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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.

(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.

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

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

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’.

Answer:

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.

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.

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

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

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’.

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.

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.

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.

(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.

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.

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.

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.

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.

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.

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.

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.)

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.

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.

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.

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.

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.

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.

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”.

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.

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.

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.

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.

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.

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.

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.

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)
= $\frac{2}{21}+\frac{4}{21}+\frac{6}{21}=\frac{12}{21}$

(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 = $\frac{3}{13}$

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:

(ii) (4x – 1)²
Solution:

(iii) 2 + x + √x³
Solution:

(iv) x – $\sqrt{x^2-1}$
Solution:

(v) $\frac{1}{x^{2 / 5}}$ + 1
Solution:

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

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:

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

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 $\overline{\mathrm{BP}}$
⇒ $\frac { AB }{ BC }$ = $\frac { AP }{ PC }$ ⇒ $\frac { 4 }{ 3 }$ = $\frac { AP }{ PC }$
∴ $\frac { AP }{ AC }$ = $\frac { 4 }{ 7 }$

(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°)

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 ~ _____ |

(△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 ~ △ ___ |

[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 = _____

(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:

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:

$\frac { AB }{ DE }$ = $\frac { 4 }{ 6 }$ = $\frac { 2 }{ 3 }$
$\frac { BC }{ EF }$ = $\frac { 6 }{ 9 }$ = $\frac { 2 }{ 3 }$
∴ $\frac { AB }{ DE }$ = $\frac { BC }{ EF }$ = 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 |

Solution:
△ABC ଓ △DAC ମଧ୍ୟ6ର
∠BAC ≅ ∠DAC (ଦର)
∠ACB ≅ ∠ACD (ସାଧାରଣ 6କାଣ)
ଥଦଣଘ୍ତ ∠ACB ≅ ∠DAC

∴ △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|

(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 ।

Solution:

Question 4.
ଦୁଇଟି ସଦୃଶ ତ୍ରିଭୁଜର ପରିସୀମା ସମାନ ହେଲେ, ପ୍ରମାଣ କର ଯେ ତ୍ରିଭୁଜ ଦୁଇଟି ସର୍ବସମ ।
Solution:

Question 5.
ଦୁଇଟି ସଦୃଶ ତ୍ରିଭୁଜର ପରିସୀମା ସମାନ ହେଲେ, ପ୍ରମାଣ କର ଯେ ତ୍ରିଭୁଜ ଦୁଇଟି ସର୍ବସମ ।
Solution:

Question 6.
ପ୍ରମାଣ କର : ଦୁଇଟି ସଦୃଶ ତ୍ରିଭୁଜର କ୍ଷେତ୍ରଫଳର ଅନୁପାତ, ଉକ୍ତ ତ୍ରିଭୁଜ ଦ୍ଵୟର
(a) ଅନ୍ତୁପ ଭରତାମାନକର 6ବିଶଇ ଦଗାନ୍ ପାତ ସହ ସମାନ |
Solution:

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:

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

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

Question 7.
△ABC ର $\overline{\mathrm{AB}}$ ଓ $\overline{\mathrm{AC}}$ ଢାଦୁ ଭଜଟି ଦିନ୍ଦୁ 6ଯପରିକି △BQP ଓ △CPQ ସମ6ଷତ୍ରଫଳ ଦିଣିସ୍ତୃ | ପ୍ରମାଣ କର ଯେ $\frac { PQ }{ BC }$ = $\frac { AP }{ AB }$ |

Solution:

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 ?

Solution:

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:

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 ,

Solution:

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:

Question 12.
ପାଣମ୍କ ଟିତ୍ରଭେ , △ABC ର ∠ABC ଏକ ସମ6କାଣ | PQRS ଏକ ଥାଯତରିତ୍ର 6ତ୍ର6କ ଯେ,
△APS ~ △QCR ~ △PQB ~ △ACB |
Solution:

Question 13.
ABCD ଟାପିକଯମ୍6ର $\overline{\mathrm{AD}}$ || $\overline{\mathrm{BC}}$ | ∠ABD ≅ ∠DCB 6ଦୁ6କ , ପ୍ତମାଣକର ପେ BD = AD. BC |
Solution:

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

∠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:

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:

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:

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 |

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 |

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:

Question 21.
ଦୁଇଟି ସଦୃଶ ତ୍ରିଭୁଜର ଅନ୍ତଃବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ ଦ୍ବୟର ଅନୁପାତ, ଉକ୍ତ ତ୍ରିଭୁଜର ଦୁଇଟି ଅନୁରୂପ ବାହୁର ଦୈର୍ଘ୍ୟର ଅନୁପାତ ସହ ସମାନ, ପ୍ରମାଣ କର ।
Solution:

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:

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 |

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:

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:

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)
⇒ $\frac { AE }{ EG }$ = $\frac { AF }{ AG }$ (∵ $\frac { AD }{ BG }$ = $\frac { AF }{ AG }$)

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:

(ii) x⁴
Solution:

(iii) x² + 1
Solution:

(iv) $\frac{1}{x}$
Solution:

(v) $\frac{1}{3 x+2}$
Solution:

(vi) $\frac{1}{x^2}$
Solution:

(vii) $\frac{x}{x+1}$
Solution:

(viii) t(t – 1)
Solution:

(ix) s² – bs + 5
Solution:

(x) √x
Solution:

$\frac{1}{\sqrt{z}+\sqrt{z}}=\frac{1}{2 \sqrt{z}}$

(xi) tan θ
Solution:

(xii) cos 2θ
Solution:

(xiii) x sin x
Solution:

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:

(ii) 2x² + x + 1 at x = 1
Solution:

(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

(iv) tan x at x = $\frac{\pi}{3}$
Solution:

(v) $\sqrt{3 x+2}$ at x = 0
Solution:

(vi) In x at x = 2
Solution:

(vii) $e^x$ at x = 1
Solution:

(viii) sin² θ at θ = $\frac{\pi}{4}$
Solution:

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) = $\frac{x+1}{x-1}$

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 –$\frac{1}{2}$

Question 4.
√x at x = 0
Solution:
Let f(x) = √x
Then f(0) = 0

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:

