:”

Immediate inferences are particularly useful for simplifying and clarifying propositions, allowing for the quick derivation of conclusions based on the structure and content of a single statement. These processes provide a direct route from a given proposition to a logically equivalent conclusion.

Mediate Inference :
In contrast to immediate inference, mediate inference involves drawing conclusions by using two or more propositions in a logical sequence. This form of inference relies on the establishment of a relationship between premises and the subsequent derivation of a conclusion. Syllogism, a fundamental structure in mediate inference, consists of three propositions: two premises and a conclusion.

1. Categorical Syllogism : A categorical syllogism is a specific form of mediate inference that involves three categorical propositions. The premises and conclusion are statements that assert or deny the inclusion or exclusion of a particular subject within a specified class. The classic example of a categorical syllogism is :
• Premise 1: All humans are mortal.
• Premise 2: Socrates is a human.
• Conclusion: Therefore, Socrates is mortal.
The conclusion follows logically from the combination of the two premises, demonstrating the process of mediate inference.

2. Hypothetical Syllogism : Hypothetical syllogism involves conditional propositions or “if-then” statements. If one proposition implies another and the second proposition implies a third, then the first proposition implies the third. For example :
• Premise 1 : If it rains, then the streets will be wet.
• Premise 2 : If the streets are wet, then people will use umbrellas.
• Conclusion : Therefore, if it rains, people will use umbrellas.
The conclusion is reached by combining the implications of the two conditional premises

3. Disjunctive Syllogism : Disjunctive syllogism involves a disjunctive proposition (an “either/or” statement). If one of the alternatives is eliminated, the other must be true. For example :
• Premise: Either it is sunny or it is raining.
• Elimination: It is not sunny.
• Conclusion: Therefore, it is raining.
The conclusion is derived by eliminating one of the alternatives presented in the initial disjunctive proposition.

In conclusion, mediate inference involves the use of multiple propositions to establish a logical relationship and draw conclusions. Categorical, hypothetical, and disjunctive syllogisms are common forms of mediate inference, providing a structured approach to reasoning and deduction. Immediate inference, on the other hand, allows for the direct derivation of conclusions from a single proposition through processes like conversion, obversion, and contraposition. Both immediate and mediate inferences are fundamental to understanding and constructing logical arguments in various fields of study.

(b) Simple conversion and conversion by limitation.
In the realm of categorical propositions, conversion is a logical operation that involves interchanging the subject and predicate terms of a given statement. Two main types of conversion are simple conversion and conversion by limitation. These techniques are employed to derive nev. propositions from existing ones, and understanding the distinctions between them is crucial for effective reasoning in formal logic.

Simple Conversion :
Simple conversion is a straightforward process that involves interchanging the subject and predicate terms of a given categorical proposition without altering the quality (affirmative or negative) of the original statement. It is applicable to both universal and particular propositions.

Universal Affirmative (A-type) :
For a universal affirmative proposition like “All S is P,” simple conversion yields “All P is S.” This maintains the original affirmation and switches the subject and predicate terms.

Universal Negative (E-type) :
In the Case of a universal negative proposition such as “No S is P,” simple conversion results in “No P is S.” The negativity of the original statement is preserved, but the subject and predicate terms are interchanged.

Particular Affirmative (I-type):
Simple conversion is not applicable to particular affirmative propositions (“Some S is P”) Attempting to convert a particular affirmative proposition using the simple method may lead to ambiguous or invalid conclusions.

Particular Negative (O-type) :
Similarly, simple conversion is not applicable to particular negative propositions (“Some S is not P”). The attempt to convert a particular negative proposition using simple conversion can result in an ambiguous or invalid statement.

Conversion by Limitation :
Conversion by limitation is a more nuanced form of conversion that involves interchanging the subject and predicate terms of a given proposition while also making adjustments to the quantity (universal or particular) and quality (affirmative or negative) of the original statement. This method is applicable to both universal and particular propositions.

Universal Affirmative (A-type) :
When applying conversion by limitation to a universal affirmative proposition “All S is P,” the result is a particular affirmative proposition, “Some P is S.” This conversion maintains the affirmation but changes the quantity from universal to particular.

Universal Negative (E-type) :
Conversion by limitation applied to a universal negative proposition “No S is P” yields a particular negative proposition, “Some non-P is non-S.” Here, the negativity is preserved, and the quantity changes from universal to particular.

Particular Affirmative (I-type) :
For a particular affirmative proposition “Some S is P,” conversion by limitation results in another particular affirmative proposition, “Some P is S.” The original affirmation is retained, and the quantity remains particular.

Particular Negative (O-type) :
Conversion by limitation applied to a particular negative proposition “Some S is not P” produces another particular negative proposition, “Some non-P is not non-S.” The negativity is preserved, and the quantity remains particular.

Distinctions :

• Quantity and Quality :
  1. Simple conversion maintains the quantity and quality of the original proposition.
  2. Conversion by limitation involves adjusting both the quantity and quality during the conversion process.
• Applicability :
  1. Simple conversion is applicable to universal affirmative and negative propositions.
  2. Conversion by limitation is applicable to both universal and particular propositions, and it allows for a more nuanced transformation.
• Resulting Proposition:
  1. Simple conversion results in a proposition with the same quantity and quality as the original statement.
  2. Conversion by limitation results in a proposition with a modified quantity while preserving the quality of the original statement.

In conclusion, while both simple conversion and conversion by limitation involve interchanging subject and predicate terms, they differ in terms of the adjustments made to quantity and quality. Simple conversion is straightforward and maintains the original quantity and quality, whereas conversion by limitation involves more nuanced adjustments, particularly in changing the quantity of the proposition. Understanding these distinctions is essential for precise and accurate reasoning in formal logic.

Group – C

Objective type Questions with Answers
I. Multiple Choice Questions with Answers :

Question 1.
An immediate inference in which the subject and the predicate are interchanged is called :
(i) Conversion
(ii) Obversion
(iii) Inversion
(iv) Nothing
Answer:
(i) Conversion

Question 2.
A term which is not distributed in the premise :
(i) can be distributed in the conclusion
(ii) cannot be distributed in the conclusion
(iii) may sometimes be distributed in the conclusion
(iv) None of these
Answer:
(ii) cannot be distributed in the conclusion

Question 3.
Which of the following is not true of immediate inference ?
(i) It’s conclusion follows from a single premise
(ii) It is a deductive inference
(iii) It is an inductive inference
(iv) Conversion, obversion, contraposition etc. are it’s types
Answer:
(iii) It is an inductive inference

Question 4.
Which of the following is called an inference?
(i) Inference is a logical phenomena
(ii) Inference is a mental phenomena
(iii) Inference is a philosophical phenomena
(iv) None of these are correct.
Answer:
(ii) Inference is a mental phenomena

Question 5.
When an inference is expressed in Language that is called what ?
(i) Argument
(ii) Proposition
(iii) Judgement
(iv) Term
Answer:
(i) Argument

Question 6.
Deductive inference is divided into which of the following?
(i) Conversion and obversion
(ii) Obversion and contraposition
(iii) Immediate and Mediate
(iv) Direct and indirect
Answer:
(iii) Immediate and Mediate

Question 7.
Which of the following is the main division of immediate inference?
(i) Conversion and obversion
(ii) Conversion, obversion, inversion and contraposition
(iii) Conversion, obversion, syllogism.
(iv) None of these are correct .
Answer:
(ii) Conversion, obversion, inversion, contraposition

Question 8.
In immediate inference, the conclusion is drawn from how many premises ?
(i) One
(ii) Two
(iii) Three
(iv) Four
Answer:
(i) One

Question 9.
Conversion is a what kind of inference?
(i) mediate
(ii) Immediate
(iii) Inductive
(iv) Both (ii) & (iii)
Answer:
(ii) immediate

Question 10.
The given premise of conversion is called vyhat?
(i) Convertend
(ii) Convert
(iii) Obvertend
(iv) Converse
Answer:
(i) Convertend

Question 11.
The conclusion of conversion is called what?
(i) Convertend
(ii) Convert
(iii) Converse
(iv) Obverse
Answer:
(iii) Converse

Question 12.
When a conclusion is drawn from more than one premise that is called what?
(i) Immediate inference
(ii) Mediate inference
(iii) Deductive inference
(iv) Inductive inference
Answer:
(ii) Mediate inference

Question 13.
Which of the following are the main classificatioq of inference?
(i) Mediate and immediate
(ii) Deductive and inductive
(iii) Conversion and obversion
(iv) Direct and indirect
Answer:
(ii) Deductive and inductive

Question 14.
What is the coversion of ‘A’ proposition?
(i) ‘A’
(ii) ‘E’
(iii) ‘T’
(iv) ‘O’
Answer:
(iii) ‘T’

Question 15.
What is the conversion of ‘E’ proposition ?
(i) ‘A’
(ii) ‘E’
(iii) ‘I’
(iv) ‘O’
Answer:
(ii) ‘E’

Question 16.
What is the conversion of T proposition?
(i) ‘A’
(ii) ‘E’
(iii) ‘T’
(iv) ‘O’
Answer:
(iii) ‘T’

Question 17.
What is the conversion of ‘O’ proposition ?
(i) ‘A’
(ii) ‘E’
(iii) ‘I’
(iv) Cannot be converted
Answer:
(iv) Cannot be converted

Question 18.
Which of the following propositions convert simply?
(i) I and O proposition
(ii) A and E proposition
(iii) E and I proposition
(iv) E and O proposition
Answer:
(iii) E and I proposition

Question 19.
Which of the following propositions convert practically ?
(i) ‘A’proposition
(ii) ‘E’Proposition
(iii) T proposition
Answer: (i) ‘A’ proposition

Question 20.
Conversion is mainly divided into
(i) Two types
(ii) Threes types
(iii) Four types
(iv) Five Types
Answer:
(i) Two types

Question 21.
Which proposition cannot be converted?
(i) ‘E’
(ii) T
(iii) ‘O’
(iv) ‘A’
Answer:
(iii) ‘O’

Question 22.
State the conversion of “All men are honest”
(i) Some honest beings are men
(ii) No men are not honest
(iii) Some men are honest
(iv) Some men are not honest
Answer:
(i) Some honest beings are men

Question 23.
State the conversion of “No men are Birds”.
(i) All men are not birds
(ii) No birds are men
(iii) Some men are not birds
(iv) No men are not birds
Answer:
(ii) No birds are men

Question 24.
State the conversion of “Some students are intelligent”.
(i) Some students are not intelligent
(ii) Some intelligent beings are students
(iii) No students are intelligent
(iv) No intelligent beings are students
Answer:
(ii) Some intelligent beings are students

Question 25.
State the conversion of “some students are not intelligent”
(i) Some intelligent beings are not students
(ii) Some students are not intelligents
(iii) All students are intelligent
(iv) None of these
Answer:
(iv) None of these

Question 26.
The given premise of obversion is called What ?
(i) Convertend
(ii) Obvertend
(iii) Obverse
(iv) Ob vert
Answer:
(ii) Obvertend

Question 27.
The conclusion of obversion is called what?
(i) Obverse
(ii) Obvertend
(iii) Obvert
(iv) ‘O’Proposition
Answer:
(i) Obverse

Question 28.
What is the obversion of ‘A’ Proposition?
(i) ‘A’proposition
(ii) ‘E’ proposition
(iii) ‘T’proposition
(iv) ‘O’proposition
Answer:
(ii) ‘E’ proposition

Question 29.
What is the obversion of ‘E’ proposition ?
(i) ‘A’proposition
(ii) ‘E’Proposition
(iii) ‘T’Proposition
(iv) ‘O’proposition
Answer:
(i) ‘A’ proposition ,

Question 30.
What is the obversion of T proposition?
(i) ‘A’Proposition
(ii) ‘E’Proposition
(iii) T Proposition
(iv) ‘O’Proposition
Answer:
(iv) ‘O’ Proposition

Question 31.
If we obvert the proposition ‘O’ then we will get which proposition?
(i) ‘A’Proposition
(ii) ‘E’Proposition
(iii) ‘T’Proposition
(iv) ‘O’Proposition
Answer:
(iii) ‘T’ Proposition

Question 32.
The fallacy of obversion is called what?
(i) Fallacy of material obversion
(ii) Fallacy of Accident
(iii) Fallacy of Accent
(iv) None of these
Answer:
(i) Fallacy of material obversion

Question 33.
When the quality and quantity of both convertend and converse are equal that is called what?
(i) Simple conversion
(ii) Partial conversion
(iii) Conversion per limitation
(iv) Material obversion
Answer:
(i) Simple conversion

Question 34.
When only the quantity of both converted and converse are differ from each other but the quality is remain same that is called what?
(i) Simple conversion
(ii) Partial conversion
(iii) Material obversion
(iv) None of these
Answer:
(ii) Partial conversion

Question 35.
What kind of obversion is the following?
Knowledge is good.
Ignorance is bad
(i) Conversion
(ii) Obversion
(iii) Fallacy of material obversion
(iv) None of these
Answer:
(iii) Fallacy of material obversion

Question 36.
State the obvert of the proposition, Some flowers are fragrance.
(i) Some flowers are fragrance
(ii) Some flowers are not-fragrance
(iii) All flowers are fragrance
(iv) No flowers are fragrance.
Answer:
(ii) Some flowers are not-fragrance

Question 37.
The other name of mediate inference is called what?
(i) Syllogism
(ii) Conversion
(iii) Obversion
(iv) Contraposition
Answer:
(i) Syllogism .

Question 38.
How the predicate of obverse is related to the predicate of obvertend?
(i) Same
(ii) Contradictory
(iii) Contrary
(iv) None of these
Answer:
(ii) Contradictory .

II. Fill in the blanks :

Question 1.
_______ is an indirect way of getting the different of knowledge.
Answer:
immediate, mediate

Question 2.
Inference is a _____ process.
Answer:
mental

Question 3.
When an inference is expressed in language is called _____
Answer:
Argument.

Question 4.
Inference is mainly divided into two types such as _____ and _____.
Answer:
Deductive, inductive

Question 5.
Deductive inference is sub divided into _____ and _____ .
Answer:
immediate, mediate

Question 6.
In an inference if the conclusion is drawn out of the only one premise that is called _____ inference.
Answer:
Immediate

Question 7.
In an inference if the conclusion is drawn from two premises that is called _____ inference.
Answer:
Mediate

Question 8.
Immediate inference is divided into _____ types.
Answer:
four

Question 9.
Immediate inference is divided into four types, such as; _____,_____,_____ and _____.
Answer:
Conversion, obversion, contraposition, inversion

Question 10.
In deductive inference we proceed from _____ .
Answer:
all to some

Question 11.
In _____ inductive inference we proceed from _____.
Answer:
some to all

Question 12.
In _____ inference the conclusion is more general than the premises.
Answer:
Inductive

Question 13.
In _____ inference the conclusion is less general than the premises.
Answer:
Deductive

Question 14.
Conversion is a kind of _____ inference.
Answer:
Immediate

Question 15.
The given premise of conversion is called _____.
Answer:
Convertend

Question 16.
The conclusion of conversion is called _____ .
Answer:
Converse

Question 17.
In conversion the quality is_____.
Answer:
Remain same

Question 18.
If the convertend is affirmative, the converse is_____.
Answer:
Affirmative

Question 19.
If the convertend is negative, the converse is _____.
Answer:
Negative

Question 20.
The converse of ‘A’ is _____.
Answer:
‘I’

Question 21.
The converse of ‘E’ is _____.
Answer:
‘E’

Question 22.
The converse of‘T’ is _____.
Answer:
‘T’

Question 23.
The converse of ‘O’ is _____.
Answer:
Impossible

Question 24.
Conversion is divided into two ways, such as _____ and _____.
Answer:
Simple, partial

Question 25.
In _____ there is a legitimate transposition of the subject and the predicate of a proposition.
Answer:
conversion

Question 26.
There is no change in _____ conversion.
Answer:
quality

Question 27.
_____ proposition cannot be converted.
Answer:
‘O’ proposition

Question 28.
If a term is not distributed in the premise, it should not be _____ in the conclusion.
Answer:
Distributed

Question 29.
Where the qualify and quantity of both convertend and converse are same that is called _____ conversion.
Answer:
Simple.

Question 30.
Where the quantity of both convertend and converse are differ but quality is same that is called _____ conversion.
Answer:
Partial

Question 31.
‘I’ gives ‘I’, ‘E’ gives ,‘E’, are the example of _____ conversion.
Answer:
Simple

Question 32.
Generally ‘A’ proposition is converted by _____
Answer:
Limitation

Question 33.
‘A’ gives ‘I’ is the example of _____ conversion.
Answer:
Partial

Question 34.
Obversion is a kind of _____ inference.
Answer:
Immediate

Question 35.
In obversion, ‘A’ gives _____ .
Answer:
‘E’

Question 36.
The obverse of ‘E’ is _____ .
Answer:
‘A’

Question 37.
The obverse of‘T’ is _____.
Answer:
‘O’

Question 38.
The obverse of ‘O’ is _____.
Answer:
‘I’

Question 39.
The given premise of obversion is called _____.
Answer:
Obvertend

Question 40.
The conclusion of obversion is called _____.
Answer:
obverse

Question 41.
In obversion the quantity is _____ .
Answer:
same

Question 42.
In obversion the quality is _____.
Answer:
change

Question 43.
The kind of obversion based on the facts of experience is called as _____.
Answer:
Material obversion

Question 44.
______ has putforth material obversion.
Answer:
Bain

Question 45.
If we violate the rule of obversion then we commit the fallacy of _______.
Answer:
Material obversion

Question 46.
Knowledge is good
∴ Ignorance is bad.
This is an example of ______.
Answer:
Material obversion

Question 47.
The other name of mediate inference is called ______.
Answer:
syllogism

III. Correct the Sentences:

Question 1.
Logic is directly concerned with inference and indirectly concerned with argument.
Answer:
Logic is directly concerned with argument and indirectly concerned with inference.

Question 2.
In deductive inference, the conclusion speaks something new than the premises.
Answer:
In inductive inference, the conclusion speaks something new than the premises.

Question 3.
In inductive inference, the conclusion does not say anything about the inference.
Answer:
In deductive inference, the conclusion does not say anything about the inference.

Question 4.
In inductive inference the conclusion is less general than the premises.
Answer:
In inductive inference the conclusion is more general than the premises.

Question 5.
In inductive inference the conclusion is less general than the premises?
Answer:
In deductive inference the conclusion is less general than the premises

Question 6.
Deductive inference is divided into two types, such as conversion and obversion.
Answer:
Deductive inference is divided into two, types, such as immediate and mediate inference.

Question 7.
Conversion, observation, inversion and contraposition are.
Answer:
Conversion, obversion, inversion and contraposition are the division of immediate inference.

Question 8.
The premise of conversion is called converse.
Answer:
The premise of conversion is called convertend.

Question 9.
The conclusion of conversion is called convertend.
Answer:
The conclusion of conversion is called converse.

Question 10.
The subject of the convertend becomes the subject of the converse and the predicate of the convertend becomes the predicate of the converse.
Answer:
The subject of the convertend becomes the predicate of the converse and the predicate of the convertend becomes the subject of the converse.

Question 11.
The quality of the convertend is the opposite quality of the converse.
Answer:
The quality of the convertend is same with the quality of the converse.

Question 12.
The term which is not distributed in the convertend is distributed in converse.
Answer:
The term which is not distributed in the convertend should not be distributed in converse.

Question 13.
If the quality of convertend and converse remains the same, it is called simple conversion.
Answer:
If the quality of convertend and converse remains the same, it is called partial conversion.

Question 14.
Conversion of ‘E’ proposition is called conversion per accidens.
Answer:
Conversion of ‘A’ proposition is called conversion per accidens.

Question 15.
The conversion of ‘A’ and ‘O’ propositions are called simple conversion.
Answer:
The conversion of ‘E’ and ‘T’ proposition are called simple conversion.

Question 16.
‘A’proposition convert to‘E’proposition.
Answer:
‘A’ proposition covert to ‘T’ proposition.

Question 17.
‘ E ’ proposition convert to ‘ A’ proposition.
Answer:
‘E’ proposition convert to ‘E’ proposition.

Question 18.
T proposition convert to ‘O’ proposition.
Answer:
T proposition convert to ‘T’ proposition.

Question 19.
‘O’ proposition covert to T proposition.
Answer:
‘O’ proposition cannot be converted.

Question 20.
Obversion is a kind of mediate inference.
Answer:
Obversion is a kind of immediate inference.

Question 21.
The premise of obversion is called obverse.
Answer:
The conclusion of obversion is called obverse.

Question 22.
The conclusion of obversion is called obvertend.
Answer:
The premise of obversion is called obvertend.

Question 23.
The quality of the obverse is the same as the quality of the obvertend.
Answer:
The quality of the obverse is the opposite of the quality of the obvertend.

Question 24.
The quantity of the obverse is the opposite of the quantity of the obvertend.
Answer:
The quantity of the obverse is the same as the quantity of the obvertend.

Question 25.
The term which is not distributed in the obvertend is distributed in the Obverse.
Answer:
The term which is not distributed in the obvertend should not be distributed in the obverse.

Question 26.
‘A’ proposition obvert to‘T’ proposition.
Answer:
‘A’ proposition obvert to ‘E’ proposition.

Question 27.
‘E ’ proposition obvert to ‘E’ proposition.
Answer:
‘E’ proposition obvert to ‘A’ proposition.

Question 28.
‘T’ proposition obvert to ‘O’ proposition.
Answer:
‘T’ proposition obvert to ‘O’ proposition.

Question 29.
‘O’ proposition obvert to ‘O’ proposition.
Answer:
‘O’ proposition obvert to ‘T’ proposition.

Question 30.
Mediate inference is otherwise called a conversion.
Answer:
Mediate inference is otherwise called a Syllogism.

