CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Long Answer Questions

Odisha State Board CHSE Odisha Class 12 Sociology Solutions Unit 1 Introducing Indian Society Long Answer Questions.

CHSE Odisha 12th Class Sociology Unit 1 Introducing Indian Society Long Answer Questions

Long Questions With Answers

Question 1.
What do you mean by Unity in diversity? Discuss the various factors promoting unity in India?
Answer:
In spite of diversity, there is unity in India. There are bonds of unity may be located in certain mechanisms of integration Herbert Rosley census commissioner in 1911, was right when he observed “Beneath the manifold diversity of physical and social type, language, custom and religion which strikes the observer and in India.

there can still be discovered a certain underlying uniformity of life from the Himalayas to cape Cambrian. The concept of one united India has always been the fascinating idea of many great thinkers of this land. Since time immemorial India has been regarded as. one district society, unity in India. In the production of certain factors. We will now describe each of them.

Geographical Unity:
From a very early time. India has been visualised as one nation covering the whole landmass from the Himalayas to the ocean on the other sides. The Aryan thinkers and other philosophers named this landmass as Bharat Varsha. The term Bharat Varsha i.e., India has always referred to thus the vast expense of the land expanding from the Himalayas in North to the cap commotion on the South and from the Brahmputra in the East to the Indus in the west. These natural boundaries give the land geographical unity.

Religious Unity:
Although India in a land of many religions, religion has been one of the bases of national unity. As Srinivas says. The concept of the unity of India is essentially a religious one Hinduism being the religion of the majority of the people of India provides a basis for unity. Religious unity of the country in expressed through the existence of pilgrimage centres of great spiritual merits in the four parts of the country. Badrinaryan in the North. Rameswaram in the South, Dwarika in the West and Puri in the East eloquently speak the religious unity in India.

Cultural Unity :
India has cultural unity. The fundamental approaches to art, philosophy and literature are typically Indian in the character. Social institutions like the caste system and the joint family are found all over India which are once again topically Indian. Caste in such a widespread social institution in India that every Indian whether a Hindu, a Muslim or a Christian finds himself a universe of caste. These common social institutions and cultural traditions faster unity among the Indians.

Political Unity :
India has also political unity in India in the product of cultural and religious unity. The idea of bringing the whole country under one central authority is not new to Indians. Unification of vast areas and populations under one rule has been the highest goal of the kings statements. The ancient Indian political concept of Chakravarti clearly refers to the idea of political unification of the whole land. Many great kings right from Dilip, Yayati, Mandhata and Yudhisthira to Chandragupta Maurya.

Ashoka and Samudra Gupta of the later period had established there in sovereignty almost all over India the country and achieved the distinction of being a Chartkravai or universal ever lord. In the past kings often waged wars on one another with another purpose than to achieve thus indicating that religious support extended to the idea of political unification of the country.

Emotional Unity:
Last but not least there is an emotional bond that binds all the inhabitants of this vast land. The very name Bharat Mata emotionally brings all Indians together. Although linguistic diversity poses a threat to emotional integration, Sanskrit being the mother of all Indian languages serves as a significant untying bond in the emotional integration of the people.

A tradition of Independence:
We have had a remarkable tradition to independence which has held us together throughout countries. One manifestation is found in the form of the jamjar system. It is a system of independence of castes. The term Jajamani refers generally to a relationship between food-producing formally and the families that supported them with goods and services. These came to be called the Jajamani relations Jajamani system as the backbone of rural economy and social order under this system each caste group within a village is expected to give certain stand and used services to the families of other castes.

The tradition of Accommodation:
We have heard of the syncretic quality of Indian culture and its remarkable quality of accommodation and tolerance. Hinduism, which is the majority religion of India has an elastic character. We know that Hinduism as not a homogeneous religion. It is not a religion having one God, one book and one temple. It is described as a federation of faiths. goes to the extent of accommodating village-level deities of the tribal faiths. Hinduism has been open religion an encompassing religion.

CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Long Answer Questions

Question 2.
Write a brief essay on the “Unity in Diversity” in Indian society. Or, Indian Society essentially presents a “Unity in Diversity”.Discuss?
Answer:
India is not only the world’s largest but also probably the most complex federal democracy. While its democratic structure protects its political unity, its federal form guarantees the harmonious co-existence of socio-cultural diversities. That is why political unity is super imposed by the constitution and cherished by the major national parties.

The world has seldom known a country like India with an age-old socio-cultural diversity which gives a unique impression of pluralism. None of the federal politics, old or new, the bourgeoisie or socialist, Imperial Prussia, Austro-Hungarian Empire, Switzerland, United States, Canada, Nigeria, Malaysia, Yugoslavia or the Soviet Union are known to encompass such a wide range of distinct sub-national identities and that too with a long historical past as is the case with India.

Viewing the unity and its rich cultural heritage intellectuals say that India presents diversity in unity while others say that “India presents unity in diversity.” It is a fact that India is the second-largest state in the world in terms of population. It is the sixth-largest state in terms of territory. All the eight major religious communities of the world have a place in this benevolent motherland.

The problems of Indian society are very in number. Conflict takes place between states for sharing of river water. Still, there is a border dispute between different states. The terrorist and secessionist activities in different parts of India poses threat to Are very unity of India. The crossing of borders from alien countries creates very many problems in Indian society. There are several diversifying factors, among Communication, Casteism, Linguism and Regionalism are notable.

Communalism in Indian society refers to feelings of rivalry based on religious differences. We all know communalism was responsible for the division of the country into India and Pakistan in 1947. Even partition could not solve this problem in India because it still left a large Muslim minority in the country. The world’s major religious groups are found in India.

They are Hindus, Muslims, Christians, Sikhs, Buddhists, Jains, Parsis and Jews. Apart from the tribes who also constitute about 7% of the total Indian population, Hindus constitute an overwhelming majority which in percentages is about four times more than the total of all the other religious communities put together. Adherence to religion and the religious system is not communalism.

The exploitation of religion is communalism. Using a religious community against other communities and against nations is communalism. So, due to religious differences, sometimes conflicts and tensions are created among people of different communities. The political parties at the centre create a vote bank in favour of them depending upon religious groups. This appeasement policy of govt creates hatred feeling among other communities.

Now, there is not uniform civil code for all the nationals of India. Though there is a provision in the constitution in favour of a common civil code. Such a critical period is dawned that the national song for every people comes into question. Some of religious groups are getting assistance from outside to disturb national development and also to disintegrate Indian society.

Casteism is another serious problem in the way of national Integration. Though democracy and caste system both oppose each other in their principle yet it is seen that casteism is gradually strengthening in modem democratic India. Casteism is a feeling that creates a favourable attitude towards one’s own caste and hatred towards other castes. Thus, casteism is blind loyalty towards one’s own caste.

In other words, when the consideration of superiority between the castes and the tendency to consider the interest of one’s own caste as opposed to other castes is called casteism. Among other causes social inequality, social distance, illiteracy conservatism, development in the means of transportation and communication etc. are responsible for casteism.

It is also seen that many political parties are formed on the basis of caste in India encouraging caste loyalties to win the elections, which creates ill- feeling among different castes and hence hampers national unity. India is not only a multi-ethnic and multi-religious but also a multi-linguistic society. India, therefore, has rightly been called a “veritable tower of babel”.

Eighteen languages are recognized as the major languages of India, linguisticians are not unanimous about the total Languages spoken in India. All the eighteen languages listed in the eighth schedule of the constitution with the exception of Sanskrit. Indian states are reorganized on the basis of language. The real problem in India is that no link language has yet evolved. Though Hindi has been constitutionally recognized as the national language to replace English, its enforcement as of India appears to be problematic.

Being motivated by a group blindly one who raises his voice against the other language is called linguism. It is an anti-linguistic attitude or feeling of people towards other languages and a positive feeling towards their own language. Regionalism is another problem which poses a problem to the federal infrastructure of Indian society. Owing to its vastness, India is divided in to a number of regions and the territory of each region is determined on the basis of language.

Regionalism as we mean is the love and loyalty towards own region. Even people of one region try to develop their own region even at the cost of other regions. At other times it is seen that this feeling does not allow people from other regions to work and settle in their region. Thus regionalism militates against nationalism and thereby impedes the process of national integration. Regionalism refers to the sub-nationalism demanding preference of the region as against the country as a whole.

In spite of these diversities, there are other diversifying factors which pose a problem for national unity. But behind all in Indian society, the idea of unity is not something new to India. The concept of one unified India has always been the fascinating idea of many great thinkers of this land.

Geographical unity:
India, though very large in size possesses geographical unity with natural boundaries. It is surrounded on one side by the great Himalayas and on the other side by the high seas. Religious thinkers political philosophers, poets, statements and kings have always conceived Bharat Varsha as expanding from the Himalayas in the north to Cape Cameron in the south and from the Brahmaputra in the east to the Indus in the west.

Religious unity:
Though India is a land of many religions, some say that the concept of the unity of India is essentially a religious one. Hinduism is the religion of the majority of people of India and provides a basis for unity. The same myths, legends and deities are shared by all the Hindus in spite of their sectarian differences. Epics like Ramayan, Mahabharat and Bhagabat Gita are read by every people.

pilgrimage centres like Badrinarayan, Rameswaram, Dwarika and Puri four comers speak the religious unity of this vast land. The rivers like Ganga, Jamuna, Godavari etc. are regarded as sacred for every Hindu. Each Indian regard each inch of this Bharat Varsha as sacred.

Cultural unity :
India possesses cultural unity which runs through every aspect of Indian social life. The art, literature, philosophy, customs, traditions etc, are typically Indian in character. Social institutions like the caste system and joint family system are found all over the country. Every Indian, whether he is a Hindu, a Muslim, a Sikh, or a Buddhist finds himself in a universe of caste. Similarly, festivals are observed all over the country with a very similar fashion.

Political unity :
Political unity is not new to India. The idea of bringing the whole country under one central authority has always been a preoccupation with great kings and statesmen in India. The concept of “Chakravarti” clearly refers to this idea of the political unification of India under one authority. The prevalence of religious practices like the Aswamedha Yajna only indicates the religious support extended to the idea of political unification of India under one central authority.

Emotional unity:
There is an emotional bond in India that binds all the inhabitants of the land. The very Name ‘Bharatmata’ brings all Indians emotionally closer to one another. From the above discussion, it is very clear that in India there is an undercurrent of unity running through the apparent diversity of race, religion, language, customs etc. India is thus a fine example of unity in diversity.

CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Long Answer Questions

Question 3.
Discuss the Religious Composition of Indian Society?
Answer:
India is a country, where the world’s almost all major religions are found. Here are Hindus, Islam, Christianity, Sikhism, Buddhism, Zoroastrianism and Animism. Again each of these main religions has a number of sects of its own. A real man of God i. e. a. true believer in religion is also expected to be a good human being, and indeed most of the time.

In Indian people seem to be more loyal to their respective religions than to their nation. This religious diversity has been a factor and a source of dis- unity and disharmony in the country. It is also a fact that religious difference was responsible for the development of the two-nation theory and the consequent partition of the country into India and Pakistan.

The partition of India neither solved the Muslim minority problem nor it has created a homogeneous population in India from a religious point of view. The religious differences in Indian society sometimes create communalism which results in loss of life and property and poses problems in the way of national integration. Let us look at the composition of Indian society from a religious point of view.

Hinduism:
Hinduism is the religion of the majority of the people of India. Some also say that Hinduism is not a religion but a way of life. Anyone can adopt the way of life provided by Hinduism, for a better life. Hinduism is the amalgamation of Indo- Aryan, Dravidian and Pre- Dravidian religious, elements. Nearly about 83% of the Indian population are Hindus.

Hindus (the Caste Hindus and scheduled caste) constitute an overwhelming majority which in percentages is about four times more than the total of all the other religious communities put together. In fact, even the aggregate and the percentage of the caste Hindus alone is singly bigger than that of all the rest. Hinduism is not a monolithic religion, it allows a number of possible conceptions of God and also a variety of alternative ways of attaining union with God.

This does not mean that there is nothing common among the Hindus. There are certain beliefs which are shared by all Hindus irrespective of the sects they belong to. For example, every Hindu believes in the immortality of the soul, transmigration of the soul, law of Karma, Dharma, Moksha, Maya etc. Hinduism as a religion of the majority of people provides a basis for national unity.

Hinduism unites more than two-thirds of the Indian population by means of common Gods, common beliefs, common festivals etc. Lingayat, Kabirpanthi, Sakta, Radhaswamy, Satnami, Brahmasamaj and a host of other sects are a number of sects of Hinduism.

Islam:
Islam is the religion of Muslims. Muslims constitute about 13% of India’s total population. In undivided, India Muslims constituted nearly 24% of the total population. Muslims living in India are more in number than in Pakistan. Today, India has the second-largest Muslim population in the world. Muslims entered India towards the end of the twelfth century A.D.

They spread and settled in India under the political patronage of Muslim rulers. The number of Muslims gradually swelled mainly due to conversions from Buddhism and Hinduism, Majority of the Indian Muslims are the descendants of converts. Therefore, they are not very different from their Hindu neighbours Muslims are mainly of two sects. They are the Sunni and the Shia.

Christianity:
Christians constitute more than 2% of the population in India. In Kerala, their population is about 25% of the state’s population. Christians are widely scattered all over the country. In northern India, Christianity is mainly confined to certain sections of the tribal population and depressed castes. There are mainly three sects in Christianity. They are

  • Romo- Syrians,
  • Roman Catholics,
  • Protestants.

Sikhism:
The Sikh population in India is around 2% which is mainly concentrated in Punjab. Sikhism was founded by Guru Nanak in the 16th century A.D. The Sikhs are ideologically nearer to the Hindus than to Muslims. The Sikhs can easily be identified by anyone, because of the five ‘K’s they always wear. The five ‘K’s are – Kesh, Kanga, Kaccha, Kera and Kripan.

Buddhism:
The founder of Buddhism is Gautam. Buddhism originated in India during the 6th century B.C. Buddhism enjoyed royal patronage for a long period beginning from the great emperor Ashoka in the 3rd century B.C. As a result, Buddhism spread not only in India but also in countries outside India. Practically Buddhism lost its influence, by the 1 Oth century A.D.

The Buddhists represent only less than 1 % of the total population. Its main creed is Ahimsa. The Buddhists are found in Maharashtra as a result of the recent conversions under the leadership of Dr Ambedkar. Buddhism has two sects, namely- Hinayana and the Mahayana.

Jainism:
Jainism was established by Lord Mahavir, in India during 6th century B.C. It represents only a small portion of the Indian population i.e. 0.45% of our population. Jainism is very close to Hinduism. Many of the Hindu doctrines are retained in it. They worship cows and enter into Hindu temples. Jains are mainly urban people and are found in the towns and cities of Punjab, U.P., Rajasthan, Gujarat and Maharashtra. Jains are divided in to three sects namely.

  • Digambaras,
  • Svetambaras,
  • Dhundias.

Zoroastrianism or Parsi:
The Parsis or the followers of Zoroaster of Zarathushtra came to India in the 7th century A.D. from Persia in order to escape the forcible conversions to Islam. They worship fire. They expose their dead on the so-called ‘towers of silence to be eaten up by vultures so that the elements- earth, fire and water are not defiled by the contact of the dead matter. They are about 0.3% in number and half of which live in the city of Bombay alone. They are mainly urban and are on the top of the economic ladder of India.

