CHSE Odisha Class 11 English Grammar Countable and Uncountable Nouns

Odisha State Board CHSE Odisha Class 11 Invitation to English 4 Solutions Grammar Countable and Uncountable Nouns Textbook Activity Questions and Answers.

CHSE Odisha 11th Class English Grammar Countable and Uncountable Nouns

SECTION – 1

Study the following sentences :

  • Red is a color.
  • Pradeep is a man of ability.
  • Nothing succeeds like success.

The underlined words, namely, red, ability, and success are Nouns.
Here, the definition of old grammar does not account for identifying a Noun in a sentence, i.e. name of a person or place or thing.
Let’s see for ourselves what a Noun in English is and whether there is any precise way of knowing (identifying) it in a sentence.
Look at the different natures and functions of a Noun.

  • Birds fly.
  • The tall dark red-haired Russian scientist will give a talk tomorrow morning.

The above-underlined words, such as, ‘Birds and scientist’ function as the ‘headword’ in the subject part of the sentences.
We can distinguish a ‘Noun ’ from other word classes if they satisfy the following criteria :

(1) These are open-class words to which new items are constantly added.
Example :
cosmonaut, astronaut, videotape, flyover, etc.

CHSE Odisha Class 11 English Grammar Countable and Uncountable Nouns

(2) Noun can function as the subject, object, complement, or adjunct in a sentence.
Example :
The man died yesterday, (subj.)
Grammarians are nasty creatures, (comp.)
I am writing this lesson now. (obj.)
The student looked up the word (obj.) in the dictionary. (adjunct)

(3) Most nouns form plurals by taking -s or ~es. e.g. -birds, books, watches, etc.
(4) Nouns can be preceded by determiners like a, an, the, one, my, two, some, a few, this, that, these, those or can go without any determiner.
(5) Many nouns can go with genitives such as -’s, -s’.
Example: A bird’s nest is destroyed.
There will be a writers’ camp soon.
(6) Nouns can be preceded by prepositions. e.g. – It is meant for birds only.
(7) Nouns can also be used directly before other nouns as modifiers, e.g. tea- stall, Art College, etc.
(8) Words quoted with plural marker -s are also sometimes used as nouns e.g., ifs and buts, ups and downs, etc.

The above properties provide a clear-cut basis for identifying words as nouns.
Countable nouns refer to ‘things’ (nouns) that can be counted (numbered) and they have got two forms, such as singular form and plural forms. For example, a pen [singular countable], and two pens [plural countable]. On the other hand, uncountable nouns do not take a plural form. These are normally used in the singular form. For example, hair, sugar, and water. [However, in exceptional cases, we, under the circumstances, use the plural form of uncountable nouns, like – water – waters, hair – hairs, etc.]

Now mark the following conversation/talking.
Mother: We have run out of rice, flour, butter, and a few other things.
Father: All right. I’ll make a shopping list. Let me get a piece of paper and a pen.

Now you need some sugar, tea, and milk, right? We also need ten kilos of rice and half a liter of cooking oil. Anything else?
Remember the criteria or markers we discussed earlier and try out yourself first to find out nouns in the above conversation/dialogue. Then check your answers with the answers given below.
Answer:
rice, flour, butter, things, list, piece, paper, pens, sugar, tea, milk, kilos, oil, etc.

Having discussed the various properties/characteristics of nouns in general, let us now try to classify the nouns. The nouns like pen, piece, kilo, etc. can be counted. So they are called Countable Nouns and nouns like rice, paper, oil, water, etc. cannot usually be counted. So these nouns are called Uncountable Nouns. So there are two groups or categories of nouns, such as Countable Nouns and Uncountable Nouns.

Countable Nouns have the following characteristics.
(i) They have two number forms, such as pen-pens, book-books, etc. (singular – plural forms).
(ii) The singular countable nouns take modifiers like an, a, the, three, fourth, etc. (called determiners) before them.
(We cannot say: There is a book on the table.)
(iii) The plural form may or may not take modifiers like the few, and many before them. (Modifiers are optional here.)
(iv) Nouns like sheep and deer are count nouns, but they have the same form for singular and plural. We call them one-form count nouns.

CHSE Odisha Class 11 English Grammar Countable and Uncountable Nouns

Uncountable Nouns have the following characteristics.
Examine the characteristics of rice the noun :
– Rice is our chief food.
– The rice in his shop is fine.
– Much rice is grown in the plains of Orissa.
– Little rice is grown in hilly areas.
The noun rice (uncountable) has the following characteristics.
(i) It has one form. It does not take -s or -es form.
(ii) It can be used without modifiers.
(iii) It can be used with the modifiers like much, and little but not with a few, or many.
(iv) It cannot be used with numerals like one, two, third, fourth, etc.
Such nouns like ‘rice’ are called uncountables.

Activity – 1:

Say whether the nouns in the list below are countables or uncountables.

1. car        2. Music     3. Cloud       4. information       5. bottle

6. chair     7. Advice    8. Loaf          9. video               10. pot

11. fruit   12. Hair      13. Girl          14. water               15. weather

16. taxi    17. Traffic  18. leather     19. furniture           20. apple

Answer:
1. countable
2. uncountable
3. countable
4. uncountable
5. countable
6. countable
7. uncountable
8. countable
9. countable/uncountable
10. countable
11. uncountable/countable
12. uncountable/countable
13. countable
14. uncountable
15. uncountable
16. countable
17. uncountable
18. uncountable
19. uncountable
20. countable

CHSE Odisha Class 11 English Grammar Countable and Uncountable Nouns

Activity — 2

Choose the correct alternatives in the sentences below.
1. There is/are usually a lot of traffic/traffics in the city during working hours.
2. The young man is looking for a work /job at the moment.
3. I want to make a list of candidates attending the meeting. Have you got a paper/ some paper?
4. I don’t want to have a bread/bread for my breakfast.
5. The girl with a blonde (golden or pale colored) hair/hairs lives next door.
6. Good accommodation/accommodations is/are not available in this city.
7. We need some meat/meats for the dinner tonight.
8. The old man could not carry the luggage/luggages.
Answer:
1. There is usually a lot of traffic in the city during working hours.
2. The young man is looking for a job at the moment.
3. I want to make a list of candidates attending the meeting. Have you got a paper
4. I don’t want to have bread for my breakfast.
5. The girl with blonde hair lives next door.
6. Good accommodation is not available in this city.
7. We need some meat for the dinner tonight.
8. The old man could not carry the luggage.

Activity – 3

Use a, an, the, somewhere required, In the blank spaces below. If no word is required, leave the space blank.
1. _________health is more precious than _________ wealth. To keep healthy, we need _________ good food, _________ sleep, and _________ exercise.
2. _________ travel gives us _________ opportunity of seeing how_________ other people live. When we go on _________ journey, we should take _________ note-book with us to make _________ notes of the names of _________ people we meet.
Answer:
1. Health is more precious than wealth. To keep healthy, we need some good food, sleep, and exercise.
2. Travel gives us an opportunity of seeing how other people live. When we go on a journey, we should take a notebook with us to make a note of the names of people we meet.

Activity – 4

There are three countable nouns in the list below. Can you find them?

gold match rubbish love
jam fun equipment knowledge
happiness wood homework food
snow progress cheese bottle
cream help cupboard wind

Answer:
Countable
match, cupboard, bottle

CHSE Odisha Class 11 English Grammar Countable and Uncountable Nouns

SECTION – 2

Read the dialogue below.
Wife: What did you buy in the market?
Husband: I bought a bottle of milk, a packet of washing powder, and a tube of toothpaste.
Wife: What about the bar of chocolate I asked you to buy?
Husband: Oh dear. I completely forgot.

Do mark now the uncountable nouns used as countable in the above dialogue between wife and husband. We made countable quantities with uncountable nouns such as milk, washing powder, and toothpaste using a bottle of milk, and a packet of washing powder. and a tube of toothpaste.

An uncountable noun does not take words like a, an or one, two before it and does not have a plural form; but we can use expressions like a piece of a drop of a bar of, etc. before it to make it countable. They are used to refer to ‘units’ of such uncountables.
For example — a ball of string, a bar of chocolate, a kilo of rice, etc.

Activity – 5

Match the items in Column ‘A’ with suitable items in Column-‘B’.
Match the items in Column ‘A’ with suitable items in Column-‘B’.
Answer:
Match the items in Column ‘A’ with suitable items in Column-‘B’ 1

Activity – 6

Imagine that you ‘went with a friend, on a week-long camping trip. You took some supplies (food, matches, candles, etc.) with you, but you find that most of the things have been used up by the third day. Ask your friend to tell you what supplies are left with him/her, and then tell your friend what things are left with you.
The following is the list of supplies that you took on the trip. You will have to decide which of these supplies have been completely used up and which are still left, and the quantities that still remain.
Follow this pattern :
A: Is there any sugar left?
B: Yes, we still have about a kilogram of sugar.
A: What about soap?
B: I’m afraid there’s no soap left.
(or)
B: Yes, we have four cakes of soap left.
List of supplies taken on the camping trip.

rice salt matchbox chewing-gum
bread sugar milk condensed milk
soap torch chocolate toothpaste
tea bandages batteries antiseptic ointment

Answer:
A: Is there any rice left?
B: No, there isn’t any rice left.
A: What about bread?
B: No, we don’t have any loaf of bread.
A: What about soap?
B: Yes, we have a cake of soap left.
A: Is there any salt left?
B: Yes, we have some pinches of salt left.
(Or)
B: I’m afraid there is no grain of salt left.
A: Is there any chewing-gum left?
B: No, there is no chewing-gum left.
A: Is there any condensed milk left?
B: No, there is no condensed milk left.
A: What about your torch?
B: Yes, we have a torch. It is still working well.
A: Is there any antiseptic ointment left?
B: I’m afraid there’s no antiseptic ointment left.
A: Is there any matches left?
B: Yes, there are five boxes of matches left.
A: What about toothpaste?
B: Yes, we have five tubes of toothpaste left.
A: Is there any chocolate left?
B: Yes, there is still a bar of chocolate left.
A: What about milk?
B: Yes, we have some pints of milk left.
A: Is there any tea left?
B: Yes, we have a packet/pound of tea left.
(Or)
B: We have fifty grams of tea left.
A: What about sugar?
B: Yes, we have two kilograms of sugar left.
A: What about batteries?
B: Yes, we still have five (pieces of) batteries left.
A: Is there any bandage left?
B: Yes, we still have four (strips of) bandages left.

CHSE Odisha Class 11 English Grammar Countable and Uncountable Nouns

SECTION – 3

Nouns that can be countable as well as uncountable.
Look at the examples mindfully.
1.
(a) Our science teacher wrote a paper on the effects of the cyclone, (a paper means here an article (essay): countable)
(b) Books are printed on paper. (uncountable)
(c) every day I read a paper, (newspaper: countable)
(d) The geography paper was difficult, (subject: countable)

2.
(a) Yesterday I bought an iron. (countable: a thing for ironing dress/clothes)
(b) Iron is stronger than wood, (metal: uncountable)
(c) He is a man of iron, (physical strength: uncountable)
In the above examples paper and iron have different meanings in their use. They are used both as countable and uncountable.

Uncountable Countable
1. There is a good deal of noise. 1. Do not make a noise.
2. Bread is our staple food. 2. What breads have you got today? (types of)
3. Do you have much difficulty with your English? 3. We have had very few difficulties so far.
4. The temple is made of stone. 4. The boys are pelting stones at the frogs.
5. There is some egg on your chin, (food) 5. There are four eggs in, the basket.
6. Eat a little more fish, (food) 6. There are fish/fishes in the pond.
7. I don’t like tea. 7. We want three teas and two coffees.
8. I don’t like your talk, (gossip) 8. He gave a nice talk on science.
9. The table is made of wood. 9. The poet loves woods. Ismail forest!
10. Beauty is to be admired. 10. Lara Dutt is a beauty.
11. He ate a whole chicken. 11. I Would like to have some more chicken.
12. Could I have a glass of water? 12. Glass is brittle.
13. Tea is grown in Assam. 13. This shop sells teas from different parts of India.


SECTION – 4

More on Quantifiers.
1. much, many, a lot of, (a) little, (a) few
Much and many are used in negatives and questions/interrogatives generally. Much is used before uncountable nouns and many with plural countable.

  • A lot of and lots of cans be used with countable and uncountables in positive (remarks) sentences.
  • A little and little can be used with uncountable nouns only.
  • Few and a few can be used with plural countable.

CHSE Odisha Class 11 English Grammar Countable and Uncountable Nouns

Activity – 7

Use much, many, a lot of, lots of in the blank spaces, where required in the sentences below. (In some blank spaces more than one alternative is possible.)
(a) I can’t come with you. I’ve got _________ work to do.
(b) He’s not got _________money, so he can’t buy that house.
(c) He is very quiet person. He doesn’t speak _________.
(d) I’m hoping to get a ticket for the match. But there aren’t _________ seats left, I hear.
(e) That car is very old. It uses _________ petrol.
(f) I haven’t got _________ time for watching sport at the moment.
(g) ______ people go me to the public meeting addressed by the Chief Minister.
(h) We didn’t visit _________ places when we were on holiday.
(i) We heard the cheapest washing machine costs 10,000 rupees. That is _______, in my opinion.
(j) My father drinks _________ water, – ten liters a day.
(k) The players haven’t won _________ medals.
(l) I take photographs but not as _________ as I used to.
Answer:
(a) much a lot of
(b) much
(c) much
(d) many
(e) a lot of
(f) much/a lot of
(g) lots of
(h) many
(i) lots of
(j) lots of
(k) many
(l) many

Activity – 8

Complete these mini dialogues with much, many, a lot of, and lots of
(a) A: Too ________ students fail in English every year.
B: Yes, and the schools aren’t doing ________ about it.
(b) A: We didn’t have ________ time to spare at the railway station.
B: No, we didn’t have ________ either.
(c) A: There were ________ people at the annual function, weren’t there?
B: Yes, we weren’t expecting so ________.
(d) A: We haven’t bad ________ rain this year, have we?
B: No, there haven’t been ________ rainy days.
(e) A: I don’t think my daughter knows ________ about people!
B: I don’t think ________ children know ________ about people!
Answer:
(a) A: many     B: much
(b) A: much     B: much
(c) A: a lot of   B: many
(d) A: much     B: many
(e) A: much     B: many, much

CHSE Odisha Class 11 English Grammar Countable and Uncountable Nouns

2. few, a few, little, a little.
We have discussed few, a few, little, and a little in Section 4.1. Do mark that a few, few are used with plural countables. They are not generally used with uncountables.
Example:
He is a man of few words, (negative meaning)
A few students passed in English last year, (positive meaning)
There is little water in the glass, (negative meaning)
I need a little help to move the box. (positive meaning)

Now, look at the following examples.
(a) He isn’t very fond of books. He has only a few books at home.
(b) This is a difficult book to read. I’ve had to look up quite a few words in the dictionary.
(c) If what you say is true, there is little we can do about it.
(d) I can’t give you an opinion now. I need a little time to -think.
In the above examples few’ means ‘not many’, ‘a few’ means ‘a small number’ and ‘little’ means ‘not much’, and ‘a little’ means ‘a small amount’.

Activity – 9

Choose the correct alternative in the sentences below.
(a) They could speak few/a few words of Assamese, but they weren’t very fluent.
(b) This is a boring little town; there’s little/a little to do here.
(c) A: Would you like some pepper in your soup?
B: Yes, please, little / a little.
(d) The mud was quite deep. They had little / a little hope of getting out.
(e) Would you like a little/little more tea? There’s still a little/little left in the pot.
(f) I don’t think Ranjan can become a scientist. He’s got little / a little intelligence.
(g) A: Have you ever been to Koraput?
B: Yes, we’ve been there few / a few times.
(h) Father will be away for / a few days next week.
(i) My brother has got a few /few friends in Delhi and he is very happy there.
(j) They won’t take much time to reach the station. There’s a little/little traffic on the road at this time of the day.
Answer:
(a) a few
(b) little
(c) a little
(d) a little
(e) a little, a little
(f) little
(g) a few
(h) a few
(i) a few
(j) a little

CHSE Odisha Class 11 English Grammar Countable and Uncountable Nouns

Activity – 10

Use a little, a lot of, few, a few, fewer, many, and much where required, in the sentences below:
I moved to this neighborhood two years ago. There seemed to be _________people in this area who were without telephones, so I expected to get a new phone quickly. I applied for one as soon as I moved into the new house. “We aren’t supplying _________ new phones in your area”, an engineer told me.” _________ people want new phones at present and the company is employing _________ engineers than last year so as to save money. A new phone won’t cost _________ money, but it will take _________ time. We can’t do anything for you before December. You need _________ patience if you are waiting for a new phone and you should have _________ friends whose phones you can use when necessary.” Fortunately, I had both. December came and went, but there was no sign of a phone. I went to the office of the telephone company to protest. They told me I would have a phone by December. I protested. “Which year ?“ the clerk asked.
Answer:
I moved to this neighborhood two years ago. There seemed to be a lot of people in this area who were without telephones, so I expected to get a new phone quickly. I applied for one as soon as I moved into the new house. “We aren’t supplying many new phones in your area”, an engineer told me. “A lot of people want new phones at present and the company is employing fewer engineers than last year so as to save money. A new phone won’t cost much money, but it will take a lot of time. We can’t do anything for you before December. You need a little patience if you are waiting for a new phone and you should have a few friends whose phones you can use when necessary.” Fortunately, I had both. December came and went, but there was no sign of a phone. I went to the office of the telephone company to protest. They told me I would have a phone by December. I protested. “Which year ?“ the clerk asked.

SECTION – 5

The articles: a/an and the
Study the following sentences.
(a) I met a beggar and an orphan. I didn’t like the beggar much, but the orphan was very nice.
(b) My brother wrote a novel and a play. I found the novel very interesting, but the play was boring.

A / An is used before singular countable nouns when the speaker or the writer does not know the person or thing. It means that when he/she wants to say about a thing or a person for the first time, he/she uses a/an. A/An is used for indefinite things or persons. So, a and an are known as indefinite articles.

When the speaker or writer speaks or writes about a person or thing (not for the first time) for the second, third time, and so on, he/she uses article the. It gives the definite or particular meaning of the noun. It is used for countable and uncountable nouns. So it is called the ‘definite article’.

CHSE Odisha Class 11 English Grammar Countable and Uncountable Nouns

Activity – 11

Use a/an or the, where required, in the blank spaces below.
(a) My uncle lives in _________ small house in _________ remote village. There is _________ beautiful garden behind _________ house. _________ garden has many rare plants.
(b) There are two bags on the table: _________ white one and _________ yellow one. _________ white one belongs to my friend but I don’t know who _________ owner of _________ yellow one is.
(c) My friend witnessed _________ accident this morning. _________ truck crashed into _________ lamp post. _________ driver of _________ truck wasn’t hurt but _________ truck was badly damaged.
(d) _________ taxi drove up to our house. _________ taxi stopped outside our house and _________ woman got out of the taxi. _________ man who was carrying _________ case in his hand also got out. With _________ case in his hand, _________ man looked like _________ salesman.
Answer:
(a) My uncle lives in a small house in a remote village. There is a beautiful garden behind the house. The garden has many rare plants.
(b) There are two bags on the table: a white one and a yellow one. The white one belongs to my friend but I don’t know who the owner of the yellow one is.
(c) My friend witnessed an accident this morning. A truck crashed into a lamp post. The driver of the truck wasn’t hurt but the truck was badly damaged.
(d) A taxi drove up to our house. The taxi stopped outside our house and a woman got out of the taxi. A man who was carrying a case in his hand also got out. With the case in his hand, the man looked like a salesman.

Activity – 12

Complete the story below by using a, an, or the where required.
_________ man walked into _________ bank in America and handed _________ note to one of the cashiers, _________ young woman. _________ woman read _________ note, which told her to give _________ man some money. Afraid that he might have _________ gun, she followed _________ instruction. _________ man then walked out of _________ building, leaving _________ note behind. However, it was not _________ successful crime. _________ man had no time to spend _________ money because he was arrested _________ same day. He had made _________ stupid mistake. He had written _________ note on _________ back of _________ envelope. On _________ other side of _________ envelope was _________ man’s name and address. This information was enough for the police to arrest the man.
Answer:
A man walked into a bank in America and handed a note to one of the cashiers, a young woman. The woman read the note, which told her to give the man some money. Afraid that he might have a gun, she followed the instruction. The man then walked out of the building, leaving the note behind. However, it was not a successful crime. The man had no time to spend the money because he was arrested the same day. He had made a stupid mistake. He had written a note on the back of the envelope. On the other side of the envelope was the man’s name and address. This information was enough for the police to arrest the man.

CHSE Odisha Class 11 English Grammar Countable and Uncountable Nouns

SECTION – 6

More on the indefinite article (a/an).
Look at the following sentences.
My friend is a doctor. He was born in America but now he lives in a small village in a tribal district of Orissa. He works in a hospital attached to a Catholic Church.
A/An is used before singular count nouns.
Example: doctor, village, hospital, etc. Here we refer to persons, people, and things. With the use of a/an article, we usually get an indefinite meaning.

Activity – 13

Complete the sentences below. (You will have to use a/an in the first blank space in each sentence.)
(a) A place where bread is made is called _________.
(b) A shelter for a horse is called _________.
(c) The home of a lion is called _________.
(d) A place where birds are kept is called _________.
(e) A person who mends water pipes is called _________.
(f) A person who tells the future by the stars is called _________.
(g) A person who treats diseases by performing operations is called _________.
(h) A book in which the events of the day are recorded is called _________.
(i) A play with a happy ending is called _________.
(j) A list of the items to be served at a meal is called _________.
Answer:
(a) a bakery
(b) a stable
(c) a den
(d) a cage
(e) a plumber
(f) an astrologer
(g) a surgeon
(h) a diary
(i) a comedy
(j) a menu

Activity – 14

Can you expand the newspaper headlines below into sentences? You will have to use a/an as well as the other words which are missing. The first one has been done for you.
(a) Blast in Billiar town. ➨ There was a blast in a town in Bihar.
(b) Bomb scare delays train. ➨ Bomb scare delays an Express train.
(c) Snake found in a fruit basket at the hotel. ➨ A snake was found in a fruit basket at a hotel.
(d) Indian ship sinks off Abu Dhabi. ➨ An Indian ship sinks off Abu Dhabi.
(e) Ten injured in the clash. ➨ Ten people were injured in a clash.
(f) Bag snatched. ➨ A bag full of gold was snatched by a robber.

CHSE Odisha Class 11 English Grammar Countable and Uncountable Nouns

Activity – 15

Who were these people?
1. Kalidas
Answer:
Kalidas was an Indian poet.

2. Newton
Answer:
Newton was a great scientist.

3. Charlie Chaplin
Answer:
Charlie Chaplin was an artist.

4. Nargis
Answer:
Nargis wasa him-star.

5. Tansen
Answer:
Tansen was a great musician.

SECTION – 7

The definite article: the
(a) My uncle bought the house next to the post office.
(b) We are not keen on meeting the people next door.
(c) He works in the garage opposite our house.
(d) The boy I met at the station last night is going to America on a scholarship.
In the above examples look at the words in italics. They tell us something about the noun that comes just before them. With the use of those situations/contexts, we will
be able to know that they identify the persons, things, or places. They speak about a definite/particular person, thing or place. So ‘the’ is usually used before them.

Activity – 16

Rewrite the sentences below, inserting the definite article the at the places.
(a) Cottage by the river has been deserted for many years.
Answer:
The cottage by the river has been deserted for many years.

(b) I bought the book you referred to yesterday.
Answer:
I bought the book you referred to yesterday.

(c) They wanted to meet old man living in the yellow house.
Answer:
They wanted to meet the old man living in the yellow house.

CHSE Odisha Class 11 English Grammar Countable and Uncountable Nouns

(d) I like tea grown in gardens of Assam.
Answer:
I like the tea grown in gardens of Assam.

(e) This is house that my father built.
Answer:
This is the house that my father built.

SECTION – 8

The definite article used for ‘unique reference’.
Mark the use of ‘the’ in the following sentences.
(a) The moon goes round the earth and the earth goes round the sun.
(b) What is the longest bridge in the world?
(c) No one can predict the end of the universe.
The above nouns such as moon, earth, sun, world, and universe are the unique nouns. Their number is one. They don’t usually have plural form. So the definite article ‘the’ is used before them. The use of a/an is possible before them.
Look at the following example.

  • The sky got dark.
  • There is only one sky in the universe.

Activity – 17

Use the in the blanks, wherever necessary.
(a) _________Prime Minister met _________ President yesterday.
(b) _________ sky is blue.
(c) I am studying History at _________ university.
(d) It’s difficult to live near _________ North Pole.
(e) _________ manufacturing industry is using _________ computers more and more.
(f) Sailors used to spend weeks crossing _________ Atlantic Sea.
(g) _________ equator runs through Africa.
(h) _________ grass is green but _________ grass in my neighbor’s garden is greener.
(i) There are hundreds of small islands in _________ Pacific Ocean.
Answer:
(a) The Prime Minister met the President yesterday.
(b) The sky is blue.
(c) I am studying at university.
(d) It’s difficult to live near the North Pole.
(e) The manufacturing industry is using computers more and more.
(f) Sailors used to spend weeks crossing the Atlantic Sea.
(g) The equator runs through Africa.
(h) Grass is green but the grass in my neighbor’s garden is greener.
(i) There are hundreds of small islands in the Pacific Ocean.

CHSE Odisha Class 11 English Grammar Countable and Uncountable Nouns

SECTION – 9

Look at the use of the in the passage below.
When we got home, everyone seemed to be busy. Brother was working in the garden. Father was reading the newspaper in the living room. Mother was making tea in the kitchen. Sister was playing with her friend on the terrace.
The use of the nouns garden, living room, kitchen, etc. is for a definite or particular thing or place. That is why the is used in the above situations.

Activity – 18

Use the where necessary.
(a) _______ teacher entered _______ classroom, went to _______ table, held _______ duster in his left hand and _______ book in his right hand. Then he cleaned _______ blackboard and started writing a passage from _______ book.
(b) Ravi went to the post office and talked to _______ man at _______ counter. He asked Ravi to meet _______ postmaster. _______ postmaster asked him to sign _______ register. Then he handed _______ parcel over to Ravi.
Answer:
(a) The teacher entered the classroom, went to the table, held the duster in his left hand and the book in his right hand. Then he cleaned the blackboard and started writing a passage from the book.
(b) Ravi went to the post office and talked to the man at the counter. He asked Ravi to meet the postmaster. The postmaster asked him to sign the register. Then he handed the parcel over to Ravi.

Activity – 19

Supply ‘the’ where necessary.
(a) A: Who’s at ______ door?
B: It is ______ postman.
(b) Will you please go to ______ market and get some butter for ______ cake I am baking?
(c) They prefer to spend their holidays in ______ country, among ______ mountains, or by ______ sea.
(d) This is ______ front room. ______ ceiling and ______ walls need repair but ______ floor is in good condition.
(e) A: Where is your sister?
B: She is in ______ kitchen at ______ moment.
Answer:
(a) A: Who’s at the door?
B: It is the postman.
(b) Will you please go to (the) market and get some butter for the cake I am baking?
(c) They prefer to spend their holidays in the country, among the mountains or by the Sea.
(d) This is the front room. The ceiling and the walls need repair but the floor is in good condition.
(e) A: Where is your sister?
B: She is in the kitchen at the moment.

CHSE Odisha Class 11 English Grammar Countable and Uncountable Nouns

SECTION – 10

Place names with and without ‘the’.
We normally use ‘the’ with the following.
(a) Seas and oceans: The Pacific Sea, the Mediterranean.
(b) Rivers : The Mahanadi, the Baitarani, the Kathajori etc.
(c) Canals: The Suez Canal, the Panama Canal, etc.
(d) Deserts The Sahara, the Thar.
(e) Island groups: The West Indies, the Canaries.
(f) Hotels, Cinemas: The Grand Hotel, The Metro Cinema.
(g) Museums, Clubs: The Salarjung Museum (Hyderabad), the Saturday Club, etc.
(h) Restaurants, Pubs: The Sultan Cafe, the Swan (Pub)

Names of ships, trains, newspapers, etc. very often have ‘the’.

  • the Queen Elizabeth (ship)
  • the Falakanama Express
  • the Times of India
  • the harmonium (any musical instrument)

We do not normally use the following place names.
(a) Continents: Europe, Asia, Africa
(b) Countries, States, Departments: India, China, Orissa
These are the single names of the Countries or States. So they do not go with the definite article ‘the’.
(c) Cities, towns, and villages: Delhi, Bhubaneswar, Astarang.
(d) Individual Islands: Long Island
(e) Lakes : Lake Chilika, Lake Ansupa
(f) Individual mountains: Mount Everest
(g) Streets, Roads: College Street, Lewis Road
(h) Names of games: Football, Cricket, Volley Ball.

We use the with place names when they include a count noun
e.g. Union, Republic, State, Kingdom, Isle, etc.
We use ‘the’ before place names containing of phrases with of:
The University of Delhi, The State of Liberty
The Bank of India, The History of India, etc.