IV. Answer the Following Questions in One Word :

Question 1.
What type of knowledge gives us inference?
Answer:
Indirect

Question 2.
What type of process in inference?
Answer:
Mental

Question 3.
How many kinds of inference are there?
Answer:
Two

Question 4.
In which inference the conclusion is more general than the premises?
Answer:
Inductive

Question 5.
In which inference the conclusion is less general than the premises?
Answer:
Deductive

Question 6.
How many types of classifications are there of deductive inference?
Answer:
Two

Question 7.
How many types of classifications are there of immediate inference?
Answer:
Four types
OD

Question 8.
What is the other name of mediate inference?
Answer:
Syllogism

Question 9.
In which inference the conclusion is drawn from only one premise?
Answer:
Immediate

Question 10.
What is the premise of conversion called?
Answer:
Convertend

Question 11.
What is the conclusion of conversion called?
Answer:
Converse

Question 12.
How many propositions does an immediate inference consist of?
Answer:
Two

Question 13.
How many propositions does the mediate inference consist of?
Answer:
Three

Question 14.
What the quality of convertend and converse?
Answer:
Remains same

Question 15.
‘A’proposition convert to which proposition?
Answer:
‘T’

Question 16.
‘E ’ proposition convert to which proposition?
Answer:
‘E’

Question 17.
‘I’ proposition convert to which proposition?
Answer:
‘T’

Question 18.
‘O’ proposition convert to which proposition?
Answer:
cannot be converted

Question 19.
Which proposition cannot be converted?
Answer:
‘O’ proposition

Question 20.
In which conversion the quantity of convertend and converse remain the same?
Answer:
Simple conversion

Question 21.
In which conversion the quantity of convertend and converse are different from each other?
Answer:
partial

Question 22.
Which proposition convert simply?
Answer:
E & I

Question 23.
Which proposition convert partially?
Answer:
‘A’

Question 24.
What the premise of obversion is called?
Answer:
Obvertend

Question 25.
What the conclusion of obversion is called ?
Answer:
Obverse

Question 26.
What the quality of the obvertend and obverse?
Answer:
Change

Question 27.
What the quantity of the obvertend and obverse?
Answer:
Same

Question 28.
What the obvert of ‘A’ proposition?
Answer:
‘E’

Question 29.
What the obvert of ‘E’ proposition ?
Answer:
‘A’

Question 30.
What the obvert of‘T’ proposition?
Answer:
‘O’

Question 31.
What the obvert of ‘O’ proposition?
Answer:
‘T’

Question 32.
What is the name of the fallacy of Obversion?
Answer:
Material Obversion

Question 33.
Can Material obversion be regarded as a form of obversion?
Answer:
No

CHSE Odisha 12th Class Logic Book Solutions (+ 2 2nd Year)

CHSE Odisha 12th Class Logic Book Solutions in English Medium

• Chapter 1 The Theory of Inference
• Chapter 2 Syllogism
• Chapter 3 Mixed Syllogism
• Chapter 4 Deductive and Inductive Fallacies
• Chapter 5 Symbolic Logic
• Chapter 6 Experimental Methods of Mill
• Chapter 7 Scientific Explanation
• Chapter 8 Naya Theory of Knowledge
• Chapter 9 Doctrine of Karma
• Chapter 10 Gandhi: Truth and Non-Violence

CHSE Odisha 12th Class Logic Book Solutions in Odia Medium

Unit 1 ଅନୁମାନ – ଅନୁମାନର ପ୍ରକାରଭେଦ – ଅବ୍ୟବହିତ ଓ ବ୍ୟବହିତ

• ଅନୁମାନ – ଅନୁମାନର ପ୍ରକାରଭେଦ – ଅବ୍ୟବହିତ ଓ ବ୍ୟବହିତ Objective Questions
• ଅନୁମାନ – ଅନୁମାନର ପ୍ରକାରଭେଦ – ଅବ୍ୟବହିତ ଓ ବ୍ୟବହିତ Short & Long Answer Questions

Unit 2 ବ୍ୟବହିତ ଅନୁମାନ ଓ ମିଶ୍ର ତ୍ରିପଦୀଯୁକ୍ତି

• ବ୍ୟବହିତ ଅନୁମାନ ଓ ମିଶ୍ର ତ୍ରିପଦୀଯୁକ୍ତି Objective Questions
• ବ୍ୟବହିତ ଅନୁମାନ ଓ ମିଶ୍ର ତ୍ରିପଦୀଯୁକ୍ତି Short Answer Questions
• ବ୍ୟବହିତ ଅନୁମାନ ଓ ମିଶ୍ର ତ୍ରିପଦୀଯୁକ୍ତି Long Answer Questions

Unit 3 ତର୍କଦୋଷ ଓ ପ୍ରତୀକାତ୍ମକ ତର୍କଶାସ୍ତ୍ର

• ତର୍କଦୋଷ ଓ ପ୍ରତୀକାତ୍ମକ ତର୍କଶାସ୍ତ୍ର Objective Questions
• ତର୍କଦୋଷ ଓ ପ୍ରତୀକାତ୍ମକ ତର୍କଶାସ୍ତ୍ର Short Answer Questions
• ତର୍କଦୋଷ ଓ ପ୍ରତୀକାତ୍ମକ ତର୍କଶାସ୍ତ୍ର Long Answer Questions

Unit 4 ମିଲ୍‌ଙ୍କ ପରୀକ୍ଷଣ ପଦ୍ଧତି, ବୈଜ୍ଞାନିକ ବ୍ୟାଖ୍ୟାନ

• ମିଲ୍‌ଙ୍କ ପରୀକ୍ଷଣ ପଦ୍ଧତି, ବୈଜ୍ଞାନିକ ବ୍ୟାଖ୍ୟାନ Objective Questions
• ମିଲ୍‌ଙ୍କ ପରୀକ୍ଷଣ ପଦ୍ଧତି, ବୈଜ୍ଞାନିକ ବ୍ୟାଖ୍ୟାନ Short & Long Answer Questions

Unit 5 ନ୍ୟାୟଙ୍କ ଜ୍ଞାନ ସିଦ୍ଧାନ୍ତ ଓ କର୍ମବାଦ

• ନ୍ୟାୟଙ୍କ ଜ୍ଞାନ ସିଦ୍ଧାନ୍ତ ଓ କର୍ମବାଦ Objective Questions
• ନ୍ୟାୟଙ୍କ ଜ୍ଞାନ ସିଦ୍ଧାନ୍ତ ଓ କର୍ମବାଦ Short & Long Answer Questions

CHSE Odisha Class 12 Logic Syllabus (+2 2nd Year) in English

Unit 1
The Theory of Inference: Classification of Inference, Conversion, Obversion, Categorical Syllogism: Structure, Figure, Moods. Rules of syllogism, Determination of valid Moods.

Unit 2
Special rules of Figures, Aristotle’s Dictum, Direct and Indirect Reduction.
Mixed Syllogism: Different forms – Hypothetical categorical, Alternative Categorical, Disjunctive Categorical, Dilemma: Forms, Refutation, Rebuttal of Dilemma.

Unit 3
Fallacy: Deductive Fallacy, Semi-logical Fallacies, Inductive Fallacies: Fallacy of Illicit Generalisation, False Analogy, Ignoratio Elenchi. Propositional Logic: Symbolic Logic and its Characteristics, Propositional Variables, Logical Constants, Propositional Connectives, Truth Functions, Construction of Truth Tables, Testing Validity by direct Truth Table Method.

Unit 4
Methods of Experimental Enquiry: Mill’s Five Experimental Methods.
Scientific Explanation: Nature of Scientific Explanation.

Unit 5
Nyaya Theory of Knowledge :Perception and Inference: Vyapti and its ascertainments.
Doctrine of karma: Niskama Karma of Bhagavad Gita, Gandian Concept of Non Violence.

CHSE Odisha Class 12 Logic Book Syllabus (+2 2nd Year)

Unit 1 ଅନୁମାନ – ଅନୁମାନର ପ୍ରକାରଭେଦ – ଅବ୍ୟବହିତ ଓ ବ୍ୟବହିତ
ଅନୁମାନ – ଅନୁମାନର ପ୍ରକାରଭେଦ – ଅବ୍ୟବହିତ ଓ ବ୍ୟବହିତ ଅନୁମାନ – ଅବ୍ୟବହିତ ଅନୁମାନ – ସମବର୍ତ୍ତନ, ବ୍ୟାବର୍ତ୍ତନ; ବ୍ୟବହିତ ଅନୁମାନ– ତ୍ରିପଦୀଯୁକ୍ତି, ଏହାର ଅବୟବାବଳୀ, ନ୍ୟାୟ-ସଂସ୍ଥାନ, ନ୍ୟାୟରୂପ, ତ୍ରିପଦୀଯୁକ୍ତିର ସାଧାରଣ ନିୟମାବଳୀ, ନ୍ୟାୟରୂପ-ମାନଙ୍କର ନିର୍ଦ୍ଧାରଣ ପ୍ରକ୍ରିୟା ।

Unit 2 ବ୍ୟବହିତ ଅନୁମାନ ଓ ମିଶ୍ର ତ୍ରିପଦୀଯୁକ୍ତି
ପ୍ରତ୍ୟେକ ସଂସ୍ଥାନର ସ୍ବତନ୍ତ୍ର ନିୟମାବଳୀ, ଆରିଷ୍ଟୋଟଲଙ୍କ ମୌଳିକ ସୂତ୍ର, ସାକ୍ଷାତ୍ ଓ ଅସାକ୍ଷାତ୍ ରୂପାନ୍ତରୀକରଣ ।
ମିଶ୍ର ତ୍ରିପଦୀଯୁକ୍ତି – ବିଭିନ୍ନ ପ୍ରକାର : ପ୍ରାକଳ୍ପିକ ନିରପେକ୍ଷ, ବିଯୋଜକ – ନିରପେକ୍ଷ, ବୈକଳ୍ପିକ — ନିରପେକ୍ଷ, ଦ୍ବିଶୃଙ୍ଗକ ନ୍ୟାୟ, ଦ୍ବିଶୃଙ୍ଗକ ଯୁକ୍ତିର ପ୍ରକାରଭେଦ, ଦ୍ବି ଶୃଙ୍ଗକ ଯୁକ୍ତିର ଆକାରଗତ ବୈଧତା, ବସ୍ତୁଗତ ସତ୍ୟାସତ୍ୟ ବିଚାର, ଦ୍ବିଶୃଙ୍ଗକ ଯୁକ୍ତିର ପ୍ରତିରୋଧ ।

Unit 3 ତର୍କଦୋଷ ଓ ପ୍ରତୀକାତ୍ମକ ତର୍କଶାସ୍ତ୍ର
ତର୍କଦୋଷ – ଅବରୋହୀ ତର୍କଦୋଷ, ଅବରୋହୀ – ଅନୁମାନ ସମ୍ପର୍କୀୟ ତର୍କଦୋଷ, ଆପାତଃ ତର୍କଦୋଷ । ଆରୋହୀ ତର୍କଦୋଷ – ଅବୈଧ ସାମାନ୍ୟକରଣ ତର୍କଦୋଷ, ଦୁଷ୍ଟ ଉପମା କିମ୍ବା ଦୁର୍ବଳ ଉପମା ତର୍କଦୋଷ, ବଳକା ତର୍କଦୋଷ, ଅବାନ୍ତର ପ୍ରସଙ୍ଗ ଦୋଷ ।
ପ୍ରତୀକାମୂକ ତର୍କଶାସ୍ତ୍ର ଏବଂ ଏହାର ବୈଶିଷ୍ଟ୍ୟ- ତର୍କବାକ୍ୟମୂଳକ ଚଳ- ତର୍କଶାସ୍ତ୍ରୀୟ ସ୍ଥିରାଙ୍କ- ସତ୍ୟଫଳନ – ସତ୍ୟସାରଣୀ- ସତ୍ୟ ସାରଣୀ ପଦ୍ଧତି । ସାକ୍ଷାତ୍ ସତ୍ୟସାରଣୀ ପଦ୍ଧତି ସାହାଯ୍ୟରେ ବୈଧତା ପରୀକ୍ଷା ।
ପୁନରୁକ୍ତିକ ତର୍କବାକ୍ୟମୂଳକ ସୂତ୍ର- ବିରୁଦ୍ଧ ତର୍କବାକ୍ୟମୂଳକ ସୂତ୍ର- ଆପାତିତ ତର୍କବାକ୍ୟମୂଳକ ସୂତ୍ର । ବିଭିନ୍ନ ଉଦାହରଣମାନଙ୍କର ସାକ୍ଷାତ୍ ସତ୍ୟ ସାରଣୀ ପଦ୍ଧତି ସାହାଯ୍ୟରେ ବୈଧତା ପରୀକ୍ଷା ।

Unit 4 ମିଲ୍ଲଙ୍କ ପରୀକ୍ଷଣ ପଦ୍ଧତି, ବୈଜ୍ଞାନିକ ବ୍ୟାଖ୍ୟାନ
ମିଲ୍‌ଙ୍କ ପରୀକ୍ଷଣ ପଦ୍ଧତି—ମିଲ୍‌ଙ୍କ ପାଞ୍ଚଟି ପରୀକ୍ଷଣ ପଦ୍ଧତି— (୧) ଅନ୍ବୟ ପଦ୍ଧତି, (୨) ବ୍ୟତିରେକ ପଦ୍ଧତି, (୩) ସଂଯୁକ୍ତ ପଦ୍ଧତି, (୪) ସହଚାରୀ ପରିବର୍ତ୍ତନ ପଦ୍ଧତି, (୫) ପରିଶେଷ ପଦ୍ଧତି ।
ବୈଜ୍ଞାନିକ ବ୍ୟାଖ୍ୟାନ– ବୈଜ୍ଞାନିକ ବ୍ୟାଖ୍ୟାନର ଲକ୍ଷଣ ।

Unit 5 ନ୍ୟାୟଙ୍କ ଜ୍ଞାନ ସିଦ୍ଧାନ୍ତ ଓ କର୍ମବାଦ
ନ୍ୟାୟଙ୍କ ଜ୍ଞାନ ସିଦ୍ଧାନ୍ତ – ପ୍ରତ୍ୟକ୍ଷ ଓ ଅନୁମାନ, ବ୍ୟାପ୍ତି ଓ ଏହାର ନିର୍ଦ୍ଧାରଣ ପ୍ରକ୍ରିୟା
କର୍ମବାଦ – ଭଗବତ୍ ଗୀତାର ନିଷ୍କାମ କର୍ମ, ଗାନ୍ଧିଜୀଙ୍କ ଅହିଂସାବାଦ

CHSE Odisha Class 12 Text Book Solutions

CHSE Odisha 12th Class Economics Book Solutions (+ 2 2nd Year)

CHSE Odisha Class 12 Economics Book Solutions in Odia Medium

Chapter 1 ଅର୍ଥଶାସ୍ତ୍ରର ସଂଜ୍ଞା, ପରିସର ଓ ବିଷୟବସ୍ତୁ

• ଅର୍ଥଶାସ୍ତ୍ରର ସଂଜ୍ଞା, ପରିସର ଓ ବିଷୟବସ୍ତୁ Objective & Short Answer Questions
• ଅର୍ଥଶାସ୍ତ୍ରର ସଂଜ୍ଞା, ପରିସର ଓ ବିଷୟବସ୍ତୁ Long Answer Questions

Chapter 2 ଅର୍ଥବ୍ୟବସ୍ଥାର ପରିଚୟ ଏବଂ ଅର୍ଥଶାସ୍ତ୍ରର କେନ୍ଦ୍ରୀୟ ସମସ୍ୟାବଳୀ

• ଅର୍ଥବ୍ୟବସ୍ଥାର ପରିଚୟ ଏବଂ ଅର୍ଥଶାସ୍ତ୍ରର କେନ୍ଦ୍ରୀୟ ସମସ୍ୟାବଳୀ Objective & Short Answer Questions
• ଅର୍ଥବ୍ୟବସ୍ଥାର ପରିଚୟ ଏବଂ ଅର୍ଥଶାସ୍ତ୍ରର କେନ୍ଦ୍ରୀୟ ସମସ୍ୟାବଳୀ Long Answer Questions

Chapter 3 ମୌଳିକ ଧାରଣା (ମାନବୀୟ ଅଭାବ, ଉପଯୋଗିତା, ଦ୍ରବ୍ୟ, ମୂଲ୍ୟ, ଦର ଓ ସମ୍ପଦ)

• ମୌଳିକ ଧାରଣା Objective Questions
• ମୌଳିକ ଧାରଣା Short Answer Questions
• ମୌଳିକ ଧାରଣା Long Answer Questions

Chapter 4 ଉପଭୋଗର ନିୟମ

• ଉପଭୋଗର ନିୟମ Objective Questions
• ଉପଭୋଗର ନିୟମ Short & Long Answer Questions

Chapter 5 ଚାହିଦା

• ଚାହିଦା Objective Questions
• ଚାହିଦା Short Answer Questions
• ଚାହିଦା Long Answer Questions

Chapter 6 ଉତ୍ପାଦନ

• ଉତ୍ପାଦନ Objective Questions
• ଉତ୍ପାଦନ Short & Long Answer Questions

Chapter 7 ପରିବ୍ୟୟ

• ପରିବ୍ୟୟ Objective Questions
• ପରିବ୍ୟୟ Short Answer Questions
• ପରିବ୍ୟୟ Long Answer Questions

Chapter 8 ଆୟ

Chapter 9 ଯୋଗାଣ

• ଯୋଗାଣ Objective & Short Answer Questions
• ଯୋଗାଣ Long Answer Questions

Chapter 10 ବଜାର

• ବଜାର Objective Questions
• ବଜାର Short Answer Questions
• ବଜାର Long Answer Questions

Chapter 11 ସମଷ୍ଟି ଅର୍ଥନୀତି

• ସମଷ୍ଟି ଅର୍ଥନୀତି Objective & Short Answer Questions
• ସମଷ୍ଟି ଅର୍ଥନୀତି Long Answer Questions

Chapter 12 ଜାତୀୟ ଆୟ

• ଜାତୀୟ ଆୟ Objective Questions
• ଜାତୀୟ ଆୟ Short Answer Questions
• Long Answer Questions

Chapter 13 କେନ୍‌ସଙ୍କ ଆୟ ନିର୍ଦ୍ଧାରଣ ତତ୍ତ୍ବ

• କେନ୍‌ସଙ୍କ ଆୟ ନିର୍ଦ୍ଧାରଣ ତତ୍ତ୍ବ Objective Questions
• କେନ୍‌ସଙ୍କ ଆୟ ନିର୍ଦ୍ଧାରଣ ତତ୍ତ୍ବ Short Answer Questions
• କେନ୍‌ସଙ୍କ ଆୟ ନିର୍ଦ୍ଧାରଣ ତତ୍ତ୍ବ Long Answer Questions

Chapter 14 ମୁଦ୍ରା

• ମୁଦ୍ରା Objective Questions
• ମୁଦ୍ରା Short & Long Answer Questions

Chapter 15 ବ୍ୟାଙ୍କ

• ବ୍ୟାଙ୍କ Objective Questions
• ବ୍ୟାଙ୍କ Short Answer Questions
• ବ୍ୟାଙ୍କ Long Answer Questions

Chapter 16 ରାଷ୍ଟ୍ରବିତ୍ତ

• ରାଷ୍ଟ୍ରବିତ୍ତ Objective Questions
• ରାଷ୍ଟ୍ରବିତ୍ତ Short Answer Questions
• ରାଷ୍ଟ୍ରବିତ୍ତ Long Answer Questions

Chapter 17 ବଜେଟ୍

CHSE Odisha Class 12 Economics Book Solutions in English Medium

Part A: Introductory Microeconomics

Unit 1 Introduction

• Chapter 1 Definition of Economics and Central Problems of An Economy
• Chapter 2 Basic Economic Concepts (Wants, Utility, Goods, Value, Price and Wealth)

Unit II Consumption and Demand

• Chapter 3 Laws of Consumption
• Chapter 4 Demand and Price Elasticity of Demand

Unit III Production

• Chapter 5 Production

Unit IV Cost, Revenue and Supply

• Chapter 6 Cost and Revenue
• Chapter 7 Supply

Unit V Market

• Chapter 8 Market

Part B: Introductory Macroeconomics

Unit VI Introduction

• Chapter 9 Meaning of Macroeconomics

Unit VII National Income

• Chapter 10 Aggregates Related to National Incorne
• Chapter 11 Theory of Income Determination

Unit VIII Money, Banking and Public Finance

• Chapter 12 Money and Banking
• Chapter 13 Public Finance

CHSE Odisha Class 12 Economics Syllabus (+2 2nd Year)

Second Year CHSE (2025-2026)
Economics Paper-II
(Elementary Micro and Macro Economics)

Part A: Introductory Micro Economics

Unit I Introduction (10 Periods, 10 Marks)

• Definition, scope, and subject matter of economics.
• Meaning of economy and central problems of an economy – scarcity and choice, what, how, and for whom to produce?
• Basic concepts – wants, utility, goods, value, price, and wealth.

Unit II Consumption and Demand (14 Periods, 15 Marks)

• Laws of consumption – marginal and total utility, law of diminishing marginal utility, the law of equimarginal utility, and conditions of consumer’s equilibrium.
• Demand – meaning and determinants, individual and market demand, demand schedule and demand curve, movement along and shifts in the demand curve.
• Price elasticity of demand – concept, determinants, measurement of price elasticity of demand; percentage and geometric methods (linear demand curve), the relation of price elasticity of demand with total expenditure.

Unit III Production (10 Periods, 10 Marks)

• Meaning of production and production function – short run and long run.
• Total, Average, and Marginal Product.
• Law of variable proportions and returns to a factor.

