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Odisha State Board CHSE Odisha Class 11 Political Science Solutions Unit 4 Constitution at Work-I Objective Questions and Answers.

CHSE Odisha 11th Class Political Science Unit 4 Constitution at Work-I Objective Questions

Multiple Choice Questions With Answers

Question 1.
How many members are there in the election commission of India?
(a) one
(b) Two
(c) Three
(d) Five
Answer:
(c) Three

Question 2.
Which article of the constitutions of India deals with Election Commission?
(a) Art. 248
(b) Art. 268
(c) Art. 348
(d) Art. 324
Answer:
(d) Art. 324

Question 3.
How many members were there in the Election Commission of India where the Indian Constitution came into force?
(a) one
(b) two
(c) three
(d) four
Answer:
(a) one

Question 4.
In which year did membership of the election commission increase?
(a) 1977
(b) 1981
(c) 1985
(d) 1989
Answer:
(d) 1989

Question 5.
Who appoints the members of the Election commission?
(a) President of India
(b) Prime minister of India
(c) Parliament of India
(d) Council ministers
Answer:
(a) President of India

Question 6.
When the members of the election commission retire from service?
(a) At the age of 60 years
(b) At the age of 65 years
(c) At the age of 62 years
(d) At the age of 70 years
Answer:
(b) At the age of 65 years

Question 7.
Which type of representation is never seen in a democracy?
(a) Propositional Representation
(b) Universal suffrage
(c) Minority representation
(d) Communal representation
Answer:
(d) Communal representation

Question 8.
To which of the following offices the Election Commission does not conduct polls?
(a) President & Vice President
(b) Loksabha & Rajya Sabha
(c) Legislative Assembly Members
(d) Members of Municipality
Answer:
(d) Members of Municipality

Question 9.
From which year electronic voting. machine (EVM) was introduced in General Election?
(a) January 15, 1990
(b) March 15, 1989
(c) April 15, 1996
(d) February 15, 1998
Answer:
(b) March 15, 1989

Question 10.
When the campaigning for election ends?
(a) 24 hours before the poll
(b) 48 hours before the poll
(c) 24 hours before the end of the poll
(d) 48 hours before the end of the poll
Answer:
(d) 48 hours before the end of the poll

Question 11.
What percentage of vote does a candidate need to secure so that his security deposit will not be forfeited?
(a) One third
(b) One fourth
(c) One fifth
(d) One-sixth of the valid votes.
Answer:
(d) One-sixth of the valid votes.

Question 12.
Who among the following is a supporter of universal suffrage?
(a) J.S. Mill
(b) II.J. Laski
(c) Sir Henry Maine
(d) Blantschli
Ans.
(b) II.J. Laski

Question 13.
Electoral reforms have been introduced on the recommendation of which commission?
(a) Savalwal Committee
(b) Sindhi Committee
(c) Dinesh Goswamy Committee
(d) Virappa moilee Committee
Answer:
(c) Dinesh Goswamy Committee

Question 14.
On whose name the bills are introduced into the Parliament?
(a) President of India
(b) Prime Minister
(c) Speaker
(d) Departmental Minister
Ans.
(a) President of India

Question 15.
What is the tenure of Lok Sabha?
(a) five years
(b) four years
(c) six years
(d) none of these
Answer:
(a) five years

Question 16.
Who nominates the 12 members of Rajyasabha
(a) President of India
(b) Chairman of Rajya Sabha
(c) Members of Rajya Sabha
(d) Prime Minister
Answer:
(a) President of India

Question 17.
Who initiates money bill in the Lok Sabha?
(a) President of India
(b) Prime Minister
(c) Finance Minister of India
(d) Speaker of Lok Sabha
Answer:
(c) Finance Minister of India

Question 18.
How many seats are reserved for SC and ST in the Lok Sabha?
(a) 79 and 40
(b) 75 and 51
(c) 34 and 22
(d) 85 and 47
Answer:
(a) 79 and 40

Question 19.
What is the term of office of a Member of the Rajya Sabha?
(a) 5 years
(b) 6 years
(c) 4 years
(d) permanent
Answer:
(b) 6 years

Question 20.
Through which motion the President can be removed from office?
(a) No – confidence motion
(b) Censure motion
(c) Impeachment
(d) Exit motion
Answer:
(c) Impeachment

Question 21.
What is the minimum age qualification for membership of LokSabha?
(a) 25 years
(b) 30 years
(c) 35 years
(d) 21 years
Answer:
(a) 25 years

Question 22.
How many states of India have bicameral legislature?
(a) five
(b) six
(c) ten
(d) seven
Answer:
(a) five

Question 23.
Which one is not a function of the legislature?
(a) Enactment of Law
(b) Amendment of the Constitution
(c) Preparation of budget
(d) Ventilation of public grievances
Answer:
(c) Preparation of budget

Question 24.
How the Legislature controls the executive?
(a) By passing a no-confidence motion
(b) By putting questions to the ministers
(c) By initiation cut motion
(d) All the above
Answer:
(d) All the above

Question 25:
Which is the most powerful organ of Govt, in India?
(a) Executive
(b) Parliament
(c) Supreme Court
(d) Bureaucracy
Answer:
(b) Parliament

Question 26.
Which is the exclusive power of Rajya Sabha?
(a) To initiate money bills
(b) To initiate proposals for the removal of the President
(c) To initiate proposals for the removal of the Vice President
(d) To initiate proposals for the removal of the comptroller Auditor-General
Answer:
(c) To initiate proposals for the removal of the Vice President

Question 27.
On whose recommendation posts under All India Services can be increased or decreased?
(a) Lok Sabha
(b) Rajya Sabha
(c) Supreme Court
(d) Cabinet Sub-Committee
Answer:
(b) Rajya Sabha

Question 28.
What constitutes the Quorum of the LokSabha?
(a) l/5th of the total membership
(b) l/6th of the total membership
(c) l/10thof-total membership
(d) l/20th of the total membership
Answer:
(c) l/10thof-total membership

