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Odisha State Board CHSE Odisha Class 12 Invitation to English 4 Solutions Grammar Tense Patterns Unit 2 The Present Perfect Textbook Activity Questions and Answers.

CHSE Odisha 12th Class English Grammar Tense Patterns Unit 2 The Present Perfect

SECTION- 1:
Examine the use of the Present Perfect in the following sentences.
(a) A: Where’s your T.V. set? I don’t see it.
B: I have sold it.
(b) A: Why are you looking so happy?
B: I have just got a job.
In the above examples, the speaker ‘B’ talks about some events beginning in the past and lasting up to the present moment (or still continuing). Though the event started in the past, it is connected to the present moment of speaking. In this sense, we used the Present Perfect Tense.

Activity — 9
Complete the sentences marked B. Use the verb in brackets, together with ‘just/already/yet’.
A: What does your wife think of your plan?
B : I __________(not tell) her yet.
A: Would you like something to eat?
B : No, thanks I __________(just/eat).
A: Is your brother here yet?
B : Yes, he __________(just/arrive).
A: What’s on TV today?
B: I don’t know. I __________(not see) the programme yet.
A: Do you know where Bidhu lives?
B : Yes, he __________(just/move) to Satyanagar.
A: Are your friends coming to the circus with us?
B : No, they __________(already/see) it.
A: When is Prakash leaving?
B : He __________(already/leave).
Answer:
1. A: What does your wife think of your plan?
B: I have not told her yet.
A: Would you like something to eat?
B: No, thanks I have just eaten.
A: Is your brother here yet?
B: Yes, he has just arrived.
A: What’s on TV today?
B: I don’t know. I have not seen the program yet.
A: Do you know where Bidhu lives?
B: Yes, he has just moved to Satyanagar.
A: Are your friends coming to the circus with us?
B: No, they have already seen it.
A: When is Prakash leaving?
B: He has already left.

Activity – 10
Study the situations suggested below and makeup sentences with “yet”, “already” or “just”.
1. You are going to Koraput next Sunday. You phone your travel agent to buy a ticket for you.
Later your father says. ‘Shall I get the ticket for you ?”
You: No, (buy) ___________________________________.
2. Alok goes to the Post Office but returns after a while. His friend asks you if he is still at the Post Office.
You: No, (come back) ___________________________________.
3. You know that one of your classmates is looking for a house. When you meet him, you want to know if he has been successful.
You: ___________________________________.
4. You visit a friend’s house after lunch. He asks if you would like something to eat.
You: ___________________________________.
5. You are doing your homework. Your brother thinks that you have finished and turns the light off. What would you tell him?
You: ___________________________________.
6. Amar goes out. Ten minutes later his friend comes and asks you if he can meet Amar.
You: ___________________________________.
Answer:
1. No, I have already bought it from a travel agent.
2. No, he has already come back.
3. Have you got a house yet?
4. Have you had your lunch yet?
5. I have not finished my homework yet.
6. Amar has iust gone out

Activity – 11
Below is a list of things that your parents have asked you to do today. You have checked the things you’ve done so far. Talk about the things you’ve already done and the things you haven’t done yet. (Two have been done for you as examples.)
1. do the washing up
2. do your homework ✓
3. wash the scooter
4. write to the brother
5. read today’s newspapers ✓
6. defrost the fridge
7. buy some fruits ✓
8. watch the news on TV
9. clean the windows ✓
10. water the plants
11. empty the dustbins ✓
12. phone uncle
Answer:
1. I haven’t done the washing up yet.
2. I have already done my homework.
3. I have not washed the scooter yet.
4. I have not written to my brother yet.
5. I have already read today’s newspaper.
6. I have not de-frosted the fridge yet.
7. I have already bought some fruit.
8. I have not watched the news on TV.
9. I have already cleaned the windows.
10. I haven’t watered the plants yet.
11. I have not emptied the dustbin yet.
12. I have not phoned my uncle yet.

Activity – 12
Anil, Anima, Mohan and Anand are talking about the places they have visited. Fill in the spaces using the information in the chart below.

	Kolkata	Koraput	Puri	Sambalpur	Shillong
Anil	Yes	No	No	Yes	No
Anima	Yes	No	Yes	Yes	No
Mohan	No	No	Yes	No	No
Anand	Yes	No	Yes	Yes	No

1. Anil __________been to Kolkata, but Mohan ___________.
2. Three people __________ been to Puri.
3. Only one person __________been to Sambalpur.
4. Mohan is the only one who __________ visited only one place.
5. Nobody __________ been to Shillong.
6. Two people __________ been to three places.
7. Anima and Mohan __________ both been to Puri, but neither __________ been to Koraput.
Answer:
1. Anil has been to Calcutta, but Mohan hasn ’t.
2. Three people have been to Puri.
3. Only one person hasn’t been to Sambalpur.
4. Mohan is the only one who has visited only one place.
5. Nobody has been to Shillong.
6. Two people have been to three places.
7. Anima and Mohan have both been to Puri, but neither haven’t been to Koraput.

Note:
Look at the difference between ‘have /has been ’ and ‘have/has gone ‘Has /have been’ means went and returned. But ‘has gone/have gone means went and not returned. The Present Perfect form of ‘go’ (has/have gone) is not used when the subject of the sentence is 7, we or you.’ The Present Perfect is used to refer to some happening in the past, for which the time of action is not given.

SECTION – 2
Look at the sentences below.
1. Hari didn’t have a beard six months ago. He has a beard now. He has grown a beard.
2. Malati was very shy. She is smart now. She has become smart.
3. She was a little baby when I last saw her. She is a young girl now. She has grown up.
When a change has taken place between now and sometime before now, we use the Present Perfect Tense.

Activity – 13
Study the situations below and make up appropriate sentences using the verbs suggested.
1. Yesterday my sister bought a pen. She can’t find it now. (lose)
______________________________________________.
2. The children were playing here some time ago. Now they are not seen, (leave)
______________________________________________.
3. My friend weighed 50 kilograms. Now he weighs 70. (gain weight)
______________________________________________.
4. The man met with an accident. Now he is not able to speak, (loses voice)
______________________________________________.
5. It was raining in the morning. Now the sky is clear. (stop)
______________________________________________.
6. The tiger attacked the man. He is dead now. (kill)
______________________________________________.
7. He had some paper with him. Now he does not have any to write on. (run out)
______________________________________________.
8. My teacher got a job in a bank. He is not coming to school anymore, (resign)
______________________________________________.
Answer:
1. She has lost it somewhere.
2. They have already left.
3. He has gained weight.
4. He has lost voice in the accident.
5. The rain has already stopped.
6. The tiger has already killed the man.
7. He has already run out of paper.
8. He has already resigned from school.

SECTION – 3
Present Perfect is often used with the following time expressions.

lately	until now	ever	for five years
not yet	never	always	over the last eight years
recently	just	so far	in the past Iwo years
in recent years	already	since 1990

Do remember that the expressions like “last year, ago, yesterday” etc. cannot be used in the Present Perfect.

Activity – 14
Rewrite the following sentences putting the words in brackets in the right place. The first one has been done for you.
1. My teacher has wanted to be a writer. (never)
Ans. My teacher has never wanted to be a writer.
2. I’ve found him helpful, (always)
______________________________________________.
3. People have misunderstood him. (often)
______________________________________________.
4. I’ve had lunch. (just)
______________________________________________.
5. Has he been to Puri? (ever)
______________________________________________.
6. Don’t panic. The police have arrested the culprit, (already)
______________________________________________.
Answer:
2. I’ve always found him helpful.
3. People have often misunderstood him.
4. I’ve Just had lunch.
5. Has he ever been to Puri?
6. Don’t panic. The police have already arrested the culprit.

Activity – 15
Imagine that you suddenly run into an old friend whom you have not met for the last five years. But he has changed so much that you can hardly recognize him. Describe the changes that have taken place in your friend.

a) _______________________________
b) _______________________________
c) _______________________________
d) _______________________________
e) _______________________________
Answer:
1. He has grown into a young man.
2. He has become strong and healthy.
3. He has grown fair and tall.
4. He has made himself smart and confident.
5. He has grown a thick beard.

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Odisha State Board BSE Odisha 6th Class English Solutions Follow-Up Lesson 3 My Story is Said Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Follow-Up Lesson 3 My Story is Said

BSE Odisha 6th Class English Follow-Up Lesson 3 My Story is Said Text Book Questions and Answers

Session 1 (ସୋପାନ- ୧):
I. Pre-Reading (ପଢ଼ିବା ପୂର୍ବରୁ):

→ Socialisation (ସାମାଜିକୀକରଣ):
→ You can think of a pre-reading activity by linking this with the main lesson or from the picture.
(ତୁମେ ଏକ ପଢ଼ିବା ପୂର୍ବବର୍ତ୍ତୀ କାର୍ଯ୍ୟ ବିଷୟରେ ଚିନ୍ତା କରିପାର – ହୁଏତ ଏହାକୁ ମୁଖ୍ୟ ବିଷୟ ସହ ସଂଯୁକ୍ତ କରି କିମ୍ବା ଛବିରୁ ।)

II. While-Reading (ପଢ଼ିବା ସମୟରେ):

• Follow the steps of the main lesson.
  (ମୁଖ୍ୟ ପାଠର ସୋପାନଗୁଡ଼ିକୁ ଅନୁସରଣ କର ।)
• Read the following chain poem translated from Odia. This poem is generally sung when a story ends.
  (ଓଡ଼ିଆରୁ ଅନୁବାଦ ହୋଇଥିବା ନିମ୍ନଲିଖ୍ ଶିକୁଳି କବିତାଟିକୁ ପଢ଼ । ଏହି କବିତାଟି ସାଧାରଣତଃ ବୋଲାଯାଏ ଯେତେବେଳେ ଗୋଟିଏ ଗପ ଶେଷ ହୋଇଯାଏ ।)

TEXT (ବିଷୟବସ୍ତୁ):

1. My story is said
The flowering plant is dead.

2. O flowering plant! why did you die ?
The black cow ate me up and made me lie.

3. O black cow! why did the plant you eat?
Because the cowherd did not well me treat.

4. O cowherd! why didn’t you well the cow treat to eat?
The daughter-in-law did not give me food.

5. O daughter-in-law ! why didn’t you give food, why?
Because my little baby did cry.

6. O little baby! why did you cry?
The ant bit me, that is why.

7. O ant! why did you bite the little child?
Under the soil I hide
And bite soft flesh when I do find.

କବିତାର ଓଡ଼ିଆ ଉଚ୍ଚାରଣ:
1. ମାଇଁ ଷ୍ଟୋରି ଇଜ୍ ସେଡ୍
ଦି ଫ୍ଲାୱାରିଙ୍ଗ୍ ପ୍ଲାଣ୍ଟ୍ ଇଜ୍ ଡେଡ୍ ।

2. ଓ ଫ୍ଲାୱାରିଙ୍ଗ୍ ପ୍ଲାଣ୍ଟ ! ହୁଏ ଡିଡ଼୍ ଇୟୁ ଡାଏ ?
ଦି ବ୍ଲାକ୍ କାଓ ଏଟ୍ ମି ଅପ୍ ଆଣ୍ଡ ମେଡ଼୍ ମି ଲାଏ ।

3. ଓ ବ୍ଲାକ୍ କାଓ ! ହ୍ଵାଏ ଡିଡ୍ ଦ’ ପ୍ଲାଣ୍ଟ୍‌ ଇୟୁ ଇଟ୍ ?
ବିକଜ୍ ଦ’ କାଓହଡ଼୍ ଡିଡ଼୍ ନଟ୍ ୱେଲ୍ ମି ଟ୍ରିଟ୍ ।

4. ଓ କାଓହଡ଼ ! ହ୍ବାଏ ଡିଡ଼ିଣ୍ଟ୍ ଇୟୁ ୱେଲ୍‌ ଦ’ କାଓ ଟ୍ରିଟ୍ ଟୁ ଇଟ୍ ?
ଦି ଡଟର୍-ଇନ୍-ଲ୍ ଡିଡ୍ ନଟ୍ ଗିଭ୍ ମି ଫୁଡ୍ ।

5. ଓ ଡଟର୍‌-ଇନ୍-ଲ୍ ! ହ୍ବାଏ ଡିଡ଼ିଣ୍ଟ୍‌ ଇୟୁ ଗିଭ୍ ଫୁଡ୍, ଦ୍ଵାଏ ?
ବିକଜ୍ ମାଇଁ ଲିଟିଲ୍ ବେବି ଡିଡ଼୍ କ୍ରାଏ ।

6. ଓ ଲିଟିଲ୍ ବେବି ! ହୁଏ ଡିଡ଼୍ ଇୟୁ କ୍ରାଏ ?
ଦି ଆଣ୍ଡ୍ ବିଟ୍ ମି, ଦ୍ୟାଟ୍ ଇଜ୍ ହୁଏ ।

7. ଓ ଆଣ୍ଡ୍ ! ହୁଏ ଡିଡ୍ ଇୟୁ ବାଇଟ୍ ଦ’ ଲିଟିଲ୍ ଚାଇଲ୍‌ଡ୍ ?
ଅଣ୍ଡର୍ ଦ’ ସିଏଲ୍ ଆଇ ହାଇଡ଼୍
ଆଣ୍ଡ୍ ବାଇଟ୍ ସଫ୍ଟ ପ୍ଲେସ୍ ସ୍ପେନ୍ ଆଈ ଡୁ ଫାଇଣ୍ଡ୍ ।

କବିତାର ସାରକଥା | ଓଡ଼ିଆ ଅନୁବାଦ:
୧।  ମୋ ଗପଟି କୁହା ସରିଲା
ଫୁଲଗଛଟି ମରିଗଲା ।

୨। ହେ ଫୁଲଗଛ ! ତୁ କାହିଁକି ମରିଗଲୁ ?
କାଳୀଗାଈଟା ମୋତେ ଖାଇଦେଲା ଏବଂ ମୋତେ ତଳେ ପକାଇ ଦେଲା ।

୩ । ହେ କାଳୀଗାଈ ! ଫୁଲଗଛଟିକୁ ତୁ କାହିଁକି ଖାଇଦେଲୁ ?
କାରଣ ଗାଈ ଜଗୁଆଳିଟା ମୋ’ର ଖାଇବା ପିଇବା ବୁଝିଲା ନାହିଁ ।

୪ । ହେ ଗାଈ ଜଗୁଆଳି ! ତୁ କାହିଁକି ଗାଈର ଭଲ ଯତ୍ନ ନେଲୁ ନାହିଁ ଓ ତାକୁ ଖାଇବାକୁ ଦେଲୁନି ?
ବୋହୂଟା ମୋତେ ଖାଦ୍ୟ ଦେଲା ନାହିଁ ।

୫ । ହେ ବୋହୂ ! ତୁ କାହିଁକି ଖାଦ୍ୟ ଦେଲୁନି, କାହିଁକି ?
କାରଣ ମୋ ଛୋଟ ଛୁଆଟି କାନ୍ଦିଲା ।

୬ । ହେ ଛୋଟ ଛୁଆ ! ତୁ କାହିଁକି କାନ୍ଦିଲୁ ?
ପିମ୍ପୁଡ଼ିଟା ମୋତେ କାମୁଡ଼ି ଦେଲା, ସେଇଥପାଇଁ।

୭ | ହେ ପିମ୍ପୁଡ଼ି ! ତୁ କାହିଁକି ଛୋଟ ଛୁଆଟିକୁ କାମୁଡ଼ି ଦେଲୁ ?
ମୁଁ ମାଟିତଳେ ଲୁଚିଥାଏ
ଏବଂ ଯେତେବେଳେ ମୁଁ ନରମ ମାଂସ ପାଏ କାମୁଡ଼ିଦିଏ ।

Session – 2 (ସୋପାନ – ୨):
III. Post-Reading (ପଢ଼ିସାରିବା ପରେ):

1. Writing (ଲେଖୁ ):
(a) Answer the following questions.
(ନିମ୍ନଲିଖୂତ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

(i) What language is this poem from ?
(କେଉଁ ଭାଷାରୁ ଏହି କବିତାଟି ଆସିଛି ?)
Answer:
This poem is from the Odia language.

(ii) When is this song normally sung?
(ସାଧାରଣତଃ କେତେବେଳେ ଏହି ଗୀତଟି ବୋଲାଯାଏ ? )
Answer:
This song is normally sung when a story ends.

(iii) Why was the flowering plant dead?
(କାହିଁକି ଫୁଲଗଛଟି ମରିଗଲା ?)
The ____________________________________dead because
_________________________________________
_________________________________________
Answer:
The flowering plant was dead because the black cow ate it up and made it lie.

(iv) Why did the cow eat up the flowering plant?
(ଗାଈ କାହିଁକି ଫୁଲଚାରାଟିକୁ ଖାଇଦେଲା ?)
Ans. The cow ate up the flowering plant because the cowherd did not treat the cow well.

Word Note (ଶବ୍ଦାର୍ଥ):
(The words/phrases have been defined mostly on contextual meanings.)
(ଶବ୍ଦ । ଖଣ୍ଡବାକ୍ୟଗୁଡ଼ିକ ଅଧିକାଂଶତଃ ପ୍ରସଙ୍ଗଗତ ଅର୍ଥ ଉପରେ ନିର୍ଭର କରି ବ୍ୟାଖ୍ୟା କରାଯାଇଛି।)

battle — fight, ଯୁଦ୍ଧ
bite — cut with teeth, କାମୁଡିବା
cowherd — one who looks after cows, ଗାଈ ଜଗୁଆଳି
kingdom — country ruled by a king, ରାଜ୍ୟ
look after — take care of, ଯତ୍ନ ନେବା, ଦେଖାଶୁଣା କରିବା
nail — metal nail for fixing horse-shoe, ଲୁହାକଣ୍ଟା
rider — (here) the person who rides a horse, ଆରୋହୀ |
shoe nail — horse shoe nail, ଘୋଡ଼ା ନାଲ
treat — caring, feeding, ଯତ୍ନ ନେବା, ଖାଇବା ପିଇବା ବୁଝିବା
daughter-in-law — the wife of the son, ବୋହୂ ବା ବଧୂ
hide — ଲୁଚାନ୍ତୁ |
bite — କାମୁଡିବା
soft — ନରମ
flesh — ମାଂସ
lie — ମିଛ କହିବା
die — ମର
dead — ମୃତ
bit — ବିଟ୍
lost — ହଜିଯାଇଛି |

Odisha State Board CHSE Odisha Class 12 Invitation to English 4 Solutions Grammar Tense Patterns Unit 7 Future Time Reference Textbook Activity Questions and Answers.

CHSE Odisha 12th Class English Grammar Tense Patterns Unit 7 Future Time Reference

SECTION – 1

The Present Progressive is used for future reference
A: What are you doing tomorrow evening?
B: I’m going to the town hall. My friends are coming. We are putting up a show for the handicapped. We are also meeting the Minister for some funds. Are you going somewhere tomorrow evening?
A: Yes, I’m going to the library. I thought I could ask you to come along.
In the dialogue above you can only see the use of more number of Present Progressive constructions/sentences.

Here are a few questions for you to answer.
1. Which period of time does the use of the Present Progressive refer to: time which is past, time which is present, or future time?
2. Are ‘A’ and ‘B’ talking about actions that they have already planned and arranged to do?
3. Is the arrangement personal (made by either A or B) or is official (made by someone else, who is in a position of authority)?
Answer:
1. The time in all the Present Progressive sentences refers to a future time (futurity).
2. Yes. In the dialogue between A and B, we see the actions have already been planned/arranged.
3. The arrangements are personal. No official planning/arrangements.
The above sentences (Present Progressive form) are based on a personal plan or decision or need or requirement. So now we can say these sentences are based on the ‘internal evidence’ of the speaker or writer.

Activity – 26
Surabhi, who is 16, wants to go on an excursion with her friends and teachers. . Her mother is worried and has a lot of questions to ask about the arrangements. Look at the hints supplied and complete the dialogue between them, using the appropriate forms of the verbs.
Mother: Who / you / go / with?
______________________
Surabhi: friends/teachers
______________________
Mother: Where / you / go?
______________________
Surabhi: Darjeeling
______________________
Mother: When / go / there?
______________________
Surabhi: next Monday
______________________
Mother: How / you / get there?
______________________
Surabhi: by bus?
______________________
Mother: Where / you / stay?
______________________
Mother: a hotel.
______________________
Answer:
Mother: Who are you going with?
Surabhi: I’m going with my friends and teachers.
Mother: Where are you going?
Surabhi: I’m (We are) going to Darjeeling.
Mother: When are you going there?
Surabhi: I’m (We are) going there next Monday.
Mother: How are you getting there?
Surabhi: I’m (We are) getting there by bus.
Mother: Where are you staying?
Surabhi: I’m (We are) staying in a hotel.
Mother: What are you doing the next Tuesday there?
Surabhi: I’m visiting a cinema with my cousin.
Mother: When are you coming back?
Surabhi: I’m (We are) coming back next Sunday.

Activity – 27
Bindu maintains a diary in which she writes down a list of the things that she plans to do during the week ahead. Here is an outline of the entries in her diary for the next week. Complete the entries, using the hints given below. Use the Present Progressive.
Monday: meet music teacher.
______________________
Tuesday: go to the cinema with cousin
______________________
Wednesday: play badminton
______________________
Thursday: see off Maya at the station.
______________________
Friday: throw a party for friends.
______________________
Saturday: visit a dentist in the evening.
______________________
Sunday: rest.
______________________
Answer:
Monday: Bindu is meeting her music teacher on Monday.
Tuesday: She is going to the cinema with her cousin.
Wednesday: She is playing badminton in the evening.
Thursday: She is seeing off Maya at the station.
Friday: She is throwing a party for her friends.
Saturday: She is visiting the dentist in the evening.
Sunday: She is having a rest on Sunday.
Or, She is resting on Sunday.

Activity – 28
You plan to visit Koraput after your examination. Mention five things that you have arranged to do there.
(a) ……………………………………………………………………………………
(b) ……………………………………………………………………………………
(c) ……………………………………………………………………………………
(d) ……………………………………………………………………………………
(e) ……………………………………………………………………………………
Answer:
(a) I’m meeting one of my school friends there.
(b) I’m attending her birthday ceremony.
(c) I’m visiting the Damanjodi project there.
(d) I’m helping her with the party.
(e) I’m visiting new places there with my friend.

SECTION – 2

The Present Simple for future time reference
Minister: Have you drawn up my tour program for the next week?
P. A.: Yes sir. You leave for Sambalpur at 7.00 a.m. on Monday morning.
Minister: When do I get there?
P. A. : You reach Sambalpur at 1.00 p.m. You halt at Angul for a few minutes, on the way Then, after lunch, you meet the Commissioner at 3.00 for a discussion.
Minister: Where do I stay in Sambalpur?
P.A.: The Guest House at the Hirakud Dam has been reserved for you, sir. Then, the next morning at 8.00, you proceed to Rourkela. The Commissioner accompanies you to Rourkela.
Minister: And when do I return to Bhubaneswar?
P. A. . On Thursday, sir.
Look at the use of the present simple in the dialogue. What is the time reference here: past time, present time, or future time?
The Present Simple Tense’ can be used for future time reference. We use this tense for future events, i.e. fixed by the calendar or an official or departmental timetable. It is used to talk about the activity which has already been planned and fixed.

Activity — 29
The following is the list of official engagements of the Chief Minister for next Monday. Write one sentence to describe each item. Use the Present Simple form of the verb given in brackets :
8.30 a.m. inaugural address, Conference on Preservation of Human Rights (deliver)
9.30 a.m. Speech on Syllabus Reform, Utkal University. (give)
1.00 p.m. Cabinet Committee meeting (preside over)
4.30 p.m. National Book Fair (inaugurate)
5.50 p.m. Parliamentary delegation from Turkey (welcome)
8.00 p.m. Dinner party in honor of the Prime Minister of Bangladesh (host)
____________________________________________
____________________________________________
____________________________________________
____________________________________________
____________________________________________
Answers
8.30 a.m. The Chief Minister delivers his inaugural address at the Conference on Preservation of Human Rights.
9.30 a.m. The Chief Minister gives his speech on Syllabus Reform, at Utkal University.
1.00 p.m. The Chíef Minister presides over the Cabinet Committee meeting.
4.30 p.m. The Chief Minister inaugurates the National Book Fair at Unit — III, Exhibition field, Bhubaneswar.
5.50 p.m. The Chief Minister welcomes the Parliamentary delegation from Turkey.
8.00 p.m. The Chief Minister hosts the Dinner party in honor of the Prime Minister of Bangladesh.