Animism :
Animism is a tribal faith. It is a very primitive religion, according to which man is believed to be surrounded by a number of impersonal ghostly powers. These powers are said to reside in rocks, rivers, trees stones etc. In India, there are about 2 5 million people who believe in Animism.

CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Long Answer Questions

Question 4.
Discuss the Racial composition of Indian Society?
Answer:
Indian society is a multi-ethnic, multi-religious, multi-linguistic and multi-racial society. Its people worship differently. Its people have different faiths and different ideologies. The Indian society, the vast population is composed of people having diverse creeds, customs and colours. Let us see the racial composition of Indian society. A.W. Green says that “A race is a large biological human grouping with a number of distinctive inherited characteristics which vary within a certain range.

Similarly, A.L. Krober says that “A race is a valid biological concept. It is a group united by heredity, a breed or genetic strain or a sub-species. The Indian sub-continent received a large number of migratory races mostly from the western and eastern directions. Observing C.B. Memoria remarks that the “Indian population contains many primitive strains of mankind not found elsewhere to the same extent.

Different scholars have described the racial composition of Indian society. The first racial classification of the Indian population on scientific lines is probably of Sir Herbert Risely. He classified the Indian population into seven types. They are

  • Turko – Iranian
  • Indo-Aryan
  • Scytho-Dravidian
  • Arya-Dravidian
  • Mongolo-Dravidian
  • Mongoloid
  • Dravidian

Turko- Iranian:
People having this strain in their blood are mainly found in Beluchisthan and Afghanistan, which are now outside the political borders of India.

Indo- Aryan:
Indo- Aryan strain is mainly found in east Punjab, Raj as than and Kashmir, especially among the people belonging to the castes of Rajput, Khatri and Jat.

Scythe- Dravidian :
Scythe – Dravidian is a mixed racial type of Scythians and Dravidians. People having this racial ancestry are said to be found in Saurashtra, Coorg, and the hilly tracts of Madhya Pradesh. It is said that the upper strata people are of Scythians while the lower strata people are dominated by Dravidians.

Arya- Dravidian :
Arya- Dravidian racial type is an admixture of Indo- Aryan and Dravidian elements. The Aryan elements is more pronounced among the upper castes, especially Brahmins. The Dravidian element is more prominent among the Harijans and other lower-caste people.

Mongol – Dravidian:
Mongols- Dravidian racial type is the intermixture of the Dravidian and Mongolian races. The Brahmins and Kayasthas of Bengal and Odisha are believed to belong to this race.

Mongoloid:
Mongoloid racial element is mainly found among the tribal people of North-East Frontier and Assam.

Dravidian:
The people of south India and Madhya Pradesh are claimed to be of this stock. Most of anthropologists are not ready to accept the racial classification of Risley. D. N. Mazumdar says that the ‘Dravidian’ like the term ‘Aryan’ is a linguistic classification and not a racial classification. A.C. Haddon gives his own classification dis- regarding the classification of Risley.

According to him the principal races are

  • The Pre- Dravidian
  • The Dravidian
  • The Indo-Aryan
  • The Indo-Alpine
  • The Mongolian

J.N. Hutton also describes about the composition of the Indian population. But it is Dr B.S. Guha, after having revised the earlier classification has presented his own list of races that are believed to have composed the Indian population. His classifications are

The Negrito:
The presence of Negrito race in India is a controversial issue among anthropologists who say the presence of Negrito race say that they are relatively in pure form are still found in Andaman Island and also among some South Indian tribal people. Keeping these facts in view the protagonists of this view believe that the earliest occupants of India were Negritos, who were later displaced by Proto- Australoid.

But the antagonists of this view say that there is no weighty evidence to prove conclusively the existence ofNegrito element in the Indian population. So it may be safely said that Negrito race though existed in past, has left little trace in India today.

The Proto- Austroloid:
Indian tribal population is by and large dominated by this racial element. The Santals, the Juangs, the Koreas, the Soares, the Parjas, the Khonds, the Chenchus are a few of the many tribes of this race.

The Mongoloid:
The Mongoloid race came to India from Northwestern China, via- Tibet. People of this racial element are mainly found in North-Eastern India. This race consists of two fundamental types. They are

Palaeo- Mongoloid:
The Palaeo- Mongoloid consists of a long-headed type and of broad-headed type.

Tibeto- Monogoloid :
The people of Sikkim and Bhutan are said to belong to the Tibeto-Mongoloid race.

The Mediterranian:
The Mediterranian race is divided into three types. They are

Palaeo – Mediterranian:
This racial type is represented by the Tamil and Telugu Brahmins of the South.

The Mediterranian:
People of this racial type are believed to be the builders of the Indus valley civilization.

Oriental:
This race is very much similar to the Mediterranean type. The Mediterranean race as a whole, once a predominant race all over India, is now mainly concentrated in the south.

Western Brachycephals:
Western- Brachycephal race entered India from the west. The Alpinoid, the Dinaric and the Armenoid are the three main types of this race.

Alpinoid:
The people of Saurashtra, Gujarat and also Bengal are said to have of this race.

Dinaric:
This strain is claimed to be found among the people of Odisha, Bengal and Coorg.

Armenoid:
The Parsis of Bombay are believed to be the true representatives of this racial type.

Nordic Race :
People belonging to the Nordic race came to India from the North and spread all over Northern India during the 2nd millennium B.C. The people of this stock are believed to have enriched Indian culture by contributing new ideas to its philosophy and literature. From the above-discussed races, the first three races namely the Proto-Austroloid and the Mongoloid and Negrito constitute the Indian tribal population, while the other three races namely the Mediterranian, the Alpo- Dinaric and the Nordic, constitute the general population of India.

From the above discussion, it is very clear that the Indian population is composed of almost all the important races of the world. Today, there is no such race in India which is completely pure. All the races have got inter- mixed with one another. The intermixture of race is thorough that even in the same family we find one brother is quite fair while the other is quite dark. So India is rightly called a “Museum of races” or a “Melting pot of races”.

CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Long Answer Questions

Question 5.
India is a ‘Spectacle of Museum of tongues’- Explain. Or. Discuss the linguistic composition of Indian society?
Answer:
India is not only a multi-ethnic and multi-religious but also a polyglot society. Apart from English which is one of the major languages of administration, law higher academic teaching and research, Journalism and inter-regional communication, eighteen languages are recognized as the major languages of India. Linguists are not unanimous about the total number of languages spoken in India. Various enumerations are given in support of different criteria.

Dr George Grierson in the ‘Linguistic Survey of India’ tests 179 major languages and 544 dialects. The Linguistic Survey of India conducted in the first decade of this century recorded that “Language changes every 20 miles in India.” While the 1931 census lists more than 200 languages dialects, the 1961 census registered 1652 spoken languages of which 1549 were indigenous to India. Of these, about 572 covered almost 90% of the total population.

The multiplicity of languages creates new social cleavages in the already divided population of India by caste and creed. In the words of A.R. Desai, “India presents a spectacle of the museum of tongues.” Some also say that India is a ‘Veritable tower of babe. ’ We all know that all the eighteen languages listed in the Eighth Schedule of the constitution with the exception of Sanskrit. By 1966 partly by design and partly by the accident of political action, the linguistic formula acquired legitimacy as the primary criterion of state re¬organization in India.

These numerous Indian languages can however be grouped into four different speech families. They are

  • Indo-Aryan
  • The Dravidian
  • TheAustric
  • The Sino- Tibetan

Indo- Aryan Languages:
Hindi is the most important language, of all the languages belonging to the Indo- Aryan speech family. The Indian constitution recognizes Hindi as the national language which eventually is to replace English as the official link language on the lingua franca of India Hindi along with other allied languages like Hindusthani, Urdu, and Punjabi is claimed to be the mother tongue of 46.3% of India’s total population.

The people inhabiting the great Indo- Gangetic plains in Northern India speak this language in one form or the other. The other Indo-Aryan languages are Assamese, Bengali, Odia, Gujarathi, Marathi, and Kashmiri. These languages are spoken by the people of the states named after these languages. Hindi, unlike Urdu, borrowed words from Sanskrit and used Devanagari as its script. Hindi and Urdu were together known as Hindustani till 1947. These two languages are very similar in their spoken form, though they greatly differ in their written form.

At present Hindi is progressively being Sanskritized by dropping Persian words and incorporating itself more and more Sanskrit words. Punjabi is the language of mainly Sikh religious minorities. Though it resembles Hindi in its spoken form, it differs from it in its written form. Urdu freely borrowed a large number of words from the Persian language. It also adopted the Persian script as its own.

Dravidian Languages:
Tamil, Telugu, Kannada and Malayalam, the four southern languages belonging to the Dravidian speech family. Tamil is the language spoken by the people of Tamil Nadu who constitute 8.2% of our population. Telugu is the language of the people of Andhra Pradesh. It represents 10.2% of the total population of India and as such form the second-largest linguistic group in India after Hindi. The people of Karnataka speak Kannada, which constitutes 4.5% of the total population. Malayalam is the language spoken by the Keralites form 4.1% of our population.

Austria – Languages:
Most of the languages and dialects are spoken by the tribes of Central India belong to the Austric speech family.

Sino- Tibetan Languages :
The languages and dialects of North- Eastern tribal population are mainly of the Sino- Tibetan speech family. Nearly 3.2% of India’s total population speak one or the other of the 23 tribal languages belonging to either the Austric or the Sino-Tibetan speech families. At present India is divided into states along linguistic lines for the simple reason that most of these major languages are localized in different parts of the country.

While each language tends to concentrate in a particular part of the country, there is much overlapping of two or more languages, especially in areas near the borders. Therefore there are linguistic minorities in each of the linguistic states as well as in big cities. Viewing such linguistic composition of Indian society an American specialist in Indian languages says that villagers in the locality may find it difficult to understand the speech of their relatives living a hundred miles away from them.

CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Long Answer Questions

Question 6.
Discuss the Tribal composition of Indian society?
Answer:
India is claimed to have the largest total tribal population compared to any other single country in the world. The tribal population in India constitutes nearly 7% of the total population. The vast tribal population in India is not a homogeneous group. Rather it is composed of different tribes. According to the scheduled tribe’s list modification order 1956, there are 414 tribes in India. All these tribes are classified into different groups on various bases like geographical location, language, race and levels of economic and social development.

1. On the basis of geographical location:
On the band of geographical or territorial location tribes are divided into three main zones, namely

The North-North Eastern zone :
Eastern Kashmir, Eastern Punjab, Himachal Pradesh, North Uttar Pradesh, Nagaland, Assam etc. constitute this zone. The Aka, Mishmi, Chulikata, and Naga are some of the tribes of this zone.

Central Zone:
Bihar, Bengal, Southern, U.P., Southern Rajasthan, Madhya Pradesh and Odishacome are under this zone. The Savara, the Khond, the Santhal, the Bhil, the Gond, the Muria and the Bangla etc. live in this zone.

Southern Zone:
The Southern zone is comprised of the four Southern states, namely Andhra Pradesh, Tamilnadu, Karnataka and Kerala. The Toda, the Chenchu, the Kedar etc. are some of the important tribes of this zone.

Classification on the basis of Language:
The languages of the tribes of the three zones belong to three different speech families namely

Sino- Tibetan linguistic group :
Most of the tribes of the North- North- Eastern zone speak Sino- Tibetan languages in one form or other.

Austro- Asiatic linguistic group:
The dialects and languages spoken by most of the central zone belong to the Austric speech family. But some of the important tribes of this region like the Gonds and the Khonds speak languages having an affinity with the Dravidian speech family.

Dravidian Linguistic groups:
Tribal people in the Southern Zone in one form on other the Dravidian languages. Telugu, Tamil, Kannada and Malayalam.

Classification on the Basis of Race:
Indian tribes are broadly classified into three racial groups. They are.

The Mongoloid:
The North- North – Eastern zone tribes, in habiting the Himalayan regions belong to the one on the other type of Mongoloid race.

The Proto- Austroloid :
The central Indian tribes are mainly found to be Proto- Austroloid in their physical features.

The Negrito:
A few tribes like the Kadar of the extreme South and the Andamanese are said to be of this racial origin.

On the basis of cultural development:
Dr Elvin has classified Indian tribes into four classes on the basis of their levels of cultural development and contact with the plains.

Class -I:
The members of the tribes of class 1 are more or less isolated, and primitive, lead a communal life and cultivate with axes.

Class – II:
The people belonging to class – II are more individualistic, less simple, less honest, more used to outside life and less occupied with axe cultivation than the members of class-1 tribes.

Class – III:
People of this category are most exposed to external influence and they represent the largest section of the Indian tribal population. These people are already on the way of losing their tribal culture, religion and social organization as a result of external influences.

Class -IV:
The Bills and Nagas are members of this class of tribes. Dr Elvin believes that these people are the representatives of the old aristocracy of the country. Madan and Mazumdar have clarified the tribal communities into three main groups exclusively on the basis of cultural distance maintained from the rural-urban groups. The first group of tribes are culturally far away from the rural-urban groups. It includes all those tribes which have not come in contact with rural-urban groups in any significant way.

The second group consists of tribes which are in active contact with and under the direct influence of the outside rural-urban groups. As a consequence, these tribal communities are experiencing discomfort and problems. The third group consists of those tribes that have come into contact with the rural-urban groups but have not suffered from it. That is to say, these tribes have been successfully acculturated into the rural-urban way of life.

The tribal Welfare- Committee has suggested a classification of Indian tribes into four classes.
Tribal Communities:
These communities are by and large confined to their forest habitat and still lead a primitive way of life.

Semi-Tribal Communities:
The communities have come out of their forest habitat to settle in the rural areas and have taken up agricultural and other allied occupations.

Acculturate Tribal Communities:
These are the communities which have migrated to urban and semi-urban centres and have taken up occupations in modem industries and other works. They also have adopted many of the cultural traits of modem society.

Totally Assimilated Tribal Communities:
These are the tribes which have totally got assimilated into the modem way of life. An attempt has been made by Madan and Mazumdar to make a classification of Indian tribes on the basis of economic development. The first category of tribes are mainly dependent on forests for their livelihood and their economy is food gathering.

The tribes of the second category are those whose economy falls midway between food gathering and primitive agriculture. The economy of the third category is mainly agriculture but is supplemented by forest produce whenever and wherever possible. Finally, there is a new economic category of tribal men which is coming up in India with the growth of Industry. This new economic category includes those tribes who have been driven out of their traditional occupations and habitat and are employed in modern industry.

CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Short Answer Questions

Odisha State Board CHSE Odisha Class 12 Sociology Solutions Unit 1 Introducing Indian Society Short Answer Questions.

CHSE Odisha 12th Class Sociology Unit 1 Introducing Indian Society Short Answer Questions

Short Answer Type Questions

Question 1.
What is Unity?
Answer:
Unity means integration where in hitherto divisive people and culture are synthesized into a united whole. It connotes a sense of oneness, a sense of oneness. It stands for the bonds which hold the members of a society together.

Question 2.
What is diversity?
Answer:
Diversity means difference, however, means something more than more difference. It means collective differences that is differences which mark off one group of people from other. These differences may be biological, religious linguistic, etc.

Question 3.
Mention two factors of diversity.
Answer:
Mention two factors of diversity

  • Racial Diversity
  • Linguistic Diversity

Question 4.
Mention two factors of Unity.
Answer:

  • Geographical Unity
  • Religious Unity

5. Define race.
Answer:
A. W. Green says A race as a large biological human grouping with a number of distinctive inherited characteristics which vary within a certain range.