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

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

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) = $\frac{|\mathrm{C}|}{|\mathrm{S}|}=\frac{4}{8}=\frac{1}{2}$

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)
= $\frac{|\mathrm{A}|}{|\mathrm{S}|}+\frac{|\mathrm{B}|}{|\mathrm{S}|}-\frac{|\mathrm{A} \cap \mathrm{B}|}{|\mathrm{S}|}$
= $\frac{50}{100}+\frac{20}{100}-\frac{10}{100}=\frac{60}{100}=\frac{3}{5}$

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) = $\frac{|\mathrm{C}|}{|\mathrm{S}|}=\frac{1}{6}$

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, R_4, 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(A^c) and P(B^c)
Solution:
P(A^c) = 1 – P(A) = 1 – $\frac{3}{8}=\frac{5}{8}$
P(B^c) = 1 – P(B) = 1 – $\frac{1}{2}=\frac{1}{2}$

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

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

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

(vi) P(A^c ∩ B)
Solution:
P(A^c ∩ 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 ∩ B^c)
Solution:
P(A ∩ B^c) = P(A – B) = P(A) – P(A ∩ B)
= $\frac{1}{3}-\frac{1}{4}=\frac{4-3}{12}=\frac{1}{12}$

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

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.

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 = $\frac{{ }^8 \mathrm{C}_1}{{ }^{10} \mathrm{C}_3}$

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) = $\frac{6 !}{6^6}$

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 = $\frac{1}{6}$

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.
x^5/3 – x^1/2
Solution:
Let y = x^5/3 – x^1/2
$\frac{d y}{d x}=\frac{5}{3} x^{\frac{2}{3}}-\frac{1}{2} x^{-\frac{1}{2}}$

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

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:

Question 12.
tan 2x + sec 2x
Solution:

Question 13.
$\frac{x^2}{x+1}-\frac{x}{1-x}$
Solution:

Question 14.
$\frac{\sqrt{x}-1}{\sqrt{x}+1}$
Solution:

Question 15.
$\frac{\tan x-\cos x}{\sin x \cos x}$
Solution:

Question 16.
$\left(\frac{x-1}{x+1}\right)^2$
Solution:

Question 17.
x³ (1 + x)(2 – x)
Solution:

Question 18.
x₃ (sin x) e^{4 ln x}
Solution:

Question 19.
$\frac{1}{\sqrt{x}}$ + x ln x³
Solution:

Question 20.
x² log₂ x + sec x
Solution:

Question 21.
$\frac{x^2-1}{x^3+1}$
Solution:

Question 22.
(x³ + 1)(3x² + 2x – 7)
Solution:

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

Question 24.
$\frac{1-\cos x}{1+\cos x}$
Solution:

Question 25.
$\frac{1-\tan x}{1+\tan x}$
Solution:

Question 26.
$\frac{\left[x^{\frac{3}{5}}-2 e^2 \ln x+\ln ^{\frac{2}{3}}\right]}{(1+x)}$
Solution:

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 + a^x
Solution:
tan² x + a^x
$\frac{d y}{d x}$ = 2 tan x. sec² x + a^x. ln a

Question 30.
sin² x + x ln x
Solution:
sin² x – x ln x

Question 31.
cos² x + e^x cos x
Solution:

Question 32.
$\frac{a^x-b^x}{x}$
Solution:

Question 33.
$\frac{e^x+e^{-x}}{x^2+1}$
Solution:

Question 34.
$\frac{\ln x}{x^2}$
Solution:

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

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

APOLLOTYRE Pivot Point Calculator

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.

Question 2.
For the same set of values as in (1) above, find the variance and standard deviation.
Solution:
x: 1, 2, 3, ….., n

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

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

(b) Values: 2  4  6 …..2n
Frequency: 1  1  1      1

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:

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

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:

∴ 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.

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:

(ii) 5, 7, 9, 10, 7, 5, 8, 9, 3
Solution:

(iii) 3, 3, 3, 4, 4, 4, 5, 5, 5
Solution:

Question 9.
Suppose the values x₁, x₂, …. x_n having frequency f₁, f₂, …. f_n respectively having mean value x̄ and variance σ². Let a be a fixed real number
x₁ + a, x₂ + a, ….. , x_n + a with frequency f₁, f₂, ….. f_n respectively will have mean value x̄ + a and variance σ.
Solution:

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:

Question 11.
Let x₁, x₂, …. x_n be a set of observations with mean value 0 and variance σ²_x and y₁, y₂, …. y_m be another set of observations with mean value 0 and variance σ²_y. Find the mean value and variance of the set of observations x₁, x₂, …. x_n , y₁, y₂, …. y_m combined.
Solution:
x₁, x₂, …. x_n be a set of observations with mean value 0 and variance σ²_x and y₁, y₂, …. y_m be another set of observations with mean value 0 and variance σ²_y

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.

(iv) Standard deviation
Standard deviation is the square root of the mean of squared deviations from the mean.
∴ Standard deviation

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:

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 12 Conic Sections Ex 12(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 12 Conic Sections Exercise 12(b)

Question 1.

Fill in the blanks by choosing the correct answer from the given ones :
(a) The equation of the directrix to the parabola x² = -6y is _____________. [y + 6 = 0, 2y – 3 = 0, y – 6 = 0, 2y + 3 = 0]
Solution:
2y – 3 = 0

(b) The eccentricity of the parabola y² = 8x is ____________. (2, 8, 0, 1)
Solution:
1

(c) The line y + x = k is tangent to the parabola y² + 12x = 0 if k = ______________. (-3, 3, 6, -6)
Solution:
3

(d) The latus rectum of the parabola (y – 2)² = 8(x + 3) is ______________. (2, 4, 8, 16)
Solution:
8

(e) The equation of tangent to the parabola x² = 6y at vertex is _______________. (x = 0, y = 0, x = $\frac{-3}{2}$, y = $\frac{-3}{2}$)
Solution:
y = 0

(f) The equation of axis of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}$ = 1 is ___________. (x = 4, y = 3, x = 0, y = 0)
Solution:
y = 0

(g) The equation of the major axis of the ellipse $\frac{(x+1)^2}{16}+\frac{(y-2)^2}{25}$ = 1 is __________. (x = 4, x = -1, y = 5, y = 2)
Solution:
x = -1