Unit IV Cost, Revenue, and Supply (12 Periods, 15 Marks)

• Cost – money and real cost, implicit and explicit cost, fixed and variable cost, Total, average, and marginal costs in the short ru,n and their relationship (simple analysis).
• Revenue – Total, average, and marginal revenue and their relationship.
• Supply – meaning and law of supply

Unit V Market (8 Periods, 10 Marks)

• Meaning and forms of market, pure and perfect competition, price determination under perfect competition, and effects of shifts in demand and supply.
• Meaning and features of monopoly, monopolistic competition, and oligopoly.

Part B: Introductory Macroeconomics

Unit VI Introduction (4 Periods, 5 Marks)

• Meaning of macroeconomics, Distinction between macro- and microeconomics, the subject matter of macroeconomics

Unit VII National Income (10 Periods, 15 Marks)

• Meaning and aggregates related to national income – GNP, NNP, GDP, and NDP at market price and factor cost.
• National disposable income (Gross and Net), Private Income, Personal income, Personal disposable income, Nominal and real national income.
• Income determination – Aggregate Demand and Supply and their components, simple Keynesian Theory of Income Determination.

Unit VIII Money, Banking, and Public Finance (12 Periods, 20 Marks)

• Meaning and Functions of Money.
• Meaning and Functions of Commercial Banks.
• Functions of the Central Bank.
• Meaning of Public Finance and Difference between public and private finance.
• Budget – Meaning and objectives, a balanced and unbalanced budget, surplus and deficit budget.

CHSE Odisha Class 12 Text Book Solutions

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 4 Matrices Ex 4(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Exercise 4(b)

Question 1.
State which of the following matrices are symmetric, skew-symmetric, both or not either:

Solution:
(i) Symmetric
(ii) Neither Symmetric nor skew-symmetric
(iii) Symmetric
(iv) Skew symmetric
(v) Both
(vi) Neither symmetric nor skew-symmetric
(vii) Skew symmetric

Question 2.
State ‘True’ or ‘False’:
(i) If A and B are symmetric matrices of the same order and AB – BA ≠ 0, then AB is not symmetric.
Solution:
True

(ii) For any square matrix A, AA’ is symmetric.
Solution:
True

(iii) If A is any skew-symmetric matrix, then A² is also skew-symmetric.
Solution:
False

(iv) If A is symmetric, then A², A³, …, Aⁿ are all symmetric.
Solution:
True

(v) If A is symmetric then A – A¹ is both symmetric and skew-symmetric.
Solution:
False

(vi) For any square matrix (A – A¹)² is skew-symmetric.
Solution:
True

(vii) A matrix which is not symmetric is skew-symmetric.
Solution:
False

Question 3.
(i) If A and B are symmetric matrices of the same order with AB ≠ BA, final whether AB – BA is symmetric or skew symmetric.
Solution:
A and B are symmetric matrices;
Thus A’ = A and B’ = B
Now (AB – BA)’ = (AB)’ – (BA)’
= B’A’ – A’B’
= BA – AB = – (AB – BA)
∴ AB – BA is skew symmetric.

(ii) If a symmetric/skew-symmetric matrix is expressed as a sum of a symmetric and a skew-symmetric matrix then prove that one of the matrices in the sum must be zero matrix.
Solution:
We know that zero matrix is both symmetric as well as skew-symmetric.
Let A is symmetric.
∴ A = A + O where A is symmetric and O is treated as skew-symmetric. If B is skew-symmetric then we can write B = O + B where O is symmetric and B is skew-symmetric.

Question 4.
A and B are square matrices of the same order, prove that
(i) If A, B and AB are all symmetric, then AB – BA = 0
Solution:
Let A, B and AB are all symmetric.
∴A’ = A, B’ = B and (AB)’ = AB
⇒ B’A’ = AB
⇒ BA = AB
⇒ AB – BA = 0

(ii) If A, B and AB are all skew symmetric then AB + BA = 0
Solution:
Let A, B and AB are all skew symmetric matrices
∴ A’ = -A, B’ = -B and (AB)’ = -AB
Now (AB)’ = -AB
⇒ B’A’ = -AB
⇒ (-B) (-A) = -AB
⇒ BA = -AB
⇒ AB + BA = 0

Question 5.
If A = \(\left[\begin{array}{rrr}
1 & 2 & 0 \\
0 & 1 & 3 \\
-2 & 5 & 3
\end{array}\right]\), then verify that A’ = \(\left[\begin{array}{ccc}
1 & 0 & -2 \\
2 & 1 & 5 \\
0 & 3 & 3
\end{array}\right]\)

(i) A+A’ is symmetric

(ii) A-A’ is skew-symmetric

Question 6.
Prove that a unit matrix is its own inverse. Is the converse true?
IfA = \(\left[\begin{array}{rrr}
0 & 1 & -1 \\
4 & -3 & 4 \\
3 & -3 & 4
\end{array}\right]\) show that A² = I and hence A= A^-1.
Solution:
No the converse is not true for example:

Question 7.
Here A is an involuntary matrix, recall the definition given earlier.
Solution:

Question 8.
Show that \(\left[\begin{array}{ll}
\mathbf{0} & \mathbf{1} \\
\mathbf{1} & \mathbf{0}
\end{array}\right]\) is its own inverse.
Solution:

Question 9.
Express as a sum of a symmetric and a skew symmetric matrix.

Solutions:

Question 10.
What is the inverse of

Question 11.
Find inverse of the following matrices by elementary row/column operation (transformations):
(i) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 5
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{cc}
4 & -2 \\
3 & 1
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{cc}
1 & 0 \\
2 & -3
\end{array}\right]\)
Solution:

(vi) \(\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\)
Solution:

Question 12.
Find the inverse of the following matrices using elementary transformation:
(i) \(\left[\begin{array}{lll}
\mathbf{0} & \mathbf{0} & 2 \\
\mathbf{0} & \mathbf{2} & \mathbf{0} \\
\mathbf{2} & \mathbf{0} & \mathbf{0}
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & 1 & 1
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{ccc}
3 & -2 & 3 \\
2 & 1 & -1 \\
4 & -3 & 2
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{lll}
1 & 1 & 2 \\
0 & 1 & 2 \\
1 & 2 & 1
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 1 & 4 \\
1 & 0 & 2
\end{array}\right]\)
Solution:

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 7 Continuity and Differentiability Ex 7(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 7 Continuity and Differentiability Exercise 7(b)

Question 1.
Differentiate from definition
(i) e^3x
Solution:

(ii) 2^x2
Solution:

(iii) In (3x + 1)
Solution:

(iv) logx⁵ (Hint : logx⁵ = \(\frac{\ln 5}{\ln x}\))
Solution:

(v) In sin x
Solution:

(vi) x² a^2x
Solution:

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 5 Determinants Ex 5(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 5 Determinants Exercise 5(b)

Question 1.
Write the number of solutions of the following system of equations.
(i) x – 2y = 0
Solution:
No solution

(ii) x – y = 0 and 2x – 2y = 1
Solution:
Infinite

(iii) 2x + y = 2 and -x – 1/2y = 3
Solution:
No solution

(iv) 3x + 2y = 1 and x + 5y = 6
Solution:
One

(v) 2x + 3y + 1 = 0 and x – 3y – 4 = 0
Solution:
One

(vi) x + y + z = 1
x + y + z = 2
2x + 3y + z = 0
Solution:
No solution

(vii) x + 4y – z = 0
3x – 4y – z = 0
x – 3y + z = 0
Solution:
One

(viii) x + y – z = 0
3x – y + z = 0
x – 3y + z = 0
Solution:
One

(ix) a₁x + b₁y + c₁z = 0
a₂x + b₂y + c₂z = 0
a₃x + b₃y + c₃z = 0
and \(\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|\) = 0
Solution:
Infinite solutions as Δ = Δ₁ = Δ₂ = Δ₃ = 0

Question 2.
Show that the following system is inconsistent.
(a – b)x + (b – c)y + (c – a)z = 0
(b – c)x + (c – a)y + (a – b)z = 0
(c – a)x + (a – b)y + (b – c)z =1
Solution:

Question 3.
(i) The system of equations
x + 2y + 3z = 4
2x + 3y + 4z = 5
3x + 4y + 5z = 6 has
(a) infinitely many solutions
(b) no solution
(c) a unique solution
(d) none of the three
Solution:
(a) infinitely many solutions

(ii) If the system of equations
2x + 5y + 8z = 0
x + 4y + 7z = 0
6x + 9y – z = 0
has a nontrivial solution, then is equal to
(a) 12
(b) -12
(c) 0
(d) none of the three
Solution:
(b) -12

(iii) The system of linear equations
x + y + z = 2
2x + y – z = 3
3x +2y + kz = 4
has a unique solution if
(a) k ≠ 0
(b) -1 < k < 1
(c) -2 < k < 2
(d) k = 0
Solution:
(a) k ≠ 0

(iv) The equations
x + y + z = 6
x + 2y + 3z = 10
x + 2y + mz = n
give infinite number of values of the triplet (x, y, z) if
(a) m = 3, n ∈ R
(b) m = 3, n ≠ 10
(c) m = 3, n = 10
(d) none of the three
Solution:
(c) m = 3, n = 10

(v) The system of equations
2x – y + z = 0
x – 2y + z = 0
x – y + 2z = 0
has infinite number of nontrivial solutions for
(a) = 1
(b) = 5
(c) = -5
(d) no real value of
Solution:
(c) = -5

(vi) The system of equations
a₁x + b₁y + c₁z = 0
a₂x + b₂y + c₂z = 0
a₃x + b₃y + c₃z =0
with has
(a) more than two solutions
(b) one trivial and one nontrivial solutions
(c) No solution
(d) only trivial solutions
Solution:
(a) more than two solutions

Question 4.
Can the inverses of the following matrices be found?
(i) \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
Solution:
|A| = 0
∴ A^-1 can not be found.

(ii) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:
∴ |A| = 4 – 6 = -2 ≠ 0
∴ A^-1 exists.

(iii) \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
Solution:
|A| = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\) = 1 – 1 = 0
∴ A^-1 does not exist.

(iv) \(\left[\begin{array}{ll}
1 & 2 \\
2 & 4
\end{array}\right]\)
Solution:
|A| = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 4
\end{array}\right]\) = 4 – 4 = 0
∴ A^-1 does not exist.

(v) \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
Solution:
|A| = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) = 1 ≠ 0
∴ A^-1 exists.

Question 5.
Find the inverse of the following:
(i) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{cc}
2 & -1 \\
1 & 3
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{cc}
4 & -2 \\
3 & 1
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{cc}
1 & 0 \\
2 & -3
\end{array}\right]\)
Solution:

(vi) \(\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\)
Solution:

(vii) \(\left[\begin{array}{cc}
i & -i \\
i & i
\end{array}\right]\)
Solution:

(viii) \(\left[\begin{array}{ll}
x & -x \\
x & x^2
\end{array}\right]\), x ≠ 0, x ≠ -1
Solution:

Question 6.
Find the adjoint of the following matrices.
(i) \(\left[\begin{array}{ccc}
1 & 1 & -1 \\
2 & -1 & 2 \\
1 & 3 & -2
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ccc}
-2 & 2 & 3 \\
1 & 4 & 2 \\
-2 & -3 & 1
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{lll}
2 & 1 & 2 \\
2 & 2 & 1 \\
1 & 2 & 2
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ccc}
1 & 3 & 0 \\
2 & -1 & 6 \\
5 & -3 & 1
\end{array}\right]\)
Solution:

Question 7.
Which of the following matrices are invertible?
(i) \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
1 & 1 & 1 \\
2 & -1 & 1
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ccc}
2 & 1 & -2 \\
1 & 2 & 1 \\
3 & 6 & 4
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{ccc}
-1 & -2 & 3 \\
2 & 1 & -4 \\
-1 & 0 & 2
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ccc}
1 & 0 & 1 \\
2 & -2 & 1 \\
3 & 2 & 4
\end{array}\right]\)
Solution:

Question 8.
Examining consistency and solvability, solve the following equations by matrix method.
(i) x – y + z = 4
2x + y – 3z = 0
x + y + z = 2
Solution:

(ii) x + 2y – 3z = 4
2x + 4y – 5z = 12
3x – y + z = 3
Solution:
Let

(iii) 2x – y + z = 4
x + 3y + 2z = 12
3x + 2y + 3z = 16
Solution:

(iv) x + y + z = 4
2x + 5y – 2x = 3
x + 7y – 7z = 5
Solution:

(v) x + y + z = 4
2x – y + 3z = 1
3x + 2y – z = 1
Solution:

(vi) x + y – z = 6
2x – 3y + z = 1
2x – 4y + 2z = 1
Solution:

(vii) x – 2y = 3
3x + 4y – z = -2
5x – 3z = -1
Solution:

(viii) x + 2y + 3z = 14
2x – y + 5z = 15
2y + 4z – 3x = 13
Solution:

(ix) 2x + 3y +z = 11
x + y + z = 6
5x – y + 10z = 34
Solution:

Question 9.
Given the matrices
A = \(\left[\begin{array}{ccc}
1 & 2 & 3 \\
3 & -2 & 1 \\
4 & 2 & 1
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and C = \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)
write down the linear equations given by AX = C and solve it for x, y, z by matrix method.
Solution:

Question 10.
Find X, if \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1 & -1 \\
2 & 1 & -1
\end{array}\right]\) X = \(\left[\begin{array}{l}
6 \\
0 \\
1
\end{array}\right]\) where X = \(\left[\begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array}\right]\)
Solution:

Question 11.
Answer the following:
(i) If every element of a third order matrix is multiplied by 5, then how many times its determinant value becomes?
Solution:
125

(ii) What is the value of x if \(\left|\begin{array}{ll}
4 & 1 \\
2 & 1
\end{array}\right|^2=,\left|\begin{array}{ll}
3 & 2 \\
1 & x
\end{array}\right|-\left|\begin{array}{cc}
x & 3 \\
-2 & 1
\end{array}\right|\) ?
Solution:
x = 6

(iii) What are the values of x and y if \(\left|\begin{array}{ll}
x & y \\
1 & 1
\end{array}\right|=2,\left|\begin{array}{ll}
x & 3 \\
y & 2
\end{array}\right|=1\) ?
Solution:
x = 5, y = 3

(iv) What is the value of x if \(\left|\begin{array}{ccc}
x+1 & 1 & 1 \\
1 & 1 & -1 \\
-1 & 1 & 1
\end{array}\right|\) = 4?
Solution:
x = 0

(v) What is the value of \(\left|\begin{array}{ccc}
\mathbf{o} & -\mathbf{h} & -\mathbf{g} \\
\mathbf{h} & \mathbf{0} & -\mathbf{f} \\
\mathbf{g} & \mathbf{f} & \mathbf{0}
\end{array}\right|\)?
Solution:
0

(vi) What is the value of \(\left|\begin{array}{l}
\frac{1}{a} 1 \mathrm{bc} \\
\frac{1}{b} 1 c a \\
\frac{1}{c} 1 a b
\end{array}\right|\)
Solution:
0

(vii) What is the co-factor of 4 in the determinant \(\left|\begin{array}{rrr}
1 & 2 & -3 \\
4 & 5 & 0 \\
2 & 0 & 1
\end{array}\right|\)
Solution:
-2

(viii)In which interval does the determinant \(\left|\begin{array}{ccc}
1 & \sin \theta & 1 \\
-\sin \theta & 1 & \sin \theta \\
-1 & -\sin \theta & 1
\end{array}\right|\) lie?
Solution:
[2, 4]

(ix) Ifx + y + z = n, what is the value of Δ = \(\left|\begin{array}{ccc}
\sin (x+y+z) & \sin B & \cos C \\
-\sin B & 0 & \tan A \\
\cos (A+B) & -\tan A & 0
\end{array}\right|\) Where A, B, C are the angles of triangle.
Solution:
0

Question 12.
Evaluate the following determinants:
(i) \(\left|\begin{array}{ccc}
14 & 3 & 28 \\
17 & 9 & 34 \\
25 & 9 & 50
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
14 & 3 & 28 \\
17 & 9 & 34 \\
25 & 9 & 50
\end{array}\right|\)
= 2\(\left|\begin{array}{ccc}
14 & 3 & 28 \\
17 & 9 & 34 \\
25 & 9 & 50
\end{array}\right|\) = 0
(∵ C₁ = C₃)

(ii) \(\left|\begin{array}{ccc}
16 & 19 & 13 \\
15 & 18 & 12 \\
14 & 17 & 11
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
16 & 19 & 13 \\
15 & 18 & 12 \\
14 & 17 & 11
\end{array}\right|\) = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1 & 1 \\
14 & 17 & 11
\end{array}\right|\)
( R₁ = R₁ – R₂, R₂ = R₂ – R₃)
= 0 (∵ R₁ = R₂)

(iii) \(\left|\begin{array}{ccc}
224 & 777 & 32 \\
735 & 888 & 105 \\
812 & 999 & 116
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
224 & 777 & 32 \\
735 & 888 & 105 \\
812 & 999 & 116
\end{array}\right|\)
= 7\(\left|\begin{array}{ccc}
32 & 777 & 32 \\
105 & 888 & 105 \\
116 & 999 & 116
\end{array}\right|\) = 0
(∵ C₁ = C₂)

(iv) \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 3 & 4 \\
3 & 4 & 6
\end{array}\right|\)
Solution:

(v) \(\left|\begin{array}{ccc}
1 & 2 & 3 \\
3 & 5 & 7 \\
8 & 14 & 20
\end{array}\right|\)
Solution:

(vi) \(\left|\begin{array}{ccc}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2
\end{array}\right|\)
Solution:

= 225 – 256 – 4(100 – 144) + 9(64 – 81)
= -31 – 4(-44) + 9(-17)
= -31 + 176 – 153 = -184 + 176
= -8

(vii) \(\left|\begin{array}{ccc}
1 & 0 & -5863 \\
-7361 & 2 & 7361 \\
1 & 0 & 4137
\end{array}\right|\)
Solution:
\(\left|\begin{array}{ccc}
1 & 0 & -5863 \\
-7361 & 2 & 7361 \\
1 & 0 & 4137
\end{array}\right|\)
= 2\(\left|\begin{array}{cc}
1 & -5863 \\
1 & 4137
\end{array}\right|\)
(expanding along 2nd column)
= 2(4137 + 5863)
= 2 × 10000 = 20000

(viii) \(\left|\begin{array}{lll}
265 & 240 & 219 \\
240 & 225 & 198 \\
219 & 198 & 181
\end{array}\right|\)
Solution:

(ix) \(\left|\begin{array}{ccc}
0 & a^2 & b \\
b^2 & 0 & a^2 \\
a & b^2 & 0
\end{array}\right|\)
Solution:

= -a² (0 –  a²) + b (b⁴ –  0) = a⁵ + b⁵

(x) \(\left|\begin{array}{ccc}
a-b & b-c & c-a \\
\boldsymbol{x}-\boldsymbol{y} & \boldsymbol{y}-\boldsymbol{z} & z-\boldsymbol{x} \\
\boldsymbol{p}-\boldsymbol{q} & \boldsymbol{q}-\boldsymbol{r} & \boldsymbol{r}-\boldsymbol{p}
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
a-b & b-c & c-a \\
x-y & y-z & z-x \\
p-q & q-r & r-p
\end{array}\right|\)
= \(\left|\begin{array}{lll}
0 & b-c & c-a \\
0 & y-z & z-x \\
0 & q-r & r-p
\end{array}\right|\) (C₁ = C₁ + C₂ + C₃)
= 0 ( ∵ C₁ = 0)

(xi) \(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\)
Solution:
\(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\)
= \(\left|\begin{array}{lll}
0 & b-c & c-a \\
0 & c-a & a-b \\
0 & a-b & b-c
\end{array}\right|\) (C₁ = C₁ + C₂ + C₃)
= 0

(xii) \(\left|\begin{array}{ccc}
-\cos ^2 \theta & \sec ^2 \theta & -0.2 \\
\cot ^2 \theta & -\tan ^2 \theta & 1.2 \\
-1 & 1 & 1
\end{array}\right|\)
Solution:

(Expanding along 3rd row)
= (-cos² θ + sec² θ) (-tan² θ – 1.2) – (sec² θ + 0.2) (cot² θ – tan² θ)
= sin² θ – 1.2 cos² θ – sec² θ tan² θ – 1.2 sec² θ – cosec² θ +  sec² θ tan² θ – 0.2 cot² θ + 0.2 tan² θ
= sin² θ – cosec² θ + 1.2 (cos² θ – sec² θ) + 0.2 (tan² θ – cot² θ) ≠ 0
The question seems to be wrong.

Question 13.
If \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+x & 1 \\
1 & 1 & 1+y
\end{array}\right|\) = 0 what are x and y?
Solution:

or, xy – 0 = 0 ⇒ xy = 0, ⇒ x = 0, or y = 0

Question 14.
For what value of x \(\left|\begin{array}{ccc}
2 x & 0 & 0 \\
0 & 1 & 2 \\
-1 & 2 & 0
\end{array}\right|\) = \(\left|\begin{array}{lll}
1 & 0 & 0 \\
2 & 3 & 4 \\
0 & 3 & 5
\end{array}\right|\)?
Solution:

Question 15.
Solve \(\left|\begin{array}{ccc}
x+a & 0 & 0 \\
a & x+b & 0 \\
a & 0 & x+c
\end{array}\right|\) = 0
Solution:
\(\left|\begin{array}{ccc}
x+a & 0 & 0 \\
a & x+b & 0 \\
a & 0 & x+c
\end{array}\right|\) = 0
or, (x – a) \(\left|\begin{array}{cc}
x+b & 0 \\
0 & x+c
\end{array}\right|\) = 0
or, (x + a) (x + b) (x + c) = 0
x = -a, x = -b, x = -c

Question 16.
Solve \(\left|\begin{array}{lll}
a+x & a-x & a-x \\
a-x & a+x & a-x \\
a-x & a-x & a+x
\end{array}\right|\) = 0
Solution:

Question 17.
Solve \(\left|\begin{array}{ccc}
x+a & b & c \\
a & x+b & c \\
a & b & x+c
\end{array}\right|\) = 0
Solution:

Question 18.
Show that x = 2 is a root of \(\left|\begin{array}{ccc}x & -6 & -1 \\ 2 & -3 x & x-3 \\ -3 & 2 x & x+2\end{array}\right|\) = 0 Solve this completely.
Solution:
Putting x = 2,

= (x – 1) (-15x + 30 – 5x² + 10x)
= (x – 1) (-5x² – 5x + 30)
= -5(x – 1) (x² + x – 6)
= -5(x – 1) (x + 3) (x – 2) = 0
⇒ x = 1 or, -3 or 2.