Question 6.
What is Racial Diversity?
Answer:
India is a museum of different races. Herbert Rosely had classified the people of India into seven racial types. These are Turko- Indian, Indo- Aryan, Scytho Dravidian Aryo – Dravidian, Mangolo Dravidian Mongolid and Dravidian.

Question 7.
What is Linguistic Diversity?
Answer:
According to A. R. Desai India presents as a spectacle of a museum of tongues. The famous linguist given dismissive noted that India has 179 languages and 544 dialects.

Question 8.
What is Religious Diversity?
Answer:
India is a land of many religions. There are also followers of various faiths particularly of Hinduism, Islam, Christianity, Buddhism, Jainism, and Zoroastrianism among others, etc.

Question 9.
Caste Diversity?
Answer:
Indis is a country of castes. The caste as peculiar to Indian society. There are more than 3,000 castes or Jati in India. These are hierarchically graded on different ways on different regions.

Question 10.
Geographical Unity?
Answer:
From very early times, India has been visualized as one nation covering the whole landmass from the Himalayan to the ocean on the other sides. The Aryan thinkers and other philosophers named this landmass as Bharatvarsha.

CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Short Answer Questions

Question 11.
Religious Unity?
Answer:
Although India as a land of many religious religions has been one of the basis of national unity. As Srinivas says. The concept of the unity of India is essentially a religious one Hinduism being the religion of the majority of the people of India provides a basis for unity.

Question 12.
Cultural Unity?
Answer:
India has cultural unity. The fundamental approaches to art, philosophy and literature and typically Indian on the character. Social institutions like the caste system and the joint family are found all over India which are once again typically Indian.

Question 13.
Political Unity?
Answer:
India has also political unity. Political unity in India as the product of cultural and religious unity. The Idea of being the whole country under one central authority as not new to Indians.

Question 14.
Emotional Unity?
Answer:
The emotional bond that binds all the inhabitants of this vast land. The very same Bharat Mata emotionally brings all Indians together. Although linguistic diversity poses a threat to emotional integration.

Question 15.
Tribal Composition of Ridia?
Answer:
The tribal population in India constitutes nearly 7% of the total population India is cleaned to have the largest total tribal population compared to any other single country on the world.

CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Short Answer Questions

Question 16.
Geographical Unity?
Answer:
From very early time India has been visualized as one nation covering the whole landmass from the Himalayans to the ocean on the other sides. The Aryan thinkers and other philosophers named this landmass as Bharat Varsha. The term Bharat Varsha i.e., India has always referred to this vast expense of the land expanding from the Himalayas in North to the cap camo rain on the south and from the Brahmaputra on the East to the Indus the west. These national boundaries given the land geographical unity.

Question 17.
Religious Unity?
Answer:
Although India is land of many religious religions has been one of the basis of national unity. As Srinivas says. The concept of the unity of India is monolithic essentially religious one Hinduism the religion of the majority of the people of India. Provides a basis for unity. It is a fact that Hinduism is not a religion, There are number seats of in Hinduism. But they have some faith in common with slight variation.

Question 18.
Cultural Unity?
Answer:
India has cultural unity. The fundamental approaches to art philosophy and literature are typically Indian in the character. Social institutions like the caste system and the joint family are found all over India which are once again topically Indian. Caste is such a social institution in India that every Indian whether a Hindu, a Muslim or a Christian finds himself a universe of caste. The common social institutions and cultural traditions foster unity among the Indians.

Question 19.
Political Unity?
Answer:
India has also political unity, political unity in India in the product of cultural and religious unity. The idea of bringing the whole country under one central authority is not new to Indians. Unification of vast area and populations under one rule has been the highest goal of the things and statesmen. The ancient Indian political unification of the whole land.

Many great things right from Dilip, Yayati, Mandhata and Yudhistira to Chandragupta Maurya, Ashoka and Samudra Gupta of later period had established theory sovereignty all most all over India the country and achieved the distinction of being a Chakravarti or universal overlord.

Question 20.
Emotional Unity?
Answer:
The emotional bond that binds all the inhabitants of this vast land. The very name Bharat Mata emotionality brings all Indians together. Although linguistic diversity poses a threat to emotional integration. Sanskrit belongs the mother of all Indian languages and serves as a significant unifying bond in the emotional integration of the people.

CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Short Answer Questions

Question 21.
What do you mean by Unity in diversity?
Answer:
Unity means integration. Where in hitherto divisive people and culture are synthesized into a united whole. It connotes a sense of oneness, a sense of we-ness. It stands for the bonds which hold the members of a society together. There is a difference between unity and uniformity, uniformity presupposes similarly, but unity does not.

Unity may be bom out of similarity. It many or many not be based in uniformity. Diversity means difference it however means something more than more difference. It means collective differences that as a difference which mark off one group of people from other.

Question 22.
Discuss various religions in India?
Answer:
Different religions have coexisted in India in past centuries. The history of our religious movements have been such that out us incorporated diverse faiths and values. Religion has been, on the one hand, the basis of national unity. On the other hand, religious diversity has been a factor of disunity and disharmony in out severity. India as a country where all the world’s major religions are found. In India are come across Hindus, Buddhists, Jains, Zeroastorms, Sikhs, Muslims and Christians etc.

Question 23.
Racial Composition?
Answer:
A race as a biological human grouping with a number of distinctive physical characteristics. A group of people are distinguished from other groups of people on the basis of physical characteristics such as completion height and shape of face etc. A such distinctive group of human beings is each called a race. These races are Aryan, Dravidian, Austroheld and Mongoloid, etc.

CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Short Answer Questions

Question 24.
Linguistic Groups?
Answer:
A says, A.R. Desai. India presents a spectacle of the museum of tongues. According to the 1951 census, there are 845 different languages and dialects spoken in India. But the actual number is estimated to be around 1000, George Grierson, in his linguistic. Survey of spoken alone in North India. While a few of these languages are spoken by crores of people a large number of them have less than a hundred thousand speakers each. All these Indian languages and dialects can be grouped into four speech families mainly.

  • Indo-Aryan,
  • Dravidian,
  • Austria, and
  • Sino – Tibetan, etc.

Question 25.
Tribal Groups?
Answer:
The tribes of India form an important part of the total population. The tribal population in India constitutes about 7% of the Indian population. India has the largest total tribal population compared to any other country in the world. The large tribal population of India is not a homogeneous group.

It is composed of a number of different tribes – our constitution recognized 212 tribes and these tribes are found in different parts of the country. According to the scheduled Tribes’ last modification order in 1956, there are 414 tribes in India. The tribes of India are classified into different groups on the basis of geographical location, language, race, and levels of socioeconomic development.

CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Objective Questions

Odisha State Board CHSE Odisha Class 12 Sociology Solutions Unit 1 Introducing Indian Society Objective Questions and Answers.

CHSE Odisha 12th Class Sociology Unit 1 Introducing Indian Society Objective Questions

Multiple Choice Questions With Answers

Question 1.
Does unity mean _____?
(a) Integration
(b) difference
(c) Uniformity
(d) Above all
Answer:
(a) Integration

Question 2.
Does diversity mean _____?
(a) Integration
(b) Difference
(c) Uniformity
(d) None of the above all
Answer:
(b) Difference

Question 3.
Which of the following forms of Diversity?
(a) Racial Diversity
(b) Linguistic Diversity
(c) Religious Diversity
(d) Above all
Answer:
(d) Above all

Question 4.
Which of the following Bonds of Unity?
(a) Geographical Unity
(b) Religious Unity
(c) Cultural Unity
(d) Above all
Answer:
(d) Above all

Question 5.
A Race is a ____?
(a) Biological human grouping
(b) A human grouping
(c) Complexion
(d) Above all
Answer:
(a) Biological human grouping

Question 6.
Who says that race is a valid Biological concept? it is a group united by heredity a breed or genetic strain or a subspecies?
(a) A.L. Krober
(b) FrazzBoas
(c) A. W. Green
(d) None of the above
Answer:
(a) A. L. Krober

Question 7.
Who says that In common parlance when we speak of race, we mean of a group of people that have certain blindly and perhaps also mental characteristics in common”?
(a) A.L. Krober
(b) FrazzBoas
(c) A. W. Green
(d) None of the above
Answer:
(b) Frazz Boas

Question 8.
Who says that “A race is a biological human grouping with a member of distinctive inherited characteristics which vary within a certain range”?
(a) A. L. Krober
(b) FrazzBoas
(c) A. W. Green
(d) Above all
Answer:
(c) A. W. Green

Question 9.
Which of the following HerbertRistey’s Racial classification?
(a) The Turko Iranian
(b) Indo Aryan
(c) saythodravidian
(d) Above all
Answer:
(d) Above all

Question 10.
Which of the following classification of A.C. Haddon?
(a) the pre-Dravidian Jungle tribes
(b) the Dravidians who are long-headed and brunette
(c) the Indo-Aryans who are fair-complexioned and long-headed
(d) Above all
Answer:
(d) Above all

CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Objective Questions

Question 11.
Which of the following is the classification of B.S Guha in the race?
(a) The Proto-Australoid
(b) Monogoloid
(c) TheTibetomongoloid
(d) Above all
Answer:
(d) Above all

Question 12.
Which of the following religious composition of India?
(a) Hinduism
(b) Islam
(c) Christians
(d) Above all
Answer:
(d) Above all

Question 13.
Which of the following are the linguistic Groups of India?
(a) Indo-Aryan
(b) Dravidriari
(c) A Rastrick
(d) Above all
Answer:
(d) Above all

Question 14.
Which of the following are the tribal groups in India?
(a) North-North Eastern zone
(b) Central Zone
(c) Southern Zone
(d) Above all
Answer:
(d) Above all

Question 15.
Which of the following the classification on the basis of language?
(a) The Sino Tibetan
(b) Austric Linguistic group
(c) The Dravidian Linguistic Group
(d) Above all
Answer:
(d) Above all

Question 16.
Which of the following the classification on the basis of race?
(a) The Mongoloid
(b) The Australoid
(c) TheNegreto
(d) Above all
Answer:
(d) Above all

Question 17.
Which of the following the classification on the basis of the socioeconomic level of Development?
(a) class-I
(b) class-II
(c) class -III
(d) Above all
Answer:
(d) Above all

Question 18.
Which of the following classification of Indian tribes?
(a) Tribal Communities
(b) Semi- Tribal Communities
(c) Accultarated Tribal Communities
(d) Above all
Answer:
(d) Above all

Question 19.
Who says that “Race refers to human stocks that are genetically distinguished to major physiological difference to a remote separation of ancestry”?
(a) Maclver
(b) A.W. Green
(c) D.N.Majumdar
(d) None of the above.
Answer:
(a) Maclver

Question 20.
Who says that “If a group of people who by their possession of a member of common physical traits can be distinguished from others, even of the members of this biological groups are widely scattered they form a race”?
(a) A. W. Green
(b) Maclver
(c) A.L. Krober
(d) D. N. Majumdar.
Answer:
(d) D.N. Majumdar

CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Objective Questions

Question 21.
Which of the following classification ofJ.H. Hutton?
(a) Negritos
(b) Proto-Australoid
(c) Pre-Dravidiran
(d) Above all
Answer:
(d) Above all

Question 22.
The 1961 census shows nearly of the Indian population as Hindusim?
(a) 86%
(b) 84%
(c) 85%
(d) 90%
Answer:
(b) 84%

Question 23.
Which of the following classification of tribes on the basis of geographical location?
(a) North-North Eastern Zone
(b) Central Zone
(c) Southern Zone
(d) Above all
Answer:
(d) Above all

Question 24.
Which of the following the classification ofB.S.Guha?
(a) The Negritos
(b) The Proto Australoids
(c) TheMongoids
(d) Above all
Answer:
(d) Above all

Question 25.
Who says that Persia as the land of light Greece the land of grace India- the land of dreams, Rome the land of empire, doubt India is the land of dreams – a dream that strives to blend inextricably her various similarities and differences?
(a) Hegel
(b) Guha
(c) Huddem
(d) None of above
Answer:
(a) Hegel

CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Objective Questions

One Word Answers

Question 1.
unity mean ________?
Answer:
Unity means integration

Question 2.
Diversity means _________?
Answer:
Diversity means difference

Question 3.
Mentions one forms of Diversity in India?
Answer:
Racial Diversity

Question 4.
Mentions one Bonds of Unity in India?
Answer:
Geographical Unity

Question 5.
Mentions one racial composition of Indian Society?
Answer:
The Tunku Iranian.

Question 6.
Mentions one religion’s composition?
Answer:
Hinduism

Question 7.
Mentions one linguistic Group?
Answer:
Dravidian Languages.

Question 8.
Mentions one tribal Group?
Answer:
North-North Eastern Zone.

Question 9.
Mentions one on the basis of language?
Answer:
The Sino-Tibetan

Question 10.
Mentions one on the basics of Race?
Answer:
The Mongoloid

CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Objective Questions

Question 11.
Mentions one classification on the basis of the socio-economic level of Development?
Answer:
The tribes of class-I

Question 12.
Mentions one of the classifications of Haddon, race?
Answer:
The Pre-Dravidian.

Question 13.
Mention one classification of the J.H. button of race?
Answer:
Mongoloids.

Question 14.
Mention one of the Dr B.S. Guha of race ________?
Answer:
The Negrito.

Question 15.
Mention one of the classification of tribes on the basis of geographical location?
Answer:
North – North – Eastern zone.

Question 16.
Mention one contributing to the unity of India?
Answer:
Religious Unity

Question 17.
Does mention 1971 census show a total languages spoken in India?
Answer:
1108

Question 18.
Who says that A race as a large biological human grouping with a number of distinctive inherited characteristics which very within a certain range?
Answer:
A. W. Green

Question 19.
Who said that Race reforms to human stocks that are genetically distinguished to major physiological differences to a remote separation of ancestry?
Answer:
Maclver

Question 20.
Who says that race as a valid biological concept? It is a group united by hereditary, a breed or genetic strain or a sub-species?
Answer:
A. L. Krober

CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Objective Questions

Question 21.
Who says that “If a group of people who by their possession of a number of common physical traits can be distinguished from others, even of the members of thus biological group are widely scattered they form a race”?
Answer:
D.N. Majumdar.

Correct The Sentences

1. Unity means difference?
Answer:
Unity means integration.

2. Diversity means integration?
Answer:
Diversity means difference.

3. India is a museum of the different region?
Answer:
India is a museum of different race?

4. A. R. Desai India presents a spectacle of the museum of tongues?
Answer:
right of all

5. A. L. Krober says The sale a large group of people is intensified by inherited physical differences?
Answer:
BiesazandBiesanz

6. A. W. Green says A race as a biological concept. It is a group united by heredity a breed or genetic strown or sub-species?
Answer:
A.L. Kroeber says a race is a biological concept It is a group united by hereditary a breed or genetic strain or subspecies.

7. Biesaz and Biesanz say in common when we speak of race, are mean of group of people that have certain bodily and perhaps also mental characteristics on common?
Answer:
Frazz Boas says in common parlance when we speak of race we mean of a group of people that have certain bodily and perhaps also mental characteristics in common.

8. A. L. Krober A race as a biological human grouping with a number or distinctive inherited characteristics which very which a certain range?
Answer:
A. W. Green A race as a biological human group with a number of distinctive inherited characteristics which vary within a certain range.

9. A race is a psychological human grouping with a number of distinctive physical characteristics?
Answer:
A race is a biological human grouping with a number of distinctive physical characteristics.