(h) The distance between the focii of the ellipse 3x² + 4y² = 1 is _____________. (1, $\frac{1}{\sqrt{3}}$, $\frac{2}{\sqrt{3}}$, $\frac{1}{2 \sqrt{3}}$
Solution:
$\frac{1}{\sqrt{3}}$

(i) The eccentricity of the ellipse $\frac{x^2}{16}+\frac{y^2}{25}$ = 1 is _______________. ($\frac{4}{5}, \frac{5}{4}, \frac{3}{5}, \frac{16}{25}$)
Solution:
$\frac{3}{5}$

(j) The line y = 2x + K is a tangent to the ellipse 5x² + y² = 5 if K = ______________. (2, 5, √3, √21)
Solution:
3

(k) The length of the latus rectum of the ellipse $\frac{(x-2)^2}{4}+\frac{(y+3)^2}{25}$ = 1 is _______________. ($\frac{4}{25}, \frac{2}{5}, \frac{5}{2}, \frac{8}{5}$)
Solution:
$\frac{8}{5}$

(l) The equation of the conjugate axis of the hyperbola $\frac{x^2}{9}-\frac{(y+2)^2}{16}$ = 1 is ____________. (x = 0, x = 3, y = -3, y = 4)
Solution:
x = 0

(m) the hyperbola $\frac{y^2}{16}-\frac{x^2}{12}$ = 1 intersects x – axis at ___________. [(0, ±4), (±2√3, 0), (2, 0), no where]
Solution:
no where

(n) The eccentricity of the hyperbola 4x² – 3y² = 1 is ____________. ($\frac{4}{3}, \frac{3}{4}, \frac{\sqrt{21}}{3}, \frac{\sqrt{7}}{2 \sqrt{3}}$)
Solution:
$\frac{\sqrt{21}}{3}$

(o) The latus rectum of the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}$ = 1 is ___________. ($\frac{16}{9}, \frac{9}{16}, \frac{1}{9}, \frac{32}{9}$)
Solution:
$\frac{32}{9}$

(p) The line y = 3x – k is a tangent to the hyperbola 6x² – 9y² = 1 if k ____________. (1, $\frac{5}{3 \sqrt{2}}, \frac{1}{\sqrt{6}}, \frac{2}{3}$)
Solution:
$\frac{5}{3 \sqrt{2}}$

Question 2.
Mention which of the following statements are true (T) and false (F) :
(a) The equation y = x² + 2x + 3 represents a parabola with its axis parallel to y – axis.
Solution:
True

(b) The latus rectum of the parabola y² = -8x is 2.
Solution:
False

(c) The eccentricity of the parabola (y – 1)² = 2(x + 3)² is $\frac{1}{3}$.
Solution:
False

(d) The line y = 3 is a tangent to the parabola (x + 2)² = 6(y – 3).
Solution:
True

(e) The equation Ax² + By² = 1 represents an ellipse with its axis parallel to x – axis, if A > B > 0.
Solution:
False

(f) The focii of the ellipse $\frac{x^2}{3}+\frac{y^2}{2}$ = 1 are the points (±1, 0).
Solution:
True

(g) The equation of the ellipse with focii at (0, ±4) and vertices (0, ±7) is $\frac{x^2}{16}+\frac{y^2}{49}$ = 1.
Solution:
False

(h) The length of the latera recta of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}$ = 1 and $\frac{(x+2)^2}{4}+\frac{(y-1)^2}{9}$ = 1 are equal.
Solution:
True

(i) The equation of the latera recta of the ellipse $\frac{(x-4)^2}{16}+\frac{(y-1)^2}{9}$ = 1 are x = 4 ± √7.
Solution:
True

(j) The line y = x + 2 is a tangent to the ellipse $\frac{x^2}{2}+\frac{y^2}{1}$ = 1.
Solution:
False

(k) The conjugate axis of the hyperbola, $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ = 1 meets the hyperbola at two points which are at distance 2b from each other.
Solution:
False

(l) The conjugate axis of the hyperbola, $\frac{(y-3)^2}{9}+\frac{(x+2)^2}{3}$ = 1 is parallel to the line x = 4.
Solution:
False

(m) The length of the transverse axis of the hyperbola with focii at (±5, 0) and vertices at (±2, 0) is 10.
Solution:
False

(n) The latera recta of the ellipse $\frac{x^2}{25}-\frac{y^2}{16}$ = 1 are same.
Solution:
True

(o) The y – axis is tangent to the hyperbola ay² – bx² = 1
Solution:
False

Question 3.
Find the equation of the parabola in each of the following cases :
(a) the vertex at (0, 0) and focus at (0, 3).
Solution:

(b) the vertex at (0, 0) and directrix x – 2 = 0.
Solution:

(c) the vertex at (6, -2) and focus at (-3, -2).
Solution:

and the parabola facing towards left.
Eqn. of the parabola is (y – k)² = -4a(x – h)
or, (y + 2)² = -4 × 9(x – 6)
= -36(x + 6)
or, (y + 2)² + 36(x – 6) = 0

(d) the vertex at (-2, 1) and focus at (-2, 4).
Solution:

(e) the length of the latus rectum is 6. and the vertex is at (0, 0), the parabola opening to the right.
Solution:
The length of the latus rectum is 6
∴ 4a = 6
The parabola opens to right and vertex at (0, 0).
∴ Eqn. of the parabola is y² = 4ax = 6x.

(f) the vertex is at (0, 0) the parabola opening to the left and passing through (-1, 2).
Solution:
Vertex at (0, 0), parabola opening to left, passing through (-1, 2).
∴ Let the eqn. of the parabola be y² = -4ax
As it passes through (-1, 2),
we have 4 = -4a (-1) or, a = 1
∴ Eqn. of the parabola is y² = 4x.

(g) the vertex at (0, 0) the parabola opens downwards, and the latus rectum of length 10.
Solution:
Vertex at (0, 0), the parabola opens downward, length of the latus rectum is 10.
∴ 4a = 10
∴ Eqn. of the parabola is
x² = -4ay or, x²  = -10y.