Question 29.
Who presides over the joint sitting of the Parliament?
(a) President of India
(b) Prime Minister of India
(c) Vice President
(d) Speaker of Lok Sabha
Answer:
(d) Speaker of Lok Sabha

Question 30.
What is the maximum strength of a Legislative Assembly?
(a) 485
(b) 450
(c) 500
(d) 600
Answer:
(c) 500

Question 31.
The Union Council of Ministers is responsible to
(a) Lok Sabha
(b) Rajya Sabha
(c) Union Parliament .
(d) President of India
Answer:
(a) Lok Sabha

Question 32.
Who prorogues the session of Parliament?
(a) Speaker of Lok Sabhas
(b) Prime Minister
(c) President
(d) Vice President
Answer:
(c) President

Question 33.
A money bill is introduced into?
(a) The Lok Sabha
(b) The Rajya Sabha
(c) Into either House of Parliament
(d) Union Cabinet
Answer:
(a) The Lok Sabha

Question 34.
Which House can initiate proposals of amendment?
(a) Lok Sabha
(b) Rajya Sabha
(c) Either Lok Sabha or Rajya Sabha
(d) State Legislative Assembly
Answer:
(c) Either Lok Sabha or Rajya Sabha

Question 35.
The Rajya Sabha is
(a) The Second Chamber of Parliament
(b) The Upper House of Parliament
(c) A Permanent Chamber
(d) All the above
Answer:
(d) All the above

Question 36.
Who acts as a link between the President and Union Parliament?
(a) Prime Minister
(b) Vice President
(c) Home Minister
(d) Speaker of Lok Sabha
Answer:
(d) Speaker of Lok Sabha

Question 37.
Who protects the privileges of members of Parliament?
(a) President
(b) Vice President
(c) Prime Minister
(d) Speaker
Answer:
(d) Speaker

Question 38.
Which is not a duty of the Speaker of Lok Sabha?
(a) To maintain discipline
(b) To interpret the rules of business of the House.
(c) To introduce bills
(d) To certify the money bills
Answer:
(c) To introduce bills

Question 39.
On whose recommendation the Governor dissolves the State Legislative Assembly?
(a) Prime Minister
(b) chief minister
(c) President
(d) State Council of Minister
Answer:
(c) President

Question 40.
On which items the State Legislature can make law?
(a) State List
(b) State List & Concurrent List
(c) Union List and State List
(d) State List and residuary matters
Answer:
(d) State List and residuary matters

Question 41.
How many S.C. and S.T. members are there in the Odisha Legislative Assembly?
(a) 22 and 34
(b) 34 and 22
(c) 32 and 24
(d) 24 and 32
Answer:
(a) 22 and 34

Question 42.
What is the strength of Odisha Vidhan Sabha?
(a) 155
(b) 165
(c) 149
(d) 147
Answer:
(d) 147

Question 43.
How many members represent Odisha in the Lok Sabha?
(a) Ten
(b) Twenty
(c) Twenty one
(d) Twenty-five
Answer:
(c) Twenty one

Question 44.
What is the term of the Legislative Assembly?
(a) five years
(b) six years
(c) one year
(d) four years
Answer:
(a) five years

Question 45.
Out of the 21 seats how many seats are reserved for SC and ST candidates representing Odisha in the Lok Sabha?
(a) 3 and 4
(b) 3 and 5
(c) 4 and 5
(d) 5 and 7
Answer:
(b) 3 and 5

Question 46.
When does the question hour start?
(a) 10 AM to 11 AM
(b) 11 AM to 12 Noon
(c) 12 Noon to 1 PM
(d) 4 PM to 5 PM
Answer:
(d) 4 PM to 5 PM

Question 47.
When the zero-hour starts?
(a) 9AMto 10 AM
(b) 10 AM to 11 AM
(c) 11 AM to 12 Noon
(d) 12 Noon to 1 PM
Answer:
(d) 12 Noon to 1 PM

Question 48.
Out of 147 members in Odisha Vidhan Sabha how many seats are reserved for SC and ST candidates?
(a) 23 and 27
(b) 21 and 25
(c) 23 and 34
(d) 27 and 39
Answer:
(c) 23 and 34

Question 49.
Who presides over the meeting of the Rajya Sabha?
(a) President of India
(b) Vice President
(c) Speaker
(d) Home Minister
Answer:
(b) Vice President

Question 50.
What is the maximum strength of Lok Sabha?
(a) 547
(b) 545
(c) 550
(d) 552
Answer:
(d) 552

Question 51.
When does the financial year start?
(a) January 1st
(b) March 1st
(c) April 1st
(d) June 1st
Answer:
(c) April 1st

Question 52.
Who certifies whether a bill is a money bill or not?
(a) President of India
(b) Speaker of Loksabha
(c) Finance Minister
(d) Prime Minister
Answer:
(b) Speaker of Loksabha

Question 53.
Who convinces the sessions of Parliament?
(a) President of India
(b) Prime Minister
(c) Speaker
(d) Secretary of Loksabha
Answer:
(a) President of India

Question 54.
In which part of the constitution there is mention of the election commission?
(a) Part-VIII
(b) Part – XI
(c) Part – XIV
(d) Part-XV
Answer:
(d) Part-XV

Question 55.
By which amendment Act the voting age of voters is reduced from 21 to 18 years?
(a) 59th
(b) 60th
(c) 61st
(d) 62nd
Answer:
(c) 61st

Answer the following questions in one word or digit

Question 1.
Who was the first Chief Election Commissioner of India?
Answer:
Sukumar Sen

Question 2.
Who is responsible for conducting a free and fair pool in India?
Answer:
Election Commission of India.

Question 3.
From which year EVMs are being used in Elections?
Answer:
14th March 1989

Question 4.
How many proposers and seconders sign the presidential nomination?
Answer:
50 and 50

Question 5.
For which posts direct election is held in India?
Answer:
Membership of Loksabha and Vidhansabha

Question 6.
For which posts indirect election is held?
Answer:
President, Vice-President’ & Rajya Sabha members.

Question 7.
From which date the voting age of voters is reduced from 21 to 18 years?
Answer:
21st March 1989.

