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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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Odisha State Board CHSE Odisha Class 12 Foundations of Education Solutions Unit 1 Contribution of Educators Long Answer Questions.

CHSE Odisha 12th Class Foundations of Education Unit 1 Contribution of Educators Long Answer Questions

Long Answer Questions With Answers

Question 1:
Discuss the life philosophy of Gandhi.
Answer:
Gandhi enunciated integral philosophy of fife. He was a naturalist, an idealist,
individualist in one. His ‘Experiment on Truth’ was the outcome of his experience and prominent philosophical activities are his concept of God, truth, doctrines of morality, non¬violence, satyagraha, labor, equality, citizenship, brotherhood of man His life was concerned with:

• His concept of truth.
• His concept of karma.
• His concept of non-violence.
• His concept of satyagraha.
• His idea of centralization.
• His idea of machine.
• His concept of the village.
• His news of morality.

1) His concept of truth – Gandhiji believed in truth to be the ultimate reality and God can be realized through truth. God is truth, and truth is God. He said truth is manifested both externally and it is expressed through the voice of God. He was the pioneer of truth and non¬violence and conquered the brutal force. One can realise god through truth.

2) His concept of Karma – In Gita, there is view on life and karma. Gandhiji was deeply influenced by Gita for a religious dedication to the service of man. Service of humanity is God. Religion is not a part from human activity. Action takes its origin from Brahma and Brahma is present in all kinds of sacrifice of service. To Gandhi, society and social service are an integral part of life and they are sacred activities.

3) His concept of non-violence – Non-violence of Gandhi was equivalent to love. His concept of non-violence retained deeply in Indian spirituality. The concept ofAhimsa or non-violence of Gandhi was a means and that an end. Man is the end of his material, mental and moral well-being and growth.

4) His concept of Satyagraha – Gandhiji’s concept of Satyagraha was dynamic aspect of non-violence and a tool that created a human context for social conflict. Truth is the end and non-violence is the means to human activities.The term ‘ Satyagraha’ is derived from the Gujarati word ‘agrapha’ which means firmness. For Gandhiji satyagraha is dynamic quality of non-violence. Satyagraha for Gandhiji’s way a truth force for acting socially and humanely.

5) His idea of decentralization – Gandhiji was against concentration of power and individualism of capitalism. He wished a kind of society where economic and social structure of decentralization on the basis of industry and agriculture. This is the productivity aim of education.

6) His idea of machine – Gandhiji was not against the machine, but he did not want it to become the master of machine. He opposed strongly machines because it created unemployment and exploitation of the poor workers by capitalists and too much dependence on man on machine. So he suggested to limit the manufacture of machine and emphasized on cottage industry and handicrafts.

7) His concept of the village – To Gandhiji, village is a small group of people, consisting a unit of society. So the village should be self-governing. He considered that it should be self¬sufficient in the matter of its vital necessities of life like food, clothing and shelter. Secondly his village was not an agricultural community, there should be a balance between agriculture and the village industries. He desired to create agro-industrial community.

8) Gandhiji’s gramraj – Village self government was the opinion of Gandhi. His idea of Gram raj or village self government means it is a complete republic independent or its neighbours are independent with other necessaries. Thus, for every villages the first concern will be no caste and hopes to abandon untouchability and create a class less society.

9) His views on morality – To Gandhiji, the end of all knowledge is the development of morality. The society and individual progress through morality, purity in thought, speech and deeds. So a social foundation of truth and purity should be established through education. To him moral education is to be imparted in schools. Morality is the best virtue of humanity. By participation in games and sports discipline in thought and action is maintained.

Question 2:
Write a note on the Educational Philosophy of Gandhi.
Answer:
As a socio-political, reformer, educationist, idealist, naturalist, social leader and practical philosopher, Gandhiji father of the Nation, the apostle ofpeace and non-violence, the champion of Freedom movement led a scheme of education of India known as ‘Basic Education’. In another way it is known as “Nayee Talim”. His educational philosophy is the potent force for social reconstruction. To him true education is “An all round drawing out of the best in child and man” – body, mind and spirit. The chieftenents ofhis educational philosophy are as follows:

• Education should be free and compulsory.
• Craft centred education.
• Self supporting education
• Emphasis on mother tongue.
• Child centredness.
• Emphasis of education on non-violence.

1) Education should be free and compulsory : Gandhi advocated free and compulsory education for 7 to 14 years and wanted to combine primary education with secondary education called it English Less Matriculation Courses. To him democracy will become successful when education will be free and compulsory. It will develop love for creative work. As India is a poor land and 60 % of population are below the poverty line so education should be free and compulsory for them.