(h) the axis is vertical and the parabola passes through the points (0, 2), (-1, 1), (2, 10).
Solution:
Axis is vertical, parabola passes through the points (0, 2), (-1, 1), and (2, 10).
Let the eqn. of the parabola be
x² + 2gx + 2fy + c = 0. As it passes through the above points,
we have 0 + 0 + 2f . 2 + c = 0
⇒ c = -4f, 1 – 2g + 2f . 1 + c = 0
⇒ c = 2g – 2f – 1 and
4 + 2g . 2 + 2f . 10 + c = 0
⇒ c = -4g – 20f – 4
∴ -4f = 2g – 2g – 1

∴ Equation of the parabola is
x² + 2gx + 2fy + c = 0
or, x² + 2x – y + 2 = 0
or, y = x² + 2x + 2

(i) the axis is horizontal and the parabola passes through the points (2, -1), (-2, -4), and (-1, 3).
Solution:
Le the eqn. of the parabola be
y² + 2gx + 2fy + c = 0
As it passes through the points (2, -1), (-2, -4), and (-1, 3)
we have 1 + 2g . 2 +2f (-1) + c = 0    …(1)
16 + 2g (-2) + 2f (-4) + c = 0     …(2)
and 9 + 2g (-1) + 2f . 3 + c = 0    …(3)
∴ From eqn. (1) → 1 + 4g – 2f + c = 0
⇒ c = 2f – Ag – 1
Eqn. (2) → 16 – 4g – 8f + c = 0
⇒ c = 8f + 4g – 16
and eqn. (3) → 9 – 2g + 6f + c = 0
⇒ c = 2g – 6f – 9
∴ – 1 + 2f – 4g = 8f + 4g – 16
or, -6f = 8g – 15 or, f = $\frac{8 g-15}{-6}$
Again -1 + 2f – 4g = 2g – 6f – 9
or, 8f = 6g – 8 or, f = $\frac{6 g-8}{8}$
∴ $\frac{8 g-15}{-6}=\frac{6 g-8}{8}$
or, 32g – 60 = -18g + 24
or, 50g = 84

(j) Vertex at (1, 3) and the directrix, x + 3 = 0.
Solution:
Vertex at (1, 3), directrix x + 3 = 0
∴ h = 1, k = 3
We have the directrix is x = h – a
∴ h – a = -3
or, a = h + 3 = 4
∴ Eqn. of the parabola is
(y – k)² = 4a(x – h)
or, (y – 3)² = 16(x – 1)

(k) Vertex at (1, -1) and the directrix y – 2 = 0
Solution:
Vertex at (2, -1), directrix y – 2 = 0
∴ h = 1, k = -1
We have the directrix is y = k – a
∴ k – a = 2
or, a = k – 2 = -1 – 2 = – 3
∴ Eqn. of the parabola is
(x – h)² = 4a(y – k)
⇒ (x – 1)² = 4 (-3) (y + 1)
⇒ (x – 1)² = -12 (y + 1)
⇒ (x – 1)² + 12 (y + 1) = 0

(l) the focus at(-2, 3) and the directrix 3x + 4y – 2 = 0.
Solution:
Focus at (-2, 3),
directrix 3x + 4y – 2 = 0.

Question 4.
Find the equation of the ellipse in each of the following cases :
(a) center at (0, 0), one vertex at (0, -5) and one end of minor axis is (3, 0).
Solution:

(d) centre at (0, 0), one vertex at (7, 0) and one end of the minor axis is (0, -5).
Solution:

(c) foci at (±5, 0), and the length of the major axis is 12.
Solution:
Foci at (±5, 0),
length of the major axis is 12
∴ c = 5, 2a = 12
∴ a = 6
∴ b² = a² – c² = 36 – 25 = 11
Equ. of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ = 1 or, $\frac{x^2}{25}+\frac{y^2}{4}$ = 1

(e) center at (5, 4) and the major axis, is of length 16 and the minor axis is of length 10.
Solution:
Centre at (5, 4), major axis 16, minor axis 10.
∴ h = 5, k = 4, 2a = 16, 2b = 10
∴ a = 8, b = 5
Eqn. of the ellipse is $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}$ = 1
or, $\frac{(x-5)^2}{64}+\frac{(y-4)^2}{25}$ = 1

(f) Centre at (-3, 3), vertex at (-3, 6), and one end of minor axis at (0, 3).
Solution:
Centre at (-3, 3), vertex at (-3, 6),
one end of minor axis (0, 3).

(g) Centre at (0, 0), axes parallel to coordinate axes, eccentricity is $\frac{1}{\sqrt{2}}$ and the minor axis is of length 5.
Solution:
(0, 0), axes parallel to coordinate axes, eccentricity is $\frac{1}{\sqrt{2}}$, and the minor axis is 5.

(h) centre at (0, 0) axes parallel to coordinate axes, eccentricity is $\frac{\sqrt{3}}{2}$ and the ellipse passing through the point (√3, $\frac{1}{2}$)
Solution:

we have 3 + 4 × $\frac{1}{4}$ = a² or, a² = 4
∴ Eqn of the ellipse is x² + 4y² = 4

(i) centre at (0, 0), one end of the major axis is (-5, 0) and eccentricity $\frac{3}{5}$
Solution:
centre at (0, 0), one end of the major axis is (-5, 0), eccentricity $\frac{3}{5}$
∴ a = 5, $\frac{c}{a}$ = $\frac{3}{5}$ or, c = 3
∴ b² = a² – c² = 25 – 9 = 16
∴ Eqn. of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ = 1 or, $\frac{x^2}{25}+\frac{y^2}{16}$ = 1

(j) axis parallel to coordinate axes, the centre at (0, 0) and the ellipse passing through (3, -2) and (-1, 3).
Solution:
Axis parallel to coordinate axes, centre at (0, 0), ellipse passing through (3, -2) and (-1, 3).
Let the eqn. of the ellipse be

(k) centre at (3, 4), axis parallel to x – axis and passing through (6, 4) and (3, 6).
Solution:
Centre at (3, 4), axis parallel to x – axis, ellipse passing through (6, 4) and (3, 6).
Let the eqn. of the ellipse be