Question 8.
What is the tenure of office of the members of the Election Commission?
Answer:
6 yrs or attaining 65 yrs which is earlier.

Question 9.
Which chamber of the legislature safeguards the interests of a federation?
Answer:
Upper Chamber

Question 10.
Which one is the primary organ of government?
Answer:
Legislature

Question 11.
The council Ministers in a parliamentary system remain answerable to which house of the legislature?
Answer:
Lower

Question 12.
From which word has the term Parliament has been derived?
Answer:
Latin word ‘Parle’

Question 13.
How many nominated members are there in the Rajya Sabha?
Answer:
12

Question 14.
How many members of Loksabha represent the Union Territories?
Answer:
13

Question 15.
Besides the Union List on which list can, the Parliament make law during normalcy?
Answer:
Concurrent List

Question 16.
By which power, the President can return an ordinary bill back to the Parliament for reconsideration?
Answer:
Suspensive Veto

Question 17.
Who gives the certificate to a money bill?
Answer:
Speaker

Question 18.
Who presides over the Rajya Sabha?
Answer:
Vice President

Question 19.
Who elects the members of the Rajya Sabha?
Answer:
Members of Vidhan Sabha

Question 20.
When was the first election to the Parliament in India held?
Answer:
1952

Question 21.
Who reads out the Oath of office and secrecy to the members of the Rajya Sabha?
Answer:
Chairman or Vice Chairman of the house

Question 22.
Which house of parliament can initiate proposals for the removal of the Vice President?
Answer:
Upper House

Question 23.
Which house of parliament initiates A confidence motion against the union council ministers
Answer:
LokSabha

Question 24:
For how many days can the Rajya Sabha detain a money bill?
Answer:
14 days.

Question 25.
Who acts is the leader of Lok Sabha?
Answer:
Prime Minister

Question 26.
Who acts as a link between the Parliament and President?
Answer:
Speaker

Question 27.
Who introduces the Annual Budget into the Lok Sabha?
Answer:
Finance Minister of India

Question 28.
For how many days the Rajya Sabha can delay an ordinary bill?
Answer:
Six Months

Question 29.
What is the hour just after the question hour is called?
Answer:
Zero hour

Question 30.
Who exercises casting vote in case of a tie?
Answer:
Speaker

Question 31.
Can there be a joint sitting for a constitutional amendment bill?
Answer:
No

Question 32.
Which constitutional authority can enhance the jurisdiction of the Supreme Court?
Answer:
Parliament

Question 33.
When the Lok Sabha is dissolved who approves the emergency proclamations?
Answer:
Rajya Sabha

Question 34.
Who Presides over the Lok Sabha during the absence of the Speaker?
Answer:
Deputy Speaker

Question 35.
What is the maximum limit of expenditure of a member of Lok Sabha during elections?
Answer:
25 lakhs

Question 36.
What is the maximum election expenditure limit of a member of Vidhan Sabha?
Answer:
15 lakhs

Question 37.
In which court of law the electoral disputes between the President and Vice President are settled?
Answer:
Supreme Court of India

Question 38.
In which court of law the electoral disputes of MPs and MLAs are settled?
Answer:
High Court

Question 39.
What is the amount of security deposit a presidential candidate must pledge?
Answer:
Rs. 15,000/

Question 40.
Who reads the oath of office of the member of parliament?
Answer:
President of India or his representative

Question 41.
What is the amount of the MP LAD fund?
Answer:
Rs.5 crores

Question 42.
What is the amount of the MLA LAD fund?
Answer:
One crore

Question 43.
How many members are there in the Rajya Sabha?
Answer:
233+12 = 245

Question 44.
When does the financial year start?
Answer:
April 1st

Question 45.
By which motion, the Union Council of ministry can be removed?
Answer:
No confidence motion.

Question 46.
Who casts the casting vote in case of a tie in the Loksabha?
Answer:
Speaker of Loksabha

Question 47.
What is the monthly salary of the Vice President?
Answer:
Rs. 1,25,000

Question 48.
What is the constitutional name of the budget?
Answer:
Annual Financial Statement

Fill in the blanks

1. Free and fair elections are indispensable for a ____ state.
Answer: Democracy.

2.The Indian constitution under Art._____ provides for an election commission.
Answer: Art. 324.

3. The membership of the election commission has increased to three from _____ general election.
Answer: Xth.

4. The tenure of office of the election commission is ____ years.
Answer: Six.

5. The members of the election commission retire at the age of ______ years.
Answer: 65.

6. The Chief Election Commissioner can be removed from office on the ground of _____.
Answer: Proved misbehavior and incapacity.

7._____ removes the members of the Election Commission?
Answer: President.

8. Seats have been reserved for the SC and STs in the Loksabha under Art. ______
Answer: Art. 330.

9. Under Art. _____ spats have been reserved for SC and ST candidates in the Vidhansabha.
Answer: Art. 332.

10. The president nominates two members from Anglo Indian Community to the Loksabha under Art._______
Answer: Art. 331.

11.______ Allots symbols to different political parties.
Answer: Election Commission.

12. Art. _____ provides for a joint sitting of the parliament.
Answer: Art. 108.

13. The Legislative Council can be created or abolished under Art. ______ of the constitution?
Answer: Art. 169.

14.Art._____ provides for consolidated fund of India.
Answer: Art. 266.

15. The constitution under Art. ______ provides for the emergency funds of India.
Answer: Art. 267.

16.______ sanctions fund out of the contingency fund of India.
Answer: President of India.

17. The parliament consists of the Loksabha. The Rajyasabha and the _____.
Answer: President of India.

18. Odisha sends ____ members to Rajyasabha.
Answer: 10.

19. The maximum strength, of Loksabha, is ____.
Answer: 552.

20. The maximum strength of the Rajya Sabha is _____.
Answer: 250.

21._____ members from Odisha represent the Parliament.
Answer: 31.