10. India is a Tower port of races?
Answer:
India as a melting pot of races.

CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Objective Questions

11. A. R. Desai presents a spectacle of the museum of tongues?
Answer:
Right

12. Animism as a Hinduism faith?
Answer:
Animism is atribal faith.

13. Joins the followers of lord Mahavira?
Answer:
Right

14. Sikhism is founded by Mahama?
Answer:
Sikhism is founded by Guru Nanak.

15. Zoroastrians are followers of Zoroaster?
Answer:
Right

16. Budhism emerged on India during the 7th century B.C Gautam?
Answer:
Buddism emerged in India during the 6th century B.C. Gautam.

17. Islam as the religion of Muslims. Did it extend in India towards the end of the 10th century A.D?
Answer:
Islam as the religion of Muslims. It extended in India towards the end of the 12th century A.D.

18. Muslim in India constitute nearly 12% of the total population?
Answer:
Muslims in India constitutes nearly 10% of the total population.

19. Christians constitute more than 5% of the Indian population?
Answer:
Christians constitute more than 2% of the Indian Population.

20. Tribal population in India constitutes about 10% of the Indian population?
Answer:
The tribal population in India constitutes about 7% of the Indian population.

CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Objective Questions

21. Constitutions recognize 210 tribes and these tribes are found in different parts of the country?
Answer:
Constitutions recognize 212 tribes and these tribes are found in different points of the country.

22. Scheduled Tribes’ last modification order 1956, therefore 414 tribes in India?
Answer:
Right

Fill In The Blanks.

1. Unity means ______.
Answer: Integration

2. Diversity means ______.
Answer: difference

3. India is a ______ of different races.
Answer: Museum

4. A. R. Desai India presents a spectacle of ______ of tongues.
Answer: Museum

5. India is a _____ of many religions.
Answer: Lord

6. India is a country of ______.
Answer: Castes

7. A race is a _____ human grouping with a number of distinctive physical characteristics.
Answer: biological

8. race is a valid ______ concept.
Answer: biological

9. _____ religions have coexisted in India in part centuries.
Answer: Different

10. Zoroastrians are followers of _______.
Answer: Zoroaster

CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Objective Questions

11. Sikhism was founded by _______.
Answer: GuruNanak

12. Jains the followers of ______.
Answer: LordMahavir

13. Buddhism emerged in India during ______ Century B.C. Gautam.
Answer: 6th

14. Christians Constitute more than _____ of the Indian population.
Answer: 2%

15. Muslim in India Constitute nearly ______ of the total population.
Answer: 10%

16. Islam is the ______ of Muslims.
Answer: Religious

17. Majority of the people of India are ______.
Answer: Hindus.

18. A.R. Desai India presents a ______ of the museum of tongues.
Answer: Spectacle

19. Tribal population in India constitutes about 7% of the ______ population.
Answer: Indian

20. Our Constitution recognizes 212 tribes and these tribes are founded in different parts of the ______.
Answer: Country

CHSE Odisha Class 12 Sociology Unit 1 Introducing Indian Society Objective Questions

21. Tribes Cast modification order 1956 there are _______ tribes in India.
Answer: 414

22. Varrier Elvira has classified Indian tribes into ______classes.
Answer: Four

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Ex 1(a)

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 1 Relation and Function Ex 1(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Exercise 1(a)

Question 1.
If A = {a,b,c,d} mention the type of relations on A given below, which of them are equivalence relations?
(i) {(a, a), (b, b)}
(ii) {(a, a), (b, b), (c, c), (d, d)}
(iii) {(a, b), (b, a), (b, d), (d, b)}
(iv) {(b, c), (b, d), (c, d)}
(v) {(a, a), (b, b), (c, c), (d, d), (a, d), (a, c), (d, a), (c, a), (c, d), (d, c)}
Solution:
(i) Symmetric and transitive but not reflexive.
(ii) Reflexive, symmetric as well as transitive. Hence it is an equivalence relation.
(iii) Only symmetric
(iv) Only transitive
(v) Reflexive, symmetric and transitive. Hence it is an equivalence relation.

Question 2.
Write the following relations in tabular form and determine their type.
(i) R = {(x, y) : 2x – y = 0] on A = {1,2,3,…, 13}
(ii) R = {(x, y) : x divides y} on A = {1,2,3,4,5,6}
(iii) R = {(x, y) : x divides 2 – y} on A = {1,2,3,4,5}
(iv) R = {(x, y) : y ≤, x ≤, 4} on A = {1,2,3,4,5}.
Solution:
(i) R = {(x, y) : 2x- y = 0} on A
= {(x, y) : y = 2x} on A
= {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10), (6, 12)}
R is neither reflexive nor symmetric nor transitive.

(ii) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1,5), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (5,5), (6, 6)}
R is reflexive transitive but not symmetric.

(iii) R = {(x, y) : x divides 2 – y} on A
= {1, 2, 3, 4, 5}
= {(x, y) : 2-y is a multiple of x}
= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 4), (3, 2), (3, 5), (4, 2), (5, 2)}
R is neither reflexive nor symmetric nor transitive.

(iv) R = {(x, y) : y ≤ x ≤ 4} on A
= {1, 2, 3, 4, 5}
= {(1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (4, 4)}
R is neither reflexive nor symmetric but transitive.

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Ex 1(a)

Question 3.
Test whether the following relations are reflexive, symmetric or transitive on the sets specified.
(i) R = {(m,n) : m-n ≥ 7} on Z.
(ii) R = {(m,n) : 2|(m+n)} on Z.
(iii) R = {(m,n) : m+n is not divisible by 3} Z.
(iv) R = {(m,n) : is a power of 5} on Z – {0}.
(v) R = {(m,n) : mn is divisible by 2} on Z.
(vi) R = {(m,n) : 3 divides m-n} on {1,2,3…,10}.
Solution:
(i) R = {{m, n) : m- n ≥ 7} on Z
Reflexive:
∀ m ∈ Z, m – m = 0 < 7
⇒ (m, m) ∉ R
Thus, R is not reflexive.
Symmetry:
Let (m, n) ∈ R
⇒ m – n ≥ 7
⇒ n – m < 7
∴ (n, m) ∉ R
⇒ R is not symmetric.
Transitive:
Let (m, n), (n, p) ∈ R
m – n ≥ 7
and n – p > 7
⇒ m – p ≥ 7
⇒ (m, p) ∈ R
⇒ R is transitive.

(ii) R = {(m, n) : 2 | (m + n)} on Z
Reflexive:
∀ m ∈ Z, m + m = 2m
which is divisible by 2.
⇒ 2 | (m + m)
⇒ (m, m) ∈ R
⇒ R is reflexive.
Symmetry:
Let (m, n) ∈ R
⇒ 2 | (m + n)
⇒ 2 | (n + m)
(n, m) ∈ R
⇒ R is symmetric.
Transitive:
Let (m, n), (n, p), ∈ R
⇒ 2 | (m + n) and 2 | (n + p)
⇒ m + n = 2k1
⇒ n + p = 2k2
⇒ m + 2n + p = 2k1 + 2k2
⇒ m + p = 2(k1 + k2 – 1)
⇒ 2 | (m + p)
⇒ (m, p) ∈ R
⇒ R is transitive.
Thus, R is an equivalence relative.

(iii) R = {(m, n) : m + n is not divisible by 3} on Z
Reflexive:
As 3 + 3 is divisible by 3
we have (3, 3) ∉ R
⇒ R is not reflexive.
Symmetric:
Let (m, n) ∈ R
⇒ m + n is not divisible by 3
⇒ n + m is not divisible by 3
⇒ (n, m) ∈ R
⇒ R is symmetric.
Transitive:
(3, 1), (1, 6) ∈ R
But (3, 6) ∉ R
⇒ R is not transitive.

(iv) R = {(m, n) : \(\frac{m}{n}\) is a power of 5} on Z – {0}
Reflexive:
∀ m ∈ Z – {0}
\(\frac{m}{m}\) = 1 = 5°
⇒ (m, m) ∈ R
⇒ R is reflexive.
Symmetric:
Let (m, n) ∈ R
\(\frac{m}{n}\) = 5k
\(\frac{n}{m}\) = 5-k
⇒ (n, m) ∈ Z
⇒ R is symmetric.
Transitive:
Let (m, n), (n, p) ∈ R
⇒ \(\frac{m}{n}\) = 5k1 , \(\frac{n}{p}\) = 5k2
⇒ \(\frac{m}{n}\) . \(\frac{n}{p}\) = 5k1 . 5k2
⇒ \(\frac{m}{p}\) = 5 k1+k2
⇒ (m, p) ∈ R
⇒ R is transitive.
Thus R is an equivalence relation.

(v) R = {(m, n) : mn is divisible by 2} on Z
Reflexive:
3 ∈ Z
3 x 3 = 9
which is not divisible by 2.
∴ (3, 3) ∉ R
⇒ R is not reflexive.
Symmetric:
Let (m, n) ∈ R
⇒ mn is divisible by 2
⇒ nm is divisible by 2
⇒ (n, m) ∈ R
⇒ R is symmetric.
Transitive:
⇒ (3, 2), (2, 5) ∈ R
⇒ But 3 x 5 = 15,
⇒ which is not divisible by 2.
⇒ (3, 5) ∉ R
R is not transitive.

(vi) R = {(m, n) : 3 divides m-n} on A = {1, 2, 3……,10}
Reflexive:
Clearly ∀ m ∈ A, m – m = 0
which is divisible by 3
⇒ (m, m) ∈ R
⇒ R is reflexive
Symmetric:
Let (m, n) ∈ R
⇒ m – n is divisible by 3
⇒ n – m is also divisible by 3
⇒ (n, m) ∈ R
⇒ R is symmetric
Transitive:
Let (m, n), (n, p) ∈ R
⇒ m – n and n – p are divisible by 3
⇒ m – n + n – p is also divisible by p.
⇒ m – p is divisible by p.
⇒ (m, p) ∈ R
⇒ R is transitive.
Thus R is an equivalence relation.

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Ex 1(a)

Question 4.
List the members of the equivalence relation defined by the following partitions on X= {1,2,3,4}. Also find the equivalence classes of 1,2,3 and 4.
(i) {{1},{2},{3, 4}}
(ii) {{1, 2, 3},{4}}
(iii) {{1,2, 3, 4}}
Solution:
(i) The equivalence relation is
R = {(1, 1), (2, 2), (3, 3), (4, 4), (3, 4), (4, 3)}
[1] = {1}, [2] = {2}, [3] = {3, 4} and [4] = {3, 4}

(ii) The equivalence relation is
R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)}
[1] = [2] = [3] = {1, 2, 3}
[4] = {4}

(iii) The equivalence relation is
R = A x A, [1] = [2] = [3] = [4] = A

Question 5.
Show that if R is an equivalence relation on X then dom R = rng R = X.
Solution:
Let R is an equivalence relation on X.
⇒ R is reflexive
⇒ (x, x) ∈ R ∀ x ∈ X
⇒ Dom R = Rng R = X

Question 6.
Give an example of a relation which is
(i) reflexive, symmetric but not transitive.
(ii) reflexive, transitive but not symmetric.
(iii) symmetric, transitive but not reflexive.
(iv) reflexive but neither symmetric nor transitive.
(v) transitive but neither reflexive nor symmetric.
(vi) an empty relation.
(vii) a universal relation.
Solution:
(i) The relation R = {(a, b), (b, a), (a, c), (c, a), (a, a), (b, b), (c, c)} defined on the set {a, b, c} is reflexive, symmetric but not transitive.
(ii) “The relation x ≤ y on z” is reflexive, transitive but not symmetric.
(iii) The relation R = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)} defined on the set {a, b, c, d} is symmetric, transitive but not reflexive.
(iv) The relation R = {(a, a), (b, b), (c, c), (a, b), (b, c)} defined on the set A = {a, b, c} is reflexive but neither symmetric nor transitive.
(v) R = {(a, b), (b, c), (a, c)} on A = {a, b, c} is transitive but neither reflexive nor symmetric.
(vi) On N the relation R= {(x, y) : x + y = – 5} is an empty relation.
(vii) On N the relation R = {(x, y) : x + y > 0} is an universal relation.

Question 7.
Let R be a relation on X, If R is symmetric then xRy ⇒ yRx. If it is also transitive then xRy and yRx ⇒ xRx. So whenever a relation is symmetric and transitive then it is also reflexive. What is wrong in this argument?
Solution:
Let R is a relation on X.
If R is symmetric then xRy ⇒ yRx
If R is also transitive then xRy and yRx ⇒ xRx
⇒ Whenever a relation is symmetric and transitive, then it is reflexive. This argument is wrong because the symmetry of R does not imply dom R = X and for reflexive xRx ∀ x ∈ X.

Question 8.
Suppose a box contains a set of n balls (n ≥ 4) (denoted by B) of four different colours (may have different sizes), viz. red, blue, green and yellow. Show that a relation R defined on B as R={(b1, b2): balls b1 and b2 have the same colour} is an equivalence relation on B. How many equivalence classes can you find with respect to R?
[Note: On any set X a relation R={(x, y): x and y satisfy the same property P} is an equivalence relation. As far as the property P is concerned, elements x and y are deemed equivalent. For different P we get different equivalence relations on X]
Solution:
On B, R = {(b1, b2) : balls b1 and b2 have the same colour}

Reflexive:
∀ b ∈ B, b and b are of same colour
⇒ (b, b) ∈ R
⇒ R is reflexive.

Symmetric:
Let (b1, b2) ∈ R
⇒ b1 and b2 are of same colour
⇒ b2 and b1 are of same colour
⇒ (b2, b1) ∈ R
⇒ R is symmetric.

Transitive :
Let (b1, b2) and (b2, b3) ∈ R
⇒ b1 and b2 are of same colour
b2 and b3 are of same colour
⇒ b1, b3 are of same colour
⇒ (b1, b3) ∈ R
⇒ R is transitive
∴ R is an equivalence relation.
As there are 4 types of balls there are 4 equivalence relations with respect to R.

Question 9.
Find the number of equivalence relations on X={1,2,3}. [Hints: Each partition of a set gives an equivalence relation.]
Solution:
Method – 1: Number of equivalence relations on a set A with | A | = n.
= The number of distinct partitions of A
= Bn
where Bn+1 = \(\sum_{k=0}^n \frac{n !}{k !(n-k) !} \mathrm{B}_k\)
with B0 = 1
Here n = 3
B1 = 1
B2 = \(\frac{1 !}{0 ! 1 !}\) B0 + \(\frac{1 !}{1 ! 1 !}\) B1
= 1 + 1 = 2
B3 = \(\frac{2 !}{0 ! 2 !}\) B0 + \(\frac{2 !}{1 ! 1 !}\) B1 + \(\frac{2 !}{2 ! 0 !}\) B2
= 1 + 2 + 2 = 5
Thus there are 5 equivalence relations.

Method – 2:
X= {1, 2, 3}
Number of equivalence relations = number of distinct partitions.
Different partitions of X are
{{1} {2}, {3}}
{{1}, {2, 3}}, {{2}, {1,3}},
{{3}, {1,2}} and {{1, 2,3}}
Thus number of equivalence relations = 5.

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Ex 1(a)

Question 10.
Let R be the relation on the set R of real numbers such that aRb iff a-b is an integer. Test whether R is an equivalence relation. If so find the equivalence class of 1 and ½ w.r.t. this equivalence relation.
Solution:
The relation R on the set of real numbers is defined as
R = {(a, b) : a – b ∈ Z}

Reflexive:
∀ a ∈ R (set of real numbers)
a – a = 0 ∈ Z
⇒ (a, a) ∈ R
⇒ R is reflexive.