2) Craft centred education : He believed in the principle of learning by doing’ of John Dewey. The basic education aimed at providing education on crafts. He introduced basic crafts like spinning and weaving, carpentry and agriculture. Introduction of crafts evoke the spirit of love for work and teach them the dignity of labour. To him the whole educative process be imparted through handicraft. They will leam the motto “work is worship”. It will develop the dignity of labour.

3) Self-supporting education: Gandhiji knew that India is a poor state and it cannot afford to educate the millions. So Gandhi suggested education to be self supporting. The concept of “Karma Yoga” and dignity of labour will help in the intellectual development. So the child should pay labour partly by binding a gap between education and life drawing upon the cultural, social and vocational potentialities. It is a measure of social reconstruction.

4) Emphasis on mother tongue: Gandhi emphasized mother tongue as the medium of instruction. To him the English system of education hinders understanding and clarity of ideas. By mother tongue, the children can express their views clearly and understand others and this would build sound foundation of education.

5) Child centredness: Child-centredness is an important feature of Basic Education which means the children should be taught to the needs, interests, and capacities of children. So the curriculum and the method of teaching are to be developed to the capacities of learners. Different crafts and subjects to be included in the curriculum to meet individual differences.

6) Education based on non-violence: A unique feature of Gandhiji’s education philosophy was the application of the law of non-violence. He wanted to build a classless society and elimination of exploitation. By the scheme of non-violence and peace, he conquered the heart of brutal forces. So his education of philosophy is based on non-violence. He wanted to create a generation that should believe in non-violence.

Question 3:
What should be the aims, curriculum and methods of teaching of Basic
Education?
Answer:
M. K. Gandhi is called an idealist, a realist, a spiritual person in one. He advocated his philosophy of education and put stress on religious education. The main aims of this philosophy are as follows :

• The utilitarian aim.
• The cultural aim.
• Harmonious development aim.
• Complete living aim.
• Character building aim.

1. The utilitarian aims : To fulfill this aim, the basic needs of human life like food, clothing and shelter and self-supporting to be imparted. The self-supporting aspects aimed at self-sufficient and education to meet one’s expenses. It is otherwise known as ‘Bread and Butter Aims ’ of education.

2. The cultural aims: Culture is essential to refine one’s personality.One should have the qualities of mind which should be reflected in one’s own conduct. Such aims helps in the transmission of culture.

3. Harmonious development aim: To Gandhi, education means “An all round draw ing out of the best in child and man, with body, mind and spirit”. To Gandhi, harmonious development means – innate and acquired powers development from social to intellectual. Basic education helps with all round development of personality of the individuals.

4. Complete living aims: To Gandhi, life is very complex. So he formulated a scheme of education which would fit the children to later life and a child to be prepared for complete living. He should learn how to support his living, social adjustment, occupation and self¬reliance.

5. Character building aims: Character building was the chief aim of basic education. To him character is the expression ofthe w’hole personality including the ethical and spiritual aspects. One should subordinate his own interest to the greater interest ofthe society, cooperate his fellow being about a new social order. Such a person is really a man of character. In his aims of education, Gandhi emphasis the building of character.

The individual character will influence the national character.
i)Curriculum: Gandhi emphasized on child centred curriculum. Education should be related to the environment of the child. He opposed English as the medium of instruction and mother tongue as the medium of instruction up to the level of matriculation. To him English hinders the clarity of thought and put a check on self-expression.My mother tongue, one can express his view correctly and understand others. It also helps in the development of nationalism and patriotism
He introduced craft as a part of curriculum and the whole process of education should be imparted through some handicrafts.
ii) Craft: Education should be given through the medium of some craft on productive work. So different handicrafts like weaving, spinning, carpentry, earthen pot building etc.
iii) Activity centred: The teaching of various subjects should be emphasized. Teaching of craft will be the center point and the teaching of all subjects should be related to craft.The co-related teaching methods are to be employed.
iv) Mother tongue: Mother tongue should be medium of instruction. It will help the children for clear expression and clear understanding, develop patriotism and nationalism.
v) Religious and moral education: For the development of personality, character religions and moral education is to be given. All should respect to all religions. Ethics of all religions is to be taught as a part of education. To Gandhi there is need of moral leaders in free India and building modem India.