Activity — 30
A travel agency offers a number of packages for tourists. Here is an outline of such a travel package.
Imagine that you are a salesman in a travel agency and that you are describing the program to a group of tourists, who want to know the details. Use complete sentences, with the Present Simple, to describe the program.
Monday → Leave Bhubaneswar for Hyderabad by Konark Express
March 2 →2 days’ sight-seeing in Hyderabad
Thursday → Board Kaveri Express for Chennai
March 5 → 2 nights in Chennai, Visit to Mahabalipuram
Saturday → Board Vrindaban Express for Bangalore
March 7 → 2 nights in Bangalore
Monday → By Deluxe bus to Mysore
March 9 → 1 night in Mysore
Tuesday → Mysore to Goa by taxi
March 10 → 2 nights in Goa
Thursdaý → Board Flight IC 765 for Bhubaneswar
March 12
Answer:
Monday → The tourists leave Bhubaneswar for Hyderabad by Konark Express.
March 2 → They go for 2 days sightseeing in Hyderabad.
Thursday → They board the Kaveri Express for Chennai.
March 5 → They stay 2 nights in Chennai and visit Mahabalipuram.
Saturday → They board the Vrindaban Express for Bangalore.
March 7 → They halt 2 nights in Bangalore.
Monday → They go by Deluxe bus to Mysore.
March 9 → They halt for 1 night in Mysore.
Tuesday → They go from Mysöre to Goa by taxi.
March 10 → They put up 2 nights in Goa.
Thursday → They board flight IC 765 for Bhubaneswar.
March 12

SECTION-3

Future time reference with ‘be going to’
1. I’m sweating already. I think it’s going to be a very hot day.
2. Be careful of that dog; it’s going to bite.
3. Mohan is eating too much; he’s going to be sick.
In the above sentences ‘be going to’ structure is used. In ‘be’ verb family we have eight verbs such as: am, is, are, was, were, be, being, and been. But in this structure for futurity (future time reference) am, is, are only these three are used.
be going to = is / am / are going to.
We can use this ‘be going to ’ structure for ‘external evidence ’ at the moment of speaking. We can better mark in the above three sentences some evidence or reason at the moment of speaking which helps the speaker to predict/believe something that is likely to happen in the near future.

Activity – 31
Complete the sentences using ‘be going to’ and the verbs in brackets.
(a) Look at those dark clouds. It’s _____________(rain).
(b) The lady is gasping for breath. I think she’s _____________(faint).
(c) My neighbor has packed up all his belongings. I think he _____________ (leave) the house.
(d) Rakesh _____________(fail) the exam. I don’t see him studying at all.
(e) Bijoy Babu _____________(lose) in the election. The voters are very unhappy with him.
Answer:
(a) Look at those dark clouds. It’s going to rain.
(b) The lady is gasping for breath. I think she is going to faint.
(c) My neighbor has packed up all his belongings. I think he is going to leave the house.
(d) Rakesh is going to fail the exam. I don’t see him studying at all.
(e) Bijoy Babu is going to lose in the election. The voters are very unhappy with

Activity – 32
Study the situation and guess what is going to happen. (One example has been given)
(a) The old man has been ill for a long time. He stopped taking food 5 days ago.
Ans. He is going to die.
(b) My friend has been reading the “matrimonial” column in the newspaper and collecting photographs of girls.
_____________________________________________________.
(c) A man is getting into the house opposite through the window. The people who live there are away on holiday.
_____________________________________________________.
(d) Do you see that man is trying to walk on the ice? His feet are sleeping.
____________________________________________________.
(e) The policeman is running after the thief and pointing his gun at him.
_____________________________________________________.
(f) The boy has taken the book from the shelf and put it on the table. He is drawing up a chair now.
_____________________________________________________.
(g) India meets Pakistan in the final match today. Sachin Tendulkar is injured and will not be able to play.
_____________________________________________________.
(h) Water has got into the boat.
_____________________________________________________.
Answer:
(b) My friend is going to marry soon.
(c) The man is going to steal from the house.
(d) The man is going to fall.
(e) The policeman is going to fire.
(f) He is going to read the book.
(g) India is going to lose the match.
(h) The boat is going to sink.

SECTION – 4

Here are some more sentences with ‘be going to’.
(a) A. Why have you bought so many books?
What are you going to do?
B. I‘m going to read all of them for my project.
(b) She has decided not to leave the house this year. She is going to stay on for another year.
(c) A. I’m going to buy a scooter
B. How‘re you going to pay for it?
A. I’m going to ask my brother to lend me some.
Look at these sentences with ‘be going to’. How are they different in meaning from the sentences in Section 3 (above)?
In this case, be going to is used to express an ‘intention’ to do something in the future. We can say: a decision or plan is based on a personal need or requirement.

Activity-33

There are a number of things that you haven’t done yet but intend to do. Answer
the questions below, using be going to as well as the words in brackets.
Friend: Have you had lunch?
You: No, but _____________. (after my friend arrives)
Friend: Have you written a letter to your father?
You: Not yet, _____________. (tomorrow)
Friend: Have you read the new novel by Vikram Seth?
You: No, _____________. (next week),
Friend: Have you watered the plants?
You: Not yet, _____________. (this afternoon)
Answer:
You: No, but I am going to have my lunch after my friend’s arrival.
Friend: Have you written a letter to your father?
You: Not yet, I am going to write a letter to my father tomorrow.
Friend: Have you read the new novel by Vikram Seth?
You: No, I am going to read the new novel by Vikram Seth next week.
Friend: Have you watered the plants?
You: Not yet, I am going to water the plants this afternoon.

Activity – 34
The members of a Youth Club have taken a vow on Gandhi Jayanti. They have promised that each of them will do at least one good deed, in memory of Mahatma Gandhi.
Can you draw up a list of the good things that the boys intend to do? Here is an example.
I’m going to plant 100 trees inside the school compound.
(a) ____________________________________________
(b) ____________________________________________
(c) ____________________________________________
(d) ____________________________________________
(e) ____________________________________________
(f) ____________________________________________
(g) ____________________________________________
Answer:
(a) I’m going to clean the rubbish in our lane.
(b) I’m going to help the poor and sick.
(c) I’m going to be friends with people of other religions.
(d) I’m going to be truthful.
(e) I’m going to fight against injustice.
(f) I’m going to be punctual.
(g) I’m going to preach equality among people.

Multiple-Choice Questions (MCQs) with Answers

Question 1.
London ___________the capital of the United Kingdom.
(A) were
(B) are
(C) is
(D) is being
Answer:
(C) is

Question 2.
It ___________since yesterday evening.
(A) was snowing
(B) had been snowing
(C) snowed
(D) has been snowing
Answer:
(D) has been snowing

Question 3.
Just as I was entering the room, the family was ___________for a party.
(A) leaving
(B) left
(C) had left
(D) going
Answer:
(A) leaving

Question 4.
The arm was so badly injured that he ___________it amputated.
(A) has to have
(B) had had
(C) had to have
(D) must have
Answer:
(C) had to have

Question 5.
I ___________ very busy lately.
(A) have been
(B) will be
(C) would be
(D) having being
Answer:
(A) have been

Question 6.
He was not thirsty because he ___________too much water.
(A) had drank
(B) had drunk
(C) had been drunk
(D) had drunken
Answer:
(B) had drunk

Question 7.
Where ___________since we last played together?
(A) were you playing
(B) had you been playing
(C) have you been playing
(D) do you play
Answer:
(C) have you been playing

Question 8.
As soon as she noticed the workmen, she asked them what they ___________.
(A) have been doing
(B) are doing
(C) have done
(D) had been done
Answer:
(C) have done

Question 9.
The bus has been traveling for days and it times it ___________.
(A) stops
(B) will stop
(C) stopped
(D) would stop
Answer:
(C) stopped

Question 10.
You need not ___________when the barrister asked you where you were when the crime was committed.
(A) be lying
(B) to tell a lie
(C) tell a lie
(D) have tell a lie
Answer:
(C) tell a lie

Question 11.
The news about the disaster ___________six weeks ago.
(A) broadcasted
(B) was broadcasting
(C) was broadcasted
(D) was broadcast
Answer:
(D) was broadcast

Question 12.
___________daily keeps people healthy and vigorous.
(A) In eating
(B) Eating
(C) Having eaten
(D) To eat
Answer:
(D) To eat

Question 13.
I cannot imagine myself ___________ car repairs in the garage.
(A) to do
(B) doing
(C) to be doing
(D) in doing
Answer:
(B) doing

Question 14.
The pianist ___________his performance to be well accepted by the audience willingly practiced his art.
(A) expect
(B) expecting
(C) expected
(D) expects
Answer:
(B) expecting

Question 15.
After our football team lost the cup final, the manager told them ___________ and not winning is the most important.
(A) playing the game
(B) the game
(C) the play
(D) the gaming
Answer:
(B) the game

Question 16.
I noticed the culprit ___________ away from the house.
(A) run
(B) would be running
(C) ran
(D) had running
Answer:
(B) would be running

Question 17.
I am convinced he ___________to school after his insolence.
(A) dare not return
(B) dare not returning
(C) dare not to return
(D) does not dare returning
Answer:
(A) dare not return

Question 18.
‘___________this week?’ ‘No, she’s on holiday.’
(A) Is Susan working
(B) Does Susan work
(C) Does work Susan
(D) none of these
Answer:
(A) Is Susan working

Question 19.
I don’t understand this sentence. What ___________?
(A) does mean this word
(B) does this word mean
(C) means this word
(D) none of these
Answer:
(B) does this word mean

Question 20.
John ___________tennis once or twice a week.
(A) is playing usually
(B) is usually playing
(C) usually plays
(D) none of these
Answer:
(C) usually plays

Question 21.
How ___________now ? Better than before?
(A) you are feeling
(B) do you feel
(C) are you feeling
(D) none of these
Answer:
(B) do you feel

Question 22.
It was a boring weekend ___________anything.
(A) I didn’t
(B) I don’t do
(C) I didn’t do
(D) none of these
Answer:
(C) I didn’t do

Question 23.
Tom ___________his hand when he was cooking the dinner.
(A) burnt
(B) was burning
(C) has burnt
(D) none of these
Answer:
(A) burnt

Question 24.
Jim is away on holiday. He ___________to Spain.
(A) is gone
(B) has gone
(C) has been
(D) none of these
Answer:
(B) has gone

Question 25.
Everything is going well. We ___________any problems so far.
(A) didn’t have
(B) don’t have
(C) haven’t had
(D) none of these
Answer:
(C) haven’t had

Question 26.
Linda has lost her passport again. It’s the second time this ___________.
(A) has happened
(B) happens
(C) happened
(D) none of these
Answer:
(A) has happened

Question 27.
You’re out of breath ___________?
(A) Are you running
(B) Have you run
(C) Have you been running
(D) none of these
Answer:
C) Have you been running

Question 28.
Where’s the book I gave you? What ___________with it?
(A) have you, done
(B) have you been doing
(C) are you doing
(D) none of these
Answer:
(A) have you, done

Question 29.
We’re good friends. We __________ each other for a long time.
(A) know
(B) have known
(C) have been knowing
(D) knew
Answer:
(B) have known

Question 30.
Sally has been working here __________.
(A) for six months
(B) since six months
(C) six months ago
(D)none of these
Answer:
(A) for six months

Question 31.
It’s two years __________ Joe.
(A) that I don’t see
(B) that I haven’t seen
(C) since I didn’t see
(D) since I saw
Answer:
(D) since I saw

Question 32.
They __________out after lunch and they’ve just come back.
(A) went
(B) have gone
(C) are gone
(D) none of these
Answer:
(A) went

Question 33.
The Chinese __________printing.
(A) invented
(B) have invented
(C) had invented
(D) none of these
Answer:
(A) invented

Question 34.
I am __________in Scotland for ten years. Now he lives in London.
(A) lived
(B) has lived
(C) has been living
(D) none of these
Answer:
(A) lived

Question 35.
The man sitting next to me on the plane was nervous because he __________before.
(A) hasn’t flown
(B) didn’t fly
(C) hadn’t flown
(D) none of these
Answer:
(C) hadn’t flown

Question 36.
__________a car when they were living in London?
(A) Had they
(B) Did they have
(C) Were they having
(D) Have they had
Answer:
(B) Did they have

Question 37.
I __________television a lot but I don’t anymore.
(A) was watching
(B) was used to watch
(C) used to watch
(D) none of these
Answer:
(C) used to watch

Question 38.
__________ tomorrow, so we can go out somewhere.
(A) I’m not working
(B) I don’t work
(C) I won’t work
(D) none of these
Answer:
(A) I’m not working

Question 39.
That hag looks heavy, __________ you with it.
(A) I’m helping
(B) I help
(C) I’ll help
(D) none of these
Answer:
(C) I’ll help

Question 40.
I think the weather __________be nice later.
(A) will
(B) shall
(C) is going to
(D) none of these
Answer:
(A) will

Question 41.
‘Ann is in hospital.’ ‘Yes, I know. __________her tomorrow.’
(A) I visit
(B) I’m going to visit
(C) I’ll visit
(D) none of these
Answer:
(B) I’m going to visit

Question 42.
We’re late. The film __________by the time we get to the cinema.
(A) will already start
(B) will be already started
(C) will already have started
(D) none of these
Answer:
(C) will already have started

Question 43.
Don’t worry __________late tonight.
(A) if I am
(B) when I am
(C) when I’ll be
(D) if I’ll be
Answer:
(A) if I am

Question 44.
The film __________by the time we get to the cinema.
(A) will start
(B) will have started
(C) will be starting
(D) will have been starting
Answer:
(B) will have started

Question 45.
Next year they __________for 25 years.
(A) will have been marrying
(B) will have been married
(C) will be marrying
(D) none of these
Answer:
(B) will have been married

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 13 Introduction To Three-Dimensional Geometry Ex 13 Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 13 Introduction To Three-Dimensional Geometry Exercise 13

Question 1.

Fill in the blanks in each of the following questions by choosing the appropriate answer from the given ones.
(a) The distance of the point P(x₀, y₀, z₀) from z – axis is [\(\sqrt{x_0^2+y_0^2}, \sqrt{y_0^2+z_0^2}, \sqrt{x_0^2+z_0^2},\)\(\sqrt{\left(x-x_0\right)^2+\left(y-y_0\right)^2}\)]
Solution:
\(\sqrt{x^2+y^2}\)

(b) The length of the projection of the line segment joining (1, 3, -1) and (3, 2, 4) on z – axis is ___________. [1, 3, 4, 5]
Solution:
5

(c) the image of the point (6, 3, -4) with respect to yz – plane is _____________. [(6, 0, -4), (6, -3, 4), (-6, -3, -4), (-6, 3, -4)]
Solution:
(-6, 3, -4)

(d) If the distance between the points (-1, -1, z) and (1, -1, 1) is 2 then z = _______________. [1, √2, 2, 0]
Solution:
1

Question 2.
(a) identify the axis on which the given points lie : (1, 0, 0), (0, 1, 0), (0, 0, 1)
Solution:
x-axis, y-axis, z-axis

(b) Identify the planes containing the points! (7, 0, 4), (2, -5, 0), (0, √2, -3)
Solution:
xz-plane, xy-plane, yz-plane.

Question 3.
(a) Determine, which of the following points have the same projection on x-axis. (2, -5, 7), (2, √2, -3), (-2, 1, 1), (2, -1, 3)
Solution:
(2, -5, 7), (2, √2, -3) and (2, -1, 3) have the same projection on x-axis.

(b) Find the projection of the point (7, -5, 3) on:
(i) xy-plane
Solution:
(7, -5, 0)

(ii) yz-plane,
Solution:
(0, -5, 3)

(iii) zx-plane
Solution:
(7, 0, 3)

(iv) x-axis
Solution:
(7, 0, 0)

(v) y-axis,
Solution:
(0, -5, 0)

(vi) z-axis.
Solution:
(0, 0, 3)

Question 4.
When do you say two lines in space are skewed? Do they intersect?
Solution:
A pair of non-co-planar lines are called skew lines. Skew lines do not intersect.

Question 5.
From the three pairs of lines given below, identify those which uniquely determine a plane:
(i) intersecting pair,
(ii) parallel pair,
(iii) a pair of skew lines.
Solution:
Out of given three pairs
(i) Intersecting pair and
(ii) Parallel pair of lines determine a plane.

Question 6.
Determine the unknown coordinates of the following points if :
(i) P(a, 2, -1)∈ yz – plane
Solution:
a = 0

(ii) Q(-1, y, 3) ∈ zx-plane
Solution:
y = 0

(iii) R(√2, -3, c) ∈ xy-plane
Solution:
c = 0

(iv) S(7, y, z) ∈ x-axis
Solution:
y = 0, z = 0

(v) T(x, 0, z) ∈ y-axis
Solution:
x = 0 , z = 0

(vi) V(a, b, -3) ∈ z-axis
Solution:
a = b = 0

Question 7.
Which axis is determined by the intersection of:
(i) xy-plane and yz-plane
Solution:
y-axis

(ii) yz-plane and zx-plane
Solution:
z-axis

(iii) zx-plane and xy-plane
Solution:
x-axis

Question 8.
Which axis is represented by a line passing through origin and normal to:
(i) xy-plane
Solution:
z-axis

(ii) yz-plane
Solution:
x-axis

(iii) zx-plane.
Solution:
y-axis

Question 9.
What are the coordinates of a point which is common to all the coordinate planes?
Solution:
Origin 0(0, 0, 0) is common to all coordinate planes.

Question 10.
If A, B, and C are projections of P(3, 4, 5) on the coordinate planes, find PA, PB, and PC.
Solution:
PA = 5, PB = 3, PC = 4.

Question 11.
(a) Find the perimeter of the triangle whose vertices are (0, 1, 2) (2, 0, 4) and (-4, -2, 7).
Solution:

(b) Show that the points (a, b, c), (b, c, a), and (c, a, b) form an equilateral triangle.
Solution:

(c) Show that the points (3, -2, 4) (1, 1, 1) and (-1, 4, -1) are collinear.
Solution:
Let A = (3, -2, 4), B = (1, 1, 1) and C = (-1, 4, -2)
D. rs of AB are < -2, 3, -3 >
D. rs of BC are < -2, 3, -3 >
As D. rs of AB is the same as d.rs of
BC it follows that A, B, and C lie on the same straight line.
So the points are collinear. (Proved)

(d) Show that points (0, 1, 2), (2, 5, 8), (5, 6, 6), and (3, 2, 0) form a parallelogram.
Solution:

(e) Show that the line segment joining (7, -6, 1) (17, -18, -3) intersect the line segment joining (1, 4, -5), (3, -4, 11) at (2, 0, 3).
Solution:

D.rs of AP are < -5, 6, 2 >
D.rs of PB are < -15, 18, 6>
i.e., < -5, 6, 2 >
Thus d.r.s. of AP are the same as the d.r.s. of PB.
So A.P.B. are collinear.
Again d.rs. of CP are < 1, -4, 8 >
D. rs. of PD are < 1, -4, 8 >
Thus d.rs. of CP and d.rs. of PD are equal.
So the points C.P.D. are collinear.
Hence line AB intersects the line CD at P.        (Proved)

(f) Find the locus of points that are equidistant from points (1, 2, 3) and (3, 2, -1).
Solution:
Let A = (1, 2, 3), B = (3, 2, -1)
Let P (x, y, z) be equidistant from A and B.
Then PA = DB
(x – 1 )² + (y – 2)² + (z – 3)²
= (x – 3)² + (y – 2)² + (z + 1)²
⇒ (x – 1)² – (x – 3)² + (y – 2)² – (y – 2)² + (z – 3)² – (z + 1)² = 0
⇒ (x – 1 + x – 3) (x – 1 – x + 3) + (z – 3 + z + 1)(z – 3 – z – 1) = 0
⇒ (2x – 4) . 2 + (2z – 2) . (- 4) = 0
⇒ x – 2 – 2z + 2 = 0
⇒ x – 2z = 0 This is the required locus.

Question 12.
(a) Find the ratio in which the line segment through (1, 3, -1) and (2, 6, -2) is divided by zx-plane.
Solution:

(b) Find the ratio in which the lines segment through (2, 4, 5), (3, 5, -4) is divided by xy-plane.
Solution:

(c) Find the coordinates of the centroid of the triangle with its vertices at (a₁, b₁, c₁), (a₂, b₂, c₂), and (a₃, b₃, c₃)
Solution:

(d) If A (1, 0, -1), B (-2, 4, -2), and C (1, 5, 10) be the vertices of a triangle and the bisector of the angle BAC, meets BC at D, then find the coordinates of the point D.
Solution:

(e) Prove that the points P(3, 2, -4), Q(5, 4, -6) and R(9, 8, -10) are collinear. Find the ratio in which point Q divides the line segment PR.
Solution:
Given that P = (3, 2, -4)
Q = (5, 4, -6), R = (9, 8, -10)
D. rs. of PQ are < 2, 2, -2 >
D. rs. of QR are < 4, 4, -4 >
i.e., < 2, 2, -2 >
Thus D.rs. of PQ and QR are the same.
So P, Q, R lie on the same straight line.
Hence P, Q, and R are collinear.     (Proved)
Let Q divides the join of PR in ratio k: 1
∴ \(\frac{9 k+3}{k+1}\) = 5, \(\frac{8 k+2}{k+1}\) = 4, \(\frac{-10 k-4}{k+1}\) = -6
⇒ k = \(\frac{1}{2}\)
Thus the ratio is 1: 2

Odisha State Board BSE Odisha 10th Class Maths Solutions Algebra Chapter 2 ଦ୍ବିଘାତ ସମୀକରଣ Ex 2(a) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Algebra Chapter 2 ଦ୍ବିଘାତ ସମୀକରଣ Ex 2(a)

Question 1.
ନିମ୍ନଲିଖ ଉକ୍ତିଗୁଡ଼ିକରେ ଥ‌ିବା ତ୍ରୁଟିକୁ ସଂଶୋଧନ କରି ଲେଖ ।
(i) x² – 4x + 4 = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟ ବାସ୍ତବ ଓ ଭିନ୍ନ ।
(ii) x² – 5x + 6 = 0 ସମୀକରଣର ପ୍ରଭେଦକ 2 ଅଟେ ।
(iii) ax? + bx – c = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟର ସମଷ୍ଟି \(\frac{c}{a}\)।
(iv) ax’ + bx + c = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟର ଗୁଣଫଳ \(\frac{b}{a}\)।
(v) 1 ଓ -1 ମୂଳ ବିଶିଷ୍ଟ ଦ୍ୱିଘାତ ସମୀକରଣଟି x² + 1 = 0 ।
(vi) x² = 0 ଦ୍ୱିଘାତ ସମୀକରଣର ମୂଳ ସମାନ ନୁହଁନ୍ତି ।
(vii) 3x² – 2x – 1 = 0 ସମୀକରଣର ମୂଳ ଦ୍ଵୟର ସମଷ୍ଟି \(– \frac{3}{2}\)।
(viii) 3x² – 2x – 1 = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟର ଗୁଣଫଳ \(\frac{1}{3}\)।
ଉ-
(i) x² – 4x + 4 = 0 ସମୀକରଣର ମୂଳଦ୍ବୟ ବାସ୍ତବ ପରିମେୟ ଓ ସମାନ ।
(ii) x² – 5x + 6 = 0 ସମୀକରଣର ପ୍ରଭେଦକ 1 ଅଟେ ।
(iii) ax² + bx – c = 0 ମୂଳଦ୍ଵୟର ସମଷ୍ଟି \(\frac{-b}{a}\)।
(iv) ax² + bx + c = 0 ସମୀକରଣର ମୂଳଦ୍ବୟର ଗୁଣଫଳ \(\frac{c}{a}\)।
(v) 1 ଓ – 1 ମୂଳବିଶିଷ୍ଟ ଦ୍ୱିଘାତ ସମୀକରଣଟି x² – 1 = 0।
(vi) x² = 0 ଦ୍ବିଘାତ ସମୀକରଣର ମୂଳଦ୍ଵୟ ସମାନ ଅଟନ୍ତି ।
(vii) 3x² – 2x – 1 = 0 ସମୀକରଣର ମୂଳଦ୍ବୟର ସମଷ୍ଟି \(\frac{2}{3}\)
(viii) 3x² – 2x – 1 = 0 ସମୀକରଣର ମୂଳଦ୍ବୟର ଗୁଣଫଳ \(\frac{-1}{3}\)।

(i) ଏଠାରେ a = 1, b = -4, c = 4
ପ୍ରଭେଦକ (D) = b² – 4ac = (-4)² – 4 × 1 × 4 = 16 – 16 = 0
∴ ମୂଳଦ୍ବୟ ବାସ୍ତବ ଓ ସମାନ ।

(ii) ଏଠାରେ a = 1, b=-5, c = 6
ପ୍ରଭେଦକ (D) = b² – 4ac = (- 5)² – 4 (1) (6) = 25 – 24 = 1