Question 19.
Evaluate \(\left|\begin{array}{ccc}
1 & a & b c \\
1 & b & c a \\
1 & c & a b
\end{array}\right|\) – \(\left|\begin{array}{lll}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2
\end{array}\right|\)
Solution:

= (a – b) (b – c) [(-a + c) – (b + c – a – b)]
= (a – b) (b – c) (-a + c – c + a) = 0

Question 20.
\(\left|\begin{array}{lll}
a & a^2-b c & 1 \\
b & b^2-a c & 1 \\
c & c^2-a b & 1
\end{array}\right|\)
Solution:

Question21.
For what value of X the system of equations
x + y + z = 6, 4x + λy – λz = 0, 3x + 2y – 4z = -5 does not possess a solution?
Solution:

= 24 – 6λ – 2λ = 24 – 8λ
when Δ = 0
We have 24 – 8λ, = 0 or, λ = 3
The system of equations does not posses solution for λ = 3.

Question 22.
If A is a 3 × 3 matrix and |A| = 2, then which matrix is represented by A × adj A?
Solution:

Question 23.
If A = \(\left[\begin{array}{cc}
0 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 0
\end{array}\right]\)
show that (I + A) (I – A)^-1 = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) where I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:

Question 24.
Prove the following:
(i) \(\left|\begin{array}{ccc}
a^2+1 & a b & a c \\
a b & b^2+1 & b c \\
a c & b c & c^2+1
\end{array}\right|\) = 1 + a² + b² + c²
Solution:

(ii) \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
a^3 & b^3 & c^3
\end{array}\right|\) = (b – c) (c – a) (a – b) (a + b + c)
Solution:

= (a – b) (b – c) (b² + bc + c² – a² – ab – b²)
= (a – b) (b- c) (c² – a² + bc – ab)
= (a – b) (b – c) {(c – a) (c + a) + b(c – a)}
= (a – b) (b – c) (c – a) (a + b + c) = R.H.S.
(Proved)

(iii) \(\left|\begin{array}{lll}
\boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} \\
\boldsymbol{b} & \boldsymbol{c} & \boldsymbol{a} \\
\boldsymbol{c} & \boldsymbol{a} & \boldsymbol{b}
\end{array}\right|\) = 3abc – a³ – b³ – c³
Solution:

= (a + b + c) {(b – c) (a – b) – (c – a)²}
= (a + b + c) (a + b + c) (ab – b² – ca + bc – c² – a² + 2ca)
= (a + b + c) (-a² – b² – c² + ab + bc + ca)
= -(a + b + c) (a² + b² + c² – ab – bc – ca)
=- (a³ + b³ + c³ – 3abc)
= 3abc – a³ – b³ – c³

(iv) \(\left|\begin{array}{lll}
b^2-a b & b-c & b c-a c \\
a b-a^2 & a-b & b^2-a b \\
b c-a c & c-a & a b-a^2
\end{array}\right|\) = 0
Solution:

= (b² – a² + bc – ac) (a – b) {(-a + b) (c – a) – (bc – ac – ab + a²)}
= (b² – a² + bc – ac) (a – b) (- ca + a² + bc – ab – bc + ac + ab – a²)
= (b² – a² + bc – ac) (a – b) × 0 = 0
= R.H.S.
(Proved)

(v) \(\left|\begin{array}{ccc}
-a^2 & a b & a c \\
a b & -b^2 & b c \\
a c & b c & -c^2
\end{array}\right|\) = 4a²b²c²
Solution:

(vi) \(\left|\begin{array}{lll}
(b+c)^2 & a^2 & b c \\
(c+a)^2 & b^2 & c a \\
(a+b)^2 & c^2 & a b
\end{array}\right|\) = (a² + b² + c² ) (a + b + c) (b – c) (c – a) (a – b)
Solution:

= (a – b) (b – c) (a² + b² + c²) (-a² – ab + bc + c²)
= (a – b) (b – c) (a² + b² + c²) {(c² – a²) + b(c – a)}
= (a² + b² + c²) (a – b) (b – c) (c – a) (c + a + b)

(vii) \(\left|\begin{array}{lll}
b+c & a+b & a \\
c+a & b+c & b \\
a+b & c+a & c
\end{array}\right|\) = a³ + b³ + c³ – 3abc
Solution:

= (a + b +c) {(a – b) (a – c) – (c – b) (b – c)}
= (a + b + c) (a² – ac – ab + bc – bc + c² + b² – bc)
= (a + b + c) (a² + b² + c² – ab – bc – ca)
= (a³ + b³ + c³ – 3abc)

(viii) \(\left|\begin{array}{ccc}
a+b+c & -c & -b \\
-c & a+b+c & -a \\
-b & -a & a+b+c
\end{array}\right|\) = 2(b + c) (c + a) (a + b)
Solution:

= -2(a + b) (b + c) (-a – b – c + b)
= 2(a + b) (b + c) (c + a)

(ix) \(\left|\begin{array}{ccc}
a x-b y-c z & a y+b x & a z+c x \\
b x+a y & b y-c z-a x & b z+c y \\
c x+a z & a y+b z & c z-a x-b y
\end{array}\right|\) = (a² + b² + c²) (ax + by + cz) (x² + y² + z²)
Solution:

Question 25.
If 2s = a + b + c show that \(\left|\begin{array}{ccc}
a^2 & (s-a)^2 & (s-a)^2 \\
(s-b)^2 & b^2 & (s-b)^2 \\
(s-c)^2 & (s-c)^2 & c^2
\end{array}\right|\) = 2s³ (s – a) (s – b) (s – c)
Solution:
Let s – a = A, s – b = B, s – c = C
A + B + C = 3s – (a + b + c)
= 3s – 2s = s
Also B + C = s – b + s – c = 2s – (b + c)
= (a + b + c) – b + c = a
Similarly C + A = b, A + B = c

= 2 ABC (A + B + C)²
[Refer Q.No.9 (xii) of Exercise 5(a)]
= 2(s – a) (s – b)(s – c) s³

Question 26.
if \(\left|\begin{array}{ccc}
x & x^2 & x^3-1 \\
y & y^2 & y^3-1 \\
z & z^2 & z^3-1
\end{array}\right|\) = 0 then prove that xyz =1 when x, y, z are non zero and unequal.
Solution:

= (x – y) (y – z) (z – x) (xyz – 1)
It is given that
(x – y) (y – z) (z – x) (xyz – 1) = 0
⇒ xyz – 1 (as x ≠ y ≠ z)

Question 27.
Without expanding show that the following determinant is equal to Ax + B where A and B are determinants of order 3 not involving x.
\(\left|\begin{array}{ccc}
x^2+x & x+1 & x-2 \\
2 x^2+3 x-1 & 3 x & 3 x-3 \\
x^2+2 x+3 & 2 x-1 & 2 x-1
\end{array}\right|\)
Solution:

Question 28.
If x, y, z are positive and are the pth, qth and rth terms of a G.P. then prove that \(\left|\begin{array}{lll}
\log x & p & 1 \\
\log y & q & 1 \\
\log z & r & 1
\end{array}\right|\) = 0
Solution:
Let the G.P. be
a, aR, aR², aR³ …..aR^n-1
p th term = aR^p-1
q th term = aR^q-1
r th term = aR^r-1
x = aR^p-1, y= aR^q-1, z = aR^r-1
log x = log a + (p – 1) log R,
log y = log a + (q – 1) log R,
log z = log a + (r – 1) log R

Question 29.
If D_j = \(\left|\begin{array}{ccc}
j & a & n(n+2) / 2 \\
j^2 & b & n(n+1)(2 n+1) / 6 \\
j^3 & c & n^2(n+1)^2 / 4
\end{array}\right|\) then prove that \(\sum_{j=1}^n\)D_j = 0.
Solution:

Question 30.
Ifa₁, a₂,……a_n are in G.P. and a_i > 0 for every i, then find the value of
\(\left|\begin{array}{ccc}
\log a_n & \log a_{n+1} & \log a_{n+2} \\
\log a_{n+1} & \log a_{n+2} & \log a_{n+3} \\
\log a_{n+2} & \log a_{n+3} & \log a_{n+4}
\end{array}\right|\)
Solution:

Question 31.
If f(x)= \(\left|\begin{array}{ccc}
1+\sin ^2 x & \cos ^2 x & 4 \sin ^2 x \\
\sin ^2 x & 1+\cos ^2 x & 4 \sin 2 x \\
\sin ^2 x & \cos ^2 x & 1+4 \sin ^2 x
\end{array}\right|\) what is the least value of f(x)?
Solution:

As minimum value of sin 2x is 0. So the minimum value of above function f(x) is 2.

Question 32.
If f_r(x), g_r(x), h_r(x), r = 1, 2, 3 are polynomials in x such that f_r(a) = g_r(a) = h_r(a) and
F(x) = \(\left[\begin{array}{lll}
f_1(x) & f_2(x) & f_3(x) \\
g_1(x) & g_2(x) & g_3(x) \\
h_1(x) & h_2(x) & h_3(x)
\end{array}\right]\) find F'(x) at x = a.
Solution:
We have

[Since f₁a) = g₁(a) = h₁(a), f₂(a) = g₂(a) = h₂(a) and f₃(a) = g₃(a) = h₃(a) So that each determinant is zero due to presence of two identical rows.]

Question 33.
If f(x) = \(\left[\begin{array}{ccc}
\cos x & \sin x & \cos x \\
\cos 2 x & \sin 2 x & 2 \cos 2 x \\
\cos 3 x & \sin 3 x & 3 \cos 3 x
\end{array}\right]\) find f'(\(\frac{\pi}{2}\)).
Solution:

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 4 Matrices Ex 4(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Exercise 4(a)

Question 1.
State the order of the following matrices.
(i) [abc]
(ii) \(\left[\begin{array}{l}
1 \\
2
\end{array}\right]\)
(iii) \(\left[\begin{array}{ll}
x & y \\
y & z \\
z & x
\end{array}\right]\)
(iv) \(\left[\begin{array}{cccc}
1 & 0 & 1 & 4 \\
2 & 1 & 3 & 0 \\
-3 & 2 & 1 & 3
\end{array}\right]\)
Solution:
(i) (1 x 3)
(ii) (2 x 1)
(iii) (3 x 2)
(iv) (3 x 4)

Question 2.
How many entries are there in a
(i) 3 x 3 matrix
(ii) 3 x 4 matrix
(iii) p x q matrix
(iv) a sqare matrix of order p?
Solution:
(i) 9
(ii) 12
(iii) pq
(iv) p²

Question 3.
Give an example of
(i) 3 x 1 matrix
(ii) 2 x 2 matrix
(iii) 4 x 2 matrix
(iv) 1 x 3 matrix
Solution:
(i) \(\left(\begin{array}{l}
a \\
b \\
c
\end{array}\right)\)
(ii) \(\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)\)
(iii) \(\left(\begin{array}{ll}
a & b \\
c & d \\
e & f \\
g & h
\end{array}\right)\)
(iv) (1, 2, 3)

Question 4.
Let A = \(\left[\begin{array}{lllll}
1 & 2 & 3 & 4 & 1 \\
4 & 5 & 6 & 1 & 2 \\
3 & 9 & 1 & 1 & 6
\end{array}\right]\)
(i) What is the order of A?
(ii) Write down the entries a₃₁, a₂₅, a₂₃
(iii) Write down A^T.
(iv) What is the order of A^T?
Solution:
A = \(\left[\begin{array}{lllll}
1 & 2 & 3 & 4 & 1 \\
4 & 5 & 6 & 1 & 2 \\
3 & 9 & 1 & 1 & 6
\end{array}\right]\)
(i) Order of A is (3 x 5)
(ii) a₃₁ = 3, a₂₅= 2, a₂₃ = 6
(iii) A^T = \(\left[\begin{array}{lll}
1 & 4 & 3 \\
2 & 5 & 9 \\
3 & 6 & 1 \\
4 & 1 & 1 \\
1 & 2 & 6
\end{array}\right]\)
(iv) Order of A^T is (5 x 3).

Question 5.
Matrices A and B are given below. Find A + B, B + A, A – B and B – A. Verify that A + B = B + A and B – A = -(A – B)
(i) A = \(\left[\begin{array}{l}
7 \\
1
\end{array}\right]\), B = \(\left[\begin{array}{c}
-6 \\
9
\end{array}\right]\)
Solution:

(ii) A = \(\left[\begin{array}{cc}
1 & 2 \\
3 & -1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
4 & 1 \\
-3 & 2
\end{array}\right]\)
Solution:

(iii) A = \(\left[\begin{array}{ll}
\frac{1}{2} & \frac{1}{4} \\
\frac{1}{3} & \frac{1}{5}
\end{array}\right]\), B = \(\left[\begin{array}{ll}
\frac{1}{3} & \frac{1}{2} \\
\frac{1}{2} & \frac{4}{5}
\end{array}\right]\)
Solution:

(iv) A = \(\left[\begin{array}{cc}
1 & a-b \\
a+b & -3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & b \\
-a & 5
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{rrr}
1 & -2 & 5 \\
-1 & 4 & 3 \\
1 & 2 & -3
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
-1 & 2 & -5 \\
1 & -3 & -3 \\
1 & 2 & 4
\end{array}\right]\)
Solution:

Question 6.
(i) Find the 2×2 matrix X
if X + \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\) = \(\left[\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right]\)
Solution:

(ii) Given
[x y z] – [-4 3 1] = [-5 1 0] derermine x, y, z.
Solution:
[x y z] – [-4 3 1] = [-5 1 0]
∴ (x y z) = (-4 3 1) + (-5 1 0) = (-9 4 1)
∴ x = -9, y = 4, z = 1

(iii) If \(\left[\begin{array}{ll}
x_1 & x_2 \\
y_1 & y_2
\end{array}\right]\) – \(\left[\begin{array}{ll}
2 & 3 \\
0 & 1
\end{array}\right]\) = \(\left[\begin{array}{ll}
3 & 5 \\
1 & 2
\end{array}\right]\) determine x₁, x₂, y₁, y₂.
Solution:

(iv) Find a matrix which when added to \(\left[\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right]\) gives \(\left[\begin{array}{ll}
4 & 1 \\
3 & 2
\end{array}\right]\)
Solution:

Question 7.
Calculate whenever possible, the following products.
(i) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{l}
2 \\
3
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
4 & 3
\end{array}\right]\)
Solution:
\(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
4 & 3
\end{array}\right]\) is impossible because number of columns of 1st ≠ number of rows of second.

(iii) \(\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\left[\begin{array}{ll}
3 & 1 \\
1 & 1
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{cc}
1 & -2 \\
-2 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4
\end{array}\right]\)
Solution:

Question 8.
If A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]\), C = \(\left[\begin{array}{ll}
2 & 2 \\
1 & 3
\end{array}\right]\)
Calculate (i) AB (ii) BA (iii) BC (iv) CB (v) AC (vi) CA
Solution:

Question 9.
Find the following products.
(i) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 3 \\
1 & 4
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{ll}
1 & 3 \\
1 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{cc}
1 & i \\
i & -1
\end{array}\right]^2\) where i = √-1
Solution:

(vi) \(\left[\begin{array}{ll}
\mathbf{0} & \mathbf{1} \\
\mathbf{1} & \mathbf{0}
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Solution:

(vii) \(\left[\begin{array}{ll}
0 & k \\
1 & 0
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Solution:

(viii) \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\)
Solution:

(ix) \(\left[\begin{array}{ll}
1 & 0 \\
0 & k
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Solution:

(x) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)
Solution:
\(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{array}\right]\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\) = \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)

Question 10.
Write true or false in the following cases:
(i) The sum of a 3 x 4 matrix with a 3 x 4 matrix is a 3 x 3 matrix.
Solution:
False

(ii) k[0] = 0, k ∈ R
Solution:
False

(iii) A – B = B – A, if one of A and B is zero and A and B are of the same order.
Solution:
False

(iv) A + B = B + A, if A and B are matrices of the same order.
Solution:
True

(v) \(\left[\begin{array}{cc}
1 & 0 \\
-2 & 0
\end{array}\right]\) + \(\left[\begin{array}{cc}
-1 & 0 \\
2 & 0
\end{array}\right]\) = 0
Solution:
True

(vi) \(\left[\begin{array}{ll}
3 & 1 \\
6 & 2
\end{array}\right]\) = 3 \(\left[\begin{array}{ll}
1 & 1 \\
2 & 2
\end{array}\right]\)
Solution:
False

(vii) With five elements a matrix can not be constructed.
Solution:
False

(viii)The unit matrix is its own transpose.
Solution:
True

Question 11.
If A = \(\left[\begin{array}{cc}
2 & 4 \\
3 & 13
\end{array}\right]\) and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) find A – α I, α ∈ R.
Solution:

Question 12.
Find x and y in the following.
(i) \(\left[\begin{array}{cc}
x & -2 y \\
0 & -2
\end{array}\right]=\left[\begin{array}{cc}
1 & -8 \\
0 & -2
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{c}
x+3 \\
2-y
\end{array}\right]=\left[\begin{array}{c}
1 \\
-3
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{c}
2 x-y \\
x+y
\end{array}\right]=\left[\begin{array}{c}
3 \\
-9
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{l}
x \\
y
\end{array}\right]+\left[\begin{array}{l}
3 \\
4
\end{array}\right]=\left[\begin{array}{c}
2 \\
-1
\end{array}\right]\)
Solution:

(v) [2x -y] + [y 3x] = 5 [1 0]
Solution:

Question 13.
The element of ith row and ith column of the following matrix is i +j. Complete the matrix.

Question 14.
Write down the matrix

Question 15.
Construct a 2 x 3 matrix having elements given by
(i) a_ij = i + j
(ii) a_ij = i – j
(iii) a_ij = i × j
(iv) a_ij = i / j
Solution:

Question 16.
If \(\left[\begin{array}{cc}
2 x & y \\
1 & 3
\end{array}\right]+\left[\begin{array}{cc}
4 & 2 \\
0 & -1
\end{array}\right]=\left[\begin{array}{ll}
8 & 3 \\
1 & 2
\end{array}\right]\)
Solution:

Question 17.
Find A such that
\(\left[\begin{array}{ccc}
2 & 3 & 4 \\
1 & 0 & -2 \\
3 & 1 & -1
\end{array}\right]+A=\left[\begin{array}{ccc}
1 & 2 & -1 \\
2 & -1 & 0 \\
1 & 3 & 2
\end{array}\right]\)
Solution:

Question 18.
If

Question 19.
What is the order of the matrix B if [3 4 2] B = [2 1 0 3 6]
Solution:
(3 4 2) B = (2 1 0 3 6)
Let A = (3 4 2), C = (2 1 0 3 6)
∴ Order of A = (1 x 3)
Order of C = (1 x 5)
∴ Order of B = (3 x 5)

Question 20.
Find A if \(\left[\begin{array}{l}
4 \\
1 \\
3
\end{array}\right]\) A = \(\left[\begin{array}{rrr}
-4 & 8 & 4 \\
-1 & 2 & 1 \\
-3 & 6 & 3
\end{array}\right]\)
Solution:

Question 21.
Find B if B² = \(\left[\begin{array}{cc}
17 & 8 \\
8 & 17
\end{array}\right]\)
Solution:

∴ a² + bc = 17, ab + bd= 8
ca + cd = 8, bc + d² = 17
∴ a² + bc = bc + d²
or, a² + d² or, a = d
or, ca + cd = ab + bd
or, cd + cd – bd + bd
or, 2cd = 2bd = 8
or, b = c and bd = 4 = cd
∴ ab + bd= 8
or, ab + 4 = 8
or, ab = 4
Again, a² + bc = 17
or, a² + b . b = 17 (∵ b = c)
or, a² + b² = 17
Also (a + b)² = a² + b² + 2ab
∴ (a + b)² = 17 + 8 = 25
or, a + b = 5
And (a – b)² = 17 – 8 = 9
or, a – b = 3
∴ a = 4, b = 1, So d = 4, c = 1
∴ B = \(\left[\begin{array}{ll}
4 & 1 \\
1 & 4
\end{array}\right]\)

Question 22.
Find x and y when

Question 23.
Find AB and BA given that:

Question 24.
Evaluate

Question 25.
If

Show that AB = AC though B ≠ C. Verify that
(i) A + (B + C) = (A + B) + C
(ii) A(B + C) = AB + AC
(iii) A(BC) = (AB)C
Solution:

Question 26.
Find A and B where

Question 27.
If A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\) and I be the 2 × 2 unit matrix find (A – 2I) (A – 3I)
Solution:

Question 28.
Verify that [AB]^T = B^TA^T where

Question 29.
Verify that A = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\) satisfies the equation x² – (a + d)x + (ad – bc)I = 0 where I is the 2 x 2 matrix.
Solution:

Question 30.
If A = \(\left[\begin{array}{rrr}
1 & 2 & 3 \\
3 & -2 & 1 \\
4 & 2 & 1
\end{array}\right]\), show that A³ – 23 A – 40 I = 0
Solution:

Question 31.