We use ‘the’ when we refer to the parts of the body.
→ She pulled me by the hair.
→ I shook her by the hand.
We use ‘the’ before ordinal or cardinal numbers.
→ Bakul is in the sixth class/class six.
→ Today is the fifteenth of August.
→ The book was published in the 1970s / in the seventies.
We use the + adjective to talk about a whole group of people, a class of people e.g. the poor, the young, the old, the weak, the injured etc.

CHSE Odisha Class 11 English Grammar Countable and Uncountable Nouns

Activity – 20

Fill in the blanks with ‘the’ where necessary.
(a) _______ Mount Everest is the highest peak in _______ Himalayas.
(b) _______ India celebrated the fiftieth anniversary of its independence in _______ year 1997.
(c) _______ United Nations is an international organization.
(d) _______ United States of America is a republic.
(e) We visited _______ Nehru Park when we were in Hyderabad.
(f) Is _______ Pacific Ocean larger than _______ Indian Ocean?
(g) We read _______ Statesman every day and _______ India Today every week.
(h) _______ University of Utkal is older than Berhampur University.
(i) _______ Grand Trunk Road is the oldest road in India.
(j) _______ Oberoi is _______ only 5-star hotel in _______ Bhubaneswar.
(k) _______ Thar is in _______ Rajasthan.
(l) _______ President visits _______ Pun tomorrow.
Answer:
(a) Mount Everest is the highest peak in the Himalayas.
(b) India celebrated the fiftieth anniversary of its independence in the year 1997.
(c) The United Nations is an international organization.
(d) The United States of America is a republic.
(e) We visited the Nehru Park when we were in Hyderabad.
(f) Is the Pacific Ocean larger than Indian Ocean?
(g) We read the Statesman every day and the India Today every week.
(h) The University of Utkal is older than Berhampur University.
(i) The Grand Trunk Road is the oldest road in India.
(j) The Oberoi is the only 5-star hotel in Bhubaneswar.
(k) The Thar is in Rajasthan.
(l) The President visits Pun tomorrow.

SECTION – 11

Compare the sentences in each of the following pairs.
1. (a) My sister goes to school every morning.
(b) Mina’s father went to the school to meet the headmaster.
2. (a) He has passed the school final examination and will go to college in July.
(b) The new cinema is very near the college.
3. (a) The injured persons have been sent to the hospital.
(b) The hospital was badly damaged by the cyclone.

CHSE Odisha Class 11 English Grammar Countable and Uncountable Nouns

In the above examples i.e. 1. (a), 2. (a), and 3. (a) ‘the’ is not used before the singular countable nouns like school, college, or hospital. Here the purpose is particular/ primary.
But in sentences 1. (b), 2. (b), and 3. (b) the singular countable nouns school, college, and hospital, do go before them with ‘the’ because the purpose is not primary. Only a few nouns can be used in this way without ‘the’, such as school, college, university, hospital, church, market, and bed.
Do mark deviation: The singular countable noun ‘office’ usually goes with the.
Now mark the following nouns.
work, home.

  • He goes to work at 10.
  • He will be at home this evening.

In the above examples work and home have some special meanings somewhat similar to the meanings of school, and college. That is why they are used without ‘the’. When nouns like a car, bus, train are used as means of travel (communication), they do not go with ‘the’. We do not normally use articles with the names of different meals i.e. breakfast, lunch, brunch, tea (afternoon meal), supper, dinner, etc.

Activity – 21

Put the into the blank, where necessary.
1. They always go to ______ church on Sunday.
2. When do you plan to go to ______ college?
3. My friend has been taken to ______ hospital.
4. The prisoner was sent to ______ jail.
5. My daughter will go to ______ school next year.
6. When do you usually have ______ lunch?
7. My father goes to ______ work at 9 a.m.
8. I’m taking these books back to ______ library.
9. The weather is too bad to go out. I’m staying at ______ home today.
10. Malati has just had a baby. We are going to ______ hospital to visit her.
11. The prisoner’s wife drove to ______ prison to meet her husband.
12. They like lying on the beach. They always spend their holidays at ______ seaside.
13. ______ bed is very expensive.
14. He went to Delhi by ______ train.
Answer:
1. They always go to church on Sunday.
2. When do you plan to go to college?
3. My friend has been taken to hospital.
4. The prisoner was sent to jail.
5. My daughter will go to school .next year.
6. When do you usually have lunch?
7. My father goes to work at 9 a.m.
8. I’m taking these books back to the library.
9. The weather is too bad to go out. I’m staying at home today.
10. Malati has just had a baby. We are going to the hospital to visit her.
11. The prisoner’s wife drove to the prison to meet her husband.
12. They like lying on the beach. They always spend their holidays at the seaside.
13. The bed is very expensive.
14. He went to Delhi by train.

CHSE Odisha Class 11 English Grammar Countable and Uncountable Nouns

Activity – 22

Put in ‘the’ where necessary.
Today Alan Broome is a world-famous actor. Forty years ago he was an unhappy child. He didn’t do well at ______ school and he never went to ______ university. His greatest enjoyment was going to ______ movies. The family lived in an unattractive industrial town in England. Their home was next to ______ railway station. Alan’s father was a sailor, and he spent months at ______ sea. He was seldom at ______ home. When he did come home, he did not do much. Sometimes he would lie all day in ______ bed. His wife had to get up at 5 o’clock to go to ______ work. Mr. Broome lost his job in ______ navy and then went to ______ prison for stealing.
Answer:
Today Alan Broome is a world-famous actor. Forty years ago he was an unhappy child. He didn’t do well at school and he never went to university. His greatest enjoyment was going to movies. The family lived in an unattractive industrial town in England. Their home was next to the railway station. Alan’s father was a sailor, and he spent months at sea. He was seldom at home. When he did come home, he did not do much. Sometimes he would lie all day in bed. His wife had to get up at 5 o’clock to go to work. Mr. Broome lost his job in the navy and then went to prison for stealing.

Activity – 23

Insert a / an and the where necessary, in the passage below.
Push metal rod through cork and then put two pins into cork. Take two more corks and push nails into them. Put pins on two glasses and move cork to right place so that it balances properly. Then you need candle and some matches. Make candle stand on saucer under one side of rod and light it. Heat that comes from candle will make metal expand. Extra length will make rod fall. Experiment shows that beat makes metals expand.
Answer:
Push metal rod through a cork and then put two pins into the cork. Take two more corks and push nails into them. Put pins on two glasses and move the cork to right place so that it balances properly. Then you a need candle and some matches. Make a candle stand on a saucer under one side of the rod and light it. Heat that comes from the candle will make the metal expand. An extra length will make the rod fall. The experiment shows that the heat makes the metals expand.

CHSE Odisha Class 11 English Grammar Story Developing

Odisha State Board CHSE Odisha Class 11 Invitation to English 4 Solutions Grammar Story Developing Textbook Activity Questions and Answers.

CHSE Odisha 11th Class English Grammar Story Developing

A. Introduction:

A story is made up of a number of events or happenings. Let us look at your own story or part of it. On Sundays, I get up at six in the morning. After a quick wash, I get into my jugging rig and go for a run. By 6.30 I am on the road. I run for half an hour. I return home and have a leisurely bathing, a luxury I cannot afford on weekdays. The bath is over, I get ready quickly. What have you done? You have described the events or your activities on a Sunday morning, in the order in which they take place. You can start with what you do first, then go on to what you do next, and so on until you come to your test activity.

B. A beginning, a middle, and an end:

Like any other piece of information, a story has a beginning, a middle, and an end – it is a complete whole. It invites you and opens the door for you; then it leads you through the plot; and finally, it shows you out at the exit, you walk out happy and satisfied and a door gently shuts behind you. The story ‘Jangled Bells’ is a case in point. It shows three stages. The transition from one stage to the next is not abrupt or sudden; it is smooth and natural.

CHSE Odisha Class 11 English Grammar Story Developing

C. The beginning of a story:

How a story begins is very important. The beginning should catch the reader’s attention and urge him to read on. It should set the scene for the action and the mood.

D. Sequence of events:

A story is the narration of action. All action occurs in time. The most natural way of narrating a story is to give the events strictly in the order in which they happened; with one event leading naturally and logically to the next. “What happened then ?” is the question the storyteller should ask himself at every stage.

E. Paragraphing a story:

The events in a story will fall into a few clusters of happenings, each cluster will have unity of time, place, and action. Each cluster can be put into a paragraph. Paragraphing a story is really a simple thing. Just remember that each paragraph tells one part of the story.

F. The Background:

The story must have a background in which the plot takes place. The background is usually set at the very beginning: It is sometimes done through descriptions of the place, the season, the time, etc. The descriptions should be rich in sensory impression — the reader should see, hear, smell, taste, and feel the atmosphere.

For example:
I was traveling across the desert with Mehmood Ah and his caravan of eighty camels and eighteen men. Ah was a dignified old man with fierce dark eyes and a white beard. His commands were the only laws that the men of the caravan knew.
How is the scene of the story set? (desert, caravan — camels, men, etc.)

CHSE Odisha Class 11 English Grammar Story Developing

G. The characters in a story:

The characters are the people in the story. The story is about ‘them — what they do. how they feel, what they say, etc. The characters must seem to be alive and doing things — not just moving through the story like puppets. They must look like people we see in real life; they must be believable. Each character should have special traits. The way he tells, the way he feels, and the way he reacts to things must be distinctly his own. Only then would he become an individual, not just one of the crowd.

H. Attitude:

She did the right thing!
That was an awful thing to do!
‘Poor woman! How she suffered at his hands?’

These are three different reactions to what someone did: approval, condemnation, and sympathy. But the action was the same; the difference is in the way the three people looked at it. It is the storyteller’s attitude that can change the story very much.

I. Dialogue:

It is possible for the storyteller to report what the characters say. But if this is done throughout he story it can become monotonous. Quoting the actual words of the characters will inject life into the story.

J. The end of a story:

A story must have a natural and definite ending. It should come to an end, not just stop suddenly. It should not leave the reader in the air – unhappily and wondering. The ending should give the feeling of completeness like the final knot on a garland of flowers.

Questions :

Question 1.
Complete a story that ends with the following paragraph :
The tail of the plane was in flames and the pilot knew he would not be able to land safely. There was another loud bang somewhere behind him. He made up his mind. He pulled the rip cord. In seconds he found himself sailing through space, the parachute billowing above him. Below him, he saw the plane crash on the field and explode like a bomb.

CHSE Odisha Class 11 English Grammar Story Developing

Answer:
Nick was bored with life. Everybody was exactly the same. He now wanted to break the monotony. It was summer vacation. He decided to go to Kashmir to enjoy its picturesque details. He arranged a plane ticket for the purpose. It was Sunday. He woke up at 6.30. The sun was shining, and the birds were singing. A gentle breeze was blowing. He got ready to catch up on his flight at 9.30. He. reached the airport one hour before the take-off time.

Nick’s mind wandered in the beautiful valley. His spirit soared. He was looking at his watch on and on. The moment he had been waiting for had come at last. The loudspeakers announced her flight. With a heart of excitement, Nick boarded the plane and sat near the window. The plane took off. Nick looked out of the window. What a beautiful sight! All of a sudden, ominous whispering among the passengers caught his attention. Then they gave loud shrieks. Nick felt terribly confused.

To his stunned disbelief, Nick noticed the plane swing violently. The tail of the plane was in flames and the pilot knew he would not be able to land safely. There was another loud bang somewhere behind him. He made up his mind. He pulled the rip cord. In seconds he found himself sailing through space, the parachute billowing above him. Below him, he saw the plane crash on the field and explore like a bomb.

Question 2.
The day was fine and the clock struck nine. I had an interview at ten. What if I got late? I was in the middle of a busy street, (continue …………)

Answer:
The day was fine and the clock struck nine. I had my interview at ten. What if I got late? I was in the middle of a busy street. The vehicles moved at a snail’s pace. I was getting more and more tensed. If I couldn’t make it, my career would be at stake. I had no other option but to wait. I checked out my watch. “God ! it was already 9.30.” My heart started hammering within my ribs. I had to do something. Just then I was reminded of a narrow lane some yards away which was a shortcut.

With much difficulty, I parked my car. But that was not the end to my problems. The traffic police stopped me for the wrong parking. When the senior officer came, I explained the situation to him. Thankfully, he understood and even helped me get there. When I reached there it was just two minutes to ten. I heared a sigh of sweet relief. The moment of truth came at last. 1 was called for the interview. It was a hectic one. I responded to the questions of the members with a great deal of confidence.

They shook hands with me. I noticed smiles on their faces at the time of my departure. A thrilling experience indeed! A month passed by. I had been waiting for the moment when my appointment would come. Sincerity never goes unrewarded. The moment I had been waiting for came at last. I got my appointment letter. Excitement was in the air. I was really on the moon. That day’s experience still lingers in my memory.

CHSE Odisha Class 11 English Grammar Story Developing

Question 3.
Provide a suitable ending to the following story.
As Sandhya was sitting on the steps at the temple two terrorists appeared. They were armed with AK-56 rifles and hand grenades. She was driven into panic at their sight. To her horror, she found them entering the temple. They fired their shots indiscriminately. The silent prayers of the devotees turned into wailing in a flash. Thousands of them ran hither and thither to save themselves from the brutal attacks of the two dreaded terrorists.

The pitiable cries of women and children moved Sandhya to tears. She was fortunate to leave that spot in a flash. I stood at a distance and was a silent spectator to the ghastly scene. They killed 29 innocent devotees and 3 children and injured 74 others. In the night-long operation to flush out the terrorists, the national security guards lost two of its commandos and the State Reserve Police two of its personnel, the terrorists were killed early the next morning.

Answer:
Whenever we meet by chance, Sandhya tells me she cannot forget the harrowing moment of that day. She still remembers when she was having a close look at the snow-white Akshardham temple, sitting on its marble steps. It is the pride of the Swaminarayan Sect. The bloodshed in the place of worship, which stands for universal peace and brotherhood has left many questions unanswered. There should be conceited efforts to eliminate terrorism for all time to come.

CHSE Odisha Class 11 English Grammar Story Developing

Question 4.
Provide a suitable beginning to the story.
At last, a foolish Brahmin passed by that way. Seeing him the tiger begged him to let him come out from the cage. He took pity on the tiger and opened the door of the cage. As soon as the door was opened the tiger came out and wanted to eat the Brahmin. The Brahmin now realized that he had acted foolishly. However, he told the tiger that he had done a good service to him, so he should not eat him. But the ungrateful tiger would not listen to his argument.

He said that he was very hungry. So he must eat him. The Brahmin was quite helpless. At this time a fox came there. He heard from the Brahmin what had happened. He wanted to decide the matter. But first of all, he must see how the tiger got into the cage and how the Brahmin helped him to get out. They agreed. The tiger then got into the cage through the open door. The fox then shut the door.

The tiger was thus again trapped in the cage. The fox now asked the Brahmin to throw the cage into the river. He called other men to help him. They heard the story and dragged the cage to the bank of the nearby river. Then they threw it into the water. The ungrateful tiger was drowned. Thus, the clever fox saved the foolish Brahmin’s life.

Answer:
Once a tiger was caught in a cage. He tried much to get out. The door of the cage was shut and the iron bars of the cage were very strong. So he could not come out. He asked the passers-by to let him out. But none dared to do so. They feared that the tiger would eat them if he could come out once.

CHSE Odisha Class 11 English Grammar Translation

Odisha State Board CHSE Odisha Class 11 Invitation to English 4 Solutions Grammar Additional Questions Textbook Activity Questions and Answers.

CHSE Odisha 11th Class English Grammar Translation

Translate the following passages into English.

Passage – 1

ସୀତା: ତୁମେ ବଜାରରୁ କ’ଣ ସବୁ କିଣିଲ ?
ଗୀତା: ମୁଁ ଗୋଟିଏ ବୋତଲ କ୍ଷୀର, ଏକ ପୁଡ଼ିଆ ଲୁଗାସଫା ପାଉଡ଼ର ଓ ଗୋଟିଏ ଦାନ୍ତଘଷା ପେଷ୍ଟ କିଣିଲି ।
ସୀତା: ଆଉ ମୁଁ ଯେଉଁ ଚକୋଲେଟ କିଣିବାପାଇଁ କହିଥିଲି ?
ଗୀତା: ମୁଁ ଦୁଃଖ । ପୁରାପୁରି ଭୁଲିଗଲି ।
Answer:
Sita : What did you buy in the market?
Rama : I bought a bottle of milk, a packet of washing powde and a tooth paste.
Sita : What aboUt the bar of chocolate I asked you to buy?
Rama : I am sorry. I completely forgot.

Passage – 2

ମୋର ବନ୍ଧୁ ଗୋଟିଏ ଦୂର ଗାଁର ଶେଷ ମୁଣ୍ଡରେ ଏକ ଛୋଟ ଘରେ ବାସ କରନ୍ତି । ସେ ଘରର ପଛପଟେ ଗୋଟିଏ ସୁନ୍ଦର ବଗିଚା ଅଛି ।ସେ ବଗିଚାରେ ଅନେକ ଦୁର୍ଲଭ ଔଷଧ୍ୟ ଗଛ ଅଛି । ତାଙ୍କ ଘର ସାମନାରେ ଜଣେ ବୈଦ୍ୟ ରହନ୍ତି । ସେ ଏହି ଔଷଧ ଗଛରୁ ଔଷଧ ପ୍ରସ୍ତୁତ କରନ୍ତି ।
Answer:
My friend lives in a small house at the end of a distant village. There is a beautiful garden behind that house. There are many rare medicinal plants in that garden. There lives a doctor in front of his house. He prepares medicine from these medicinal plants.

CHSE Odisha Class 11 English Grammar Translation

Passage – 3

ଆଜି ସକାଳେ ମୋର ବନ୍ଧୁ ଏକ ଦୁର୍ଘଟଣା ଦେଖ‌ିଲେ । ଗୋଟିଏ ଆଲୋକ ସ୍ତମ୍ଭରେ ଗୋଟିଏ ଟ୍ରକ ବାଡ଼େଇ ହୋଇଗଲା । ଟ୍ରକ ଚାଳକ ଆହତ ହୋଇନଥିଲେ । କିନ୍ତୁ ଟ୍ରକଟି ବହୁତ କ୍ଷତିଗ୍ରସ୍ତ ହୋଇଥିଲା । ସେ ଜାଗାରେ ବହୁତ ଲୋକ ଜମା ହୋଇଥିଲେ । ପୋଲିସ ଆସିବା ଦେଖୁ ସମସ୍ତେ ପଳାଇଗଲେ ।
Answer:
Today morning my friend witnesses an accident. A truck crashed against a lamp post. The truck driver was not injured. But the truck was badly damaged. Many people had gathered on that spot. On seeing the police, they all fled away.

Passage – 4

ଶିକ୍ଷକ ଶ୍ରେଣୀଗୃହରେ ପ୍ରବେଶ କଲେ ଓ ତାଙ୍କ ଟେବୁଲ ପାଖକୁ ଗଲେ । ତାଙ୍କ ବାମ ହାତରେ ଗୋଟିଏ ଡଷ୍ଟର ଓ ଡାହାଣ ହାତରେ ଗୋଟିଏ ବହି ଥିଲା । ତା’ପରେ ସେ କଳାପଟାକୁ ସଫା କଲେ ଓ ସେ ବହିରୁ ଗୋଟିଏ ଅନୁଚ୍ଛେଦ ଲେଖିଲେ । ଆମକୁ ସେ ଅନୁଚ୍ଛେଦକୁ ଇଂରାଜୀରେ ଭାଷାନ୍ତର କରିବାକୁ କହିଲେ । ସେ ଅନୁଚ୍ଛେଦଟି ଭାଷାନ୍ତର କରିବା ସହଜ ନ ଥିଲା ।
Answer:
The teacher entered into the classroom and went to the table. He held a duster in his left hand and a book in his right hand. Then he cleaned the blackboard and started writing a passage from the book. He asked us to translate that passage in to English. It was not easy to translate the passage.

Passage – 5

ଚା ତିଆରି କରିବାପାଇଁ ଗୋଟିଏ ସସ୍‌ପ୍ୟାନ୍‌ରେ କିଛି ପାଣି, କିଛି କ୍ଷୀର ଓ ଅଳ୍ପକିଛି ଚା’ ନେବା ଆବଶ୍ୟକ । ସେ ସବୁକୁ ଭଲରୂପେ ଫୁଟାଇବା ଦରକାର । ଗୋଟିଏ ଚା’ ଛଣାଦ୍ଵାରା ଚା’କୁ ଛାଣିବା ଦରକାର । ସେଥ‌ିରେ ସ୍ଵାଦ ମୁତାବକ ଚିନି ମିଶାଇବା ଆବଶ୍ୟକ । ଗୋଟିଏ କପରେ କିଛି ଚା ଓ ଗୋଟିଏ ପିଆଲାରେ କିଛି ବିସ୍କୁଟ ନେଇ ଅତିଥିଙ୍କୁ ଦିଅନ୍ତି । ସେ ଖୁସି ହେବେ ।
Answer:
In order to prepare tea, it is necessary to put some water, some milk and a little tea in a saucepan. They need to be boiled well. Tea needs filtration with a strainer. It is necessary to mix sugar with that according to one’s taste. Give your guest a cup of tea and a few biscuits on a plate. He will be glad.

Passage – 6

ଗୋଟିଏ ସୋଲଠିପି ଭିତରେ ଗୋଟିଏ ଲୁହାଛଡ଼ ଭର୍ତ୍ତି କର । ସେଇ ସୋଲଠିପିରେ ଦୁଇଟି ପିକଣ୍ଟା ମଧ୍ୟ ଭର୍ତ୍ତି କର । ଦୁଇଟି ଗିଲାସକୁ ଓଲଟାଇ ରଖ । ତା’ ଉପରେ ଆଉ ଗୋଟିଏ ଲୁହାଛଡ଼ ରଖ । ସୋଲଠିପିକୁ ଦ୍ଵିତୀୟ ଛଡ଼ ଉପରେ ସତର୍କତାର ସହ ରଖ । ପ୍ରଥମ ଲୁହାଛଡ଼ଟି ଠିପିର ଉଭୟ ପାର୍ଶ୍ଵରେ ସମତୁଲ ରହିବା ଦରକାର । ଗୋଟିଏ ମହମବତୀ ଜାଳ । ତା’ ଦ୍ଵାରା ପ୍ରଥମ ଲୁହାଛଡ଼ର ଗୋଟିଏ ପାର୍ଶ୍ଵକୁ ଗରମ କର । ମହମବତୀର ଉତ୍ତାପ ଯୋଗୁଁ ଲୁହାଛଡ଼ର ସମ୍ପ୍ରସାରଣ ହେବ । ସୋଲଠିପିଟି ଗୋଟିଏ ପାର୍ଶ୍ଵକୁ ଢଳିପଡ଼ିବ । ଏଥୁରୁ ଆମେ ଜାଣିପାରୁଛୁ ଯେ ଧାତବ ପଦାର୍ଥ ଉତ୍ତାପ ଯୋଗୁଁ ସମ୍ପ୍ରସାରିତ ହୁଏ ।
Answer:
Push an iron rod through a cork. Put two pins into that cork too. Put two glasses upside down. Keep another iron rod on it. Keep that cork carefully on the second rod. The first iron rod needs to be kept in balance on both sides of the cork. Bum a candle. Heat one side of the first iron rod with it. The iron rod will expand as a result of the candle’s heat. That cork will lean towards one side. From this we know that heat makes a metal expand.

CHSE Odisha Class 11 English Grammar Translation

Passage – 7

ଆମେ ଶୁକ୍ରଗ୍ରହକୁ ସନ୍ଧ୍ୟାତାରା ବା କୁଆଁତାରା ରୂପେ ଜାଣୁ । କିନ୍ତୁ ଏହା ଏକ ତାରା ନୁହେଁ । ଏହା ସୌରମଣ୍ଡଳର ଏକ ଗ୍ରହ । ପୃଥ‌ିବୀ ଅପେକ୍ଷା ଶୁକ୍ରଗ୍ରହ ସୂର୍ଯ୍ୟର ନିକଟତର । କେତେକ ବୈଜ୍ଞାନିକ ବିଶ୍ଵାସ କରନ୍ତି ଯେ ଶୁକ୍ରଗ୍ରହ ଏକ ପ୍ରକାଣ୍ଡ ମରୁଭୂମି । କିନ୍ତୁ ଅନ୍ୟମାନେ ଭାବନ୍ତି ଯେ ଏହା ପାଣି ଓ ଜଳୀୟବାଷ୍ପରେ ପୂର୍ଣ୍ଣ ।
Answer:
We know the planet Venus as the evening star or the morning star. But this is not a star. This is a planet of the solar system. The planet Venus is nearer to the sun than the earth. Some scientists believe that the planet Venus is a vast desert. But others think that it is full of water and water vapor.

Passage – 8

ମୁଁ ସବୁଦିନ ସକାଳ ୫ଟାରେ ଶଯ୍ୟା ତ୍ୟାଗ କରେ । ସକାଳେ ତିନିଘଣ୍ଟା ପଢ଼େ । ୧୦ଟାବେଳେ କଲେଜକୁ ଯାଏ । କଲେଜରୁ ଫେରି ଜଳଖୁଆ ଖାଏ ଓ ଫୁଟବଲ୍ ଖେଳିବାକୁ ଯାଏ । ସନ୍ଧ୍ୟାବେଳେ ଚାରିଘଣ୍ଟା ପଢ଼େ । ଆଜିକାଲି ମୋର ସାଙ୍ଗ ମାନେ ଫୁଟବଲ୍‌ ଖେଳୁ ନାହାନ୍ତି । ତେଣୁ ମୁଁ କ୍ରିକେଟ୍ ଖେଳୁଛି । ଆପଣମାନେ ଜାରିଥ‌ିବେ, କ୍ରିକେଟ୍ ଭାରତର ସବୁଠାରୁ ଲୋକପ୍ରିୟ ଖେଳ ।
Answer:
I get up every day at 5 in the morning. I read for three hours in the morning. I go to college at 10. Returning from college, I take tiffin and go to play football. I read for four hours in the evening. Nowadays, my friends are not playing football. So I am playing cricket. You know cricket is the most popular game of India.

Passage – 9

ଏ ବର୍ଷ ମୁଁ ଦିଲ୍ଲୀ ବିଶ୍ବବିଦ୍ୟାଳୟରେ ଇଂରାଜୀ ପଢୁଛି । ଗୋଟିଏ ଘରୋଇ ଗ୍ରନ୍ଥାଗାରରେ ମୁଁ ସାମୟିକ ଭାବରେ କାମ କରୁଛି । ଭାଗ୍ୟବଶତଃ ମୁଁ ଏହି ଚାକିରୀଟି ପାଇଛି । ମୋତେ ସକାଳୁ ଉଠିବାକୁ ପଡ଼ୁଛି । ଗ୍ରନ୍ଥାଗାର ୧୦ଟାବେଳେ ଖୋଲେ ସ ୭ଟା ବେଳେ ବନ୍ଦ ହୁଏ । କିନ୍ତୁ ମୁଁ ୩ଟାରୁ ୭ଟା ପର୍ଯ୍ୟନ୍ତ କାମ କରେ । କାମଟି ଆମୋଦଦାୟକ, କାରଣ ଲୋକମାନେ ମୋ ପାଖକୁ ଆସନ୍ତି ଓ ମୋର ସାହାଯ୍ୟ ମାଗନ୍ତି । ତେଣୁ ମୁଁ ବିଭିନ୍ନ ବିଷୟରେ ବହୁତ କଥା ଶିଖୁଛି । ବହିରେ ଚିହ୍ନ ଦେବାପାଇଁ ଲୋକମାନେ ଅଦ୍ଭୁତ ଜିନିଷସବୁ ବ୍ୟବହାର କରନ୍ତି । ଥରେ ମୁଁ ଗୋଟିଏ ଫଟୋ ପାଇଥିଲି । ତା ପଛରେ ଲେଖାଥିଲା, ‘ମୁଁ ତୁମକୁ ଭଲପାଏ ।’’
Answer:
This year I am reading English at Delhi University. I am doing a part-time job in a private library. Fortunately, I have got this job. I have to get up in the morning. The library opens at 10 and closes at 7. But I work from 3 to 7. The work is interesting because people come to me and ask me to help. So I am learning a lot about different subjects. People use strange things as bookmarks. Once I found a photograph. On its back were the words, “I love you”.

CHSE Odisha Class 11 English Grammar Translation

Passage – 10

ବହୁତ କାମ କରିବାକୁ ବାପା ମୋତେ କହିଥିଲେ । ମୁଁ ମୋର ଘରପାଠ୍ୟ ସାରିଦେଇଛି । ବଜାରରୁ ପରିବା କିଣି ଆଣିଛି । ଆଜିକାର ଖବରକାଗଜ ପଢ଼ିସାରିଛି । କିନ୍ତୁ ସ୍କୁଟରକୁ ଏ ପର୍ଯ୍ୟନ୍ତ ଧୋଇନାହିଁ । ଆଉ ବଗିଚାର ଗଛଗୁଡ଼ିକରେ ପାଣି ଦେଇନାହିଁ ।
Answer:
Father told me to do a lot of work. I have finished my homework. I have bought vegetables from the market. I have read today’s newspaper. But I have not washed my scooter yet. I have also not watered the plants in the garden.