(l) Centre at (-2, 1) axis parallel to y – axis, eccentricity is $\frac{\sqrt{7}}{4}$ and the ellipse passing through (-2, 5).
Solution:
Centre at (-2, 1), axis parallel to y – axis, eccentricity is $\frac{\sqrt{7}}{4}$ passing through (-2, 5)

Question 5.
Obtain the equation of a hyperbola in each of the following cases :
(a) foci at (±4, 0) and vertices (±2, 0).
Solution:
Here c = 4, a = 2
∴ b² = c² – a² = 16 – 4 = 12
∴ Eqn. of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}$ = 1 or, $\frac{x^2}{4}-\frac{y^2}{12}$ = 1

(b) foci at (0, ±72) and vertices (0, ±1).
Solution:

(c) centre at (0, 0) transverse axis along x – axis of length 4, and focus at (2√5,0).
Solution:
Here 2a = 4, c = 2√5
∴ a =2
∴ b² = c² – a² = 20 – 4 = 16
∴ Eqn. of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}$ = 1 or, $\frac{x^2}{4}-\frac{y^2}{16}$ = 1

(d) centre at (0, 0), the conjugate axis along x – axis of length 6 and eccentricity 2.
Solution:
Here 2b = 6 ⇒ b = 3, $\frac{c}{a}$ = 2
or, c = 2a
We have a² + b² = c²
or, a² + 9 = 4a²
or, 3a² = 9 or, a² = 3
∴ Eqn. of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}$ = 1 or, $\frac{x^2}{3}-\frac{y^2}{9}$ = 1

(e) foci at (±2√3, 0) and eccentricity √3.
Solution:
Here c = 2√3, $\frac{c}{a}$ = √3
or, a = $\frac{c}{\sqrt{3}}$ = 2
∴ b² = c² – a² = 12 – 4 = 8
∴ Eqn. of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}$ = 1 or, $\frac{x^2}{4}-\frac{y^2}{8}$ = 1

(f) centre at (0, 0) transverse axis is along y – axis, the distance between the foci is 14 and the distance between the vertices is 12.
Solution:
Here 2c = 14, 2a = 12
∴ c = 7, a = 6
∴ b² = c² – a² = 49 – 36 = 13
∴ Eqn. of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}$ = 1 or, $\frac{x^2}{36}-\frac{y^2}{13}$ = 1

(g) centre (1, -2), transverse axis parallel to the x-axis of length 6 and conjugate axis of length 10.
Solution:
Centre (1, -2), transverse axis parallel to x – axis of length 6, the conjugate axis of length 10.
∴ 2a = 6, 2b= 10
∴ a = 3, b = 5, h = 1, k = -2
∴ Eqn. of the hyperbola $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}$ = 1 or, $\frac{(x-1)^2}{9}-\frac{(y-2)^2}{25}$ = 1

(h) Centre (2, -3), eccentricity $\frac{5}{3}$ and hyperbola passing through (5, -3).
Solution:
Here h = 2, k = -3, $\frac{c}{a}=\frac{5}{3}$ or, c = $\frac{5 a}{3}$ we have c² = a² + b²

(i) centre at origin, axis perpendicular to y-axis and the hyperbola passes through the points (3, -2) and (5, -7).
Solution:
Let the eqn. of the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}$ = 1
As the hyperbola passes through the points (3, -2) and (5, -7)
we have $\frac{9}{a^2}-\frac{4}{b^2}$ = 1     …..(1)

or, 45x² – 16y² = 341.

(j) The transverse axis parallel to the y-axis, the hyperbola passes through the points ($\frac{11}{3}$, 0), (1, 2) and its centre is the intersection of the lines. x + y – 6 = 0, 4x – y + 1 = 0.
Solution:
Solving x + y – 6 = 0
$\frac{4 x-y+1=0}{5 x=5 \quad \text { or, } x=0}$
∴ y = 6 – x = 6 – 1 = 5
∴ Centre of the hyperbola is at (1, 5).

Question 6.
Reducing to standard form, obtain the coordinates of the vertex, focus, endpoints of the latus rectum, the length of latus return, the equation of axis and directrix of the following parabolas:
(a) y² – 4x + 4y – 1 = 0
Solution:
y² – 4x + 4y – 1 = 0
or, y² + 4y = 4x + 1
or, y² + 4y + 4 = 4x + 5
or, (y + 2)² = 4(x + $\frac{5}{4}$)
or, (y + 2)² = 4 × 1 × (x + $\frac{5}{4}$) standard from

Length of the latus rectum
= 4a = 4 × 1 = 4
Eqn. of the axis is y = k or, y = -2
Eqn. of the directrix is x = h – a
or, $\frac{-5}{4}$ – 1 $\frac{-9}{4}$
or, 4x + 9 = 0

(b) 2x² – 4y + 6x – 3 = 0
Solution:
2x² – 4y + 6x – 3 = 0
or, 2x² + 6x = 4y + 3
or, 2(x² + 3x) = 4(y + $\frac{3}{4}$)

(c) x² + x + y + 1 = 0
Solution:
x² + x + y + 1 = 0

and (0, -1)
Length of the latus rectum = 4a = 1.
Eqn. of the axis x = h or x = –$\frac{1}{2}$
or, 2x + 1 = 0
Eqn. of the directrix is y = k – a or, y = $\frac{-3}{4}+\frac{1}{4}$
or, y = –$\frac{1}{2}$ or, 2y + 1 = 0

(d) y² + 14y – 3x + 1 = 0
Solution:
y² + 14y – 3x + 1 = 0
or, y² + 14y = 3x – 1
or, y² + 2.y.1 + 49 = 3x – 1 + 49
or, (y + 7)² = 3x + 48 = 3 (x + 16)
or, (y + 7)² = 4 × $\frac{3}{4}$ (x + 16)
This is the standard form

y = k or, y = -7
or, y + 7 = 0
Eqn of the directrix is x = h – a
or, x = -16 = $\frac{3}{4}$ = $\frac{-67}{4}$
or, 4x + 67 = 0