Question 32.
If A and B are matrices of the same order and AB = BA, then prove that
(i) A² – B² = (A – B) (A + B)
(ii) A² + 2AB + B² = (A + B)²
(iii) A² – 2AB + B² = (A – B)²
Solution:
(i) (A – B) (A + B)
= A² + AB – BA – B²
= A² + AB – AB- B²(∵ AB = BA)
= A² – B²
(ii) (A + B)² = (A + B) (A + B)
= A² + AB + BA + B²
= A² + AB + AB + B² (∵ AB = BA)
= A² + 2AB + B²
(iii) (A – B)² = (A – B) (A – B)
= A² – AB – BA + B²
= A² – AB – AB + B² (∵ AB = BA)
= A² – 2AB + B²

Question 33.
If α and β are scalars and A is a square matrix then prove that
(A – αI) . (A – βI) = A² – (α + β) A + αβI, where I is a unit matrix of same order as A.
Solution:
(A – αI) (A – βI)
= A² – AβI – αIA + αβI²
= A² – βAI – αA + αβI
(∵ IA = A, I² = I)
= A² – βA – αA + αβI) (∵ AI = A)
= A² – (α + β) A + αβI

Question 34.
If α and β are scalars such that A = αβ + βI, where A, B and the unit matrix I are of the same order, then prove that AB = BA.
Solution:
We have A = αβ + βI
AB (αβ + βI) B
= α βB + βI B
= α βB + βB = (α + I) βB
= βB (α + 1)
(∵ Scalar mltiβlication is associative)
= Bβ (α + 1)
= Bβα + Bβ = Bαβ + BIβ
(∵ BI = B)
= B (αβ + βi) = BA
AB = BA
(proved)

Question 35.

Question 36.

Question 37.

Question 38.

Question 39.

Question 40.

Question 41.

Question 42.

Question 43.

	Men	Women	Children
Family A →	4	6	2
Family B →	2	2	4

	Family B
	Calory	Proteins
Men	2400	45
Women	1900	55
Children	1800	33

Solution:
The given informations can be written in matrix form as

∴ Calory requirements for families A and B are 24600 and 15800 respectively and protein requirements are 576 gm and 332 gm respectively.

Question 44.
Let the investment in first fund = ₹x and in the second fund is ₹(50000-x)
Investment matrix A=[x  50000-x]

⇒ 300000 – x = 278000
⇒ x = 22000
∴ He invests ₹22000 in first bond and ₹28000 in the second bond.

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Exercise 3(b)

Question 1.
Maximize Z = 5x₁+ 6x₂
Subject to: 2x₁ + 3x₂ ≤ 6
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraint as equation, we get 2x₁ + 3x₂ = 6
Step – 2 Let us draw the graph

x1	3	0
x2	0	0

Step – 3 Clearly (0,0) statisfies 2x₁ + 3x₂ ≤ 6
The shaded region is the feasible region with vertices 0(0,0), A(3,0), B(0,2).
Step – 4

Corner point	Z = 5x1+ 6x2
0(0.0)	0
A(3,0)	15 → maximum
B(0,2)	12

Z is maximum at A (3,0)
∴ The solution of LPP is x₁ = 3, x₂ = 0
Z_max = 15

Question 2.
Minimize: Z = 6x₁ + 7x₂
Subject to: x₁ + 2x₂ ≥ 4
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraint as equation we get x₁ + 2x₂ = 0
Step – 2 Let us draw the graph of x₁ + 2x₂ = 4

x1	0	4
x2	2	0

Step – 3 Clearly 0(0,0) does not satisfy
x₁ + 2x₂ > 4, x₁ > 0, x₂ > 0 is the first quadrant.
The feasible region is the shaded region with vertices A(4, 0), B(0, 2).
Step – 4 Z (4, 0) = 24
Z (0, 2) = 14 → minimum
Step – 5 As the feasible region is unbounded we cannot immediately decide Z is minimum at B (0, 2).
Let us draw the half-plane 6x₁ + 7x₂ < 14

x1	0	3.5
x2	2	-1

As this half-plane has no point common with the feasible region, we have Z is minimum for x₁= 0, x₂ = 2 and the minimum value of Z = 14.

Question 3.
Maximize Z = 20x₁+ 40x₂
Subject to: x₁ + x₂ ≤ 1
6x₁ + 2x₂ ≤ 3
x₁, x₂ ≥ 0.
Solution:
Step – 1 Treating the constraints as equations
x₁ + x₂ = 1    …. (1)
6x₁ + 2x₂ = 3   …. (2)
x₁, x₂ ≥ 0
Step – 2 Let us draw the graph:
Table – 1

x1	0	1
x2	1	0

Table – 2

x1	0	0.5
x2	1.5	0

Step – 3 As (0, 0) satisfies both the inequations the shaded region is the feasible region.
Step – 4 Solving
x₁ + x₂ = 1
6x₁ + 2x₂ = 3
we have x₁ = ¼ x₂ = ¾
The vertices are O(0, 0), A(0.5, 0), B(0,1) and C(¼, ¾)
Now Z(O) = 0
Z(A) = 10
Z(B) = 40
Z(C) = 20 × ¼ + 40 × ¾ = 35
∴ Z attains maximum at B for x₁= 0, x₂ = 1
Z_max = 40

Question 4.
Minimize: Z = 30x₁ + 45x₂
Subject to: 2x₁ + 6x₂ ≥ 4
5x₁ + 2x₂ ≥ 5
x₁, x₂ ≥ 0
Solution:
Step – 1 Consider the constraints as equations
2x₁ + 6x₂ = 4
5x₁ + 2x₂ = 5
Step – 2
Table – 1

x1	2	-1
x2	0	1

Table – 2

x1	1	0
x2	0	2.5

Step – 3 Clearly 0(0,0) does not satisfy 2x₁ + 6x₂ ≥ 4 and 5x₁ + 2x₂ ≥ 5.
Thus the shaded region is the feasible region.
Solving the equations we get
x₁ = \(\frac{11}{13}\), x₂ = \(\frac{5}{13}\).
∴ The vertices are A(2, 0)
B(\(\frac{11}{13}\), \(\frac{5}{13}\)) and C(0, \(\frac{5}{2}\)).
Step – 4 Z(A) = 60
Z(B) = \(\frac{555}{13}\) → minimum
Z(C) = \(\frac{225}{2}\)
Step – 5 As the feasible region is unbounded we cannot immediately decide Z is minimum at B(\(\frac{11}{13}\), \(\frac{5}{13}\))
Let us draw the half plane
30x₁ + 45x₂ < \(\frac{555}{13}\)

x1	\(\frac{11}{13}\)	0
x2	\(\frac{5}{13}\)	\(\frac{27}{39}\)

As this half plane and the feasible region has no point in common we have Z is minimum for x₁ = \(\frac{11}{13}\), x₂ = \(\frac{5}{13}\), and Z_min = \(\frac{555}{13}\)

Question 5.
Maximize: Z = 3x₁+ 2x₂
Subject to: -2x₁ + x₂ ≤ 1
x₁ ≤ 2
x₁+ x₂ ≤ 3
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
-2x₁ + x₂ = 1        …..(1)
x₁ = 2                   …..(2)
x₁+ x₂ = 3            …..(3)
Step – 2 Let us draw the lines.
Table – 1

x1	0	-1
x2	1	-1

Table – 2

x1	2	2
x2	0	1

Table – 3

x1	0	3
x2	3	0

Step – 3 (0, 0) satisfies all the constraints and x₁, x₂ > 0 is the 1st quadrant the shaded region is the feasible region.

Step – 4 Solving -2x₁ + x₂ = 1
x₁+ x₂ = 3
we have 3x₁ = 2
⇒ x₁ = \(\frac{2}{3}\), x₂ = 3 – \(\frac{2}{3}\) = \(\frac{7}{3}\)
From x₁+ x₂ = 3 and x₁ = 2 we have x₁ = 2, x₂ = 1
∴ The vertices are 0(0, 0), A(2, 0), B(2, 1), C(\(\frac{2}{3}\), \(\frac{7}{3}\)), D(0, 1)
Z(0) = 0, Z(A) = 6, Z(B) = 8, Z(C) = 3.\(\frac{2}{3}\) + 2.\(\frac{7}{3}\) = \(\frac{20}{3}\), Z(D) = 2
Z is maximum at B.
∴ The solution of given LPP is x₁ = 2, x₂ = 1, Z_(max) = 8.

Question 6.
Maximize: Z = 50x₁+ 60x₂
Subject to: x₁ + x₂ ≤ 5
x₁+ 2x₂ ≤ 4
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x₁ + x₂ = 5     ….(1)
x₁+ 2x₂ = 4    ….(2)
Step – 2 Let us draw the graph
Table – 1

x1	5	5
x2	0	0

Table – 2

x1	4	0
x2	0	2

Step – 3 0(0,0) satisfies x₁ + x₂ ≤ 5 and does not satisfy x₁+ 2x₂ ≤ 4
Thus the shaded region is the feasible region.
Step – 4 The corner points are A(4,0), B(5,0), C(0,5) , D(0,2)
∴

Corner point	z = 50x1+ 60x2
A(4,0)	200
B (5,0)	250 → maximum
C(0,5)	300
D(0,2)	120

Z is maximum for x₁ = 0, x₂ = 5, Z_(max) = 300.

Question 7.
Maximize: Z = 5x₁+ 7x₂
Subject to: x₁ + x₂ ≤ 4
5x₁+ 8x₂ ≤ 30
10x₁+ 7x₂ ≤ 35
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get,
x₁ + x₂ = 4           …. (1)
5x₁+ 8x₂ = 30      …. (2)
10x₁+ 7x₂ = 35    …. (3)
Step – 2 Let us draw the graph
Table – 1

x1	4	0
x2	0	4

Table – 2

x1	6	2
x2	0	2.5

Table – 3

x1	0	3.5
x2	5	0

Step – 3 0(0,0) satisfies all the constraints.
Thus the shaded region is the feasible region.
From (1) and (2) we get (\(\frac{2}{3}\), \(\frac{10}{3}\))
From (1) and (3) we get
x₁ = \(\frac{7}{3}\), x₁ = \(\frac{5}{3}\)
∴ The corner points are 0(0,0), A(\(\frac{7}{2}\), 0), B(\(\frac{7}{3}\), \(\frac{5}{3}\)), C(\(\frac{2}{3}\), \(\frac{10}{3}\)), D(0, \(\frac{15}{4}\))
Step – 4

Corner point	z = 5x1+ 7x2
0(0,0)	0
A(\(\frac{7}{2}\), 0)	\(\frac{35}{2}\)
B(\(\frac{7}{3}\), \(\frac{5}{3}\))	\(\frac{70}{3}\)
C(\(\frac{2}{3}\), \(\frac{10}{3}\))	\(\frac{80}{3}\)
D(0, \(\frac{15}{4}\))	\(\frac{105}{4}\)

Z attains its maximum value \(\frac{80}{3}\) for x₁ = \(\frac{2}{3}\) and x₂ = \(\frac{10}{3}\).

Question 8.
Maximize: Z = 14x₁ – 4x₂
Subject to: x₁ + 12x₂ ≤ 65
7x₁ – 2x₂ ≤ 25
2x₁+ 3x₂ ≤ 10
x₁, x₂ ≥ 0
Also find two other points which maximize Z.
Solution:
Step – 1 Treating the constraints as equations we get
x₁ + 12x₂ = 65   …. (1)
7x₁ – 2x₂ = 25    …. (2)
2x₁ + 3x₂ = 10   …. (3)
Step – 2 Let us draw the graph
Table – 1

x1	65	5
x2	0	5

Table – 2

x1	5	10
x2	5	22.5

Table – 3

x1	5	2
x2	0	2

Step – 3 Clearly 0(0,0) satisfies x₁ + 12x₂ ≤ 65 and 7x₁ – 2x₂ ≤ 25 but does not satisfy 2x₁+ 3x₂ ≤ 10. Thus shaded region is the feasible region.
Equation (1) and (2) meet at (5, 5).
From (2) and (3)

∴ The corner points of the feasible region are A(0, \(\frac{10}{3}\)), B(\(\frac{19}{5}\), \(\frac{4}{5}\)), C(5, 5), D(0, \(\frac{65}{12}\)).
Step – 4

Corner point	z = 14x1 – 4x2
A(0, \(\frac{10}{3}\))	\(\frac{-40}{3}\)
B(\(\frac{19}{5}\), \(\frac{4}{5}\))	50 → maximum
C(5, 5)	50 → maximum
D(0, \(\frac{65}{12}\))	\(\frac{65}{3}\)

Z is maximum for x₁ = \(\frac{19}{5}\), x₂ = \(\frac{4}{5}\) or x₁ = 5, x₂ = 5 and Z_max = 50
There is no other point that maximizes Z.

Question 9.
Maximize: Z = 10x₁ + 12x₂ + 8x₃
Subject to: x₁ + 2x₂ ≤ 30
5x₁ – 7x₃ ≤ 12
x₁ + x₂ + x₃ = 20
x₁, x₂ ≥ 0
[Hints: Eliminate x₃ from all expressions using the given equation in the set of constraints, so that it becomes an LPP in two variables]
Solution:
Eliminating x₃ this LPP can be written as Maximize Z = 2x₁ + 4x₂ + 160
Subject to: x₁ + 2x₂ ≤ 30
5x₁ – 7x₃ ≤ 12
x₁, x₂ ≥ 0
Step – 1 Treating the consraints as equations we get
x₁ + 2x₂ = 30    …..(1)
5x₁ – 7x₃ = 12   …..(2)
Step – 2 Let us draw the graph
Table – 1

x1	30	0
x2	0	15

Table – 2

x1	8	1
x2	8	20

Step – 3 Clearly 0(0,0) satisfies x₁ + 2x₂ ≤ 30 and does not satisfy 12x₁ + 7x₂ ≤ 152
∴ The shaded region is the feasible region.

Step – 4

Z is maximum for x₁ = 30, x₂ = 0 and Z_max = 220

Question 10.
Maximize: Z = 20x₁ + 10x₂
Subject to: x₁ + 2x₂ ≤ 40
3x₁ + x₂ ≥ 30
4x₁+ 3x₂ ≥ 60
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equalities we have:
x₁ + 2x₂ = 40   ….(1)
3x₁ + x₂ = 30   ….(2)
4x₁+ 3x₂ = 60  ….(3)
Step – 2 Let us draw the graph

Step – 3 (0, 0) satisfies x₁ + 2x₂ ≤ 40 and does not satisfy 3x₁ + x₂ ≥ 30 and 4x₁+ 3x₂ ≥ 60, x₁, x₂ ≥ 0 is the first quadrant.
∴ The shaded region is the feasible region.
Step – 4 x₁ + 2x₂ = 40 and 3x₁ + x₂ = 30

∴ The vetices are A(15, 0), B(10, 0), C(4, 18) and D(6, 12)
Z(A) = 300, Z(B) = 800
Z (C) = 20 x 4 + 10 x 18 = 260
Z (D) = 120 + 120 = 240
Z attains minimum at D(6 ,12).
∴ The required solution x₁ = 6, x₂ =12 and Z_min = 240

Question 11.
Maximize: Z = 4x₁ + 3x₂
Subject to: x₁ + x₂ ≤ 50
x₁ + 2x₂ ≥ 80
2x₁+ x₂ ≥ 20
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
x₁ + x₂ ≤ 50    ….(1)
x₁ + 2x₂ ≥ 80  ….(2)
2x₁+ x₂ ≥ 20   ….(3)
Step – 2 Let us draw the graph

Step – 3 (0, 0) satisfies x₁ + x₂ < 50, x₁ + 2x₂ < 80 but does not satisfy
2x₁ + x₂ > 20, x₁ > 0, x₂ > 0 is the 1st quadrant.
Hence the shaded region is the feasible region.
Step – 4 x₁ + x₂ = 50
x₁ + 2x₂ = 80
=> x₂ = 30, x₁ = 20
The vertices of feasible region are
A(10, 0), B(50, 0), C(20, 30), D (0, 40) and E (0, 20)

Point	Z = 4x1 + 3x2
A(10,0)	40
5(50,0)	200
C(20,30)	170
D(0,40)	120
E(0,120)	60

Question 12.
Optimize: Z = 5x₁ + 25x₂
Subject to: -0.5x₁ + x₂ ≤ 2
x₁ + x₂ ≥ 2
-x₁+ 5x₂ ≥ 5
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
-0.5x₁ + x₂ = 2   ….(1)
x₁ + x₂ = 2         ….(2)
-x₁+ 5x₂ = 5      ….(3)
Step – 2 Let us draw the graph.

Step – 3 (0, 0) satisfies -0.5x₁ + x₂ ≤ 2, but does not satisfy x₁ + x₂ ≥ 2 and -x₁+ 5x₂ ≥ 5, x₁ > 0, x₂ > 0 is the 1st quadrant.
The shaded region is the feasible region with vertices A(\(\frac{5}{6}\), \(\frac{7}{6}\)) and B(0, 2).
Step – 4 Z can be made arbitrarily large.
∴ Problem has no maximum.
But Z(A) = \(\frac{100}{3}\), Z(B) = 50
Z is minimum at A(\(\frac{5}{6}\), \(\frac{7}{6}\)).
But the feasible region is unbounded.
Hence we cannot immediately decide, Z is minimum at A.
Let us draw the half plane
5x₁ + 25x₂ < \(\frac{100}{3}\)
⇒ 3x₁ + 15x₂ < 20
As there is no point common to this half plane and the feasible region.
we have Z is minimum for x₁ = \(\frac{5}{6}\), x₂ = \(\frac{7}{6}\) and the minimum value = \(\frac{100}{3}\)

Question 13.
Optimize: Z = 5x₁ + 2x₂
Subject to: -0.5x₁ + x₂ ≤ 2
x₁ + x₂ ≥ 2
-x₁+ 5x₂ ≥ 5
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
-0.5x₁ + x₂ = 2   ….(1)
x₁ + x₂ = 2         ….(2)
-x₁+ 5x₂ = 5      ….(3)
Step – 2 Let us draw the graph.

Step – 3 The shaded regian is feasible region which is unbounded, thus Z does not have any maximum.

As Z can be made arbitrarily large, the given LPP has no maximum.
Z is minimum at B (0, 2). But we cannot immediately decide, Z is minimum at B.
Let us draw the half plane 5x₁ + 2x₂ < 4

x1	0	4/5
x2	2	0

As there is no point common to this half plane and the feasible region,
we have Z is minimum for x₁ = 0, x₂ = 2 and the minimum value of Z = 4.

Question 14.
Optimize: Z = -10x₁ + 2x₂
Subject to: -x₁ + x₂ ≥ -1
x₁ + x₂ ≤ 6
x₂ ≤ 5
x₁, x₂ ≥ 0
Solution:
Step – 1 Treating the constraints as equations
-x₁ + x₂ = -1     ….(1)
x₁ + x₂ = 6        ….(2)
x₂ = 5                ….(3)
Step – 2 Let us draw the graph
Table – 1

x1	1	0
x2	0	-1

Table – 2

x1	6	0
x2	0	1

Step – 3 Clearly 0(0,0) satisfies all the constraints.
Thus the shaded region is the feasible region.
The vertices are 0(0,0) , A(1,0), B(\(\frac{7}{2}\), \(\frac{5}{2}\)) ,C(1, 5) and D (0, 5)
Step – 4 Z(O) = 0
Z(A) = -10
Z(B) = – 30
Z(C) = 0
Z(D) = 10
∴ Z is maximum for x₁= 0, x, = 5 and Z_(max) = 10
Z is minimum for x₁ = \(\frac{7}{2}\)  x₂ = \(\frac{5}{2}\) and Z_(min) = -30

Question 15.
Solve the L.P.P.s obtained in Exercise 3(a) Q.1 to Q. 9 by graphical method.
(1) Maximise: Z = 1500x + 2000y
Subject to: x + y < 20
x + 2y < 24
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x + y = 20
x + 2y = 24
Step – 2 Let us draw of graph.

Step – 3 Clearly 0(0,0) satisfies all the constraints.
Thus the shaded region is the feasible region.
From (1) and (2) we get
y = 14
x = 16
With vertices 0(0, 0), A(20, 0), B(16, 4), C(0, 12).
Step – 4 Z(0) = 0
Z(A) = 30,000
Z(B) = 32,000 → Maximum
Z(C) = 24000
Z is maximum for x = 16, y = 4 with Z = 32000
To get maximum profit he must keep 16 sets of model X and 4 sets of model Y.
Maximum profit = 1500 × 16 + 2000 × 4 = ₹32,000

(2) Maximize: 15x + 10y
Subject: x + 3y ≤ 600
2x + y ≤ 480
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
2x +3y = 600
2a + y = 480
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints.
The corner point are 0(0, 0), A (240, 0) B(210, 60),C(0, 200)
Step – 4 Z(0) = 6
Z(A) = 3600
Z(B) = 3150 + 600
= 3750 → maximum
Z(C) = 2000
Thus Z is maximum for x = 210 and y = 60
and Z_(max) = 3750

(3) Maximize: Z = 20x + 30y
Subject to: x + 2y ≤ 10
x + y ≤ 6
x ≤ 4
x, y ≥ 0.
Solution:
Step – 1 Treating the constraints as equations we get
x + 2y = 10       …(1)
x + y = 6           …(2)
x = 4
Step – 2 Let us draw the graph

Step – 3 As 0(0,0) satisfies all the constraints the shaded region is the feasible region.
Solving (1) and (2) we get x = 2, y = 4.
The vertices and 0(0, 0) , A(4, 0), B(4, 2), C(2, 4), D (0, 5).
Step – 4 Z(0) =0
Z(A) = 80
Z (B) =140
Z(C) = 1 60 → maximum
Z (D) = 150
∴ Z is Maximum when x = 2, y = 4 and Z_(max) = 160

(4) Maximize: Z = 15x + 17y
Subject to: 4x + 7y ≤ 150
x + y ≤ 30
15x + 17y > 300
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
4x + 7y = 150      ….(1)
x + y = 30            ….(2)
15x + 17y = 300  ….(3)
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints.
4x + 7y ≤ 150, x + y ≤ 30, but does not satisfy 15x + 17y ≥ 300.
∴ The shaded region is the feasible region.
From (1) and (2) we get

∴ Z is maximum for x = 20. y = 10 and Z_(max) = 470

(5) Maximize: Z = 2x + 4y
Subject to: 3x + 2y ≤ 10
2x + 5y ≤ 15
5x + 6y ≤ 21
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
3x + 2y = 10  …(1)
2x + 5y = 15  …(2)
5x + 6y = 21  …(3)
Step – 2 Let us draw the graph

Step – 3 As 0(0,0) satisfies all the constraints the shaded region is the feasible region.
From (1) and (3) we get

From (2) and (3) we get

Step-4 Z(O) = 0

(6) Maximize: Z = 1000x + 800y
Subject to: x + y ≤ 5
2x + y ≤ 9
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x + y = 5    ….(1)
2x + y = 9  ….(2)
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints.
∴ Thus the shaded region is the feasible region.
From (1) and (2) we get x = 4, y = 1.
∴ The vertices are A(0, 0), A(4.5, 0), B(4, 1) and C(0, 5).
Step – 4 Z(0) =0
Z (A) = 4500
Z (B) = 4800 → Maximum
Z (C) = 4000
Z is maximum for x = 4 and y = 1, Z_(max) = 4800

(7) Minimize: Z = 4960 – 70x – 130y
Subject to: x + y ≤ 12
x + y ≥ 6
x ≤ 8
y ≤ 8
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x + y = 12   ….(1)
x + y = 6     ….(2)
x = 8           ….(3)
y = 4           ….(4)
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints except x + y > 6.
The shaded region is the feasible region.
The vertices are A(6, 0), B(8, 0), C(8, 4), D(4, 8), E(0, 8) and F(0, 6).
Step – 4 Z (A) = 4540
Z (B) = 4400
Z (C) = 3880
Z (D) = 3640 → Minimum
Z (E) = 3920
Z (F) = 4180
∴ Z is maximum for x = 4 and y = 8 and Z_(min) = 3640.