Passage – 11

ମଧୁ ମୋର ପୁରାତନ ବନ୍ଧୁ । କାଲି ରେଳଷ୍ଟେସନରେ ହଠାତ୍ ତା ସହ ଦେଖା ହେଲା । ଗତ ପାଞ୍ଚବର୍ଷ ଧରି ମୁଁ ତାକୁ ଦେଖୁ ନ ଥିଲି । ତା’ର ବହୁତ ପରିବର୍ତ୍ତନ ହୋଇଯାଇଛି । ସେ ବହୁତ ଡେଙ୍ଗା ହୋଇଯାଇଛି । ତା’ର ସ୍ଵର ଗମ୍ଭୀର ହୋଇଯାଇଛି । ସେ ଆଜିକାଲି ପ୍ରାଞ୍ଜଳ ଭାବରେ ଇଂରାଜୀ କହୁଛି । ତା’ର ବ୍ୟକ୍ତିତ୍ଵ ଉନ୍ନତ ହୋଇଯାଇଛି । ଏବେ ସେ ପାଖ ସହରରେ ପାଠ ପଢୁଛି ।
Answer:
Madhu is my old friend. I met him suddenly at the railway station yesterday. I had not seen him for the last five years. He has changed a lot. He has become very tall. His voice has become grave. Nowadays, he speaks English fluently. His personality. has developed. Now he is reading in a nearby town.

Passage – 12

ତୁମେ : ହଇରେ, ତୁମ କେବେ ଘୋଡ଼ା ଚଢ଼ିଛୁ ?
ମଧୁ : ହଁ ଚଢ଼ିଛି ।
ତୁମେ: ତୁ କେବେ ଚଢ଼ିଲୁ ?
ମଧୁ : ଗତ ଖରା ଛୁଟିରେ ।
ତୁମେ : କେମିତି ଲାଗିଲା ?
ମଧୁ : ଭୟଙ୍କର ଲାଗିଲା ।
ତୁମେ : କାହିଁକି ? କ’ଣ ହେଲା କି ?
ମଧୁ : ମୁଁ ଘୋଡ଼ା ଉପରୁ ଖସି ପଡ଼ିଲି ।
Answer:
You : Hello, have you ever ridden a horse?
Madhu : Yes, I have.
You : When did you ride?
Madhu : I rode one last summer.
You : What was it like?
Madhu : Oh, it was terrible.
You : Why? What happened?
Madhu : I fell off the horse.

CHSE Odisha Class 11 English Grammar Translation

Passage – 13

ଆମେ କାଲି ଅନୀଲର ଘରକୁ ଯାଇଥିଲୁ । ତା ଦ୍ୱାରରେ ଠକ୍‌ ଠକ୍ କଲୁ କିନ୍ତୁ କୌଣସି ଉତ୍ତର ପାଇଲୁ ନାହିଁ । କେହି ଦ୍ଵାର ଖୋଲିଲେ ନାହିଁ କି କାହାର ସ୍ଵର ଶୁଭୁ ନ ଥିଲା । ବୋଧହୁଏ ସେ ବାହାରକୁ ଯାଇଥିଲା କିମ୍ବା ଆମକୁ ସାକ୍ଷାତ କରିବାକୁ ଚାହୁଁ ନ ଥିଲା । ଆଗରୁ ସେ ବହୁତ ମେଳାପୀ ଥିଲା । ତା’ର କ’ଣ ହୋଇଛି କି ?
Answer:
We went to Anil’s house yesterday. We knocked at his door but we didn’t get any response. Nobody opened the door nor spoke anything. He either had gone out or did not want to meet us. He was very sociable before. What has happened to him?

Passage – 14

ଆମେ ଗାଁରେ ଅବୋଲକରା ନାମରେ ଗୋଟିଏ ଭିକାରୀ ଅଛି । ଗତକାଲି ସେ ଆମ ଦୁଆର ବାଡ଼େଇଲା । ବକତେ ଖାଇବାକୁ ମାଗିଲା । ତା’ ବଦଳରେ ସେ ନାଚିଲା ଓ ଗୀତଟିଏ ଗାଇଲା । ମୁଁ ତାକୁ ବକତେ ଖାଇବାକୁ ଦେଲି । ଖାଇବା ଶେଷରେ ସେ ଗୋଟିଏ ରସଗୋଲା ମାଗିଲା । ସେ କହିଲା ଯେ ସେ ଦିନ ତା’ର ଜନ୍ମଦିନ । ସେ ମାସକୁ ଥରେ ପ୍ରତ୍ୟେକ ଘରକୁ ବୁଲି ଆସେ । ସବୁଦିନ ତା’ର ଜନ୍ମଦିନ ଥାଏ ।
Answer:
There is a beggar Abolkara by name in our village. He banged our door yesterday. He asked us for a meal. He danced and sang a song in return for this. I offered him a meal. He asked for a rasgola after the meal. He said that that day was his birthday. He calls at every house once a month. His birthday comes every day.

Passage – 15

ମା : ତୁମେ ଆର ମାସରେ ଷ୍ଟଡିଟୁର ଯାଉଛ କି ?
ସୁରଭି : ହଁ ମା, ଆମେ ଦାର୍ଜିଲିଂ ଯାଉଛୁ ।
ମା : ତୁମ ସାଥ୍‌ରେ ଆଉ କିଏ ଯାଉଛନ୍ତି ?
ସୁରଭି : ମୋ ସାଥ୍‌ରେ ସାଙ୍ଗମାନେ ଓ ଶିକ୍ଷକମାନେ ଯାଉଛନ୍ତି ।
ମା : ତୁମେ ଦାର୍ଜିଲିଂ କିପରି ଯିବ ?
ସୁରଭି : ଆମେ ବସ୍‌ରେ ଯିବୁ।
ମା : ତୁମେ ସେଠାରେ କେଉଁଠି ରହିବ ?
ସୁରଭି : ହୋଟେଲରେ ।
Answer:
Mother : Are you going on the studio next month ?
Surabhi : Yes mother, we are going to Darjling.
Mother : Who else are going with you ?
Surabhi : Friends and teachers are accompanying me.
Mother : How will you go to Darjleeng ?
Surabhi : We shall go by bus.
Mother : Where will you stay there ?
Surabhi : We shall stay at the hotel.

CHSE Odisha Class 11 English Grammar Translation

Passage – 16

ମନ୍ତ୍ରୀ : ଆଗାମୀ ସପ୍ତାହରେ ମୋ ଗସ୍ତର କାର୍ଯ୍ୟସୂଚୀ ଠିକ୍ କରିଛ ?
ବ୍ୟକ୍ତିଗତ ସଚିବ : ହଁ ଆଜ୍ଞା, ସୋମବାର ୭ଟାବେଳେ ଆପଣ ସମ୍ବଲପୁର ଯିବାପାଇଁ ବାହାରିବେ ।
ମନ୍ତ୍ରୀ : ସମ୍ବଲପୁରରେ କେତେବେଳେ ପହଞ୍ଚିବି ?
ବ୍ୟକ୍ତିଗତ ସଚିବ : ଆପଣ ୧୧ଟାବେଳେ ପହଞ୍ଚେ । ବାଟରେ ଅନୁଗୁଳଠାରେ କିଛି ସମୟପାଇଁ ଅଟକିବେ । ସମ୍ବଲପୁରରେ ମଧ୍ୟାହ୍ନ ଭୋଜନ ସରିଲାପରେ ଆପଣଙ୍କ ସହ କମିଶନର ଦେଖା କରିବେ ଓ କେତେକ ସମସ୍ୟା ବିଷୟରେ ଆଲୋଚନା କରିବେ ।
ମନ୍ତ୍ରୀ : ମୁଁ ସମ୍ବଲପୁରରେ କେଉଁଠି ରହିବି ?
ବ୍ୟକ୍ତିଗତ ସଚିବ : ହୀରାକୁଦ ବନ୍ଧ ପାଖରେ ଥ‌ିବା ଅତିଥ୍‌ଭବନ ଆପଣଙ୍କପାଇଁ ସଂରକ୍ଷିତ ହୋଇଛି । ତା’ ପରଦିନ ସକାଳ ୮ଟାରେ ଆପଣ ରାଉରକେଲା ଅଭିମୁଖେ ଯାତ୍ରା ଆରମ୍ଭ କରିବେ । କମିଶନର ଆପଣଙ୍କ ସାଥ୍‌ରେ ଯିବେ ।
Answer:
Minister : Have you drawn up my tour programme for the next week ?
Personal Secretary : Yes, sir. You leave for Sambalpur at 7.00 a.m. on Monday morning.
Minister : When do I get there ?
Personal Secretary: You reach Sambalpur at 11 a.m. You halt at Angul for a few minutes on the way. Then, after lunch, you meet the commissioner for a discussion on some problems.
Minister: Where do I stay in Sambalpur?
Personal Secretary: The guest house at the Hirakud Dam has been reserved for you, Sir. The next morning at 8.00 you set out for Rourkela. The commissioner accompanies you.

Passage – 17

ମାତୃପକ୍ଷୀ : ତୁ କିଛି ଚାଉଳ ଆଣିପାରିବୁ କି ?
ଛୁଆ ଚଢ଼େଇ : ନାଁ, ମା, ମୁଁ ପାରିବି ନାହିଁ।
ମାତୃପକ୍ଷୀ : ଯେତିକି ଚାଉଳ ଅଛି ତା’କୁ କୁଟିପାରିବୁ କି ?
ଛୁଆ ଚଢ଼େଇ : ନାଁ, ମୋ ହାତ ଦରଜ ହୋଇଛି ।
ମା : ତୁ ବୁଲି ଲଗାଇ ପାରିବୁ କି ?
ଛୁଆ ଚଢ଼େଇ : ନାଁ, ମୋ ଆଖିରୁ ଲୁହ ଗଡ଼ିବ ?
ମା ତା’ ହେଲେ ପିଠା କେମିତି ଖାଇବୁ ?
ଛୁଆ ଚଢ଼େଇ : ହଁ ତା’କୁ ମୁଁ ଖାଇପାରିବି ।
Answer:
Mother bird : Can you bring some rice?
Birdie : No mother, I can’t.
Mother : Can you pound the rice that we have?
Birdie : No, my hand aches.
Mother : Can you light the hearth?
Birdie : No, tears will roll down my eyes.
Mother :. Then how will you eat cake?
Birdie : Yes, I can eat it.

CHSE Odisha Class 11 English Grammar Translation

Passage – 18

ସ୍କୁଲରେ ପଢ଼ିବାବେଳେ ଏ ଝିଅଟି ବହୁତ ପରିଶ୍ରମ କରୁଥିଲା । ସେ ସକାଳ ୫ଟାରେ ଉଠୁଥିଲା । ଘଣ୍ଟାଏ ଗୀତ ଶିଖୁଥିଲା । ଦୁଇ ଘଣ୍ଟା ପଢ଼ୁଥିଲା । ଦିନ ୧୦ଟାରେ ସ୍କୁଲ ଯାଉଥିଲା । ତା’ ପରେ ଗୋଟିଏ ଦୋକାନରେ ଘଣ୍ଟାଏ କାମ କରୁଥିଲା । ସନ୍ଧ୍ୟାରେ ସାନଭାଇକୁ ପାଠ ପଢ଼ାଉଥିଲା । ତା’ ପରେ ରୋଷେଇ କରୁଥିଲା। ଡେରିରେ ଶୋଇବାକୁ ଯାଉଥିଲା । ଏବେ ସେ ଜଣେ ପ୍ରସିଦ୍ଧ ବୈଜ୍ଞାନିକ ।
Answer:
This girl would work hard when she was a school student. She would get up at 5 a.m. She would learn music for an hour. She would read for two hours. She would go to school at 10 a.m. Then she would work in a shop for an hour. She would teach her younger brother in the evening. After that she would cook. She would go to bed late. Now she is a famous scientist.

Passage – 19

ଆଲେକ୍ସଜାଣ୍ଡାର୍ : ମୁଁ ଆପଣଙ୍କ ସହ କିପରି ବ୍ୟବହାର କରିବା ଉଚିତ ?
ପୁର : ଯେପରି ଗୋଟିଏ ରାଜା ଅନ୍ୟ ଜଣେ ରାଜାଙ୍କ ସହ ବ୍ୟବହାର କରିବା କଥା ।
ଆଲେକ୍ସଜାଣ୍ଡାର୍ : ଆପଣ ଜଣେ ସାହସୀ ଲୋକ । ଆପଣ ମୋର ବନ୍ଧୁ ହେବେ କି ?
ପୁର : ଗୋଟିଏ ସର୍ଭରେ ।
ଆଲେକ୍ସଜାଣ୍ଡାର୍ : ଆପଣଙ୍କ ସର୍ଭ କ’ଣ ?
ପୁର : ମୋ ରାଜ୍ୟ ସ୍ଵାଧୀନ ରହିବା ଉଚିତ । ଆପଣ ମୋତେ ଜଣେ ସମାସ୍କନ୍ଧ ଭଳି ବ୍ୟବହାର କରିବା ଉଚିତ ।
Answer:
Alexander : How should I behave with (treat) you?
Puru : As a king should behave with (treat) another king?
Alexander : You are a brave man. Will you be my friend?
Puru : On one condition.
Alexander : What is your condition?
Puru : My kingdom should remain independent. You should treat me as an equal.

Passage – 20

ଅଶୋକ: ବାହାରେ ଏ ବହି ପ୍ୟାକେଟ୍ କିଏ ରଖିଛି ?
ଅଳକା : ଡାକବାଲା ନିଶ୍ଚୟ ଏଇଟାକୁ ଛାଡ଼ି ଯାଇଥବ ।
ଅଶୋକ : ସେ କାହିଁକି ଘଣ୍ଟି ବଜାଇଲା ନାହିଁ ?
ଅଳକା : ସେ ନିଶ୍ଚୟ ବଢାଇଥ । ବୋଧହୁଏ ମୁଁ ଶୁଣିପାରିଲି ନାହିଁ ।
ଅଶୋକ ତୁମେ ଅଧିକ ମନୋଯୋଗୀ ହେବା ଉଚିତ । ବହି ପୁଡ଼ିଆଟି ହଜିଯାଇଥା’ନ୍ତା ।
Answer:
Ashok : Who has kept this book packet outside?
Alka : The postman must have left it.
Ashok : Why didn’t he ring thè calling bell?
Alka : He must have rung. I could not hear it.
Ashok : You should be more attentive. The book packet would have been lost.

CHSE Odisha Class 11 English Grammar Translation

Passage – 21

ମୁଁ ଭୁବନେଶ୍ଵର ଭୋର ୩ଟାବେଳେ ପହଞ୍ଚିବି । କିନ୍ତୁ କାର୍ଯ୍ୟାଳୟସବୁ ୧୦ଟା ବେଳକୁ କାମ ଆରମ୍ଭ କରନ୍ତି । ମୁଁ ରାତିସାରା ନିଶ୍ଚୟ ଅନିଦ୍ରା ହୋଇଥବି । ତେଣୁ କିଛି ଘଣ୍ଟା ବିଶ୍ରାମ ନେବା ଦରକାର । ଦିନକ ଭିତରେ କାମ ସରିବନି । ତେଣୁ ଦୁଇ ଦିନପାଇଁ ଗୋଟିଏ ହୋଟେଲରେ ରହିଯିବା ଉଚିତ ହେବ ।
Answer:
I shall reach Bhubaneswar at 3 a.m. But offices start functioning from 10 a.m. I shall have kept awake throughout the night. So I need to take rest for some hours. The work will not be finished in a day’s time. So it will be better to stay at a hotel for two days.

Passage – 22

ଆମ ଗାଁ ପାଖରେ ଗୋଟିଏ ନଈ ବହିଯାଇଛି । ତା’ କୂଳରେ ଅନେକ ବଡ଼ ବଡ଼ ଗଛ ଅଛି । ଆମେ ସେ ନଈ କୂଳରେ ଖେଳୁ । ଗଛମୂଳେ ବସି ଗପସପ କରୁ । ମୁଁ ସେଇ ନଈରେ ପହଁରା ଶିଖିଛି । ସେଇ ନଈ ପାରହୋଇ ମୁଁ ସ୍କୁଲ ଯାଉଥୁଲି । ଏବେ ସେଇ ନଈ ଉପରେ ଗୋଟିଏ ନୂଆ ପୋଲ ତିଆରି ହୋଇଛି । ମୋ ସାନଭାଇ ଏବେ ସାଇକେଲରେ ସ୍କୁଲକୁ ଯାଉଛି ।
Answer:
A river flows by our village. There are many big trees on its bank. We play on the river bank. We gossip at the foot of the tree. I have learnt swimming in that river. I was going to school crossing the river. A new bridge has been built on the river. My younger brother is now going to school on his bicycle.

Passage – 23

ମୁଁ ଆଗାମୀ ସପ୍ତାହରେ ଆପଣଙ୍କୁ ଦେଖା କରିବି । ଦୟାକରି ସେ ପର୍ଯ୍ୟନ୍ତ ମୋ କୁକୁରର ଯତ୍ନ ନେଉଥ‌ିବେ । ତା’କୁ ବାହାରକୁ ଛାଡ଼ିବେ ନାହିଁ । ଅନ୍ୟ କୁକୁରମାନେ ତା’କୁ ଦେଖ‌ିଲେ ଭୁକିବେ । ତା’କୁ କାମୁଡ଼ିଦେଇ ପାରନ୍ତି । ସେ ଏତେ ଭୟାଳୁ ଯେ ଭୟରେ ଅଣାୟତ୍ତ ହୋଇଯାଇପାରେ । ତା’ ପରେ ତା’କୁ ସମ୍ଭାଳିବା କଷ୍ଟ ହୋଇପଡ଼ିବ ।
Answer:
I shall call on you next week. Please, look after my dog till that day. Don’t allow him to go out. Other dogs will bark at him. They may bite him. He is so timid that he becomes uncontrollable out of fear. Then it will be difficult to control him.

Passage – 24

ଆଜିକାଲି ମୁଁ ଗୋଟିଏ ଘରୋଇ କାରଖାନାରେ କାମ କରୁଛି । ଗତ ଅଗଷ୍ଟ ମାସରେ ମୁଁ ସେ କାମପାଇଁ ଦରଖାସ୍ତ କରିଥିଲି ଓ ନଭେମ୍ବରରେ କାମ ଆରମ୍ଭ କଲି । ମୁଁ ବିଶେଷ ରୋଜଗାର କରେ ନାହିଁ; ଏପରିକି ମୋ ଅଫିସ ପୋଷାକ ମୁଁ ନିଜେ କିଣିଛି । ବୋଧହୁଏ ମୁଁ ଏବେ ଦରମା ବଢ଼ାଇବାପାଇଁ ଅନୁରୋଧ କରିବା ଉଚିତ୍ । କିନ୍ତୁ ପ୍ରକୃତରେ ମୁଁ ପଇସାପାଇଁ ଏତେ ବ୍ୟସ୍ତ ନୁହେଁ । କାମ ମୋପାଇଁ ଅଧିକ ଗୁରୁତ୍ଵପୂର୍ଣ୍ଣ । ଆଶା, ଆପଣ ମୋ ସହିତ ଏକମତ
ହେବେ ।
Answer:
At present, I am working in a private firm. I applied for that job last August and joined in November. I don’t earn much; even I myself have bought my office uniform. Of course, I should ask for my pay hike. But, in reality, I am not so worried about money. Work is more important to me. Hope, you will agree with me.

CHSE Odisha Class 11 English Grammar Translation

Passage – 25

ଅପରିଚିତ : ଆଜ୍ଞା, ଟିକିଏ ଶୁଣିବେ ? ଲୁଇସ୍ ରୋଡ଼କୁ କିପରି ଯିବାକୁ ହୁଏ ମୋତେ ବତାଇପାରିବେ କି ?
ଯଦୁ : ହଁ ନିଶ୍ଚୟ । ଏଇ ରାସ୍ତାରେ ଆଗ ଛକ ପର୍ଯ୍ୟନ୍ତ ଯାଆନ୍ତୁ । ସେ ଛକଟି ଡେଇଁ ଯାଆନ୍ତୁ ଏବଂ ଆଗେଇ ଚାଲନ୍ତୁ । ତା’ ପର ଛକ ପାଖରେ ଡାହାଣକୁ ବୁଲି ଯାଆନ୍ତୁ । ପ୍ରଥମ ବାମପଟ ରାସ୍ତା ନିଅନ୍ତୁ । ଅଳ୍ପ କିଛି ବାଟ ପରେ ଲୁଇସ୍ ରୋଡ଼ରେ ପହଞ୍ଚିଯିବେ । ବାଟ ଭୁଲିଗଳେ କାହାକୁ ପଚାରନ୍ତୁ । କିନ୍ତୁ ଏ ରାସ୍ତା ପାଇବା ଖୁବ୍ ସହଜ।
ଅପରିଚିତ : ଧନ୍ୟବାଦ ।
Answer:
Stranger: Excuse me. Could you tell me how to get to Lewis Road, please?
Jadu: Yes, certainly. Go along this road till you reach the next square. Cross that square and continue going. Turn right there. Take the first turning to the left. After walking for a while you reach Lewis Road. Ask someone if you get lost. But it is quite easy to find it.
Stranger : Thank you.

Passage – 26

ଗୋଟିଏ ଅଣ୍ଡା ଭାଙ୍ଗ ଏବଂ ତା’କୁ ଗୋଟିଏ ତାଟିଆରେ ରଖ । ତା’କୁ ଫେଣ୍ଟି ଫେଣ୍ଟି ମିଶାଅ । ସୁଆଦ ମୁତାବକ ଲୁଣ ଓ ଗୋଲମରିଚ ମିଶାଅ । ପିଆଜକୁ ଛୋଟ ଛୋଟ କରି କାଟି ମିଶାଅ । ଅଳ୍ପ ଟିକିଏ ଲହୁଣୀ ବା ତେଲ ନେଇ ତାଉଆକୁ ଗରମ କର । ସେଥ‌ିରେ ଅଣ୍ଡାକୁ ଢାଳ । ଆମ୍‌ଲେଟ୍‌ ଯେପରି ଜଳି ନ ଯାଏ ସେଥ୍ୟପ୍ରତି ଲକ୍ଷ୍ୟ ରଖ । ଗୋଟିଏ ପାଖ ଭାଜିହେବା ସଙ୍ଗେ ସଙ୍ଗେ ତାହାକୁ ଓଲଟାଇ ଦିଅ । ଦୁଇ ମିନିଟ୍ ପରେ ଓମ୍‌ଲେଟ୍ ପ୍ରସ୍ତୁତ ହୋଇଯିବ ।
Answer:
Break an egg and put it in a bowl. Mix it after beating. Add salt and pepper to taste. Chop an onion to pieces and mix them. Heat a little butter or oil in a frying pan. Then pour the mixture and make sure that the omelette does not burn. Turn it over as soon as the bottom is fried. Omelette is ready after two minutes.

CHSE Odisha Class 11 English Grammar Translation

Passage – 27

A : ଆପଣ ଯାହା ଭାବିଛନ୍ତି ତାହା କ’ଣ ଏହି ପ୍ରକୋଷ୍ଠରେ ଅଛି ?
X : ହଁ ନିଶ୍ଚୟ ।
B : ତାହାର ଆକାର, ଆୟତନ ଓ ରଙ୍ଗ କ’ଣ ?
X : ସେ ପଦାର୍ଥର ଆକାର, ଆୟତନ ଓ ରଙ୍ଗ ନ ଥାଏ।
C : ତାହା କ’ଣ ଆମରି ଭାରି ଦରକାରୀ ?
X : ତାହା ବିନା ଆମେ ବଞ୍ଚୁପାରିବାନି ।
D : ଆପଣ କ’ଣ ପବନ ବିଷୟରେ ଭାବିଛନ୍ତି ?
X : ଆପଣ ପୁରା ଠିକ୍ । ଏଥରକ ଆପଣଙ୍କ ପାଳି ।
Answer:
A : Does the thing you think exist in this room?
X : Yes, certainly.
B : What is its size, area and colour?
X : That object doesn’t have size, area or colour.
C : Is that very much necessary for us?
X : We cant live without it.
D : Have you thought about air?
X : You are absolutely right. Now it is your turn.

CHSE Odisha Class 11 Math Solutions Chapter 16 Probability Ex 16(b)

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 16 Probability Ex 16(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 16 Probability Exercise 16(b)

Question 1.
A school has six classes 1, 2, 3, 4, 5, and 6. Classes 2, 3, 4, 5, and 6 each have the same number of students, but there is twice this number in class 1. If a student is selected at random from the school, what is the probability that he(she) will be in
Solution:
A school has six classes 1, 2, 3, 4, 5, and 6. Classes 2, 3, 4, 5, and 6 each have the same number of students, but there is twice this number in class 1.
Let the number of students in class 2, 3, 4, 5, and 6 is x each and the number of students in class 1 is 2x.
∴ The total number of students = 7x.
A student can be chosen from 7x students in 7xC1 = 7x ways.
∴ |S| = 7x.

(i) class 1
Solution:
Probability that the student belongs to class 1, is \(\frac{2 x}{7 x}=\frac{2}{7}\)

(ii) class 2
Solution:
Probability that the student belongs to class 2, is \(\frac{x}{7 x}=\frac{1}{7}\)

Question 2.
Let a die be weighed in such a way that the probability of getting a number n is proportional to n.
Solution:
Let a die be weighed in such a way that the probability of getting a number n is proportional to n.
Let the constant of proportionality be k.
∴ P(n) = nk so that P(1) = k
P(2) = 2k, P(3) = 3k,…. P(6) = 6k
∴ P(1) + P(2) +…….+ P(6) = 1
or, k + 2k +……..+ 6k = 1
or, 21 k = 1 or, k = \(\frac{1}{21}\)

(i) Find the probability of each elementary event.
Solution:
P(1) = \(\frac{1}{21}\), P(2) = \(\frac{2}{21}\), P(3) = \(\frac{3}{21}\), P(4) = \(\frac{4}{21}\), P(5) = \(\frac{5}{21}\) , P(6) = \(\frac{6}{21}\)

(ii) Find the probability of getting an even number in a single roll of the die.
Solution:
= P(2) + P(4) + P(6)
= \(\frac{2}{21}+\frac{4}{21}+\frac{6}{21}=\frac{12}{21}\)

CHSE Odisha Class 11 Math Solutions Chapter 16 Probability Ex 16(b)

(iii) Find the probability of getting a prime number in a single roll of the die.
Solution:
= P(1) + P(3) + P(5)
= \(\frac{1}{21}+\frac{3}{21}+\frac{5}{21}=\frac{9}{21}\)

(iv) Find the probability of getting a prime number in a single roll of a die.
Solution:
Probability of getting a prime number = P(3) + P(5) + P(2)
= \(\frac{3}{21}+\frac{5}{21}+\frac{2}{21}=\frac{10}{21}\)

Question 3.
Five boys and three girls are playing in a chess tournament. All boys have the same probability p of winning the tournament and all the girls have the same probability q of winning. If p = 2q, find the probability that
(i) a boy wins the tournament.
(ii) a girl wins the tournament.
Solution:
5 boys and 3 girls are playing a chess tournament. All boys have the same probability P of winning the tournament and all the girls have the
same probability q of winning.
We have P(B) =p, P(G) = q.
As there are 5 boys and 3 girls,
we have 5p + 3q = 1
Now putting p = 2q,
we have 10q + 3q = 1
or, q = \(\frac{1}{13}\) ∴ p = 2q = \(\frac{2}{13}\)
∴ P(B) = 5p = \(\frac{10}{13}\),
∴ P(G) = 3p = \(\frac{3}{13}\)

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

Odisha State Board CHSE Odisha Class 11 Invitation to English 1 Solutions Chapter 2 The Legend behind a Legend Textbook Exercise Questions and Answers.

CHSE Odisha 11th Class English Solutions Chapter 2 The Legend behind a Legend

CHSE Odisha Class 11 English The Legend behind a Legend Text Book Questions and Answers

UNIT – I
Gist with Glossary

Gist:
The writer walks down memory lane. Exactly 25 years ago, he had spent two days and two nights with Khairi, the tigress of Jashipur, and a collection of wild animals of Saroj and Nihar. A news item on the latest exploits of Khairi evoked his interest to visit the place. He came to know that Saroj Raj Chaudhury was taking care of Khairi. He wrote a letter to him and met with a prompt response. He was filled with great joy. Mr. Chaudhury invited him to Khairi-Jashipur, giving him particular directions on how to reach there. The writer apprised him of when and how he would visit him.