Question 7.
Reducing to standard form, obtain the coordinates of the centre, the foci, the vertices, the endpoints of the minor axis, the endpoints of latera recta, the equation of the directrices and the eccentricity of the following ellipses :
(a) 3x² + 4y² + 6x + 8y – 5 = 0
Solution:
3x² + 4y² + 6x + 8y – 5 = 0
or, 3(x² + 2y) + 4(y² + 2y) = 5
or, 3(x² + 2x + 1 – 1) + 4(y² + 2y + 1 – 1) = 5
or, 3(x + 1)² – 3 + 4(y + 1)² – 4 = 5
or, 3(x + 1)² + 4(y + 1)² = 12
or, $\frac{(x+1)^2}{4}+\frac{(y+1)^2}{3}$ = 1 is the standard form.
∴ h = -1, k = -1, a² = 4, b² = 3
∴ a = 2, b = √2, c = $\sqrt{a^2-b^2}$ = 1
Centre at (h, k) = (-1, -1), the vertices of (h±a, k)
= (-1±2, -1) = (1, -1)
and (-3, -1) foci at (h±c, k)
= (-1±1, -1) = (-2, -1) and (0, -1)
The endpoints of minor axis are (h, k±b) = (-1, -1±√3)
Endpoints of latera recta are (h + c, k±$\frac{b^2}{a}$) and (h – c, k±$\frac{b^2}{a}$)

(b) 4x² + 8y² + 4x + 24y – 13 = 0
Solution:
4x² + 8y² + 4x + 24y – 13 = 0
or, 4(x² + x) + 8(y² + 3y) = 13

(c) 2x² + 3y² – 12x + 24y + 60 = 0
Solution:
2x² + 3y² – 12x + 24y + 60 = 0
or, 2(x² – 6x) + 3(y² + 8y) = -60
or, 2(x² – 2.x.3 + 9 – 9) + 3(y² + 2.y.4 + 16 – 16 = -60
or, 2(x – 3)² – 18 + 3 (y + 4)² – 48 = -60
or, 2 (x – 3)² + 3 (y + 4)² = 6
or, $\frac{(x-3)^2}{3}+\frac{(y+4)^2}{2}$ = 1 is the standard form.
∴ h = 3, k = -4, a² = 3, b² = 2
∴ a = √3, b = √2, c = $\sqrt{a^2-b^2}$ = 1
Centre at (h, k) = (3, -4),
Foci at (h±c, k)
= (3±1, -4) = (4, -4) and (2, -4)
Vertices at (h±a, k) = (3±√3, -4)
The endpoints of the minor axis are (h, k±b) = (3, -4±√2)

(d) 9x² + 4y² + 36x – 8y + 4 = 0
Solution:
9x² + 4y² + 36x – 8y + 4 = 0
or, 9(x² + 4x) + 4(y² – 2y) = -4
or, 9(x² + 4x + 4 – 4) + 4(y² – 2y + 1 – 1) = -4
or, 9(x + 2)² – 36 + 4(y – 1)² – 4 = -4
or, 9(x + 2)² + 4(y – 1)² = 36
or, $\frac{(x+2)^2}{4}+\frac{(y-1)^2}{9}$ = 1, This is the standard form
∴ h = -2, k = 1, a² = 9, b² = 4
∴ a = 3, b = 2, c = $\sqrt{a^2-b^2}$ = √5
∴ Centre at (h, k) = (-2, 1)
Foci at (h, k±c) = (-2, 1±√5)
Vertices at (h, k±a)
= (-2, 1±3)
Endpoints of minor axis are (h±b, k) = (-2±2, 1)
The endpoints of latera recta are (h ± $\frac{b^2}{a}$, k ± c) = (-2 ± $\frac{4}{3}$, 1 ± √5)
End of directrices arty = k ± $\frac{a^2}{c}$
or y = 1 ± $\frac{9}{\sqrt{5}}$
Eccentricity is $\frac{c}{a}$ = $\frac{\sqrt{5}}{3}$.

Question 8.
Reducing to standard form, obtain the coordinates of the centre, the vertices, the foci, the endpoints of the conjugate axis, the endpoints of latera recta, the equation of directrices and the eccentricity of the following hyperbola :
(a) x² – 2y² – 6x – 4y + 5 = 0
Solution:
x² – 2y² – 6x – 4y + 5 = 0
or, x² – 6x – 2(y² + 2y) =-5
or, x² – 2.x.3 + 9 – 9 – 2(y² + 2y + 1 – 1) = -5
or, (x – 3)² – 2(y + 1)² + 2 = 4
or, (x – 3)² – 2(y + 1)² = 2
or, $\frac{(x-3)^2}{2}-\frac{(y+1)^2}{1}$ = 1 the standard form
∴ h = 3, k = -1, a² = 2, b² = 1
∴ a = √2 , b = 1, c $\sqrt{a^2+b^2}$ = √3
∴Centre at (h, k) = (3, -1)
Vertices (h±a,  k) = (3±√2, -1)
Foci at (h±c, k) = (3±√3 , -1)
The endpoints of a conjugate axis are (h, k±b) = (3, -1±1)
Endpoints of latera recta are (h ± c, k ± $\frac{b^2}{a}$) and (3±√3, -1 ± $\frac{1}{\sqrt{2}}$)
Eqn. of directrices are x = h ± $\frac{a^2}{c}$ = 3 ± $\frac{2}{\sqrt{3}}$ or, x = 3 ± $\frac{2}{\sqrt{3}}$
Eccentricity = $\frac{c}{a}=\frac{\sqrt{3}}{\sqrt{2}}$

(b) 9y² – 4x² – 90y + 189 = 0
Solution:
9y² – 4x² – 90y + 189 = 0
or, 9(y² – 10y) – 4x² = -189 .
or, 9(y² – 2.y.5 + 25 – 25) – 4x² = -189
or, 9(y – 5)² – 225 – 4x² = -189
or, 9(y – 5)² – 4x² = 225 – 189 = 36
or, $\frac{(y-5)^2}{4}-\frac{(x)^2}{9}$ = 1 the standard form
h = 0, k = 5, a² = 4, b² = 9
a = 2 , b = 3, c $\sqrt{a^2+b^2}$ = √13
Centre at (h, k) = (0, 5)
Vertices (h, k±a) = (0, 5±2)
Foci at (h , k±c) = (0, 5±√13)
The endpoints of a conjugate axis are (h±a, k) = (0±2, 5) = (±2, 5)
Endpoints of latera recta are (h ± $\frac{b^2}{a}$, k ± c) = (0 ± $\frac{9}{2}$, 5 ± √13)
Eqn. of directrices are y = k ± $\frac{a^2}{c}$ = 5 ± $\frac{4}{\sqrt{13}}$
Eccentricity = $\frac{c}{a}$ = $\frac{\sqrt{13}}{2}$