Question 4:
Discuss the essential features of Basic Education. Explain the gift of Basic education to education.
Answer:
The essential features of Basic education include:

• Free and compulsory education.
• Purposeful activity-centred education.
• Emphasis on mother tongue.
• Self-supporting education.
• Primary importance for village.
• New cooperative regime.
• Dignity of labor.
• Co-operative work.
• Integrated teaching.
• Educates body, mind, and spirit.

1. Free and compulsory education: Basic Education implies a free and compulsory education for all children between the age of 7 to 14 years. It will reduce the disparity among the children.

2. Purposeful activity centred education: Basic Education centers round some purposeful activity or useful and productive craft training which supports self supporting aims of education.

3. Mother tongue: The medium of instruction should be mother tongue of the child. By this the child can express his views fluently and understand others’ views. It also inculcates the spirit of nationalism and patriotism.

4. Self-supporting: Basic Education is aimed at self-supporting. It followed the principles of learning by doing. They earn from their craft work as well, so as to cover their expenses. Thus craft has both educational and economic value.

5. Primary importance for the village: Basic education was primarily devised for the village. Gandhiji say, “In discussing the question of primary education, I have neither to deliberately confirmed myself to the village, as it is to villages that the bulk of Indians population resides’. To tackle successfully the question of village is to solve the problem for the cities also.

6. New cooperative regime: Basic education aims at bringing about a new cooperative regime in place of the present in-human regime based on exploitation and violent forces.

7. Dignity of labor: Basic education curriculum inculcates the virtue of dignity of labor, a keen sense of discipline, and a great sense of responsibility. Labour-centred education reduces disparity among pupils. It helps with self-employment.

8. Cooperative work: In the scheme of Basic education both the teachers and pupils work for community development and social progress.

9. Integrated teaching: In Basic education, all the subjects are taught in an integrated way. All the instructions are co-related. It seeks to develop the child as a whole. The child is taught with co-related teaching methods of all subjects like math, general science, social sciences, language, drawing and paintings etc. It helps the harmonious development of the child.

10. Educates body, mind, and spirit: Basic education is meant to educate the body, mind and spirit with an unique relation among them. It seeks to develop the child as a whole.

Question 5:
Discuss the causes of downfall of Basic education.
Answer:
Inspite of the merits of Basic education, the scheme was criticised by the richer classes, educationists and suffer from a number of limitations. After the death of Gandhiji the scheme was given a death-blow. The most important reasons of the failure of Basic Education are as follows :

• The Unclear Concept.
• Emphasis on Idealistic view.
• Emphasis on Economic aspects.
• Compact Area approach. ‘
• Absence of Text Books.
• Lack of Trained Skilled Teachers and Textbooks.
• Faculty timetable.
• Costliness of Basic Education.
• Opposition of Traditionalists.
• Matriculation of Minus English.
• Lack of Research.

1) The Unclear Concept: In that period most of the educationists the common people, education administrators as well were not clear about the concept of Basic Education. They were confused of craft education and mechanical education. The common people could understand nothing. In that period, there was no provision of propagation and no mass media system to highlight the program. So the concept of the scheme could not touch the common people and got failure.

2. Emphasis on Idealistic Approach: The scheme laid stress on some idealistic practices like manual work. Such scheme of education was not accepted by the British and the intellectuals because they educated people do not appreciate that their children could do any manual labour. They sent their children to public schools and English medium schools. So the confusion is created.

3. Emphasis on Economic Aspects: In Basic Education, too much emphasis was given on economic aspect. The craft centredness was not accepted by the intellectuals as well as educationists. The productivity activity of self supporting aspect exploits the child labour and craft work puts emphasis on economic aspect. The students become money minded. The guardians felt that their children were turned into laborers, so they opposed the basic ideals of craft.

4. Compact Area Approach: Basic schools were opened in some specific areas, especially in rural areas but not in town areas. The scheme was worked out in a limited area on an experimental basis. So compact area approach was a major cause of the failure of such education system.

5. Absence of Text Books: Basic scheme of education emphasized on craft education. It was a mechanical education. Text books were riot emphasized and no text book was prepared as there were no writers to prepare textbooks on craft training. It was another major causes of the downfall of Basic Education.

6. Lack of Qualified teachers: It was a new type of education like craft. The traditional teachers failed to understand the new pattern of education and the curriculum prepared by Dr. Jakir Hussain. Qualified trained teachers were not available as it was mechanical teaching. There was no provision of teachers’ training. So lack of qualified trained teachers, the Basic Scheme of Education led its downfall.