Glossary:
bandit queen : queen of the robbers (ଦସ୍ୟୁରାଣୀ)
menagerie : a collection of wild animals (ବଣ୍ୟଜନ୍ତୁ ମାନ ଙ୍କ ସଂଗ୍ରହାଳୟ)
exploits : some unusual thing that someone does that you think is brave, exciting or entertaining
domesticated : an animal trained to live with or work for humans (ଗୃହପାଳିତ)
struck : occurred (ମନେପଡ଼ିଲା)
gruff : unfriendly and cruel (ନିଷ୍ଠୁର ସ୍ବଭାବସମ୍ପନ୍ନ)
tough : strict and severe (ଶୃଙ୍ଖଳିତ)
brooked no nonsense : tolerated only important and necessary things (ବାଜେ କଥା ସହ୍ୟ କରିପାରନ୍ତି ନାହିଁ)
suffered no feels : did not tolerate stupidity in others (ନିର୍ବୋଧତାକୁ ପ୍ରଶ୍ରୟ ଦିଅନ୍ତି ନାହିଁ)
itnerant articles : articles published in different magazines (ବିଭିନ୍ନ ପତ୍ରିକାରେ ପ୍ରକାଶିତ ଅନେକ ଲେଖା )
shot in the dark : a hopeful attempt (ଏକ ଆଶାପୂର୍ଣ୍ଣ ଉଦ୍ୟମ )
utter : great (ମାତ୍ରାତ୍ଵିକ)
delight : pleasure (ଆନନ୍ଦ)
precise : exact (ନିର୍ଦ୍ଦିଷ୍ଟ)
instructions : directions (ନିର୍ଦ୍ଦେଶ )

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

Think it out:
Question 1.
Who is Khairi?
Answer:
Khairi is a legendary tigress of Jashipur.

Question 2.
How did the writer come to know about Khairi?
Answer:
The writer came to know about Khairi when he had read a small news item in The Statesman that threw light on the latest exploits of the tigress in the Similipal forests of Odisha.

Question 3.
Who was the foster father of Khairi?
Answer:
Saroj Raj Chaudhury was the foster father of Khairi.

Question 4.
Which State does the writer belong to?
Answer:
The writer belongs to Odisha.

Question 5.
What did he learn about Saroj Raj Chaudhury as a person?
Answer:
He learnt that Saroj Raj Chaudhury tolerated only important and necessary things, but not stupidity in others.

Question 6.
How did he contact Mr. Chaudhury?
Answer:
He contacted Mr. Chaudhury by writing a letter to him after getting his address.

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

Question 7.
Why did he refer some of his articles to Mr. Chaudhury?
Answer:
He referred some of his articles to Mr. Chaudhury in the hope of getting his response.

Question 8.
Did Mr. Chaudhury reply to the author’s letter? What did he write?
Ans.
Yes, Mr. Chaudhury replied to the author’s letter. The former wanted the latter to inform him in advance of the manner and the time of his arrival.

UNIT-II
Gist with Glossary

Gist:
The writer reached Khairi-Jashipur by an overcrowded bus. It was 4 a.m. He was soon provided with food and shelter. In other words, he was accorded fabulous hospitality, thanks to Saroj Chaudhury. Terror seized him when he heard the clear voice of the Tiger just outside the door. It did not last long before the bearer met him to serve hot tea and biscuit and assured him of the presence of Khairi who was making loving inquiries about the new guest in the house. He met Saroj Chaudhury, ‘a frail man in his fifties, slightly balding on the top; the latter greeted the former in a polite manner after asking Jambu, the bear, to get down because the animal holding on to Chaudhury’s waist. They conversed with each other. In Mr. Chaudhury, the writer found a humble and careful man. One of the most caring persons he had ever met in his life was Mr.

Glossary:
semidarkness: half-darkness (ଅର୍ଥ ଅନ୍ଧକାର)
click : sound (ଶବ୍ଦ)
detailed : was given minute instructions (ସମ୍ପୂର୍ଣ୍ଣ ସୂଚନା ପାଇଥିଲେ )
escort : guide (ପଥ ପ୍ରଦର୍ଶନକାରୀ)
with a start : in fear (ଭୟଚକିତ ହୋଇ)
terror struck : the writer was seized with fear (ଭୟ ବିହ୍ବଳିତ)
sloth : lazy behaviour (ଆଳସ୍ୟ ସ୍ବଭାବସମ୍ପନ୍ନ )
frail : weak (ଦୁର୍ବଳ)
slightly : a little (ଅତି ଅଳ୍ପ)
gruff : rude, unfriendly
no-nonsense : doing things quickly and effectively without worring too much about people’s fear (ଲୋକଙ୍କ ମନ୍ତବ୍ୟ ପ୍ରତି ଉଦାସୀନ)
humane : caring people and animals ( ମାନବ ପ୍ରତି ସମ୍ବେଦନଶୀଳ)

Think it out:
Question 1.
How did the writer come to Bhubaneswar?
Answer:
The writer came to Bhubaneswar by train.

Question 2.
How did he go to Jashipur from Bhubaneswar?
Answer:
He went to Jashipur from Bhubaneswar in an over-packed bus.

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

Question 3.
How did the forest guard receive him?
Answer:
The forest guard received him in a very cordial manner. He guided the writer, took him to the guest house, and made him stay in the guest house, assuring him that the water was in the jug.

Question 4.
Why was he terror-struck?
Answer:
He was terror-struck because he heard the clear voice of the Tiger just outside his door.

Question 5.
What did the bearer tell him about Khairi?
Answer:
The bearer told him that Khairi was trying to know about the new guest in the house. There was a ring of friendliness about Khairi.

Question 6.
What was Mr. Chaudhury doing when the writer met him?
Answer:
When the writer met Mr. Chaudhury, he was sitting on a large chair.

Question 7.
How did Mr. Chaudhury greet the author?
Answer:
Mr. Chaudhury warmly greeted the author.

Question 8.
What was the name of the bear?
Answer:
The name of the bear was Jambu.

Question 9.
What was the physical appearance of Mr. Chaudhury?
Answer:
Mr. Chaudhury was a weak man in his fifties with a little baldness on the top.

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

Question 10.
What kind of man did the author find Mr. Chaudhury to be?
Answer:
In the author’s estimation, Mr. Chaudhury was friendly and responsive. Besides, he was caring to the core.

UNIT – III
Gist with Glossary

Gist:
This part begins with Mr. Chaudhury narrating a wonderful story to the writer. His house turned into a habitat for different species of wild animals. He conducted an experiment to exaggerate that the animals could exist with each other if they were together from childhood. He brought Khairi and one of the most dangerous snakes, krait close to her. The writer marked Khairi’s reaction – it was one of fondness for the strange creature. Whenever the krait got too close to Khairi, he would pull it by its back.

Once he became inattentive and was bitten by a krait. Some of its poison entered his blood and therefore, he was now a permanent patient of low blood pressure. As soon as Mr. Chaudhury finished this wonderful story Jambu tried to give the writer his bear hug, but the former’s stem warning prevented him from doing so. Mr. Chaudhury has a well-knit joint family that was living inside the compound. It comprised a mongoose, a pangolin, wild cat twins, a country dog, and a blind Hyena, each having a name.

Glossary:
emerged : became known (ଜଣାପଡ଼ିଲା)
sips : drink (something) by taking small mouthfuls (ଅଳ୍ପ ଅଳ୍ପ ପିଇବା)
debunked : exaggerated
co-exist : live together (ଏକତ୍ର ବାସ କରିବା)
infancy : childhood (ପିଲାଦିନ)
unmindful : inattentive (ଅମନୋଯୋଗୀ)
tied : bound (ବାନ୍ଧିଥିଲେ )
tourniquet : a piece of cloth bound tightly on an arm or leg to stop bleeding
hypo-glycaemia : condition of having a very low blood pressure (ନିମ୍ନ ରକ୍ତଚାପ)
astonishing : wonderful (ଆଶ୍ଚର୍ଯ୍ୟଜନକ)
take a fancy : to start liking someone (ଜଣକୁ ଭଲ ପାଇବାକୁ ଆରମ୍ଭ କରିବା)
hug : to put ones arms around someone to show love (ଆଲିଙ୍ଗନ)
dissuade : prevent (ବାଧା ଦେବା)
fondness : affection (ସ୍ନେହ, ଶ୍ରଦ୍ଧା)
progressed : advanced (ଆଗେଇ ଚାଲିଲା)
intennittent : occurring occasionally (ସାମୟିକଭାବେ ଘଟୁଥିବା)
crackle : making shout sharp sounds (ସ୍ଵଚ୍ଛ ଅବଶିଷ୍ଟ ଉଚ୍ଚ ଶବ୍ଦ)
veritable : real (ବାସ୍ତବ)

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

Think it out:
Question 1.
What theory did Mr. Chaudhury prove wrong?
Answer:
The theory that Mr. Chaudhury proved wrong was that the different species of wild animals cannot co-exist unless they are together from childhood.

Question 2.
What was his first story about?
Answer:
The first story was about how his house became the habitat of different species of wild animals who all come at different stages of their lives. It also dealt with their coexistence and the close relationship between Khairi and krait, a dangerous snake.

Question 3.
Why is it so unique and amazing?
Answer:
It is so unique and amazing because, during this experiment, Khairi showed her reaction to the presence of krait, the most poisonous snake. Khairi was curious to know more about the krait as it was a stranger to her.

Question 4.
Why did Mr. Chaudhury allow Khairi to come near the krait?
Answer:
Mr. Chaudhury allowed Khairi to come near a krait to know how they dealt with each other.

Question 5.
What was Khairi’s reaction to the presence of the krait?
Ans.
Khairi’s reaction to the presence of a krait was one of curiosity to know more about the latter.

Question 6.
How did the experiment affect him?
Answer:
The experiment made him a permanent patient of low blood pressure.

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

Question 7.
What did the bear try to do with the writer?
Answer:
The bear tried to put his arms around the writer to show his love.

Question 8.
What prevented the bear from doing so?
Answer:
Mr. Chaudhury’s strict ‘no’ prevented the bear from doing so.

Question 9.
What kind of family did Mr. Chaudhury have?
Answer:
Mr. Chaudhury had a genuine joint family that comprised a mongoose, a pangolin, wild cat twins, a country dog, and a blind Hyena, each having a name.

Question 10.
What was his relationship with different animals?
Answer:
His relationship with different animals was quite familiar. His act of naming each of them is a case in point.

UNIT – IV

Gist:
Khairi was brought to Saroj Raj Chaudhury as a two-month cub who was hungry and confused. Veteran forester and instinctive lover of wildlife as he was, Saroj imitated the sounds of a mother tigress. It worked wonderfully. The tiger cub’s confidence was restored. Saroj became nostalgic. He recollected his birthday when his mother had presented him with a gun, with which he, as a young man, shot wildlife in a carefree manner. But, soon he realized that it was a mistake and happiness lies in the conservation of these harmless beautiful animals. As the Director of Project Tiger, Saroj was the first to introduce the Tiger Tracing Method of tiger census. For the night, both camped at a guest house deep in the jungle. Never before had the writer experienced one night in a magnificent wooden structure with rooms and a bath that stood 15 feet high from the ground.

Glossary:
cub : (here) a young tigress (ବାଘଛୁଆ, ଛୁଆ ବାଘୁଣୀ)
famished : very hungry (କ୍ଷୁଧାଉଁ)
confused : disturbed (ବିବ୍ରତ ହେଲା)
snarls: making angry sounds in one’s throat and showing teeth
veteran : experienced (ଅଭିଜ୍ଞ, ଦକ୍ଷ)
handle : deal with (ଆୟତ୍ତ କରିବା)
firmly : determindedly (ଦୃଢ଼ ଭାବରେ )
anchored : restored (ଶାନ୍ତ ହେଲା)
the legends : (here) Saroj Raj Chaudhury and Khairi (କିମ୍ବଦନ୍ତୀ )
tag along : accompanied someone (ବ୍ୟକ୍ତିବିଶେଷଙ୍କ ସହ ଯାତ୍ରା
snaked (v) : moved in or had a senes of long curves (ଅଙ୍କାବଙ୍କା ରାସ୍ତାରେ ଗଲା)
amidst: in the midst of (ମଝିରେ)
lush foliage: leaves of trees growing luxuriously
abandon : in an uncontrolled way (ଅବିଚାରିତ ଭାବେ)
conserving: preventing land, water, etc. from being damaged (ସଂରକ୍ଷଣ କରିବା)
wanton harm: reckless harm
authority : (here) Saroj Choudhury (କର୍ତ୍ତୃପକ୍ଷ )
pugmarks: the mark of the footprint of an animal (ପଶୁର ପାଦଚିହ୍ନ)
distinctive: very clear
meticulously: carefully attending to every detail
functional: practical and simple
build on stilts: build on one of a set of posts
a top: at the top of (ଉପରିଭାଗରେ )
magnificent : very beautiful (ଖୁବ୍‌ ସୁନ୍ଦର)

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

Think it out:
Question 1.
How and when did Mr. Chaudhury come across Khairi?
Answer:
Mr. Chaudhury came across Khairi when 12 Kharia tribals of Similipal brought her to him when she was a two-month cub. It was on October 5, 1974.

Question 2.
In what condition did he And it?
Answer:
He found it in a state of hunger and confusion.

Question 3.
How did he manage the hungry and confused cub?
Answer:
He managed the hungry and confused cub by copying the sounds of the mother tigress. Within minutes, she became firmly confident.

Question 4.
How did he treat wildlife in a young age?
Answer:
He treated wildlife uncontrollably in a young age.

Question 5.
What did he say about his change of attitude towards wildlife to the author?
Answer:
He said to the author that his change of attitude towards wildlife took place because of his realization that there was greater happiness in safeguarding the beautiful wild animals that do not cause reckless harm to man.

Question 6.
What was his contribution to the Tiger Project?
Answer:
His contribution to the Tiger Project was the introduction of the Tiger Tracing Method of tiger census. As a result, one can measure the pugmarks of each animal distinctly and record their characteristics very carefully.

Question 7.
What was the guest house like?
Answer:
The guest house was a wooden structure with simple rooms and a bath. It was built on one of the set of posts and was at a height of 15 feet.

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

Question 8.
What new experience did the author have in the Tiger Reserve area?
Answer:
The author experienced for the first time, one night stay in the Tiger Reserve area that was unique, especially on a very beautiful ‘machan’.

UNIT – V
Gist with Glossary

Gist :
The writer revisited Khairi-Jashipur after three months. His interest to know more about Khairi gained momentum. Besides Mr. Chaudhury’s old acquaintances, he caught sight of a young python. He focused on Mr. Chaudhury whose quest and passion was amazing. Meanwhile, Mr. Chaudhury got a message from the World Wildlife Fund. He went to New Delhi by air for an urgent meeting. The writer saw him off at the Dum Dum Airport, Calcutta. It was his last meeting with Mr. Chaudhury, a legend behind a legend. Khairi and Mr. Chaudhury are no more.

Glossary:
gracious: kind
in addition to besides
python: a very large snake that kills animals for food by wrapping itself around them and crushing them (ଅଜଗର ସାପ )
quest : search (ଅନ୍ଵେଷଣ)
legend: someone who very many people know about and admire (କିମ୍ବଦନ୍ତୀ ପୁରୁଷ)
unique : extraordinary (ଅସାଧାରଣ)
due: worth (ଯୋଗ୍ୟ)

Think it out:
Question 1.
After what interval of time did the writer visit Mr. Chaudhury for the second time?
Answer:
After three months, the writer visited Mr. Chaudhury for the second time.

Question 2.
What new addition to the Chaudhury family did he find there?
Answer:
The new addition to Chaudhury’s family he found there was an eight-foot-long young python.

Question 3.
Why did he get less time to interact with Mr. Chaudhury this time?
Answer:
He got less time to interact with Mr. Chaudhury because at that moment Mr. Chaudhury got a wireless message from the World Wildlife Fund to attend an important meeting at New Delhi.

Question 4.
Who died first, Khairi or Mr. Chaudhury?
Answer:
Khairi died first.

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

Question 5.
Who are the two legends the writer talks about?
Answer:
The two legends the writer talks about are Khairi and Mr. Saroj Raj Chaudhury.

Question 6.
Is the text more about Mr. Chaudhury or Khairi?
Answer:
The text throws much light on Mr. Chaudhury, yet Khairi does not lag far behind.

Question 7.
Can you guess now why the title of the text is “The Legend behind the Legend”?
Answer:
The title is aptly justified because the legendary passionate and instinctive lover of wildlife has been instrumental in transforming a two-month-old tiger baby into a legend. She is Khairi.

Post-Reading Activities:

I. Arranging in Order
Provided below are some events from the lesson. These are not in order. Arrange them in order as they occur in the lesson by putting numbers within the brackets provided against the items. One is done for you.
(a) Khairi played with the krait. ( )
(b) Khairi was brought to Mr. Chaudhury. ( )
(c) The writer sees Mr. Chaudhury off at Dum Dum Airport. (8)
(d) The writer reads a news item about Khairi. ( )
(e) Chaudhury writes a letter to the writer. ( )
(f) The writer reaches Jashipur by bus. ( )
(g) He stays with Chaudhury in a camp guest house in the forest. ( )
(h) The writer saw a python as a pet of Chaudhury. ( )
Answer:
(a) Khairi played with the krait. (6)
(b) Khairi was brought to Mr. Chaudhury. (5)
(c) The writer sees Mr. Chaudhury off at Dum Dum Airport. (8)
(d) The writer reads a news item about Khairi. (1)
(e) Chaudhury writes a letter to the writer. (2)
(f) The writer reaches Jashipur by bus. (3)
(g) He stays with Chaudhury in a camp guest house in the forest. (4)
(h) The writer saw a python as a pet of Chaudhury. (7)

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

II. Note-making
Notes-making helps you to develop your reading and writing skills. This lesson has, you know, two major themes – Khairi and Mr. Chaudhury. You have to read the lesson, make notes (in words and phrases) on these two, and then use these points to write about them. One has been done below on Khairi as a model. Make notes on Mr. Chaudhury.

Notes on Khairi
Para 1
→ Tigress of Jashipur
→ Made the forest famously
→ Domesticated tigress in the Similipal forest of Odisha
→ Writer reads a news item about Khairi
Para 5
→ Khairi roared to welcome the writer
→ He was terror struck
Para 9
→ On October 5, 1974, Khairi was brought as a cub, hungry, confused
Para 15
→ Chaudhury manages her imitating the sounds of a mother tigress.
The death of Khairi was followed by the death of Chaudhury.
With the help of these notes, write a paragraph on Khairi.
Khairi:
Khairi was a domesticated tigress. She made the Simmilipal forest and Jashipur famous. The writer read about her from a news item. Then he visited Jashipur to see Khairi and Chaudhury who had kept Khairi as a pet. He came to know from Mr. Chaudhury that Khairi was brought to Chaudhury on October 5, \91A by twelve Kharia tribals. It was then a small cub, hungry and confused. Chaudhury handled the cub well by imitating the sounds of a tigress. She grew up as a domesticated tigress under the loving care of Chaudhury. But she did not live long. Chaudhury also died soon after the death of Khairi.

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

Now write notes (from the lesson) and develop the notes into a write-up on Mr. Chaudhury.

Notes on Mr. Chaudhury
Para 3
→ Mr. Chaudhury was a very responsible person.
Para 4
→ Hospitable to the core
Para 6
→ A weak man in his fifties, slightly balding on the top
→ Nice and down-to-earth
→ One of the most humane beings
Para 7
→ His experiment concerning the co-existence of wild animals in the presence of Khairi and the krait
→ The latter is a dangerous snake
→ Noticed Khairi’s reaction of fondness
→ A permanent patient with low blood pressure
Para 8
→ Strict
→ Possessed a joint family of animals in his house
Para 9
→ An officer of the Indian Forest Service
→ Veteran forester and instinctive lover of wildlife
→ Showed skill in handling hungry and confused two-month-old tiger cub
→ Imitated the sounds of a mother tigress
Para 11
→ An authority on the tiger and Director of Project Tiger in India
→ Introduced the Tiger Tracing Method of tiger census
Para 13
→ Gracious
Para 14
→ Journey to New Delhi for an important meeting
Para 15
→ The writer’s last meeting with him in the Dum Dum Airport, Calcutta
→ A legend behind a legend
→ He is no more.
Mr. Chaudhury
Mr. Chaudhury, responsible, hospitable, and down-to-earth, was a weak person in his fifties. He was one of the most humane beings. He made an experiment by effecting contact between Khairi and the krait, one of the most dangerous snakes, to know about the co-existence of different wild animals. Khairi’s reaction, he noticed, was one of great love. Mr. Chaudhury, a veteran forester and an instinctive lover of wildlife possessed a joint family of different animals in his house. He was strict. He showed his skill in handling the hungry and confused two-month-old tiger cub by imitating the sounds of a mother tigress. Mr. Chaudhury, an authority on the tiger and Director of Project Tiger in India, was the first to introduce the Tiger Tracing Method of tiger census. The writer saw such a graceful legend off in Dum Dum Airport on his way to New Delhi, for the last time. Mr. Chaudhury is no more.

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

III. Doing with words Collocation
(a) Collocation in expressions means which words go with which other words. Collocations are fixed expressions. For example sweet dreams, daydreams, bad dreams, pipe dreams, hard-earned money, public money, extra money, and tax-payers money.
chapter 2
Answer:
Chapter 2.1
(b) Find out five collocations from the test. (Example: to take a fancy).
Answer:
brooked no nonsense
suffered no fools
terror-struck
no-nonsense
hypo-glycaemia

(c) Which word in each line does not collocate with the headword?
(i) a theory: come up with, do, debunk, build
(ii) a debate: open, listen to, join in, find
(iii) legend: fresh, famous, well-known, sports
(iv) veteran: soldier, idealist, activist, man
(v) gracious: welcome, hospitality, building, smile
Answer:
(i) do
(ii) find
(iii) fresh
(iv) man
(v) building

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

CHSE Odisha Class 11 English The Legend behind a Legend Important Questions and Answers

I. Short Answer Type Questions with Answers

1. Read through the extract and answer the questions that follow.
Khairi made the entire forest where she lived famously. She was not a bandit queen but Khairi, the tigress of Jashipur. It was exactly 25 years ago when I spent two days and two nights with Khairi and the menagerie of Saroj and Nihar, I had read a small news item in The Statesman about the latest exploits of a domesticated tigress in the Similipal forests of Odisha. Suddenly, it struck me that this was happening in my own State. I thought, “why not attempt to experience it myself ?” I spoke to N.S. Ayyangar, a senior journalist in Berhampur, and a few other elders.

I was told that Khairi was under the care of a rather gruff and tough man called Saroj Raj Chaudhury who brooked no nonsense and suffered no fools. I got his address and wrote asking if I could visit him. For good measure, I referred to a few itinerant articles I had written for Indian magazines. It was a shot in the dark and I did not really expect to hear from him. But, to my utter delight, I got a letter within a week inviting me to Khairi-Jashipur, giving precise instructions about how to reach there. Mr. Chaudhury also asked me to let him know in advance how and when I was reaching. I gave him a date and said I would be taking a bus from Bhubaneswar on a particular night.

Questions :
(i) Why was the forest made famous?
(ii) “Why not attempt to experience it myself ?” What does ‘It’ refer to?
(iii) Why did the writer contact N.S. Ayyanger and a few others? What was the result?
(iv) Explain the expression “It was a shot in the dark.”.
(v) Suggest a suitable title to the extract.

Answers :
(i) The forest was made famous by Khairi, the tigress of Jashipur because it was her abode.
(ii) ‘It’ refers to the writer’s curiosity to be aware of the latest exciting things carried by a tigress in the Similipal forests of Odisha. She had been to live with and work for humans.
(iii) The writer contacted N.S. Ayyanger and a few other seniors to know about Khairi. He learned that Saroj Raj Chaudhury, who was rather an unfriendly and severe man, took care of the tigress.
(iv) The writer’s reference to a few articles he had written for Indian magazines was a sort of hopeful attempt to see Khairi and Saroj Raj Chaudhury’s positive response to his letter.
(v) The Writer’s Quest of Khairi

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

2. Read through the extract and answer the questions that follow.
I packed my bag, took the train, and boarded the overcrowded bus from Bhubaneswar. I arrived sometime before 4.00 a.m. wondering where to go in that semidarkness. To my utter surprise, within a minute there was the click of boots and a voice welcoming me to Khairi-Jashipur. The Forest Guard, detailed to escort me, took me to the guest house, put me in my room, and assured me that water was in the jug; I could sleep as long as I wanted and Saab would see me as soon as I was ready.

I think I had an hour of blissful sleep. I woke up with a start when I heard the unmistakable voice of the Tiger just outside my door. I was terror-struck. Within minutes, a bearer came to the room with hot tea and biscuits. He smiled at the expression on my face and assured me that it was only Khairi outside the door, making friendly inquiries about the new guest in the house. I finished my tea, had a quick shower and went to the main house.

Saroj Raj Chaudhury was sitting on a large chair. There was a sloth bear behind him, holding on to his waist and making gurgling sounds. He said, “Get down, Jambu, get down’’ and rose to greet me – a frail man in his fifties, slightly balding on the top. We got talking. I didn’t find a gruff and rough no-nonsense man. What I found was one of the most humane human beings I had ever met in my life.

Questions :
(i) Describe the bus the writer had boarded.
(ii) When did he reach Jashipur?
(iii) Describe the treatment according to the writer in Jashipur.
(iv) How did a bearer react to the writer’s terror-stricken face?
(v) Throw light on Jambu.

Answers :
(i) The bus the writer had boarded was packed with passengers beyond its capacity.
(ii) The writer reached Jashipur sometime before 4 a.m. It was half-dark.
(iii) As soon as the writer reached Jashipur, the Forest Guard led him to the guest house, and showed him his room. He assured the writer of the presence of a jug filled with water. Later a bearer provided him with hot tea and biscuits in his room.
(iv) A bearer reacted smilingly to the writer’s terror-stricken face.
(v) The bear Jambu was seen in a lazy mood, behind Saroj Raj Chaudhury, and holding on to his waist making gurgling sounds.

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

3. Read through the extract and answer the questions that follow.
Here is the first story that emerged from this very unusual man between sips of coffee: “As you will see, I have different species of wild animals in this house. They all came in at different stages of their lives. I have debunked the theory that they cannot co-exist unless they are together from infancy. One thing I wanted to experiment with was the reaction of a young tiger to a snake. One day, when Khairi was much younger, we found a baby krait in the house. As you know, the krait is one of the most poisonous snakes.

I was noting Khairi’s reaction to its presence. Khairi was curious to know more about this strange new creature. Every time the krait got too close to Khairi, I would pull it back by its tail. This went on for some time. At some point, I must have been a little unmindful. It turned around and bit me. I immediately tied a tourniquet above that and got the poison out. I saw the doctor as soon as possible. Luckily it was a baby. Still, some of the poison got into my bloodstream and as a result, I am now a permanent patient of hypo-glycemia.”

By the time he finished this astonishing story, Jambu took a fancy to me and climbed behind to give me his bear hug. A stern ‘no’ from Saroj was enough to dissuade him from this expression of fondness. As the day progressed, between our conversations and the intermittent crackle on the VHF wireless set by which he was giving instructions to his men in the forests, I got to know a veritable joint family that was living inside the compound – a mongoose, a pangolin, wild cat twins, a country dog, and a blind Hyena. Each had a name.

Questions :
(i) When did Saroj narrate the first story to the writer?
(ii) What is the theory that Mr. Chaudhury has exaggerated?
(iii) ‘It turned round and bit me.’ What does ‘It’ refer to?
(iv) How was Mr. Saroj Chaudhury’s story?
(v) Throw light on Jambu.

Answers :
(i) Between sipping coffee, Saroj narrated the first story to the writer.
(ii) The theory that Mr. Chaudhury has exaggerated is that different species of wild animals cannot co-exist unless they are together from childhood.
(iii) ‘It’ refers to a krait, one of the most poisonous snakes.
(iv) Mr. Chaudhury’s story was wonderful.
(v) The bear Jambu was a picture of fondness. He started liking the writer climbing behind. The creature was interested to give a bear hug, but in vain, because of Mr. Chaudhury’s strict ‘no’.

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

4. Read through the extract and answer the questions that follow.
Khairi’s story started on October 5, 1974, when 12 Kharia tribals of Similipal brought a two-month-old tiger cub to Saroj Raj Chaudhury, an officer of the Indian Forest Service. Saroj noticed that it was a female – famished and confused. His first experience of what was to become his passion in life was angry snarls and scratching claws. But, the veteran forester and instinctive lover of wildlife knew how to handle a hungry, angry cub. He imitated the sounds of a mother tigress.

“Within minutes, her confidence was firmly anchored in the fostering human,” is how he recalled those first few minutes between the legends. Early the next morning, Saroj started his inspection of the Tiger Reserve area. I tagged along in the jeep that snaked through a narrow road in the woods amidst lush foliage. “My mother gave me a gun for my eighth birthday. As a young man, I shot wildlife with abandon. But soon, I realized that there is greater happiness in conserving these beautiful animals that do no wanton harm to man” is one of the things he told me about his life during that long travel.

At that time, he was an authority on the tiger and Director of Project Tiger in India. Saroj introduced the Tiger Tracing Method of tiger census where the pugmarks of each animal with distinctive measurements and characteristics are meticulously recorded. For the night, we camped at a guest house deep in the jungle. It was a wooden structure with functional rooms and a bath. It was built on stilts and stood a good 15 feet above the ground. I experienced for the first time, one night atop a magnificent machan.

Questions :
(i) What picture of Saroj Chaudhury do you get in the 1st para of the extract?
(ii) When did he go to inspect the Tiger Reserve area?
(iii) Describe his journey to this place.
(iv) When did Mr. Chaudury’s realization concerning the conservation of wildlife come?
(v) What was the writer’s experience of staying one night at the guest house a top?