(8) Minimize: Z = 16x + 20y
Subject to x + 2y ≥ 10
x + y ≥ 6
3x + y ≥ 8
x, y ≥ 0
Solution:
Step – 1 Treating the constraints as equations we get
x + 2y = 10  ….(1)
x + y = 6      …(2)
3x + y = 8    …(3)
Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints. Thus the shaded region is the feasible region.
From (1) and (2) we get y = 4, x = 2.
From (2) and (3) we get x = 1, y = 5.
The vertices are A(10, 0), B(2, 4), C(1, 5), D(0, 8).
Step – 4 Z (A) = 160
Z (B) = 112 → Minimum
Z (C) =116
Z (D) = 160
As the region is unbounded, let us draw the half plane Z < Z_(min)
⇒ 16x + 20y < 112
⇒ 4x + 5y < 28

x1	7	0
x2	0	5.6

There is no point common to the shaded region and the half plane 4x + 5y ≤ 28 other than B(2, 4).
∴ Z is minimum for x = 2, y = 4 and Z_(min) = 112.

(9) Minimize: Z = (512.5)x + 800y
Subject to: 5x + 4y = 40
x ≤ 7
x ≤ 3
x, y ≥ 0
Solution:
Step – 1 Let us draw the graph of
5x + 4y = 40
x = 7, y = 3

x1	8	0
x2	0	10

Step – 1 Let us draw the graph of
5x + 4y = 40
x = 7, y = 3
Step – 2 The line segment AB is the feasible region.
Step – 3 Z (A) = 3587.5 + 1000 = 4587.5
Z (B) = 2870 + 2400 = 5270
Clearly Z is minimum for
x = 7, y = \(\frac{5}{4}\) and Z_(min) = 4587.5

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 2 Inverse Trigonometric Functions Ex 2 Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Exercise 2

Question 1.
Fill in the blanks choosing correct answer from the brackets:
(i) If A = tan^-1 x, then the value of sin 2A = ________. (\(\frac{2 x}{1-x^2}\), \(\frac{2 x}{\sqrt{1-x^2}}\), \(\frac{2 x}{1+x^2}\))
Solution:
\(\frac{2 x}{1+x^2}\)

(ii) If the value of sin^-1 x = \(\frac{\pi}{5}\) for some x ∈ (-1, 1) then the value of cos^-1 x is ________. (\(\frac{3 \pi}{10}\), \(\frac{5 \pi}{10}\),\(\frac{3 \pi}{10}\))
Solution:
\(\frac{3 \pi}{10}\)

(iii) The value of tan^-1 x (2cos\(\frac{\pi}{3}\)) is ________. (1, \(\frac{\pi}{4}\), \(\frac{\pi}{3}\))
Solution:
\(\frac{\pi}{4}\)

(iv) If x + y = 4, xy = 1, then tan^-1 x + tan^-1 y = ________. (\(\frac{3 \pi}{4}\), \(\frac{\pi}{4}\), \(\frac{\pi}{3}\))
Solution:
\(\frac{\pi}{2}\)

(v) The value of cot^-1 2 + tan^-1 \(\frac{1}{3}\) = ________. (\(\frac{\pi}{4}\), 1, \(\frac{\pi}{2}\))
Solution:
\(\frac{\pi}{4}\)

(vi) The principal value of sin^-1 (sin \(\frac{2 \pi}{3}\)) is ________. (\(\frac{2 \pi}{3}\), \(\frac{\pi}{3}\), \(\frac{4 \pi}{3}\))
Solution:
\(\frac{\pi}{3}\)

(vii) If sin^-1 \(\frac{x}{5}\) + cosec^-1 \(\frac{5}{4}\) = \(\frac{\pi}{2}\), then the value of x = ________. (2, 3, 4)
Solution:
x = 3

(viii) The value of sin (tan^-1 x + tan^-1 \(\frac{1}{x}\)), x > 0 = ________. (0, 1, 1/2)
Solution:
1

(ix) cot^-1 \(\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right]\) = ________. (2π – \(\frac{x}{2}\), \(\frac{x}{2}\), π – \(\frac{x}{2}\))
Solution:
π – \(\frac{x}{2}\)

(x) 2sin^-1 \(\frac{4}{5}\) + sin^-1 \(\frac{24}{25}\) = ________. (π, -π, 0)
Solution:
π

(xi) if Θ = cos^-1 x + sin^-1 x – tan^-1 x, x ≥ 0, then the smallest interval in which Θ lies is ________. [(\(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\)), [0, \(\frac{\pi}{2}\)), (0, \(\frac{\pi}{2}\)])
Solution:
(0, \(\frac{\pi}{2}\)]

(xii) sec² (tan^-1 2) + cosec² (cot^-1 3) = ________. (16, 14, 15)
Solution:
15

Question 2.
Write whether the following statements are true or false.
(i) sin^-1 \(\frac{1}{x}\) cosec^-1 x = 1
Solution:
False

(ii) cos^-1 \(\frac{4}{5}\) + tan^-1 \(\frac{2}{3}\) = tan^-1 \(\frac{17}{6}\)
Solution:
True

(iii) tan^-1 \(\frac{4}{3}\) + cot^-1 (\(\frac{-3}{4}\)) = π
Solution:
True

(iv) sec^-1 \(\frac{1}{2}\) + cosec^-1 \(\frac{1}{2}\) = \(\frac{\pi}{2}\)
Solution:
False

(v) sec^-1 (-\(\frac{7}{5}\)) = π – cos^-1 \(\frac{5}{7}\)
Solution:
True

(vi) tan^-1 (tan 3) = 3
Solution:
False

(vii) The principal value of tan^-1 (tan \(\frac{3 \pi}{4}\)) is \(\frac{3 \pi}{4}\)
Solution:
False

(viii) cot^-1 (-√3) is in the second quadrant.
Solution:
True

(ix) 3 tan^-1 3 = tan^-1 \(\frac{9}{13}\)
Solution:
False

(x) tan^-1 2 + tan^-1 3 = – \(\frac{\pi}{4}\)
Solution:
False

(xi) 2 sin^-1 \(\frac{4}{5}\) = sin^-1 \(\frac{24}{25}\)
Solution:
False

(xii) The equation tan^-1 (cotx) = 2x has exactly two real solutions.
Solution:
True

Question 3.
Express the value of the foilowing in simplest form.
(i) sin (2 sin^-1 0.6)
Solution:
sin (2 sin^-1 0.6)

(ii) tan (\(\frac{\pi}{4}\) + 2 cot^-1 3)
Solution:

(iii) cos (2 sin^-1 x)
Solution:

(iv) tan (cos^-1 x)
Solution:

(v) tan^-1 (\(\frac{x}{y}\)) – tan^-1 \(\frac{x-y}{x+y}\)
Solution:

(vi) cosec (cos^-1 \(\frac{3}{5}\) + cos^-1 \(\frac{4}{5}\))
Solution:

(vii) sin^-1 \(\frac{1}{\sqrt{5}}\) + cos^-1 \(\frac{3}{\sqrt{10}}\)
Solution:

(viii) sin cos^-1 tan sec √2
Solution:
sin cos^-1 tan sec √2
= sin cos^-1 tan sec \(\frac{\pi}{4}\)
= sin cos^-1 1 = sin 0 = 0

(ix) sin (2 tan^-1 \(\sqrt{\frac{1-x}{1+x}}\))
Solution:

(x) tan \(\left\{\frac{1}{2} \sin ^{-1} \frac{2 x}{1+x^2}+\frac{1}{2} \cos ^{-1} \frac{1-y^2}{1+y^2}\right\}\)
Solution:

(xi) sin cot^-1 cos tan^-1 x.
Solution:

(xii) tan^-1 \(\left(x+\sqrt{1+x^2}\right)\)
Solution:

Question 4.
Prove the following statements:
(i) sin^-1 \(\frac{3}{5}\) + sin^-1 \(\frac{8}{17}\) = cos^-1 \(\frac{36}{85}\)
Solution:

(ii) sin^-1 \(\frac{3}{5}\) + cos^-1 \(\frac{12}{13}\) = cos^-1 \(\frac{33}{65}\)
Solution:

(iii) tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{13}\) = tan^-1 \(\frac{2}{9}\)
Solution:
L.H.S = tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{13}\)

(iv) tan^-1 \(\frac{1}{2}\) + tan^-1 \(\frac{1}{5}\) + tan^-1 \(\frac{1}{8}\) = \(\frac{\pi}{4}\)
Solution:

(v) tan ( 2tan^-1 \(\frac{1}{5}\) – \(\frac{\pi}{4}\) ) + \(\frac{7}{17}\) = 0
Solution:

Question 5.
Prove the following statements:
(i) cot^-1 9 + cosec^-1 \(\frac{\sqrt{41}}{4}\) = \(\frac{\pi}{4}\)
Solution:

(ii) sin^-1 \(\frac{4}{5}\) + 2 tan^-1 \(\frac{1}{3}\) = \(\frac{\pi}{2}\)
Solution:

(iii) 4 tan^-1 \(\frac{1}{5}\) – tan^-1 \(\frac{1}{70}\) + tan^-1 \(\frac{1}{99}\) = \(\frac{\pi}{4}\)
Solution:

(iv) 2 tan^-1 \(\frac{1}{5}\) + sec^-1 \(\frac{5 \sqrt{2}}{7}\) + 2 tan^-1 \(\frac{1}{8}\) = \(\frac{\pi}{4}\)
Solution:

(v) cos^-1 \(\frac{12}{13}\) + 2 cos^-1 \(\sqrt{\frac{64}{65}}\) + cos^-1 \(\sqrt{\frac{49}{50}}\) = cos^-1 \(\frac{1}{\sqrt{2}}\)
Solution:


(vi) tan² cos^-1 \(\frac{1}{\sqrt{3}}\) + cot² sin^-1 \(\frac{1}{\sqrt{5}}\) = 6
Solution:
tan² cos^-1 \(\frac{1}{\sqrt{3}}\) + cot² sin^-1 \(\frac{1}{\sqrt{5}}\)
= tan² tan^-1 √2 + cot² cot^-1 (2)
= 2 + 4 = 6

(vii) cos tan^-1 cot sin^-1 x = x.
Solution.

Question 6.
Prove the following statements:
(i) cot^-1 (tan 2x) + cot^-1 (- tan 2x) = π
Solution:

(ii) tan^-1 x + cot^-1 (x + 1) = tan^-1 (x² + x + 1)
Solution:

(iii) tan^-1 (\(\frac{a-b}{1+a b}\)) + tan^-1 (\(\frac{b-c}{1+b c}\)) = tan^-1 a – tan^-1 c.
Solution:
tan^-1 (\(\frac{a-b}{1+a b}\)) + tan^-1 (\(\frac{b-c}{1+b c}\))
= tan^-1 a – tan^-1 b + tan^-1 b – tan^-1 c
= tan^-1 a – tan^-1 c.

(iv) cot^-1 \(\frac{p q+1}{p-q}\) + cot^-1 \(\frac{q r+1}{q-r}\) + cot^-1 \(\frac{r p+1}{r-p}\) = 0
Solution:

(v)

Solution:

Question 7.
Prove the following statements:
(i) tan^-1 \(\frac{2 a-b}{b \sqrt{3}}\) + tan^-1 \(\frac{2 b-a}{a \sqrt{3}}\) = \(\frac{\pi}{3}\)
Solution:

(ii) tan^-1 \(\frac{1}{x+y}\) + tan^-1 \(\frac{y}{x^2+x y+1}\) = tan^-1 \(\frac{1}{x}\)
Solution:

(iii) sin^-1 \(\sqrt{\frac{x-q}{p-q}}\) = cos^-1 \(\sqrt{\frac{p-x}{p-q}}\) = cot^-1 \(\sqrt{\frac{p-x}{x-q}}\)
Solution:

(iv) sin² (sin^-1 x + sin^-1 y + sin^-1 z) = cos² (cos^-1 x + cos^-1 y + cos^-1 z)
Solution:

(v) tan (tan^-1 x + tan^-1 y + tan^-1 z) = cot (cot^-1 x + cot^-1 y + cot^-1 z)
Solution:

Question 8.
(i) If sin^-1 x + sin^-1 y + sin^-1 z = π, show that x\(\sqrt{1-x^2}\) + x\(\sqrt{1-y^2}\) + x\(\sqrt{1-z^2}\) = 2xyz
Solution:
Let sin^-1 x = α, sin^-1 y = β, sin^-1 z = γ
∴ α + β + γ = π
∴ x = sin α, y = sin β, z = sin γ
or, α + β = π – γ
or, sin(α + β) = sin(π – γ) = sin γ
and cos(α + β) = cos(π – γ) = – cos γ

(ii) tan^-1 x + tan^-1 y + tan^-1 z = π show that x + y + z = xyz.
Solution:

(iii) tan^-1 x + tan^-1 y + tan^-1 z = \(\frac{\pi}{2}\). Show that xy + yz + zx = 1
Solution:

or, 1 – xy – yz – zx = 0
⇒ xy + yz + zx = 1

(iv) If r² = x² +y² + z², Prove that tan^-1 \(\frac{y z}{x r}\) + tan^-1 \(\frac{z x}{y r}\) + tan^-1 \(\frac{x y}{z r}\) = \(\frac{\pi}{2}\)
Solution:

(v) In a triangle ABC if m∠A = 90°, prove that tan^-1 \(\frac{b}{a+c}\) + tan^-1 \(\frac{c}{a+b}\) = \(\frac{\pi}{4}\). where a, b, and c are sides of the triangle.
Solution:
L.H.S. tan^-1 \(\frac{b}{a+c}\) + tan^-1 \(\frac{c}{a+b}\)

Question 9.
Solve
(i) cos (2 sin^-1 x) = 1/9
Solution:

(ii) sin^-1 x + sin^-1 (1 – x) = \(\frac{\pi}{2}\)
Solution:
sin^-1 x + sin^-1 (1 – x) = \(\frac{\pi}{2}\)
or, sin^-1 (1 – x) = \(\frac{\pi}{2}\) – sin^-1 x = cos^-1 x
or, sin^-1 (1 – x) = sin^-1 \(\sqrt{1-x^2}\)
or, 1 – x = \(\sqrt{1-x^2}\)
or, 1 + x² – 2x = 1 – x²
or, 2x² – 2x  = 0
or, 2x (x – 1) = 0
∴ x = 0 or, 1

(iii) sin^-1 (1 – x) – 2 sin^-1 x = \(\frac{\pi}{2}\)
Solution:
sin^-1 (1 – x) – 2 sin^-1 x = \(\frac{\pi}{2}\)
⇒ – 2 sin^-1 x = \(\frac{\pi}{2}\) – sin^-1 (1 – x)
⇒ cos^-1 (1 – x)
⇒ cos (– 2 sin^-1 x) = 1 – x      ….. (1)
Let sin^-1 Θ ⇒ sin Θ
Now cos (– 2 sin^-1 x) = cos (-2Θ)
= cos 2Θ = 1 – 2 sin² Θ = 1 – 2x²
Using in (1) we get
1 – 2x² = 1 – x
⇒ 2x² – x = 0 ⇒ x (2x – 1) = 0
⇒ x = 0, ½, But x = ½ does not
Satisfy the given equation, Thus x = 0.

(iv) cos^-1 x + sin^-1 \(\frac{x}{2}\) = \(\frac{\pi}{6}\)
Solution:

(v) tan^-1 \(\frac{x-1}{x-2}\) + tan^-1 \(\frac{x+1}{x+2}\) = \(\frac{\pi}{4}\)
Solution:

(vi) tan^-1 \(\frac{1}{2 x+1}\) + tan^-1 \(\frac{1}{4 x+1}\) = tan^-1 \(\frac{2}{x^2}\)
Solution:

(vii) 3 sin^-1 \(\frac{2 x}{1+x^2}\) – 4 cos^-1 \(\frac{1-x^2}{1+x^2}\) + 2 tan^-1 \(\frac{2 x}{1-x^2}\) = \(\frac{\pi}{3}\)
Solution:

(viii) cot^-1 \(\frac{1}{x-1}\) + cot^-1 \(\frac{1}{x}\) + cot^-1 \(\frac{1}{x+1}\) = cot^-1 \(\frac{1}{3x}\)
Solution:

(ix) cot^-1 \(\frac{1-x^2}{2 x}\) =  cosec^-1 \(\frac{1+a^2}{2 a}\) – sec^-1 \(\frac{1+b^2}{1-b^2}\)
Solution:

(x) sin^-1 \(\left(\frac{2 a}{1+a^2}\right)\) + sin^-1 \(\left(\frac{2 b}{1+b^2}\right)\) = 2 tan^-1 x
Solution:

(xi) sin^-1 y – cos^-1 x = cos^-1 \(\frac{\sqrt{3}}{2}\)
Solution:

(xii) sin^-1 2x + sin^-1 x = \(\frac{\pi}{3}\)
Solution:

Question 10.
Rectify the error ifany in the following:
sin^-1 \(\frac{4}{5}\) + sin^-1 \(\frac{12}{13}\) + sin^-1 \(\frac{33}{65}\)
Solution:

Question 11.
Prove that:
(i) cos^-1 \(\left(\frac{b+a \cos x}{a+b \cos x}\right)\) = 2 tan^-1 \(\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{x}{2}\right)\)
Solution:

(ii) tan \(\left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} \frac{a}{b}\right)\) + tan \(\left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} \frac{a}{b}\right)\)
Solution:

(iii) tan^-1 \(\sqrt{\frac{x r}{y z}}\) + tan^-1 \(\sqrt{\frac{y r}{y x}}\) + tan^-1 \(\sqrt{\frac{z r}{x y}}\) = π where r = x + y +z.
Solution:

Question 12.
(i) If cos^-1 (\(\frac{x}{a}\)) + cos^-1 (\(\frac{y}{b}\)) = Θ, prove that \(\frac{x^2}{a^2}\) – \(\frac{2 x}{a b}\) cos Θ + \(\frac{y^2}{b^2}\) = sin² Θ.
Solution:

(ii) If cos^-1 (\(\frac{x}{y}\)) + cos^-1 (\(\frac{y}{3}\)) = Θ, prove that 9x² – 12xy cos Θ + 4y² = 36 sin² Θ.
Solution:

(iii) If sin^-1 (\(\frac{x}{a}\)) + sin^-1 (\(\frac{y}{b}\)) = sin^-1 (\(\frac{c^2}{a b}\)) prove that b²x² + 2xy \(\sqrt{a^2 b^2-c^4}\) a²y² = c²
Solution:

(iv) If sin^-1 (\(\frac{x}{a}\)) + sin^-1 (\(\frac{y}{b}\)) = α prove that \(\frac{x^2}{a^2}\) + \(\frac{2 x y}{a b}\) cos α + \(\frac{y^2}{b^2}\) = sin² α
Solution:

(v) If sin^-1 x + sin^-1 y + sin^-1 z = π prove that x² + y² + z² + 4x²y²z² = 2 ( x²y² + y²z² + z²x² )
Solution:

Question 13.
Solve the following equations:
(i) tan^-1 \(\frac{x-1}{x+1}\) + tan^-1 \(\frac{2 x-1}{2 x+1}\) = tan^-1 \(\frac{23}{36}\)
Solution:

(ii) tan^-1 \(\frac{1}{3}\) + tan^-1 \(\frac{1}{5}\) + tan^-1 \(\frac{1}{7}\) + tan^-1 x = \(\frac{\pi}{4}\)
Solution:

(iii) cos^-1 \(\left(x+\frac{1}{2}\right)\) + cos^-1 x+ cos^-1 \(\left(x-\frac{1}{2}\right)\) = \(\frac{3 \pi}{2}\)
Solution:

(iv) 3tan^-1 \(\frac{1}{2+\sqrt{3}}\) – tan^-1 \(\frac{1}{x}\) = tan^-1 \(\frac{1}{3}\)
Solution:

Odisha State Board CHSE Odisha Class 12 Invitation to English 4 Solutions Grammar Fundamental Rules for Correcting Grammatical Errors Textbook Activity Questions and Answers.