(iii) ax² + bx – c = 0 ମୂଳଦ୍ଵୟର ସମଷ୍ଟି \(\frac{-b}{a}\)।
(iv) ax² + bx + c = 0 ମୂଳଦ୍ବୟର ଗୁଣଫଳ \(\frac{c}{a}\)।
(v) ମୂଳବିଶିଷ୍ଟ ଦ୍ୱିଘାତ ସମୀକରଣ, x² – (α + ß) x + αß = 0
⇒ x² – { 1 + (- 1)} x + 1 (-1) = 0 ⇒ x² – 1 = 0
(vi) x = 0 ଦ୍ୱିଘାତ ସମୀକରଣ ମୂଳଦ୍ଵୟ ସମାନ ଅଟନ୍ତି । (କାରଣ ପ୍ରଭେଦକ (D) = 0)
(vii) ଏଠାରେ a = 3, b = -2, c = -1 ମୂଳଦ୍ଵୟ ସମାନ = \(\frac{-b}{a}\) = \(\frac{-(-2)}{a}\) = \(\frac{2}{3}\)
(viii) 3x² – 2x-1=0 ଏଠାରେ a = 3, b = -2, c = 1 ମୂଳଦ୍ବୟର ଗୁଣଫଳ \(\frac{c}{a}\) = \(\frac{-1}{3}\)

Question 2.
ନିମ୍ନଲିଖୂ ପ୍ରଶ୍ନମାନଙ୍କର ସଂକ୍ଷିପ୍ତ ଉତ୍ତର ଦିଅ ।
(i) ଗୋଟିଏ ଦ୍ବିଘାତ ସମୀକରଣର ମୂଳଦ୍ଵୟ 3 ଓ -5 ହେଲେ, ସମୀକରଣଟି ନିରୂପଣ କର ।
(ii) mx² – 2x + (2m – 1) = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟର ଗୁଣଫଳ 3 ହେଲେ, m ର ମାନ ନିରୂପଣ କର ।
(iii) x² – px + 2 = 0 ସମୀକରଣର ଗୋଟିଏ ମୂଳ 2 ହେଲେ, p ର ମାନ ନିରୂପଣ କର ।
(iv) 4x² – 2x + c = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟ ଏକ ଓ ଅଭିନ୍ନ ହେଲେ, c ର ମାନ ନିରୂପଣ କର ।
(v) 5x² + 2x + k = 0 ସମୀକରଣର ଗୋଟିଏ ମୂଳ – 2 ହେଲେ, k ର ମାନ ନିରୂପଣ କର ।
(vi) x² – kx + 6 = 0 ସମୀକରଣର ଗୋଟିଏ ମୂଳ 3 ହେଲେ, k ର ମାନ ନିରୂପଣ କର ।
(vii) 2x² + kx + 3 = 0 ସମୀକରଣର ଦୁଇଟି ମୂଳ ବାସ୍ତବ ଓ ସମାନ ହେଲେ, k ର ମାନ ନିରୂପଣ କର ।
ସମାଧାନ :
(i) ଗୋଟିଏ ଦ୍ଵିଘାତ ସମୀକରଣର ମୂଳଦ୍ଵୟ 3 ଓ – 5 । ଅର୍ଥାତ୍ α = 3 ଓ ß = – 5
x² – (α + ß) x + αß = 0
⇒ x² – (3 – 5) x + (3)(-5) = 0 ⇒ x² + 2x – 15 = 0

(ii) mx² – 2x + (2m – 1) = 0 ମୂଳଦ୍ଵୟର ଗୁଣଫଳ = 3
ଏଠାରେ a = m, b = – 2, c = 2m – 1
ମୂଳଦ୍ବୟର ଗୁଣଫଳ = \(\frac{c}{a}\) ⇒ 3 \(\frac{2m-1}{m}\)
⇒ 3m = 2m – 1 ⇒ 3m – 2m = 1 ⇒ m = -1
∴ ଦତ୍ତ ସମୀକରଣର ମୂଳଦ୍ଵୟର ଗୁଣଫଳ 3 ହେଲେ m = -1 ହେବ ।

(iii) x² – px + 2 = 0 ସମୀକରଣର ଗୋଟିଏ ମୂଳ 2।
⇒ 2² – p(2) + 2 =0 ⇒ 4 – 2p + 2 = 0 ⇒ 6 = 2p ⇒ p = \(\frac{6}{2}\) = 3
∴ pର ମାନ = 3

(iv) 4x² – 2x + c = ଫିର ମୂଳଦ୍ବୟ ଏକ ଓ ଅଭିନ୍ନ । ଏଠାରେ a = 4, b = – 2, c = c
ପ୍ରଭେଦକ (D) = 0 ⇒ b² – 4ac = 0 ⇒ (-2)² – 4 (4) c = 0 ⇒ 4 – 16c = 0
⇒ -16c = -4 = c = – \(\frac{-4}{16}=\frac{1}{4}\)
∴ c = \(\frac{1}{4}\)
∴ ଦତ୍ତ ସମୀକରଣର ମୂଳଦ୍ଵୟ ଏକ ଏବଂ ଅଭିନ୍ନ ହେଲେ c ର ମାନ \(\frac{1}{4}\) ହେବ।

(v) 5x² + 2x + k = 0 ସମୀକରଣର ଗୋଟିଏ ମୂଳ – 2 ।
⇒ 5(- 2)² + 2(- 2) + k = 0 ⇒ 20 – 4 + k = 0 ⇒ 16 + k = 0 ⇒ k = -16
∴ ‘k’ ର ମାନ -16 ପାଇଁ ଦତ୍ତ ସମୀକରଣର ଗୋଟିଏ ମୂଳ -2 ହେବ ।

(vi) x² – kx + 6 = 0 ସମୀକରଣର ଗୋଟିଏ ମୂଳ 3।
⇒ 3² – 3k + 6 = 0 ⇒ 9 – 3k + 6 = 0 ⇒ -3k = -15 ⇒ \(\frac{-15}{-3}=5\)
∴ ଦତ୍ତ ସମୀକରଣର ଗୋଟିଏ ମୂଳ 3 ହେଲେ k ର ମାନ 5 ହେବ ।

(vii) 2x² + kx + 3 = 0, ଏଠାରେ a = 2, b = k, c = 3
ସମୀକରଣର ମୂଳଦ୍ୱୟ ବାସ୍ତବ ଓ ସମାନ ହେଲେ b² – 4ac = 0
⇒ k² – 4 (2) (3) = 0 ⇒ k² – 24 = 0 ⇒ k² = 24 ⇒ k = ±√24 ⇒ k = ± 2√6
∴ ଦତ୍ତ ସମୀକରଣର ମୂଳଦ୍ଵୟ ବାସ୍ତବ ଓ ସମାନ ହେଲେ k ର ମାନ ±2√6 ହେବ ।

Question 3.
ପ୍ରତ୍ୟେକ ପ୍ରଶ୍ନପାଇଁ ଥ‌ିବା ସମ୍ଭାବ୍ୟ ଉତ୍ତରଗୁଡ଼ିକ ମଧ୍ୟରୁ ଠିକ୍ ଉତ୍ତରଟି ବାଛି ଲେଖ ।
(i) ନିମ୍ନଲିଖ୍ ସମୀକରଣମାନଙ୍କ ମଧ୍ୟରୁ କେଉଁଟି x ରେ ଏକ ଦ୍ଵିଘାତ ସମୀକରଣ ?
(a) x² – x – 12 = 0
(b) x² + 12 = 3
(c) x + = x²
(d)x (x – 1 ) (x + 5) = 0

(ii) 7x² – 9x + 2 = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟର ସ୍ବରୂପ କ’ଣ ?
(a) ବାସ୍ତବ ସଂଖ୍ୟା ଓ ପରସ୍ପରଠାରୁ ପୃଥକ୍
(b) ବାସ୍ତବ ସଂଖ୍ୟା ଏବଂ ଏକ ଓ ଅଭିନ୍ନ
(c) ବାସ୍ତବ ହେବେ ନାହିଁ
(d) ଏଥୁରୁ କେଉଁଟି ନୁହେଁ

(iii) ନିମ୍ନଲିଖ ମଧ୍ୟରୁ କେଉଁଟି – 6 ଓ 8 ମୂଳ ବିଶିଷ୍ଟ ଦ୍ବିଘାତ ସମୀକରଣ ?
(a) (x + 6) (x+8)= 0
(b) (x + 6) (x – 8) = 0
(c) (x – 6) (x + 8) = 0
(d) (x – 6) (x – 8) = 0

(iv) 3x² + 2√5 x – 5 = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟ α ଓ ß ଓ ହେଲେ αß ର ମୂଲ୍ୟ କେତେ ?
(a) 3
(b) 2√5
(c) \(\frac{2√5}{3}\)
(d) \(\frac{-5}{3}\)

(v) 4x – 2x + 2 = 0 ସମୀକରଣର ମୂଳଦ୍ବୟ α ଓ ß ହେଲେ α + ß ର ମୂଲ୍ୟ କେତେ ?
(a) \(\frac{1}{16}\)
(b) 4
(c) \(\frac{1}{2}\)
(d) -8

(vi) 4x + 3x + 7 = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟ α ଓ ß ହେଲେ \(\frac{1}{α}+\frac{1}{ß}\) ର ହେଲେ କେତେ ?
(a) \(\frac{3}{7}\)
(b) \(– \frac{3}{7}\)
(c) \(\frac{7}{3}\)
(d) \(– \frac{7}{3}\)

(vii) ଗୋଟିଏ ଦ୍ବିଘାତ ସମୀକରଣର ମୂଳଦ୍ଵୟର ସମଷ୍ଟି ଓ ଗୁଣଫଳ ଯଥାକ୍ରମେ 4 ଓ \(\frac{5}{2}\) ହେଲେ ନିମ୍ନଲିଖ ମଧ୍ୟରୁ କେଉଁଟି ?
(a) 2x² + 8x + 5 = 0
(b) 2x² – 8x + 5 = 0
(c) 2x² + 8x – 5=0
(d) 2x² – 8x – 5 = 0
ଉ-
(i) x² – x – 12 = 0
(ii) ବାସ୍ତବ ସଂଖ୍ୟା ଓ ପରସ୍ପରଠାରୁ ପୃଥକ୍
(iii) (x +6) (x – 8) = 0
(iv) \(\frac{-5}{3}\)
(v) \(\frac{1}{2}\)
(vi) \(– \frac{3}{7}\)
(vii) 2x² – 8x + 5 = 0

Question 4.
ନିମ୍ନଲିଖୂତ ଦ୍ବିଘାତ ସମୀକରଣମାନଙ୍କୁ ପୂର୍ବ ବର୍ଗରେ ପରିଣତ କରି ସମାଧାନ କର ।
(i) x² + x – 6 = 0
(ii) 2x² – 9x + 4 = 0
(iii) 14x² + x – 3 = 0
(iv) 3x² – 32x + 12=0
(v) x² + 2px – 3qx – 6pq = 0
(vi) √3x² + 10x + 8√3 = 0
(vii) 25x² + 30x + 7 = 0
(viii) 3a²x² + 8abx + 4b² = 0 (a ≠ 0)
(ix) x² + ax + b = 0
(x) x² + bx = a² – ab
ସମାଧାନ :
(i)

(ii)

(iii)

(iv)

(v) x² + 2px – 3qx – 6pq = 0

(vi) √3x² + 10x + 8√3 = 0

(vii) 25x² + 30x + 7 = 0

(viii) 3a²x² + 8abx + 4b² = 0 (a ≠ 0)

(ix) x² + ax + b = 0


(x) x² + bx = a² – ab

Question 5.
ଦ୍ଵିଘାତ ସୂତ୍ର ପ୍ରୟୋଗ କରି ନିମ୍ନଲିଖ ସମୀକରଣମାନଙ୍କର ବୀଜ ବା ମୂଳ ନିରୂପଣ କରି ।
(i) 4x² – 11x + 6 = 0
(ii) (2x – 1)(x – 2) = 0
(iii) x² – (1+√2)x+ 2 = 0
(iv) a(x² + 1) = x(a2 + 1), a ≠ 0
(v) 6x² + 11x + 3 = 0
(vi) 2x² + 41x – 115 = 0
(vii) 12x² +x – 6 = 0
(viii) (6x + 5)(x – 2) = 0
(ix) 15x² – x – 8 = 0
(x) (x + 5)(x – 5) = 39
ସମାଧାନ :
(i) 4x² – 11x + 6 = 0

(ii) (2x – 1)(x – 2) = 0
⇒ 2x² – 4x -x + 2 = 0 ⇒ 2x² – 5x + 2 = 0
ଏଠାରେ a = 2, b = -5, c = 2

(iii) x² – (1+√2)x+ 2 = 0
ଏଠାରେ a = 1, b = – (1+√2), c = √2

(iv) a(x² + 1) = x(a² + 1) ⇒ a(x² + 1) – x(a² + 1) = 0
⇒ ax² + a – x(a² + 1) = 0 ⇒ ax² – x(a² + 1) + a = 0
ଏଠାରେ a = a, b = -(a² + 1), c = a

(v) 6x² + 11x + 3 = 0
ଏଠାରେ a = 6, b = 11, c = 3

(vi) 2x² + 41x – 115 = 0
ଏଠାରେ a = 2, b = 41, c = -115

(vii) 12x² +x – 6 = 0
ଏଠାରେ a = 12, b = 1, c = -6

(viii) (6x + 5)(x – 2) = 0 ⇒ 6x² -12x + 5x – 10 = 0 ⇒ 6x² -7x – 10 = 0
ଏଠାରେ a = 6, b = -7, c = -10

(ix) 15x² – x – 8 = 0
ଏଠାରେ a = 15, b = -1, c = -8

(x) (x + 5)(x – 5) = 39 ⇒ x² – 25 = 39 ⇒ x² – 25 – 39 = 0 ⇒ x² – 64 = 0
ଏଠାରେ a = 1, b = 0, c = -64
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{0 \pm \sqrt{0^2-4(1) 64}}{2(1)}=\frac{0 \pm \sqrt{256}}{2}=\frac{\pm 16}{2}=\pm 8\)
⇒ x = 8 ବ। x = -8
∴ ସମୀକରଣର ମୂଳଦ୍ଵୟ 8 ଓ -8 ।

Question 6.
ଯଦି 4x² – 13x + k = 0 ସମୀକରଣର ଗୋଟିଏ ମୂଳ ଅପରଟିର 12 ଗୁଣ ହେଲେ kରେ ମୂଲ୍ୟ ନିରୂପଣ କର ।
ସମାଧାନ :
ମନେକର 4x² – 13x + k = 0 ସମୀକରଣର ଗୋଟିଏ ମୂଳ । ଏବଂ ଅନ୍ୟଟି 12α ।
ଏଠାରେ a = 4, b = 13 6 c = k
∴ ମୂଳଦ୍ବୟର ସମଷ୍ଟି = \(\frac{-b}{a}\) ⇒ α + 12α = \(\frac{-(-13)}{4}\) ⇒ 13α = \(\frac{13}{4}\) ⇒ α = \(\frac{13}{4}\) × \(\frac{1}{13}\) = \(\frac{1}{4}\)
ମୂଳଦ୍ବୟର ଗୁଣଫଲ = \(\frac{-c}{a}\) ⇒ α.12α = \(\frac{k}{4}\) ⇒ 12α² = \(\frac{k}{4}\) ⇒ 12 × (\(\frac{1}{4}\))² = \(\frac{k}{4}\) ⇒ \(\frac{12}{16}\) = \(\frac{k}{4}\) ⇒ 16k = 48 ⇒ k = 3
∴ kର ମାନ 3 ହେଲେ ସମୀକରଣର ଗୋଟିଏ ମୂଳ ଅପରଟିର 12 ଗୁଣ ହେବ ।

Question 7.
x² – 5x + p = 0 ସମୀକରଣର ଗୋଟିଏ ମୂଳ ଅପରଟି ଅପେକ୍ଷା 3 ଅଧିକ ହେଲେ pର ମୂଲ୍ୟ ନିରୂପଣ କର ।
ସମାଧାନ :
ମନେକର x² – 5x + p = 0 ସମୀକରଣର ଗୋଟିଏ ମୂଳ α ଓ ଅନ୍ୟଟି α + 3
∴ ମୂଳଦ୍ବୟର ସମଷ୍ଟି = α + α + 3 = \(\frac{-(-5)}{1}\)
⇒ 2α + 3 = 5 ⇒ 2α = 5 – 3 = 2 ⇒ α = \(\frac{2}{2}\) = 1
α ର ମାନ ସମୀକରଣରେ ବସାଇଲେ 1² – 5 (1) + p = 0
⇒ 1 – 5 + p = 0 ⇒ p – 4= 0 ⇒ p = 4
∴ p ର ମାନ 4 ହେଲେ ଗୋଟିଏ ମୂଳ ଅପରୂଟି ଅପେକ୍ଷା 3 ଅଧିକ ହେବ ।

Question 8.
ଯଦି 2x² – 5x + 3 = 0 ସମୀକରଣର ମୂଳଦ୍ୱୟ α ଓ ß ହୁଏ ତେବେ α²ß + αß²ର ମୂଲ୍ୟ ନିରୂପଣ ‘କର ।
ସମାଧାନ :
2x² – 5x + 3= 0 ସମୀକରଣର ମୂଳଦ୍ୱୟ α ଓ ß।
ମୂଳଦ୍ଵୟର ସମଷ୍ଟି = α + ß = \(\frac{-(-5)}{2}\) ⇒ α + ß = \(\frac{5}{2}\)
ମୂଳଦ୍ବୟର ଗୁଣଫଲ = αß = \(\frac{3}{2}\)
∴ α²ß + αß² = αß (α + ß)\(\frac{3}{2}\)(\(\frac{5}{2}\)) = \(\frac{15}{4}\)

Question 9.
ଯଦି 2x² – 6x + 3 = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟ α ଓ ß ହୁଏ, ତେବେ (α + 1) (ß + 1) ର ମୂଲ୍ୟ ନିରୂପଣ କର ।
ସମାଧାନ :
2x² – 6x + 3= 0 ସମୀକରଣର ମୂଳଦ୍ୱୟ α ଓ ß।
ମୂଳଦ୍ବୟର ସମଷ୍ଟି = α + ß = \(\frac{-(-6)}{2}\) = \(\frac{6}{2}\) = 3
ମୂଳଦ୍ବୟର ଗୁଣଫଲ = αß = \(\frac{3}{2}\)
∴ (α + 1) (ß + 1) = αß + α + ß + 1 = \(\frac{3}{2}\) + 3 + 1 = \(\frac{3+6+2}{2}\) = \(\frac{11}{2}\)
∴ (α + 1) (ß + 1) = \(\frac{11}{2}\)

Question 10.
ଯଦି 2x² – (p + 1) x + p – 1 = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟର ଅନ୍ତର ଓ ଗୁଣଫଳ ସମାନ ହେଲେ pର ମାନ ନିରୂପଣ କର ।
ସମାଧାନ :
ମନେକର 2x² – (p + 1) x + (p – 1) = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟ α ଓ ß।
ବୀଜଦ୍ଵୟର ସମଷ୍ଟି α + ß = \(\frac{-{-(p+1)}}{2}\) = \(\frac{p+1}{2}\) ଏବଂ αß = \(\frac{p-1}{2}\)
ପ୍ରଶାନୁସାରେ ବୀଜଦ୍ଵୟର ଅନ୍ତର, ବୀଜଦ୍ୱୟର ଗୁଣଫଳ ସହ ସମାନ ।
∴ α – ß = aß ⇒ (α – ß)² = (aß)² ⇒ (α + B)² – 4aß = (αß)²
⇒ (\(\frac{p+1}{2}\))² – 4(\(\frac{p-1}{2}\)) ⇒ (\(\frac{p-1}{2}\))² = (\(\frac{p+1}{2}\))² – (\(\frac{p-1}{2}\))² = 4 (\(\frac{p-1}{2}\))
⇒ p . 1 = 2p – 2 ⇒ p = 2
∴ ଦତ୍ତ ସମୀକରଣର ମୂଳଦ୍ଵୟର ଅନ୍ତର ଓ ଗୁଣଫଳ ସମାନ ହେଲେ p ର ମାନ 2 ହେବ ।

Question 11.
ଯଦି 5x² – 3x – 2 = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟ α ଓ ß ହୁଏ, ତେବେ ପ୍ରମାଣ କର ଯେ α³ + ß³ = \(\frac{117}{125}\)
ସମାଧାନ :
5x²-3x-2=0 ସମୀକରଣର ମୂଳଦ୍ଵୟ α ଓ ß।
ମୂଳଦ୍ବୟର ସମଷ୍ଟି = α + ß = \(\frac{-(-3)}{5}\) = \(\frac{3}{5}\), ମୂଳଦ୍ବୟର ଗୁଣଫଲ = αß = \(\frac{-2}{5}\)
∴ α³ + ß³ = (α + ẞ)³ – 3αß (α + B)
= (\(\frac{3}{5}\))³ – 3(\(\frac{3}{5}\))(\(\frac{-2}{5}\)) = \(\frac{27}{125}+\frac{18}{25}\) = \(\frac{27+90}{125}\) = \(\frac{117}{125}\)
∴ α³ + ß³ = \(\frac{117}{125}\) (ପ୍ରମାଣିତ)

Question 12.
ଯଦି 5x² + 17x + 6 = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟ α ଓ ß ହୁଏ ତେବେ \(\frac{1}{α²}+\frac{1}{ß²}\) ର ମୂଲ୍ୟ ନିରୂପଣ କର ।
ସମାଧାନ :
5x² + 17x + 6 = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟ α ଓ ß।
∴ ମୂଳଦ୍ବୟର ସମଷ୍ଟି = α + ß = \(\frac{-17}{5}\), ମୂଳଦ୍ବୟର ଗୁଣଫଲ = αß = \(\frac{6}{5}\)
\(\frac{1}{α²}+\frac{1}{ß²}\) = \(\frac{ß²+α²}{α²ß²}\) = \(\frac{(α+ß)²-2αẞ}{(αß)²}\)
=\(\frac{\left(\frac{-17}{5}\right)^2-2\left(\frac{6}{5}\right)}{\left(\frac{6}{5}\right)^2}=\frac{\frac{289}{25}-\frac{12}{5}}{\frac{36}{25}}\) = \(\frac{289-60}{25}\) × \(\frac{25}{36}\) = \(\frac{229}{36}\)
∴ \(\frac{1}{α²}+\frac{1}{ß²}\) ର ମୂଲ୍ୟ \(\frac{229}{36}\) ।

Question 13.
x² – 8x + 16p = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟ α ଓ ß ହେଲେ \(\frac{αß}{α + ß}\) ପରିପ୍ରକାଶକୁ p ମାଧ୍ୟମରେ ପ୍ରକାଶ କର ।
ସମାଧାନ :
x² – 8x + 16p = 0 ମୂଳ ଦ୍ଵୟ α ଓ ß।
∴ ମୂଳଦ୍ବୟର ସମଷ୍ଟି = α + ß = \(\frac{-(-8)}{1}=8\), ମୂଳଦ୍ବୟର ଗୁଣଫଲ = αß = \(\frac{16p}{1}=16p\)
∴ \(\frac{αß}{α + ß}\) = \(\frac{16p}{8}\) = 2p

Question 14.
ଯଦି x² – 2 (5 + 2m) x + 3 (7 + 10m) = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟ ଏକ ଓ ଅଭିନ୍ନ ହୁଏ, m ର ମୂଲ୍ୟ ନିରୂପଣ କର ।
ସମାଧାନ :
x² – 2 (5 + 2m) x + 3 (7 + 10m) = 0 ସମୀକରଣର ମୂଳଦ୍ବୟ ଏକ ଓ ଅଭିନ୍ନ ।
ପ୍ରଭେଦକ (D) = 0
ଏଠାରେ a = 1, b = – 2 (5+ 2m), c = 3 (7 + 10m)
D = b² – 4ac = 0
⇒ {-2(5+ 2m)}² – 4 × 1 × 3 (7 + 10m) = 0
⇒ 4 (25 + 20m + 4m²) – 4 × 3 (7 + 10m) = 0
⇒ 100+ 80m + 16m² – 12 (7 + 10m) = 0
⇒ 16m² + 80m + 100 – 84 – 120m = 0 16m² – 40m + 16 = 0 ⇒ 2m² – 5m+2=0 2m² – 4m – m + 2=0
⇒ 2m (m – 2) – 1 (m – 2) = 0 ⇒ (m – 2) (2m – 1) = 0
m – 2 = 0 | 2m – 10 ⇒ m = 2 | m =
∴ m ର ମାନ 2 କିମ୍ବା \(\frac{1}{2}\) ହେଲେ x² – 2 (5 + 2m) x + 3 (7 + 10m) = 0 ସମୀକରଣର ମୂଳଦ୍ବୟ ଏକ ଓ ଅଭିନ୍ନ ହେବେ ।