Answers :
(i) In the first para of the extract, we learn that Saroj Chaudhury, an officer of the Indian Forest Service, was a veteran forester and instinctive lover of wildlife. The way he handled the two-year-old tiger cub is a case in point.
(ii) Early the next morning of his arrival, he went to visit the Tiger Reserve area.
(iii) In the course of his journey, Saroj Chaudhury in the jeep went through a narrow road in the woods amidst leaves of trees growing luxuriantly.
(iv) Mr. Chaudhury’s realization concerning the conservation of wildlife came when he, as a young man, had shot wildlife in a carefree life.
(v) The writer’s experience of staying one night at the guest house atop was very beautiful.

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

II. Multiple Choice Questions (MCQs) with Answers
Choose the correct option.

Unit – I
The text
Khairi made ………………. night.

Question 1.
Who was the tigress of Jashipur?
(a) Nhairi
(b) Khairi
(c) Bhairi
(d) Shairi
Answer:
(b) Khairi

Question 2.
Where did Khairi live?
(a) Bhitarkanika
(b) Similipal forests
(c) Kanchanjanga
(d) National Zoo, Kolkata
Answer:
(b) Similipal forests

Question 3.
Who was N. S. Ayyangar ?
(a) a politician
(b) a senior journalist
(c) a zoologist
(d) a veterinary doctor
Answer:
(b) a senior journalist

Question 4.
Khairi was under the care of:
(a) N. S. Ayyangar
(b) Hariharan
(c) Saroj Raj Chaudhury
(d) Nihar Raj Chaudhury
Answer:
(c) Saroj Raj Chaudhury

Question 5.
Khairi was a :
(a) domesticated tigress
(b) wild tigress
(c) zoo tigress
(d) tigress of a circus
Answer:
(a) domesticated tigress

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

Question 6.
Who had invited the author to Khairi- Jashipur?
(a) Nihar Raj Chaudhury
(b) Saroj Raj Chaudhury
(c) N. S. Ayyangar
(d) Forest ranger
Answer:
(b) Saroj Raj Chaudhury

Unit – II
The text
I packed my bag, ……………… met in my life.

Question 7.
Who escorted the author to the guest house from the bus stop at Khairi- Jashipur?
(a) Forest Ranger
(b) Forest Guard
(c) Forest Guide
(d) none of the above
Answer:
(b) Forest Guard

Question 8.
What was Khairi doing outside the guest house when the author was inside the room?
(a) making friendly enquiries
(b) making fun
(c) searching for enemies if any
(d) none of the above
Answer:
(a) making friendly enquiries

Question 9.
Who was Jambu?
(a) a monkey
(b) a bear
(c) a sloth bear
(d) a donkey
Answer:
(c) a sloth bear

Unit – III
The text
Here is the first story……………..had a name.

Question 10.
The name of the snake living in Mr. Chaudhury’s house was :
(a) cobra
(b) Python
(c) krait
(d) rattlesnake
Answer:
(c) krait

Question 11.
Mr. Chaudhury was bitten by a snake and as a result, he became a permanent patient of
(a) diabetics
(b) hypo-glycaemia
(c) leukaemica
(d) high blood pressure
Answer:
(b) hypo-glycaemia

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

Question 12.
How did Mr. Chaudhury instruct his men in the forests?
(a) on the computer internet
(b) on the VHF wireless
(c) on the T.V.
(d) by the mobile phone
Answer:
(b) on the VHF wireless

Question 13.
What did Mr. Saroj Chaudhury do when he was bitten by a krait, the poisonous snake?
(a) immediately cut the biting place
(b) immediately applied for medicine
(c) immediately tied a tourniquet
(d) met a doctor
Answer:
(c) immediately tied a tourniquet

Unit – IV
The text
Khairi’s story…….. magnificent machan.

Question 14.
When did Khairi come to Mr. Chaudhury’s hands?
(a) Nov. 5, 1974
(b) Dec. 5, 1974
(c) Oct. 5, 1974
(d) Sept. 5, 1974
Answer:
(c) Oct. 5, 1974

Question 15.
Who got the baby tiger and handed it over to Mr. Chaudhury?
(a) 12 Kharia tribals
(b) 12 Gonda tribals
(c) 12 Santhal tribals
(d) none of them
Answer:
(a) 12 Kharia tribals

Question 16.
Who was Saroj Raj Chaudhury?
(a) Indian Police Service officer
(b) Indian Foreign Service officer
(c) Indian Forest Service officer
(d) a low cadre forest official
Answer:
(c) Indian Forest Service officer

Question 17.
What do you mean by the word ‘famished’?
(a) very tired
(b) very smart
(c) very careful
(d) very hungry
Answer:
(d) very hungry

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

Question 18.
How did Mr. Choudhury handle and consoled the hungry and angry cub?
(a) by giving it milk to drink
(b) by showing her the picture of a mother tigress
(c) by making the sounds of a mother tigress
(d) by leaving it all alone
Answer:
(c) by making the sounds of a mother tigress

Question 19.
Where does lie the greater happiness of Mr. Chaudhury?
(a) in killing animals
(b) in conserving animals
(c) in destroying animals’ habitat
(d) none of the above
Answer:
(b) in conserving animals

Question 20.
What do you mean by the word ‘wanton harm’?
(a) no harm
(b) less harm
(c) reckless harm
(d) secret harm
Answer:
(c) reckless harm

Question 21.
Who was then an authority on the tiger and Director of Project Tiger in India?
(a) Nihar Raj Chaudhury
(b) Bhasker Raj Chrudhury
(c) Saroj Raj Chaudhury
(d) N. S. Ayyangar
Answer:
(c) Saroj Raj Chaudhury

Question 22.
Which parts of speech is the word ‘meticulously’?
(a) noun
(b) verb
(c) adverb
(d) adjective
Answer:
(c) adverb

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

Question 23.
Which according to the author was a magnificent machan?
(a) a guest house deep in the jungle
(b) the building where Saroj Chaudhury lived
(c) the house where Khairi and other animals lived
(d) none of the above
Answer:
(a) a guest house deep in the jungle

Unit – V
The text
I went to ……………… 25 years hence.

Question 24.
When did the author come to Khairi- Jashipur again?
(a) after six months
(b) after four months
(c) after three months
(d) after seven months
Answer:
(c) after three months

Question 25.
What was the purpose of the second visit of the author to Khairi-Jashipur?
(a) to know more about Mr. Chaudhury
(b) to know more about Khairi
(c) to see the place again
(d) to enjoy the nature
Answer:
(b) to know more about Khairi

Question 26.
Which new animal do he saw in his second visit?
(a) a pangolin
(b) a blind hyena
(c) a rattlesnake
(d) a young python
Answer:
(d) a young python

Question 27.
What did the author search about in his second visit?
(a) about Saroj Chaudhury and his passion
(b) about World Wildlife Fund
(c) about the animals
(d) none of the above
Answer:
(a) about Saroj Chaudhury and his passion

Question 28.
Where came a wireless message to the Chaudhury?
(a) Department of Forest
(b) World Wildlife Fund
(c) Central Government
(d) none of the above
Answer:
(b) World Wildlife Fund

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

Question 29.
To whom the author has described a legend behind a legend?
(a) Khairi
(b) Mr. Saroj Chaudhury.
(c) Nihar Raj Chaudhury
(d) none of the above
Answer:
(b) Mr. Saroj Chaudhury.

Question 30.
Where did the author meet Mr. Chaudhury last time?
(a) at New Delhi
(b) in Dum Dum Airport
(c) at Mumbai
(d) at Jashipur
Answer:
(b) in Dum Dum Airport

Question 31.
Who made the forest famous?
(a) Mr. Chaudhury
(b) Khairi
(c) author
(d) tourist
Answer:
(b) Khairi

Introducing the Author:
Hariharan Balakrishnan excels in the art of writing articles on wildlife. He is also a columnist.

About the Topic:
‘The Legend Behind A Legend’, as the title suggests, deals with two legends: Mr. Saroj Raj Chaudhury and Khairi; the former was an authority on tiger and Director of Project Tiger, and the latter a magnificent famous tigress. Mr. Chaudhury was a foster father of Khairi. The writer showers accolade on both legends.

Summary:
The writer takes us back to the Similipal forests of Odisha, which served as the habitat of Khairi, the tigress of Jashipur. The place carved out a name for itself, thanks to this tigress. She was not a queen of the robbers. She was the queen of Jashipur. The writer goes down memory lane. 25 years have elapsed since he had spent two days and two nights with Khairi and other wild animals of Saroj and Nihar. A news item on the latest exploits of Khairi evoked his interest to visit the place. He came to know that Saroj Raj Chaudhury was taking care of Khairi.

He wrote a letter to him and met with a prompt response. He was filled with great joy. Mr. Chaudhury invited him to Khairi – Jashipur, giving him particular directions on how to reach there. The writer apprised him of when and how he would visit him. This was his meeting. The writer left Berhampur by train. He reached in Jashipur before 4 a.m. by an overpacked bus. The Forest Guard was present there to cordially welcome him. He made all comfortable arrangements for the writer. He spent an hour of sound sleep.

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

Terror gripped him when he heard the clear voice of the Tiger just outside the door. Within minutes, the bearer came to his room and served him with hot tea and biscuits. He smiled at the panic-stricken expression of his face. He assured the writer of the presence of Khairi who was making loving enquiries about the new guest in the house. He met Saroj Chaudhury, a frail man in his fifties, slightly balding on the top; the latter greeted the former in a polite manner after asking Jambu, the bear, to get down, because the animal holding on to Chaudhury’s waist.

They conversed with each other. In Mr. Chaudhury, the writer found a humble and careful man. One of the most caring persons he had ever met in his life was Mr. Chaudhury. While sipping tea, Mr. Chaudhury narrated a story to the writer. He drew the latter’s attention to the presence of different species of wild animals in his house. They all had not come at the same time. He exaggerated the theory that they could not exist with each other unless they were together from childhood. He conducted an experiment to ascertain the truth.

He brought Khairi and one of the most dangerous snakes, krait. The writer marked Khairi’s reaction – it was one of fondness for the strange creature. Whenever the krait goes too close to Khairi, Mr. Chaudhury would pull it by back by its tail. Once he became inattentive and was bitten by the krait. Some of its poison entered his blood and therefore, he was now a permanent patient with low blood pressure. As soon as Mr. Chaudhury finished this wonderful story, Jambu tried to give the writer his bear hug, but the former’s stern warning prevented him from doing so.

Mr. Chaudhury has a well-knit joint family that was living inside the compound. It comprised a mongoose, a pangolin, wild cat twins, a country dog, and a blind Hyena, each having a name. Khairi’s story dates back to October 5, 1974, when twelve Khaira tribals of Similipal brought a two-month-old tiger cub to Saroj Raj Chaudhury who was an IFS officer. He found the female cub in a state of hunger and confusion. He managed it by imitating the sounds of a mother tigress. Her anger and disturbance vanished at once.

She was in her element. Saroj became nostalgic. He recollected his birthday when his mother had presented him with a gun, with which he, as a young man, shot wildlife in a carefree manner. But, soon he realized that it was a mistake and happiness lies in the conservation of these harmless beautiful animals. As the Director of Project Tiger, Saroj was the first to introduce the Tiger Tracing Method of tiger census. For the night, both camped at a guest house deep in the jungle. Never before had the writer experienced spending one night in a magnificent Wooden structure with rooms and baths that stood 15 feet high from the ground.

It is three months since the writer last visited Khairi-Jashipur. Then he went there again for the second time to know more about Khairi. Saroj welcomed him in his characteristic gracious manner. This time the writer saw a python as a pet of Chaudhury. His passion for wildlife knew no end. In a response to a message from the World Wildlife Fund, he went to New Delhi by plane with a view to attending an important meeting. In just over three months, Khairi passed away. Saroj Raj Chaudhury did not survive for long. The topic comes to a close with the writer saluting Saroj Chaudhury. He is worthy of it.

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

ସାରାଂଶ:
“The Legend Behind A Legend’ ବିଷୟଟି ଦୁଇଟି କିମ୍ବଦନ୍ତୀକୁ ଆଧାର କରି ରଚିତ । ବାଘମାନଙ୍କ ଉପରେ ନିଜର ଦକ୍ଷତା ହାସଲ କରିପାରିଥିବା Project Tigerର ନିର୍ଦ୍ଦେଶକ Mr. Saroj Raj Chaudhury ଏବଂ ପ୍ରସିଦ୍ଧ ମହାବଳ ବାଘୁଣୀ ‘ଖଇରୀ’ର ଜୀବନୀ ଉପରେ ଏହା ପର୍ଯ୍ୟବସିତ । ସରୋଜ ରାଜ ଚୌଧୁରୀ ଖଇରୀର ପାଳିତ ପିତା ଭଳି ଥିଲେ । ଶିମିଳିପାଳ ଜଙ୍ଗଲର ପ୍ରସିଦ୍ଧ ବାଘୁଣୀ ଥିଲା ଖଇରୀ । ସେ ଥୁଲା ଯଶିପୁରର ମହାବଳ ବାଘୁଣୀ । ଲେଖକ ସ୍ମୃତିଚାରଣ କରି କହନ୍ତି ଯେ ୨୫ ବର୍ଷ ପୂର୍ବେ ସେ ଦୁଇ ଦିନ ଓ ଦୁଇ ରାତି ଖଇରୀ ସହିତ ସରୋଜ ଓ ନିହାରଙ୍କ ବନ୍ୟଜନ୍ତୁ ସଂଗ୍ରହାଳୟରେ ସମୟ ଅତିବାହିତ କରିଥିଲେ । ଲେଖକ ଖଇରୀର ଅଦ୍ଭୁତ କାର୍ଯ୍ୟକଳାପ ବିଷୟରେ ‘The Statesman’ର ଏକ ଖବରରୁ ଜାଣିପାରିଥିଲେ ଏବଂ ସେଇ ସ୍ଥାନକୁ ବୁଲିଯିବା ପାଇଁ ଆଗ୍ରହ ପ୍ରକାଶ କରିଥିଲେ ।

ସେ ଜାଣିବାକୁ ପାଇଲେ ଯେ, ଖଇରୀ ସରୋଜ ରାଜ ଚୌଧୁରୀଙ୍କ ତତ୍ତ୍ଵାବଧାନରେ ପାଳିତ ହେଉଛି । ସେଥ‌ିପାଇଁ ଲେଖକ ସେଠାକୁ ଯିବାର ଅନୁମତି ମାଗି ସରୋଜଙ୍କ ପାଖକୁ ପତ୍ର ଲେଖିଲେ । ସରୋଜ ଅତି ଖୁସିରେ ଲେଖକଙ୍କୁ ନିମନ୍ତ୍ରଣ ପତ୍ର ଲେଖୁଲେ ଏବଂ ଯିବାପାଇଁ ସମସ୍ତ ତଥ୍ୟ ସହିତ ତାରିଖ ମଧ୍ୟ ଜଣାଇଲେ । ଭୋର ୪ଟା ସମୟରେ ଲେଖକ ଏକ ଜନଗହଳିପୂର୍ଣ୍ଣ ବସ୍‌ରେ ବସି ଖଇରୀ-ଯଶିପୁରରେ ପହଞ୍ଚିଲେ । ତାଙ୍କୁ ଉତ୍ତମ ଆତିଥ୍ୟ ସତ୍କାର କରାଗଲା । ଖାଦ୍ୟ ଓ ବିଶ୍ରାମ ପାଇଁ ସବୁପ୍ରକାରର ବ୍ୟବସ୍ଥା କରାଯାଇଥିଲା । ଲେଖକ ବିଶ୍ରାମ ନେଉଥ‌ିବାବେଳେ ଦ୍ଵାରଦେଶର ବାହାରେ ବାଘୁଣୀର ଗର୍ଜନ ଶୁଣି ଭୟଭୀତ ହୋଇଯାଆନ୍ତି । ଏହାର ଅଳ୍ପ ସମୟ ପରେ ଜଣେ ବ୍ୟକ୍ତି ଚା’ ଓ ବିସ୍କୁଟ ଧରି ଆସିଲେ ଏବଂ କହିଲେ ଖଇରୀ ବାଘୁଣୀ ବାହାରେ ଥାଇ ଆନନ୍ଦରେ ଗର୍ଜନ କରି ନୂତନ ଅତିଥିଙ୍କୁ ସ୍ଵାଗତ କରୁଛି ।

ତା’ପରେ ଲେଖକ ସରୋଜଙ୍କୁ ସାକ୍ଷାତ କଲେ । ଦୁର୍ବଳ ଶରୀରଧାରୀ ପଚାଶ ବର୍ଷୀୟ ଏହି ବ୍ୟକ୍ତିଜଣକ ଟିକେ ଚନ୍ଦା ଥିଲେ । ସେ ଲେଖକଙ୍କୁ ଅତି ଭାବରେ ସ୍ଵାଗତ କଲେ ଏବଂ ତାଙ୍କ ଅଣ୍ଟାକୁ ଧରି ଠିଆ ହୋଇଥିବା ଭାଲୁକୁ ବସିବାକୁ କହିଲେ । ପରସ୍ପର କଥାବାର୍ତ୍ତା ହେଲେ । ଲେଖକ ସରୋଜ ରାଜ ଚୌଧୁରୀଙ୍କଠାରେ ଭଦ୍ର ଯେଉଁ ବିନମ୍ର ସ୍ଵଭାବ ଦେଖିଥିଲେ, ସେ ତାଙ୍କ ଜୀବନରେ ଅନ୍ୟ କାହାଠାରେ ଦେଖିନଥିଲେ । କଥାବାର୍ଭା ସମୟରେ ଶ୍ରୀଯୁକ୍ତ ଚୌଧୁରୀ ମହାଶୟେ କହିଲେ ତାଙ୍କ ଘର କିଭଳି ଭାବେ ବିଭିନ୍ନ ଜୀବଜନ୍ତୁମାନଙ୍କର ଏକ ଆଶ୍ରୟସ୍ଥଳୀ ହୋଇଯାଇଛି । ଜୀବଜନ୍ତୁମାନଙ୍କୁ ଯଦି ଶୈଶବାବସ୍ଥାରୁ ଏକାଠି ରଖାଯାଏ, ସେମାନେ ପରସ୍ପର ସହିତ ମିଳିମିଶି ରହିପାରିବେ ବୋଲି ସେ କହିଥିଲେ । ଏହାର ପରୀକ୍ଷଣ ନିମନ୍ତେ ସେ ଖଇରୀ ନିକଟରୁ ତାଙ୍କ ଅଗଣାରୁ ଧରା ଯାଇଥିବା

ଏକ ବିଷଧର ନାଗସାପକୁ ଆଣିଥିଲେ । ସେତେବେଳେ ଖଇରୀ ଖୁବ୍ ଛୋଟ ଥିଲା । ଖଇରୀର ସେହି ସାପ ପ୍ରତି ପ୍ରତିକ୍ରିୟାକୁ ଲେଖକ ଲକ୍ଷ୍ୟ କରିଥିଲେ । ଏଇ ବିଷଧର ସାପ ପ୍ରତି ଖଇରୀର ଆଗ୍ରହ ପ୍ରକାଶ ପାଇଥିଲା । ଯେତେବେଳେ ଏହି ସାପଟି ଖଇରୀ ଆଡ଼କୁ ଆଗ୍ରସର ହେବାକୁ ଚାହୁଁଥିଲା, ଲେଖକ ତାକୁ ଲାଞ୍ଜ ଧରି ପଛକୁ ଟାଣି ଦେଉଥିଲେ । ଥରେ ସେ ଅନ୍ୟମନସ୍କ ହୋଇ ଏହି ବିଷଧର ନାଗସାପକୁ ଟାଣିଦେବା ସମୟରେ, ସେ ଲେଖକଙ୍କୁ କାମୁଡ଼ି ଦେଇଥିଲା । ତାଙ୍କ ରକ୍ତରେ କିଛି ବିଷ ପ୍ରବାହିତ ହୋଇଯାଇଥିଲା । ତା’ପରଠାରୁ ସେ ନିମ୍ନ ରକ୍ତଚାପ ରୋଗରେ ପୀଡ଼ିତ ହୋଇଥିଲେ । ଏହି ସମୟରେ ଭାଲୁ ଲେଖକଙ୍କୁ କୁଣ୍ଢାଇବାକୁ ଆସୁଥିଲା । କିନ୍ତୁ ସରୋଜଙ୍କ କଡ଼ା ନିର୍ଦ୍ଦେଶ ହେତୁ ସେ ସେଥୁରୁ ନିବୃତ୍ତ ରହିଲା ।

ସରୋଜଙ୍କର ଏହି ଯୌଥ ପରିବାରରେ ସମସ୍ତେ ଯଥା ନେଉଳ, ଗୋଧ, ପକ୍ଷୀ, ସାପ, ଦେଶୀ କୁକୁର, ବାଘ, ଭାଲୁ ସମସ୍ତେ ଉତ୍ତମ ବୁଝାମଣାରେ ଏକ ଶୃଙ୍ଖଳିତ ଜୀବନଯାପନ କରୁଥିଲେ । ଖଇରୀ ଯେତେବେଳ ଏକ ୨ ମାସର ବାଘଛୁଆ ଥିଲା, ସେତେବେଳେ ଶିମିଳିପାଳର ୧୨ ଜଣ ଖରିଆ ଆଦିବାସୀ ତାକୁ ସରୋଜଙ୍କ ନିକଟକୁ ଆଣିଥିଲେ । ସେ ଥିଲେ ଜଙ୍ଗଲ ବିଭାଗର ଜଣେ ଅଧିକାରୀ ଥିଲେ । ଏହି ବାଘଛୁଆଟି ଥୁଲା ଭୋକିଲା ଓ ବିବ୍ରତ । ସେ ରାଗି ଗର୍ଜନ କରୁଥିଲା ଓ ତା’ର ପଞ୍ଝାକୁ ଘୋଷାରୁଥିଲା । ସରୋଜ ଏକ ମା’ ବାଘର ସ୍ଵରକୁ ଅନୁକରଣ କରିଥିଲେ ଯାହାକି ଖଇରୀକୁ ପୋଷା ମନେଇବା ପାଇଁ ଚମତ୍କାର ଭାବେ କାର୍ଯ୍ୟ କରିଥିଲା । ସେଥ‌ିରେ ସେ ସଫଳ ହୋଇଥିଲେ ଏବଂ ଖୁସିରେ ବିଭୋର ହୋଇଯାଇଥିଲେ । ତା’ ପରଦିନ ସକାଳୁ ସରୋଜ ଲେଖକଙ୍କୁ ନେଇ ଏକ ଜିପ୍‌ରେ ବ୍ୟାଘ୍ର ସଂରକ୍ଷଣ ଅଞ୍ଚଳ ବୁଲିବାକୁ ଗଲେ ।

ତାହା ଥିଲା ଏକ ସରୁ ଅଣଓସାରିଆ ରାସ୍ତା । ବାଟରେ ସରୋଜ ଅତୀତର ସ୍ମୃତିଚାରଣ କରି କହିଲେ, ତାଙ୍କୁ ଜନ୍ମଦିନରେ ତାଙ୍କ ମା’ ଏକ ବନ୍ଧୁକ ଉପହାର ଦେଇଥିଲେ ଯାହାଦ୍ଵାରା ଜଣେ ଯୁବକଭାବେ ସେ ବନ୍ୟଜନ୍ତୁମାନଙ୍କୁ ମୁକ୍ତଭାବେ ଶିକାର କରିଥିଲେ । ତା’ପରେ ସେ ଅନୁଭବ କଲେ ଯେ ପଶୁମାନଙ୍କୁ ମାରିବା ଏକ ଅପରାଧ। ଆନନ୍ଦ ଥାଏ ଏହି ନିରୀହ ସୁନ୍ଦର ଜୀବଜନ୍ତୁମାନଙ୍କୁ ସଂରକ୍ଷଣ କରିବାରେ । ବ୍ୟାଘ୍ର ପ୍ରକଳ୍ପର ନିର୍ଦ୍ଦେଶକ ଭାବେ ସେ ପ୍ରଥମେ ବ୍ୟାଘ୍ର ଗଣନା ପ୍ରଣାଳୀ ପ୍ରଚଳନ କରିଥିଲେ । ସେହି ରାତ୍ରିରେ ଉଭୟେ ଘଞ୍ଚ ଜଙ୍ଗଲ ମଧ୍ୟରେ ଥିବା ଏକ ଅତିଥୁ ଗୃହରେ ରାତ୍ରିଯାପନ କରିଥିଲେ । ଏହା ପୂର୍ବରୁ ଲେଖକ ଭୂମିଠାରୁ ୧୫ ଫୁଟ ଉଚ୍ଚରେ କାଠରେ ନିର୍ମିତ ଏକ ଘରେ ରହିବାର ଆନନ୍ଦ କେବେ ଅନୁଭବ କରିନଥିଲେ । ସେହିଘରେ କେତେଗୁଡ଼ିଏ କୋଠରୀ ସହ ଏକ ଗାଧୁଆଘର ଥିଲା। ତିନିମାସ ପରେ ଲେଖକ ପୁନର୍ବାର ସେହି ଖଇରୀ ଓ ଯଶିପୁରକୁ ବୁଲିବାକୁ ଗଲେ ।

CHSE Odisha Class 11 English Solutions Chapter 2 The Legend behind a Legend

ଖଇରୀ ବିଷୟରେ ଅଧିକ ଜାଣିବାର ଇଚ୍ଛା ତାଙ୍କୁ ସେଠାକୁ ପୁନର୍ବାର ଟାଣି ନେଇଥିଲା । ଶ୍ରୀ ଚୌଧୁରୀ ତାଙ୍କୁ ସ୍ଵାଗତ କଲେ ଏବଂ ଭାରି ଖୁସି ହେଲେ । ସେ ସେଠାରେ ଅନ୍ୟ ଜୀବମାନଙ୍କ ସହିତ ଏକ ଆଠ ଫୁଟ ଲମ୍ବର ଏକ ଅଜଗର ସାପ ଦେଖିଲେ । ତା’ପରେ ସେ ଚୌଧୁରୀ ଜୀବଜନ୍ତୁମାନଙ୍କୁ ସଂରକ୍ଷଣ କରିବାର ନିଶା ବିଷୟରେ ଜାଣିବାକୁ ଆଗ୍ରହ ପ୍ରକାଶ କଲେ । ଏହି ସମୟରେ ଏକ ଜରୁରୀ ସଭାରେ ଯୋଗ ଦେବାପାଇଁ ବିଶ୍ବ ବନ୍ୟଜନ୍ତୁ ପାଣ୍ଠି ତରଫରୁ ଏକ ତାରବାର୍ତ୍ତା ଆସି ପହଞ୍ଚିଲା। ସେ କଲିକତାର ଦମ୍ଦମ୍ ଉଡ଼ାଜାହାଜ ପଡ଼ିଆରୁ ବିମାନ ଯୋଗେ ଦିଲ୍ଲୀ ଯାତ୍ରା କଲେ । ଏହା ଥିଲା ଚୌଧୁରୀଙ୍କ ସହିତ ଲେଖକଙ୍କର ଶେଷ ସାକ୍ଷାତ । ଏହାର ତିନି ମାସ ପରେ ଖଇରୀର ମୃତ୍ୟୁ ଘଟିଥିଲା । ତା’ପରେ ଖଇରୀ ବିନା ଚୌଧୁରୀ ମଧ୍ୟ ବେଶି ଦିନ ବଞ୍ଚୁରି ନ ଥିଲେ ।

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(e)

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 14 Limit and Differentiation Ex 14(e) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Exercise 14(e)

Question 1.

Find derivatives of the following functions from the definition :
(i) 3x2 – \(\frac{4}{x}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(e)

(ii) (4x – 1)2
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(e) 1

(iii) 2 + x + √x3
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(e) 2

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(e)

(iv) x – \(\sqrt{x^2-1}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(e) 3

(v) \(\frac{1}{x^{2 / 5}}\) + 1
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(e) 4

Question 2.
(i) cos (ax + b)
Solution:
Let y = cos (ax + b)
Then \(\frac{d y}{d x}\) = -sin (ax + b) × \(\frac{d}{d x}\) (ax + b) by chain rule.
= -sin(ax + b). a = -a sin (ax + b)

(ii) x2 sin x
Solution:
Let y = x2 sin x
Then \(\frac{d y}{d x}=\frac{d}{d x}\) (x2). sin x + x2 \(\frac{d}{d x}\)
[ ∴ \(\frac{\mathrm{d}}{\mathrm{dx}}(u \cdot v)=\frac{d u}{d x} \cdot v+u \cdot \frac{d v}{d x}\)
= 2x sin x + x2 cos x

(iii) \(\sqrt{\tan x}\)
Solution:
Ley y = \(\sqrt{\tan x}\) = \((\tan x)^{\frac{1}{2}}\)
Then \(\frac{d y}{d x}=\frac{1}{2}(\tan x)^{-\frac{1}{2}} \times \frac{d}{d x}\)(tan x)
= \(\frac{1}{2 \sqrt{\tan x}}\) sec2 x.