7. Faculty Time Table: In Basic Education much more was invested or devoted for craft work and other subjects were neglected. In a Basic School 2/3rd. of the time was utilized in craft work. On the timetable, academic subjects were taught after craft work. So academic subjects were neglected. Agricultural students become tired to their academic work. So the faulty timetable was an obstacle of spreading Basic Education.

8. Costliness of Basic Education: As Basic Education needed equipment, so more initial cost was required to purchase craft equipment. There were no funds to meet such expenses and the Govt, could not afford it so it led its downfall.

9. Opposition of Traditionalists: The dream of Gandhi to build a classless society was strongly opposed by the higher class people. So the traditionalists and conservatives were afraid that the new social order would upset their position and so they strongly opposed the system of education.

10. Matriculation Minus English Course: Gandhiji emphasized that English should not be taught to the students in matriculation stages. But the richer classes opposed this and did not prefer to admit their children in such a school. So the strength of the school reduced day by day.

11. Lack of Research: No research activities were encouraged and no research centers were set up. So lack of research, newer methods of teaching, and techniques the Basic Scheme of Education led its downfall.

Question 6:
Discuss the aims of education of Satyabadi School.
Or Explain Gopabandhu’s Educational thoughts.
Or Discuss the contribution of Gopabandhu to the present education.
Answer:
Gopabandhu believed in universal education. The organizers of national Education League opined that everyone has the right of being educated, just as the rays of sun and moon. One shared equally by the people. Pandit Nilakantha Das emphasized classless society education. The people of Odisha are poor and they cannot afford for the education of their children. So Gopabandhu proposed an education that should be ideal, forced, and inexpensive. In 1909 he set up a School at Satyabadi named as Open Air Schooling based on the ideals of Gopabandhu.

The aims, main tenents and educational thoughts of Gopabandhu are discussed below.
1. In-expensive Education: Gopbandhu was hilly aware of that India being a poor country and Odisha a poor province it cannot pay for education of the entire population. So the cost of education to be reduced. He experimented with the groove school on the lines of ancient Gurukul System with no school building and tuition fee payable by the students should be minimum. They should lead a simple and austre life in the school hostel.

2. Idealistic Education – Gopabandhu believed that it is not a costly school building but the idealistic and dedicated teachers who can make a good school were appointed. It is remarked that a school does not consist of only buildings, chairs, and tables, there must be well-educated, sincere and idealistic teachers. No education worthwhile can be imparted without good and efficient teachers.

3. Practical Education – Gopabandhu was cirtical about the prevailing system of education which does not equip the individual to meet the requirement of life. He wanted education to be practical which should make the students economically independent.
To him the present system of education failed to prepare our students for the struggle of life. So they should be taught crafts to maintain their livelihood and they should be taught physical exercise, industry and agriculture etc.

4. Religious and Moral Education: Gopabandhu believed in the all round development of the personality of an individual through education. The students must be taught craft skills in order to capable them to earn their living. There is need of civilized and cultured individuals. So there is need of religious and moral instructions for morality.

5. Social Service and National Integration: Gopabandhu did not stress on individuality. To him, individual is a part of society. So education would enable the individuate to perform the social functions as efficiently as successfully as possible. For him education is a preparation for a life of dedicated social service in the society or nation.Through this education, he wanted to bring about emotional and national integration. In the community dinner in his hostel, every day, high or low sit together and take meals.

6. Women’s Education – Gopabandhu was revolutionary in nature. He knew that in the backward and traditional society of Odisha,a great deal of courage is essential to advocate women’s education. To him, women are the wealth of the family as well as the wealth of the nation. They are the Goddess of family life. If women are educated, they will take care of their children.

7. Mother tongue as the Medium of Instruction – Gopabandhu emphasized on the importance of the mother tongue in the education of a child. It is essential to develop his mental powers. It is also essential to realize the intellectual cultural and spiritual aims of education So education should be imparted through the mother tongue. By mother tongue, the creative powers is to be developed.

Question 7:
Discuss the main features of Groove School ?
Answer:
The main basic principles in which the groove school grew up includes the following:

• Open Air School.
• Free Education.
• Ideal Teachers.
• All round development of Personality.
• Teaching Craft Skills.
• New Methods of Teaching.
• Community dinner and Cultural programmes.
• Emphasis on co-curricular activities.

Importance on Mother’s tongue.

1. Open Air School: Gopabandhu knew well that Indians cannot afford and spend the amount of money in constructing school building and to reduce the expenditure in education without reducing the educational standards the attempts been made opening a groove school on the lines of old Gurukul system Thus, it is an Open Air School similar to Shantiniketan.

2. Free Education: Education was free and minimum fee was charged. The attempts to set up school, an old Gurukul system and the students will lead a simple austere life.