Symmetric:
Let (a, b) ∈ R
⇒ a – b ∈ Z
⇒ b – a ∈ Z
⇒ (b, a) ∈ R
⇒ R is symmetric.

Transitive:
Let (a, b), (b, c) ∈ R
⇒ a – b and b – c ∈ Z
⇒ a – b + b – c ∈ Z
⇒ a – c ∈ Z
⇒ (a, c) ∈ R
⇒ R is transitive.
Thus R is an equivalence relation.
[1] = {x ∈ R : x -1 ∈ Z} = Z
\(\begin{aligned}
{\left[\frac{1}{2}\right] } &=\left\{x \in \mathrm{R}: x-\frac{1}{2} \in \mathrm{Z}\right\} \\
&=\left\{x \in \mathrm{R}: x=\frac{2 k+1}{2}, k \in \mathrm{Z}\right\}
\end{aligned}\)

Question 11.
Find the least positive integer r such that
(i) 185 ∈ [r]7
(ii) – 375 ∈ [r]11
(iii) -12 ∈ [r]13
Solution:
(i) 185 ∈ [r]7
⇒ 185 – r = 7k, k ∈ z and r < 7
⇒ r = 3
(ii) – 375 ∈ [r]7
⇒ – 375 – r = 11k, k ∈ z and r < 11
⇒ r = 10
(iii) – 12 ∈ [r]13
⇒ – 12 – r = 13k, k ∈ z and r < 13
⇒ r= 1

Question 12.
Find least non negative integer r such that
(i) 7 x 13 x 23 x 413 r (mod 11)
(ii) 6 x 18 x 27 x (- 225) = r (mod 8)
(iii) 1237(mod 4) + 985 (mod 4) = r (mod 4)
(iv) 1936 x 8789 = r (mod 4)
Solution:
(i) 7 x 13 x 23 x 413 ≡ r (mod 11)
Now 7 x 13 ≡ 3 mod 11
23 ≡ 1 mod 11
413 ≡ 6 mod 11
∴ 7 x 13 x 23 x 413 ≡ 3 x 1 x 6 mod 11
≡ 18 mod 11
≡ 7 mod 11
∴ r = 7

(ii) 6 x 18 x 27 x – 225 ≡ r (mod 8)
Now 6 x 18 ≡ 108 = 4 mod 8
27 ≡ 3 mod 8
– 225 ≡ 7 mod 8
⇒ 6 x 18 x 27 x – 225 ≡ 4 x 3 x 7 mod 8
≡ 84 mod 8
≡ 4 mod 8
∴ r = 4

(iii) 1237 (mod 4) + 985 (mod 4) r (mod 4)
Now 1237 ≡ 1 mod 4
985 ≡ 1 mod 4
⇒ 1237 (mod 4) + 985 (mod 4)
≡ (1 + 1) mod 4
≡ 2 mod 4
⇒ r = 2

(iv) 1936 x 8789 ≡ r (mod 4)
1936 x 8789 ≡ 0 mod 4
∴ r = 0

Question 13.
Find least positive integer x satisfying 276x + 128 ≡ (mod 7)
[Hint: 276 ≡ 3, 128 ≡ 2 (mod 7)]
Solution:
Now 128 ≡ 2 mod 7
Now 176 x + 128 ≡ 4 mod 7
⇒ 176 x ≡ (4 – 2) mod 7
⇒ 176 x ≡ 2 mod 7
176 x x ≡ 2 mod 7,
But 276 ≡ 3 mod 7
Thus x = 3.

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Ex 1(a)

Question 14.
Find three positive integers xi, i =1, 2, 3 satisfying 3x ≡ 2 (mod 7)
[Hint: If X1 is a solution then any member of [X1] is also a solution]
Solution:
3x ≡ 2 mod 7
Least positive value of x ≡ 3
Each member of [3] is a solution
∴ x = 3, 10, 17 …..

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Ex 1(b)

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 1 Relation and Function Ex 1(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Exercise 1(b)

Question 1.
Let X={x, y} and Y={u, v}. Write down all the functions that can be defined from X to Y. How many of these are (i) one-one (ii) onto and (ii) one-one and onto?
Solution:
The functions from X = {x, y} to y = {u, v} are:
f1 = {(x, u), (y, v)}
f2 = {(x, v), (y, u)}
f3 = {(x, u), (y, u)}
f4 = {(x, v), (y, v)}

Out of these 4 functions there are:
(i) 2 one-one functions
(ii) 2 onto functions
(iii) 2 one-one and onto function.

Question 2.
Let X and Y be sets containing m and n elements respectively.
(i) What is the total number of functions from X to Y.
(ii) How many functions from X to Y are one-one according as men, m > n and m = n?
Solution:
If | x | = m and | y | = n then
(i) Number of functions =nm
(ii) If m < n then number of one-one functions = nPm.
If m > n then number of one-one functions = 0
If m = n then number of one-one functions = m!

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Ex 1(b)

Question 3.
Examine each of the following functions if it is
(i) injective (ii) surjective, (iii) bijective and (iv) none of the three
(a) f : R → R, f(x) = x²
(b) f : R → [-1, 1], f(x) = sin x
(c) f : R+ → R + , f(x) = x + 1/x
where R+ = {x ∈ R : x > 0}
(d) f : R → R, f(x) = x³ + 1
(e) f : (-1, 1) → R, f(x) = \(\frac{x}{1-x^2}\)
(f) f : R → R, f(x) = [x] = the greatest integer ≤ x.
(g) f : R → R, f(x) = | x |
(h) f : R → R, f(x) = sgn x
(i) f : R → R, f = idR = the identity function on R.
Solution:
(a) f : R → R, f(x) = x²
for x1, x2 ∈ R
Let f(x1) = f(x2)
⇒ x1² = x2²
⇒ x = ± x2
∴ f is not one-one.
Hence f is not injective or bijective.
Rng f = [0, ∞) ≠ R
∴ f is not surjective.

(b) f : R → [-1, 1], f(x) = sin x
For x1 , x2 ∈ R
let f(x1) = f(x2) ⇒ sin x1 = sin x2 
⇒ x1 = nπ + (- 1)n  x2
⇒ x1 = x2 (not always)
∴ f is not injective and also not bijective.
But f is onto, as ∀ y ∈ [-1, 1]
there is a x ∈ R such that f(x) = sin x.
i.e., f is surjective.

(c) f : R+ → R+ , f(x) = x + \(\frac{1}{x}\)
f(2) = 2 + \(\frac{1}{2}\) = \(\frac{5}{2}\)
f(\(\frac{1}{2}\)) = \(\frac{1}{2}\) + 2 = \(\frac{5}{2}\)
f(2) = f(\(\frac{1}{2}\))
but 2 ≠ \(\frac{1}{2}\)
∴ f is not injective (one-one).
Again, for 1 ∈ R+ (domain)
⇒ there is no x ∈ R+(Dom)
such that x + \(\frac{1}{2}\) = 1
∴ f is not onto.

(d) f : R → R, f(x) = x³ + 1
for x1, x2 ∈ R
Let f(x1) = f(x2)
⇒ x13 = x23
⇒ x1 = x2
∴ f is injective.
Let f(x) = y ⇒ y = x³ + 1
⇒ x³ = y – 1
⇒ x = (y – 1)1/3 which exists ∀ y ∈ R
∴ f is onto.
∴ f is bijective.

(e) f : (-1, 1) → R, f(x) = \(\frac{x}{1-x^2}\)
for x1, x2 ∈ (-1, 1)
Let f(x1) = f(x2) ⇒ \(\frac{x_1}{1-x_1^2}\) = \(\frac{x_2}{1-x_2^2}\)
⇒x1 – x1 x22 = x2 – x12 x2
⇒ x1 – x2 + x12 x2 – x1 x22 = 0
⇒ (x1 – x2) (1 + x1 x2) = 0
⇒ x1 = x2 (for x1 x2 ∈ (-1, 1) x1 x2 ≠ -1)
∴ f is injective.
Again let y = \(\frac{x}{1-x^2}\) ⇒ y – x²y = x
⇒ x²y + x – y = 0
⇒ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{-1 \pm \sqrt{1+4 y^2}}{2 y}\) ∉ (-1, 1) ∀ y ∈¸.
∴f is surjective.
∴f is bijective.

(f) f : R → R
f(x) = [x]
f(1.2) = f(1.5) =1
∴ f is not injective.
Rng f = Z ⊂ R
∴ f is not surjective.
∴ Hence it is not also bijective.

(g) f : R → R
f(x) = | x |
As f(-1) = f(1) = 1
∴ f is not injective.
Again Rng f = [0, ∞) ⊂ R
⇒ f is not surjective.
Thus f is not bijective.

(h) f : R → R
f(x) = Sgn (x) = \(\left\{\begin{array}{cc}
1, & x>0 \\
0, & x=0 \\
-1, & x<0
\end{array}\right.\)
As f(1) = f(2) = 1
We have f is not injective.
Again Rng f = {- 1, 0, 1} ≠ R
∴ R is not surjective.
⇒ R is not bijective.

(i) f : R → R
f = idx
∴ f(x) = x
for x1, x2 ∈ R
Let f(x1) = f(x2) where x2 x2 ∈ R
⇒ x1 = x2
∴ f is one-one.
Again Rng f = R (codomain)
∴ f is onto.
Thus f is a bijective function.

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Ex 1(b)

Question 4.
Show that the following functions are injective.
(i) f(x) = sin x on \(\left[0, \frac{\pi}{2}\right]\)
(ii) f(x) = cos x [0, π]
(iii) fix) = loga x on (0, ∞), (a > 0 and a ≠ 1)
(iv) f(x) = ax on R. (a > 0 and a ≠ 1)
Solution:
(i) f(x) = sin x . \(\left[0, \frac{\pi}{2}\right]\)
for α, β ∈ \(\left[0, \frac{\pi}{2}\right]\)
Let f(α) = f(β) ⇒ sin α = sin β
⇒ α = β, as α, β ∈ \(\left(0, \frac{\pi}{2}\right)\) and no other values of α is possible.
∴ f is one-one.

(ii) f(x) = cos x, [0, π]
for α, β ∈ [0, π]
Let f(α) = f(β) ⇒ cos α = cos β
⇒ α = β, since α, β ∈ [0, π] and cos x is +ve in 1st quadrant and -ve in 2nd quadrant.
∴ f is one-one.

(iii) f(x) = loga x [1, ∞]
for α, β ∈ [1, ∞]
Let f(α) = f(β)
⇒ loga α = loga β ⇒ α = β
∴ f is one-one.

(iv) f(x) = ax, (a > 0), x ∈ R
for x1, x2 ∈ R
Let f(x1) = f(x2) ⇒ ax1 = ax2
⇒ x1 = x2
∴ f is one-one.

Question 5.
Show that functions f and g defined by f(x) = 2 log x and g(x) = log x2 are not equal even though log x2 = 2 log x.
Solution:
f(x) = 2 log x
g(x) = log x2
Dom f(x) = (0, ∞)
Dom g(x) = R – {0}
As Dom f(x) ≠ Dom g(x) we have f(x) ≠ g(x), though log x2 = 2 log x

Question 6.
Give an example of a function which is
(i) Surjective but not injective.
(ii) injective but not surjective.
(iii) neither injective nor surjective.
(iv) bijective
Solution:
(i) f(x) = sin x
from R → [-1, 1]
which is surjective but not injective.

(ii) f(x) = \(\frac{x}{2}\) from Z → R
is injective but not surjective.

(iii) f : (-1, 1) → R, f(x) = \(\frac{x}{1-x^2}\)
is neither surjective nor injective.
[Refer Q. No. 3(e)].

(iv) f(x) = x3 + 1, f : R → R
is bijective.
[Refer No. 3 (d)].

Question 7.
Prove that the following sets are equivalent:
{1, 2, 3, 4, 5, 6,…}
{2, 4, 6, 8, 10,…}
{1, 7, 5, 7, 9,…}
{1, 4, 9, 16, 25,…}
Solution:
Let A = {1, 2, 3, 4, 5, 6, ……..}
B = {2, 4, 6, 8, 10, …….}
C = {1, 3, 5, 7, 9, ……}
D = {1, 4, 9, 16, 25, ……}
Let f : A → B defined as f(x) = 2x
Clearly f is bijective.
There is a one-to-one correspondence between A and B.
⇒ A and B are equivalent.
Let g : A → C defined as g(x) = 2x – 1
Clearly f is bijective.
⇒ There is a one-to-one correspondence between A to C
∴ A and C are equivalent.
Let h : A → D defined as h(x) = x2.
Clearly h is bijective.
⇒ There is a one-to-one correspondence between A to D.
⇒ A and D are equivalent.
Thus A, B, C and D are equivalent.

Question 8.
Let f = {(1, a), (2, b), (3, c), (4, d)} and g = {{a, x), (b, x), (c, y), (d, x)}. Determine gof and fog if possible. Test whether fog = gof.
Solution:
f = {(1, a), (2, b), (3, c), (4, d)}
g = {{a, x), (b, x), (c, y), (d, x)}
gof (1) = g(a) = x
gof (2) = g(b) = x
gof (3) = g(c) = y
gof (4) = g(a) = x
∴ gof = {(1, x), (2, x), (3, y), (4, x)} Here fog is not defined.

Question 9.
Let f = {(1, 3), (2, 4), (3, 7)} and g = {(3, 2), (4, 3), (7, 1)}. Determine gof and fog if possible. Test whether fog = gof.
Solution:
f = {(1, 3), (2, 4), (3, 7)}
g = {(3, 2), (4, 3), (7, 1)}
We have fog (3) = f (g(3)) = f(2) = 4
fog (4) = f (g(4)) = f(3) = 7
fog (7) = f (g(7)) = f(1) = 3
∴  fog = {(3, 4), (4, 7), (7, 3)}
Again gof (1) = g (f(1)) = g(3) = 2,
gof (2) = g (f(2)) = g(4) = 3,
gof (3) = g (f(3)) = g(7) = 1
∴ gof = {(1, 2), (2, 3), (3, 1)}
∴ gof ≠ fog
So the composition of functions is not necessarily commutative.

Question 10.
Let f(x) =√x and g(x) = 1 – x2.
(i) Find natural domains of f and g.
(ii) Compute fog and gof and find their natural domains.
(iii) Find natural domain of h(x) = 1 – x.
(iv) Show that h = gof only on R0 = {x ∈ R : x ≥ 0} and not on R.
Solution:
Let f(x) = √x, g(x) = 1 – x2
(i) ∴ Dom f = R+ U{0}, Dom g = R

(ii) fog (x) = f (g(x))
= f (1 – x2) = \(\sqrt{1-x^2}\)
∴ fog (x) exists when 1 – x2 ≥ 0
⇒ x2 ≤ 1 ⇒ -1 ≤ x ≤ 1 i.e., x ∈ [-1, 1]
∴ Dom fog = [-1, 1]
Again gof (x) = g (f(x))
= g( √x ) = 1 – ( √x )2 = 1 – x
∴ Dom gof = R0 = (0, ∞)

(iii) Domain of h(x) = 1 – x is R.