Question 15.
(i) ଯଦି a = b = c ହୁଏ, ତେବେ ପ୍ରମାଣ କର ଯେ, ସମୀକରଣ
(x – a)(x – b) + (x – b)(x – c) + (x – c)(x – a) = 0 ର ମୂଳଦ୍ୱୟ ବାସ୍ତବ ଏବଂ ଏକ ଓ ଅଭିନ୍ନ ।
(ii) ଯଦି a + b + c = 0 ଏବଂ a, b, c ∈ Q ହୁଏ, ତେବେ ପ୍ରମାଣ କର ଯେ
(b + c – a) x² + (c + a – b) x + (a + b – c) = 0 ସମୀକରଣର ପରିମେୟ ହେବେ ।
ସମାଧାନ :
(i) a = b = c ହେଲେ, (x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0
⇒ (x – a)² + (x – a)² + (x – a)² = 0
⇒ 3 (x – a)² = 0 ⇒ (x – a)² = 0 ⇒ x² – 2ax + a² = 0
ଏଠାରେ ପ୍ରଭେଦକ (D) = b² – 4ac = (-2a)² – 4(1) a² = 4a² – 4a² = 0
∴ D = 0 ହେଲେ, ମୂଳଦ୍ୱୟ ବାସ୍ତବ ଏବଂ ଏକ ଓ ଅଭିନ୍ନ ।

(ii) ଯଦି a+b+c=0, a, b, c ∈ Q
(b + c – a) x² + (c + a – b) x + (a + b – c) = 0
a + b + c = 0
⇒ b + c = a, c + a = – b, a + b = -c
(b + c – a) x² + (c + a – b) x + (a + b – c) = 0
⇒ -2ax² + (-2b) x + (- 2c) = 0 ⇒ 2 (ax² + bx + c) = 0
⇒ ax² + bx + c = 0
ପ୍ରଭେଦକ (D) = b² – 4ac = (-b)² – 4ac = (a + c)² – 4ac = (a – c)²

Question 16.
ଗୋଟିଏ ଦ୍ଵିଘାତ ସମୀକରଣର ମୂଳଦ୍ଵୟର ସମଷ୍ଟି 3 ଓ ମୂଳଦ୍ବୟର ବର୍ଗର ସମଷ୍ଟି 29 ହେଲେ, ସମୀକରଣଟି ନିରୂପଣ କର ।
ସମାଧାନ :
ମନେକର ସ୍ଵିଘାତ ସମୀକରଣର ମୂଳଦ୍ଵୟ α ଓ ß ।
ମୂଳଦ୍ବୟର ସମଷ୍ଟି = α + ß = 3
ମୂଳଦ୍ଵୟର ବର୍ଗର ସମଷ୍ଟି = α² + ß² = 29
⇒ (α + ß)² – 2 aẞ = 29 ⇒ 3² – 2αß = 29
⇒9 – 2αẞ = 29 ⇒ – 2αß = 29 – 9 = 20
αß = \(\frac{-20}{2}\) = -10
ସମୀକରଣଟି x² – (a + ẞ) x + αß = 0
⇒ x² – 3x + (-10) = 0 ⇒ x² – 3x – 10 = 0

Question 17.
ଯଦି 2x² – 4x + 2 = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟ α ଓ ß ହୁଏ, ତେବେ ପ୍ରମାଣ କର ଯେ \(\frac{α}{ß}+\frac{ß}{α}\) + 4(\(\frac{1}{α}+\frac{1}{ß}\)) + 2αß = 12
ସମାଧାନ :
2x² – 4x + 2 = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟ α ଓ ß ।
∴ ମୂଳଦ୍ଵୟର ସମଷ୍ଟି α + ß = \(\frac{-(-4)}{2}\) = -2 ଏବଂ ମୂଳଦ୍ଵୟର ଗୁଣଫଲ αβ = \(\frac{2}{2}\) = 1
L.H.S. = \(\frac{α}{ß}+\frac{ß}{α}\) + 63(\(\frac{1}{α}+\frac{1}{ß}\)) – 2αß
= \(\frac{ß²+α²}{αß}\) + 63\(\frac{(α+ß)}{(αß)}\) – 2αß
= \(\frac{(α+ß)²-2αß}{(αß)}\) + 4\(\frac{(α+ß)}{(αß)}\) + 2αß
= \(\frac{(2)²-2×1}{1}\) + \(\frac{4(2)}{1}\) + 2 × 1
= 2 + 8 + 2 = 12 = R.H.S.

Question 18.
(i) ଯଦି ax² + bx + c = 0 ସମୀକରଣର ଗୋଟିଏ ମୂଳ ଅପରଟିର 4 ଗୁଣ ହୁଏ, ତେବେ ପ୍ରମାଣ କର ଯେ 4b² = 25ac।
(ii) ଯଦି x² – px + q = 0 ସମୀକରଣର ଗୋଟିଏ ମୂଳ ଅପରଟିର 2 ଗୁଣ ହୁଏ, ତେବେ ପ୍ରମାଣ କର ଯେ 2p² = 9q ।
ସମାଧାନ :
(i) ମନେକର ax² + bx + c = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟ α ଏବଂ 4α।
∴ ମୂଳଦ୍ଵୟର ସମଷ୍ଟି α + 4α = \(\frac{-b}{a}\) ⇒ 5α = \(\frac{-b}{a}\) ⇒ α= \(\frac{-b}{5a}\) .. (i)
ମୂଳଦ୍ଵୟର ଗୁଣଫଲ α.4α = \(\frac{c}{2}\)
⇒ \(4 α^2=\frac{c}{a} \Rightarrow 4 \times (\frac{-b}{5a})^2=\frac{c}{a}\) [(i)]
⇒ \(\frac{4b²}{25a²}=\frac{c}{a} \Rightarrow 4ab²=25a²c \Rightarrow 4b²=25 ac\) (ପ୍ରମାଣିତ)

(ii) ମନେକର x² – px + q = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟ α ଏବଂ 2α।
∴ ମୂଳଦ୍ଵୟର ସମଷ୍ଟି α + 2α = \(\frac{-(-p)}{1}\) ⇒ 3α = p ⇒ α= \(\frac{p}{3}\) .. (i)
ମୂଳଦ୍ଵୟର ଗୁଣଫଲ α.2α = \(\frac{q}{1}\) = 2α² = q
⇒ 2 × (\(\frac{p}{3}\))³ = q ⇒ \(\frac{2p²}{3}\) = q ⇒ 2p² = 9q (ପ୍ରମାଣିତ)

Question 19.
(i) ଯଦି 41x² – 2 (5a + 4b) x + (a² + b²) = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟ ସମାନ ହୁଅନ୍ତି, ତେବେ ପ୍ରମାଣ କର ଯେ, \(\frac{a}{b}=\frac{5}{4}\)।
(ii) ଯଦି x² + px + q = 0 ସମୀକରଣର ବୀଜଦ୍ଵୟର ସମଷ୍ଟି ସେମାନଙ୍କର ବର୍ଗର ସମଷ୍ଟି ସହ ସମାନ ଯେ, 2q = p (p + 1)।
(ii) ଯଦି x² + px + q = 0 ସମୀକରଣର ଗୋଟିଏ ବୀଜ ଅନ୍ୟଟିର ବର୍ଗ ହୁଏ, ତେବେ ଦର୍ଶାଅ ଯେ p³ + q² + q = 3pq।
ସମାଧାନ :
(i) 41x² – 2 (5a + 4b) x + (a² + b²) = 0 ସମୀକରଣର ମୂଳଦ୍ଵୟ ସମାନ।
ଏଠାରେ a = 41, b = -2(5a + 4b), c = (a² + b²)
ପ୍ରଭେଦକ (D) = b² – 4ac = 0
⇒ {-2 (5a + 4b)}² – 4 (41) (a² + b²) = 0
⇒ 4 (25a² + 40ab + 16b²) – 4 (41a² + 41b²) = 0
⇒(25a² + 40ab + 16b² – 41a² – 41b²) = 0
⇒(-16a² + 40ab – 25b²) = 0
⇒(16a² – 40ab + 25b²) = 0
⇒(4a)² – 2.4a.5b + (5b)² = 0
⇒ (4a – 5b)² = 0
⇒4a – 5b = 0 [∴ a² – 2ab + b² = (a – b)²]
⇒ 4a = 5b ⇒ \(\frac{a}{b}=\frac{5}{4}\)।

(ii) x² + px + q = 0 ସମୀକରଣର ବୀଜଦ୍ଵୟର ସମଷ୍ଟି ସେମାନଙ୍କର ବର୍ଗର ସମଷ୍ଟି ସହ ସମାନ ।
ମନେକର ସମୀକରଣର ବୀଜଦ୍ଵୟ α ଓ ß।
ବୀଜଦ୍ୱୟର ସମଷ୍ଟି α + ß = – p ଏବଂ ବୀଜଦ୍ବୟର ଗୁଣଫଳ αß = q
କିନ୍ତୁ ଦତ୍ତ ଅଛି α + ß = α² + ß²
⇒ α + ß = α² + ß² – 2αß ⇒ – p = (- p)² – 2q
⇒ 2q = p² + p ⇒ 2q = p (p + 1) (ପ୍ରମାଣିତ)

(iii) x + px + g = 0 ସମୀକରଣର ଗୋଟିଏ ବୀଜ ଅପରଟିର ବର୍ଗ ।
ମନେକର ସମୀକରଣର ବୀଜଦ୍ଵୟ α ଓ α²।
ବୀଜଦ୍ୱୟର ସମଷ୍ଟି α + α² = – p
ବୀଜଦ୍ବୟର ଗୁଣଫଳ α.α² = q ⇒ α³ = q
α + α² = – p = (α + α²)³ = -p³
⇒ a³ + α⁶ + 3α · a² (α + α²) = – p³ ⇒ q + q² + 3q (- p) = – p³
⇒ q² + q – 3pq = -p³ ⇒ p³ + q² + q = 3pq (ପ୍ରମାଣିତ)

Question 20.
ଯଦି a(b – c) x² + b(c – a) x + c (a – b) = 0 ସମୀକରଣର ବୀଜଦ୍ଵୟ ସମାନ ହୁଏ, ତେବେ ଦର୍ଶାଅ ଯେ, \(\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\)
ସମାଧାନ :
a(b – c) x² + b(c – a) x + c (a – b) = 0 ସମୀକରଣର ବୀଜଦ୍ଵୟ ସମାନ ।
ସମୀକରଣର ପ୍ରଭଦେକ (D) = 0
⇒ {b (c – a)}² – 4a (bc) c (a – b) = 0
⇒ b² (ca)² – 4a (b -c) c (a – b) = 0
⇒(bc – ab)² – 4 a(b – c) · c (a – b) = 0
⇒ (ab – ca + ca – bc)² – 4a (b- c) · c (a – b) = 0
⇒ {a (b-c) + c (a – b)}² – 4a (b – c) c (a – b) = 0
⇒ {a (bc) -c (a – b)}² = 0 ⇒ (ab – ac- ca + bc) = 0
⇒ ab + bc = 2 ca ⇒ \(\frac{ab}{abc}+\frac{bc}{abc}=\frac{2ca}{abc}\) (abc ଦ୍ବାରା ଉଭୟ ପାର୍ଶ୍ଵକୁ ଭାଗ କଲେ)
⇒ \(\frac{1}{c}+\frac{1}{a}=\frac{2}{b}\) (ପ୍ରମାଣିତ)

Odisha State Board BSE Odisha 10th Class Maths Solutions Algebra Chapter 1 ସରଳ ସହସମୀକରଣ Ex 1(b) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Algebra Chapter 1 ସରଳ ସହସମୀକରଣ Ex 1(b)

Question 1.
ପ୍ରତିକଳ୍ପନ ପ୍ରଣାଳୀରେ ନିମ୍ନଲିଖ୍ ସହସମୀକରଣ ଦ୍ଵୟର ସମାଧାନ କର ।
(i) x + y – 8 = 0, 2x – 3y – 1 = 0
(ii) 3x + 2y – 5 = 0, x – 3y – 9 = 0
(iii) 2x – 5y + 8 = 0, x – 4y + 7 = 0
(iv) 11x + 15y + 23 = 0, 7x – 2y – 20 = 0
(v) ax + by – a + b = 0, bx – ay – a – b = 0
(vi) x + y – a = 0, ax + by – b² = 0
ସମାଧାନ ପ୍ରଣାଳୀ :
(i) ସହସମୀକରଣଦ୍ଵୟ ମଧ୍ୟରୁ ଗୋଟିକରୁ ‘x’ କିମ୍ବା ‘y’ର ମାନ ନେଇ ଯଥାକ୍ରମେ y କିମ୍ବା x ମାଧ୍ୟମରେ ପ୍ରକାଶ କରାଯାଏ ।
(ii) x କିମ୍ବା yର ମାନକୁ ଅନ୍ୟ ସମୀକରଣରେ ପ୍ରୟୋଗ କରି ଯଥାକ୍ରମେ y କିମ୍ବା x ର ମାନ ନିର୍ଣ୍ଣୟ କରାଯାଉ ।
(iii) ଉକ୍ତ ନିର୍ଣ୍ଣୟ ମାନକୁ (y କିମ୍ବା x) ନେଇ ଯେକୌଣସି ଗୋଟିଏ ସମୀକରଣରେ ପ୍ରୟୋଗ କରି ଅନ୍ୟଟିର ମାନ ନିର୍ଣ୍ଣୟ କରାଯାଏ ।
ସମାଧାନ :
(i) x + y – 8 = 0 …….(i) ଏବଂ
2x – 3y – 1 = 0 …….(ii)
ସମୀକରଣ (i)କୁ ବିଚାର କରି yକୁ x ମାଧ୍ୟମରେ ପ୍ରକାଶ କରାଯାଉ ।
x + y – 8 = 0 ⇒ y = 8 – x ……..(iii)
y ର ମାନକୁ ସମୀକରଣ (ii) ରେ ପ୍ରୟୋଗ କଲେ ପାଇବା 2x – 3(8 – x) – 1 = 0
⇒ 2x – 24 + 3x – 1 = 0 ⇒ 5x = 25 ⇒ x = \(\frac{25}{5}\) = 5
x ର ମାନ ସମୀକରଣ (iii)ରେ ପ୍ରୟୋଗ କଲେ, y = 8 – x = 8 – 5 = 3
∴ ନିର୍ଦେୟ ସମୀକରଣଦ୍ଵୟର ସମାଧାନ (x, y) = (5, 3) ଅଟେ ।

(ii) 3x + 2y – 5 = 0 …….(i) ଏବଂ
x – 3y – 9 = 0 …….(ii)
ସମୀକରଣ (i) ରୁ 3x = 5 – 2y ⇒ x = \(\frac{5-2y}{3}\) …….(iii)
‘x’ ର ମାନ ସମୀକରଣ (ii) ରେ ପ୍ରୟୋଗ କଲେ, \(\frac{5-2y}{3}\) – 3y – 9 = 0
⇒ \(\frac{5-2y-9y-27}{3}\) ⇒ 0 = -22 – 11y = 0
⇒ 11y = -22 = y = \(\frac{-22}{11}\) = -2
yର ମାନ ସମୀକରଣ (iii)ରେ ପ୍ରୟୋଗ କଲେ,
x = \(\frac{5-2y}{3}\) = \(\frac{5-2(-2)}{3}\) = \(\frac{5+4}{3}\) = \(\frac{9}{3}=3\)
∴ ନିର୍ଦେୟ ସମୀକରଣଦ୍ଵୟର ସମାଧାନ (x, y) = (3, -2) ଅଟେ ।

(iii) 2x – 5y + 8 = 0 …….(i) ଏବଂ
x – 4y + 7 = 0 …….(ii)
ସମୀକରଣ (i) ରୁ ବିଚ।ର କରି x କୁ y ମାଧ୍ୟମରେ ପ୍ରକାଶ କରାଯାଉ ।
2x – 5y + 8 = 0 ⇒ 2x = 5y – 8 ⇒ x = \(\frac{1}{2}\)(5y – 8) …….(iii)
‘x’ ର ମାନ ସମୀକରଣ (ii) ରେ ପ୍ରୟୋଗ କଲେ, \(\frac{1}{2}\)(5y – 8) – 4y + 7 = 0
⇒ \(\frac{5y-8-8y+14}{2}\) = 0
⇒ -3y + 6 = 0 ⇒ y = \(\frac{-6}{-3}\) = 2
yର ମାନ ସମୀକରଣ (iii)ରେ ପ୍ରୟୋଗ କଲେ, x = \(\frac{1}{2}\)(5 × 2 – 8) = \(\frac{1}{2}\) × 2 = 1
∴ ନିର୍ଦେୟ ସମୀକରଣଦ୍ଵୟର ସମାଧାନ (x, y) = (1, 2) ଅଟେ ।

(iv) 11x + 15y + 23 = 0 …….(i) ଏବଂ
7x – 2y – 20 = 0 …….(ii)
ସମୀକରଣ (i) ରୁ ବିଚ।ର କରି y କୁ x ମାଧ୍ୟମରେ ପ୍ରକାଶ କରାଯାଉ ।
11x + 15y + 23 = 0 ⇒ 15y = -11x – 23
⇒ y = \(\frac{1}{15}\)(-11x – 23) …….(iii)
‘y’ ର ମାନ ସମୀକରଣ (ii) ରେ ପ୍ରୟୋଗ କଲେ, 7x – \(\frac{2}{15}\)(-11x – 23) = 20
⇒ \(\frac{105x+22x+46}{15}\) = 20
⇒ 127x = 300 – 46 ⇒ x = \(\frac{254}{127}\) = 2
x ର ମାନକୁ ସମୀକରଣ (iii) ରେ ପ୍ରୟୋଗ କଲେ, y = latex]\frac{1}{15}[/latex](-22 – 23) = \(\frac{1}{15}\) × -45 = -3
∴ ନିର୍ଦେୟ ସମାଧାନ (x, y) = (2, -3) ।

(v) ax + by – a + b = 0 …….(i) ଏବଂ
bx – ay – a – b = 0 …….(ii)
ସମୀକରଣ (i) ରୁ ବିଚ।ର କରି y କୁ x ମାଧ୍ୟମରେ ପ୍ରକାଶ କରାଯାଉ ।
ax + by – a + b = 0 ⇒ by = -ax + a – b
⇒ y = \(\frac{1}{b}\)(-ax + a – b) …….(iii)
y ର ମାନ ସମୀକରଣ (ii) ରେ ପ୍ରୟୋଗ କଲେ, bx – \(\frac{a}{b}\)(-ax + a – b) = 0
⇒ \(\frac{b^2x+a^2x-a^2 +ab-ab-b^2}{b}\) = 0
⇒ x(a² + b²) = a² + b² ⇒ x = \(\frac{a^2+b^2}{a^2+b^2}\) = 1
x ର ମାନକୁ ସମୀକରଣ (iii) ରେ ପ୍ରୟୋଗ କଲେ, y = latex]\frac{1}{b}[/latex](-a + a – b) ⇒ y = \(\frac{-b}{b}\) = -1
∴ ନିର୍ଦେୟ ସମାଧାନ (x, y) = (1, -1) ଅଟେ।

(vi) x + y – a = 0 …….(i) ଏବଂ
ax + by – b² = 0 …….(ii)
ସମୀକରଣ (i) ରୁ ବିଚ।ର କରି y କୁ x ମାଧ୍ୟମରେ ପ୍ରକାଶ କରାଯାଉ ।
x + y – a = 0 ⇒ y = a – x …….(iii)
y ର ମାନ ସମୀକରଣ (ii) ରେ ପ୍ରୟୋଗ କଲେ, ax + b(a – x) – b² = 0
⇒ ax + by – bx – b² = 0 ⇒ ax – bx = b² – ab
⇒ x(a – b) = -b(a – b) [a = b ତେଣୁ a – b ≠ 0]
ତେଣୁ x = -b
‘x’ ର ମାନକୁ ସମୀକରଣ (iii) ରେ ପ୍ରୟୋଗ କଲେ, y = a + b
∴ ନିର୍ଦେୟ ସମାଧାନ (x, y) = (-b, a+b)।

Question 2.
ଅପସାରଣ ପ୍ରଣାଳୀରେ ନିମ୍ନଲିଖ ସହ ସମୀକରଣମାନଙ୍କର ସମାଧାନ କର ।
(i) x – y – 3 = 0, 3x – 2y – 1 = 0
(ii) 3x + 4y = 10, 2x – 2y = 2
(iii) 3x – 5y – 4 = 0, 9x = 2y – 1
(iv) 0.4x – 1.5y = 6.5, 0.3x + 0.2y = 0.9
(v) √2x + √3y = 0, √5x + √2y = 0
(vi) ax + by = 0, x + y – c = 0 (a+b ≠ 0)
ସମାଧାନ ପ୍ରଣାଳୀ :
(i) ସମୀକରଣଦ୍ବୟରୁ ‘x’ ଅପସାରଣ କରାଯାଇ y ର ମାନ ନିର୍ଣ୍ଣୟ କରାଯାଏ ।
(ii) y ର ମାନକୁ ଯେକୌଣସି ସମୀକରଣରେ ପ୍ରୟୋଗ କରି x ର ମାନ ନିର୍ଣ୍ଣୟ କରାଯାଏ ।
(iii) ସେହିପରି ସମୀକରଣଦ୍ଵୟରୁ yକୁ ଅପସାରଣ କରାଯାଇ ‘x’ର ମାନ ନିର୍ଣ୍ଣୟ କରି ଏହାକୁ ଯେକୌଣସି ସମୀକରଣରେ ପ୍ରୟୋଗ କରି ‘y’ର ମାନ ନିର୍ଣ୍ଣୟ କରାଯାଇ ପାରିବ ।
ସମାଧାନ :
(i) x – y – 3 = 0 ……… (i) ଏବଂ
3x – 2y – 1 = 0 ………. (ii)

x ର ମାନକୁ ସମୀକରଣ (iii) ରେ ପ୍ରୟୋଗ କଲେ,
x – y – 3 = 0 ⇒ 4 – y – 3 = 0 ⇒ -y + 1 = 0 ⇒ y = 1
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (4, 1) ଅଟେ।

(ii) 3x + 4y = 10 ……… (i) ଏବଂ
2x – 2y = 2 ………. (ii)

x ର ମାନକୁ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ,
3x + 4y = 10 ⇒ 2 + 4y = 10 ⇒ 4y = 10 – 6 ⇒ y = \(\frac{4}{4}\) = 1
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (2, 1) ଅଟେ।

(iii) 3x – 5y – 4 = 0 ……… (i) ଓ
9x = 2y – 1 ⇒ 9x – 2y + 1 = 0 ………. (ii)

x ର ମାନକୁ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ,
3x – 5y – 4 = 0
⇒ 3 × (\(– \frac{1}{3}\)) – 5y – 4 = 0
⇒ -5y – 5 = 0 ⇒ y = \(\frac{5}{-5}\) = -1
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (\(– \frac{1}{3}\), -1) ଅଟେ।

(iv) 0.4x – 1.5y = 6.5 ……… (i) ଓ
0.3x + 0.2y = 0.9 ………. (ii)

y ର ମାନକୁ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ,
0.4x – 1.5y = 6.5 ⇒ 0.4x + 4.5 = 6.5
⇒ 0.4x = 6.5 – 4.5 ⇒ 1.4x = 2 ⇒ x = \(\frac{2}{0.4}\) = 5
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (5, -3) ଅଟେ।

(v) √2x + √3y = 0 ……… (i) ଓ
√5x + √2y = 0 ………. (ii)

c₁ = c₂ = 0 ଓ a₁b₂ – a₂b₁ = 0 ହେଲେ ସମୀକରଣଦ୍ଵୟର ସମାଧାନଟି (0, 0) ଅଟେ ।
ଏଠାରେ c₁ = c₂ = 0 ଏବଂ \(\frac{\sqrt{2}}{\sqrt{5}} \neq \frac{\sqrt{3}}{\sqrt{2}}\)
ତେଣୁ ସହସମୀକରଣ ଦ୍ଵୟର ସମାଧାନ (0, 0) ଅଟେ ।
ବିକଳ୍ପ ପ୍ରଣାଳୀ :

‘y’ ର ମାନକୁ ସମୀକରଣ (i)ରେ ପ୍ରୟୋଗ କଲେ,
√2x + √3 × 0 = 0 ⇒ √2x = 0 ⇒ x = 0
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (0, 0) ଅଟେ।