3. Ideal Teachers: Gopabandhu put high premium on the quality of teachers. Gopabandhu aimed at ideal education and teachers should be ideal sacrificing in nature. The teachers of Satyabadi school had voluntarily given up the pleasures of life for the purpose of rendering service to the community.

4. All round development of Personality: Satyabadi system of education aimed at the development of all the aspects of the personality. He emphasized character building inculcation of social values, virtues qualities of good citizenship, patriotism, brotherhood and spirituality. For the development of good qualities various co-curricular activities, debates, excursions, physical exercises, games and prayer assemblies are essential.

5. Teaching Craft Skills: The system of education strongly opposed the system of education bookish and aimed at imparting education of art and craft skills. Then the children will be able to prepare for the future, learn the struggle for life. Craft education may enable to earn their livelihood.

6. New Method of Teaching: Satyabadi Vana Vidyalaya was a residential school and the teachers and students stay together, in the hostels. The Headmaster, asks the teachers to submit their class notes to him for supervision. The teachers and the Headmasters, sit together and discuss mutually the course. The programs like debate, excursion etc. are continued.

7. Community dinner and Cultural program: The elected Secretary (from the boarders) manage the hostel The school and hostel for self-discipline democratic management and ideal student life. The students and teachers take dinner together. It teaches them community living.

8. Emphasis on co-curricular activities: To bring an all around development of personalities of the students various types of co-curricular activities were arranged in the Satyabadi School, such as (i) Literary activities (ii) Debates and games (iii) Prayer assembly (iv) Physical exercises (v) Literary magazines etc. Debates are aimed at oratoral abilities of the students. Physical exercises brings about the development of character, discipline and virtues. Prayer assembly develops the discipline and moral instructions.

9. Importance of Mother tongue: Mother tongue emphasized on the medium of instruction. It helps in understanding and developing nationalism, and patriotism and the child can express his views clearly and understand others.

Question 8:
Explain, the causes of downfall of Open Air Schooling or Satyabadi Vana Vidyalaya.
Answer:
Like the downfall of Basic Education of Gandhi, for some reasons Satyabadi Vana Vidyalaya led its downfall. The chief causes are:

• Lack of local support and unclear concept.
• Economic condition.
• Lack of help of the Govt.
• Participation of Gopabandhu in Indian National Movement.
• The death of Utkalmani.

1. Lack of local support and unclear concept – Satyabadi Vana Vidyalaya was surrounded by conservative villages. Conservative people and Brahmin society. Such conservatives did not appreciate the idealistic education system of Gopabandhu. Pandit Nilakantha Das wanted to build a classless society education which was a stroke to the conservatives.

The conservatives opposed this and they do not like to read their children with other backward-class children. They wanted to continue the superstitions.Lack of local support the enrolment of children is reduced.To oppose Gopabandhu’s system of education in 1912 on March 22, the conservatives fired the school and the main property of school library was destroyed. Such an incident shook the strength of the school.

2. Non-cooperation of the Govt.: It was not a traditional education system nor followed the British system of education and opposed the older beliefs, superstitions and conservatism. It is filled with idealism, patriotism, nationalism and social service programs.It aimed at creating leader for the state. So the British opposed it. No Govt, grant lias come and the application for Govt. Recognition was canceled. The school was transferred under Calcutta University in 1916 and then to Patna University in 1917, but it could not get the Govt. help.

3. Economic Condition : The third cause of the downfall of Open Air Schooling was the economic condition of the school. In 1921 Vana Vidyalaya was transferred into a National School. But the Govt, grant only two hundred rupees per year was refused and cut off very relationship from Patna University. In 1923, it turned to a National College and become autonomous. But it did not exist long. It suffered from the financial crisis. The teachers demanded to merge the college with the University. For this Gopabandhu remained away from the college. But the colleagues failed to continue the college for which it led its downfall.

4. Participation in Non-cooperation Movement: In 1920, Utkalamani joined in the Non-cooperation movement. He invested all his time and resources in the Freedom struggle. He found no time and resources to invest in groove school. Thus, the school led its downfall

5. The Death of Utkalamani: The day before the World famous Car Festival on June 17, 1928, Gopabandhu died in an immature death. The glorious chapter of Odia’s history’ came to an end. The next generation did not give any importance to the institution and just five years after the death of Utkalamani, the other Satyabadi Panchasakhas died one after another, the school was neglected and led its downfall.