22. The members of Rajya Sabha are ____ elected.
Answer: Indirectly.

23. The minimum age of a member of Lok Sabha is _____ years.
Answer: 25 years.

24. A member of the Rajya Sabha must be above _____ years of age.
Answer: 30.

25. One-third members of the Rajya Sabha retire every ____ years
Answer: Two.

26. Rajya Sabha can detain an ordinary bill for not more than ______.
Answer: Six months.

27. Money bills are introduced into the ______.
Answer: Lok Sabha.

28.No Money Bill can be introduced into the Parliament without ______.
Answer: Prior permission of the President.

29.The Lok Sabha represents ___SC and ___ST candidates.
Answer: 79 & 40.

30. ______ is the Presiding Officer of the Lok Sabha.
Answer: Speaker.

31. The Lok Sabha is summoned for at least a ____ year.
Answer: Twice.

32. The ______ casts his vote in case of a tie.
Answer: Speaker.

33. So far National Emergency has been declared in India for _____ times.
Answer: Three.

34. The Lok Sabha can be dissolved by the _____ on the advice, of the-Prime Minister.
Answer: President.

35. The _____ can initiate proposals for the removal of the Vice-President.
Answer: Rajya Sabha.

36.No confidence motion is introduced only in the ______ house of parliament.
Answer: Lower.

37. Indian Parliament is ______.
Answer: Bicameral.

38._____ is the federal chamber of the Union Parliament.
Answer: Rajya Sabha.

39. The last National Emergency in India is declared on the ground of ______.
Answer: Internal disturbance.

40.______ is the final authority to decide on the disqualification of a member.
Answer: Speaker.

41. The President nominates ______ members from the Anglo-Indian Community to the Lok Sabha if the said community has not been adequately represented in the House.
Answer: Two.

42. The Indian President exercises only _____ veto power.
Answer: Suspensive.

43. The Quorum of the Lok Sabha is ______ of the total membership.
Answer: One-tenth.

44._____ was the only Prime Minister of India who never faced the Parliament.
Answer: Charan Singh.

45._____ members of the Rajya Sabha are elected.
Answer: 238.

46. The Union Territories represent ______ members to the Upper House.
Answer: Six.

47. Proposals of the constitutional amendment are initiated by _______ house of parliament.
Answer:
Either.

48. Under Art _____the Rajya Sabha can pass a resolution and empower the Parliament to create or abolish All India Services.
Answer: 249.

49. A legislative proposal before final approval is known as _______.
Answer: Bill.

50. All money bills are _______ bills.
Answer: Public.

51. The President of India can not withhold assent to a ______bill.
Answer: Money.

52. There is a bicameral legislature in ______ states of India.
Answer: Five.

53. The state legislature in Orissa is ______.
Answer: Unicameral.

54. The second chamber of the State Legislature in India is known as _____.
Answer: Vidhan Parishad.

55. The Presiding officer of the state legislative council is known as ______.
Answer: Chairman.

56. A member of the Legislative Council must be above years of age.
Answer: 30.

57. The legislative Council can delay an ordinary bill for a maximum period of ______.
Answer: Four months.

58. A member of the State Legislative Assembly must be not less than _____ years of age.
Answer: 25.

59. The Speaker of the Legislative Assembly determines the agenda in consultation with the ______.
Answer: Chief Minister.

60. The maximum strength of the Odisha Legislative Assembly is ______.
Answer:147.

61.Out of 147 members _____ seats have been reserved for the SC and ______seats for ST candidates.
Answer:23 and 34.

62. Provision for the creation of abolition of the Legislative Council has been mentioned under Art. ______ of the Constitution.
Answer:169.

63. The members of the State Legislative Council are ______ elected.
Answer: Indirectly.

64. The second chamber of the state legislature can retain money bills for ______ days.
Answer:14 days.

65._____ legislature is suitable for a small & poor country.
Answer: Unicameral.

66. The term of the Indian Parliament is ______.
Answer:5 years.

67. There are ______ Scheduled Caste and Scheduled Tribe members in the Lok Sabha from Odisha.
Answer:3 and 5.

68. There ____ SC and _____ ST members in the Lok Sabha.
Answer:79 and 40.

69. The Speaker of Lok sabha gets a monthly salary of _____ rupees.
Answer:55,000.

70. Odisha Vidhansabha is _____ in structure.
Answer: Unicameral.

71. The second chamber of the state legislature is called ______.
Answer: Vidhan Parishad

72. A _______ legislature is useful for a federal state.
Answer: Bicameral.

73. Under Art. ______ the Rajya Sabha can recommend the creation of posts for All India Services.
Answer:312.

74. The Legislative Council should not have less than _____ members.
Answer:40.

75. Money bills are introduced into the Lok Sabha as per Art. ______.
Answer:109 .

76. Railway budget is being introduced separately from the year_______.
Answer:1925.

77. Art. ______ provides for a vote on Account.
Answer:116.

78.Art._____ of the constitution deals with state legislature.
Answer:168.

79. The minimum age required for membership in Vidhansabha is _____.
Answer:25 yrs.

80. Universal franchise is extended under Art. ______.
Answer: Art. 326.

81. Anybody contesting for membership in Lok Sabha has to deposit _______ rupees as a security deposit.
Answer:10,000.

82. Anybody contesting for a seat in the Vidhan Sabha has to deposit Rs. _______ as security.
Answer:5000.

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 3 Relations And Functions Ex 3(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 3 Relations And Functions Exercise 3(b)

Question 1.
Let A = {a, b, c }, |B| = {1, 2}
(a) Determine all the relations from A to B and determine the domain, range, and inverse of each relation.
(b) Determine all the relations from B to A.
(c) Is there any relationship that is both a relation from A to B and B to A? How many?
(d) Of all the relations from A to B, identify which relations are many ones, one-many, and one-one and represent this diagrammatically.
Solution:
(a) A = {a, b, c}, B = {1, 2}
∴ A × B = {(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)}
∴ |A × B| = 6
∴ |P(A × B)| = 2⁶ = 64
∴ There are 64 relations from A to B as any subset of A × B. The domain of these relations is any subset of A. The inverse of these relations is any subset of B × A.
(b) There are 64 relations from B to A as any sub-set of B x A is a relation from B to A.
(c) Φ is the only relation that is from A to B and from B to A.
(d) Some many-one relations are {(a, 1), (b, 1), (c, 1), (b, 2) (c, 2)}, {(a, 2), (b, 2), (c, 2)}.