(iv) We have proved in (ii) that gof (x) = 1 – x.
∴ h(x) = gof (x) ⇒ h = gof only when x ∈ R0 as dom f is R0 = [0, ∞]

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Ex 1(b)

Question 11.
Find the composition fog and gof and test whether fog = gof when f and g are functions on R given by the following:
(i) f(x) = x3 + 1, g(x) = x2 – 2
(ii) f(x) = sin x, g(x) = x5
(iii) f(x) = cos x, g(x) = sin x2
(iv) f(x) = g(x) = (1 – x3)1/3
Solution:
(i) f(x) = x3 + 1, g(x) = x2 – 2
∴ fog (x) = f (g(x)) = f(x2 – 2)
= (x2 – 2)3 + 1
gof (x) = g (f(x)) = g(x3 + 1)
= (x3 + 1)2 – 1
fog ≠ gof

(ii) f(x) = sin x, g(x) = x5
∴ fog (x) = f (g(x)) = f(x5) = sin x5
∴ gof (x) = g (f(x))
= g(sin x) = (sin x)5 = sin5 x
fog ≠ gof

(iii) f(x) = cos x, g(x) = sin x2
∴ fog (x) = f (g(x))
= f( sin x2) = cos (sin x2)
and gof (x) = g (f(x)) = g(cos x)
= sin (cos x)2 = sin (cos2 x)
fog ≠ gof

(iv) f(x) = g(x) = (1 – x3)1/3
fog (x) = f (g(x))
= (1 – (g(x))3)1/3
= [1 – (1 – x3)]1/3 = x
gof (x) = g (f(x))
= [1 – (f(x))3]1/3 = x
⇒ fog = gof

Question 12.
(a) Let f be a real function. Show that h(x) = f(x) + f(-x) is always an even function and g(x) = f(x) – f(-x) is always an odd function.
(b) Express each of the following function as the sum of an even function and an odd function:
(i) 1 + x + x2 , (ii) x2, (iii) ex, (iv) ex + sin x
Solution:
(a) We have h(x) = f(x) + f(-x)
∴ h(-x) = f(-x) + f(x) = h(x)
∴ h is always an even function.
Further, g(x) = f(x) – f(-x)
∴ g(-x) = f(-x) – f(x)
= – [f(x) – f(-x)] = – g(x).
∴ g is always an odd function.

(b) (i) Let f(x) = 1 + x + x2
∴ f(-x) = 1 – x + x2
∴ g(x) = \(\frac{f(x)+f(-x)}{2}\)
= \(\frac{1+x+x^2+1-x+x^2}{2}\)
= x2 + 1 and
g(-x) = (-x)2 + 1 = x2 + 1
∴ g is an even function.
h(x) = \(\frac{f(x)-f(-x)}{2}\)
= \(\frac{\left(1+x+x^2\right)-\left(1-x+x^2\right)}{2}\) = x
h(-x) = -x = -h(x)
⇒ h is an odd function.
∴ f(x) = g(x) + f(x)
where g is even and h is odd.

(ii) Let f(x) = x2
So that f(-x) = (-x)2 = x2
∴ g(x) = \( \frac{f(x)+f(-x)}{2}\) = \( \frac{x^2+x^2}{2}\) = x2
g(-x) = g(x)
∴ g is an even function.
and h(x) = \( \frac{f(x)-f(-x)}{2}\) = \(\frac{x^2-x^2}{2}\) = 0
h(x) = 0 is both even and odd.
∴ f(x) = g(x) = f(x),
where g is even and h is odd.

(iii) Let f(x) = ex
f(-x) = e-x
g(x) = \( \frac{e^x+e^{-x}}{2}\)
g(-x) = g(x)
g is an even function.
and h(x) = \( \frac{e^x-e^{-x}}{2}\)
h(-x) = \( \frac{e^{-x}-e^x}{2}\) = -h2(x)
⇒ h is an odd function.
∴ f(x) = g(x) + h(x),
where g is even and h is odd.

(iv) Let f(x) = ex + sin x
f(-x) = e-x + sin (-x) = e-x –  sin x
∴ g(x) = \( \frac{f(x)+f(-x)}{2}\)
= \( \frac{e^x+\sin x+e^{-x}-\sin x}{2}\)
= \( \frac{e^x+e^{-x}}{2}\) and h(x) = \( \frac{f(x)-f(-x)}{2}\)
= \( \frac{e^x+\sin x-e^{-x}+\sin x}{2}\)
= \( \frac{e^x-e^{-x}+2 \sin x}{2}\)
∴ f(x) = g(x) + h(x)
where g is even and g is odd.

Question 13.
Let X = {1, 2, 3, 4} Determine whether f : X → X defined as given below have inverses.
Find f-1 if it exists:
(i) f = {(1, 4), (2, 3), (3, 2), (4, 1)}
(ii) f = {(1, 3), (2, 1), (3, 1), (4, 2)}
(iii) f = {(1, 2), (2, 3), (3, 4), (4, 1)}
(iv) f = {(1, 1), (2, 2), (2, 3), (4, 4)}
(v) f = {(1, 2), (2, 2), (3, 2), (4, 2)}
Solution.
(i) x = {1, 2, 3, 4}
f is bijective. Hence f-1 exists.
f-1 = {(4, 1), (3, 2), (2, 3), (1, 4)}

(ii) f(2) = f(3) = 1
⇒ f is not injective
∴ f is not invertible.

(iii) f is bijective. Hence f-1 exists.
f-1 = {(2,1 ), (3, 2), (4, 3), (1, 4)}

(iv) f is not a function as
f(2) = 2 and f(2) = 3

(v) f is not injective hence not invertible.

Question 14.
Let f : X → Y.
If there exists a map g : Y → X such that gof = idx and fog = idy, then show that
(i) f is bijective and (ii) g = f-1
[Hint: Since idx is a bijective function, gof = idx is bijective. By Theorem 2(iv) f is injective. Similarly fog is bijective ⇒ f is surjective by Theorem 2(iii)]
Solution:
Let f : x → y and g : y – x
where gof = idx and fog = idy
we know that idx and idy are bijective functions.
⇒ gof and fog are both bijective functions.
⇒ f is a bijective function.

(ii) As f is bijective (by (i)) we have f-1 exists.
and f-1 : y → x where f-1of = idx and fof-1 = idy
But g : y → x with gof = idx and fog = idy
∴ g = f-1

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Ex 1(b)

Question 15.
Construct an example to show that f(A ∩ B) ≠ f(A) ∩ f(B) where A ∩ B ≠ Ø
Solution:
Let f(x) = cos x.
Let A = \(\left\{0, \frac{\pi}{2}\right\}\), B = \(\left\{\frac{\pi}{2}, 2 \pi\right\}\).
∴ f(A) = \(\left\{\cos 0, \cos \frac{\pi}{2}\right\}\)
= {1, 0} = {0, 1}
∴ f(B) = \(\left\{\cos \frac{\pi}{2}, \cos 2 \pi\right\}\) = {0, 1}
∴ f(A) ∩ f(B) = {0, 1}
Again,
A ∩ B = \(\left\{\frac{\pi}{2}\right\}\) and f(A ∩ B) = cos \(\frac{\pi}{2}\) = {0}
∴ f(A ∩ B) ≠ f(A) ∩ f(B)

Question 16.
Prove that for any f : X → Y, foidx = f = idyof.
Solution:
Let f : X → Y, so that y = f(x), x ∈ X.
∴ foidx = fof-1 of (x) = fof-1 (f(x))
= f(x) = y (∵ idx = fof-1‍)
Again, (idyof)(x) = (fof-1) of (x)
= (fof-1)(y) = f (f-1(y)) = f(x) = y  ….(2)
∴ From (1) and (2)
we have foidx = f = idyof

Question 17.
Prove that f : X → Y is surjective iff for all B ⊆ Y, f (f-1(B)) = B.
Solution:
Let f : X → Y is surjective.
i.e. for all y ∈ Y, ∃ a x ∈ X such that
y = f(x).
∴ x = f-1(y) ⇔ f(x)
= f (f-1(y)) ∈ f (f-1(B)).
for y = B ⊂ Y ⇔ y ∈ f (f-1(B)).
∴ y ∈ f (f-1(B)) ⇔ y ∈ B
∴ f (f-1(B)) = B

Question 18.
Prove that f : X → Y is injective iff f (f-1(A)) = A for all A ⊆ X.
Solution:
f : X → Y is injective.
Let x ∈ A ⇔ f(x) ∈ f(A) (∵ f is injective)
⇔ x ∈ f (f-1(A))
∴ A = f (f-1(A)) for all A ⊆ X.

Question 19.
Prove that f : X → Y is injective iff for all subsets A, B of X, f(A ∩ B) = f(A) ∩ f(B).
Solution:
f : X → Y is injective.
Let A and B are subsets of X.
Let f(x) ∈ f(A ∩ B)
⇔ x ∈ A ∩ B ⇔ x ∈ A ∧ x ∈ B
⇔ f(x) ∈ f(A) ∧ f(x) ∈ f(B) (∵ f is injective)
⇔ f(x) = f(A) ∩ f(B)
∴ f(A ∩ B) = f(A) ∩ f(B)

Conversely, suppose that
f(A ∩ B) = f(A) ∩ f(B)
Let f is not injective.
The if f(x) ∈ f(A ∩ B) ⇔ x ∈ A ∩ B
⇔ x ∈ A ∧ ⇔ x ∈ B
≠ f(x) ∈ f(A) ∧ ⇔ f(x) ∈ f(B)
⇔ f(x) ∈ f(A) ∩ f(B)
∴ f(A ∩ B) = f(A) ∩ f(B) is false.
so f must be injective.

Question 20.
Prove that f : X → Y is surjective iff for all A ⊆ X, (f(A))‘ ⊆ f(A‘), where A‘ denotes the complement of A in X.
Solution:
f : X → Y is surjective.
Then for all y ∈ Y ∃ x ∈ X
such that f(x) = y.
Let y ∈ [f(A)]‘ ⇒ y ∉ f(A)
⇒ f(x) ∉ f(A) ⇒ x ∉ A ⇒ x ∈ A‘
⇒ f(x) ∈ f(A‘) ⇒ y ∈ f(A‘)
∴ [f(A)]‘ ⊂ f(A‘)

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Ex 1(c)

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 1 Relation and Function Ex 1(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Exercise 1(c)

Question 1.
Show that the operation ∗ given by x ∗ y = x + y – xy is a binary operation on Z, Q and R but not on N.
Solution:
The operation ∗ given by
x ∗ y = x + y – xy
Clearly for all x, y ∈ Z
x + y – xy ∈ Z
⇒ x ∗ y ∈ Z
∴ ∗ is a binary operation on Z.
For all x, y ∈ Q
x + y – xy ∈ Q
⇒ x ∗ y ∈ Q
⇒ ∗ is a binary operation on Q for all x, y, ∈ R.
x + y – xy ∈ R
⇒ x ∗ y ∈ R
⇒ ∗ is a binary operation on R
Again 3, 4 ∈ N.
3 + 4 – 3 x 4 = 7 – 12 = – 5 ∉ N
i.e., x, y ∈ N
≠ x ∗ y ∈ N
∴ ∗ is not a binary operation on N.

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Ex 1(c)

Question 2.
Determine whether the following operations as defined by ∗ are binary operations on the sets specified in each case. Give reasons if it is not a binary operation.
(i) a ∗ b = 2a + 3b on Z.
(ii) a ∗ b = ma – nb on Q+ where m and n ∈ N.
(iii) a ∗ b = a + b (mod 7) on {0, 1, 2, 3, 4, 5, 6}
(iv) a ∗ b = min {a, b} on N.
(v) a ∗ b = GCD {a, b} on N.
(vi) a ∗ b = LCM {a, b} on N.
(vii) a ∗ b = LCM {a, b} on {0, 1, 2, 3, 4……, 10}
(viii) a ∗ b = \(\sqrt{a^2+b^2}\) on Q+
(ix) a ∗ b =a × b (mod 5) on {0, 1, 2, 3, 4}.
(x) a ∗ b = a2 + b2 on N.
(xi) a ∗ b = a + b – ab on R – {1}.
Solution:
(i) For all a, b ∈ Z
2a + 3b ∈ Z
⇒ a ∗ b ∈ Z
∗ is a binary operation on Z.

(ii) Let a = 1, b = 2
m = 1, n = 3
ma – nb = 1 – 6 = – 5 ∉ Q+
∴ a, b ∈ Q+ ≠ a ∗ b ∈ Q+
⇒ ∗ is not a binary operation on Q+

(iii) a ∗ b = a + b (mod 7) ∈ {0, 1, 2, 3, 4, 5, 6}
for a, b ∈ 7
∗ is a binary operation on the given set.

(iv) a, b ∈ N ⇒ min {a, b} ∈ N
∴ a ∗ b ∈ N
⇒ ∗ is a binary operation on N.

(v) for all a, b ∈ N, GCD [a, b] ∈ N
⇒ a ∗ b ∈ N
⇒ ∗ is a binary operation on N.

(vi) for all a, b ∈ N, LCM {a, b} ∈ N
⇒ a ∗ b ∈ N
⇒ ∗ is a binary operation on N.

(vii) Let A = {0, 1, 2, ….. 10}
4, 5 ∈ A but 4 ∗ 5 = LCM {4, 5}
= 20 ∉ A
⇒ ∗ is not a binary operation on A.

(viii) for all a, b ∈ Q+
a ∗ b = \(\sqrt{a^2+b^2}\) ∉ Q+
⇒ ∗ is not a binary operation on Q+.

(ix) For all a, b ∈ {0, 1, 2, 3, 4}
a ∗ b = a × b (mod 5) ∈ {0, 1, 2, 3, 4}
∴ ∗ is a binary operation on the given set.

(x) for all a, b ∈ N, a * b = a2 + b2 ∈ N
∴ ∗ is a binary operation on N.

(xi) For all a, b ∈ R – {1}
a ∗ b = a + b – ab ∈ R – {1}
∴ ∗ is a binary operation on R – {1}

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Ex 1(c)

Question 3.
In case ∗ is a binary operation in Q2 above, test whether it is (i) associative
(ii) commutative, Test further if the identity element exists and the inverse element for any element of the respective set exists.
Solution:
(i) On Z the binary operation is
a ∗ b = 2a + 3b
Commutative:
b ∗ a = 2b + 3a ≠ a ∗ b
∴ ∗ is not commutative.

Associative:
(a ∗ b) ∗ c = (2a + 3b) ∗ c
= 2 (2a + 3b) + 3c
= 4a + 6b + 3c
a ∗ (b ∗ c) = a ∗ (2b + 3c)
= 2a + 3 (2b + 3c)
= 2a + 6b + 9c
As (a ∗ b) ∗ c ≠ a ∗ (b ∗ c)
∗ is not associative.

Existance of identity:
Let e is the identity
∴ e ∗ a = a
⇒ 2e + 3a = a
⇒ e = -2a / 2 = -a
which depends on a.
∴ Identity element does not exist.

(iii) A = (0, 1, 2, 3, 4, 5, 6}
Commutative:
a ∗ b = a + b (mod 7)
= The remainder obtained when a + b is divided by 7.
b ∗ a = b + a (mod 7) = a + b (mod 7)
∴ ∗ is commutative.

Associative:
(a ∗ b) ∗ c = {a + b (mod 7)} ∗ c
= a + b + c (mod 7)
= The remainder obtained if a + b + c is divided by 7.
a ∗ (b ∗ c) = a ∗ {b + c (mod 7)}
= a + b + c (mod 7)
= The remainder obtained if a + b + c is divided by 7.
∴ (a ∗ b) ∗ c = a + (b ∗ c)
∴ ∗ is associative.