Question 9:
Discuss the educational philosophy of Sri Aurobindo.
Answer:
Sri Aurobindo dedicated his whole life for society and education to provide contribution to travel towards divine perfection and to express the power, harmony, beauty and joy of self-realization.By education, he means that which will offer the tools whereby one can live for the divine for the country for oneself and for others. His principle of the philosophy of education, the awareness of man as a spiritual being.

Integral Education – Aurovindo’s integral education means integral growth of the individual’s personality. The function of education is to study the mind of individuals, people, nature, the universe. He put more emphasis to study the mind.

The human mind consists of four layers. They are :
1. Chitta – the storehouse of memory.
2. Manas – the sixth sense like – sight, sound, taste, smell, touch.
3. Budhi – thought or intellect.
4. Truth – Satya
Integral education, attempts to the integral development of physical being, vital beings, psychic being and mental being about a transformation of man into a spiritual being.

Aurobindo’s philosophy of Education are:
1. Education of the Physical being: To Aurovindo beauty is the ideal of physical life such as (i) To discipline and control the physical function (ii) For harmonious development of the body and physical movements (iii) Rectification of defects (iv) To awaken the body consciousness one has to undertake physical exercises. The factors of spiritual discipline, service, bhakti, and yoga are essential physical education. Restlessness is the important aspect to control the body and to achieve. Concentration physical education also controls sex drives or impulses. Emphasis is given on games and sports which an renew physical and higher forms of energy, develop tolerance, self control, friendliness etc.

2. Education of the Vital Being : It helps in the development of character. Vital education emphasizes on the vital being of man by which the man be able to understand the inner and outer world. It develops self observation. Vital being means utilization of the sense organs which help us to receive knowledge. The senses like a sight hearing, small, touch, taste and mind should be trained. Vital education also aimed at the aesthetic personality development.

3. Education of the Mental Being: Mental being education emphasizes on mental science and concentration. The mother says to silence the mind, one has to take the help of classical yoga. By yoga, one can acquire mastery of the mind. So knowledge of the education of mental being which helps is the gradual liberation from ignorance.
Mental education has three-fold functions:
i)To gather old knowledge.
ii) To discover new knowledge.
iii) To develop the capacity use and apply the knowledge acquired.
Through the application of knowledge, the pupil develops cognition, ideas, intelligence and mental perception. For this man becomes the source of knowledge.

4. Education for Psyche Being: Psyche being is the psychological centre of mind. The function of education is to enable man to become conscious of the psychological centre which is key to an integral personality. Psyche education is to enable an individual to see his soul grow in freedom. It supports the vital, physical, and mental being. When an individual develops psyche consciousness, he understands life and himself.

To him the psyche being is a spiritual personality put forward by the soul in its evaluation. The business of education is to develop the capacity of psychological being towards a realization of potentialities. Psyche being is possible through yoga or tapasya of love. As a result of this yoga one can attain liberation from suffering.As a result four-fold approval like the physical, the mental, the psyche and love in the ii idi v ideal student, the man gets liberation from the material world, desires, ignorance and suffering.
A tota 1 spiritual education is the goal of education and spiritual transformation of man is the goal o f integral education.

Question 10:
Discuss the philosophical thoughts of Rousseau.
Answer:
Rousseau’s philosophical thoughts are related to Naturalism and Negative Education.

1) His Naturalism
a) God has made all humans good but when they come in contact with society they become spoiled. In order to change them again into good, we should bring them back to nature.
b) In the beginning of human civilization, man was happy and good, but now he is unhappy. If he goes back to nature, he will again be happy and good. Thus, a child be educated and developed according to his natural tendencies. Society and school has no role to play in the process.
c) He did not like old values and traditions of the society. According to him, social relations can be brought about by destroying these values.
d) There are three main forms of naturalism such as Social, Psychological and Physical. In his social naturalism, he devises education as a method to mold the society. He opined that we could not become a man and citizens at the same time. Out of two options, we should become a man only. Thus, the individuality of man is honored by him. Psychological naturalism meant that the child should be given chance to develop on the basis of his inner feeling and natural tendencies and experiences gained from contact with others are harmful and unnatural.By physical naturalism, Rousseau means that child should be given chance to come in contact with birds, animals, and other physical objects of nature and learn in the process. This learning will make him free from evils.
e) He opposes the organization of education in social foundations. Thus he opposed school/education in the formal sense and advocated an individual basis of education.
f) He opposed too all sorts of habit formation in the child. It is because this can make a child traditional.
Thus, Rousseau’s Naturalism is fully of many unnatural and impracticable ideas and he himself realized it but he put three ideas with such a force that is influenced not only the society of Europe but also the educational system of the period.