Question 2.
Are the following sets related?
(i) Φ from A to B.
(ii) A × B from A to B.
(iii) A × Φ from A to Φ.
(iv) Φ × B from Φ to B.
(v) Φ × Φ from Φ to Φ.
(vi) Φ × C from A to B.
(vii) Φ × Φ from A to B.
Determine the domain range and inverse of each of the relations mentioned above
Solution :
(i) Φ from A to B is a relation.
(ii) A × B from A to B is a relation.
(iii) A × Φ from A to Φ is a relation.
(iv) Φ × B from Φ to B is a relation.
(v) Φ × Φ from Φ to Φ is a relation.
(vi) Φ × C from A to B is a relation.
(vii) Φ × Φ from A to B is a relation.
∴ Domain of Φ i.e. D_Φ = Φ
Range of Φ i.e., R_Φ = Φ
Similarly, D_{A × B} = A, R_{A × B} = β
D_{A × Φ} = Φ, R_{A × Φ} = Φ
D _{Φ × B} = Φ = Φ, R _{Φ × B} = Φ
D _{Φ × Φ} = Φ, R _{Φ × C} = Φ
D _{Φ × C} = Φ, R _{Φ × C} = Φ
D _{Φ × Φ} = Φ, R _{Φ × Φ} = Φ
The inverse of the above relations is Φ, B × A, Φ × A, B × Φ, Φ × Φ, C ×  Φ, and Φ × Φ respectively.

Question 3.
Express the following relations on A to B in each case in tabular form :
(i) A = {n ∈ N : n ≤ 10}, B = N
f = {(x, y) ∈ A × B : y = x²}
Solution:
A = {n ∈ N : n ≤ 10}
= {1, 2, 3,…..10}, B = N
∴ B = {1, 2, 3}
∴ f ={(x, y) ∈ A × B : y = x²}
= {(1, 1), (2, 4), (3, 9)…..(10, 100)}

(ii) A = B = R
∴ f = {(x, y) : x² + y² = 1 and |x – y| = 1}
Solution:
A = B = R
∴ f = {(x, y) : x² + y² = 1 and |x – y| = 1}
={(0, 1) (1, 0), (-1, 0), (0, -1)}

(iii) (1, 2, 3, 4), B = {1, 2, 3, 4, 5}
f = {x, y) : 2 divides 3x+y}
Solution:
A = {1, 2, 3, 4}, B = {1, 2, 3, 4, 5}
∴ f = {(x, y) : 2 divides 3x+y}
={(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (3, 1), (3, 3), (3,5), (4, 2), (4, 4)}

Question 4.
A and B are non-empty sets such that |A| = m, |B| = n. How many relations can be defined from A to B ? (Remember that the number of relations is the number of subsets of (A × B).
Solution:
|A| = m, |B| = n
⇒ |A × B| = mn
A relation is a subset of A to B
∴ Number of relations from A to B
= Number of subsets of A × B
= 2^mn (∴ |A × B| = mn)

Question 5.
Give an example of a relation f such that
(i) dom f – rng f (ii) dom f ⊂ rng f
(iii) dom f ⊃ rng f
(iv) f ∪ f^-1 = Φ
(v) f = f^-1
(vi) f ∪ f^-1 ≠ Φ
Solution:
Let A = { 1, 2, 3} = B
(i) Let f = {(x, y) ∈ A × B : x = y}
∴ Dom f = {1, 2, 3} = Range f

(ii) Let f = {(1, 1), (1, 2), (2, 3)}
on A = (1, 2, 3}
∴ Dom f = {1, 2} ⊂ { 1, 2, 3} = Range f

(iii) Do yourself

(iv) Let f = Φ
∴ f^-1 = Φ = f ∪ f^-1 = Φ

(v) Let f = {(x, y) ∈ A × B; x² + y² = 1}, where A = B = {1, – 1, 0}
= {(1, 0), (0, 1), (-1, 0), (0, -1)}
f^-1 = {(0, 1) (1, 0), (0, -1), (-1, 0)}
=f

(vi) Let f = {(1, 3), (3, 1)} on A = { 1, 2, 3}
∴ f^-1 = {3, 1), (1, 3)},
so that f ∩ f^-1 = Φ.

Question 6.
Let R = {(a, a³) I a is a prime number less than 10}
Fine (i) R, (ii) dom R, (iii) rng R (iv) R^-1 (v) dom R^-1 (vi) rng R^-1
Solution:
R = {(a, a³)} a is a prime number less than 10}
(i) R = {(2, 8), (3, 27), (5, 125), (7, 343)}
(ii) dom R = {2, 3, 5, 7}
(iii) rng R = {8, 27, 125, 343}
(iv) R^-1 = {(8, 2), (27, 3), (125, 5), (343, 7)}
(v) Dom R^-1 = {8, 27, 125, 343} = rng R
(vi) rng R^-1 = {2, 3, 5, 7} = dom R

Question 7.
Let A = {1, 2, 3, 4, 5, 6} and Let R be a relation on A defined by R = {(a, b)} a divides b
Find (i) R, (ii) dom R, (iii) rng R (iv) R^-1, (v) Dom R^-1 (vi) rng R^-1
Solution:
A = {1, 2, 3, 4, 6}
R on A is defined by
R = {(a, b) | a divides b}
(i) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6) (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}
(ii) dom R = {1,2, 3, 4, 6} = A
(iii) rng R = {1, 2, 3, 4, 6} = A
(iv) R^-1 = {(1, 1), (2, 1), (3, 1), (4, 1), (6, 1), (2, 2) (4, 2), (6, 2), (3, 3), (6, 3), (4, 4), (6, 6)}
(v) dom R^-1 = {1, 2, 3, 4, 6} = A
(vi) rng R^-1 = {l, 2, 3, 4, 6} = A

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 4 Trigonometric Functions Ex 4(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Exercise 4(a)

Question 1.
State which of the following is positive.
(i) cos 271°
Solution:
cos 271° is + ve as 271° lies in 4th quadrant.