Existance of identity:
Let e is the identity
⇒ e ∗ a = a ∗ e = a
⇒ e + a mod 7 = a
⇒ e = 0
∴ 0 is the identity.

Existance of inverse:
Let a-1 = the inverse of a
⇒ a ∗ a-1 = a-1 ∗ a = e = 0
⇒ a + a-1 (mod 7) = 0
⇒ a + a-1 is divisible by 7.
1-1 = 6, 6-1 = 1
2-1 = 5, 5-1 = 2
3-1 = 4, 4-1 = 3

(iv) a ∗ b = min {a, b} on N.
Commutative:
a ∗ b = min {a, b}
b ∗ a – min {b, a} = a ∗ b
∴ ∗ is commutative.
Associative:
(a ∗ b) ∗ c = min {a, b} ∗ c
= min {a, b, c}
a ∗ (b ∗ c)= a ∗ min {b, c}
= min {a, b, c}
⇒ a ∗ (b ∗ c) = (a ∗ b) ∗ c
∴ ∗ is associative.

Existance of Identity:
Let e is the identity
∴ For all a ∈ N
e ∗ a = a ∗ e = a
⇒ min {e, a} = a
No such element exists in N.
∴ ∗ has no identity element on N.

(v) a ∗ b = GCD {a, b} on N.
b ∗ a = GCD {b, a} = GCD {a, b} = a ∗ b
∴ ∗ is commutative.
Associative:
(a ∗ b) ∗ c = GCD {a, b} ∗ c
= GCD {a, b, c}
a ∗ (b ∗ c) = a ∗ GCD {b, c}
= GCD {a, b, c}
⇒ (a ∗ b) ∗ c = a ∗ (b ∗ c)
⇒ ∗ is associative.

Existance of Identity:
Let e is the identity
∴ a ∗ e = e ∗ a = a
⇒ GCD {e, a} = a
No such element exists in N
⇒ ∗ has no indentity element.

(vi) a ∗ b = LCM {a, b} on N
Commutative:
a ∗ b = LCM {a, b}
= LCM {b, a}
= b ∗ a
∴ ∗ is commutative.

Associative
(a ∗ b) ∗ c = LCM {a, b} ∗ c
= LCM {a, b, c}
a ∗ (b ∗ c) =» a ∗ LCM {b, c}
= LCM {a, b, c}
⇒ (a ∗ b) ∗ c = a ∗ (b ∗ c)
∴ ∗ is associative.

Existance of Identity:
Let e is the identity
∴ e ∗ a = a ∗ e = a
⇒ LCM {e, a} = a
⇒ e – 1
∴ 1 is the identity element.

Existance of inverse:
Let a-1 is the inverse of a
⇒ a * a-1 = e = 1
⇒ LCM [a, a-1} = 1
a = a-1 = 1
Only 1 is invertible with 1-1 = 1.

(ix) a ∗ b = a × (mod 5) on {0, 1, 2, 3, 4}

Commutative:
a ∗ b = a x b (mod 5)
= Remainder on dividing a x b by 5
= Remainder on dividing b x a by 5
= b x a (mod 5)
= b x a
∴ ∗ is commutative.

Associative:
(a ∗ b) ∗ c=a x b (mod 5) ∗ c
= a x b x c (mod 5)
a and a ∗ {b ∗ c} = a ∗ {b x c (mod 5)}
= a x b x c (mod 5)
∴ (a ∗ b) ∗ c -=a ∗ (b ∗ c)
⇒ ∗ is associative.

Existance of identity:
Let e is the identity
∴ For all a ∈ {0, 1, 2, 3, 4}
a ∗ a = e ∗ a = a
a × e (mod 5) = a
⇒ e = 1
∴ 1 is the identity element.

Existance of inverse:
Let a-1 is the inverse of a
∴ a ∗ a-1= a-1 ∗ a = e = 1
⇒ a x a-1 (mod 5) = 1
⇒ 1-1 = 1
2-1 = 3, 3-1 = 2, 4-1 = 4
0 has no inverse.

(x) a ∗ b = a2 + b2 on N.
Commutative:
a ∗ b = a2 + b2
b ∗ a = b2 + a2 = a2 + b2 = a ∗ b
∴ ∗ is commutative.

Associative:
(a ∗ b) ∗ c = (a2 + b2) ∗ c
= (a2 + b2)2 + c2
a ∗ (b ∗ c) = a ∗ (b2 + c2)
= a2 + (a2 + b2)2
(a ∗ b) ∗ c ≠ a ∗ (b ∗ c)
∴ ∗ is not associative.

Existance of Identity:
Let e is the identity
a ∗ e = e ∗ a = a
⇒ a2 + e2 = a
⇒ e = \( \sqrt{a-a^2}\) which depends on a
∴ Identity does not exist.

(xi) a ∗ b = a + b – ab on R – {1}
Commutative:
a ∗ b = a + b – ab
b ∗ a = b + a – ba
a ∗ b = b ∗ a
∴ ∗ is commutative.

Associative:
a ∗ (b ∗ c) = a ∗ (b + c – bc)
= a + (b + c – bc) – a (b + c – bc)
= a + b + c – bc – ab – ac + abc
(a ∗ b) ∗ c = (a + b – ab) ∗ c
= a + b – ab + c – (a + b – ab) c
= a + b + c – ab – bc – ca + abc
∴ (a ∗ b) ∗ c = a ∗ (b ∗ c)
⇒ ∗ is associative.

Existance of Identity:
Let e is the identity
∴ e ∗ a = a ∗ e = a
⇒ a + e – ae = a
⇒ e (1 – a) = 0
⇒ e = 0 ( a ≠ 1)
∴ 0 is the identity.

Existance of inverse:
Let a-1 is the inverse of a
⇒ a ∗ a-1 = a-1 ∗ a = e
⇒ a + a-1 – aa-1 = 0
⇒ a-1 (1 – a) = – a
⇒ a-1 = \(\frac{a}{a-1}\) for a ∈ R – {1}

CHSE Odisha Class 12 Math Solutions Chapter 1 Relation and Function Ex 1(c)

Question 4.
Construct the composition table/multiplication table for the binary operation ∗ defined on {0, 1, 2, 3, 4} by a ∗ b = a × b {mod 5). Find the identity element if any. Also find the inverse elements of 2 and 4.
[This operation is called multiplication moduls 5 and denoted by x5. In general, on a finite subset of N, xm denotes the operation of multiplication modulo m where m is a fixed positive integer].
Solution:
A = {0. 1, 2, 3, 4}
a ∗ b = a × b mod 5

0 1 2 3 4
0 0 0 0 0 0
1 0 1 2 3 4
2 0 2 4 1 3
3 0 3 1 4 2
4 0 4 3 2 1

As 3rd row is identical to the first row we have 1 is the identity clearly 2-1 = 3 and 4-1 = 4.

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(a)

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

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

Question 1.
A merchant sells two models X and Y of TV with cost price ₹25000 and ₹50000 per set respectively. He gets a profit of ₹1500 on model X and ₹2000 on model Y. The sales cannot exceed 20 sets in a month. If he cannot invest more than 6 lakh rupees, formulate the problem of determining the number of sets of each type he must keep in stock for maximum profit.
Solution:
To get maximum profit let x TVs of model X and Y TVs of model Y must be kept in stock.
∴ Total profit = Z = 1500x + 2000y which is to be maximum.
According to the question, the sales cannot exceed 20 sets i.e. x + y ≤ 20.
Again total investment does not exceed 6 lakh.
25000x + 50000y ≤ 600000
⇒ x + 2y ≤ 24
∴ The LPP is:
maximise: Z = 1500x + 2000y
subject to: x + y ≤ 20
x + 2y ≤ 24
x, y ≥ 0

Question 2.
A company manufactures and sells two models of lamps L1 and L2, the profit being ₹15 and ₹10 respectively. The process involves two workers W, and W2 who are available for this kind of work 100 hours and 80 hours per month respectively, W1 assembles L1 in 20 and L2 in 30 minutes. W2 paints L1 in 20 and L2 in 10 minutes. Assuming that all lamps made can be sold, formulate the LPP for determining the production figures for maximum profit.
Solution:
Let x units of L1 and y units of L2: are to be produced to get maximum profit.
Total profit = Z = 15x + 10y
According to the question
20x + 30y ≤ 600
and 20x + 10y ≤ 480
=> 2x + 3y ≤ 600
2x + y ≤ 480
.-. The LPP is maximize: Z = 15x + 10y
subject to: 2x + 3y ≤ 600
2x + y ≤ 480
x, y ≥ 0.

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(a)

Question 3.
A factory uses three different resources for the manufacture of two different products, 20 units of the resource A, 12 units of B and 16 units of C being available. One unit of the first product requires 2, 2 and 4 units of the resources and one unit of the second product requires 4, 2 and 0 units of the resources taken in order. It is known that the first product gives a profit of ₹20 per unit and the second ₹30 per unit. Formulate the LPP so as to earn maximum profit.
Solution:
Let to earn maximum profit the factory produces x units of first product and y units of the second product.
The given data can be summarised as below:

Resource A Resource B Resource C Profit per unit in ₹
Product – I 2 2 4 20
Product – II 4 2 0 30
Availability 20 12 16

Total profit = 20x + 30y
which is to be maximum.
According to the question
2x + 4y ≤ 20
2x + 2y ≤ 12
4x + 0y ≤ 16
.-. The LPP is
maximize: Z = 20x + 30y
subject to: x + 2y ≤ 10
x + y ≤ 6
x ≤ 4
x, y ≥ 0.

Question 4.
A man plans to start a poultry farm by investing at most ₹3000. He can buy old hens for ₹80 each and young ones for ₹140 each, but he cannot house more than 30 hens. Old hens lay 4 eggs per week and young ones lay 5 eggs per week, each egg being sold at ₹5. It costs ₹5 to feed an old hen and ₹8 to feed a young hen per week. Formulate his problem determining the number of hens of each type he should buy so as to earn a profit of more than ₹300 per week.
Solution:
Let to get maximum profit he has to purchase x old hens and y young hens.
Total cost = 80x + 140y ≤ 3000
⇒ 4x + 7y ≤ 150
Total number of hens x + y ≤ 30
Number of eggs per week = 4x + 5y
Total income per week = 20x + 25y
Total cost to feed per week = 5x + 8y
∴ Weekly profit = 15x + 17y
∴ 15x + 17y > 300
and also total profit = z = 15x + 17y is to be maximum.
∴ The LPP is
maximize: Z = 15x + 17y
subject to: 4x + 7y ≤ 150
x + y ≤ 30
15x + 17y ≥ 300
x, y ≥ 0.

Question 5.
An agro-based company produces tomato sauce and tomato jelly. The quantity of material, machine hour, labour (man-hour) required to produce one unit of each product and the availability of raw material ore given in the following table:

Sauce Jelly availability
Man-hour 3 2 10
Machine hour 1 2.5 7.5
Raw material 1 1.2 4.2

Assume that one unit of sauce and one unit of jelly each yield a profit of ₹2 and ₹4 respectively. Formulate the LPP so as to yield maximum profit.
Solution:
Let the company produces x units of sauce and y units of jelly.
Total profit = 2x + 4y to be maximum.
Man hour = 3x + 2y ≤ 10
Machine hour = x + 2.5y ≤ 7.5
⇒ 2x + 5y ≤ 15
Raw material = x + 1.2y ≤ 4.2
⇒ 5x + 6y ≤ 21
∴ The LPP is
maximize: Z = 2x + 4y
subject to: 3x + 2y ≤ 10
2x + 5y ≤ 15
5x + 6y ≤ 21
x, y ≥ 0.

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(a)

Question 6.
(Allocation Problem.) A farmer has 5 acres of land on which he wishes to grow two crops X and Y. He has to use 4 cart loads and 2 cartloads of manure per acre for crops X and Y respectively. But not more than 18 cartloads of manure is available. Other expenses are ₹200 and ₹500 per acre for the crops X and Y respectively. He estimates profit from crops X and Y at the rates ₹1000 and ₹800 per acre respectively. Formulate the LPP as to how much land he should allocate to each crop for maximum profit.
Solution:
Let x acres are allocated for crop X and y acres for crop Y.
Total profit = 100x + 800y to be maximum.
According to the question
x + y ≤ 5
Manure = 4x + 2y ≤ 1 8 ⇒ 2x + y ≤ 9
∴ The LPP is
maximize: Z = 1000x + 800y
subject to: x + y ≤ 5
2x + y ≤ 9
x, y ≥ 0.

Question 7.
(Transportation Problem) A company has two factories at locations X and Y. He has to deliver the products from these factories to depots located at three places A, B and C. The production capacities at X and Y are respectively 12 and 10 units and the requirements at the depots are 8, 8 and 6 units respectively. The cost of transportation from the factories to the depots per unit of the product is given below.

(Cost in ₹)
To → A B C
From X 210 160 250
Y 170 180 140

The company has to determine how many units of product should be transported from each factory to each depot so that the cost of transportation is minimum. Formulate this LPP.
Solution:
Let x units are transported from X to A and y units from X to B. The transportation matrix is
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(a) Q.7

Total cost of transportation
= 210x + 160y + 250 (12 – x – y) +170 (8 – x) +180 (8 – y) + 140 (x + y – 6)
= 4960 – 70x – 130y to be minimum.
Now all costs of transportation are ≥ 0.
∴ x ≥ 0, y ≥ 0.
12 – x – y ≥ 0 ⇒ x + y ≤ 12
8 – x ≥ 0 ⇒ x ≤ 8
8 – y ≥ 0 ⇒ y ≤ 8
x + y – 6 ≥ 0 x + y ≥ 6
∴ The LPP is
Minimize: Z = 4960 – 70x – 130y
subject to: x + y ≤ 12
x ≤ 8
y ≤ 8
x + y ≥ 6
x, y ≥ 0

Question 8.
(Diet Problem) Two types of food X and Y are mixed to prepare a mixture in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. These vitamins are available in one kg of food as per the table given below.

Vitamins
food A B C
X 1 2 3
Y 2 2 1

One kg. of food X cost ₹16 and one kg. of food Y costs ₹20. Formulate the LPP so as to determine the least cost of the mixture containing the required amount of vitamins.
Solution:
Let x units of food X and y units of food Y are to be mixed to prepare the mixture.
Cost of the mixture = 16x + 20y to be minimum.
According to the question
Vitamin A content = x + 2y ≥ 10
Vitamin B content = 2x + 2y ≥ 12
Vitamin C content = 3x + y ≥ 8
∴ The LPP is
minimize: Z = 16x + 20y
subject to x + 2y ≥ 10
x + y ≥ 6
3x + y ≥ 8
x, y ≥ 0.

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(a)

Question 9.
Special purpose coins each weighing 10gms are to be manufactured using two basic metals M1 and M2 and a mix of other metals M3. M1, M2 and M3 cost ₹500, ₹800 and ₹50 per gram respectively. The strength of a coin demands that not more than 7gm. of M1 and a minimum of 3 gm of M1 should be used. The amount of M3 in each coin is maintained at 25% of that of M1. Since the demand for the coin is related to its price, formulate the LPP to find the minimum cost of a coin.
Solution:
Let x gm of M1 and y g of M2 are used to make the coin. According to the demand of the coin \(\frac{x}{4}\) g of M3 is to be mixed.
Cost of the coin = 500x + 800y + \(\frac{50 x}{4}\)
= (512.5)x + 800y
which is to be minimum.
Again weight of the coin = 10g
⇒ x + y + \(\frac{x}{4}\) = 10
⇒ 5x + 4y = 40
According to the question x ≤ 7, y ≤ 3.
Thus the L.P.P is
minimize: Z = (512.5)x + 800y
subject to: 5x + 4y = 40
x ≤ 7
x ≤ 3
x, y ≥ 0.