(iv) cot x2
Solution:
Let y = cot x2
Then \(\frac{d y}{d x}=-{cosec}^2 x^2 \times \frac{d}{d x}\left(x^2\right)\)
= – cosec2 x2. 2x
= -2x. cosec2 x2

(v) cosec 3x
Solution:
Let y = cosec 3x
Then \(\frac{d y}{d x}\) = -3 cosec 3x . cot 3x

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(e)

Question 3.
(i) √x sin x
Solution:
Let y = √x sin x
Then \(\frac{d y}{d x}=\frac{d}{d x}\)(√x) sin x + √x. \(\frac{d}{d x}\)(sin x)
= \(\frac{1}{2 \sqrt{x}}\) sin x + √x. cos x

(ii) \(\sqrt{x^2+1}\)cos x
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(e) 5

(iii) tan x – x2 – 2x
Solution:
Let y = tan x – x2 – 2x
\(\frac{d y}{d x}\) = sec2 x – 2x – 2

BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 ଜ୍ୟାମିତିରେ ସାଦୃଶ୍ୟ Ex 1(c)

Odisha State Board BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 ଜ୍ୟାମିତିରେ ସାଦୃଶ୍ୟ Ex 1(c) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Geometry Chapter 1 ଜ୍ୟାମିତିରେ ସାଦୃଶ୍ୟ Ex 1(c)

Question 1.
ବନ୍ଧନୀ ମଧ୍ଯରୁ ଠିକ୍ ଉତ୍ତର ବାଛି ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର :
(i) △ABC ଓ △DEF ମଧ୍ୟ 6ର , m∠A = m∠D, m∠B = m∠E, AB = 3 ସେ.ମି., , BC = 5 ସେ.ମି.,ଏବଂ DE = 7.5 ସେ.ମି. ହେଲେ,, EF : _____ ସେ.ମି., (10, 10.5, 12, 12.5)
Solution:
12.5
Hint:
△ABC ~ △DEF ⇒ \(\frac { AB }{ DE }\) = \(\frac { BC }{ EF }\) ⇒ \(\frac { 3 }{ 7.5 }\) = \(\frac { 5 }{ EF }\) ⇒ EF = 12.5 ସେ.ମି. |

(ii) △ABC ରେ AB = 5 6 ସେ.ମି., BC = 7 ସେ.ମି., CA = 8 6 ସେ.ମି.; △PQR ରେ PQ = 10 ସେ.ମି., QR = 14 ସେ.ମି. । PR = _____ ସେ.ମି. ହେଲେ, △ABC ଓ △PQR ସଦୃଶକୋଣୀ ହେବେ । (12, 16, 20, 24)
Solution:
16
Hint:
△ABC ~ △PQR ⇒ \(\frac { AB }{ PQ }\) = \(\frac { BC }{ QR }\) ⇒ \(\frac { AC }{ PR }\)

(iii) △ABC ଓ △POR ମଧ୍ଯରେ ∠B ≅ ∠Q | △ABC ର AB = 8 ସେ.ମି. ଏବଂ BC = 12 ସେ.ମି. । A POR ର PQ = 12 ସେ.ମି. ଏବଂ QR = 18 ସେ.ମି. । △ABC ର କ୍ଷେତ୍ରଫଳ 48 ବର୍ଗସେ.ମି. ହେଲେ △PQR ର କ୍ଷେତ୍ରଫଳ = _____ ସେ.ମି. ହେଲେ, (84, 96, 104, 108)
Solution:
108
Hint:
△ABC ~ △PQR
∴ \(\frac { △ABC ର 6ସ୍ତୃତ୍ତ୍ଵଫଳ }{ △PQR ର 6ସ୍ତୃତ୍ତ୍ଵଫଳ }\) = \(\frac{\mathrm{AB}^2}{\mathrm{PQ}^2}\)
⇒ \(\frac { 48 }{ △PQR ର 6ସ୍ତୃତ୍ତ୍ଵଫଳ }\) = \(\frac { 4 }{ 9 }\)
⇒ △PQR ର 6ସ୍ତୃତ୍ତ୍ଵଫଳ = 108 ଦ . ସେ.ମି.

(iv) △ABC ଓ ∠ABC ର ସମଦ୍ୱିଖଣ୍ଡକ \(\overline{\mathrm{AC}}\) କୁ P ଦିନ୍ଦୁ6ର ଛେଦ କରେ | AB = 12 ସେ.ମି. ଓ BC = 9 ସେ.ମି. ହେବେ , AP : AC _____ | (4 : 3, 3 : 4, 7 : 4, 4 : 7)
Solution:
4 : 7
Hint:
∠B ର ସମଦ୍ୱିଖଣ୍ଡନ \(\overline{\mathrm{BP}}\)
⇒ \(\frac { AB }{ BC }\) = \(\frac { AP }{ PC }\) ⇒ \(\frac { 4 }{ 3 }\) = \(\frac { AP }{ PC }\)
∴ \(\frac { AP }{ AC }\) = \(\frac { 4 }{ 7 }\)

BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 ଜ୍ୟାମିତିରେ ସାଦୃଶ୍ୟ Ex 1(c)

(v) ଦୁଇଟି ସମବାହୁ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳର ଅନୁପାତ 16 : 25 ହେଲେ, ସେହି ତ୍ରିଭୁଜ ଦ୍ଵୟର ଅନୁରୂପ ଯୋଡ଼ାର ଦୈର୍ଘ୍ୟର ଅନୁପାତ _____ | (4 : 5, 2 : 5, 5 : 4, 5 : 2)
Solution:
4 : 5
Hint:
ଦୁକଟି ସମବାହୁ ତିଦୁକର 6ସ୍ତୃତ୍ରଫଳର ଅନ୍ନପାଜର ଦାହୁଦୂଯର ତାଦଣପୁର ବ୍ରଣ ଅନ୍ମଣ ପର ସହ ସମାନ |

(vi) ପାଣ୍ଡଷ୍ଟ ଚିତ୍ରରେ , m∠B = 50°, m∠BDC = 100° ଓ △DBC ~ △CBA ହେଲେ , m∠ACD ______ | (60°, 70°, 80°, 90°)
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 1
Solution:
70°
Hint:
△DBC ~ △CBA
⇒ m∠BDC = m∠ACB = 100° କକ୍ମ m∠BCD = 30°
∴ m∠ACD = 70°

(vii) ପାଣ୍ଡଷ୍ଟ ଚିତ୍ରରେ , △ABE ଓ △ACD ର 6ସ୍ତ୍ ତ୍ରଫଳ ସମାଜ 6 ଦୃ6କ , △BOC ~ _____ |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 2
(△ADE, △DOB, △EOD, △OEC)
Solution:
△EOD
Hint:
△ABE 6ସ୍ତୃତ୍ତ୍ଵଫଳ = D ACD 6ସ୍ତୃତ୍ତ୍ଵଫଳ ⇒ △BDE 6ସ୍ତୃତ୍ତ୍ଵଫଳ = △DEC 6ସ୍ତୃତ୍ତ୍ଵଫଳ ⇒ \(\overline{\mathrm{DE}}\) || \(\overline{\mathrm{BC}}\) ⇒ △BOC ~ △EOD

(viii) ପାଶ୍ଚଣ୍ଠ ଚିତ୍ରରେ △ABC ର \(\overline{\mathrm{AE}}\) ଓ \(\overline{\mathrm{BD}}\) ଯଥାକୃ6ମ \(\overline{\mathrm{BC}}\) ଓ \(\overline{\mathrm{AC}}\) ପୃତି ଦିପତାତ ଶାସ୍ତ୍ର ଦିନ୍ଦରୁ କମ , 6ତ6ଦ △BEM ~ △ ___ |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 3
[BEA, ABD, BDC, AEC]
Solution:
△AEC
Hint:
m∠EBM = m∠EAC ଏବଂ m∠MEB = m∠AEF
⇒ △BEM ~ △AEC

(ix) ପାସ୍ତସ୍ଥ ଚିତ୍ର6ର BC ରପରିସ୍ଥ D ଏକ ଦନ୍ଦୁ |
∠ADC ≅ ∠BAC ତ୍ରଫଳ ,
CB. CD = _____
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 4
(AC2 , AB2 , AD . AB, AD. AC)
Solution:
AC2
Hint:
△ABC ~ △DAC ⇒ \(\frac { CB }{ AC }\) = \(\frac { AC }{ CD }\)
⇒ CB . CD = AC2

(x) △ABC ରେ ∠BAC ର ସମଦିଖଣ୍ଡକ BC କୁ M ଦିନ୍ଦୁ6ର ଛେଦକ6ର | AB : AC = 2 : 3 ଏବଂ BC = 15 ସେ.ମି. ହେଲେ, , BM = _____ ସେ.ମି. | (6, 9, 10, 12)
Solution:
6
Hint:
\(\frac { AB }{ AC }\) = \(\frac { BM }{ MC }\) (∠BAC ର ସମଦ୍ୱିଗଣ୍ଡକ \(\overline{\mathrm{AM}}\))
⇒ \(\frac { 2 }{ 3 }\) = \(\frac { BM }{ MC }\)
⇒ BC = BM + MC ⇒ 15 = 2x + 3x ⇒ x = 3, BM = 6

Question 2.
(i)△ABC ରେ AB = 2.5 ସେ.ମି., BC = 2 ସେ.ମି., AC = 3.5 ସେ.ମି. ଏବଂ △PQR 66 PQ = 5 ସେ.ମି. QR = 4 ସେ.ମି. , PR = 7 ସେ.ମି. | m∠A = x° ଓ m∠Q = y° ଛେଦକ, m∠B, m∠C, m∠P ଓ m∠R ହେଲେ କୁର |
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 6
ଏO|6ର \(\frac { AB }{ PQ }\) = \(\frac { 2.5 }{ 5 }\) = \(\frac { 1 }{ 2 }\) , \(\frac { BC }{ QR }\) = \(\frac { 2 }{ 4 }\) = \(\frac { 1 }{ 2 }\) ଓ \(\frac { AC }{ PR }\) = \(\frac { 3.5 }{ 7 }\) = \(\frac { 1 }{ 2 }\)
⇒ \(\frac { AB }{ PQ }\) = \(\frac { BC }{ QR }\) = \(\frac { AC }{ PR }\) ⇒ △ABC ~ △PQR
⇒ m∠A = m∠P = x° , m∠B = m∠Q = y°
m∠C = 180° – (x – y)° = m∠R |

(ii) △ABC ଓ △DEF 68 ∠B ≅ ∠E, AB = 4 ସେ.ମି., BC = 6 ସେ.ମି., EF = 9 ସେ.ମି. ଓ DE = 6 ସେ.ମି. | △ABC ର ଶ୍ରେତ୍ରଫଳ 20 ଦଗ ସେ.ମି.ଦ୍ରେଭେ , DEF ର ଶ୍ରେତ୍ରଫଳ ନିଗ୍ରଯ କର |
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 7
\(\frac { AB }{ DE }\) = \(\frac { 4 }{ 6 }\) = \(\frac { 2 }{ 3 }\)
\(\frac { BC }{ EF }\) = \(\frac { 6 }{ 9 }\) = \(\frac { 2 }{ 3 }\)
∴ \(\frac { AB }{ DE }\) = \(\frac { BC }{ EF }\) = ଓ ∠B ≅ ∠E
⇒ △ABC ~ △DEF
⇒ \(\frac { △ABC ର ଶ୍ରେତ୍ରଫଳ }{ △DEF ର ଶ୍ରେତ୍ରଫଳ }\) = \(\frac{\mathrm{AB}^2}{\mathrm{DE}^2}\) = \(\frac { 4 }{ 9 }\)
⇒ \(\frac { 20 ବଗ 6ସ.ମି. }{ △DEF ର 6ଘ୍ତତ୍ରଫଳ }\) = \(\frac { 4 }{ 9 }\)
⇒ △DEF ର 6ଘ୍ତତ୍ରଫଳ = \(\frac { 20 × 9 }{ 4 }\) ଚ୍ଚଗ ସେ.ମି. = 45 ଦଗ6ସ.ମି. |

(iii) ଦୁଇଟି ସଦୃଶ ତ୍ରିଭୁଜ ମଧ୍ୟରୁ ପ୍ରଥମଟିର କ୍ଷେତ୍ରଫଳ ଦ୍ବିତୀୟଟିର କ୍ଷେତ୍ରଫଳର 9 ଗୁଣ ହେଲେ, ତ୍ରିଭୁଜ ଦୁଇଟିର ଅନୁରୂପ ବାହୁଦ୍ୱୟର ଅନୁପାତ ନିର୍ଣ୍ଣୟ କର ।
Solution:
ତ୍ରିଭୁଜଦ୍ୱୟର କ୍ଷେତ୍ରଫଳର ଅନୁପାତ = 9 : 1 = ତ୍ରିଭୁଜଦ୍ଵୟର ଅନୁରୂପ ବାହୁ ।
⇒ ତ୍ରିଭୁଜଦ୍ଵୟର ଅନୁରୂପ ବାହୁ ଦ୍ୱୟର ଦୈର୍ଘ୍ୟର ବର୍ଗାନୁପାତ = \(\sqrt{9^2}\) : \(\sqrt{1^2}\) = 3 : 1

(iv) ପାଣ୍ସ୍ଟଣ ଚିତ୍ର6ର , ∠BAC ≅ ∠DAC , AC = 12 6ସ.ମି. ଓ BC = 15 6ସ.ମି. | △ADC ର ଷ୍ଟେତୃଫକ 32 ଦ.6ସ.ମି. 6ଦୃ6କ , △ABD ର ଷ୍ଟେତୃଫକ ଚିତ୍ର6ର କର |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 5
Solution:
△ABC ଓ △DAC ମଧ୍ୟ6ର
∠BAC ≅ ∠DAC (ଦର)
∠ACB ≅ ∠ACD (ସାଧାରଣ 6କାଣ)
ଥଦଣଘ୍ତ ∠ACB ≅ ∠DAC
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 8
∴ △ABC ~ △DAC (6କା – 6କା – 6କା ସାଧାରଣ)
⇒ \(\frac { △ABC ର ଷ୍ଟେତୃଫକ }{ △DAC ର ଷ୍ଟେତୃଫକ }\) = \(\frac { △ABC ର ଷ୍ଟେତୃଫକ }{ 32 ପରିସାମା }\) = \(\frac{\mathrm{BC}^2}{\mathrm{AC}^2}\) = \(\frac{15^2}{12^2}\)
⇒ △ABC ର ଷ୍ଟେତୃଫକ = (\(\frac { 225 }{ 144 }\) × 32 ) ପରିସାମା = 50 ପରିସାମା |
△ABD ର ଷ୍ଟେତୃଫକ = △ABC ର ଷ୍ଟେତୃଫକ – △ADC ର ଷ୍ଟେତୃଫକ = 50 ପରିସାମା – 32 ପରିସାମା = 18 ଦ.6ସ.ମି.

(v) △ABC ର AB = 5 6ସ.ମି., BC = 7 6ସ.ମି.. ଓ CA = 9 6ସ.ମି. | △PQR ~ △ABC ଏବଂ △PQR ର ପରିସାମା 63 6ସ.ମି. ହୋ6କ , PQ, QR ଓ PR କିଣ୍ଡଯ କର |
Solution:
△PQR ~ △ABC (ଦଉ)
△ABC ର ପରିସାମା = (5 + 7 + 9 ) 6ସ.ମି. = 21 6ସ.ମି.
\(\frac { △PQR ର ଷ୍ଟେତୃଫକ }{ △ABC ର ଷ୍ଟେତୃଫକ }\) = \(\frac { 63 }{ 21 }\) = 3 (△PQR ର ପରିସାମା = 63 6ସ.ମି. )
⇒ \(\frac { △PQR ର ଷ୍ଟେତୃଫକ }{ △ABC ର ଷ୍ଟେତୃଫକ }\) = \(\frac{\mathrm{PQ}+\mathrm{QR}+\mathrm{PR}}{\mathrm{AB}+\mathrm{BC}+\mathrm{AC}}\) = \(\frac { PQ }{ AB }\) = \(\frac { QR }{ BC }\) = \(\frac { PR }{ AC }\)
⇒ 3 = \(\frac { PQ }{ 5 }\) = \(\frac { QR }{ 7 }\) = \(\frac { PR }{ 9 }\)
∴ PQ = 5 × 3 6ସ.ମି. = 15 6ସ.ମି. , QR = 7 × 3 6ସ.ମି. = 21 6ସ.ମି. PR = 9 × 3 6ସ.ମି. = 27 6ସ.ମି. |

(vi) △ABC ~ △PQR ; AB = 5 6ସ.ମି. , BC = 12 6ସ.ମି. ., AC = 13 6ସ.ମି. ଓ QR = 8 6ସ.ମି. ସମଦ୍ୱିଗଣ୍ଡକ △PQR ର ଷ୍ଟେତୃଫକ ଚିତ୍ର6ର କର |
Solution:
△ABC ~ △PQR(ଦଉ)
\(\frac { △ABC ର ଷ୍ଟେତୃଫକ }{ △PQR ର ଷ୍ଟେତୃଫକ }\) = \(\frac{\mathrm{BC}^2}{\mathrm{QR}^2}\) = \(\frac { 144 }{ 64 }\) = \(\frac { 9 }{ 4 }\)
(ତ୍ରିଭୁଜଦ୍ୱୟର କ୍ଷେତ୍ରଫଳର ଅନୁପାତ, ସେମାନଙ୍କର ଅନୁରୂପ ବାହୁଦ୍ୱୟର ଦୈର୍ଘ୍ୟର ଅନୁପାତ ସହ ସମାନ ।)
ଆମେ ଜାଣୁ 52 + 122 = 132 ଅର୍ଥାତ୍ ABC ଏକ ସମକୋଣୀ ତ୍ରିଭୁଜ । m∠ABC = 90°
∴△ABC ର ଷ୍ଟେତୃଫକ = \(\frac { 1 }{ 2 }\) × 5 × 12 6ସ.ମି = 30 6ସ.ମି2
\(\frac { 30 6ସ.ମି^2 }{ △PQR ର ଷ୍ଟେତୃଫକ }\) = \(\frac { 9 }{ 4 }\)
⇒ △PQR ର ଷ୍ଟେତୃଫକ = \(\frac { 30 × 4 }{ 9 }\) 6ସ.ମି2 = 13\(\frac { 1 }{ 3 }\) 6ସ.ମି2|

BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 ଜ୍ୟାମିତିରେ ସାଦୃଶ୍ୟ Ex 1(c)

(vii) △ABC ~ △PQR | △ABC ପରିସୀମା 60 ସେ.ମି. ଓ କ୍ଷେତ୍ରଫଳ 81 ବର୍ଗ ସେ.ମି. ଏବଂ △PQR ର ପରିସୀମା 80 ସେ.ମି. ହେଲେ, ଏହାର କ୍ଷେତ୍ରଫଳ କେତେ ?
Solution:
△ABC ~ △PQR (ଦଉ)
\(\frac { △ABC ର ପରିସୀମା }{ △PQR ର ପରିସୀମା }\) = \(\frac { 60 ସେ.ମି. }{ 80 ସେ.ମି. }\) = \(\frac { 3 }{ 4 }\)
⇒ \(\frac { AB }{ PQ }\) = \(\frac { 3 }{ 4 }\)
∴ \(\frac { △ABC ର କ୍ଷେତ୍ରଫଳ }{ △PQR ର କ୍ଷେତ୍ରଫଳ }\) = \(\frac{3^2}{4^2}\) = \(\frac { 9 }{ 16 }\)
⇒ \(\frac { 81 30 6ସ.ମି^2 }{ △PQR ର କ୍ଷେତ୍ରଫଳ }\) = \(\frac { 9 }{ 16 }\) ⇒ △PQR ର କ୍ଷେତ୍ରଫଳ = \(\frac { 81 × 16 }{ 9 }\) 6ସ.ମି2 = 144 6ସ.ମି2 .

Question 3.
ପ୍ତମାଣ କର 6ଯ କୁଲଟି ସହଶ ତ୍ରରୁଜର
(a) ଅନୁରୂପ ଉଚ୍ଚତାମାନଙ୍କର ଦୈର୍ଘ୍ୟ, ଉକ୍ତ ତ୍ରିଭୁଜ ଦ୍ୱୟର ଅନୁରୂପ ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ ସହ ସମାନୁପାତୀ ।
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 9
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 10

Question 4.
ଦୁଇଟି ସଦୃଶ ତ୍ରିଭୁଜର ପରିସୀମା ସମାନ ହେଲେ, ପ୍ରମାଣ କର ଯେ ତ୍ରିଭୁଜ ଦୁଇଟି ସର୍ବସମ ।
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 13

Question 5.
ଦୁଇଟି ସଦୃଶ ତ୍ରିଭୁଜର ପରିସୀମା ସମାନ ହେଲେ, ପ୍ରମାଣ କର ଯେ ତ୍ରିଭୁଜ ଦୁଇଟି ସର୍ବସମ ।
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 14

Question 6.
ପ୍ରମାଣ କର : ଦୁଇଟି ସଦୃଶ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳର ଅନୁପାତ, ଉକ୍ତ ତ୍ରିଭୁଜ ଦ୍ଵୟର
(a) ଅନ୍ତୁପ ଭରତାମାନକର 6ବିଶଇ ଦଗାନ୍ ପାତ ସହ ସମାନ |
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 15
ଦଭ : △ABC ~ △PQR , A <-> P, B <-> Q ଓ C <-> R
\(\overline{\mathrm{AD}}\) ⊥ \(\overline{\mathrm{BC}}\) ଓ \(\overline{\mathrm{PS}}\) ⊥ \(\overline{\mathrm{QR}}\)
ପ୍ତ।ମାଣୟ : \(\frac { △ABCରଶ୍ରେତ୍ରଫଳ }{ △PQRରଶ୍ରେତ୍ରଫଳ }\) = \(\frac{\mathrm{AD}^2}{\mathrm{PS}^2}\)
ପ୍ତମାଣ : △ABD ଓ △PQR ରେ ∠ABD ≅ ∠PQS (∵ ∠ABC ≅ ∠PQR)
∠ADB ≅ ∠PSQ (ଇବ6ପ୍ ସମ6କାଣ)
ଅତ୍ଣିପ୍ରତି ∠BAD ≅ ∠QPS
△ABD ଓ △PQS (6କା. 6କା. 6କା. ସାଦ୍ଶ୍ୟ)
⇒ \(\frac { AB }{ PQ }\) = \(\frac { AD }{ PS }\) (ସାଦ୍ୱଣଦର ଫଲ୍ଲା)
△ABC ~ △PQR ⇒ \(\frac { AB }{ PQ }\) = \(\frac { BC }{ QR }\) = \(\frac { AC }{ PR }\)
∴ \(\frac { AB }{ PQ }\) = \(\frac { BC }{ QR }\) = \(\frac { AC }{ PR }\) = \(\frac { AD }{ PS }\)
\(\frac { △ABCରଶ୍ରେତ୍ରଫଳ }{ △DEFରଶ୍ରେତ୍ରଫଳ }\) = \(\frac{\mathrm{AB}^2}{\mathrm{PQ}^2}\) = \(\frac{\mathrm{BC}^2}{\mathrm{QR}^2}\) = \(\frac{\mathrm{AC}^2}{\mathrm{PR}^2}\) = \(\frac{\mathrm{AD}^2}{\mathrm{PS}^2}\) (ପ୍ତମାଣିତ)

(b) ଅନୁରୁପ 6ଲାଣ-ସମଦ୍ଦିଖଣନମାନକର 6ବିଣ୍ୟର 6ବଣ୍ୟର ତାଗାନ୍ପାତ ସହ ପାପନ |
Solution:

BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 16
(c) ଅନୁରୁପ 6ଲାଣ-ସମଦ୍ଦିଖଣନମାନକର 6ବିଣ୍ୟର 6ବଣ୍ୟର ତାଗାନ୍ପାତ ସହ ପାପନ |
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 17
(d) ଅନୁରୁପ 6ଲାଣ-ସମଦ୍ଦିଖଣନମାନକର 6ବିଣ୍ୟର 6ବଣ୍ୟର ତାଗାନ୍ପାତ ସହ ପାପନ |
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 18

Question 7.
△ABC ର \(\overline{\mathrm{AB}}\) ଓ \(\overline{\mathrm{AC}}\) ଢାଦୁ ଭଜଟି ଦିନ୍ଦୁ 6ଯପରିକି △BQP ଓ △CPQ ସମ6ଷତ୍ରଫଳ ଦିଣିସ୍ତୃ | ପ୍ରମାଣ କର ଯେ \(\frac { PQ }{ BC }\) = \(\frac { AP }{ AB }\) |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 19
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 20

BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 ଜ୍ୟାମିତିରେ ସାଦୃଶ୍ୟ Ex 1(c)

Question 8.
ନିମ୍ନ ଚିତ୍ରରେ \(\overline{\mathrm{AB}}\) ଓ \(\overline{\mathrm{CD}}\) ର 6ଛଦଦିନ୍ଦୁ O |
(a) AO. OD = BO. OC 6ହକେ , ପୃମାଣ କର ଯେ △AOC ~ △BOD |
(b) CO. OD = AO. OB 6ହକେ , ପୃମାଣ କର ଯେ △AOC ~ △DOB |
(c) ପୃଦତରା 6କରି ପେଣ \(\overline{\mathrm{AC}}\) ଓ \(\overline{\mathrm{DB}}\) ସମାତୃର 6ଦୃ6ଦ ?
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 21
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 22

Question 9.
ABCD ଟ୍ରାପିଜିଯମ୍ବର \(\overline{\mathrm{AB}}\) || \(\overline{\mathrm{DC}}\) | କଣ୍ଡ \(\overline{\mathrm{AC}}\) s \(\overline{\mathrm{BD}}\) ପରଘରକୁ O ଦିନ୍ଦୁ6ର ଛେଦ କରତି | AO = 3 6ପ.ମି. ଏବଂ OC = 5 6ପ.ମି. | △AOB ର ଘେତ୍ରଫଳ 36 ଦ. 6ପ.ମି. ହେଲେ , △COD ର ଘେତ୍ରଫଳ କିଣ୍ଡଯ କର |
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 23
ABCD ଟ୍ଟାପିଳିଯମ୍ନରେ \(\overline{\mathrm{AB}}\) || \(\overline{\mathrm{DC}}\) |
\(\overline{\mathrm{AC}}\) ଓ \(\overline{\mathrm{BD}}\) ର ଛେଦଦିନ୍ଦୁ O |
∠ABO ≅∠ODC (ଏକାନ୍ତର 6କାଣ)
∠BAO ≅∠OCD (ଏକାନ୍ତର 6କାଣ)
∠AOB ≅∠COD (ପ୍ତତାପ 6କାଣ)
⇒ △AOB ~ △COD (6କା. 6କା. 6କା. ସାଦ୍ଶ୍ୟ)
⇒ \(\frac { △AOBରଶ୍ରେତ୍ରଫଳ }{ △CODରଶ୍ରେତ୍ରଫଳ }\) = \(\frac{\mathrm{AO}^2}{\mathrm{OC}^2}\)
⇒ \(\frac { 36 ଦଗ ସେ.ମି. }{ △CODରଶ୍ରେତ୍ରଫଳ }\) = \(\frac { 9 }{ 25 }\)
⇒ △COD ର ଶ୍ରେତ୍ରଫଳ = \(\frac { 36 × 25 }{ 9 }\) ଦଶ6ସ.ମି. = 100 ଦଶ6ପ.ମି. |

Question 10.
କିମ୍ନ ଚିତ୍ର6ର △ABC ଓ △DBC ଭଉଯ ଏକ ରମି \(\overline{\mathrm{BC}}\) ଭପରିଷ୍ଟ | \(\overline{\mathrm{AC}}\) ଓ \(\overline{\mathrm{BD}}\) ର 6ଛବଦିହୁ O 6ହ6ଲ ,
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 24
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 25

Question 11.
ପ୍ରମାଣ କର ଯେ ଏକ ତ୍ରିଭୁଜର ବାହୁମାନଙ୍କର ମଧ୍ୟବିନ୍ଦୁର ସଂଯୋଜକ ରେଖାଖଣ୍ଡମାନଙ୍କ ଦ୍ୱାରା ତ୍ରିଭୁଜଟି ଯେଉଁ ଚାରୋଟି ତ୍ରିଭୁଜରେ ପରିଣତ ହୁଏ, ସେମାନେ ସର୍ବସମ ଓ ପ୍ରତ୍ୟେକ ମୂଳ ତ୍ରିଭୁଜ ସହ ସଦୃଶ । ପୁନଶ୍ଚ ପ୍ରମାଣ କର ଯେ ଉତ୍ପନ୍ନ ହୋଇଥିବା ପ୍ରତ୍ୟେକ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳ, ମୂଳତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳର ଏକ ଚତୁର୍ଥାଂଶ ।
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 26

Question 12.
ପାଣମ୍କ ଟିତ୍ରଭେ , △ABC ର ∠ABC ଏକ ସମ6କାଣ | PQRS ଏକ ଥାଯତରିତ୍ର 6ତ୍ର6କ ଯେ,
△APS ~ △QCR ~ △PQB ~ △ACB |
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 27

BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 ଜ୍ୟାମିତିରେ ସାଦୃଶ୍ୟ Ex 1(c)