(v) ax + by = 0 ……… (i) ଏବଂ
x + y – c = 0 ………. (ii)

‘x’ ର ମାନକୁ ସମୀକରଣ (ii)ରେ ପ୍ରୟୋଗ କଲେ, y = c – x = c – \(\frac{bc}{b-a}\) =
= \(\frac{bc-ca-bc}{b-a}\) = \(\frac{-ca}{b-a}\) = \(\frac{ca}{a-b}\)
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (\(\frac{bc}{b-a}\), \(\frac{ca}{a-b}\)) ଅଟେ।

Question 3.
ବଜ୍ରଗୁଣନ ପ୍ରଣାଳୀରେ ନିମ୍ନଲିଖୂତ ସହ ସମୀକରଣମାନଙ୍କର ସମାଧାନ କର ।
(1) x + 2y + 1 = 0, 2x – 3y – 12 = 0
(ii) 2x + 5y = 1, 2x + 3y = 3
(iii) x + 6y + 1 = 0, 2x + 3y + 8 = 0
(iv) \(\frac{x}{a}+\frac{y}{b}\) = a+b, \(\frac{x}{a^2}+\frac{y}{b^2}\) = 2
(v) x + 6y + 1 = 0, 2x + 3y + 8 = 0
(vi) 4x – 9y = 0, 3x + 2y – 35 = 0
ବଜ୍ରଗୁଣନ ପୃତ୍ର : \(\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{x}{a_1b_2-a_2b_1}\)
ଯେଉଁଠାରେ a₁b₂ = a₂b₁ ≠ 0
ସମାଧାନ :
(i) x + 2y + 1 = 0 ……… (i) ଏବଂ
2x – 3y – 12 = 0 ………. (ii)
ସମୀକରଣମାନଙ୍କର a₁b₂ – a₂b₁ = (1)(-3) – (2)(2) = -7 ≠ 0
ତେଣୁ ସମାଧାନ ସମ୍ଭବ।
ବଜ୍ରଗୁଣନ ପ୍ରଣାଳୀ ଅବଲମୂନରେ।

⇒ \(\frac{x}{-21}=\frac{y}{14}=\frac{1}{-7}\)
⇒ x = \(\frac{-21}{-7}\) ଓ y = \(\frac{14}{-7}\)
⇒ x = 3 ଓ y = -2
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (3, -2) ଅଟେ।

(ii) 2x + 5y = 1 ……… (i) ଏବଂ
2x + 3y = 3 ………. (ii)
ସମୀକରଣ (i) ଓ (ii) ଦ୍ବୟରୁ 2x + 5y = 1, 2x + 3y = 3
ସମୀକରଣମାନଙ୍କର a₁b₂ – a₂b₁ = (2)(3) – (2)(5) = -4 ≠ 0
ତେଣୁ ସହସମୀକରଣ ଦ୍ଵୟର ସମାଧାନ ସମ୍ଭବ ।
ବଜ୍ରଗୁଣନ ପ୍ରଣାଳୀ ଅବଲମୂନରେ।

⇒ \(\frac{x}{15-(-3)}=\frac{y}{-2-(-6)}=\frac{1}{6-10}\)
⇒ \(\frac{x}{15+3}=\frac{y}{-2+6}=\frac{1}{6-10}\)
⇒ \(\frac{x}{-12}=\frac{y}{4}=\frac{1}{-4}\)
⇒ x = \(\frac{-12}{-4}=3\) ଓ y = \(\frac{4}{-4}=-1\)
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (3, -2) ଅଟେ।

(iii) x + 6y + 1 = 0 ……… (i) ଏବଂ
2x + 3y + 8 = 0 ………. (ii)
ଏଠାରେ a₁b₂ – a₂b₁ = (1)(3) – (2)(6) = 3- 12 = -9 ≠ 0
ତେଣୁ ଦତ୍ତ ସହସମୀକରଣ ଦ୍ଵୟର ସମାଧାନ ସମ୍ଭବ ।
ବଜ୍ରଗୁଣନ ପ୍ରଣାଳୀ ଅବଲମୂନରେ

⇒ \(\frac{x}{48-3}=\frac{y}{2-86}=\frac{1}{3-12}\)
⇒ \(\frac{x}{45}=\frac{y}{-6}=\frac{1}{-9}\)
⇒ x = \(\frac{-45}{-9}=-5\) ଓ y = \(\frac{-6}{-9}=\frac{2}{3}\)
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (-5, \(\frac{2}{3}\)) ଅଟେ।

(iv) \(\frac{x}{a}+\frac{y}{b}\) = a+b ……… (i)
\(\frac{x}{a^2}+\frac{y}{b^2}=2\) ………. (ii)
ସମୀକରଣ (i) ଓ (ii) ରୁ \(\frac{x}{a}+\frac{y}{b}-(a+b)=0\), \(\frac{x}{a^2}+\frac{y}{b^2}-2=0\)

∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (a², b²) ଅଟେ।

(v) x + 6y + 1 = 0 ……… (i)
2x + 3y + 8 = 0 ………. (ii)
ଏଠାରେ a₁ = 1
b₁ = 6
c₁ = 1
a₂ = 5
b₂ = 3
c₂ = 8
ପୁନଶ୍ଚ \({a_1}{a_2}=\frac{1}{2}\), \({b_1}{b_2}=\frac{6}{3}=2\)
\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), ତେଣୁ ସହ-ସମୀକରଣଦ୍ୱୟର ଅନନ୍ୟ ସମାଧାନ ରହିବ ।
b₁c₂ – b₂c₁ = 6 × 8 – 3 × 1 = 48 – 3 = 45
c₁a₂ – c₂a₁ = 1 × 2 – 8 × 1 = 2 – 8 = -6
a₁b₂ – a₂b₁ = 1 × 3 – 2 × 6 = 3 – 12 = -9

⇒ \(\frac{x}{45}=\frac{y}{-6}=\frac{1}{-9}\)
⇒ x = \(\frac{-45}{-9}=-5\) ଓ y = \(\frac{-6}{-9}=\frac{2}{3}\)
∴ ସମାଧାନ (x, y) = (-5, \(\frac{2}{3}\) ) ।

(vi) 4x – 9y = 0 ……… (i)
3x + 2y – 35 = 0 ………. (ii)
ଏଠାରେ a₁ = 4
b₁ = -9
c₁ = 0
a₂ = 3
b₂ = 2
c₂ = -35
ତେଣୁ ସହ-ସମୀକରଣଦ୍ୱୟର ଅନନ୍ୟ ସମାଧାନ ରହିବ ; \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) (\(\frac{4}{3} \neq \frac{-9}{2}\))
b₁c₂ – b₂c₁ = (-9)(-35) – 2 × 0 = 315
c₁a₂ – c₂a₁ = 0 × 3 – (-35) × 4 = 140
a₁b₂ – a₂b₁ = 4 × 2 – 3 × (-9) = 35 ≠ 0

⇒ \(\frac{x}{315}=\frac{y}{140}=\frac{1}{35}\)
⇒ \(\frac{x}{9}=\frac{y}{4}=1\) ⇒ x = 9 ଓ y = 4
∴ ସମାଧାନ (x, y) = (9, 4) ।

Question 4.
ନିମ୍ନଲିଖ୍ ସହସମୀକରଣମାନଙ୍କ ସମାଧାନ କର ।
(i) \(\frac{2}{x}+\frac{3}{y}=17, \frac{1}{x}+\frac{1}{y}=7(x \neq 0, y \neq 0)\)
(ii) \(\frac{5}{x}+6 y=13, \frac{3}{x}+20 y=35(x \neq 0)\)
(iii) \(2 x-\frac{3}{y}=9,3 x+\frac{7}{y}=2(y \neq 0)\)
(iv) 4x + 6y = 3xy, 8x + 9y = 5xy (x ≠ 0, y ≠ 0)
(v) (a – b)x + (a + b)y = a² – 2ab – b², (a + b)x+(a + b)y = a² + b²
(vi) \(\frac{2}{x}+\frac{3}{y}=2\), ax – by = a² – b²
(vii) \(\frac{5}{x+y}-\frac{2}{x-y}+1=0, \frac{15}{x+y}+\frac{7}{x-y}-10=0\)
(viii) \(\frac{xy}{x+y}=\frac{6}{5}, \frac{xy}{x+y}=6(x+y \neq 0, x-y \neq 0)\)
(ix) 6x + 5y = 7, x + 3y + 1 = 2 (x + 6y – 1)
(x) \(\frac{x+y-8}{2}=\frac{x+2 y-14}{3}=\frac{3 x+y-12}{11}\)
(xi) \(\frac{x+y}{2}-\frac{x-y}{3}=8, \frac{x+y}{3}+\frac{x-y}{4}=11\)
(xii) \(\frac{x}{a}=\frac{y}{b}, ax + by=a^2+b^2 \)
ସମାଧାନ :
(i) \(\frac{2}{x}+\frac{3}{y}=17\) ⇒ \(\frac{2}{x}+\frac{3}{y}-17=0\) …….(1)
\(\frac{1}{x}+\frac{1}{y}=7\) ⇒ \(\frac{1}{x}+\frac{1}{y}-7=0\) …….(2)
ଏଠାରେ \(\frac{1}{x}=u\) ଏବଂ \(\frac{1}{x}=v\) ନେଲେ ଦତ୍ତ ସମୀକରଣଦ୍ବୟ
2u + 3v – 17 = 0, ଏବଂ u + v – 7 = 0 ହେବ ।
ଏଠାରେ a₁ = 2
b₁ = 3
c₁ = -17
a₂ = 1
b₂ = 1
c₂ = -7
\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) ହୋଇଥିବାରୁ ସମୀକରଣ ଦ୍ଵୟର ଅନନ୍ୟ ସମାଧାନ ରହିବ ।
b₁c₂ – b₂c₁ = (3)(-7) – 1(-17) = -21 – 17 = -4
c₁a₂ – c₂a₁ = (-17) × 1 – (-7) × 2 = -17 + 14 = -3
a₁b₂ – a₂b₁ = 2 × 1 – 1 × 3 = 2 – 3 = -1 (≠ 0)

⇒ \(\frac{u}{-4}=\frac{v}{-3}=\frac{1}{-1}\) ⇒ u = 4 ଏବଂ v = 3
⇒ \(\frac{1}{x}=4\) ଏବଂ \(\frac{1}{y}=3\) ⇒ x = \(\frac{1}{4}\) ଏବଂ y = \(\frac{1}{3}\)
∴ ସମାଧାନ (x, y) = (\(\frac{1}{4}\), \(\frac{1}{3}\)) ।

‘x’ ର ମାନ ସମୀକରଣରେ (ii) ପ୍ରୟୋଗ କଲେ y = \(\frac{1}{3}\) ହେବ ।

(ii) \(\frac{5}{x}+6 y=13\) ⇒ \(\frac{5}{x}+6 y-13=0\) ……..(1)
\(\frac{3}{x}+20 y=35\) ⇒ \(\frac{3}{x}+20 y-35=0\) ……….(2)

ସମୀକରଣ (1) ରେ x = \(\frac{41}{25}\)

⇒ \(\frac{125}{41}+6 y-13=0 \Rightarrow 6 y=13-\frac{125}{41}=\frac{533-125}{41}\)
⇒ \(6 y=\frac{408}{41} \Rightarrow y=\frac{408}{41} \times \frac{1}{6}=\frac{68}{41}\)
∴ ସମାଧାନ (x, y) = (\(\frac{41}{25}\), \(\frac{68}{41}\)) ।

(iii) \(2 x-\frac{3}{y}=9,\) …….(i)
\(3 x+\frac{7}{y}=2\) …….(ii)

‘x’ ର ମାନ ସମୀକରଣରେ (ii) ପ୍ରୟୋଗ କଲେ 2 × 3 – \(\frac{3}{y}\) = 9 ⇒ – \(\frac{3}{y}\) = 9 – 6
⇒ 3y = -3 ⇒ y = -1
∴ ସମାଧାନ (x, y) = (3, -1)।

(iv) 4x + 6y = 3xy ………..(i)
8x + 9y = 5xy ………. (ii)
ସମୀକରଣ (i) ଓ (ii) ର ଉଭୟ ପାର୍ଶ୍ଵକୁ xy ଦ୍ବାରା ଭାଗକଲେ,
\(\frac{4}{y}+\frac{6}{x}=3\) …..(iii)
\(\frac{8}{y}+\frac{9}{x}=5\)
ମନେକର \(\frac{1}{x}=u\) ଓ \(\frac{1}{y}=v\)। ତେଣୁ ସମୀକରଣ (iii) ଓ (iv) ରୁ
6u + 4v = 3 ……(v), 9u + 8v = 5 ……..(vi)

ସମୀକରଣ (v) ରେ u = \(\frac{1}{3}\) ବସୀକଲେ, 6 × \(\frac{1}{3}\) + 4v = 3 ⇒ 4v = 1 ⇒ v = \(\frac{1}{4}\)
u = \(\frac{1}{3}\) ⇒ \(\frac{1}{x}\) = \(\frac{1}{3}\) ⇒ x = 3, v = \(\frac{1}{4}\) ⇒ \(\frac{1}{y}\) = \(\frac{1}{4}\) ⇒ y = 4
∴ ସମାଧାନ (x, y) = (3, 4)।

(v) (a – b)x + (a + b)y = a² – 2ab – b² ……..(i)
(a + b)x+(a + b)y = a² + b² ……….(ii)
ସମୀକରଣ (ii) କୁ ସମୀକରଣ (i) ରୁ ବିପ୍ରୟୋଗ କଲେ,
x(a + b) – x(a + b) = -2ab – 2b²
⇒ x(a + b – a – b) = -2ab – 2b²
⇒ -2bx = -2b(a + b) ⇒ x = a + b
‘x’ ର ମାନ ସମୀକରଣ (ii) ରେ ପ୍ରୟୋଗ କଲେ, (a+b)(a+b) + (a+b) y = a² + b²
⇒ a² + b² + 2ab + (a + b) y = a² + b²
⇒ (a + b) y = -2ab ⇒ y = \(\frac{-2ab}{a+b}\)
∴ ସମାଧାନ (x, y) = (a+b, \(\frac{-2ab}{a+b}\))।

(vi) \(\frac{2}{x}+\frac{3}{y}=2\) ⇒ \(\frac{bx+ay}{ab}=2\)
⇒ bx + ay = 2ab ……(1), ax – by = a² – b² ………(2)

⇒ x(a² + b²) = a³ + ab² ⇒ x(a² + b²) = a(a² + b²) ⇒ x = a
ସମୀକରଣ (1) ରେ x = a ସ୍ଥାପନ କଲେ, b.a + ay = 2ab ⇒ ay = ab
⇒ y = b
∴ ସମାଧାନ (x, y) = (a, b)।

(vii) \(\frac{5}{x+y}-\frac{2}{x-y}+1=0\) ……..(i) ଏବଂ
\(\frac{15}{x+y}+\frac{7}{x-y}-10=0\) ……..(ii)
\(\frac{1}{x+y}=a\) ଏବଂ \(\frac{1}{x-y}=b\) ହେଲେ
ସମୀକରଣଦ୍ଵୟ 5a – 2b + 1 = 0 ………(iii) ଏବଂ 15a + 7b – 10 = 0 ……….(iv)

⇒ x – y = 1 …….. (v)
b ର ମାନ ସମୀକରଣ (ii) ରେ ପ୍ରୟୋଗକଲେ,
5a – 2 × 1 + 1 = 0
⇒ 5a – 1 = 0 ⇒ 5a = 1 ⇒ a = \(\frac{1}{5}\)
⇒ \(\frac{1}{x+y}=\frac{1}{5}\) ⇒ x + y = 5 ……..(iv)
ସମୀକରଣ (v) ଓ (vi) ରୁ ୟୋଗକଲେ x + y + x – y = 5 + 1
⇒ 2x = 6 ⇒ x = 3
∴ ନିଶ୍ଚେୟ ସମାଧାନ (x, y) = (3, 2)।

(viii) \(\frac{xy}{x+y}=\frac{6}{5}\) …….(i)
\(\frac{xy}{x+y}=6\) ……..(ii)

⇒ \(\frac{1}{y}=\frac{2}{6}\) ⇒ \(\frac{1}{y}=\frac{1}{3}\) ⇒ y = 3
∴ ସମାଧାନ (x, y) = (2, 3)।

(ix) 6x + 5y = 7x + 3y + 1 = 2 (x + 6y – 1)
⇒ 6x + 5y = 7x + 3y + 1 ⇒ x – 2y + 1 = 0 ……(i)
ପୁନଶୃ 7x + 3y + 1 = 2 (x + 6y – 1)
⇒ 7x + 3y + 1 = 2x + 12y – 2 ⇒ 5x – 9y + 3 = 0 …..(ii)

ସମୀକରଣ (i) ରେ x = 3 ସ୍ଥାପନ କଲେ, 3 – 2y + 1 = 0
⇒ 2y = 4 ⇒ y = 2
∴ ସମାଧାନ ପେଟ୍ (x, y) = (3, 2)।

(x) \(\frac{x+y-8}{2}=\frac{x+2 y-14}{3}=\frac{3 x+y-12}{11}\)
⇒ \(\frac{x+y-8}{2}=\frac{x+2 y-14}{3}\) ⇒ 3(x + y – 8) = 2(x + 2y – 14)
⇒ 3x + 3y – 24 = 2x + 4y – 28 ⇒ x – y = -4 ……..(i)
ପୁନଶୃ \(\frac{x+2 y-14}{3}=\frac{3 x+y-12}{11}\) ⇒ 11(x + 2y – 14) = 3(3x + y – 12)
⇒ 11x + 22y – 154 = 9x + 3y – 36 ⇒ 2x + 19y = 118 ……….(ii)

ସମୀକରଣ (i) ରେ x = 2 ସ୍ଥାପନ କଲେ, 2 – y = -4 ⇒ y = 6
∴ ସମାଧାନ (x, y) = (2, 6)।

(xi) \(\frac{x+y}{2}-\frac{x-y}{3}=8\) ⇒ \(\frac{3(x+y)-2(x-y)}{6}=8\)
⇒ 3x + 3y – 2x + 2y = 48 ⇒ x + 5y = 48 ……..(i)
ପୁନଶୃ \(\frac{x+y}{3}+\frac{x-y}{4}=11\) ⇒ \(\frac{4(x+y)+3(x-y)}{12}=11\)
⇒ 4x + 4y + 3x – 3y = 132 ⇒ 7x + y = 132 ……..(ii)

ସମୀକରଣ (i) ରେ y = 6 ସ୍ଥାପନ କଲେ, x + 5 × 6 = 48 ⇒ x = 18
∴ ସମାଧାନ (x, y) = (18, 6)।

(xii) \(\frac{x}{a}=\frac{y}{b}\) ⇒ bx = ay ⇒ bx – ay = 0 ……..(i)
ଏବଂ \(ax + by=a^2+b^2 \) ………(ii)

x ର ମାନ ସମୀକରଣ (i) ରେ କଲେ, ba – ay = 0 ⇒ ay = ab ⇒ y = b

ବିକଳ୍ପ ସମାଧାନ :
\(\frac{x}{a}=\frac{y}{b}=k\) (ମନେକର) x = ak, y = bk
ax + by = a² + b² = a.ak + b.bk = a² + b²
k (a² + b²) = a² + b² ⇒ k = 1
∴ x = ak = a . 1 = a; y = bk = b . 1 = b

Question 5.
ନିମ୍ନଲିଖତ ଡିଟରମିନାଣ୍ଟର ମୂଲ୍ୟ ନିର୍ଣ୍ଣୟ କର ।
(i) \(\left|\begin{array}{ll}
2 & 5 \\
6 & 0
\end{array}\right|\)
(ii) \(\left|\begin{array}{ll}
2 & -1 \\
3 & 2
\end{array}\right|\)
(iii) \(\left|\begin{array}{ll}
0 & 4 \\
5 & -1
\end{array}\right|\)
(iv) \(\left|\begin{array}{ll}
\frac{1}{2} & 1 \\
\frac{3}{4} & \frac{1}{5}
\end{array}\right|\)
ସମାଧାନ :
(i) \(\left|\begin{array}{ll}
2 & 5 \\
6 & 0
\end{array}\right|\) = 2(0) – 6(5) = 0 – 30 = -30

(ii) \(\left|\begin{array}{ll}
2 & -1 \\
3 & 2
\end{array}\right|\) = 2 × 2 – 3 (-1) = 4 + 3 = 7

(iii) \(\left|\begin{array}{ll}
0 & 4 \\
5 & -1
\end{array}\right|\) = 0(-1) – 5 × 4 = 0 – 20 = -20

(iv) \(\left|\begin{array}{ll}
\frac{1}{2} & 1 \\
\frac{3}{4} & \frac{1}{5}
\end{array}\right|\) = \((\frac{1}{2})(\frac{1}{5})-(\frac{3}{4})(1)=\frac{1}{10}-\frac{3}{4}=\frac{2-15}{20}=\frac{-13}{20}\)

Question 6.
Cramer ଙ୍କ ନିୟମ ପ୍ରୟୋଗ କରି ନିମ୍ନ ସହସମୀକରଣମାନଙ୍କର ସମାଧାନ କର ।
(i) 2x + 3y = 5, 3x + y = 4
(ii) x + y = 3, 2x + 3y = 8
(iii) x – y = 0, 2x + y = 3
(iv) 2x – y = 3, x – 3y = -1
ସମାଧାନ :
(i) \(\Delta=\left|\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right|=\left|\begin{array}{ll}
2 & 3 \\
3 & 1
\end{array}\right|=2 \times 1-3 \times 3=2-9=-7\)
ଏଠାରେ ∆ ≠ 0 ତେଣୁ ସମୀକରଣର ସମାଧାନ ସମ୍ଭବ ।
\(∆_x=\left|\begin{array}{ll}
-c_1 & b_1 \\
-c_2 & b_2
\end{array}\right|=\left|\begin{array}{ll}
5 & 3 \\
4 & 1
\end{array}\right|=5 \times 1-4 \times 3=5-12=-7\)
\(∆_y=\left|\begin{array}{ll}
a_1 & -c_1 \\
a_2 & -c_2
\end{array}\right|=\left|\begin{array}{ll}
2 & 5 \\
3 & 4
\end{array}\right|=2 \times 4-5 \times 3=8-15=-7\)
x = \(\frac{∆_x}{∆}=\frac{-7}{-7}=1\), y = \(\frac{∆_y}{∆}=\frac{-7}{-7}=1\)
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (1, 1)

(ii) \(\Delta=\left|\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right|=\left|\begin{array}{ll}
1 & 1 \\
2 & 3
\end{array}\right|=1 \times 3-2 \times 1=3-2=1\)
ଏଠାରେ ∆ ≠ 0 ତେଣୁ ସମୀକରଣର ସମାଧାନ ସମ୍ଭବ ।
\(∆_x=\left|\begin{array}{ll}
-c_1 & b_1 \\
-c_2 & b_2
\end{array}\right|=\left|\begin{array}{ll}
3 & 1 \\
8 & 3
\end{array}\right|=3 \times 3-8 \times 1=9-8=1\)
\(∆_y=\left|\begin{array}{ll}
a_1 & -c_1 \\
a_2 & -c_2
\end{array}\right|=\left|\begin{array}{ll}
1 & 3 \\
2 & 8
\end{array}\right|=1 \times 8-2 \times 3=8-6=2\)
x = \(\frac{∆_x}{∆}=\frac{1}{1}=1\), y = \(\frac{∆_y}{∆}=\frac{2}{1}=2\)
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (1, 2)

(iii) \(\Delta=\left|\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right|=\left|\begin{array}{ll}
1 & -1 \\
2 & 1
\end{array}\right|=1 \times 1-2 \times -1=1+2=3\)
∆ ≠ 0 ତେଣୁ ସହ-ସମୀକରଣର ସମାଧାନ ସମ୍ଭବ ।
\(∆_x=\left|\begin{array}{ll}
-c_1 & b_1 \\
-c_2 & b_2
\end{array}\right|=\left|\begin{array}{ll}
0 & -1 \\
3 & 1
\end{array}\right|=0 \times 1-3 \times -1=0+3=3\)
\(∆_y=\left|\begin{array}{ll}
a_1 & -c_1 \\
a_2 & -c_2
\end{array}\right|=\left|\begin{array}{ll}
1 & 0 \\
2 & 3
\end{array}\right|=1 \times 3-2 \times 0=3-0=3\)
x = \(\frac{∆_x}{∆}=\frac{3}{3}=1\), y = \(\frac{∆_y}{∆}=\frac{3}{3}=1\)
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (1, 1)