2) Negative Education
In the 17th. century, Europe, man was considered bad by nature. So efforts were made to change the nature of man by imparting religious education.
Rousseau went contrary to this believing infallibility of man and proposed the idea of Negative Education. By this education he means not teaching truth or virtues to a child but shielding his heart from evils and mind from errors. The feature of his negative education are given below:
a) Nothing against the interest, attitude on age of the child should be taught. He should be given full freedom to choose his own curriculum.
b) The education of a child should be based on his natural tendencies and stages of development by using different organs and senses of the child. Mind should be less taxed for the purpose. Mind should be inactive toll the child develops discretion power in him thus he emphasized, the training of senses by keeping the mind idle.
c) Child should be protected from outside environment to keep him alien to vices. In this way, there will not be any need to impart knowledge of virtues to the child. Virtues may be taught of the later stage of life.
d) The child should not be taught anything at all, especially from books. Small children should learn from nature itself.
e) Thus by Negative Education, Rousseau opposed not only the mental development of the individuals/child but he also opposed moral and spiritual development.

Question 11:
Discuss the different phases of education of Emile as supported by Rousseau. Curriculum, methods of teaching and discipline.
Answer:
Rousseau’s programme of education for Emile is devised into different phases and Rousseau had divided the whole programme of education curriculum on the basis of the development stages of human child.
(i) Infancy (from 00 to 5 years): At this stage instead of giving the child controlled information of various subjects it is better to pay attention to the development of child’s body and his sense organs. He should be allowed to play with whatever things he likes. The dresses should be made for free movement. His toys should be cheap ordinary and natural like leaves, plants flowers, stems etc. According to Rousseau Emile should be given a negative Education druing infancy.

(ii) Childhood (5 to 12 years): Rousseau opposed the use of textbooks during this period. The child should be given chance to leam everything through observation and experiences. These experiences develop sense organs, which will lead to the development of mind and power of reasoning. He should be allowed full freedom to use his sense, the sense organs, the eyes to be trained to measure height, weight, and distance.

(iii) Adolescence (12-15 years): During this period natural curiosity of the adult develops. So he should be taught natural science and languages, mathematics, music, painting and social services. He should use maps in learning geography.

(iv) Adulthood (15 – 20 years): The organs become active with a maturity of mind and intellect he should leam social and moral education like objects of social services.
His Discipline: Rousseau advocated complete freedom, left free to the environment.
Self discipline is learnt in the process of experiences. There is no scope ofgiving any punishment to the child but natural process of feed back in stills discipline in him.

Question 12:
Dsicuss the life philosophy of Rousseau. Give his educational aims and ‘Self Education’ and curriculum.
Answer:
Life Philosophy of Rousseau: Jean Jacques Rousseau (1712 -1778) lost his mother just after his birth and his father brought him up but he could not look after him. Rousseau fell to prey to all sorts of bad habits. He originally belonged to Geneva and the beauty of that place lured him to use it and it made him a careless man.
Rousseau got matured. He hated the society for its evils. Evil customs and artificial traditions and tried to retun it on the line of nationalism and wished everybody shun the evil society and live with nature.

Out of five books, “The progress of arts and science”, “New Heloise”, Confessions, ‘Social Contract’ and ‘Emile’ are unique and the last two books ‘Social Contract’ and ‘Emile’ brought great name and fame to him.His ideas of education can be seen in Emile. It is a novel whose hero is Emile and the heroine is Sofia. In his book, Rousseau keeps Emile away from society and culture and leaves him under the guidance of an ideal teacher in the natural environment and Emile is educated in an atmosphere of natural beauty. There are five chapters in the book. The first chapter describes education of Emile is in fancy, childhood, adolescence and adulthood and the last chapter deals with the education of Emile’s wife Sofia.

People reacted against his book “Emile very sharply. France and Switzerland banned it end it was put into the fire at many place in Europe consequently. Rousseau had to leave France for England in 1766.After 11 years of exile, he returned to France and wrote his last book “Confession”. Rousseau’s philosophical thoughts are related to Naturalism and Negative Education.

Educational Aims: Rousseau emphasized on the following aims of education, l) We are bom weak and we want strength. We are poor and we want help. Whatever we do, not have is to be given by education is the ultimate aim of education.

• To establish harmony between man, object and nature.
• Child attains pleasure by using his organs and senses and by applying his strength. So the aims of education is to develop his various innate powers by helping him in his natural activities.
• Books are not end of education. They are only means and the child is the end. The aim of education is to develop the child to be fullest for a complete and happy life.