Question 13.
ABCD ଟାପିକଯମ୍6ର \(\overline{\mathrm{AD}}\) || \(\overline{\mathrm{BC}}\) | ∠ABD ≅ ∠DCB 6ଦୁ6କ , ପ୍ତମାଣକର ପେ BD = AD. BC |
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 28
ଦର : ABCD ଟ୍ରାପିଚ୍ଚିଯମ୍6ର \(\overline{\mathrm{AD}}\) || \(\overline{\mathrm{BC}}\)
ଏଣ ∠ABC ≅ ∠DCB |
ସ୍ତ।ମାଶ୍ୟ : BC2 = AD. BC
ପ୍ତମାତ : \(\overline{\mathrm{AD}}\) || \(\overline{\mathrm{BC}}\)
⇒ ∠ADB ≅∠DCB (ଏକାତ୍ରଉ 6କାଣ)
∠ADB ≅∠DCB (ଦଭ)
⇒ △ABD ~ △DCB (6କା. 6କା. 6କା. ସାଦ୍ଶ୍ୟ)
⇒ \(\frac { BD }{ BC }\) = \(\frac { AD }{ BD }\) (ପlଦୃଶ୍ୟର ଫଳା)
⇒ BD2 = BC. AD

Question 14.
ନପ୍ ଟିତ୍ର6ର \(\overline{\mathrm{AB}}\) || \(\overline{\mathrm{DC}}\) | △ADO ~ △BCO ବ୍ରେକେ , ପ୍ରମାଣ କର AD = BC |
(ସୂଚନା : ପ୍ରଶ୍ନ 5ରେ ପ୍ରମାଣିତ ତଥ୍ୟକୁ ବ୍ୟବହାର କର ।)
Solution:
ଦର : ABCD ଟ୍ରାପିଚ୍ଚିଯମ୍6ର \(\overline{\mathrm{AB}}\) || \(\overline{\mathrm{DC}}\) , △ADO ~ △BCO |
ସ୍ତ।ମାଶ୍ୟ : AD = BC
ପାଣମ୍କ : \(\overline{\mathrm{AB}}\) || \(\overline{\mathrm{DC}}\)
⇒ △ABD ର ସେତ୍ରଫଳ – △ABC ର ସେତ୍ରଫଳ
⇒ △ABD ର ସେତ୍ରଫଳ – △AOB ର ସେତ୍ରଫଳ
= △ABC ର ସେତ୍ରଫଳ – △AOB ର ସେତ୍ରଫଳ
⇒ △ADO ର ସେତ୍ରଫଳ – △BOC ର ସେତ୍ରଫଳ |
ପ୍ନନଣ୍ଡ , △ADO ~ △BCO (ଜଉ)
⇒ △ADO ≅ △BCO
(∵ ଦୁଇଟି ସଦୃଶ ତ୍ରିଭୁଜ ସମକ୍ଷେତ୍ରଫଳ ବିଶିଷ୍ଟ ହେଲେ ସେ ଦୁଇଟି ସର୍ବସମ ହେବେ । )
⇒ AD = BC

Question 15.
△ABC ର \(\overline{\mathrm{AB}}\) ଓ \(\overline{\mathrm{AC}}\) ବାହୁ ଉପରେ ଯଥାକ୍ରମେ X ଓ Y ବିନ୍ଦୁ ଅବସ୍ଥିତ ଯେପରିକି \(\overline{\mathrm{XY}}\) || \(\overline{\mathrm{BC}}\)ପ୍ରମାଣ କର ଯେ, △ABC ର ମଧ୍ୟମା \(\overline{\mathrm{AD}}\) , \(\overline{\mathrm{XY}}\) କୁ ସମତ୍ତିଖଣ୍ଡ କରେ ।
Solution:
ଦର : △ABC ର \(\overline{\mathrm{AB}}\) ଓ \(\overline{\mathrm{AC}}\) ବାହୁ ଉପରେ ଯଥାକ୍ରମେ X ଓ Y ବିନ୍ଦୁ ଅବସ୍ଥିତ ଯେପରିକି \(\overline{\mathrm{XY}}\) || \(\overline{\mathrm{BC}}\) | \(\overline{\mathrm{AD}}\) ତ୍ରିରୁ ଜର ଏକ ମଧ୍ୟମା | \(\overline{\mathrm{AD}}\) ଓ \(\overline{\mathrm{XY}}\) ର ଛେଦ ଦିନ୍ଦ୍ର O |
ସ୍ତ।ମାଶ୍ୟ : OX = OY
ପାଣମ୍କ : △AXO ଓ △ABD ଦର
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 29
∠AXO ≅ ∠ABD (ର୍ଥନ୍ମରୁପ 6କାଣ) (∵ \(\overline{\mathrm{OX}}\) || \(\overline{\mathrm{BD}}\)
∠AOX ≅ ∠ADB (ର୍ଥନ୍ମରୁପ 6କାଣ) (∵ \(\overline{\mathrm{OX}}\) || \(\overline{\mathrm{BD}}\)
⇒ △AXO ≅ △ABD (6କା. 6କା. 6କା. ସାଦ୍ଶ୍ୟ)
⇒ \(\frac { AO }{ AD }\) = \(\frac { OX }{ DB }\) (ସାଦ୍ୱଶ୍ୟର ସଂଳା)
ସେଦିପରି △AYO ~ △ACD ⇒ \(\frac { AO }{ AD }\) = \(\frac { OY }{ DC }\)
⇒ \(\frac { OX }{ DB }\) = \(\frac { OY }{ DC }\) ⇒ OX = OY (∵DB = DC ଦଇ)

Question 16.
△ABC ରେ \(\overline{\mathrm{AD}}\) ଏକ ମଧ୍ୟମା ଏବଂ \(\overline{\mathrm{AD}}\) ର ମଧ୍ୟବିନ୍ଦୁ E | \(\overrightarrow{\mathbf{B E}}\) ରଶ୍ମି \(\overline{\mathrm{AC}}\) କୁ X ବିନ୍ଦୁରେ ଛେଦକଲେ, ପ୍ରମାଣ କର ଯେ BE = 3EX |
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 30

Question 17.
△ABC 6ର \(\overline{\mathrm{AD}}\) ⊥\(\overline{\mathrm{BC}}\) ଏବଂ AD2 = BD. CD ହେଲେ, ପ୍ରମାଣକର ଯେ
(i) ∠BAC ଏକ ସମକୋଣ,
(ii) △ABD ର କ୍ଷେତ୍ରଫଳ ଓ △CADର କ୍ଷେତ୍ରଫଳ AB2 ଓ AC2 ସହ ସମାନୁପାତୀ
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 31

Question 18.
△ABC ଓ △DEF 6ର m∠A = m∠D, m∠B = m∠E | \(\overline{\mathrm{BC}}\) ଓ \(\overline{\mathrm{EF}}\) ଉ ମଧ୍ୟ ଦିନ୍ଦୁ ଯଥାକୁ ମେ X ଓ Y ହେଲେ , ପ୍ରମାଣ କର ଯେ
(i) △AXC ~ △DYF (ii) △AXB ~ △DYF |
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 32
ସ୍ତ।ମାଶ୍ୟ : (i) △AXC ~ △DYF (ii) △AXB ~ △DYF
ପାଣମ୍କ : △ABC ଓ △DEF 6ର m∠A = m∠D, ଓ m∠B = m∠E
⇒ △ABC ~ △DEF (6କା. 6କା. 6କା. ସାଦ୍ଶ୍ୟ)
⇒ \(\frac { AB }{ DE }\) = \(\frac { BC }{ EF }\) = \(\frac { AC }{ DF }\)
⇒ \(\frac { AB }{ BC }\) = \(\frac { DE }{ EF }\) ⇒ \(\frac { AB }{ 2BX }\) = \(\frac { DE }{ 2EY }\) (∵ X, \(\overline{\mathrm{BC}}\) ର ମଧ୍ୟଦିନ୍ଦୁ ଓ Y, \(\overline{\mathrm{EY}}\) ରମଧ୍ୟଦିନ୍ଦୁ )
⇒ \(\frac { AB }{ BX }\) = \(\frac { DE }{ EY }\)
ର୍ଥତ୍ରଗତ ∠ABX ≅ △DEY (∵ m∠B = m∠E)
⇒ △AXB ~ △DYE
ସେହପରି ପ୍ରମାଣ କରାଯାଇପାରିବ △AXC ~ △DYE |

BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 ଜ୍ୟାମିତିରେ ସାଦୃଶ୍ୟ Ex 1(c)

Question 19.
ପାଶ୍ୱମ୍ ଟି ତୃଭେ △ABC ର \(\overline{\mathrm{AB}}\) ଉପରିସ୍କ Q ଏକ ଦିନ୍ଦୁ , \(\overline{\mathrm{QR}}\) || \(\overline{\mathrm{BC}}\) 6ପପରିକି A-R-C, \(\overline{\mathrm{DR}}\) || \(\overline{\mathrm{QC}}\) ରମଧ୍ୟଦିନ୍ଦୁ A-D-B | ପ୍ରମାଣକର ମେ AQ2 = AD × AB |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 33
Solution:
ଦର : △ABC ର \(\overline{\mathrm{AB}}\) ଭପରିସ୍ଥ Q ଏକ ଦିନ୍ଦୁ , \(\overline{\mathrm{QR}}\) || \(\overline{\mathrm{BC}}\) 6ଯପରିକି
A-R-C ଏବଂ \(\overline{\mathrm{DR}}\) || \(\overline{\mathrm{QC}}\) 6ଯପରିକି A-D-B |
ସ୍ତ।ମାଶ୍ୟ : AQ2 = AD × AB
ପାଣମ୍କ : △AQC ରେ \(\overline{\mathrm{DR}}\) || \(\overline{\mathrm{QC}}\)
⇒ \(\frac { AR }{ AC }\) = \(\frac { AD }{ AQ }\) …(i)
△ABC ରେ \(\overline{\mathrm{QR}}\) || \(\overline{\mathrm{BC}}\) ⇒ \(\frac { AR }{ AC }\) = \(\frac { AQ }{ AB }\) …(ii)
(i) ଓ (ii ) ହ \(\frac { AQ }{ AB }\) = \(\frac { AD }{ AQ }\) ⇒ AQ2 = AD × AB

Question 20.
ପାଶଙ୍କ ଚିତ୍ର 6ର \(\overline{\mathrm{AB}}\) || \(\overline{\mathrm{CD}}\) || \(\overline{\mathrm{EF}}\) ଏର୍ଦ \(\overline{\mathrm{AF}}\) ଓ
\(\overline{\mathrm{BE}}\) ପରସ୍ତରକୁ C ବିନ୍ଦୁ 6ର 6ଛଦ କରନ୍ତି | ପ୍ରମାଣ କର ସେ EF × BD = DF × AB |
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 34

Question 21.
ଦୁଇଟି ସଦୃଶ ତ୍ରିଭୁଜର ଅନ୍ତଃବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ ଦ୍ବୟର ଅନୁପାତ, ଉକ୍ତ ତ୍ରିଭୁଜର ଦୁଇଟି ଅନୁରୂପ ବାହୁର ଦୈର୍ଘ୍ୟର ଅନୁପାତ ସହ ସମାନ, ପ୍ରମାଣ କର ।
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 35

Question 22.
A-P-B ଓ A-Q-B 6ହାଲେ ଏଇ \(\frac { AP }{ PB }\) = \(\frac { AQ }{ QB }\) 6ହାଲେ , ପ୍ତମାଣା କର ମେ P ଓ Q ଥିରିନ୍ନ |
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 36

Question 23.
ପାଣମ୍ଠ ଚିତ୍ରରେ △ABC ର ∠ABC ଏକ ଶୁକ6କାଣ | A ରୁ \(\overrightarrow{\mathbf{B C}}\) ପ୍ରତ ଅଳିତ ଲମୂର ପାଦ ଦିନ୍ଦୁ D | ଯଦି AD2 = BD. DC ହୁଏ , ପ୍ରମାଣ କର ଯେ ∠BAD ଓ ∠CAD ପରମର ଅନୁତ୍ପରକ |
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 37
Solution:
ଦର : △ABC ରେ ∠ABC ତ୍ପଳ6କାଣ | \(\overline{\mathrm{AD}}\) ⊥ \(\overrightarrow{\mathbf{CB}}\) , AD2 = BD. DC
ପ୍ରାମାଣ୍ୟ: m∠BAD + m∠CAD = 90°
ପ୍ରାମାଣ : AD2 = BD. DC (ଦର)
⇒ \(\frac { AD }{ BD }\) = \(\frac { DC }{ AD }\)
ଅନ୍ତଗତ ∠ADB ≅ ∠ADC (ପ୍ତ6ତ୍ୟକ ସମ6ଳାଣ)
⇒ △ADB ~ △CDA
⇒ ∠BAD ≅ ∠ACD
△ADC ରେ m∠ACD + m∠CAD = 90°
⇒ m∠BAD + m∠CAD = 90° (∵ ∠BAD ≅ ∠ACD)

Question 24.
△ABC ର \(\overline{\mathrm{AB}}\) ଓ \(\overline{\mathrm{AC}}\) ଉପରେ ଯଥାକ୍ରମେ X ଓ Y ବିନ୍ଦୁ ଅବସ୍ଥିତ, ଯେପରିକି \(\overline{\mathrm{XY}}\) || \(\overline{\mathrm{BC}}\) ଟ୍ରାପିଜିୟମ୍ XBCY ର କ୍ଷେତ୍ରଫଳ, △AXY ର କ୍ଷେତ୍ରଫଳର ଆଠଗୁଣ ହେଲେ, AX : BX ନିର୍ଣ୍ଣୟ କର ।
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 38

Question 25.
ABCD ଏକ ସ|ମାନ୍ତରିକ ଚିତ୍ର | \(\overrightarrow{\mathbf{AG}}\) ରଣ , \(\overline{\mathrm{BD}}\) , \(\overline{\mathrm{CD}}\) ଓ \(\overrightarrow{\mathbf{BC}}\) କୁ ଯଥାକ୍ତ6ମ E, F ଓ G ଦିନ୍ଦୁ 6ର 6ରଦରକା , ତ୍ପମାଣ କର ଯେ AE : EG = AF : AG |
Solution:
BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 Img 39
ଦର୍ : ABCD ଏକ ସ|ମାନ୍ତରିକ ଚିତ୍ର | \(\overrightarrow{\mathbf{AG}}\) ରଣ , \(\overline{\mathrm{BD}}\) , \(\overline{\mathrm{CD}}\) ଓ \(\overrightarrow{\mathbf{BC}}\) କୁ ଯଥାକ୍ତ6ମ E, F ଓ G ଦିନ୍ଦୁ 6ର 6ରଦରକା
ପ୍ରାମାଣ୍ୟ: AE : EG = AF : AG
ପ୍ରାମାଣ : △ABG ରେ \(\overline{\mathrm{CF}}\) || \(\overline{\mathrm{BA}}\) |
⇒ \(\frac { BC }{ BG }\) = \(\frac { AF }{ AG }\)
⇒ \(\frac { AF }{ AG }\) = \(\frac { BC }{ BG }\) = ⇒ \(\frac { AD }{ BG }\) (∵ AD = BC)
ପୁନଶ୍ଚ △AED ଓ △GEB ରେ ∠DAE ≅ ∠EGB (ଏଲାନ୍ତର ଲୋଣ)
∠AED ≅ ∠GEB (ପୃତାପ ଲୋଣ)
⇒ △AED ~ △GEB (କୋ . କୋ .ସାହଣ୍ୟ)
⇒ \(\frac { AD }{ BG }\) = \(\frac { AE }{ EG }\) (ସାଦୃଶ୍ୟର ସକା)
⇒ \(\frac { AE }{ EG }\) = \(\frac { AF }{ AG }\) (∵ \(\frac { AD }{ BG }\) = \(\frac { AF }{ AG }\))

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d)

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 14 Limit and Differentiation Ex 14(d) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Exercise 14(d)

Question 1.

Find the derivative of the following functions ‘an initio’, that is, using the definition.
(i) 2x3
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d)

(ii) x4
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 1

(iii) x2 + 1
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 2

(iv) \(\frac{1}{x}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 3

(v) \(\frac{1}{3 x+2}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 4

(vi) \(\frac{1}{x^2}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 5

(vii) \(\frac{x}{x+1}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 6

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d)

(viii) t(t – 1)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 7

(ix) s2 – bs + 5
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 8

(x) √x
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 9
\(\frac{1}{\sqrt{z}+\sqrt{z}}=\frac{1}{2 \sqrt{z}}\)

(xi) tan θ
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 10

(xii) cos 2θ
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 11

(xiii) x sin x
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 12

Question 2.
Find the derivative of the following function from the definition at the indicated points. Test whether the following functions are differentiable at the indicated points. If so find the derivative.
(i) x4 at x = 2
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 13

(ii) 2x2 + x + 1 at x = 1
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 14

(iii) x3 + 2x2 – 1 at x = 0
Solution:
Let x3 + 2x2 – 1
Then \(\left.\frac{d y}{d x}\right]_{x=0}\) = \(\lim _{h \rightarrow 0}\left[\frac{\left(h^3+2 h^2-1\right)-(-1)}{h}\right]\)
= \(\lim _{h \rightarrow 0}\) (h2 + 2h) = 0

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d)

(iv) tan x at x = \(\frac{\pi}{3}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 15

(v) \(\sqrt{3 x+2}\) at x = 0
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 16

(vi) In x at x = 2
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 17

(vii) \(e^x\) at x = 1
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 18

(viii) sin2 θ at θ = \(\frac{\pi}{4}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 19

Question 3.
\(\frac{x+1}{x-1}\) at x = -1
Solution:
We know that a function f(x) is differentiable at a point
x = c if (i) L.H.D. exists
(ii) R.H.D. exists
(iii) L.H.D. = R.H.D
Let f(x) = \(\frac{x+1}{x-1}\)
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 20
Thus L.H.D. and R.H.D. both exist and L.H.D. = R.H.D.
Hence f(x) is differentiable at x = -1 and the derivative is –\(\frac{1}{2}\)

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d)

Question 4.
√x at x = 0
Solution:
Let f(x) = √x
Then f(0) = 0
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 21

Question 5.
f(x) = \(\left\{\begin{array}{r}
1-x, x \leq \frac{1}{2} \\
x, x>\frac{1}{2}
\end{array} \text { at } x=\frac{1}{2}\right.\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 22

Question 6.
f(x) = \(\left\{\begin{array}{r}
\sin \frac{1}{x}, x \neq 0 \\
0, x=0
\end{array}\right.\) at x = 0
Solution:
f(0) = 0
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 23

Question 7.
f(x) = \(\left\{\begin{array}{r}
x^2 \sin \frac{1}{x^{\prime}}, x \neq 0 \\
0, x=0
\end{array}\right.\) at x = 0
Solution:
f(0) = 0
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(d) 24

CHSE Odisha Class 11 Math Solutions Chapter 16 Probability Ex 16(a)

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 16 Probability Ex 16(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 16 Probability Exercise 16(a)

Question 1.
A coin was tossed twice. Find the probability of getting.
(i) exactly one head
Solution:
A coin is tossed twice.
∴ S = {HH, HT, TH, TT}, |S| = 4
Let A be the event of getting exactly one head.
∴ A = {HT, TH} ⇒ |A| = 2
∴ P(A) = \(\frac{|\mathrm{A}|}{|\mathrm{S}|}=\frac{2}{4}=\frac{1}{2}\)

(ii) at least one head
Solution:
Let B be the event of getting at least one head.
∴ B = {HT, TH, HH}
∴ |B| = 3
∴ P(B) = \(\frac{|\mathrm{B}|}{|\mathrm{S}|}=\frac{3}{4}\)

(iii) at most one head
Solution:
Let C be the events of getting at most one head
∴ C = {HT, TH, TT} ⇒ |C| = 3
∴ P(C) = \(\frac{|C|}{|S|}=\frac{3}{4}\)

Question 2.
A coin is tossed three times. Find the probability of getting.
Solution:
A coin is tossed three times.
∴ S = {HHH, HTT, HTH, THH, TTH, THT, HHT, TTT}
∴ |S| = 8

(i) all heads
Solution:
Let A be the event of getting all heads.
∴ A = {HHH} ⇒ O(A) = 1
∴ P(A) = \(\frac{|\mathrm{A}|}{|\mathrm{S}|}=\frac{1}{8}\)

(ii) at most 2 heads
Solution:
Let B be the event of getting at most 2 heads.
∴ B = {HTT, HTH, THH, TTH, THT, HHT, TTT} ⇒ |B| = 7
∴ P(B) = \(\frac{|\mathrm{B}|}{|\mathrm{S}|}=\frac{7}{8}\)

(iii) at least 2 heads.
Solution:
Let C be the event of getting at least 2 heads.
∴ C = {HTH, THH, HHT, HHH} ⇒ |C| = 4
∴ P(C) = \(\frac{|\mathrm{C}|}{|\mathrm{S}|}=\frac{4}{8}=\frac{1}{2}\)

CHSE Odisha Class 11 Math Solutions Chapter 16 Probability Ex 16(a)

Question 3.
List all possible outcomes when a die is rolled twice or a pair of dice is rolled once. Then find the probability that
Solution:
A die is rolled twice
∴ S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ |S| = 36

(i) Sum of points is 10
Solution:
Let A be the event of getting the sum 10.
∴ A = {(4, 6), (5, 5), (6, 4)} ⇒ |A| = 3
∴ P(A) = \(\frac{3}{16}=\frac{1}{12}\)

(ii) sum of points is at least 10
Solution:
Let B be the event of getting the sum at least 10.
∴ B = {(4, 6), (5, 5), (6, 4) (5, 6) (6, 5) (6, 6)} ⇒ |B| = 6
∴ P(B) = \(\frac{|\mathrm{B}|}{|\mathrm{S}|}=\frac{3}{36}=\frac{1}{6}\)

(iii) sum of points is at most 10.
Solution:
Let C be the event of getting the sum 11 or 12.
∴ C = {(5, 6), (6, 5), (6, 6)} ⇒ |C| = 3
The C is the event of getting the sum at most 10.
∴ P(C’) = 1 – P(C’) =  1 – \(\frac{3}{36}=\frac{33}{36}\)

Question 4.
A die rolled twice. Find the probability that the result of the first roll exceeds the result of the second roll by
Solution:
A die rolled twice S = \(\left\{\begin{array}{llllll}
1 & 2 & 3 & 4 & 5 & 6 \\
1 & 2 & 3 & 4 & 5 & 6
\end{array}\right\}\)
∴ |S| = 36

(i) 3
Solution:
Let A be the event of getting the 1st roll exceeds the result of the 2nd roll by 3.
∴  A = {(4, 1), (5, 2), (6, 3)} ⇒ |A| = 3
∴ P(A) = \(\frac{|\mathrm{A}|}{|\mathrm{S}|}=\frac{3}{36}=\frac{1}{12}\)

(ii) at least 3
Solution:
Let B be the event of getting the 1st roll exceeds the result of the second roll by at least 3.
∴ B = {(4, 1), (5, 2), (6, 3), (5,1), (6, 2), (6, 1)}
∴ P(B) = \(\frac{|\mathrm{B}|}{|\mathrm{S}|}=\frac{6}{36}=\frac{1}{6}\)

(iii) at most 3
Solution:
Let A be the event of getting the 1st roll exceeds the result of the 2nd roll by 4 or 5.
∴ A ={(5, 1), (6, 2), (6, 1)}
P(A’) = 1 – P(A) = 1 – \(\frac{|A|}{|S|}\)
= 1 – \(\frac{3}{36}=\frac{33}{36}=\frac{11}{12}\)

Question 5.
A card is selected from 100 cards numbered 1 to 100. If a card is selected at random, find the probability that the number on the card is
Solution:
A card is selected from 100 cards numbered 1 to 100.
∴ |S| = 100

(i) divisible by 5
Solution:
Let A be the event of getting the card whose number is divisible by 5.
∴ A = {5, 10, 15, 20, ….. 10} ⇒ |A| = 20
∴ P(A) = \(\frac{|\mathrm{A}|}{|\mathrm{S}|}=\frac{20}{100}=\frac{1}{5}\)

(ii) divisible by 2
Solution:
Let B be the event of getting the card whose number is divisible by 2.
∴ B = {2, 4, 6, 8,…., 100} ⇒ |B| = 50
∴ P(B) = \(\frac{|B|}{|S|}=\frac{50}{100}=\frac{1}{2}\)

(iii) divisible by both 2 and 5
Solution:
If a number is divisible by both 2 and 5 then it is divisible by 10. Let A be a such an event.
∴ A ={10, 20, 30,……,100} ⇒ |A| = 10
∴ P(A) = \(\frac{|\mathrm{A}|}{|\mathrm{S}|}=\frac{10}{100}=\frac{1}{10}\)

(iv) divisible by either 2 or 5.
Solution:
Let A be the event of getting the number divisible by 2 and B be the event of getting the number divisible by 5.
∴ A = {2, 4, 6,……… 100}
B = {5, 10, 15, 20, ……, 100}
∴ A ∩ B = {10, 20, 30, ….., 100} ⇒ A ∩ B = 10
∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{|\mathrm{A}|}{|\mathrm{S}|}+\frac{|\mathrm{B}|}{|\mathrm{S}|}-\frac{|\mathrm{A} \cap \mathrm{B}|}{|\mathrm{S}|}\)
= \(\frac{50}{100}+\frac{20}{100}-\frac{10}{100}=\frac{60}{100}=\frac{3}{5}\)

CHSE Odisha Class 11 Math Solutions Chapter 16 Probability Ex 16(a)

Question 6.
Eight persons stand in a line at random. What is the probability that two person X and Y don’t stand together?
Solution:
Eight persons stand in a line at random.
∴ |S| = 8!
Let A be the event that two persons X and Y stand together. Considering X and Y as one person, the total number of persons is 7, who can stand in 7 ! × 2 ways.
∴ |A| = 2 × 7 !
∴ P(A) = \(\frac{|\mathrm{A}|}{|\mathrm{S}|}=\frac{2 \times 7 !}{8 !}=\frac{1}{4}\)
∴ P(A’) = 1 – P(A) = 1 – \(\frac{1}{4}=\frac{3}{4}\)

Question 7.
What is the probability that four aces appear together when a pack of 52 cards is shuffled completely?
Solution:
Let A be the event of getting four aces appearing together. Then considering the four aces as one card, the total number of cards is 49, which can be shuffled in 49! × 4! ways.
∴ P(A) = \(\frac{|\mathrm{A}|}{|\mathrm{S}|}=\frac{4 ! \times 49 !}{52 !}\)

Question 8.
If 8 persons are to sit around a table, what is the probability that X and Y don’t sit together?
Solution:
If 8 persons sit in a round table then the number of ways is (8 – 1)!
∴ |S| = (8 – 1)! = 7!
Let A be the event of getting that X and Y sit together. Considering X and Y as one person, the total number of persons is 7, who can be sit in 2 × 6 ! ways.
∴ P(A) = \(\frac{|\mathrm{A}|}{|\mathrm{S}|}=\frac{2 ! \times 6 !}{7 !}=\frac{2}{7}\)
P(A’) = 1 – P(A) = 1 – \(\frac{2}{7}=\frac{5}{7}\)

Question 9.
A die is rolled three times. Find the probability that the numbers obtained are in strictly increasing order.
Solution:
A die is rolled three times.
|S| = 63 = 216
Let A be the event of getting the numbers in strictly increasing order.
A = {(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 3, 4), (1, 3, 5), (1, 3, 6), (1, 4, 5), (1, 4, 6), (1, 5, 6), (2, 3, 4), (2, 3, 5), (2, 3, 6), (2, 4, 5), (2, 4, 6), (2, 5, 6), (3, 4, 5), (3, 5, 6), (4, 5, 6), (3, 4, 6)} ⇒ |A| = 20
∴ P(A) = \(\frac{|\mathrm{A}|}{|\mathrm{S}|}=\frac{20}{216}\)

Question 10.
Three phonorecords are removed from their jackets, played with, and then returned to the jackets at random. Find the probability that
Solution:
Three phonorecords are removed from their jackets, played with, and returned to the jackets at random. Let the records be numbered 1, 2, and 3, and let their jackets be similarly numbered 1, and 2,3. The number of ways in which the records can be put in their jackets is 3! = 6.
S = \(\left\{\left(\begin{array}{lll}
1 & 2 & 3 \\
1 & 2 & 3
\end{array}\right),\left(\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right),\left(\begin{array}{lll}
1 & 2 & 3 \\
3 & 1 & 2
\end{array}\right),\left(\begin{array}{lll}
1 & 2 & 3 \\
1 & 3 & 2
\end{array}\right),\right.\)
\(\left.\left(\begin{array}{lll}
1 & 2 & 3 \\
3 & 2 & 1
\end{array}\right),\left(\begin{array}{lll}
1 & 2 & 3 \\
2 & 1 & 3
\end{array}\right)\right\}\)

(i) none of the records goes to the right jacket
Solution:
Let A be the event that none of the records goes to the right jacket.
∴ A = \(\left\{\left(\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right),\left(\begin{array}{lll}
1 & 2 & 3 \\
3 & 1 & 2
\end{array}\right)\right\}\)
∴ P(A) = \(\frac{|\mathrm{A}|}{|\mathrm{S}|}=\frac{2}{6}=\frac{1}{3}\)