(c) 49x² – 4y² – 98x + 48y – 291 = 0
Solution:
49x² – 4y² – 98x + 48y – 291 = 0
or, 49(x² – 2x) – 4(y² – 12y)= 291
or, 49(x² – 2x + 1 – 1) – 4(y² – 2.y.6 + 36 – 36) = 291
or, 49(x – 1 )² – 49 – 4(y – 6)² + 144 = 291
or, 49(x – 1)² – 4(y – 6)²
= 291 – 144 + 49 = 196
or, $\frac{(x-1)^2}{4}$ – $\frac{(y-6)^2}{49}$ = 1
h= 1, k = 6, a² = 4, b² = 49
a = 2, b = 7, c = $\sqrt{a^2+b^2}$ = √53
Centre at (h, k) = (1, 6)
Vertices at (h±a, k) = (1±2, 6)
Foci at (h±c, k) = (1±√53, 6)
Endpoints of latera recta are (h ± c, k ± $\frac{b^2}{a}$) = (1 ±√53, 6 ± $\frac{49}{2}$)
Eccentricity = $\frac{c}{a}=\frac{\sqrt{53}}{2}$
Endpoints of conjugate axis are (h, k±b) = (1, 6±7) = (1, -1) and (1, 13)

(d) 3x² – 2y² – 4y – 26 = 0
Solution:
3x² – 2y² – 4y – 26 = 0
or, 3x² – 2(y² + 2y) = 26
or, 3x² – 2(y² + 2y + 1 – 1) = 26
or, 3x² – 2(y + 1)² + 2 = 26
or, 3x² – 2(y + 1)² = 24
or, $\frac{x^2}{8}-\frac{(y+1)^2}{12}$ = 1 Standard form
h= 0, k = -1, a² = 8, b² = 12
a = 2√2, b = 2√3, c = $\sqrt{a^2+b^2}$ = √20 = 2√3
Centre at (h, k) = (0, -1)
Vertices at (h±a, k) = (±2√2, -1)
Foci at (h±c, k) = (±2√5, -1)

Question 9.
Prove that the equation of the parabola whose vertex and focus are at distances α and β from the origin on the x-axis respectively is y² = 4(β – α)(x – α)
Solution:

Question 10.
Find the locus of the point of trisection of a double ordinate of the parabola y² = 4ax.

Question 11.
(a) Prove that a double ordinate of the parabola y² = 4ax of length 8a subtends a right angle at its vertex.
Solution:

(b) Find the angle which a double ordinate of length 2a subtends at its vertex and focus.
Solution:

Question 12.
(a) Obtain the equations of the tangent and normal of the parabola y² = 4ax at a point where the ordinate is equal to three times the abscissa.
Solution:
Putting y = 3x in the parabola equation.
we have y² = 4ax
or, 9x² = 4ax
or, x = $\frac{4 a}{9}$
∴ y = 3x = $\frac{4 a}{3}$
∴ The point of contact is ($\frac{4 a}{9}$, $\frac{4 a}{3}$)

(b) Find the equation of tangents and normals to the parabola y² = 4ax at the ends of its latus rectum.
Solution:
The endpoints of the latus rectum of the parabola y² = 4ax are (a, 2a) and (a, -2a).
∴ Eqn. of the tangent at (a, 2a) is yy₁ = 2a (x + x₁)
or, 2ay = 2a (x + a)
or, y = x + a
or, x – y + a = 0
Eqn. of the tangent at (a, -2a) is -2ay = 2a(x + a).
or, -y = x + a
or, x + y + a = 0
Eqn. of the normal at (a, 2a) is y – y₁ = $\frac{-y_1}{2 a}$ (x – x₁)
or, y + 2a = $\frac{+2 a}{2 a}$ (x – a)
or, y – 2a = -x + a
or, x – y – 3a = 0

(c) Find the equations of tangents and normals to the parabola y² = 4ax at the points, where it is cut by the line y = 3x – a.
Solution:
Solving y = 3x – a, y² = 4ax
∴ (3x – a)² = 4ax
or, 9x² + a² – 6ax – 4ax = 0
or, 9x² – 10ax + a² = 0
or, 9x² – 9ax – ax + a² = 0
or, 9x(x – a) – a(x – a) = 0
or, (x – a) (9x – a) = 0
∴ x = a, x = $\frac{a}{9}$
∴ y = 3x – a = 3a – a = 2a

(d) Show that the tangent to the parabola y² = 4ax at the point (a’, b’) is perpendicular to the tangent at the point ($\frac{a^2}{a^{\prime}}, \frac{-4 a^2}{b^{\prime}}$).
Solution:

∴ The product of their slopes = -1
∴ The two tangents are perpendicular to each other.

(e) A tangent to the parabola y² = 8x makes an angle of 45° with the line 3x – y + 5 = 0. Find the equation and the point of contact.
Solution:

or, 4a + 2y + 2 = 0 or, 2x + y + 1 = 0

(f) Prove that, for all values of k, the line y = k(x + a) + $\frac{a}{k}$ is a tangent to the parabola y² = 4a(x + a).
Solution:

(g) Obtain the condition that the line lx + my + n = 0 will touch the parabola y² = 4ax
Solution:
We have lx + my + n = 0
or, y = $\frac{-l x-n}{m}$
Now putting the value of y in the parabola Eqn., we have y² = 4ax.
or, $\frac{-l x-n}{m}$2 = 4ax
or, l²x² + n² + 2lnx = 4am²x
or, l²x² + x(2ln – 4am²) + n² = 0
a’ = l², b’ = 2ln – 4am², c’ = n²
As the line is a tangent to the parabola,
we have b^’2 – 4a’c’ = 0.
or, (2ln – 4am²)² – 4l²x² = 0
or, 4l²n² + 16a²m⁴ – 16alnm² – 4l²n² = 0
or, 16am²(am² – ln) = 0
or, am² – ln = 0
or, am² = ln

(h) Prove that the line 4x – 2y – 1 = 0 touches the parabola whose focus is at (0, 0) and the directrix is the line y = 2x – 1.
Solution:
According to the definition,

Question 13.
(a) If (-2, 0) and (2, 0) are the two vertices of a triangle with a perimeter of 16, then obtain the locus of the third vertex.
Solution:

or, 8x² + 9y² – 288 = 0
∴ This is the locus of the 3rd vertex of the triangle.