(iii) \(\Delta=\left|\begin{array}{ll}
a_1 & b_1 \\
a_2 & b_2
\end{array}\right|=\left|\begin{array}{ll}
2 & -1 \\
1 & -3
\end{array}\right|=2 \times (-3)-1 \times (-1)=-6+1=-5\)
∆ ≠ 0 ତେଣୁ ସହ-ସମୀକରଣର ସମାଧାନ ସମ୍ଭବ ।
\(∆_x=\left|\begin{array}{ll}
-c_1 & b_1 \\
-c_2 & b_2
\end{array}\right|=\left|\begin{array}{ll}
3 & -1 \\
-1 & -3
\end{array}\right|=3 \times -3-(-1) \times -1=-9-1=-10\)
\(∆_y=\left|\begin{array}{ll}
a_1 & -c_1 \\
a_2 & -c_2
\end{array}\right|=\left|\begin{array}{ll}
2 & 3 \\
1 & -1
\end{array}\right|=2 \times -1-1 \times 3=-2-3=-5\)
x = \(\frac{∆_x}{∆}=\frac{-10}{-5}=1\), y = \(\frac{∆_y}{∆}=\frac{-5}{-5}=1\)
∴ ନିର୍ଣ୍ଣେୟ ସମାଧାନ (x, y) = (2, 1)

Odisha State Board BSE Odisha 10th Class Hindi Solutions Poem 1(c) तुलसीदास के दोहे Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Hindi Solutions Poem 1(c) तुलसीदास के दोहे

प्रश्न और अभ्यास (ପ୍ରଶ୍ନ ଔର୍ ଅଭ୍ୟାସ)

1. निम्नलिखित प्रश्नों के उत्तर दो-तीन वाक्यों में दीजिए।
(ନିମ୍ନଲିଖୂ ପ୍ରଶ୍ନୋ କେ ଉତ୍ତର ଦୋ-ତୀନ୍ ୱାର୍କୋ ମେଁ ଦୀଜିଏ।)
(ନିମ୍ନଲିଖ ପ୍ରଶ୍ନର ଉତ୍ତର ଦୁଇ-ତିନୋଟି ବାକ୍ୟରେ ଦିଅ।)

(क) कठोर वचन का क्यों परिहार करना चाहिए?
(କଠୋର୍ ବଚନ୍ କା ର୍ଯ୍ୟା ପରିହାର୍ କର୍‌ନା ଚାହିଏ?)
उत्तर:
कठोर वचन सबको दुःख पहुँचाता है। परिवेश को अशान्त कर देता है। कठोर वचन से दूसरों को पीड़ा पहुचती है इसलिए कठोर वचन को परिहार करके मीठे वचन बोलना चाहिए।

(ख) मीठे वचन से क्या लाभ होता है?
(ମୀଠ ବଚନ୍ ସେ କ୍ୟା ଲାଭ୍ ହୋତା ହୈ ?)
उत्तर:
मीठे वचन सबको प्रिय होते हैं। मीठी वाणी से हम सबको अपने वश में कर सकते हैं। मीठी वाणी से चारों ओर शांति बनी रहती है। सबको सुख मिलता है।

(ग) सन्तोष धन के सामने कौन-कौन से धन धूरि के बराबर माने जाते हैं?
(ସନ୍ତୋଷ୍ ଧନ୍ କେ ସାମ୍‌ନେ କୌନ୍-କୌନ୍ ସେ ଧନ୍ ଧୂରି କେ ବରାବର୍ ମାନେ ଜାତେ ହୈ ?)
उत्तर:
सन्तोष धन के सामने गोधन, गजधन, बाजीधन, रतनधन आदि धन धूरि के बराबर माने जाते हैं। क्योंकि इस प्रकार के धन से सुख शांति नहीं मिलती। मन चिंतित रहता है।

(घ) रोष या गुस्से के समय क्या नहीं खोलना चाहिए और क्यों?
(ରୋଷ୍ ୟା ଗୁସ୍‌ କେ ସମୟ କ୍ୟା ନହୀ ଖୋଲ୍‌ନା ଚାହିଏ ଔର୍ କ୍ୟା?)
उत्तर:
रोष या गुस्से के समय जीभ नहीं खोलनी चाहिए। क्योंकि क्रोध में मनुष्य कड़वी बातें बोल जाता है। कड़वी बातें तलवार से भी अधिक घाव कर देती है।

(ङ) मीठे वचन की तुलना वशीकरण मन्त्र से क्यों की गई है?
(ମୀଠ ବଚନ୍ କୀ ତୁଲନା ବଶୀକରଣ୍ ମନ୍ତ୍ର ସେ କୈ କୀ ଗଈ ହୈ ?)
उत्तर:
मीठे वचन की तुलना वशीकरण मन्त्र से की गई है क्योंकि मीठे वचन से हम सबको अपने वश में कर सकते हैं। मीठे वचन सबको प्रिय होते हैं, इससे सबको शांति और सुख मिलता हैं।

(च) हमें सोच विचार कर क्यों बोलना चाहिए?
(ହର୍ମେ ସୋଚ୍ ବିଚାର୍ କର୍ ଜ୍ୟୋ ବୋଲନା ଚାହିଏ ?)
उत्तर:
हमें सोच विचार कर हमेशा बोलना चाहिए। क्योंकि क्रोध में मनुष्य कड़वी बातें बोल जाता है। ये कड़वी बातें तलवार से भी अधिक घाव करती है। इसका प्रहार सीधे हृदय और मन पर होता है। मधुर वचन का परिणाम मधुर होता है।

निम्नलिखित अवतरणों का आशय दो-तीन वाक्यों में स्पष्ट कीजिए।
(ନିମ୍ନଲିଖ୍ ଅବତରର୍ଡୋ କା ଆଶୟ ଦୋ-ତୀନ୍ ୱାର୍କୋ ମେଁ ସ୍ପଷ୍ଟ କୀଜିଏ ।)
(ତଳଲିଖ୍ ଅବତରଣଗୁଡ଼ିକର ଆଶୟ ଦୁଇ-ତିନି ବାକ୍ୟରେ ସ୍ପଷ୍ଟ କର ।)

(क) तुलसी मीठे वचन ते, सुख उपजत चहुँओर।
(ତୁସୀ ମୀଠେ ବଚନ୍ ତେ, ସୁଖ୍ ଉପଜତ୍ ଚହୁଁଓର ।)
उत्तर:
इस पंक्ति में तुलसीदास यह बतलाते हैं कि मीठ वचन से सबको सुख मिलता है। चारों ओर शांति बनी रहती है। मीठे वचन सबको प्रिय होते हैं।

(ख) जब आवे सन्तोष धन, सब धन धूरि समान।
(ଜବ୍ ଆୱେ ସନ୍ତୋଷ୍ ଧନ୍, ସବ୍ ଧନ୍ ଧୂରି ସମାନ୍ ।)
उत्तर:
इस पंक्ति में कवि ने यह कहा है कि सन्तोष धन के सामने सब धन धूल के समान है। क्योंकि इस प्रकार के धन से सुख शांति नहीं मिलती, मन चिंतित रहता है।

(ग) रोष न रसना खोलिए, बरु खोलिओ तलवारि।
(ରୋସ୍ ନ ରସ୍‌ନା ଖୋଲିଏ ବରୁ ଖୋଲିଓ ତଲବାରି ।)
उत्तर:
गुस्से में जीभ नहीं खोलनी चाहिए। क्योंकि गुस्से में मनुष्य आपे से बाहर हो जाता है और कड़वी बातें बोल जाता है। ये कड़वी बाते दिल और मन को घायल करके अधिक कष्ट देती है।

3. निम्नलिखित प्रश्नों के उत्तर एक-एक वाक्य में दीजिए।
(ନିମ୍ନଲିଖତ୍ ପ୍ରକ୍ଷ୍ନୌ କେ ଉତ୍ତର୍ ଏକ୍-ଏକ୍ ୱାର୍କୋ ମେଁ ଦୀଜିଏ ।)
(ତଳଲିଖ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଗୋଟିଏ-ଗୋଟିଏ ବାକ୍ୟରେ ଦିଅ ।)

(क) किससे चारों ओर सुख उपजता है?
(କିସ୍‌ ଚାର୍ରେ ଓର ସୁଖ୍ ଉପଚ୍ଚତା ହୈ ?)
उत्तर:
मीठे वचन से चारों ओर सुख उपजाता है।

(ख) वशीकरण का मंत्र क्या है?
(ବଶୀକରଣ୍ କା ମନ୍ତ୍ର କ୍ୟା ହୈ ?)
उत्तर:
मीठे वचन वशीकरण का मंत्र है।

(ग) हमें क्या परिहार करना या छोड़ना चाहिए?
(ହର୍ମେ କ୍ୟା ପରିହାର୍ କର୍‌ନା ୟା ଛାଡୁନା ଚାହିଏ?)
उत्तर:
हमें कटु वचन को परिहार करना या छोड़ना चाहिए।

(घ) कवि ने सन्तोष की तुलना किस से की है?
(କବି ନେ ସନ୍ତୋଷ୍ କୀ ତୁଲନା କିସ୍ ସେ କୀ ହୈ ?)
उत्तर:
कवि ने सन्तोष की तुलना धन से की है।

(ङ) कब रसना नहीं खोलनी चाहिए?
(କବ୍ ରସ୍‌ନା ନହୀ ଖୋଲ୍‌ନୀ ଚାହିଏ?)
उत्तर:
अधिक गुस्से में रसना नहीं खोलनी चाहिए।

(च) किस धन के सामने सारे धन तुच्छ माने जाते हैं ?
(କିସ୍ ଧନ୍ କେ ସାମ୍‌ ସାରେ ଧନ୍ ତୁଚ୍ଛ ମାନେ ଜାତେ ହେଁ ?)
उत्तर:
संतोष धन के सामने सारे धन तुच्छ माने जाते हैं।

(छ) सन्तोष धन के सामने सब धन किसके समान होते हैं?
(ସନ୍ତୋଷ୍ ଧନ୍ କେ ସାମ୍‌ ସବ୍ ଧନ୍ କିସ୍କେ ସମାନ୍ ହୋତେ ହେଁ ?)
उत्तर:
सन्तोष धन के सामने सब धन धूल के समान होते हैं।

(ज) विचार करके वचन कहने से क्या होता है?
(ବିଚାର୍ କର୍‌କେ ବଚନ୍ କହନେ ସେ କ୍ୟା ହୋତା ହୈ ?)
उत्तर:
विचार करके वचन कहने से उसका परिणाम मधुर होता है।

भाषा-ज्ञान (ଭାଷା-ଜ୍ଞାନ)

1. निम्नलिखित शब्दों के विपरीत शब्द लिखिए । ( 160 61 ।)
(ନିମ୍ନଲିଖ୍ ଶବ୍ଦଗୁଡ଼ିକର ବିପରୀତ ଶବ୍ଦ ଲେଖ ।)
मीठा, सुख, कठोर, छोड़ना, समान, खोलना
उत्तर:
मीठा – कड़वा/खट्टा
कठोर – मृदु/कोमल
समान – असमान
सुख – दुःख
छोड़ना – पकड़ना
खोलना – बंद करना

2. निम्नलिखित शब्दों के समानार्थक शब्द लिखिए।
(ନିମ୍ନଲିଖତ୍ ଶବ୍ଦା କେ ସମାନାର୍ଥକ ଶବ୍ଦ ଲିଖିଏ।)
(ନିମ୍ନଲିଖ୍ ଶବ୍ଦଗୁଡ଼ିକର ସମାନାର୍ଥୀ ଶବ୍ଦ ଲେଖ )
वचन, सुख, कठोर, उपजना, गो, गज, बाजि
उत्तर:
वचन – वाणी/बात
कठोर – निर्दयी
गो – गाय/गऊ
सुख – आनन्द
उपजना – पैदा होना/जन्म होना
गज – हाथी

3. निम्नलिखित शब्दों के प्रयोग से सार्थक वाक्य बनाइए।
(ନିମ୍ନଲିଖତ୍ ଶବ୍ଦା କେ ପ୍ରୟୋଗ୍ ସେ ସାର୍ଥକ ବାକ୍ୟ ବନାଇଏ ।)
(ନିମ୍ନଲିଖ୍ ଶବ୍ଦଗୁଡ଼ିକର ପ୍ରୟୋଗ କରି ସାର୍ଥକ ବାକ୍ୟ ଗଠନ କର।)
वसीकरण, कठोर, गोधन, सन्तोष, तलवार
उत्तर:
वसीकरण – मीठे वचन तो वशीकरण मंत्र के समान हैं।
कठोर – राहुल अत्यन्त कठोर स्वभाव का है।
गोधन – यशोदा गोधन की कसम खाकर कहती हैं कि कृष्ण ही उनका पुत्र है।
संन्तोष – सन्तोष रूपी धन के सामने बाकी सारा धन तुच्छ है।
तलवार – तलवार शरीर पर घाव करती है मगर कड़वी बातें दिल पर घाव कर देती हैं।

4. निम्नलिखित शब्दों के शुद्ध रूप लिखिए।
(ନିମ୍ନଲିଖତ୍ ଶବ୍ଦା କେ ଶୁଦ୍ଧ ରୂପ ଲିଖିଏ।)
(ନିମ୍ନଲିଖ୍ ଶବ୍ଦଗୁଡ଼ିକର ଶୁଦ୍ଧ ରୂପ ଲେଖ।)
चहुँओर, वसीकरण, धूरि, तरवारि, परिनाम
उत्तर:
चहुँओर – चारों ओर
तरवारि – तलवार
वसीकरण – वशीकरण
धूरि – धूल/धूलि
परिनाम – परिणाम

5. निम्नलिखित शब्दों के साथ करण कारक ‘से’ चिह्न का प्रयोग करके वाक्य बनाइए:
(ନିମ୍ନଲିଖ୍ ଶବ୍ଦା କେ ସାଥ୍ କରଣ୍ କାରକ୍ ‘ସେ’ ଚିହ୍ନ କା ପ୍ରୟୋଗ କର୍‌କେ ବାକ୍ୟ ବନାଇଏ)
(ନିମ୍ନଲିଖତ ଶବ୍ଦଗୁଡ଼ିକ ସହିତ କରଣ କାରକ ‘ସେ’ ଚିହ୍ନର ପ୍ରୟୋଗ କରି ବାକ୍ୟ ଗଠନ କର )
वचन, मंत्र, धन, तलवार
उत्तर:
वचन – मीठे वचन से सबको सुख मिलता है।
मंत्र – वशीकरण मंत्र से सभी को वश में किया जा सकता है।
धन – धन से सुख नहीं मिलता।
तलवार – तलवार से मत खेलो।

Very Short & Objective type Questions with Answers

A. निम्नलिखित प्रश्नों के उत्तर एक वाक्य में दीजिए।

1. तुलसीदास का जन्म कब और कहाँ हुआ?
उत्तर:
तुलसीदास का जन्म सन् 1532 में उत्तर प्रदेश के राजापुर में हुआ।

2. तुलसीदास के गुरु कौन थे?
उत्तर:
श्री नरहरि दास तुलसीदास के गुरु थे

3. तुलसी जी किस-किस भाषा में लिखते थे?
उत्तर:
तुलसी जी अवधी और ब्रजभाषा में लिखते थे।

B. निम्नलिखित प्रश्नों के उत्तर एक शब्द / एक पद में दीजिए।

1. संतोष-धन आ जाता है तो बाकी सब धन किसके समान हो जाते हैं?
उत्तर:
धूल के

2. किसका परिणाम हितकर होता है?
उत्तर:
वशीकरण का

3. तुलसी ने श्रेष्ठ धन किसे कहा है?
उत्तर:
संतोष

4. तुलसीदास क्या परिहार करने को कहते हैं?
उत्तर:
कठोर वचन

5. जो वाणी सुनने में मधुर लगती है, उसका परिणाम क्या होता है?
उत्तर:
हितकर

6. किस अवस्था में रसना नहीं खोलनी चाहिए?
उत्तर:
गुस्से के समय

7. कवि तुलसीदास के अनुसार वशीकरण मंत्र का अर्थ क्या है?
उत्तर:
मीठी वाणी से सबको वश में करना

8. रोष के समय क्या नहीं खोलनी चाहिए ?
उत्तर:
रसना

9. हमे किस प्रकार की वाणी बोलनी चाहिए?
उत्तर:
मीठी

10. ‘परिनाम’ का अर्थ क्या है?
उत्तर:
परिणाम

C. रिक्त स्थानों की पूर्ति कीजिए।

1. विचार करके वचन बोलने से परिणाम ……………… होता है।
उत्तर:
हितकर

2. ………………… सी बात तलवार से अधिक घाव करती है।
उत्तर:
कड़वी

3. कड़वी बातें. ……………. घायल करती है।
उत्तर:
मन को

4. वशीकरण मंत्र ……………….. है।
उत्तर:
मधुर वाणी

5. तुलसी जी के गुरु ………………… है।
उत्तर:
रामानन्द

6. ………………… में रसना नहीं खोलनी चाहिए।
उत्तर:
अधिक गुस्से

7. तुलसी ने सन्तोष की तुलना …………………. से की है।
उत्तर:
धन

8. ………………. से चारों ओर सुख उपजता है।
उत्तर:
मिठे वचन

9. ………………. के सामने सारे धन तुछ माने जाते है।
उत्तर:
संन्तोष धन

10 ‘बरु’ का अर्थ है…………….. ।
उत्तर:
बल्कि

D. ठिक् या भूल लिखिए।

1. संतोष धन के आने से सारे धन धूल के समान हो जाते हैं।
उत्तर:
ठिक्

2. तुलसी के अनुसार विचार करके घूमना चाहिए।
उत्तर:
भूल

3. तलवार से रसना घातक होती है।
उत्तर:
ठिक्

4. चुटकुले चारों ओर सुख उपजाने में सहायक हैं।
उत्तर:
भूल

5. मीठे वचनों से सुख मिलता है।
उत्तर:
ठिक्

6. विचार करके कहानी कहना जरुरी है।
उत्तर:
भूल

7. ‘रामचरित मानस’ तुलसीदास ने लिखा है।
उत्तर:
भूल

8. बाजिधन के सामने सारे धन तुच्छ हैं
उत्तर:
ठिक्

9. ‘बाजी’ का अर्थ हाथी है।
उत्तर:
भूल

10. विनय पत्रिका तुलसीदास की रचना है।
उत्तर:
ठिक्

11. कवि ने संन्तोष की तुलना धन से की है।
उत्तर:
ठिक्

Multiple Choice Questions (mcqs) with Answers

सही उत्तर चुनिए : (MCQS)

1. संतोष-धन आ जाता है तो बाकी सब धन किसके समान हो जाते हैं?
(A) धूल के
(B) बाजि के
(C) रतन के
(D) प्राण के
उत्तर:
(A) धूल के

2. किसका परिणाम हितकर होता है?
(A) कटु वचन का
(B) रत्न-धन का
(C) वशीकरण का
(D) मधुर वचन का
उत्तर:
(C) वशीकरण का

3. तुलसी ने श्रेष्ठ धन किसे कहा है?
(A) रत्न-धन को
(B) राम – रत्न को
(C) संतोष-धन को
(D) सोने-चाँदीको
उत्तर:
(C) संतोष-धन को

4. तुलसीदास क्या परिहार करने को कहते हैं?
(A) कुसंग
(B) चिंत
(C) कामना
(D) कठोर वचन
उत्तर:
(D) कठोर वचन

5. जो वाणी सुनने में मधुर लगती है, उसका परिणाम क्या होता है?
(A) हितकर
(B) अहितकर
(C) भयानक
(D) खुशामद
उत्तर:
(A) हितकर

6. चहुँओर सुख उपजाता है।
(A) मीठे वचन से
(B) कटुवचन से
(C) अल्प वचन से
(D) धीमे वचन से
उत्तर:
(A) मीठे वचन से

7. तुलसीदास के अनुसार सबसे बड़ा धन है।
(A) गोधन
(B) गज धन
(C) संतोष धन
(D) रतन धन
उत्तर:
(C) संतोष धन

8. किस अवस्था में रसना नहीं खोलनी चाहिए।
(A) गुस्से के समय
(B) शान्ति के समय
(C) खाते समय
(D) राते समय
उत्तर:
(A) गुस्से के समय

दोहे (ଦୋହେ)

(i) तुलसी मीठे बचन ते, सुख उपजत चहुँओर।
वसीकरण यह मंत्र है, परिहरु बचन कठोर॥
ତୁଲସୀ ମୀଠେ ବଚନ୍ ତେ, ସୁଖ୍ ଉପଜତ ଚହୁଁଓର।
ବସୀକରଣ ୟହ ମନ୍ତ୍ର ହୈ, ପରିହରୁ ବଚନ୍ କଠୋର୍॥

हिन्दी व्याख्या:
तुलसीदास कहते हैं कि मीठे वचन सबको प्रिय होते हैं। मधुर वचन से हम सबको अपने वश में कर सकते हैं। मीठे वचन से चारों ओर शान्ति बनी रहती है। सबको सुख मिलता है। लेकिन कड़वी वाणी से सबको दुःख पहुँचता है। मीठे वचन तो वशीकरण मंत्र के समान है। इसलिए कड़वा वचन न बोलकर मीठे वचन बोलना चाहिए।

ଓଡ଼ିଆ ଅନୁବାଦ:
ମିଠା କଥା ସମସ୍ତଙ୍କୁ ଭଲ ଲାଗେ। ମିଠା କଥା କହି ଆମେ ସମସ୍ତଙ୍କୁ ଆପଣେଇ ପାରିବା। ମିଠା କଥାଦ୍ଵାରା ସବୁଆଡ଼େ ଶାନ୍ତି ଲାଗି ରହିଥାଏ। ସମସ୍ତଙ୍କୁ ସୁଖ ମିଳେ। କିନ୍ତୁ କଟୁକଥା ସମସ୍ତଙ୍କ ମନକୁ ଆଘାତ କରେ ଓ ଦୁଃଖ ଦିଏ। ମଧୁର ବଚନ ବଶୀକରଣ ମନ୍ତ୍ର ପରି। ଏଣୁ ଆମକୁ କଟୁ କଥା ନ କହି ମିଠା କଥା କହିବା ଦରକାର।

(ii) गोधन, गजधन, बाजिधन और रतनधन खान।
जब आवे सन्तोष धन सब धन धूरि समान॥
ଗୋଧନ୍, ଗଜଧନ୍, ବାଜିଧନ୍ ଔର୍ ରତନଧନ୍ ଖାନ୍।
ଜବ୍ ଆୱେ ସନ୍ତୋଷ୍ ଧନ୍, ସବ୍‌ ଧନ୍ ଧୂରି ସମାନ୍॥

हिन्दी व्याख्या :
कवि कहते हैं कि साधारणतः हमारी धारणा है कि जिसके पास पर्याप्त गाय-भैंस, हाथी और घोड़े होते हैं या धन रत्न आदि होते हैं, वह इस संसार में सबसे अधिक धनी है। लेकिन कवि के अनुसार ये सारे धन होते हुए भी अगर मन में सन्तोष नहीं है तो ये सबकुछ मूल्यहीन होता है। सन्तोष धन के सामने ये सब धन धूल के समान तुच्छ हैं। क्योंकि इस प्रकार के धन से सुख शान्ति नहीं मिलती। मन चिंतित रहता है।

ଓଡ଼ିଆ ଅନୁବାଦ:
ସାଧାରଣତଃ ଆମେ ଭାବିଥାଉ ଯେ ଯାହା ପାଖରେ ଗାଈ, ମଇଁଷି, ହାତୀ-ଘୋଡ଼ା ବା ଧନରତ୍ନ, ହୀରା-ମୋତି ଅଛି, ସେ ଜଣେ ବଡ଼ ଧନୀ ବ୍ୟକ୍ତି। କିନ୍ତୁ ତୁଳସୀ ଦାସଙ୍କ ଅନୁସାରେ ଏତେ ସବୁ ଧନ ଥାଇ ମଧ୍ୟ ଯଦି ତା’ମନରେ ଶାନ୍ତି ନାହିଁ ତାହାହେଲେ ସେ ଧନ ସବୁ ତା’ପାଇଁ ମୂଲ୍ୟହୀନ। ସନ୍ତୋଷ ରୂପୀ ଧନ ପାଖରେ ଏସବୁ ଧନରତ୍ନ ଧୂଳି ସଙ୍ଗେ ସମାନ ଓ ତୁଚ୍ଛ। ଏପରି ଧନ ସମ୍ପଦରେ ସୁଖ ଶାନ୍ତି ମିଳେ ନାହିଁ। ମନରେ ଚିନ୍ତା ଲାଗି ରହିଥାଏ।