(ii) just one record goes to the right jacket.
Solution:
Let A be the event that none of the records goes to the right jacket.
∴ A = \(\left\{\left(\begin{array}{lll}
1 & 2 & 3 \\
1 & 3 & 2
\end{array}\right),\left(\begin{array}{lll}
1 & 2 & 3 \\
3 & 2 & 1
\end{array}\right),\left(\begin{array}{lll}
1 & 2 & 3 \\
2 & 1 & 3
\end{array}\right)\right\}\)
∴ P(A) = \(\frac{|\mathrm{A}|}{|\mathrm{S}|}=\frac{3}{6}=\frac{1}{2}\)

(iii) just two records go to the right jackets.
Solution:
Let B be the event that just two records goes to the right jackets. When two records goes to the right jackets, then it is obvious that the 3rd jacket must go to the right jacket.
∴ B = Φ
∴ P(B) = 0

(iv) all three of them go to the right jackets.
Solution:
Let C be the event that all 3 of them go to the right jackets.
∴ C = \(\left\{\left(\begin{array}{lll}
1 & 2 & 3 \\
1 & 2 & 3
\end{array}\right)\right\}\)
∴ P(C) = \(\frac{|\mathrm{C}|}{|\mathrm{S}|}=\frac{1}{6}\)

CHSE Odisha Class 11 Math Solutions Chapter 16 Probability Ex 16(a)

Question 11.
Four records are taken out of their jackets, played and returned to the jackets at random. Find the probability that
Solution:
Four records are taken out of their jackets, played and returned to the jackets at random.
∴ The number of ways in which the records can be put is 4!
∴ |S| = 24

(i) none of the records goes into the right jacket.
Solution:
Let the records and jackets be denoted as R1, R2, R3, R4, and J1, J2, J3, J4, respectively.
Considering \(\left(\begin{array}{llll}
\mathrm{R}_1 & \mathrm{R}_2 & \mathrm{R}_3 & \mathrm{R}_4 \\
\mathrm{~J}_2 & \mathrm{~J}_1 & \mathrm{~J}_4 & \mathrm{~J}_3
\end{array}\right),\left(\begin{array}{llll}
\mathrm{R}_1 & \mathrm{R}_2 & \mathrm{R}_3 & \mathrm{R}_4 \\
\mathrm{~J}_2 & \mathrm{~J}_4 & \mathrm{~J}_1 & \mathrm{~J}_3
\end{array}\right)\)
\(\left(\begin{array}{llll}
\mathrm{R}_1 & \mathrm{R}_2 & \mathrm{R}_3 & \mathrm{R}_4 \\
\mathrm{~J}_2 & \mathrm{~J}_3 & \mathrm{~J}_4 & \mathrm{~J}_1
\end{array}\right)\)
∴ When R1 be put in J2, there are 2 such cases. Similarly when R1 be put in J3 and J4 the number of such cases is 3 each.
∴ The total number of ways in which none of the records goes to the right jackets is 3 × 3 = 9.
∴ Its probability = \(\frac{15}{24}=\frac{5}{8}\)

(ii) at least one record is put in the right jacket.
Solution:
The number of ways in which at least one record goes to the right jacket, i.e. 1, 2, 3 of 4 records goes to the right jacket is 24 – 9 = 15.
∴ ItS probability = \(\frac{15}{24}=\frac{5}{8}\)

Question 12.
Let A and B be events with P(A) = \(\frac{3}{8}\), P(B) = \(\frac{1}{2}\) and P(A ∩ B) = \(\frac{1}{4}\). Find
(i) P(A ∪ B)
Solution:
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{3}{8}+\frac{1}{2}-\frac{1}{4}=\frac{3+4-2}{8}=\frac{5}{8}\)

(ii) P(Ac) and P(Bc)
Solution:
P(Ac) = 1 – P(A) = 1 – \(\frac{3}{8}=\frac{5}{8}\)
P(Bc) = 1 – P(B) = 1 – \(\frac{1}{2}=\frac{1}{2}\)

(iii) P(Ac ∪ Bc)
Solution:
P(Ac ∪ Bc) = P(A ∩ B)c = 1 – (A ∩ B)
= 1 – \(\frac{1}{4}=\frac{3}{4}\)

(iv) P(Ac ∩ Bc)
Solution:
P(Ac ∩ Bc) = P(A ∪ B)c = 1 – (A ∪ B)
= 1 – \(\frac{5}{8}=\frac{3}{8}\)

(v) P(A ∩ Bc)
Solution:
P(A ∩ Bc)
= P(A – B) = P(A) – P(A ∩ B)
= \(\frac{3}{8}-\frac{1}{4}=\frac{3-2}{8}=\frac{1}{8}\)

(vi) P(Ac ∩ B)
Solution:
P(Ac ∩ B)
= P(B – A) = P(B) – P(A ∩ B)
= \(\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\)

Question 13.
Let A and B be the events with P(A) = \(\frac{1}{3}\) P(A ∪ B) = \(\frac{3}{4}\) and P(A ∩ B) = \(\frac{1}{4}\), Find
(i) P(A)
Solution:
P(A) = \(\frac{1}{3}\)

(ii) P(B)
Solution:
we have
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
or, \(\frac{3}{4}=\frac{1}{3}\) +P(B) – \(\frac{1}{4}\)
or, P(B) = \(\frac{3}{4}+\frac{1}{4}-\frac{1}{3}=1-\frac{1}{3}=\frac{2}{3}\)
P(A’) = 1 – P(A) = 1 – \(\frac{1}{3}=\frac{2}{3}\)

(iii) P(A ∩ Bc)
Solution:
P(A ∩ Bc) = P(A – B) = P(A) – P(A ∩ B)
= \(\frac{1}{3}-\frac{1}{4}=\frac{4-3}{12}=\frac{1}{12}\)

(iv) P(A ∪ Bc)
Solution:
P(A ∪ Bc) = 1 – P(A ∪ Bc)c
= 1 – P(Ac ∪ B) = 1 – P(B – A)
= 1 – P(B) + P(A ∩ B)
= 1 – \(\frac{2}{3}+\frac{1}{4}=\frac{12-8+3}{12}=\frac{7}{12}\)

CHSE Odisha Class 11 Math Solutions Chapter 16 Probability Ex 16(a)

Question 14.
There are 20 defective bulbs in a box of 100 bulbs. If 10 bulbs are chosen at random what is the probability that
Solution:
There are 20 defective bulbs in a box of 100 bulbs. If 10 bulbs are chosen at random.

(i) there are just 3 defective bulbs
Solution:
|S| = 100C10
The number of defective bulbs is 20 so that the number of non-defective bulbs is 80.
Let A be the event of getting defective bulbs.
∴ |A| = 20C3 × 80C7
∴ P(A) = \(\frac{|\mathrm{A}|}{|\mathrm{S}|}=\frac{{ }^{20} \mathrm{C}_3 \times{ }^{80} \mathrm{C}_7}{{ }^{100} \mathrm{C}_{10}}\)

(ii) there are at least 3 defective balls.
Solution:
Let B be the event of getting at least 3 defective bulbs.
∴ B’ is the event of getting at most 2 defective bulbs.
CHSE Odisha Class 11 Math Solutions Chapter 16 Probability Ex 16(a)

Question 15.
A pair of dice is rolled once. Find the probability that the maximum of the two numbers
Solution:
A pair of dice is rolled once.
∴ S = \(\left\{\begin{array}{llllll}
1 & 2 & 3 & 4 & 5 & 6 \\
1 & 2 & 3 & 4 & 5 & 6
\end{array}\right\}\)
∴ |S| = 62 = 36

(i) is greater than 4
Solution:
A be the event of getting the maximum of two numbers greater than 4.
|A| = 20
∴ P(A) = \(\frac{20}{36}\)

(ii) is 6.
Solution:
Let A be the event of getting the maximum of two numbers is 6.
∴ A ={(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)}
∴ P(A) = \(\frac{|\mathrm{A}|}{|\mathrm{IS}|}=\frac{15}{36}\)

Question 16.
4 girls and 4 boys sit in a row. Find the probability that
Solution:
4 girls and 4 boys sit in a row.
∴ |S| = 8 !

(i) the four girls are together
Solution:
Let A be the event that 4 girls are together. Considering 4 girls as one, the total number of children is 5 which can be sit in 5! × 4! ways.
∴ |A| = 5! × 4!
∴ P(A) = \(\frac{\mid \mathrm{Al}}{|\mathrm{S}|}=\frac{5 ! \times 4 !}{8 !}\)

(ii) the boys and girls sit in alternate seats.
Solution:
When the boys and girls sit in alternate positions. So the arrangement can be as follows:
BC1BC1BC1BC1
C1BC1BC1BC1B
∴ The total number of ways = 2(4! × 4!)
∴ Its probability = \(\frac{2 \times 4 ! \times 4 !}{8 !}\)

Question 17.
A committee of 3 is to be chosen from among 10 people including X and Y. Find the probability that
Solution:
A committee of 3 is to be chosen from among 10 people including X and Y.
∴ |S| = 10C3

(i) X is the committee
Solution:
Let A be the event that X is in the committee. So we have chosen 2 persons from 9 persons in 9C2 ways
∴ |A| = 9C2
∴ P(A) = \(\frac{|\mathrm{A}|}{|\mathrm{S}|}=\frac{{ }^9 \mathrm{C}_2}{{ }^{10} \mathrm{C}_3}\)

(ii) X or Y belongs to the committee
Solution:
Let B be the event that X or Y belongs to the committee,
When X is in the committee, its probability = \(\frac{{ }^9 C_2}{{ }^{10} C_3}\)
When Y is the in the committee, its probability = \(\frac{{ }^9 C_2}{{ }^{10} C_3}\)
When X and Y both are in the committee, its probability = \(\frac{8 \mathrm{C}_1}{{ }^{10} \mathrm{C}_3}\)
∴ Probability that X or Y is in the committee
= \(\frac{{ }^9 \mathrm{C}_2+{ }^9 \mathrm{C}_2-{ }^8 \mathrm{C}_1}{{ }^{10} \mathrm{C}_3}=\frac{2 \times{ }^9 \mathrm{C}_2-{ }^8 \mathrm{C}_1}{{ }^{10} \mathrm{C}_3}\)

(iii) X and Y belong to the committee.
Solution:
When X and Y are both in the committee, we have to choose 1 person from 8 persons in 8C1 ways.
∴ Its probability = \(\frac{{ }^8 \mathrm{C}_1}{{ }^{10} \mathrm{C}_3}\)

CHSE Odisha Class 11 Math Solutions Chapter 16 Probability Ex 16(a)

Question 18.
A class consists of 25 boys and 15 girls. If a committee of 6 is to be chosen at random, find the probability that
Solution:
A class consists of 25 boys and 15 girls. A committee of6 is to be chosen at random.
∴ |S| = 40C6

(i) all members of the committee are girls.
Solution:
Let A be the event of getting all members of the committee are girls.
∴ |A| = 40C6
∴ P(A) = \(\frac{|\mathrm{A}|}{|\mathrm{S}|}=\frac{{ }^{15} \mathrm{C}_6}{{ }^{40} \mathrm{C}_6}\)

(ii) all members of the committee are boys.
Solution:
If all members of the committee are boys, then its probability = \(\frac{{ }^{25} \mathrm{C}_6}{{ }^{40} \mathrm{C}_6}\)

(iii) there are exactly 3 boys in the committee.
Solution:
Let A be the event of getting exact 3 boys in the committee.
∴ |A| = 25C3 × 15C3
∴ P(A) = \(\frac{{ }^{25} \mathrm{C}_3 \times{ }^{15} \mathrm{C}_3}{{ }^{40} \mathrm{C}_6}\)

(iv) there are exactly 4 girls in the committee.
Solution:
Let B the event of getting exactly 4 girls in the committee.
∴ |B| = 15C4 × 25C2
∴ P(B) = \(\frac{{ }^{15} \mathrm{C}_4 \times{ }^{25} \mathrm{C}_2}{{ }^{40} \mathrm{C}_6}\)

(v) there is at least one girl in the committee.
Solution:
Let C be the event of getting at least one girl in the committee.
∴ C’ is the event of getting no girl in the committee.
∴ |C’| = 25C6 ∴ P|C’| = \(\frac{\left|\mathrm{C}^{\prime}\right|}{|\mathrm{S}|}\)
∴ P(C) = 1 – P(C’) = 1 – \(\frac{{ }^{25} \mathrm{C}_6}{{ }^{40} \mathrm{C}_6}\)

Question 19.
There are 20 boys and 10 girls in the class. If a committee of 6 is to be chosen at random having at least 2 boys and 2 girls, find the probability that
Solution:
There are 20 boys and 10 girls in the class. A committee of 6 is to be chosen at random having at least 2 boys and 2 girls.

(20) Boys (10) girls
2 4
3 3
4 2

∴ |S| = (20C2 × 10C4) + (20C3 × 10C3) + (20C4 × 10C2)

(i) there are 3 boys in the committee.
Solution:
When there are 3 boys in the committee, its probability = \(\frac{{ }^{20} \mathrm{C}_3 \times{ }^{10} \mathrm{C}_3}{|\mathrm{~S}|}\)

(ii) there are 4 boys in the committee.
Solution:
When there are 4 boys in the committee, its probability = \(\frac{{ }^{20} \mathrm{C}_4 \times{ }^{10} \mathrm{C}_2}{|\mathrm{~S}|}\)

Question 20.
There are 120 students in a class who have opted for the following MIL. English 20, Oriya 70, Bengali 30. If a student is chosen at random, find the probability that the student is studying.
Solution:
There are 120 students in a class who have opted for the English 20, Oriya 70, Bengali 30.
∴ |S| = 120.
Let Be be the event of getting Bengali and E be the event of getting English.

(i) Bengali or English
Solution:
Since B and E are mutually exclusive events.
P(B ∪ E) = P(B) + P(E)
= \(\frac{|\mathrm{B}|}{|\mathrm{S}|}+\frac{|\mathrm{E}|}{|\mathrm{S}|}=\frac{30}{120}+\frac{20}{120}=\frac{50}{120}\)

(ii) neither Bengali nor English.
Solution:
Let A be the event of getting neither Bengali nor English, i.e. A is the event of getting Odia only.
∴ P(A) = \(\frac{|\mathrm{A}|}{|\mathrm{S}|}=\frac{70}{120}\)

Question 21.
Sometimes, probability of an event A is expressed as follows. We say that odds in favour of A are x toy if P(A) = \(\frac{x}{x+y}\). Similarly, we say that odds against A are x to y if P(A) = \(\frac{y}{x+y}\). Find P(A) and P(A)c if
Solution:
Odds in favour of A are x to y if P(A) = \(\frac{x}{x+y}\)
Odds against A are x to y if P (A) = \(\frac{y}{x+y}\)

(i) odds in favour of A are 2 to 5.
Solution:
P(A) = \(\frac{x}{x+y}\) = \(\frac{2}{2+5}=\frac{2}{7}\)
and P(A’) = 1 – P(A) = 1 – \(\frac{2}{7}=\frac{5}{7}\)

(ii) odds against A are 4 to 3.
Solution:
P(A) = \(\frac{3}{4+3}=\frac{3}{7}\)
P(A’) = 1 – P(A) = 1 – \(\frac{3}{7}=\frac{4}{7}\)

Question 22.
Six dice are rolled. Find the probability that all six faces show different numbers.
Solution:
Six dice are rolled once.
∴ |S| = 66
Let A be the event that all six faces show different numbers.
∴ |S| = 6!
∴ P(A) = \(\frac{6 !}{6^6}\)

CHSE Odisha Class 11 Math Solutions Chapter 16 Probability Ex 16(a)

Question 23.
There are 60 tickets in a bag numbered 1 through 60. Ifa ticket is picked at random, find the probability that the number on it is divisible by 2 or 5 and is not divisible by any of the numbers 3, 4, 6.
Solution:
There are 60 tickets in a bag numbered 1 through 60. A ticket is to be chosen, whose number is divisible by 2 or 5 and is not divisible by 3, 4, 6.
∴ |S| = 6 !
Let A be the event of getting the numbers divisible by 2 but not divisible by 3, 4, 6.
B be the event of getting* the numbers divisible by 5 but not divisible by 3, 4, 6.
∴ A = {2, 10, 14, 22, 26, 34, 38, 46, 50, 58}
B = {5, 10, 25, 35, 50, 55}
A ∩ B = {10, 50}
∴ P(A ∪ B) = P(A) + P (B) – P (A ∩ B)
= \(\frac{10+6-2}{60}=\frac{14}{60}\)

Question 24.
Compute P (A Δ B) in terms of P (A), P (B) and P (A ∩ B) where A Δ B denotes the symmetric difference of A and B.
Solution:
P (A Δ B) = P[(A – B)∪ (B – A)]
= P (A – B) + P (B – A) as (A – B) n (B – A) = Φ
= P[A – (A ∩ B)]+ P[B – (A ∩ B)]
= P(A) – P(A ∩ B) + P(B) – P(A ∩ B)
= P(A) + P(B) – 2P(A ∩ B)

Question 25.
Three volumes of a book and five volumes of another book are placed at random on a book shelf. Find the probability that all volumes of both the books will be found together.
Solution:
Three volumes of a book and five volumes of another book are placed at random on a book shelf.
∴ |S| = 8 !
When all volumes of both the books will find together, then considering the volumes as one book each, we have the total number of books is 2, which can be arranged in 2 × 3! × 5!
∴ Its probability = \(\frac{2 \times 3 ! \times 5 !}{8 !}\)

Question 26.
2 black cards and 2 red cards are lying face down on the table, If you guess their colours, find the probability that you get
Solution:
2 black cards and 2 red cards are lying face down on the table.
∴ We can guess their colours in \(\frac{4 !}{2 ! 2 !}\) = 4 ways
Cards kept face down as:
A student can guess:

(a) B B R R
(b) B R B R
(c) B R R B
(d) R R B B
(e) R B R B
(f) R B B R

(i) none of them right
Solution:
A student can guess none of them right only in case (d).
∴ Its probability = \(\frac{1}{6}\)

(ii) two of them right
Solution:
A student can guess two of them right in (b), (c), (d), (f).
∴ Its probability = \(\frac{4}{6}=\frac{2}{3}\)

(iii) all four of them right
Solution:
The student can guess all 4 of them right in (a) only.
∴ Its probability = \(\frac{1}{6}\)

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f)

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 14 Limit and Differentiation Ex 14(f) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Exercise 14(f)

Differentiate.

Question 1.
x8 + x7
Solution:
Let  y = x8 + x7
Then \(\frac{d y}{d x}\) = 8x7 + 7x6

Question 2.
x5/3 – x1/2
Solution:
Let y = x5/3 – x1/2
\(\frac{d y}{d x}=\frac{5}{3} x^{\frac{2}{3}}-\frac{1}{2} x^{-\frac{1}{2}}\)

Question 3.
x3 – 5x
Solution:
Let y = x3 – 5x
Then \(\frac{d y}{d x}\) = 3x2 – 5

Question 4.
√x + \(\frac{1}{\sqrt{x}}-\sqrt[3]{x^2}\)
Solution:
Let y = √x + \(\frac{1}{\sqrt{x}}-\sqrt[3]{x^2}\)
= \(x^{\frac{1}{2}}+x^{-\frac{1}{2}}-x^{\frac{2}{3}}\)
⇒ \(\frac{d y}{d x}=\frac{1}{2} x^{\frac{-1}{2}}-\frac{1}{2} x^{\frac{-3}{2}}-\frac{2}{3} x^{\frac{-1}{3}}\)

Question 5.
x2 + 2x – sin x + 5
Solution:
x2 + 2x – sin x + 5
\(\frac{d y}{d x}\) = 2x + 2 – cos x

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f)

Question 6.
\(\frac{1}{2} x^{\frac{1}{2}}+\frac{1}{3} x^{\frac{1}{3}}\)
Solution:
\(\frac{1}{2} x^{\frac{1}{2}}+\frac{1}{3} x^{\frac{1}{3}}\)
\(\frac{d y}{d x}=\frac{1}{4} x^{\frac{-1}{2}}+\frac{1}{9} x^{\frac{-2}{3}}\)

Question 7.
ax2 + b tan x + ln x3
Solution:
ax2 + b tan x + ln x3
\(\frac{d y}{d x}\) = 2ax + b sec2 x + \(\frac{3}{x}\)

Question 8.
√x(√x + 1)
Solution:
Let y = √x(√x + 1) = \(x+x^{\frac{1}{2}}\)
\(\frac{d y}{d x}=1+\frac{1}{2} x^{\frac{-1}{2}}\)

Question 9.
(x – 1)2
Solution:
Let y = (x – 1)2
Then \(\frac{d y}{d x}\) = 2(x – 1)

Question 10.
(x2 – x + 2)2
Solution:
Let y = (x2 – x + 2)2
\(\frac{d y}{d x}\) = 2(x2 – x + 2) × \(\frac{d}{d x}\)(x2 – x + 2)
= 2(x2 – x + 2)(2x – 1)

Question 11.
x sin x – \(\frac{e^x}{1+x^2}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f)

Question 12.
tan 2x + sec 2x
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f) 1

Question 13.
\(\frac{x^2}{x+1}-\frac{x}{1-x}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f) 2

Question 14.
\(\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f) 3

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f)

Question 15.
\(\frac{\tan x-\cos x}{\sin x \cos x}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f) 4

Question 16.
\(\left(\frac{x-1}{x+1}\right)^2\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f) 5

Question 17.
x3 (1 + x)(2 – x)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f) 6

Question 18.
x3 (sin x) e4 ln x
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f) 7

Question 19.
\(\frac{1}{\sqrt{x}}\) + x ln x3
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f) 8

Question 20.
x2 log2 x + sec x
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f) 9

Question 21.
\(\frac{x^2-1}{x^3+1}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f) 10

Question 22.
(x3 + 1)(3x2 + 2x – 7)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f) 11

Question 23.
cot x – sec x – log10 x
Solution:
\(\frac{d y}{d x}\) = -cosec 2 x – sec x. tan x – \(\frac{1}{x} \log _{10} e\)

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f)

Question 24.
\(\frac{1-\cos x}{1+\cos x}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f) 12

Question 25.
\(\frac{1-\tan x}{1+\tan x}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f) 13

Question 26.
\(\frac{\left[x^{\frac{3}{5}}-2 e^2 \ln x+\ln ^{\frac{2}{3}}\right]}{(1+x)}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f) 14

Question 27.
cosec x + cot x
Solution:
Let y = cosec x + cot x
\(\frac{d y}{d x}\) = -cosec x. cot x – cosec 2 x

Question 28.
tan2 x + sec2 x
Solution:
Let y = tan2 x + sec2 x
\(\frac{d y}{d x}\) = 2 tan x. \(\frac{d}{d x}\)(tan x) + 2 sec x \(\frac{d}{d x}\)(sec x)
= 2 tan x. sec2 x + 2 sec2 x. tan x
= 4 sec2 x tan x

Question 29.
tan2 x + ax
Solution:
tan2 x + ax
\(\frac{d y}{d x}\) = 2 tan x. sec2 x + ax. ln a

Question 30.
sin2 x + x ln x
Solution:
sin2 x – x ln x
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f) 15

Question 31.
cos2 x + ex cos x
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f) 16

Question 32.
\(\frac{a^x-b^x}{x}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f) 17

Question 33.
\(\frac{e^x+e^{-x}}{x^2+1}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f) 18

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f)

Question 34.
\(\frac{\ln x}{x^2}\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f) 19

Question 35.
Show that f(x) = \(\left\{\begin{array}{l}
x \sin \frac{1}{x}, x \neq 0 \\
0, x=0
\end{array}\right.\) is not differentiable x = 0
Solution:
Differentiability
CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(f) 20

CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15 Textbook Exercise Questions and Answers.

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CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Exercise 15

Question 1.
If the values observed are 1, 2, …..,n each with frequency 1, find
(i) the mean value
Solution:
Mean of 1, 2, 3, ….. n
= \(\frac{1+2+3 \ldots . . .+n}{n}=\frac{n(n+1)}{2 n}=\frac{n+1}{2}\)

(ii) the mean deviation from the mean separately for two cases when n is odd and when n is even.
Solution:
If n is even, let n = 2m.
CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15

Question 2.
For the same set of values as in (1) above, find the variance and standard deviation.
Solution:
x: 1, 2, 3, ….., n
CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15 1

CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15

Question 3.
From the table below, find the mean value and the variance.
(a) Values: 1  2  3 ….. n
Frequency: 1  2  3 …. n
Solution:
x: 1  2  3 ….. n
y: 1  2  3 …. n
CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15 2

Question 4.
From the table below, find the mean and the variance.
Solution:
(a) Values: 1  2  5 ….. (2n – 1)
Frequency: 1  1  1        1
CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15 3

(b) Values: 2  4  6 …..2n
Frequency: 1  1  1      1
CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15 4

Question 5.
From the table below, calculate the mean and the variance
\(\text { Values } \quad \mathbf{0} \quad 1 \quad 2 \ldots \quad r \ldots n\)
\(\text { Frequency: } \quad{ }^n \mathbf{C}_0{ }^n \mathbf{C}_1{ }^n \mathbf{C}_2{ }^n \mathbf{C}_r \ldots . .{ }^n \mathbf{C}_n\)
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15 5

CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15

Question 6.
From the following table calculate the mean, mean deviation from the mean, and variance.

Marks Number of students
30-35 5
35-40 7
40-45 8
45-50 20
50-55 16
55-60 12
60-65 7
65-70 5

Solution:

C. I f Mid value (x) d = x – A
30-35 5 32.5 -15
35-40 7 37.5 -10
40-45 8 42.5 -5
45-50 20 47.5 0
50-55 16 52.5 5
55-60 12 57.5 10
60-65 7 62.5 15
65-70 5 67.5 20
∑f = 80

Let A (working mean) = 47.5, i = 5

u = d/i fu u2 fu2
-3 -15 9 45
-2 -14 4 28
-1 -8 1 8
0 0 0 0
1 16 1 16
2 24 4 48
3 21 9 63
4 20 16 80
∑fu = 44 ∑fu2 = 44

CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15 6

Question 7.
In a soccer league, two teams A and B have the following records
A: Goals scored: 0  1  2  3  4
Number of matches: 11 18 8 6 2
B: Goals scored: 0  1  2  3  4  5
Number of matches: 5 20 10 6 3 1
Which team is more consistent? Which is a better team.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15 7
∴ The mean of B is more than that of A, so B is the better team. A is more consistent as its variance is less than that of B.

CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15

Question 8.
The coefficient of variation is defined as \(\sigma / \bar{x}\), that is the standard deviation divided by the mean value. Find the coefficient of variation c.v. for each of the following sets of observations.
(i) 2, 3, 4, 2, 5, 7, 8, 9
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15 8

(ii) 5, 7, 9, 10, 7, 5, 8, 9, 3
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15 9

(iii) 3, 3, 3, 4, 4, 4, 5, 5, 5
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15 10

Question 9.
Suppose the values x1, x2, …. xn having frequency f1, f2, …. fn respectively having mean value x̄ and variance σ2. Let a be a fixed real number
x1 + a, x2 + a, ….. , xn + a with frequency f1, f2, ….. fn respectively will have mean value x̄ + a and variance σ.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15 11

Question 10.
Find the mean and deviation from the mean and the standard deviation of a, a + d, a + 2d, …. , a + 2nd assume that d > 0.
Solution:
CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15 13
CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15 14

Question 11.
Let x1, x2, …. xn be a set of observations with mean value 0 and variance σ2x and y1, y2, …. ym be another set of observations with mean value 0 and variance σ2y. Find the mean value and variance of the set of observations x1, x2, …. xn , y1, y2, …. ym combined.
Solution:
x1, x2, …. xn be a set of observations with mean value 0 and variance σ2x and y1, y2, …. ym be another set of observations with mean value 0 and variance σ2y
CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15 15

CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15

Question 12.
Find which group of the following data is more dispersed :

Range 10-20 20-30 30-40 40-50 50-60
(Group A)
Frequency
5 1 3 2 1
(Group A)
Frequency
1 3 2 3 1

Solution:
Let us find the mean and standard deviation for the given two distributions.
(i) Mean deviation about mean
M. D = \(\frac{1}{N} \sum_{i=1}^n f_i\left|x_i-\bar{x}\right|\)

(ii) Mean deviation about median
M. D = \(\frac{1}{N} \sum_{i=1}^n f_i\left|x_i-M\right|\)

(iii) variance
Variance is the mean of squared deviations from the mean.
CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15 16

(iv) Standard deviation
Standard deviation is the square root of the mean of squared deviations from the mean.
∴ Standard deviation
CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15 17

Question 13.
The price of land per square meter and that of gold per ten grams over five consecutive years is given below. Decide, which price maintains better stability. [Hint: Stability ⇔ Consistency]

Price of land/Sq.meter(₹) 1500 2500 2600 3000 4000
Price of gold/10 gms(₹) 2500 2600 2750 2900 2850

Solution:
CHSE Odisha Class 11 Math Solutions Chapter 15 Statistics Ex 15 18