(ii) sec 73°
Solution:
sec 73° is + ve as sec +ve in the 1st quadrant.

(iii) sin 302°
Solution:
sin 302° is- ve as sin is -ve in the 4th quadrant

(iv) cosec 159°
Solution:
cosec 159° is + ve as 159° lies in 2nd quadrant and cosec is +ve there.

(v) sec 199°
Solution:
sec 199° is – ve as 199° lies in the 3rd quadrant and sec is -ve there.

(vi) cosec 126°
Solution:
cosec 126° is + ve as cosec is +ve in 2nd quadrant.

(vii) cos 315°
Solution:
cos 315° is +ve as 315° lies in 4th quadrant and cos is +ve there.

(viii) cot 375°
Solution:
cot 375° is +ve as 375° lies in 1st quadrant.

Question 2.
Express the following as trigonometric ratios of some acute angles.
(i) sin 1185°
Solution:
sin 1185° = sin\(\left(13 \frac{\pi}{2}+15^{\circ}\right)\)
=(- 1) ^{\(\frac{13-1}{2}\)} cos 15° = cos 15°

(ii) tan 235°
Solution:
tan 235° = tan (180° + 45°) = tan 45°

(iii) sin (- 3333°)
Solution:
sin (-3333°) – -sin 3333°
= – sin\(\left(37 \frac{\pi}{2}+3^{\circ}\right)\)
= – (- 1) ^{\(\frac{27-1}{2}\)} cos 3° =- cos 3°

(iv) cot (- 3888°)
Solution:
cot (-3888°) = – cot 3888°
= – cot\(\left(43 \frac{\pi}{2}+18^{\circ}\right)\)
= – (- tan 18°) = tan 18°

(v) tan 458°
Solution:
tan 458° = tan\(\left(5 \frac{\pi}{2}+8^{\circ}\right)\) = – cot 8°

(vi) cosec (- 60°)
Solution:
cosec (- 60°) = – cosec 60°

(vii) cos 500°
Solution:
cos 500° = cos\(\left(5 \frac{\pi}{2}+50^{\circ}\right)\)
= – (-1) ^{\(\frac{5+1}{2}\)} sin 55° – sin 50°

(viii)sec 380°
Solution:
sec 380° = sec (360° + 20°)
= sec 20°

Question 3.
Find the domain of tangent and cotangent functions.
Solution:
Domain of tan x is R – \(\left\{\frac{(2 n+1) \pi}{2}, n \in Z\right\}\) as tangent is not defined for
x = \(\frac{(2 n+1) \pi}{2}\)
The domain of cot x is R – {nπ, n ∈ Z} as cotangent is not defined for x = nπ.

Question 4.
Determine the ranges of sine and cosine functions.
Solution:
The maximum and minimum values of sine and cosine are 1 and -1, respectively.
∴ Ranges of sine and cosine are [-1, 1].

Question 5.
Find a value of A when cos 2A = sin 3A
Solution:
cos 2A = sin 3A = cos (90° – 3A)
or, 2A = 90° – 3A
or, 5A = 90° or, A = 18°

Question 6.
Find the value of
cos 1°. cos 2° …..cos 100°
Solution:
cos 1° cos 2° …..cos 100°
= 0 as cos 90° is there which is zero.

Question 7.
Find the value of
cos 24° + cos 5° + cos 175° + cos 204° + cos 300°
Solution:
cos 24° + cos 5° + cos 175° + cos 204° + cos 300°
= cos 24° + cos 5° + cos (180° – 5°) + cos (180° + 24°) + cos (360°- 60°)
= cos 24° + cos 5° – cos 5° – cos 24° + cos 60° = cos 60° = 1/2

Question 8.
Evaluate
tan\(\frac{\pi}{20}\).tan\(\frac{3 \pi}{20}\).tan\(\frac{5 \pi}{20}\).tan\(\frac{7 \pi}{20}\).tan\(\frac{9 \pi}{20}\)
Solution:
tan\(\frac{\pi}{20}\).tan\(\frac{3 \pi}{20}\).tan\(\frac{5 \pi}{20}\).tan\(\frac{7 \pi}{20}\).tan\(\frac{9 \pi}{20}\)
= tan 9° tan 27° tan 45° tan 63° tan 81°
= tan 9°. tan 27°. 1 tan (90° – 27°). tan (90° – 9°)
= tan 9° tan 27° cot 27° cot 9°
= (tan 9°. cot 9°) x (tan 27°. cot 27°)
=1 × 1=1

Question 9.
Show that
\(\frac{\sin ^3\left(180^{\circ}+\mathbf{A}\right) \cdot \tan \left(360^{\circ}-\mathbf{A}\right) \sec ^2\left(180^{\circ}-\mathbf{A}\right)}{\cos ^2\left(90^{\circ}+\mathbf{A}\right){cosec}^2 A \cdot \sin \left(180^{\circ}-A\right)}\) = tan³ A
Solution:
L.H.S

= tan³A      (R.H.S)

Question 10.
If A = cos² θ + sin⁴ θ then prove that for all values of θ, 3/4 ≤ A ≤ 1.
Solution:
A = cos² θ + sin⁴ θ =1 – sin² θ sin⁴ θ
or, sin⁴ θ – sin² θ + (1 – A) = 0 …(1)
Eqn. (I) is quadratic in sin² θ.
∴ sin² θ = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
\(=\frac{1 \pm \sqrt{1-4(1-\mathrm{A})}}{2 \times 1}\)
Where a=1, b = – 1, c = 1 – A
∴ sin² θ = \(\frac{1 \pm \sqrt{4 A-3}}{2}\)
We know that sin20 is not negative and lies in [0, 1]
So, \(\sqrt{4 \mathrm{~A}-3}\) ≤ 1
⇒ 4A – 3 ≤ 1 ⇒ 4A ≤ 4 ⇒ A ≤ 1  …(2)
Again, since sin² θ is real,
b² – 4ac must be +ve
i.e., 4A – 3 ≥ 0 ⇒ A ≥ 3/4
∴ From (2) and (3),
We have 3/4 ≤ A ≤ 1          (Proved)