(iii) रोष न रसना खोलिए, बरु खोलिओ तरबार।
सुनत मधुर परिनाम हित, बोलिओ बचन बिचारि॥
ରୋସ୍ ନ ରସନା ଖୋଲିଏ, ବରୁ ଖୋଲିଓ ତରବାରି।
ସୁନତ ମଧୁର୍ ପରିନାମ ହିତ, ବୋଲିଓ ବଚନ୍ ବିଚାରି॥

हिन्दी व्याख्या:
कवि तुलसीदास कहते हैं कि जब क्रोध अधिक हो तो जीभ नहीं खोलनी चाहिए। क्रोध में मनुष्य कड़वी बातें बोल जाता है। ये कड़वी बातें तलवार से भी अधिक घाव करती है। कड़वी बात का प्रहार सीधे दिल और दिमाग पर होता है। तलवार तो शरीर पर घाव करती है, लेकिन कड़वी बातें दिल और दिमाग को घायल करके अधिक कष्ट देती हैं।

ଓଡ଼ିଆ ଅନୁବାଦ :
ଯେତେବେଳେ ରାଗ ଅଧ୍ବକ ବଢ଼ିଚାଲେ ସେତେବେଳେ ଜିଭକୁ ନିୟନ୍ତ୍ରଣ ରଖିବାକୁ ପଡ଼ିବ। ରାଗରେ ମଣିଷ କଟୁ କଥା କହି ଚାଲିଥାଏ। ଏହି କଟୁ କଥା ଖଣ୍ଡାଠାରୁ ମଧ୍ୟ ଅଧିକ ଆଘାତ କରିଥାଏ। କଟୁ କଥାର ପ୍ରହାର ସିଧା ମନ ଓ ହୃଦୟ ଉପରେ ପଡ଼ିଥାଏ। ଖଣ୍ଡାର ଚୋଟରେ ଶରୀରରେ କ୍ଷତ ହୋଇଥାଏ। କିନ୍ତୁ କଟୁ କଥା ମନ ଓ ହୃଦୟକୁ ଆଘାତ କରେ ଏବଂ ଅଧିକ କଷ୍ଟ ଦିଏ।

शबनार: (ଶରାର୍ଥି)

मीठे – मीठा/मधुर (ମିଠା/ମଧୁର )।

उपजत – उपजना/पैदा होना (ଜନ୍ମହେବା/ଉପୁଜିବା)।

चहुँओर – चारों ओर (ଚାରିଆଡ଼େ)।

परिहरु – त्यागना/परित्याग करना (ତ୍ଯାଗ କରିବା)।

गोधन – गाय रूपी धन (ଗୋଧନ)।

बाजि – घोड़ा (ଘୋଡ଼ା)

ते – से/द्वारा (ଦ୍ଵାରା)।

वसीकरण – वशीभूत (ହାତି)।

गज – हाथी (ରତ୍ନ)।

रतन – रत्न (ବଶୀଭୂତ)।

खान – भंड़ार (ଭଣ୍ଡାର)।

धूरि – धूल यूरिकि)। (ଧୂଳି)।

रसना – जीभ (ଜିଭ)।

खोलिओ – खोले (ଖୋଲିବା)।

परिनाम – परिणाम (ପରିଣାମ)।

सुनत – सुनकर (ଶୁଣି)।

विचारि – विचार करके (ବିଚାର କରି)।

आवे – आए (ଆସେ)।

रोष – गुस्सा (ରାଗ)।

बरु – बल्कि ( ବରଂ)।

तरवारि – तलवार (ଖଣ୍ଡା)।

हित – मंगल (ହିତମଙ୍ଗଳ)।

बोलिअ – बोलो (କୁହ)।

कवि परिचय

भक्त कवि तुलसी दास का जन्म सन् 1532 में उत्तर प्रदेश के राजापुर में हुआ था और देहांत सन् 1633 में। पितामाता के स्नेह से वंचित होकर बचपन में उनको बड़ा कष्ट उठाना पड़ा। सौभाग्य से गुरु नरहरिदास ने उनकी बड़ी मदद की। तुलसी रामभक्त थे और यौवन काल में ही साधु बन गये। रामानंद उनके गुरु थे। वे हिन्दी और संस्कृत के बड़े पंड़ित थे। उस समय मुगलों का शासन था। देश की सामाजिक और धार्मिक परिस्थितियाँ अस्तव्यस्त थीं।

तुलसी दास ने रामचरित मानस लिखकर लोगों के सामने निष्कपट जीवन और आचरण का उदाहरण रखा। आज भी यह देश का अत्यंत लोकप्रिय ग्रंथ है। जन साधारण उसे बड़े चाव से पढ़ते हैं। दुःखी, निराश तथा भक्त लोगों को रामचरितमानस पढ़कर सुख शान्ति मिलती है। विनयपत्रिका, कवितावली, दोहावली, गीतावली आदि उनके अनेक ग्रंथ हैं। वे अवधी और ब्रजभाषा दोनों में लिखते थे।

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 10 Sequences and Series Ex 10(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 10 Sequences and Series Exercise 10(b)

Question 1.
Expand in ascending power of x.
(i) 2^x
Solution:

(ii) cos x
Solution:

(iii) sin x
Solution:

(iv) \(\frac{x e^{7 x}-e^{-x}}{e^{3 x}}\)
Solution:

(v) \(\boldsymbol{e}^{e^x}\) up to the term containing x⁴
Solution:

Question 2.
If x = y + \(\frac{y^2}{2 !}+\frac{y^3}{3 !}\) + ….. then show that y = x – \(\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}\) +….
Solution:

Question 3.
Find the value of \(x^2-y^2+\frac{1}{2 !}\left(x^4-y^4\right)+\frac{1}{3 !}\left(x^6-y^6\right)\) + ….
Solution:

Question 4.
Show that
(i) 2\(\left(\frac{1}{3 !}+\frac{2}{5 !}+\frac{3}{7 !}+\ldots\right)=\frac{1}{e}\)
Solution:

(ii) \(\frac{9}{1 !}+\frac{19}{2 !}+\frac{35}{3 !}+\frac{57}{4 !}+\frac{85}{5 !}\) + …. = 12e – 5
Solution:

(iii) \(1+\frac{1+3}{2 !}+\frac{1+3+3^2}{3 !}+\ldots=\frac{1}{2}\left(e^3-e\right)\)
Solution:

(iv) \(\frac{1.3}{1 !}+\frac{2.4}{2 !}+\frac{3.5}{3 !}+\frac{4.6}{4 !}\) + …. = 4e
Solution:
t_n for L.H.S.
= \(\frac{n(n+2)}{n !}=\frac{n^2+2 n}{n !}\)

(v) \(\frac{1}{1.2}+\frac{1.3}{1.2 .3 .4}+\frac{1.3 .5}{1.2 .3 .4 .5 .6}\) + …. = √e – 1
Solution:

Question 5.
Prove that
(i) loge(1 + 3x + 2x²) = 3x – \(\frac{5}{2}\)x² + \(\frac{9}{3}\)x³ – \(\frac{17}{4}\)x⁴ + …..,|x| < \(\frac{1}{2}\)
Solution:

(ii) log_e(n + 1) – log_e(n – 1) = 2 \(\left[\frac{1}{n}+\frac{1}{3 n^3}+\frac{1}{5 n^5}+\ldots\right]\)
Solution:
There is a printing mistake in the question. The correct question is log_e(n + 1) – log_e(n – 1)

(iii) log_e(n + 1) – log_en = 2 \(\left[\frac{1}{2 n+1}+\frac{1}{3(2 n+1)^3}+\frac{1}{5(2 n+1)^5}+\ldots\right]\)
Solution:

(iv) log_em – log_en = \(\frac{m-n}{m}+\frac{1}{2}\left(\frac{m-n}{m}\right)^2\)
Solution:

= – [log n – log m]
= log m – log n = L.H.S.

(v) log_ea – log_eb = \(2\left[\frac{a-b}{a+b}+\frac{1}{3}\left(\frac{a-b}{a+b}\right)^3\right.\) \(\left.+\frac{1}{5}\left(\frac{a-b}{a+b}\right)^5+\ldots\right], a>b\)
Solution:

(vi) log_en = \(\frac{n-1}{n+1}+\frac{1}{2} \cdot \frac{n^2-1}{(n+1)^2}\)\(+\frac{1}{3} \cdot \frac{n^2-1}{(n+1)^3}\) + …..
Solution:

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Ex 9(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 9 Binomial Theorem Exercise 9(b)

Question 1.
²ⁿC₀ + ²ⁿC₂ + …… + ²ⁿC_2n = 2^2n-1 and ²ⁿC₁ + ²ⁿC₃ + ….. + ²ⁿC_2n-1 = 2^2n-1
Solution:
We know that
(1 + x)²ⁿ = ²ⁿC₀ + ²ⁿC₁x + ²ⁿC₂x² + …..+ ²ⁿC_2nxⁿ …(1)
Putting x = 1
We get putting x = – 1  we get
∴ (²ⁿC₀ + ²ⁿC₂ + ²ⁿC₄ + ….. + ²ⁿC_2n) – (²ⁿC₁ + ²ⁿC₃ + ….. + ²ⁿC_2n-1) = 0
∴ ²ⁿC₀ + ²ⁿC₂ + ….. + ²ⁿC_2n
= ²ⁿC₁ + ²ⁿC₃ + ….. + ²ⁿC_2n-1
= \(\frac{2^{2 n}}{2}\) = 2^2n-1

Question 2.
Find the sum of
(i) C₁ + ²C₂ + ³C₃ + ….. + ⁿC_n
Solution:
C₁ + ²C₂ + ³C₃ + ….. + ⁿC_n

(ii) C₀ + ²C₁ + ³C₂ + ….. + ⁽ⁿ⁺¹⁾C_n Hint: write (C₀ + C₁ + ….. + C_n) + (C₁ + ²C₂ + ….. + ⁿC_n) use (5) and exercise 1.

Question 3.
Compute \(\frac{(1+k)\left(1+\frac{k}{2}\right) \ldots\left(1+\frac{k}{n}\right)}{(1+n)\left(1+\frac{n}{2}\right) \ldots\left(1+\frac{n}{k}\right)}\)
Solution:

Question 4.
Show that
(i) C₀C₁ + C₁C₂ + C₂C₃ + ….. + C_n-1C_n = \(\frac{(2 n) !}{(n-1) !(n+1) !}\)
Solution:

(ii) C₀C_r + C₁C_r+1 + C₂C_r+2 + ….. + C_n-rC_n = \(\frac{(2 n) !}{(n-r) !(n+r) !}\) Hint : Proceed as in Example 13. Compare the coefficient of a^n-1 to get (i) and the coefficient of a^n-r to get (ii)
Solution:


∴ C₀C_r + C₁C_r+1 + C₂C_r+2 + ….. + C_n-rC_n = \(\frac{(2 n) !}{(n-r) !(n+r) !}\)

(iii) 3C₀ – 8C₁ + 13C₂ – 18C₃ + ….. + (n+1)^th term = 0
Solution:
3C₀ – 8C₁ + 13C₂ – 18C₃ + ….. + (n+1)^th term
= 3C₀ – (5 + 3)C₁ + (10 + 3) C₂ – (15 + 3) C₃ + …
= 3(C₀ – C₁ + C₂ …) + 5 (- C₁ + 2C₂ – 3C₃ …..)
But (1 – x)ⁿ =C₀ – C₁x + C₂x² – C₃x³ + … + (-1)ⁿ xⁿC_n … (1)
Putting x = 1 we get
C₀ – C₁ + C₂ ….. + (-1)ⁿC_n
and differentiating (1) we get n(1 – x)^n-1 (- 1)
= – C₁ + 2C₂x – 3C₃x² +…..
Putting x = 1 we get
– C₁ + 2C₂ – 3C₃ + ….. = 0
∴ 3C0 – 8C1 +13C2 …… (n+1) terms = 0

(iv) C₀n² + C₁(2 – n)² +C₂(4 – n)² + ….. + C_n(2n – n)² = n.2ⁿ
Solution:
C₀n² + C₁(2 – n)² +C₂(4 – n)² + ….. + C_n(2n – n)² = n.2ⁿ
= n²(C₀ + C₁ + + C_n) + (2²C₁ + 4²C₂ + ….. + (2n)²C_n) – 4n (C₁ + 2C₂ + 3C₃ + ….+ nC_n)
= n² . 2ⁿ + 4n(n+l)2^n-2 – 4 n . n . 2^n-1
= 2^n-1 (2n² + 2n² + 2n – 4n²)
= 2n . 2^n-1 = n2ⁿ

(v) C₀ – 2C₁ +3C₂ + ….. + (-1)ⁿ(n+1)C_n = 0
Solution:

(vi) C₀ – 3C₁ +5C₂ + ….. + (2n+1)C_n = (n+1)²ⁿ
Solution:

Question 5.
Find the sum of the following :
(i) C₁ – 2C₂ + 3C₃ + ….. + n(-1)^n-1C_n 
Solution:

(ii) 1.2C₂ + 2.3C₃ + ….. + (n-1)nC_n
Solution:

(iii) C₁ + 2²C₂ + 3²C₃ + ….. + n²C_n
Solution:

(vi) C₀ – \(\frac{1}{2}\)C₁ + \(\frac{1}{3}\)C₂ + ….. + (-1)ⁿ \(\frac{1}{n+1}\)C_n
Solution:

Question 6.
Show that
(i) C₁² + 2C₂² +3C₃² + ….. + nC_n² = \(\frac{(2 n-1) !}{\{(n-1) !\}^2}\)
Solution:

(ii) C₂ + 2C₃ + 3C₄ + ….. + (n – 1)C_n = 1 + (n – 2)2^n-1
Solution:

Question 7.
C₁ – \(\frac{1}{2}\)C₂ + \(\frac{1}{3}\)C₃ + ….. +(-1)ⁿ⁺¹ \(\frac{1}{n}\)C_n = 1 + \(\frac{1}{2}\) + …. + \(\frac{1}{n}\)
Solution:

Question 8.
C₀C₁ + C₁C₂ + ….. + C_n-1C_n = \(\frac{2^n \cdot n \cdot 1 \cdot 3 \cdot 5 \ldots(2 n-1)}{(n+1)}\)
Solution:

Question 9.
The sum \(\frac{1}{1 ! 9 !}+\frac{1}{3 ! 7 !}+\ldots+\frac{1}{7 ! 3 !}+\frac{1}{9 ! 1 !}\) can be written in the form \(\frac{2^a}{b !}\) find a and b.
Solution:

Question 10.
(a) Using binominal theorem show that 1⁹⁹ + 2⁹⁹ + 3⁹⁹ + 4⁹⁹ + 5⁹⁹ is divisible by 5 (Regional Mathematical Olympiad, Orissa – 1987)
Solution:
1⁹⁹ + 2⁹⁹ + 3⁹⁹ + 4⁹⁹ + 5⁹⁹
= 1 + (5 – 3)⁹⁹ + 3⁹⁹ + (5 – 1)⁹⁹ + 5⁹⁹
= 1 + (5⁹⁹ – ⁹⁹C₁5⁹⁸.3¹ + ⁹⁹C₂5⁹⁷.3² – …3⁹⁹) + 3⁹⁹ – (1 – ⁹⁹C₁5¹ + ⁹⁹C₂5² – …  5⁹⁹) + 5⁹⁹
= (3 × 5⁹⁹ – ⁹⁹C₁5⁹⁸.3¹ + ⁹⁹C₂5⁹⁷.3² – ….. + ⁹⁹C⁹⁸5¹.3⁹⁸) + (⁹⁹C₁5¹ – ⁹⁹C₂5² + …. – ⁹⁹C₉₈5⁹⁸) ….(1) which is divisible by 5 as each term is a multiple of 5.

(b) Using the same procedure show that 1⁹⁹ + 2⁹⁹ + 3⁹⁹ + 4⁹⁹ + 5⁹⁹ is also divisible by 3 so that it is actually divisible by 15.
Solution:
From Eqn. (1) above, it is clear that each term within the 1st bracket is divisible by 3 and the terms in the 2nd bracket are divisible by 99 and hence divisible by 3.
Each term in Eqn. (1) is divisible by 3. As it is divisible by 3 and 5, it is divisible by 3 × 5 = 15

Odisha State Board BSE Odisha 6th Class English Solutions Follow-Up Lesson 2 Bamboo Curry Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Follow-Up Lesson 2 Bamboo Curry

BSE Odisha 6th Class English Follow-Up Lesson 2 Bamboo Curry Text Book Questions and Answers

Session – 1 (ସୋପାନ – ୧):
I. Pre-Reading (ପଢ଼ିବା ପୂର୍ବରୁ):

→ You have read the Odia folk-tale “The Foolish Son-in-Law”. In other languages, there are similar stories. Let us read a similar Santal folk tale “Bamboo Curry”.
(ତୁମେ ଓଡ଼ିଆ ଲୋକକଥା ‘ନିର୍ବୋଧ ଜ୍ଵାଇଁ’’ ପଢ଼ିସାରିଛ । ଅନ୍ୟ ଭାଷାମାନଙ୍କରେ, ସେହି ଏକାପରି ଗପସବୁ ଅଛି । ଆସ ଆମେ ଏକାପରି ଏକ ସାନ୍ତାଳ ଲୋକକଥା ‘ବାଉଁଶ ତରକାରି’’ ପଢ଼ିବା ।)

II. While-Reading (ପଢ଼ିବା ସମୟରେ):
Text (ବିଷୟବସ୍ତୁ):

SGP-1:
Read paragraphs 1-2 silently and answer the questions that follow.
(ଅନୁଚ୍ଛେଦ ୧ – ୨ କୁ ନୀରବରେ ପଢ଼ ଏବଂ ପରବର୍ତୀ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

1. Once a foolish Santal son-in-law went to his in-law’s place. His mother-in-law cooked delicious dishes for her son-in-law. One of the dishes was a curry made out of the bamboo shoot. The son-in-law liked it very much and asked his mother-in-law, “Mother, the curry is extremely delicious. What is the curry made from ?” Instead of answering his question, she pointed at the bamboo door. The son-in-law asked, “Is it from bamboo ?” “Yes son, the curry is made from bamboo and is, therefore, called “Bamboo Curry”.

2. Next day, the son-in-law was about to leave for his home. The bamboo curry came to his mind. He thought of cooking bamboo curry at home. But they did not have bamboo. So he carried home the bamboo door of his in-laws’ house.

ଓଡ଼ିଆ ଅନୁବାଦ :
(୧). ଏକଦା (ଥରେ) ଜଣେ ନିର୍ବୋଧ (ମୂର୍ଖ) ସାନ୍ତାଳ ଜ୍ଵାଇଁ ତା’ର ଶ୍ୱଶୁର ଘରକୁ ଗଲା । ତା’ର ଶାଶୁ ତା’ର ଜ୍ଵାଇଁ ପାଇଁ ସୁସ୍ୱାଦୁ ଖାଦ୍ୟ ରୋଷେଇ କଲା । ଖାଦ୍ୟଗୁଡ଼ିକ ମଧ୍ୟରୁ ଗୋଟିଏ ଥିଲା ବାଉଁଶ ଗଜାରେ ତିଆରି ହୋଇଥାଏ ତରକାରି । ଜ୍ଵାଇଁ ଜଣଙ୍କ ଏହାକୁ ବହୁତ ପସନ୍ଦ କଲା ଏବଂ ତା’ର ଶାଶୁକୁ ପଚାରିଲା, ‘‘ମାଆ, ତରକାରି ବହୁତ ସୁଆଦିଆ ହୋଇଛି । ତରକାରି କେଉଁଥିରେ ତିଆରି ହୋଇଛି ?’’ ତା’ର ପ୍ରଶ୍ନର ଉତ୍ତର ଦେବା ପରିବର୍ତ୍ତେ, ସେ ବାଉଁଶ ଦରଜା (ତାଟି) ଆଡ଼କୁ ଆଙ୍ଗୁଳି ନିର୍ଦ୍ଦେଶ କରି ଠାରିଲେ । ଜ୍ଵାଇଁ ପଚାରିଲା, ‘ଏହା ବାଉଁଶରେ ତିଆରି ହୋଇଛି ?’’ ‘ହଁ ପୁଅ, ତରକାରିଟି ବାଉଁଶରେ ତିଆରି ହୋଇଛି ଏବଂ ଏଥିପାଇଁ ‘ବାଉଁଶ ତରକାରି’’ କୁହାଯାଏ ।’’
(୨) ତା’ପରଦିନ, ଜ୍ଵାଇଁ ତାଙ୍କ ଘରକୁ ବାହାରୁଥିଲେ । ତାଙ୍କର ମନକୁ ବାଉଁଶ ତରକାରି କଥା ଆସିଲା । କଥା ଭାବିଲେ । କିନ୍ତୁ ସେମାନଙ୍କର ବାଉଁଶ ନଥିଲା । (ତାଟି)କୁ ତା’ ନିଜ ଘରକୁ ବୋହିନେଲା । ସେ ତାଙ୍କ ନିଜ ଘରେ ବାଉଁଶ ତରକାରି କରିବା ତେଣୁ ସେ ତାଙ୍କ ଶ୍ଵଶୁର ଘରର ବାଉଁଶ ଦରଜା

Comprehension Questions : (ବୋଧମୂଳକ ପ୍ରଶ୍ନବଳୀ)

Question 1.
Who went to his father-in-law’s house?
(କିଏ ତା’ର ଶ୍ୱଶୁର ଘରକୁ ଗଲା ?)
Answer:
A foolish Santal son-in-law went to his father-in-law’s house.

Question 2.
What curry did his mother-in-law cook?
(ତା’ର ଶାଶୁ କି ତରକାରି ତିଆରି କଲା ?)
Answer:
His mother-in-law cooked “Bamboo Curry”.

Question 3.
Did he like it?
(ଏହାକୁ ସେ ପସନ୍ଦ କଲା କି ?)
Answer:
Surely (ନିଶ୍ଚୟ), he liked it very much.

Question 4.
Why did he carry home the bamboo door of his in-law’s house?
(ସେ କାହିଁକି ଘରକୁ ଶ୍ୱଶୁର ଘରର ବାଉଁଶ ଦରଜା ବୋହିନେଲା ?)
Answer:
He carried home the bamboo door of his in-laws’ house because he thought of cooking bamboo curry at home, but they did not have any bamboo.

Session – 2 (ସୋପାନ – ୨):
SGP-2:

• Read paragraphs 3-4 silently and answer the questions that follow.
  (ଅନୁଚ୍ଛେଦ ୩ – ୪ କୁ ନୀରବରେ ପଢ଼ ଏବଂ ପରବର୍ତ୍ତୀ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

3. Reaching home, he asked his wife to prepare the bamboo curry. He helped his wife in chopping the dry bamboo sticks. But the dry bamboo pieces did not get boiled. The pieces remained as hard and stiff as before. He asked his wife to put more water and boil.
4. That evening his in-laws came to their son-in-law’s house. The son-in-law offered them the bamboo curry. The in-laws laughed at their foolish son-in-law. They told him, “The bamboo curry is made from soft bamboo shoots and not from dry bamboo pieces”.

ଓଡ଼ିଆ ଅନୁବାଦ :
(୩) ଘରେ ପହଞ୍ଚିସାରି, ସେ ତା’ର ସ୍ତ୍ରୀକୁ ବାଉଁଶ ତରକାରି ପ୍ରସ୍ତୁତ କରିବାକୁ କହିଲା । ସେ ତା’ର ସ୍ତ୍ରୀକୁ ସେହି ଶୁଖୁଲା ବାଉଁଶ କାଠିଗୁଡ଼ିକୁ ଖଣ୍ଡ ଖଣ୍ଡ କରି କାଟିବାରେ ସାହାଯ୍ୟ କଲା । କିନ୍ତୁ ସେହି ଶୁଖୁ ବାଉଁଶ ଖଣ୍ଡଗୁଡ଼ିକ ସିଝିଲା ନାହିଁ । ସେହି ବାଉଁଶ କାଠି ଖଣ୍ଡଗୁଡ଼ିକ ପୂର୍ବପରି କଠିନ (ଟାଣ) ହୋଇ ରହିଲା । ସେ ତା’ର ସ୍ତ୍ରୀକୁ ଅଧିକ ପାଣି ଦେଇ ସିଝାଇବାକୁ କହିଲା ।
(୪) ସେହିଦିନ ସନ୍ଧ୍ୟାରେ ତା’ର ଶାଶୁ-ଶ୍ଵଶୁର ସେମାନଙ୍କର ଜ୍ୱାଇଁ ଘରକୁ ଆସିଲେ । ଜ୍ଵାଇଁ ଜଣଙ୍କ ତାଙ୍କୁ ବାଉଁଶ ତରକାରି ଖାଇବାକୁ ଦେଲେ । ଶାଶୁ-ଶ୍ଵଶୁର ତାଙ୍କର ନିର୍ବୋଧ (ବୋକା) ଜ୍ୱାଇଁ ଆଡ଼କୁ ଅନାଇ ହସିଲେ । ସେମାନେ ତାକୁ କହିଲେ, ‘ବାଉଁଶ ତରକାରି କୋମଳ (ନରମ) ବାଉଁଶ ଗଜାରେ ତିଆରି ହୁଏ କିନ୍ତୁ ଶୁଖିଲା ବାଉଁଶ କାଠି ଖଣ୍ଡଗୁଡ଼ିକରେ ନୁହେଁ ।

Comprehension Questions : (ବୋଧମୂଳକ ପ୍ରଶ୍ନବଳୀ)

Question 1.
Who did he ask to prepare bamboo curry?
(ସେ କାହାକୁ ବାଉଁଶ ତରକାରି ପ୍ରସ୍ତୁତ କରିବାକୁ କହିଲା ?)
Answer:
He asked his wife to prepare bamboo curry.