CHSE Odisha 12th Class English Grammar Fundamental Rules for Correcting Grammatical Errors

Question 1.
In English Grammar verbs always agree with their subject in persons and numbers.
Example:
(a) I am a teacher.
subject – I
verb – am
(b) He is a student.
subject – He
verb – is
(c) They are farmers.
subject – They
verb – are
So while correcting the errors first underline the subject and verb of each sentence and see whether the verb agrees with its subject.

Question 2.
Prepositions always follow either a noun or an adjective or a verb. So find out preposition in each sentence and see their appropriate use.
Example:
(a) He is jealous of my success.
Adjective – jealous
Preposition – of
(b) The boy always depends on his father’s help.
Adjective – depends
Preposition – on
(c) He has confidence in me.
Noun- Confidence
Preposition – in

Question 3.
To find out errors in sentences articles play important role. In some sentences: there is wrong use of articles i.e., wrong use of ‘a’, ‘an’, ‘the’. So go through the use of articles meticulously and get it corrected.
Example:
A hermit live in a small cottage in a remote village. There is an orchard behind the cottage. The orchard has many rare trees.

Question 4.
Sometimes the use of quantifiers like ‘much’, ‘many’, ‘a lot of, ‘little’, ‘a little’, ‘few’ and ‘a few’ is wrongly used in sentences. It should be corrected carefully.
Example:
(a) They don’t find much time to work in the garden.
Uncountable – time
(b) Is there much water in the pond?
Uncountable – water

From the above examples it is found that ‘much’ is used before uncountable nouns in negative and interrogative sentences.
Example:
(c) He doesn’t have many shirts.
PI. Count. – shirts
(d) Does he have many books in his library.
PI. Count. – books

From the above examples it is found that ‘many’ is used before plural countable noun in negative and interrogative sentences.
Example:
(e) The mechanic repairs a lot of bikes everyday.
PI. Count. – bikes
(f) He has got a lot of money.
Uncountable – money

From the above examples it is found that ‘a lot of can be used with plural countable nouns as well as uncountable nouns in affirmative sentences.

Question 5.
Some verbal expressions like would rather and had better always go with the Bare Infinitive form of a non-finite verbs.
Example:
(a) You would rather go than stay here.
(going, go, gone, to go)
(Choose the correct alternative)
(b) They had better do their duty in time.
(doing, done, do, to do)
(Choose the correct alternative)

Question 6.
An imperative sentence beginning with “Let” can be followed by objective case of a pronoun.
Example:
(a) (i) Let he do whatever he wants to do. (Incorrect)
(ii) Let him do whatever he wants to do. (Correct)
(b) (i) Let you and I solve the problem. (Incorrect)
(ii) Let you and me solve the problem. (Correct).
(c) (i) Let there be no secret between you and we. (Incorrect)
(ii) Let there be no secret between you and us. (Correct)

Pronouns
Subjective case	Objective Case
I	Me
We	Us
You	You
He	Him
She	Her
It	It

Question 7.
The expression “I wish” always goes with “I were” i.e., past simple or “had + past participle” i.e., past perfect.
Example:
(a) (i) I wish I am the king. (Incorrect).
(ii) I wish I were the king. (Correct)
(b) (i) I wish I know your name. (Incorrect)
(ii) I wish I knew /had known your name. (Correct)

Question 8.
When the singular pronouns of first, second and third person go together their order should be “2nd person, third person and first person” i.e., “you + he and I”.
Example: (a) (i) I, you and he are friends. (Incorrect)
(ii) You, he and I are friends. (Correct)
(b) (i) You, they and we must work together. (Incorrect)
(ii) We, you and they must work together. (Correct)

Question 9.
Verbs like absent, apply, acquit, enjoy, resign pride, avail always go with reflexive pronouns (self/selves forms).
I – myself
he – himself
she – herself
you – yourself (singular) yourselves (plural)
we – ourselves
they – themselves
child/animal/ object (singular) – itself.

Example:
(a) (i) He prides on his money. (Incorrect)
(ii) He prides himself on his money. (Correct)

(b) (i) He availed of the chance. (Incorrect)
(ii) He availed himself of the chance. (Correct)

(c) (i) The children are expected to behave in the class. (Incorrect)
(ii) The children are expected to behave themselves in the class. (Correct)

(d) (i) The man resigned to the will of God. (Incorrect)
(ii) The man resigned himself to the will of God. (Correct)

(e) (i) My sister absented from her chemistry class yesterday. (Incorrect)
(ii) My sister absented herself from her chemistry class yesterday. (Correct)

Question 10.
The idiomatic expressions like “with a view to”, “look forward to”, “used to”, “habituated to”, and “objected to” always go with the “V+ing” form of a non-finite verb.
Example:
(a) (i) Sulipta is working hard with a view to win the match. (Incorrect)
(ii) Sulipta is working hard with a view to i. (Correct)
(b) (i) I look forward to see the doctor next month. (Incorrect).
(ii) I look forward to seeing the doctor next month. (Correct)
(c) (i) He is used to get up late. (Incorrect)
(ii) He is used to getting up late. (Correct)

Question 11.
(a) Some nouns look singular but they are originally plural. These nouns are people, police, public, cattle, swine, geese, teeth, oxen, mice and they can take plural verbs after them.
(b) The + Adjective and ‘The + Nationality word’ used as the subjects can take plural verbs.
Example:
(i) The police are on duty.
(ii) The people are shouting.
(iii) The cattle are grazing.
(iv) The poor are helpful.
(v) The wounded have been admitted in the hospital.
(vi) The English are a nation of traders.
(vii) The Japanese are industrious.
(viii) The geese look hungry.
(ix) The swine are the lovers of filth.
(x) When the cat is away, mice dance in joy.
(xi) The blind are the most unfortunate.

Question 12.
Grammatical expressions like “It is no use”, “There is no use” and “There is no point” always go with “-ing” form of non-finite verbs.
Example:
(i) It is no use attending classes today.
(ii) There is no point talking with her.

Question 13.
Some verbs tell us about our feelings, emotions, opinions, relations or about a permanent state. Such verbs are called stative verbs. These verbs are usually used in the simple present form.
(a) Verbs of possession: have, own, possess, belong to, contain, consist.
(b) Verbs of liking/disliking: like, dislike, love, hate, prefer, admire and want.
(c) Verbs of perception: see, hear, smell, taste, feel.
(d) Verbs of thinking: think, believe, understand, know.
(e) Verbs of mental activity: hope, forget, remember.
(f) Verbs of appearance: appear, seem, look (like) resemble.
(g) Some other verbs: depend, weight, cost, measure, sound.

Example:
(i) He belongs to Cuttack.
(ii) A year consists of six seasons.
(iii) Lipsa hates telling a lie.
(iv) Honey tastes sweet.
(v) Rose smells sweet.
(vi) Batu loves this girl.
(vii) Mama resembles her mother.
(viii) He knows me.
(ix) Empty vessel sounds much.
(x) I never believe in ghosts.

Question 14.
Adverb like No sooner, hardly/seldom, little, under no circumstances, nowhere and conjunctions like neither, not only etc. are used with inversion if they are used at the beginning of the sentences, i.e., Adv. + Aux. Verb + Sub. + Main verb.
Example:
(a) (i) No sooner the bell rang than the students came out of the class. (Incorrect)
(ii) No sooner did the bell rang than the students came out of the class. (Correct)

(b) (i) Hardly father had come out of the house when it began to rain. (Incorrect)
(ii) Hardly had father come out of the house when it began to rain. (Correct)

(c) (i) Little he understands my problem. (Incorrect)
(ii) Little does he understand my problem. (Correct)

(d) (i) Under no circumstances I shall compel you to do this work. (Incorrect)
(ii) Under no circumstances shall I compel you to do this work. (Correct)

(e) (i) Not only the robbers captured the city but also destroyed it. (Incorrect)
(ii) Not only did the robbers captured the city but also destroyed it. (Correct)

(f) (i) Neither the girl can dance nor sing. (Incorrect)
(ii) Neither can the girl dance nor sing. (Correct)

Question 15.
Some nouns look like plural but originally they are singular and take singular verbs. News, physics, politics, economics, measles, mathematics, diabetes, gymnastics, mumps, rabies, itches, scabies, electronics, athletics, sports, billiards, hurdles, cards.
Example:
(i) Mathematics is my favourite subjects.
(ii) Diabetes is a disease.
(iii) Gymnastics is good for health.
(iv) Billiards is my favourite indoor game.
(v) Physics is a tough subject.
(vi) The new has been greeted with cheers.
(viii)Politics is dirty game.

Question 16.
When the pronoun is the object of a verb or a preposition, it should be in the objective case.
Example:
(a) (i) These books are for you and I. (Incorrect)
(ii) These books are for you and me. (Correct)
(b) (i) Between he and I there is an understanding. (Incorrect)
(ii) Between he and me, there is an understanding. (Correct)

Question 17.
Subject form of the pronouns like (I/he/she/ you/they/we, etc.) is used with the sentences taking than.
Example:
(a) (i) He is taller than me. (Incorrect)
(ii) He is taller than I. (Correct)
(b) (i) I love you more than him. (Incorrect)
(ii) I love you more than he (Correct)

Question 18.
To avoid repetition of a noun, we use that for singular noun and those for the plural noun.
Example:
(a) (i) The climate of Simla is better than Darjeeling. (Incorrect)
(ii) The climate of Simla is better than that of Darjeeling. (Correct)
(b) (i) The roads of Bhubaneswar are wider than Cuttack. (Incorrect)
(ii) The roads of Bhubaneswar are wider than those of Cuttack. (Correct)

Question 19.
‘Each other’, the reciprocal pronoun is used in speaking of two persons/things, ‘one another’ for more than two.
Example:
(a) (i) Jack and Jill love one another. (Incorrect)
(ii) Jack and Jill love each other. (Correct)
(b) (i) The street dogs barked at each other. (Incorrect)
(ii) The street dogs barked at one another. (Correct)

Question 20.
The expressions like each (of), everyone (of), neither of (of the two), either of (of the two), the number of, one of etc. always take singular verb after them.
Example:
(a) (i) Each boy were given a prize. (Incorrect)
(ii) Each boy was given a prize. (Correct)

(b) (i) Each man and each woman have been sent to the workfield. (Incorrect)
(ii) Each man and each woman has been sent to the workfield. (Correct)

(c) (i) Everyone of the cars look attractive. (Incorrect)
(ii) Everyone of the cars looks attractive. (Correct)

(d) (i) Either of the books have a lot of pictures. (Incorrect)
(ii) Either of the books has a lot of pictures. (Correct)

(e) (i) Neither of the girls were called to appear the test. (Incorrect)
(ii) Neither of the girls was called to appear the test. (Correct)

(f) (i) The number of books stolen are forty. (Incorrect)
(ii) The number of books stolen is forty. (Correct)

(g) (i) One of the chairs have a broken leg. (Incorrect)
(ii) One of the chairs has a broken leg. (Correct)

(h) (i) One of my brothers are in the Indian army. (Incorrect)
(ii) One of my brothers is in the Indian army. (Correct)

Question 21.
The name of the shops (books/ opticals) always takes singular verb after them.
Example:
(a) (i) The Books and Books stand on the main road, Bhubaneswar. (Incorrect)
(ii) The Books and Books sstands on the main road, Bhubaneswar. (Correct)
(b) (i) The Giri Opticals have a good name in the whole town. (Incorrect)
(ii) The Giri Opticals has a good name in the whole town. (Correct)
(c) (i) ‘Gulliver’s Travels’ were written by Jonathan Swift. (Incorrect)
(ii) ‘Gulliver’s Travels’ was written by Jonathan Swift. (Correct)

Question 22.
After same or such, relative pronoun as or that is used.
Example:
(a) (i) This isn’t such a good book with I expected. (Incorrect)
(ii) This isn’t such a good book as I expected. (Correct)
(b) (i) This is the same beggar who came to our house last week. (Incorrect)
(ii) this is the same beggar that came to our house yesterday. (Correct)
(c) (i) My problem is the same which yours. (Incorrect)
(ii) My problem is the same as yours. (Correct)

Question 23.
We use relative pronoun that in the noun phrase on antecedent have the + superlative degree, ordinal (like first/second/ tenth /last, etc.) something/ anything/ all/nothing/ somebody etc. (Indefinite pronouns)
Example:
(a) (i) This is the best film which I have ever seen. (Incorrect)
(ii) This is the best film that I have ever seen. (Correct)
(b) (i) Love is something which money can’t buy. (Incorrect)
(ii) Love is something that money can’t buy. (Correct)
(c) (i) She is the most beautiful girl which has ever lived. (Incorrect)
(ii) She is the most beautiful girl that has ever lived. (Correct)
(d) (i) This is all which is yours. (Incorrect)
(ii) This is all that is yours. (Correct)
(e) (i) The second train which left for Puri just now, has faced an accident. (Incorrect)
(ii) The second train that left for Puri just now, has faced an accident. (Correct)

Question 24.
Comparative degree is used for two persons things/ animals and superlative degree for more than two.
Example:
(a) (i) It is the best of the two books. (Incorrect)
(ii) It is better of the two books. (Correct)
(b) (i) He is the better of the three boys. (Incorrect)
(ii) He is the best of the three boys. (Correct)
(c) (i) Which is the best: bread or butter? (Incorrect)
(ii) Which is better: bread or butter? (Correct)
(d) (i) Out of these two watches, which is the best? (Incorrect)
(ii) Out of these two watches, which is better? (Correct)

Question 25.
We use Fewer/Few/ A Few to denote number and less for quantity.
Example:
(a) (i) There are no less than twenty boys in this class. (Incorrect)
(ii) There are no fewer than twenty boys in this class. (Correct)
(b) (i) He takes no fewer than one litre of milk. (Incorrect)
(ii) He takes no jess than one litre of milk, (milk – u.n.) (Correct)

Question 26.
When comparative degree is used in the superlative sense, it is followed by ‘any other’.
Example:
(a) (i) Akram is better than any bowler. (Incorrect)
(ii) Akram is better than any other bowler. (Correct)
(b) (i) Raman is better than any student in the class. (Incorrect)
(ii) Raman is better than any other student in the class. (Correct)

Question 27.
Comparative Adjectives like senior, junior, superior, inferior take to after them instead of than’.
Example:
(a) (i) My father is senior than you in service. (Incorrect)
(ii) My father is senior to you in service. (Correct)
(b) (i) I am junior than her. (Incorrect)
(ii) I am junior to her. (Correct)
(c) (i) This book is inferior than that book. (Incorrect)
(ii) This book is inferior to that book. (Correct)

Question 28.
When expression of measurement, amount, and quantity are used as adjectives, the nouns occuring after the hyphen (-) are always singular.
Example:
(a) (i) The Prime Minister went on a three day state visit to America. (Incorrect)
(ii) The Prime Minister went on a three day state visit to America. (Correct)
(b) (i) A seven-members jury decided the case. (Incorrect)
(ii) A seven-members jury decided the case. (correct)
(c) (i) I came upon a hundred-rupees note. (Incorrect)
(ii) I came upon a hundred-rupee note. (Correct)

Question 29.
We usually use ‘but’ (not ‘than’) after ‘else’.
Example:
(a) (i) It is nothing else than pride. (Incorrect)
(ii) It is nothing else but pride. (Correct)
(b) (i) Call me anything else than a thief. (Incorrect)
(ii) Call me anything else but a thief. (Correct)

Question 30.
‘Very’ is used with adjectives and adverbs in the positive degree with precedent and much is used with Adjectives and adverbs in the comparative degree with past participle.
Example:
(a) (i) She is very slower than Reena. (Incorrect)
(ii) She is much slower (c.d) than Reena. (Correct)
(b) (i) You are very older than me. (Incorrect)
(ii) You are much older (c.d.) than me. (Correct)
(c) (i) The policeman was walking much slowly. (Incorrect)
(ii) The policeman, was walking very slowly. (Correct)

Question 31.
The verb ‘know’ is followed by ‘how to + infinitive’.
Example:
(a) (i) The professor knows to teach the topic. (Incorrect)
(ii) The professor knows how to teach the topic. (Correct)
(b) (i) An expert pilot knows to land the flight in hostile weather. (Incorrect)
(ii) An expert pilot knows how to land the flight in hostile weather. (Correct)

Question 32.
When a noun or pronoun is placed before a gerund (verb + ing working like a noun is called ‘gerund’), it (that noun/pronoun) should be put in the ‘possessive case’, (my/his/her/our/their/teacher’s/one’s etc)
Example:
(a) (i) Please excuse me being late. (Incorrect)
(ii) Please excuse my being late. (Correct)
(b) (i) I would like you stopping smoking. (Incorrect)
(ii) I would like your stopping smoking. (Correct)
(c) (i) The professor disliked us coming late. (Incorrect)
(ii) The professor disliked our coming late. (Correct)

Question 33.
Certain verbs looking alike have different meanings. They are as follows :
Example:
(a) (i) He hanged the lamp on the wall. (Incorrect)
(ii) He hung the lamp on the wall. (Correct)
(b) (i) The criminal was hung for murder. (Incorrect)
(ii) The criminal was hanged for murder. (Correct)
(c) (i) The hens have laid no eggs today. (Incorrect)
(ii) The hens have lain no eggs today. (Correct)
(d) (i) Let me lay on the bed. (Incorrect)
(ii) Let me lay on the bed. (Correct)
Look at the list of following verbs with their past and perfect or past participle forms.

Present	Past	P.P. / Perfect
lie (rest/lie down)	lay	lain
lay (place, arrange/deposit)	laid	laid
lie (to tell a lie)	lied	lied
hang (put up)	hung	hung
hang (execute the order of death sentence)	hanged	hanged
flow (water)	flowed	flowed
fly (bird)	flew	flown
flee (run away).	fled	fled
bear (put up with/tolerate)	bore	borne
bore (make a hole/make tied)	bored	bored
find (to discover)	found	found
found (establish)	founded	founded
fall (drop in the ground)	fell	fallen
fell (cut down a tree)	felled	felled
feel (sensitize)	felt	felt
fill (to pour till the end)	filled	filled
awake (intransitive verb)	awoke	awoke
awake (transitive verb)	awaked	awaked

Question 34.
Indirect questions have the usual Wh-word + subject + verb order.
Example:
(a) (i) Tell me where are you going. (Incorrect)
(ii) Tell me where you are going. (Correct)
(b) (i) Father asked the servant where had he gone. (Incorrect)
(ii) Father asked the servant where he had gone. (Correct)
(c) (i) Can you tell me why does the girl cry bitterly. (Incorrect)
(ii) Can you tell me why the girl cries bitterly. (Correct)
(d) (i) The gentleman asked the station master when was the next train. (Incorrect)
(ii) The gentleman asked the station master when next train was. (Correct)

Question 35.
The past tense in the Principal Clause or Main Clause is used with the same past tense in the Subordinating or Dependent Clause.
Example:
(a) (i) She knew that I am coming. (Incorrect)
(ii) She knew that I was coming. (Correct)
(b) (i) Father said that he won’t go to office that day. (Incorrect)
(ii) Father said that he wouldn’t go to office that day. (Correct)
(c) (i) The doctor asked the patient if he (the patient) has taken medicine at regular
intervals. (Incorrect)
(ii) The doctor asked the patient if he had taken medicine at regular intervals.(Correct)

Question 36.
It is time + subject + past tense verb.
Example:
(a) (i) It is time you go to bed. (Incorrect)
(ii) It is time you went to bed. (Correct)
(b) (i) It is time the doctor operates the patient. (Incorrect)
(ii) It is time the doctor operated the patient. (Correct)

Question 37.
After imperatives (order, advice, request) we use won’t you? (to invite people to do things) and will you/would you/could you? (to tell people to do things).
Example:
(a) (i) Do sit down, will you? (Incorrect)
(ii) Do sit down, won’t you? (Correct)
(b) (i) Give me sufficient time, won’t you? (Incorrect)
(ii) Give me sufficient time, will you? (Correct)
(c) (i) Shut up, can you? (Incorrect)
(ii) Shut up, can’t you? (Correct)

Question 38.
Certain verbs in English are not used in their progressive forms. They are :
Example:
(a) (i) I amn’t seeing you these days. (Incorrect)
(ii) I don’t see you these days. (Correct)

(b) (i) This bag is containing a lot of story books. (Incorrect)
(ii) This bag is contains a lot of story books. (Correct)

(c) (i) Are you appearing disappointed? (Incorrect)
(ii) Do you appear disappointed? (Correct)

(d) (i) Why is the girl hating me so much? (Incorrect)
(ii) Why does the girl hate me so much? (Correct)

(e) (i) Honey is tasting sweet. (Incorrect)
(ii) Honey tastes sweet. (Correct)

(f) (i) Rose is smelling sweet. (Incorrect)
(ii) Rose smells sweet. (Correct)

(g) (i) We are not believing in ghost. (Incorrect)
(ii) We don’t believe in ghost. (Correct)

(h) (i) I am loving this girl. (Incorrect)
(ii) I love this girl. (Correct)

Question 39.
Would rather + subject takes past simple tense of the verb.
Example:
(a) (i) I would rather you resign the job. (Incorrect)
(ii) I would rather you resigned the job. (Correct)
(b) (i) I would rather he leaves this place. (Incorrect)
(ii) I would rather he left this place. (Correct)

Some incorrect words/expressions and their correct use.