Question 10.
A company produces three types of cloth A, B and C. Three kinds of wool, say red, green and blue are required for the cloth. One unit length of type A cloth needs 2 metres of red and 3 metres of blue wool; one unit length of type B cloth needs 3 metres of red, 2 metres of green and 2 metres of blue wool and one unit length of type C cloth needs 5 metres of green and 4 metres of blue wool. The firm has a stock of only 80 metres of red, 100 metres of green and 150 metres of blue wool. Assuming that income obtained from one unit length of cloth is ₹30, ₹50 and ₹40 of types A, B and C respectively, formulate the LPP so as to maximize income.
Solution:
Let x units of cloth A, y units of cloth B and z units of cloth C are to be produced from the available materials to get the maximum income. The given data can be summarised as:

Red wool Green wool Blue wool Income
Cloth A 2 3 30
Cloth B 3 2 2 50
Cloth C 5 4 40
Availability 80 100 150

Total Income = 30x + 50y + 40z
which is to be maximum.
According to the question
2x + 3y ≤ 80
2y + 5z ≤ 100
3x + 2y + 4z ≤ 150
Thus the L.P.P. is
minimize: Z = 30x + 50y + 40z
subject to: 2x + 3y ≤ 80
2y + 5z ≤ 100
3x + 2y + 4z ≤ 150
x, y, z ≥ 0

Question 11.
A person wants to decide the constituents of a diet which will fulfil his daily requirements of proteins, fats and carbohydrates at minimum cost. The choice is to be made from three different types of food. The yields per unit of these foods are given in the following table.

food yield/unit cost/unit
Protein Fat Carbonate
f1 3 2 6 45
f2 4 2 3 40
f3 8 7 7 85
Minimum Requirement 100 200 800

Formulate the LPP.
Solution:
Let the diet constitues x units of f1, y units of f2 and z units of f3.
Total cost = 45x + 40y + 85z, which is to be minimum.
According to the question
3x + 4y + 8z ≥ 1000
2x + 2y + 7z ≥ 200
6x + 3y + 7z ≥ 800
Thus the LPP is
Minimize: Z = 45x + 40y + 85z
Subject to: 3x + 4y + 8z ≥ 1000
2x-+2y + 7z ≥ 200
6x + 3y + 7z ≥ 800
x, y, z ≥ 0

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

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

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.1
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
Zmax = 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

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.2
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.

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

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

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.3
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
Zmax = 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 Zmin = \(\frac{555}{13}\)

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.
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.5
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.

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

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

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.6
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

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.8
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)
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.8.1
∴ 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 Zmax = 50
There is no other point that maximizes Z.

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

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

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.9
Step – 3 Clearly 0(0,0) satisfies x1 + 2x2 ≤ 30 and does not satisfy 12x1 + 7x2 ≤ 152
∴ The shaded region is the feasible region.
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.9.1
Step – 4
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.9.2
Z is maximum for x1 = 30, x2 = 0 and Zmax = 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
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.10
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
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.10.1

∴ 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 Zmin = 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
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.11
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

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

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.
D:\BSE Odisha.guru\Image\CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.12.png
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.
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.13
Step – 3 The shaded regian is feasible region which is unbounded, thus Z does not have any maximum.
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.13(1)
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

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.14
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

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

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.
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.15(1)
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
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.15(2)
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
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.15(3)
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
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.15(4)
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
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.15(4.1)
∴ Z is maximum for x = 20. y = 10 and Z(max) = 470

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

(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
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.15(5)
Step – 3 As 0(0,0) satisfies all the constraints the shaded region is the feasible region.
From (1) and (3) we get
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.15(5.1)
From (2) and (3) we get
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.15(5.2)
Step-4 Z(O) = 0
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.15(5.3)

(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
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.15(6)
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
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.15(7)

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.

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

(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
CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.15(8)
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

CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Q.15(9)
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

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

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:
\(\frac{\pi}{3}\)

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

(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

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

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)
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.3(1)

(ii) tan (\(\frac{\pi}{4}\) + 2 cot-1 3)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.3(2)

(iii) cos (2 sin-1 x)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.3(3)

(iv) tan (cos-1 x)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.3(4)
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.3(4.1)

(v) tan-1 (\(\frac{x}{y}\)) – tan-1 \(\frac{x-y}{x+y}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.3(5)

(vi) cosec (cos-1 \(\frac{3}{5}\) + cos-1 \(\frac{4}{5}\))
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.3(6)

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

(vii) sin-1 \(\frac{1}{\sqrt{5}}\) + cos-1 \(\frac{3}{\sqrt{10}}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.3(7)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.3(9)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.3(10)

(xi) sin cot-1 cos tan-1 x.
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.3(11)

(xii) tan-1 \(\left(x+\sqrt{1+x^2}\right)\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.3(12)

Question 4.
Prove the following statements:
(i) sin-1 \(\frac{3}{5}\) + sin-1 \(\frac{8}{17}\) = cos-1 \(\frac{36}{85}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.4(1)

(ii) sin-1 \(\frac{3}{5}\) + cos-1 \(\frac{12}{13}\) = cos-1 \(\frac{33}{65}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.4(2)

(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 \(\frac{1}{13}\)
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.4(3)

(iv) tan-1 \(\frac{1}{2}\) + tan-1 \(\frac{1}{5}\) + tan-1 \(\frac{1}{8}\) = \(\frac{\pi}{4}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.4(4)

(v) tan ( 2tan-1 \(\frac{1}{5}\) – \(\frac{\pi}{4}\) ) + \(\frac{7}{17}\) = 0
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.4(5)
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.4(5.1)

Question 5.
Prove the following statements:
(i) cot-1 9 + cosec-1 \(\frac{\sqrt{41}}{4}\) = \(\frac{\pi}{4}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.5(1)

(ii) sin-1 \(\frac{4}{5}\) + 2 tan-1 \(\frac{1}{3}\) = \(\frac{\pi}{2}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.5(2.1)
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.5(2.2)

(iii) 4 tan-1 \(\frac{1}{5}\) – tan-1 \(\frac{1}{70}\) + tan-1 \(\frac{1}{99}\) = \(\frac{\pi}{4}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.5(3.1)
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.5(3.2)

(iv) 2 tan-1 \(\frac{1}{5}\) + sec-1 \(\frac{5 \sqrt{2}}{7}\) + 2 tan-1 \(\frac{1}{8}\) = \(\frac{\pi}{4}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.5(4.1)
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.5(4.2)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.5(5.1)
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.5(5.2)
(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

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

(vii) cos tan-1 cot sin-1 x = x.
Solution.
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.5(7)

Question 6.
Prove the following statements:
(i) cot-1 (tan 2x) + cot-1 (- tan 2x) = π
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.6(1)

(ii) tan-1 x + cot-1 (x + 1) = tan-1 (x2 + x + 1)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.6(2)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.6(4)

(v)
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.6(5.1)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.6(5.2)
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.6(5.3)

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:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.7(1)

(ii) tan-1 \(\frac{1}{x+y}\) + tan-1 \(\frac{y}{x^2+x y+1}\) = tan-1 \(\frac{1}{x}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.7(2.1)
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.7(2.2)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.7(3.1)
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.7(3.2)

(iv) sin2 (sin-1 x + sin-1 y + sin-1 z) = cos2 (cos-1 x + cos-1 y + cos-1 z)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.7(4)

(v) tan (tan-1 x + tan-1 y + tan-1 z) = cot (cot-1 x + cot-1 y + cot-1 z)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.7(5)

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
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.8(1)

(ii) tan-1 x + tan-1 y + tan-1 z = π show that x + y + z = xyz.
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.8(2)

(iii) tan-1 x + tan-1 y + tan-1 z = \(\frac{\pi}{2}\). Show that xy + yz + zx = 1
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.8(3)
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:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.8(4)

(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 \(\frac{c}{a+b}\)
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.8(5)

Question 9.
Solve
(i) cos (2 sin-1 x) = 1/9
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.9(1)

(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.

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

(iv) cos-1 x + sin-1 \(\frac{x}{2}\) = \(\frac{\pi}{6}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.9(4)

(v) tan-1 \(\frac{x-1}{x-2}\) + tan-1 \(\frac{x+1}{x+2}\) = \(\frac{\pi}{4}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.9(5)

(vi) tan-1 \(\frac{1}{2 x+1}\) + tan-1 \(\frac{1}{4 x+1}\) = tan-1 \(\frac{2}{x^2}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.9(6)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.9(7)

(viii) cot-1 \(\frac{1}{x-1}\) + cot-1 \(\frac{1}{x}\) + cot-1 \(\frac{1}{x+1}\) = cot-1 \(\frac{1}{3x}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.9(8)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.9(9)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.9(10)

(xi) sin-1 y – cos-1 x = cos-1 \(\frac{\sqrt{3}}{2}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.9(11)

(xii) sin-1 2x + sin-1 x = \(\frac{\pi}{3}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.9(12)

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:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.10

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:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.11(1)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.11(2.1)
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.11(2.2)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.11(3)

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:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.12(1.1)
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.12(1.2)

(ii) If cos-1 (\(\frac{x}{y}\)) + cos-1 (\(\frac{y}{3}\)) = Θ, prove that 9x2 – 12xy cos Θ + 4y2 = 36 sin2 Θ.
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.12(2)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.12(3)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.12(4)

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

(v) If sin-1 x + sin-1 y + sin-1 z = π prove that x2 + y2 + z2 + 4x2y2z2 = 2 ( x2y2 + y2z2 + z2x2 )
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.12(5)

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:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.13(1)

(ii) tan-1 \(\frac{1}{3}\) + tan-1 \(\frac{1}{5}\) + tan-1 \(\frac{1}{7}\) + tan-1 x = \(\frac{\pi}{4}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.13(2)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.13(3.1)
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.13(3.2)

(iv) 3tan-1 \(\frac{1}{2+\sqrt{3}}\) – tan-1 \(\frac{1}{x}\) = tan-1 \(\frac{1}{3}\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Ex 2 Q.13(4)

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

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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.1
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

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

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
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.5

(ii) A-A’ is skew-symmetric
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.5(2)

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

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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.6

Question 7.
Here A is an involuntary matrix, recall the definition given earlier.
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.7

Question 8.
Show that \(\left[\begin{array}{ll}
\mathbf{0} & \mathbf{1} \\
\mathbf{1} & \mathbf{0}
\end{array}\right]\) is its own inverse.
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.8

Question 9.
Express as a sum of a symmetric and a skew symmetric matrix.
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.9
Solutions:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.9(1)

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.9(3)
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.9(4)
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.9(5)
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.9(6)
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.9(7)

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

Question 10.
What is the inverse of
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.10

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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.11(1)

(ii) \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.11(2)

(iii) \(\left[\begin{array}{cc}
4 & -2 \\
3 & 1
\end{array}\right]\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.11(3)

(iv) \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.11(4)

(v) \(\left[\begin{array}{cc}
1 & 0 \\
2 & -3
\end{array}\right]\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.11(5)

(vi) \(\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.11(6)

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

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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.12(1)

(ii) \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & 1 & 1
\end{array}\right]\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.12(2)

(iii) \(\left[\begin{array}{ccc}
3 & -2 & 3 \\
2 & 1 & -1 \\
4 & -3 & 2
\end{array}\right]\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.12(3)

(iv) \(\left[\begin{array}{lll}
1 & 1 & 2 \\
0 & 1 & 2 \\
1 & 2 & 1
\end{array}\right]\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.12(4)

(v) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 1 & 4 \\
1 & 0 & 2
\end{array}\right]\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(b) Q.12(5)

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

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

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

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 AT.
(iv) What is the order of AT?
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) AT = \(\left[\begin{array}{lll}
1 & 4 & 3 \\
2 & 5 & 9 \\
3 & 6 & 1 \\
4 & 1 & 1 \\
1 & 2 & 6
\end{array}\right]\)
(iv) Order of AT 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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.5(1)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.5(2)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.5(3)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.5(4)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.5(5)

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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.6(1)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.6(3)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.6(4)

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

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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.7(1)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.7(3)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.7(4)

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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.8

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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.9(1)

(ii) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.9(2)

(iii) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 3 \\
1 & 4
\end{array}\right]\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.9(3)

(iv) \(\left[\begin{array}{ll}
1 & 3 \\
1 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.9(4)

(v) \(\left[\begin{array}{cc}
1 & i \\
i & -1
\end{array}\right]^2\) where i = √-1
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.9(5)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.9(6)

(vii) \(\left[\begin{array}{ll}
0 & k \\
1 & 0
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.9(7)

(viii) \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\)
Solution:
D:\BSE Odisha.guru\Image\CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.9(8).png

(ix) \(\left[\begin{array}{ll}
1 & 0 \\
0 & k
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.9(9)

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

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

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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.11

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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.12(1)

(ii) \(\left[\begin{array}{c}
x+3 \\
2-y
\end{array}\right]=\left[\begin{array}{c}
1 \\
-3
\end{array}\right]\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.12(2)

(iii) \(\left[\begin{array}{c}
2 x-y \\
x+y
\end{array}\right]=\left[\begin{array}{c}
3 \\
-9
\end{array}\right]\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.12(3)

(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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.12(4)

(v) [2x -y] + [y 3x] = 5 [1 0]
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.12(5)

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

Question 13.
The element of ith row and ith column of the following matrix is i +j. Complete the matrix.
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.13

Question 14.
Write down the matrix
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.14

Question 15.
Construct a 2 x 3 matrix having elements given by
(i) aij = i + j
(ii) aij = i – j
(iii) aij = i × j
(iv) aij = i / j
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.15

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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.16

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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.17

Question 18.
If
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.18

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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.20

Question 21.
Find B if B2 = \(\left[\begin{array}{cc}
17 & 8 \\
8 & 17
\end{array}\right]\)
Solution:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.21
∴ 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 = \(\left[\begin{array}{ll}
4 & 1 \\
1 & 4
\end{array}\right]\)

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

Question 22.
Find x and y when
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.22

Question 23.
Find AB and BA given that:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.23

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.23(2)

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.23(3)

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.23(4)

Question 24.
Evaluate
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.24(1)

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.24(2)

Question 25.
If
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.25
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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.25.1

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.25(1)

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.25(2)

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.25(3)

Question 26.
Find A and B where
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.26

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 = BTAT where
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.28.1

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.28.2

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

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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.29

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:
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.30

Question 31.
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.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)

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

Question 35.
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.35

Question 36.
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.36

Question 37.
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.37

Question 38.
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.38(1)

CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.38(2)

Question 39.
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.39

Question 40.
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.40

Question 41.
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.41
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.41(1)

Question 42.
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.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
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.43
∴ Calory requirements for families A and B are 24600 and 15800 respectively and protein requirements are 576 gm and 332 gm respectively.

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

Question 44.
Let the investment in first fund = ₹x and in the second fund is ₹(50000-x)
Investment matrix A=[x  50000-x]
CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Ex 4(a) Q.44
⇒ 300000 – x = 278000
⇒ x = 22000
∴ He invests ₹22000 in first bond and ₹28000 in the second bond.