(b) A point in a plane is such that the sum of its distances from point (2, 2) and (6, 2) is 12. Find the locus of the point.
Solution:

or, x² + 484 – 44x
= 9x² + 9y² – 108x – 36y + 360
or, 8x² + 9y² – 64x – 36y – 124 – 0 which is the locus of the point P.

(c) Obtain the equation of the ellipse which has its centre at the origin, a focus at (2, 0) and the corresponding directrix is the line 2x = 7. Calculate the length of the latus rectum.
Solution:
Centre at (0, 0), focus at (2, 0),
directrix is 2x = 7

(d) Find the equation of the ellipse which has its centre at (-1, 4), eccentricity $\frac{1}{\sqrt{3}}$ and the ellipse passes through the point (3, 2).
Solution:

Question 14.
(a) Find the equation of the tangent and normal to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}$ = 1 at the point ($\frac{8}{3}$, √5)
Solution:

(b) Find the equation of the tangent and normals to the ellipse 2x² + 3y² = 6 at the endpoints of the latera recta.
Solution:
2x² + 3y² = 6

(c) Prove that the line y = 2x + 5 is a tangent to the ellipse 9x² + 4y² = 36 and find the point of contact.
Solution:
Putting y = 2x + 5 in the ellipse eqn.
we have 9x² + 4y² = 36
or, 9x² + 4(2x + 5)² = 36
or, 9x² + 4(4x² + 25 + 20x) = 36
or, 9x² + 16x² + 100 + 80x – 36 = 0
or, 25x² + 80x + 64 = 0     …..(1)
∴ b² – 4ac = (80)² – 4 × 25 × 64
= 6400 – 6400 = 0
∴ The line y = 2x + 5
touches the ellipse 9x² + 4y² = 36.
Now from eqn. (1),
we have (5x + 8)² = 0

(d) Find the equation of the tangent to the ellipse 4x² + 5y² = 20 which are parallel to the line x – y = 2
Solution:
4x² + 5y² = 20

(e) Find the equation of the tangent to the ellipse 4x² + 9y² = 1, which are perpendicular to 2ax + y – 1 = 0.
Solution:
4x² + 9y² = 1

(f) Prove that the line x cos α + y sin α = p touches the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ = 1, if p² = a² cos² α + b² sin² α.
Solution:

(g) Prove that the product of the distances of the foci from any tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ = 1 is equal to b²
Solution:
The tangent at (x₁, y₁) to the ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text { is } \frac{x x_1}{a^2}+\frac{y y_1}{b^2}=1$

Question 15.
(a) Find the equation of the hyperbola which has it foci at (0, 0) and (0, 4) and which passes through the point (12, 9).
Solution:
Foci at (0, 0) and (0, 4)
∴ Centre at (0, 2)

(b) Find the equation of the hyperbola with foci at (±3, 0) and directrices x = ± 2.
Solution:
Foci at (±3, 0) and directrices x = ±2.
∴ c = 3, $\frac{a^2}{c}$ = 2, a² = 2c = 6
∴ b² = c² – a² = 9 – 6 = 3
∴ Eqn. of the ellipse $\frac{x^2}{a^2}-\frac{y^2}{b^2}$ = 1
or, $\frac{x^2}{6}-\frac{y^2}{3}$ = 1 or, x² – 2y² = 6

(c) Find the foci and latus rectum of the hyperbola whose transverse and conjugate axes are 6 and 4 and whose centre is at (0, 0).
Solution:
The transverse axis is 6, the conjugate axis is 4, the centre is at (0, 0).
∴ 2a = 6, 2b = 4
∴ a = 3, b = 2
∴ c = $\sqrt{a^2+b^2}=\sqrt{9+4}=\sqrt{13}$
The foci are (±c, 0) = (±√13 , 0)
∴ Length of the latus rectum = $\frac{2 b^2}{a}=\frac{2 \times 4}{3}=\frac{8}{3}$

Question 16.
(a) Find the equation of tangent and normal to the hyperbola x² – 6y² = 3 at the point (-3, -1).
Solution:

(b) Find the equation of the tangent to the hyperbola 4x² – 11y² = 1 which is parallel to the straight line 20x – 33y = 13.
Solution:

(c) Find the equation of tangent to hyperbola 9x² – 16y² = 144 which are perpendicular to the line 2x + 3y = 4.
Solution:
9x² – 16y² = 144
or, 18x – 32y$\frac{d y}{d x}$ = 0

(d) Prove that the line x + y + 2 = 0 touches the hyperbola 3x² – 5y² = 30 and find the point of contact . Find also the equation of normal at the point.
Solution:
x + y + 2 = 0
or, y = -x – 2
putting y = -x – 2 in the hyperbola equation we have 3x² – 5y² = 30
or, 3x² – 5 (-x – 2)² = 30
or, 3x² – 5 (x² + 4 + 4x) = 30
or, 3x² – 5x² – 20 – 20x = 30
or, 2x² + 20x + 50 = 30
or, x² + 10x + 25 = 0
b² – 4ac = 100 – 4 × 25 = 0
The line x + y + 2 = 0 touches the hyperbola 3x² – 5y² = 30
Now solving x² + 10x + 25 = 0
we have x = – 5.
y = – x – 2 = 5 – 2 = 3
The point of contact is (-5, 3).
Eqn. of the normal is
y – y₁ = m (x – x₁)
or, y – 3 = 1 (x + 5)
or, x – y + 8 = 0

(e) Prove that the line x cos α + y sin α = p touches the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}$ = 1, if p² = a² cos² α – b² sin² α.
Solution:

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