Incorrect – Correct
arm (weapon) – arms
blotting – blotting paper
boarding – boarding school
bowel – bowels
foundation – foundations
arrear – arrears
furnitures – furniture
luggages – luggage
earning – earnings
breads – pieces/slices/loaves of bread
equipments – equipment
informations – information
gentries – gentry
machineries – machinery/machines
poetries – poetry
sceneries – scenery
traffics – traffic
scissor – scissors
trouser – trousers
a coward man – a coward/ a cowardly man
a miser man – a miser/ a miserly man
a man of letter – a man of letters(literate)
arrear bill – arrears bill
birth date – date of birth
cousin brother/sister – cousin
custom duty – customs duty
famous criminal – notorious criminal
in the campus – on the campus
in the committee – on the committee
in leave – on leave
in holiday – on holiday
tennis field – tennis court
Cheque of Rs. 200/- – Cheque for Rs. 200/-
no place (bus/train etc.) – no room (bus/train etc.)
today morning/ – this morning/afternoon/
afternoon/evening – evening
this night – tonight
two dozens pens – two dozen pens
saving bank – savings bank
make noise – make a noise
tell lie – tell a lie/tell lies
white hair – grey hair
miles after miles – mile after mile
in hurry – in a hurry
bad in studies – bad at studies
strong/weak of/ good in – strong/weak in/ good at
what to speak of – not to speak of
with bag and baggage – bag and baggage
with heart and soul – heart and soul
with tooth and nail (Completely) – tooth and nail
with black and blue (mercilessly/beat) – black and blue
clever in figure works – clever at figure works
build a home – build a house
cut jokes – crack jokes
cut the pencil – mend the pencil
cook bread – bake bread
describe about – describe
discuss about – discuss
attack on – attack
give a speech – deliver a speech
goodbye – bid goodbye
eat the poor – feed the poor
give order – give orders
make a goal – score a goal
rise the lid – raise the lid
see the pulse – feel the pulse
to have headache – to have a headache
to have temperature – to have a temperature
speak a lie – tell a lie
make prayers – say prayers
ladies bicycle – ladies’ bicycle
women’s college – womens’ college
family members – members of the family
passing marks – pass marks
decrease fear – allay fear
in class tenth – in class ten/in the tenth class
lecture (person) – lecturer
out of orders – out of order (technically defect)
deny request/invitation – refuse request/invitation
refuse stealing/lying – deny stealing/lying
refuse – reject/not to accept
deny – not acknowledge/tell that something is untrue.
a furniture – a piece of furniture
a luggage – a piece of luggage
an information – a piece of information.
a wood – a piece of wood
a grass – a blade of grass
a chocolate – a bar of chocolate
a paper – a sheet of paper
a news – a piece of news/some news
a work – a piece of work / some work
a bread – a slice of bread
a meat – a piece of meat
a toothbrush – a stick of toothbrush
a toothpaste – a tube of toothpaste

Rewrite the following passage, correcting all the grammatical errors in it:

Question 1.
The streets crowd by traffic even before the daybreak. Taxis have been bringing tired people from the airport and the railway station to hotels. They hope sleeping for a few hours before their busy day in the big city is beginning. Trucks are bringing fresh fruits and vegetables in the city. Ships loaded with food and fuel tie up in the dock. Towards morning the streets are quieter, and they are never deserted in the big city.
Answer:
The streets are crowded by traffic even before the daybreak. Taxis bring tired people from the airport and the railway station to the hotels. They hope to sleep for a few hours before their busy day in the big city begins. Trucks bring fresh fruits and vegetables to the city. Ships loaded with food and fuel tied up at the dock. Towards the morning the streets are quieter, and they are never deserted in the big city.

Question 2.
Every country in the world want rapid economic development today. Some economists tells us that it is possible to remove poverty and make everyone prosperous, provide we adopt the right economic policies. The key to prosperity, we are also told, lies in rapid and large-scale industrialization: setting up more factories which will churn out an endless stream of consumer good – products designed to make life more pleasant; motor cars to carried us in comfort and at high speed along smooth super highways; air conditioners to keeps us cool in summer, television sets which will keep us informed as well as entertain and so on. It is believed that as more and more consumers buy the goods that these factories will produce, more and more workers find employment in them; and as their levels of income rise, they will, in their turn create a farther demand for yet more goods. In this way, everyone becomes rich. There are no limits to economical growth and prosperity.
Answer:
Every country in the world wants rapid economic development today. Some economists tell us that it is possible to remove poverty and make everyone prosperous, provided we adopt the right economic policies. The key to prosperity, we are also told, lies in rapid and large-scale industrialization: setting up more factories which will churn out an endless stream of consumer goods – products designed to make life more pleasant; motor cars to carry us in comfort and at high speed along smooth super highways.

air conditioners to keep us cool in summer, television sets which will keep us informed as well as entertained and so on. It is believed that as more and more consumers buy the goods that these factories will produce, more and more workers find employment in them; and as their level of income rises, they will, in their turn create a further demand for yet more goods. In this way, everyone becomes rich. There are no limits to economical growth and prosperity.

Question 3.
Thrift means regulating expenses by such a way that there might be some saving in income. There could be no hard and fast standards for what shall be one’s expenditure and saving. It varied according to one’s circumstances. The rich man may neglect the duty of saving in special occasions because he has power to make up for this neglect. And people with limited income need thrift the most. It gives them strength by relieving them of anxieties for the future. Good housewives in poor families are found to lay off something from daily expenses.
Answer:
Thrift means regulating expenses in such a way that there might be some saving in income. There should be no hard and fast standard for what could be one’s expenditure and saving. It varies according to one’s circumstances. The rich man may neglect the duty of saving on special occasions because he has the power to make up for this neglect. And people with limited income need thrift the most. It gives them strength by relieving them from anxieties for future. Good housewives in poor families are found laying off something from daily expenses.

Question 4.
Preschools have offered good basic education as well as help the child in becoming much independent but confident. Parents may rely in preschools for all-round development of the children. The pre-primary education of the child generally began in home by parents and grandparents. But the picture changes rapidly.
Answer:
Preschools offer good basic education as well as help the child in becoming much independent and confident. Parents rely on preschools for an all-round development of their children. The pre-primary education of the child begins at home by parents and grandparents. But the picture has changed rapidly.

Question 5.
Once a lion was enjoy a nap in his den. A mouse came out in its hole in the den. It start frisking about. In so doing it leap upon the lion’s face. The lion’s sleep disturbs. He wake up furious. He caught the mouse and had been killed it, but the mouse entreated “Your Majesty, I humbly beg your pardon, I’m a poor and little subject of you. But a tiny creature as 1 am, I shall be of some help to you in time. So, please let me go.” The lion laughed aloud for this, but he released the mouse all the same.
Answer:
Once a lion was enjoying a nap in his den. A mouse came out of its hole in the den. It started frisking about. In so doing it leapt upon the lion’s face. The lion’s sleep was disturbed. He woke up furiously. He caught the mouse and was about to kill it. But the mouse entreated “Your Majesty, I humbly beg your pardon, I’m a poor and little subject of yours. But a tiny creature as I am, I shall be of some help to you in time. So, please let me go.” The lion laughed aloud for this and he released the mouse all the same.

Question 6.
Man is at first like an animal. His power then rested only in physical strength. But in this respect he was no match to many beasts. So he has to live in constant fear of them. In course of time came knowledge and it gave him the power to get mastery the entire animal kingdom. He invented weapons with which he not only scared off beasts but even killed them. He learnt how to hunt beasts like a deer which may run more faster than he did Now man could achieve things which were considered impossible then.
Answer:
Man was at first like an animal. His power then rested only on physical strength. But in this respect he was no match for many beasts. So he had to live in constant fear of them. In course of time came knowledge and it gave him a power to get mastery upon the entire animal kingdom. He invented weapons with which he not only scared off beasts but also killed them. He learnt how to hunt beasts like a deer who could run faster than he. Now man has achieved things which were considered impossible then.

Question 7.
The foodbazar took the entire responsibility in sending a farmer’s produce to the customer, without depending on a chain of middlemen. The private company with its vast resources could set in cold storages, acquire refrigerated trucks to transport the produce to cities. In the end of the food chain there are airconditioned supermarkets where consumers can buy the produce at good condition. A kilo of tomato which a customer will buy for 10 rupees may be available for only 7 rupees in a supermarket. Don’t imagine that any private company will do all this out from charity or love towards the farmers. In time food chains might come on away from the cities and closer to the farms.
Answer:
The foodbazar is taking the entire responsibility of sending a farmer’s produce to the customer, without depending on a chain of middlemen. The private company with its vast resources can set up cold storages and acquire refrigerated trucks to transport the produce to cities. At the end of the food chain there are airconditioned supermarkets where consumers can buy the produce in good condition. A kilo of tomato which a customer will buy for 10 rupees can be available for only 7 rupees in a supermarket. We can’t imagine that any private company will do all this out of charity or love for the farmers. With time food chains may come away from the cities and closer to the farms.

Question 8.
A big fat hen lived of a farmyard in a village. A dove also lived in a large tree besides the same farmyard. Soon both of them became good friend. They use to meet every evening to share their thoughts. Oneday, the hen see by a fox. He enters the farmyard secretly and was able to catch the hen. The clever fox put the hen in a sack, carried at his back and the hen started crying aloud in protest. The dove heard the crying of the hen from the top of the tree. She at once realised that the hen fell in danger. The dove thought off an idea to save her friend. She went ahead and lay on the path motionless as of she was dead. The fox put his sack down at the wayside and went near the dove. In the meantime, the hen managed to escape, and hide behind a bush. When the fox was about to catch the ‘dead’ dove, he surprised to see that the dove quickly flew away. In the evening the dove met the hen at the farmyard and the two began to laugh for the foolishness of the fox.
Answer:
A big fat hen lived in a farmyard in a village. A dove also lived in a large tree beside the same farmyard. Soon both of them became good friends. They used to meet every evening to share their thoughts. Oneday, the hen was seen by a fox. He entered the farmyard secretly and was able to catch the hen. The clever fox put the hen in a sack, carried on his back and the hen started crying aloud in protest. The dove heard the cry of the hen from the top of the tree.

She at once realised that the hen had fallen into danger. The dove thought of an idea to save her friend. She went ahead and lay on the path motionless as if she was dead. The fox put his sack down on the wayside and went near the dove. In the meantime, the hen managed to escape, and hid behind a bush. When the fox was about to catch the ‘dead’ dove, he was surprised to see that the dove quickly flew away. In the evening the dove met the hen at the farmyard and the two began to laugh for the foolishness of the fox.

Question 9.
The food bazaar is taking entire responsibility in sending the farmer’s produce to the consumer. The private company with its vast resource, may set up cold storages, acquire fleets of refrigerated trucks to transport the produce into cities and even construct roads for speedy transportation. In the end of the food chain, there have been air-conditioned supermarkets where consumers could buy produce of high quality, in good condition, at comparatively reasonable prices, in clean and hygienic surrounding. A kilo of tomatoes which a customer could buy from a vegetable-vendor for ten rupees must be available, weighed but neatly packed for only t 7.50 in a super-market. Out of that amount, the farmer is likely to have got at least ? 3.50 a much higher price than he would get if he would sell his produce to a middleman.
Answer:
The food bazaar is taking an responsibility in sending a farmer’s produce to the consumer. The private company with its vast resources, can set up cold storages, acquire a fleet of refrigerated trucks to transport the produce to the cities and even construct roads for speedy transportation. At the end of the food chain, there are air-conditioned supermarkets where consumers can buy produce of high quality, in good condition, in a comparatively reasonable prices, in a clean and hygienic surrounding.

A kilo of tomato which a customer can buy from a vegetable-vendor for ten rupees may be available, weighed but neatly packed for only t 7.50 in a super-market. Out of that amount, the farmer is likely get at least ? 3.50 a much higher price than he would get if he would sell his produce to a middleman.

Question 10.
Few years ago I was spending a week at Port Blair. The week had been over but I was in the airport ready for leaving when I discovered, in my dismay, that I forgot one of my suitcase in my hotel. Quickly, I jumped into a taxi and had explained my situation to the taxi driver. We sped away in the direction of my hotel. Suddenly the taxi driver slowed down so that he would talk with the driver of a truck moving through a road next to us. The truck contained live chickens.
Answer:
A few years ago I was spending a week in Port Blair. As the week was over but 1 was in the airport ready to leave when I discovered, to my dismay, that I had forgotten one of my suitcases in my hotel. Quickly, I jumped into a taxi and explained my situation to the taxi driver. We sped away in the direction of my hotel. Suddenly the taxi driver slowed down so that he might talk to the driver of a truck moving through the road next to us. The truck carried live chichens.

Question 11.
Rahul and Ramesh were best friend studying in school. Rahul was good in sports but poor in studies. Ramesh is good at both. At any competition, Ramesh always managed to win. As a result, Rahul became jealous with Ramesh. The sports day was near. Rahul and Ramesh were practising for it. Rahul was not sure if he score a win over Ramesh. Slowly they talked less among themselves. Ramesh asked him the reason many time but Rahul always put him of with a excuse or other. One day before the sports event, Rahul hit at a plan to defeat Ramesh. He went to the ground before anyone has arrived there. He dug a small pit on the path where Ramesh was supposed to run. Then he covered it with leaves and went back to his classroom. When the race started. Ramesh was ahead of Rahul but after sometimes he stepped on the pit and fell down. Rahul took over him. He was very happy as his plan worked and he had own the first prize. Soonafter, Rahul realized his fault and begged forgiveness.
Answer:
Rahul and Ramesh were best friends when studying in school. Rahul was good at sports but poor in studies. Ramesh was good at both. In any competition, Ramesh always managed to win. As a result, Rahul became jealous of Ramesh. The sports day was near. So Rahul and Ramesh were practising for it. Rahul was not sure if he would score a win over Ramesh. Gradually they talked less between themselves. Ramesh asked him the reason many a time but Rahul always put him off with an excuse or other.

One day before the sports event, Rahul hit upon a plan to defeat Ramesh. He went to the ground before anyone had arrived there. He dug a small pit on the path where Ramesh was supposed to run. Then he covered it with leaves and went back to his classroom. When the race started. Ramesh was ahead of Rahul but after sometime he stepped on the pit and fell down. Rahul took over him. He was very happy as his plan worked and he had won the first prize. Soonafter, Rahul realized his fault and begged forgiveness.

Question 12.
Generally the word ‘superstition’ was used as the term of disgrace or reproach to senseless belief based in ignorant fear. We look down to the uncivilized people because they worshipped the different aspect of nature and are filled in awe by things that are simple enough for us. Many Hindus considered a sneeze, the cry of a lizard as ominous. For the Westerners, number 13 is so unlucky that it can be avoided by all means. If the cat crosses their path they call up their tour. Indeed, every community is subject to some superstition and we could not conceive for any community being completely free of it.
Answer:
Generally the word ‘superstition’ is used as a term of disgrace or reproach to senseless belief based on ignorant fear. We look down upon the uncivilized people because they worshipped different aspects of nature and were fdled with awe by the things that are simple enough for us.

Many Hindus consider a sneeze, the cry of a lizard as ominous. For the Westerners, number 13 is so unlucky that it should be avoided by all means. If the cat crosses their path they call off their tour. Indeed, every community is subject to some superstition and we can not conceive of any community completely free from it.

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 9 Integration Ex 9(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Exercise 9(a)

Question 1.
(i) ∫2 dx
Solution:
∫2 dx = 2x + C

(ii) ∫3x² dx
Solution:
∫3x² dx = 3 \(\frac{x^3}{3}\) + C = x³ + C

(iii) ∫4x³ dx
Solution:
∫4x³ dx = 4 \(\frac{x^4}{4}\) + C = x⁴ + C

(iv) ∫x⁵ dx
Solution:
∫x⁵ dx = \(\frac{x^6}{6}\) + C

(v) ∫x³¹ dx
Solution:
∫x³¹ dx = \(\frac{x^{32}}{32}\) + C

(vi) ∫\(\left(2 \sqrt{x}+\frac{3}{\sqrt{x}}\right)\) dx
Solution:

(vii) ∫\(\frac{1}{x \sqrt{x}}\) dx
Solution:

(viii) ∫\(\left(x^{\frac{4}{7}}+\frac{1}{x^{\frac{1}{3}}}\right)\) dx
Solution:

(ix) ∫\(\frac{4}{x}\) dx
Solution:
∫\(\frac{4}{x}\) dx = 4 ln |x| + C

(x) ∫\(\left(\frac{3}{x^2}+\frac{1}{x^{12}}\right)\) dx
Solution:

(xi) ∫(x² + √x)² dx
Solution:

(xii) ∫(x + 3) (2 – x) dx
Solution:
∫(x + 3) (2 – x) dx
= ∫(2x + 6 – x² – 3x) dx
= ∫(-x² – x + 6) dx
= –\(\frac{1}{3}\)x³ – \(\frac{1}{2}\)x² + 6x + C

(xiii) ∫\(\frac{(\sqrt{x}+2)^2}{x^4}\) dx
Solution:

(xiv) ∫\(\frac{(x+\sqrt{x})(2 x+1)}{x^2}\) dx
Solution:

Question 2.
(i) ∫cos dx
Solution:
∫cos dx = sin x + C

(ii) ∫\(\frac{d x}{\cos ^2 x}\)
Solution:
∫\(\frac{d x}{\cos ^2 x}\) = ∫sec² x dx = tan x + C

(iii) ∫\(\frac{d x}{1-\cos ^2 x}\)
Solution:
∫\(\frac{d x}{1-\cos ^2 x}\) = ∫\(\frac{d x}{\sin ^2 x}\)
= ∫cosec² x dx
= -cot x + C

(iv) ∫\(\frac{\sin x}{\cos ^2 x}\) dx
Solution:
∫\(\frac{\sin x}{\cos ^2 x}\) dx = ∫\(\frac{\sin x}{\cos x} \cdot \frac{1}{\cos x}\) dx
= ∫sec x . tan x dx = sec x + C

(v) ∫\(\frac{2 \cos x}{1-\cos ^2 x}\) dx
Solution:

(vi) ∫\(\frac{1-\sin ^3 x}{\sin ^2 x}\) dx
Solution:
∫\(\frac{1-\sin ^3 x}{\sin ^2 x}\) dx = ∫(cosec² x – sin x) dx
= -cot x + cos x + C

(vii) ∫\(\frac{\sin ^2 x}{1+\cos x}\) dx
Solution:

(viii) ∫\(\frac{\cos 2 x}{\cos x+\sin x}\) dx
Solution:

(ix) ∫\(\frac{\cos ^4 x-\sin ^4 x}{\cos x-\sin x}\) dx
Solution:

(x) ∫\(\frac{\cos 2 x}{\sin ^2 x \cdot \cos ^2 x}\) dx
Solution:

(xi) ∫\(\frac{a^2 \sin ^2 x+b^2 \cos ^2 x}{\sin ^2 2 x}\) dx
Solution:

(xii) ∫(tan x + cot x)² dx
Solution:

(xiii) ∫\(\frac{1-\cos 2 x}{1+\cos 2 x}\) dx
Solution:

(xiv) ∫sec² x . cosec² x dx
Solution:

(xv) ∫\(\frac{\sin ^3 x+b \cos ^3 x}{\sin ^2 x \cos ^2 x}\) dx
Solution:

(xvi) ∫\(\sqrt{1+\sin 2 x}\) dx
Solution:

(xvii) ∫\(\sqrt{1-\cos 2 x}\) dx
Solution:

(xviii) ∫\(\sqrt{1+\cos 2 x}\) dx
Solution:
∫\(\sqrt{1+\cos 2 x}\) dx = ∫√2 cos x dx
= √2 sin x + C

(xix) ∫\(\frac{\cos 3 x \cos 2 x+\sin 3 x \sin 2 x}{1-\cos ^2 x}\) dx
Solution:

(xx) ∫(a cot x + b tan x)² dx
Solution:
∫(a cot x + b tan x)² dx
= ∫{a² cot² x + b² tan² x + 2ab cot x – tanx} xdx
= ∫(a² cot² x + b² tan² x + 2ab) dx
= a² ∫cot² x dx + b² ∫tan² x dx + 2ab ∫dx
= a² ∫(cosec² x – 1) dx + b² ∫(sec² x – 1) dx + 2ab x + C
= a² (-cot x – x) + b² (tan x – x) + 2ab x + C
= b² tan x – a² cot x – x (a² + b² – 2ab) + C
= b² tan x – a² cot x – (a – b)² x + C
= b² tan x – a² cot x + C

Question 3.
(i) ∫(e^x + 2) dx
Solution:
∫(e^x +2) dx = ∫e^x dx + 2 ∫dx
= e^x + 2x + C

(ii) ∫3^x dx
Solution:
∫3^x dx = ∫\(\frac{3^x}{\ln 3}\) + C

(iii) ∫a^x+2 dx
Solution:
∫a^x+2 dx = ∫a^x a² dx
= a² . \(\frac{a^x}{\ln a}\) + C
= \(\frac{a^{x+2}}{\ln a}\) + C

(iv) ∫a^3x dx
Solution:
∫a^3x dx = \(\frac{1}{3} \frac{a^{3 x}}{\ln a}\) + C

(v) ∫\(\frac{e^{2 x}+1}{e^x}\) dx
Solution:
∫\(\frac{e^{2 x}+1}{e^x}\) dx = ∫(e^x + e^-x) dx
= e^x – e^-x + C

Question 4.
(i) ∫\(\left(\frac{5}{\sqrt{1-x^2}}+\frac{7}{1+x^2}\right)\) dx
Solution:

(ii) ∫\(\frac{3 x^2}{x^2+1}\) dx
Solution:

(iii) ∫\(\frac{x^6}{x^2+1}\) dx
Solution:

(iv) ∫\(\frac{x^4+x^3+x^2+x+2}{x^2+1}\) dx
Solution:

(v) ∫\(\left(\sqrt{1-x^2}+\frac{x^2}{\sqrt{1-x^2}}\right)\) dx
Solution:

(vi) ∫\(\frac{x^2+\sqrt{x^2+1}}{x^3 \sqrt{x^2-1}}\) dx
Solution:

Question 5.
Find the unique antiderivative F(x) of f(x) = 2x² + 1, whese F(o) = -2.
Solution:
F(x) = ∫f(x) dx = ∫(2x² + 1) dx
= \(\frac{2}{3}\)x³ + x + C
But F(0) = -2
∴ -2 = C
Thus F(x) = \(\frac{2}{3}\)x³ + x – 2