Question 2.
How did he help his wife ?
(କିପରି ସେ ତା’ ସ୍ତ୍ରୀକୁ ସାହାଯ୍ୟ କଲା ?)
Answer:
He helped his wife in chopping the dry bamboo sticks.

Question 3.
When the bamboo did not boil what did he ask his wife to do?
(ଯେତେବେଳେ ବାଉଁଶ ଖଣ୍ଡଗୁଡ଼ିକ ସିଝିଲା ନାହିଁ, ସେ ତା’ସ୍ତ୍ରୀକୁ କ’ଣ କରିବାକୁ କହିଲା ?)
Answer:
When the bamboo pieces did not boil, he asked his wife to put more water and boil.

Question 4.
Who came to his house?
(କିଏ ଜ୍ଵାଇଁ ଘରକୁ ଆସିଲେ ?)
Answer:
His in-laws came to his house.

Question 5.
Where from is the bamboo curry made?
(କେଉଁଥୁରୁ ବାଉଁଶ ତରକାରି ପ୍ରସ୍ତୁତ ହୁଏ ?)
Answer:
The bamboo curry is made from soft bamboo shoots.

Session – 3 (ସୋପାନ – ୩):
III. Post-Reading (ପଢ଼ିସାରିବା ପରେ):

1. Writing (ଲେଖ୍) :
(a) Answer the following questions.
(ନିମ୍ନଲିଖ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

(i) What is the story about ?
(ଗପଟି କାହା ଉପରେ ଆଧାରିତ ?)
Answer:
The story is about a foolish Santal son-in-law.

(ii) What curry did the mother-in-law prepare?
(କେଉଁ ତରକାରି ଶାଶୁ ତିଆରି କଲେ ?)
Answer:
The mother-in-law prepared “Bamboo Curry”.

(iii) Is the son-in-law foolish ? Why ?
(ଜ୍ଵାଇଁଟି ନିର୍ବୋଧ ଥିଲା କି ? କାହିଁକି ?)
Yes, the ____________ because he asked ____________ out of dry bamboo.
Answer:
Yes, the son-in-law was very foolish because he asked his wife to prepare bamboo curry out of dry bamboo.

Word Note (ଶବ୍ଦାର୍ଥ):
(The words / phrases have been defined mostly on contextual meanings.) (ଶବ୍ଦ । ଖଣ୍ଡବାକ୍ୟଗୁଡ଼ିକ ଅଧ୍ଵଂଶତଃ ପ୍ରସଙ୍ଗଗତ ଅର୍ଥ ଉପରେ ନିର୍ଭର କରି ବ୍ୟାଖ୍ୟା କରାଯାଇଛି ।)

ashamed – feeling shame, ଲଜ୍ଜା ଅନୁଭବ
bamboo curry – a kind of dish (food) made out of young bamboo plants ବାଉଁଶ ଗଜା ତରକାରି
bamboo shoot- new-young bamboo plants, ବାଉଁଶ ଗଜା
chopping- cutting into small pieces, ଖଣ୍ଡ ଖଣ୍ଡ କରି କାଟିବା ସ୍ବାଦିଷ୍ଟ,ସୁସ୍ବାଦୁ
delicious- tasty (food), ସ୍ଵାଦିଷ୍ଟ, ସୁସ୍ୱାଦୁ
dishes- food items, curry, ତରକାରି, ସ୍ଵାଦିଷ୍ଟ ବ୍ୟଞ୍ଜନ
folk-tale- popular story of a community, କଥୁତଳ୍ପ
gentlest- very kind (behaviour) ବହୁତ ଦୟାଳୁ (ଆଚରଣ)
heaved a great sign of relief- feel relieved,ଆରାମ ଅନୁଭବ କର |
high sounding words- difficult words, କଠିନ ବା ବଡ଼ ବଡ଼ ଶବ୍ଦ
impolite- not good behaviour, rude, ଭଲ ବ୍ୟବହାର ନୁହେଁ
lamb- young sheep, ଛୋଟ ମେଣ୍
offered- gave, served, ଦେଲେ, (ଖାଦ୍ୟ) ପରିବେଷଣ କଲେ
piled high – kept (things) in a heap, ଗଦା, ସ୍ତୂପ
plucking- collecting (from a tree) ସଂଗ୍ରହ (ଏକ ଗଛରୁ)
pointed – showed hand towards (bamboo door), ହାତ ଦେଖାଇ ନିର୍ଦ୍ଦେଶିତ କଲେ
preferred – chose, ପସନ୍ଦ କଲେ ବା ଆଗ୍ରହ ଦେଖାଇଲେ
quack – a self claimed ignorant practitioner, ଶଠ ବଇଦ | ଠକ ବଇଦ
smeared- spread something (substance) on body, ବୋଳି ହୋଇଗଲା, ଲାଗିଗଲା
stiff- hard, କଠିନ
thought of- got an idea, ଗୋଟିଏ ଉପାୟ ଚିନ୍ତା କଲେ
thrashed- beat, ମାଡ଼ିଦେଲେ, ପିଟିଲେ

Odisha State Board BSE Odisha 6th Class English Solutions Follow-Up Lesson 5 The Crab and the Fox Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Follow-Up Lesson 5 The Crab and the Fox

BSE Odisha 6th Class English Follow-Up Lesson 5 The Crab and the Fox Text Book Questions and Answers

Session – 1 (ସୋପାନ – ୧):
I. Pre-Reading (ପଢ଼ିବା ପୂର୍ବରୁ):

→ The teacher finds an activity to introduce the topic. S/he may use the pictures in the text for the purpose.
(ବିଷୟକୁ ଉପସ୍ଥାପନ କରିବାପାଇଁ ଶିକ୍ଷକ ଏକ କାର୍ଯ୍ୟ ସ୍ଥିରକରିବେ । ସେ (ପୁ/ସ୍ତ୍ରୀ) ବିଷୟବସ୍ତୁରେ ଥିବା ଛବିଗୁଡ଼ିକୁ ଏହି ଉଦ୍ଦେଶ୍ୟରେ ବ୍ୟବହାର କରିପାରନ୍ତି ।)

II. While-Reading (ପଢ଼ିବା ସମୟରେ):
→ Follow the three steps-teacher’s reading aloud two times followed by silent reading by the students.
(ତିନୋଟି ପର୍ଯ୍ୟାୟକୁ ଅନୁସରଣ କର – ଶିକ୍ଷକଙ୍କର ଦୁଇଥର ବଡ଼ପାଟିରେ ପଢ଼ିବା ଓ ଶିକ୍ଷାର୍ଥୀମାନେ ନୀରବରେ ତାଙ୍କ ପରେ ପଢ଼ିବା ।)

TEXT (ବିଷୟବସ୍ତୁ)

• Read the poem silently and answer the questions that follow.
  (କବିତାଟିକୁ ନୀରବରେ ପଢ଼ ଏବଂ ପରବର୍ତ୍ତୀ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

It was a very cool night
And there was no crab in sight.
The fox looked for one

But there was none.
“Where did they go?”
Not even one in sight!
They must be in their holes
If I’m right.”

Thinking so he got ready
(The hunger too made him greedy)
To go in search of a crab hole
And he straightened his tail like a pole.
Inside the hole his bushy tail he pushed
The crab, he thought, it slightly missed.
He waited long for a sweet pull
But for long there was none at all.
Finally, he pulled out his tail
But the crab was inside he could smell.
So he changed his plan and called “Brother Crab,
Let’s some song and dance have.
The weather calls for such merrymaking
What is life without dancing and singing ?”
The crab well understood
The fox’s real mood.
Thought he to himself ‘Am I a fool ?’

And answered from his hole :
“Who is going to sing and dance
In such weather cool?
I’ll rather eat and sleep well
in my cozy little hole.”

କବିତାଟିର ଓଡ଼ିଆ ଉଚ୍ଚାରଣ :
ଇଟ୍ ୱାଜ୍ ଏ ଭେରି କୁଲ୍ ନାଇଟ୍
ଆଣ୍ଡ୍ ଦେୟାର୍ ୱାଜ୍ ନୋ କ୍ରାବ୍‌ ଇନ୍ ସାଇଟ୍ ।
ଦି ଫକ୍ସ ଲୁକ୍‌ ଫର୍ ୱାନ୍
ବଟ୍ ଦେୟାର ୱାଜ୍ ନନ୍ ।
‘‘ହୋୟାର୍ ଡିଡ଼୍ ଦେ ଗୋ ?’’
ନଟ୍ ଇଭିନ୍ ୱାନ୍ ଇନ୍ ସାଇଟ୍ !
ଦେ ମଷ୍ଟ ବି ଇନ୍ ଦେୟାର୍ ହୋଲ୍‌ସ୍
ଇଫ୍ ‘ଆଇ’ମ୍ ରାଇଟ୍ ।’’
ଥଙ୍କିଙ୍ଗ୍ ସୋ ହି ଗଟ୍ ରେଡ଼ି
(ଦି ହଙ୍ଗର୍ ଠୁ ମେଡ଼୍ ହିମ୍ ଗ୍ରିଡ଼ି)
ଟୁ ଗୋ ଇନ୍ ସର୍ଚ ଅଫ୍ ଏ କ୍ରାବ୍ ହୋଲ୍
ଆଣ୍ଡ ହି ଷ୍ଟେଟେଡ୍ ହିଜ୍ ଟେଲ୍ ଲାଇକ୍ ଏ ପୋଲ୍ ।
ଇସାଇଡ୍ ଦି’ ହୋଲ୍ ହିଜ୍ ବୁସି ଟେଲ୍ ହି ପୁସ୍‌
ଦି କ୍ରାବ୍, ହି ଥଟ୍, ଇଟ୍ ସ୍ଲାଇଟ୍‌ଲି ମିସିଡ୍ ।
ହି ୱେଟେଡ଼ ଲଙ୍ଗ୍ ଫର୍ ଏ ସୁଇଟ୍ ପୁଲ୍
ବଟ୍ ଫର୍ ଲଙ୍ଗ୍ ଦେୟାର୍ ୱାଜ୍ ନନ୍ ଆଟ୍ ଅଲ୍ ।
ଫାଇନାଲି ହି ପୁଲ୍‌ ଆଉଟ୍ ହିଜ୍ ଟେଲ୍
ବଟ୍ ଦି’ କ୍ରାବ୍ ୱାଜ୍ ଇନ୍‌ସାଇଡ୍ ହି କୁଡ଼ ସ୍କେଲ୍ ।
ସୋ ହି ଚେଞ୍ଜେଡ଼୍ ହିଜ୍ ପ୍ଲାନ୍ ଆଣ୍ଡ କଲ୍‌ ‘ବ୍ରଦର୍‌ କ୍ରାନ୍’’,
ଲେଟ୍ସ ସମ୍ ସଙ୍ଗ୍ ଆଣ୍ଡ୍ ଡ୍ୟାନ୍ସ ହାଭ୍ ।
ଦି ୱେଦର୍ କଲସ୍ ଫର୍ ସବ୍ ମେରିମେକିଙ୍ଗ୍
ଦ୍ଵାଟ୍ ଇଜ୍ ଲାଇଫ୍ ଉଇଦାଉଟ୍ ଡ୍ୟାନ୍‌ସିଙ୍ଗ୍ ଆଣ୍ଡ ସିଙ୍ଗିଙ୍ଗ୍ ?’’
ଦି କ୍ରାବ୍‌ ୱେଲ୍ ଅଣ୍ଡରଷ୍ଟୁଡ୍
ଦି ଫକ୍ସ’ସ୍ ରିଅଲ୍ ମୁଡ଼ ।
ଥଟ୍ ହି ଟୁ ହିମ୍‌ସେଲୁ ‘ଆମ୍ ଆଇ ଏ ଫୁଲ୍ ?’
ଆଣ୍ଡ ଆନ୍‌ସର୍‌ଡ୍‌ ଫ୍ରମ୍ ହିଜ୍ ହୋଲ୍ :
“ହୁ ଇଜ୍ ଗୋଇଙ୍ଗ୍ ଟୁ ସିଙ୍ଗ୍ ଆଣ୍ଡ୍ ଡ୍ୟାନ୍ସ,
ଇନ୍ ସଚ୍ ଏ ୱେଦର୍ କୁଲ୍ ?
ଆଇ’ଲ୍ ନ୍ୟାଦର୍ ଇଟ୍‌ ଆଣ୍ଡ ସ୍କ୍ରିପ୍ ୱେଲ୍
ଇନ୍ ମାଇଁ କୋଜି ଲିଟିଲ୍ ହୋଲ୍ ।’’

କବିତାର ଓଡ଼ିଆ ଅନୁବାଦ:
ଏହା ଏକ ବହୁତ ଶୀତଳ ରାତ୍ରି ଥିଲା
ଏବଂ କୌଣସି କଙ୍କଡ଼ା ଦୃଷ୍ଟିରେ ପଡୁନଥିଲା ।
କୋକିଶିଆଳ ଗୋଟାଏକୁ ଖୋଜୁଥିଲା
କିନ୍ତୁ ସେଠାରେ ଗୋଟିଏ ବି ନଥିଲା ।
‘ସେମାନେ କୁଆଡ଼େ ଗଲେ?’
ଏପରିକି ଗୋଟିଏବି ହେଲେ
ଦୃଷ୍ଟିରେ ପଡୁନାହାନ୍ତି !
ନିଶ୍ଚୟ ସେମାନେ ଥିବେ ସେମାନଙ୍କର ଗାତଗୁଡ଼ିକ ଭିତରେ
ଯଦି ମୁଁ ଠିକ୍ କହୁଥାଏ ।’
ଭାବି ସେ ପ୍ରସ୍ତୁତ ହୋଇଗଲେ |
(କ୍ଷୁଧା ମଧ୍ଯ ତାକୁ ଲୋଭୀ କରିଦେଲା)
ଅନ୍ଵେଷଣରେ ଯିବା ପାଇଁ ଗୋଟିଏ କଙ୍କଡ଼ା ଗାତ
ଏବଂ ସେ ସିଧା ବା ସଳଖ କରିଦେଲା ତା’ର ଲାଞ୍ଜକୁ ଗୋଟିଏ ଖୁଣ୍ଟ ପରି !
ଗାତ ଭିତରକୁ ତା’ର ବୁଦାଳିଆ ଲୋମଶ ଲାଞ୍ଚକୁ ସେ ଠେଲିଦେଲା
ସେ ଭାଙ୍ଗିଲା, କଙ୍କଡ଼ାଟିରେ ବାଜିବାରେ ଟିକେ ଭୁଲ୍ ହେଲା ।
ସେ ବହୁତ ସମୟ ଅପେକ୍ଷା କଲା ଗୋଟିଏ ମଧୁର ଟଣା ପାଇଁ
କିନ୍ତୁ ଦୀର୍ଘ ସମୟ ଧରି ସେପରି କିଛି ହେଲା ନାହିଁ ।
ଶେଷରେ ସେ ତା’ର ଲାଞ୍ଜକୁ ବାହାରକୁ ଟାଣି ଆଣିଲା
କିନ୍ତୁ କଙ୍କଡ଼ାଟି ଭିତରେ ଥ‌ିବାର ବାସନା ସେ ବାରିପାରିଲା ।
ତେଣୁ ସେ ତା’ର ଯୋଜନାକୁ ବଦଳାଇ ଦେଲା ଏବଂ ଡାକିଲା ‘କଙ୍କଡ଼ା ଭାଇ’’,
ଆସ ଆମେ କିଛି ଗୀତ ଏବଂ ନାଚ କରିବା ।
ଏପରି ମଉଜ କରିବାକୁ ପାଗ ଡାକୁଛି
ନାଚ ଓ ଗୀତ ବିନା ଜୀବନର ଅର୍ଥ କ’ଣ ଅଛି ?
କଙ୍କଡ଼ା ଭଲ ଭାବରେ ବୁଝିଗଲା
ସେ ନିଜକୁ ନିଜେ ଭାବିଲା, ‘ମୁଁ କ’ଣ ଏତେ ବୋକା ?”’
କୋକିଶିଆଳର ପ୍ରକୃତ ମନୋବୃତ୍ତି !
ସେ ନିଜକୁ ନିଜେ ଭାବିଲା, ‘ମୁଁ କ’ଣ ଏତେ ବୋକା ?”’
ଏବଂ ତା’ର ଗାତ ମଧ୍ୟରେ ଥାଇ ଉତ୍ତର ଦେଲା :
‘‘କିଏ ଯାଉଛି ଗୀତ ଗାଇବାକୁ ଓ ନାଚିବାକୁ
ଏପରି ଏକ ଶୀତଳ ପାଗରେ ?
ମୁଁ ବରଂ ଖାଇବି ଓ ଶୋଇବି ଭଲ ଭାବରେ
ମୋ’ର ଆରାମଦାୟକ ଛୋଟ ଗାତ ମଧ୍ଯରେ ।’’

Comprehension Questions (ବୋଧମୂଳକ ପ୍ରଶ୍ନବଳୀ):

Question 1.
What is the story about?
(ଗପଟି କେଉଁ ବିଷୟରେ ଲେଖାଯାଇଛି ?)
Answer:
The story is about a crab and a fox.

Question 2.
What did the fox look for?
(କୋକିଶିଆଳ କ’ଣ ଖୋଜୁଥୁଲା ?)
Answer:
The fox looked for a crab.

Question 3.
Did he find one?
(ସେ ଗୋଟିଏ ହେଲେ କଙ୍କଡ଼ା ପାଇଲା କି ?)
Answer:
He did not find one.

Question 4.
Where did he push his tail? Why?
(ସେ ତା’ର ଲାଞ୍ଜକୁ କେଉଁଠାକୁ ଠେଲିଲା ? କାହିଁକି ?)
Answer:
He pushed his tail inside the hole. Because he wanted to pull the crab out of its hole.

Question 5.
How could he know that the crab was inside?
(କଙ୍କଡ଼ା ଭିତରେ ଅଛି ବୋଲି ସେ କିପରି ଜାଣିପାରିଲା ? )
Answer:
He could smell that the crab was inside the hole.

Question 6.
What was his new plan?
(ତା’ର ନୂଆ ଯୋଜନାଟି କ’ଣ ଥିଲା ?)
Answer:
His new plan was to make friends with the crab and to call him to come out of its hole for enjoying the fine weather.

Question 7.
Did the new plan work? Why?
(ନୂଆ ଯୋଜନାଟି କାମ କଲା କି ? କାହିଁକି ?)
Answer:
No, the new plan did not work well. Because the crab was cunning enough to the fox’s real mood.

Question 8.
Did the crab understand the intention of the fox?
(କୋକିଶିଆଳର ଉଦ୍ଦେଶ୍ୟକୁ କଙ୍କଡ଼ାଟି ବୁଝିପାରିଥିଲା କି ?)
Answer:
Yes, the crab understood the intention of the fox.

Question 9.
Did the crab come out of her hole?
(କଙ୍କଡ଼ାଟି ତା’ର ଗାତ ବାହାରକୁ ଆସିଥିଲା କି ?)
Answer:
No, the crab did not come out of her hole.

Question 10.
Who is clever?
(କିଏ ଚତୁର ଥିଲା ?)
Answer:
The crab was clever.

Session – 2 (ସୋପାନ – ୨):
III. Post-Reading (ପଢ଼ିସାରିବା ପରେ):

1. Writing (ଲେଖୁବା):
(a) Answer the following questions.
(ନିମ୍ନଲିଖ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

Question (i).
What is the story about?
(ଗପଟି କେଉଁ ବିଷୟରେ ଲେଖାଯାଇ ଅଛି ?)
Answer:
The story is about a crab and a fox.

Question (ii).
What did the fox look for?
(କୋକିଶିଆଳଟି କ’ଣ ଖୋଜୁଥୁଲା ?)
Answer:
The fox looked for a crab.

Question (iii).
Where did the fox push his tail?
(କୋକିଶିଆଳଟି ତା’ର ଲାଞ୍ଜକୁ କେଉଁଆଡ଼କୁ ଠେଲିଲା ?)
Answer:
The fox pushed his tail inside the crab hole.

Question (iv).
Who is clever?
(କିଏ ଚତୁର ଅଟେ ?)
Answer:
The crab is clever.

(b) Write the story by filling in the gaps:
(ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କରି ଗପଟିକୁ ଲେଖ ।)
(Question with Answer)
Once there lived a ____________ and a ____________. The fox looked for ____________. He____________ his tail inside the ____________. But the crab did not catch his ____________. The fox _____________his plan. He sang a song and called the crab to come out. But the ____________. She said, “Am I _________. I’ll _____________in my ____________.”
Answer:
Once there lived a crab and a fox. The fox looked for a crab. He pushed his tail inside the crab hole. But the crab did not catch his tail. The fox changed his plan. He sang a song and called the crab to come out. But the crab did not come out. She said, “Am I a fool ?” I’ll rather eat and sleep well in my cozy little hole.”

WORD NOTE (ଶବ୍ଦାର୍ଥ):
(The words/phrases have been defined mostly on contextual meanings.)
(ଶବ୍ଦ । ଖଣ୍ଡବାକ୍ୟଗୁଡ଼ିକ ଅଧ୍ଯକାଂଶତଃ ପ୍ରସଙ୍ଗଗତ ଅର୍ଥ ଉପରେ ନିର୍ଭର କରି ବ୍ୟାଖ୍ୟା କରାଯାଇଛି ।)

dog — କୁକୁର
cat — ବିଲେଇ
thin — ପତଳା |
fat — ମେଦ
neither — ଏହା ନୁହେଁ କିମ୍ବା ତାହା ନୁହେଁ
pet — ଗୃହପାଳିତ ପଶୁ
mat — ଆସନ (ମସିଣା)
claimed — ଦାବି କଲେ
chased — ଗୋଡ଼ାଇଲା
retire — ବିଶ୍ରାମ ନେବା ବା ଶୋଇବା
hither — here, ଏଠାରେ
owner’s — ମାଲିକଙ୍କର,
cursed — ଅଭିଶାପିତ,
fate — ଭାଗ୍ୟ,
left — ବାମ,
As — ଯେପରି,
someone — କେହି ଜଣେ
pack — ପ୍ୟାକ୍ କରନ୍ତୁ |
sack — ଅଖା ବସ୍ତା
Hey – ହେ
gunny bag – ଛୋଟ ଅଖାଥଳି
grey — ଧୂସର ରଙ୍ଗ
wish — ଇଚ୍ଛା
religious — ଧାର୍ମିକ
carry — ବହନ କର
obey — ମାନ
nanny — ନାନୀ
funny — ମଜାଳିଆ
thought — ଭାବିଲା
rush — ଭିଡ଼, ଜନଗହଳି
Miss — ମିସ୍
hate — ଘୃଣା କରିବା
always — ସର୍ବଦା
late — ବିଳମ୍ବ, ଡେରି
cosy — ଉଷୁମ ଓ ଆରାମଦାୟକ
merry making — ହସଖୁସିରେ ତିଆରି
straightened — ସିଧା
fox — ଠେକୁଆ
crab — କଙ୍କଡ଼ା
greedy — ଲୋଭୀ
pole — ଖୁଣ୍ଟ

busy — ବୁଦାଳିଆ, ଲୋମଶ
pushed — ଠେଲି ହୋଇଗଲା
slightly — ଅଳ୍ପ ଟିକିଏ
missed — ମିସ୍
pull — ଟାଣନ୍ତୁ
smell — ଗନ୍ଧ ବା ବାସନା ଠଉରାଇବା
weather — ପାଣିପାଗ
understood — ବୋଧଗମ୍ୟ
real — ବାସ୍ତବ
mood — ମନ
himself – ତା’ ନିଜକୁ ନିଜେ
fool — ମୂର୍ଖ
hole — ଗାତ
cool — ଶୀତଳ
rather — ବରଂ
well — କୂପ