Odisha State Board BSE Odisha 10th Class Hindi Solutions Poem 1(b) सूरदास के पद Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Hindi Solutions Poem 1(b) सूरदास के पद

प्रश्न और अभ्यास (ପ୍ରଶ୍ନ ଔର୍ ଅଭ୍ୟାସ)

1. निम्नलिखित प्रश्नों के उत्तर दो-तीन वाक्यों में दीजिए।
(ନିମ୍ନଲିଖ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦୁଇ-ତିନି ବାକ୍ୟରେ ଦିଅ ।)

(क) कृष्ण यशोदा से क्या शिकायत करते है और क्यों?
(କୃଷ୍ଣ ଯଶୋଦା ସେ କ୍ୟା ଶିକାୟତ୍ କରନେ ହୈ ଔର୍ କେଁ ?)
उत्तर:
कृष्ण यशोदा से यह शिकायत करते हैं कि- माँ! मुझे बलराम भैया चिढ़ाते हैं। वे कहते हैं कि मैं खरीद कर लाया गया हुँ। जसुमति ने तुम्हें जन्म नहीं दिया है। इसलिए मैं उनके साथ खेलने नहीं जाता।

(ख) बलराम कृष्ण से क्या पूछते हैं?
(ବଳରାମ କୃଷ୍ଣ ସେ କ୍ୟା ପୁଛତେ ହୈ ?)
उत्तर:
बलराम कृष्ण से पूछते हैं कि – कौन तेरे माता पिता हैं? नंद गोरे हैं और यशोदा गोरी हैं, क्यों काला है?

(ग) यशोदा किसकी कसम खाती हैं और क्या कहती हैं?
(ଯଶୋଦା କିସ୍‌ କସମ୍ ଖାତୀ ହୈ ଔର୍ କ୍ୟା କହତୀ ହୈ ?)
उत्तर:
यशोदा गोधन की कसम खाती हैं और कहती हैं कि ‘मैं तेरी माता हूँ और तू मेरा पुत्र है। यह बलराम चुगलखोर है और जन्म से शरारती है।

(घ) चुटकी देकर ग्वाल-बालक क्यों नाचते हैं?
(ଚୁଟକୀ ଦେକର୍ ଗ୍ୱାଲା-ବାଳକ ଜ୍ୟୋ ନା ହେଁ ?)
उत्तर:
कृष्ण को बलराम के कहने पर कि नंद गोरे हैं, यशोदा गोरी हैं मगर तू काला क्यों है? तुझे माता यशोदा ने खरीद कर लाया है। यही सब बातें सुनकर कृष्ण को चिढ़ाने के लिए ग्वाल बालक चुटकी बजाकर नाचते हैं।

2. निम्नलिखित पदों के अर्थ दो-तीन वाक्यों में स्पष्ट कीजिए:
(ନିମ୍ନଲିଖତ୍ ପଦୌ କେ ଅର୍ଥ ଦୋ-ତୀନ୍ ୱାର୍କୋ ମେଁ ସ୍ପଷ୍ଟ କୀଜିଏ: )
(ନିମ୍ନଲିଖ୍ ପଦଗୁଡ଼ିକରେ ଅର୍ଥ ଦୁଇ-ତିନି ବାକ୍ୟରେ ସ୍ପଷ୍ଟ କର: )

(क) पुनि-पुनि कहत कौन है माता, कौन है तुमरो तात।
(ପୁନି-ପୁନି କହତ କୌନ୍ ହୈ ମାତା, କୌନ୍ ହୈ ତୁମରେ ତାତ ।)
उत्तर:
कृष्ण की बाललीला का वर्णन करते हुए कवि ने उसका सुंदर वर्णन किया है। कृष्ण बलराम चिढ़ाने के लिए बारबार कहते हैं कि तुम्हारी माता कौन हैं और पिता कौन हैं? तुम बाबानंद और माता यशोदा के पुत्र नहीं हो। तुम्हें माता यशोदा ने खरीद कर लाया है।

(ख) सूर स्याम मोहि गोधन की सौं हौं माता तू पूत।
(ସୂର୍ ଶ୍ୟାମ ମୋହି ଗୋଧନ କୀ ସୌ ହେଁ ମାତା ତୂ ପୂତ ।)
उत्तर:
जब बालक कृष्ण के मुख पर यशोदा गुस्सा देखती है और कृष्ण के मुख से गुस्सैली वाणी सुनती हैं तो वह प्रसन्न हो जाती हैं। फिर कृष्ण से कहती हैं कि बलराम चुगलखोर है और शरारती है। वह गोधन की कसम खाकर कहती हैं कि कृष्ण ही उनका पुत्र है और वह कृष्ण की माता है।

(ग) तनक – तनक चरननि सौं, नाचत, मनहिं मनहिं रिझावत।
(ନିମ୍ନଲିଖ୍ ପଦଗୁଡ଼ିକରେ ଅର୍ଥ ଦୁଇ-ତିନି ବାକ୍ୟରେ ସ୍ପଷ୍ଟ କର: )
उत्तर:
इस पंक्ति में कवि ने कहा है कि कृष्ण अपने आप कुछ गा रहे हैं। वे गाते-गाते अपने नन्हे चरणों से नाचते हैं और मन-ही-मन खुश हो रहे हैं।

(घ) कबहुँ चितै प्रतिबिम्ब खंभ में, लौनी – लिए खबावत।
(କବହୁ ଚିତୈ ପ୍ରତିବିମ୍ବ ଖମ୍ବ ମେଁ, ଲୌନୀ-ଲିଏ ଖବାୱତ୍।)
उत्तर:
बालक कृष्ण घर के भीतर जाकर थोड़ा मक्खन हाथ में लेकर खाते हैं और कुछ मक्खन अपने मुँह पर भी लगा देते हैं। कभी खम्भे में अपना प्रतिबिंब देखकर उसे भी कुछ मक्खन खिलाते हैं।