His Self-Education:
i) He opposes strongly the imposition of ideals and morals into the mind of the child from
outside. Children should learn these things through activities. It is because the children are more interested in activities rather than sitting idle and hearing lectures. At that stage, helps enough power of assimilate between construction and destruction. His only concern is to bring about change will form through any activity.
ii) Body can become strong through physical exericse and mind also become strong through self study. In self-education the child can proceed further according to his own physical and mental capacities.
iii) Only that knowledge gets retained for a longer period, which is learned from self-experiences. We should accept the experiences of others only after our own wisdom.
iv) Blind fellowship is not accepted at all. A child should not learn a thing because he has been asked to do so but he should learn oniyit in the process of his self study.
v) Special emphasis is to be given on the physical development of the child.
Curriculum
Rousseau was against the fixed curriculum, the child should be educated through activities and first-hand experiences.During infancy positive instructions to be imparted with good health training of senses and cultivation of natural habits. At the stage of childhood provision of in parting physical education through a set of gymnastics and the exercises training of senses.

At the stage of boyhood, the chief intellect is to be trained through teaching of good sense of physical senses, language, mathematics, manual works, social relations, music and drawing.At adolescence morality of the individual is to be trained through teaching of good education and by activity method and occupation. Moral education subjects are : history, religion, aesthetics, physical culture and sex education etc. are included in the curriculum at the adolescence stage.

Question 13:
Discuss John Dewey’s contribution to educational thought and practice.
Answer:
John Dewey is one of the greatest educationists of the modem age. He has revolutionized the exponent of pragmatism.
Meaning of Education to John Dewey: John Dewey considered individuality as the aim of education. According to him, all education proceeds by the participation of the individual in social consciousness of the race. The educative process, according to John Dewey has two aspects – Psychological and Sociological because every individual is a psychological as well as social being. So no aspect can be neglected in the process of education.

1.  Education as life: According to John Dewey education is a process of living. So the school experiences should have a resemblance with life outside the school and the school functions as a society in miniature form.
2.  Education as Growth: Education is a continuous process that adds to child’s experience resulting in the growth and development of the child. It helps the child to grow to its full extent.
3.  Education as Reconstruction of Experience: Every generation inherits certain experiences from its previous generation. These experiences are to be modified according to the changing situations through education.
4. Education as a Social Process: Man is a social being and education socializes to human child. Right education helps the child make suitable adjustments with his social environment.John Dewey and Aims of Education: John Dewey is of the opinion that education process no aim beyond itself, it is its own end. There are no fixed aim of education to John Dewey.

Because, human life changes with the changes in time, place and situations and education should cater to the changing nature of human life. Education should therefore aim at cultivating a dynamic and adaptable mind in the child so that he can suitably adjust with his changing environment. Education should also create new value for the child. Education should aim at inculcating democratic values and attitudes in the child.

John Dewey and Curriculum: Curriculum according to John Dewey should reflect the child’s social life and social activities. It should be flexible, and changeable and it should take into consideration the child’s interests and educative experiences. As per example the curriculum at the primary stages should be based on the fourfold interests of the small child, interests in conservation and communication interests in inquiry interests in making change or construction and interest in artistic expression. Hence subjects like reading, writing, counting, handwork and drawing etc. are to be included at this stage.

The curriculum at higher students of education must have provisions for the environment and reorganization of past experiences. It just stimulates the learner to acquire new experiences and new ideas to they learned ones. John Dewey has also put emphasis on curriculum. This means each subject should be linked with the other and each should also related to the day life of the child.

John Dewey and Method of Teaching: John Dewey put emphasis on learning by doing. He stressed project method and this method is meant for the child. Earning was to be emphasized on teaching. The students should be free to carry on their whole planning and activity. Learning should be incidental and an outcome of the purposeful activity.

John Dewey and Discipline: John Dewey believed in the theory of free discipline and self-discipline. Discipline imposed from outside is directive with eyes of the John Dewey. Discipline should be social in character. The natural impulses of the child should be channelized in a socially desirable way. The main function of school discipline is to cultivate in the children’s social attitude, habits and ideal conduct through cooperative activity.

John Dewey and Role of the Teacher: In John Dewey’s system of education the child occupies the central position. The teacher is the friend, philosopher and guide of students. The teacher has to observe the pupil’s planning, encourage their activity and provide them the necessary opportunity and environment. John Dewey’s teacher is free to frame his own curriculum and carry on the administration ofhis own school.

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