3. निम्नलिखित प्रश्नों के उत्तर एक-एक वाक्य में दीजिए।
(ନିମ୍ନଲିଖୂ ପ୍ରଶ୍ନ କେ ଉତ୍ତର୍ ଏକ୍- ଏକ୍ ୱାକ୍ୟ ମେଁ ଦୀଜିଏ । (ନିମ୍ନଲିଖ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଗୋଟିଏ-ଗୋଟିଏ ବାକ୍ୟରେ ଦିଅ ।)

(क) बाललीला (पद) के रचयिता कौन हैं?
(ବାଳଲୀଳା (ପଦ) କେ ରଚୟିତା କୌନ୍ ହୈ ।)
उत्तर:
बाललीला (पद) के रचयिता सूरदास हैं।

(ख) कौन कहते हैं कि तुझे मोलकर लाया गया है?
(କୌନ୍ କହତେ ହୈ କି ତୁଝେ ମୋଲ କର ଲାୟା ଗୟା ହୈ?)
उत्तर:
बलराम कहते हैं कि तुम्हें मोल कर लाया गया है।

(ग) बलराम पुनः पुनः क्या कहते हैं?
(ବଳରାମ ପୁନଃ ପୁନଃ କ୍ୟା କହତେ ହୈ ?)
उत्तर:
बलराम पुनः पुनः कहते हैं कि कौन तुम्हारी माता हैं और कौन तुम्हारे पिता हैं।

(घ) ग्वाल बालक किस तरह हँसते हैं?
(ସ୍ଵାଲେ ବାଳକ କିସ୍ ତରହ ହଁସତେ ହେଁ ?)
उत्तर:
ग्वाल बालक चुटकी बजाकर नाचते हैं और हँसते हैं।

(ङ) माँ यशोदा ने किसे मारना सिखा है?
(ମାଁ ଯଶୋଦା ନେ କିସେ ମାର୍‌ନା ସିଖା ହୈ ?)
उत्तर:
माँ यशोदा ने कान्हा को मारना सीखा है।

(च) कौन दाऊ पर नहीं खीझती है?
(କୌନ୍ ଦାଊ ପର ନହୀ ଶୀଝାତୀ ହୈ ?)
उत्तर:
माता यशोदा दाऊ पर नहीं खीझती हैं।

(छ) स्याम शरीर का अर्थ क्या है?
(ଶ୍ୟାମ୍ ଶରୀର କା ଅର୍ଥ କ୍ୟା ହୈ ?)
उत्तर:
स्याम शरीर का अर्थ है ‘काला शरीर’ या श्याम मरंग का शरीर।

(ज) ‘जनमत ही को धूत’ का अर्थ क्या है?
(‘ଜନମତ୍‌ ହୀ କୋ ଧୂତ୍’ କା ଅର୍ଥ କ୍ୟା ହୈ ?)
उत्तर:
‘जनमत ही को धूत’ का अर्थ है जन्म से शरारती।

(झ) माँ यशोदा किसकी सौगंध खाती हैं?
(ମାଁ ଯଶୋଦା କିସ୍‌ ସୌଗନ୍ଧା ଖାତି ହେଁ ?)
उत्तर:
माँ याशोदा गोधन की सौगंध खाती हैं।

(अ) कृष्ण किसे माखन खिलाते हैं?
(ମାଁ ଯଶୋଦା କିସ୍‌ ସୌଗନ୍ଧା ଖାତି ହେଁ ?)
उत्तर:
कृष्ण खंभे में अपने प्रतिबिंब को देखकर उसे माखन खिलाते हैं।

(ट) यशोमति क्या देखकर हर्षित हो जाती हैं?
(ଯଶୋମତି କ୍ୟା ଦେଖ୍କର୍ ହର୍ଷିତ୍ ହୋ ଜାତୀ ହୈ ?)
उत्तर:
यशोमती कृष्ण की बाललीला को देखकर हर्षित हो जाती हैं।

भाषा-ज्ञान (ଭାଷା-ଜ୍ଞାନ)

प्रश्न 1.
निम्नलिखित शब्दों के तत्सम रूप लिखिए:
(ନିମ୍ନଲିଖ୍ ଶବ୍ଦଗୁଡ଼ିକର ତତ୍‌ସମ ରୂପ ଲେଖ)
मोल, सौं, पूत, तनक, धूत, बाँह, मैया, खिझायो, गैयनि, कजरी
उत्तर:
मोल – मूल्य
पूत – पुत्र
धूत – दुष्ट
मैया – माता
गैयनि – गौ
सौं – शपथ
तनक – क्षुद्र
बाँह – हस्त
खिझायो – उपहास
कजरी – श्याम

प्रश्न  2.
निम्नलिखित शब्दों के समानार्थी शब्द लिखिए :
(ନିମ୍ନଲିଖ୍ ଶବ୍ଦଗୁଡ଼ିକର ସମାନାର୍ଥୀ ଶବ୍ଦ ଲେଖ ।)
नित, चरन, कर, बाँह, रिस, तात, जात, बदन, स्याम, धूत, सौं, पूत, शरीर, खीझें, चबाई
उत्तर:
नित – हमेशा/सदैव
चरन – पैर / पाँव
कर – हाथ
बाँह – हाथ
रिस – गुस्सा
तात – पिता
जात – जन्म
बदन – मुँह / चेहरा
स्याम – काला
धूत – धूर्त / शरारती / दुष्ट
सौं – सौगंध/शपथ/कसम
पूत – पुत्र/तनय/बेटा
शरीर – काया/देह
खीझे – गुस्सा
चबाई – चुगलखोर/निंदक

Very Short & Objective type Questions with Answers

A. निम्नलिखित प्रश्नों के उत्तर एक वाक्य में दीजिए।

प्रश्न  1.
भक्तिकाल के सर्वश्रेष्ठ कृष्ण-भक्त कवि किसे माना जाता है?
उत्तर:
भक्तिकाल के सर्बश्रेष्ठ कृष्ण-भक्त कवि सूरदास को माना जाता है।

प्रश्न  2.
सूरदास की प्रमुख प्रामाणिक रचन क्या है?
उत्तर:
सूरदास के प्रमुख प्रामाणिक रचना ‘सूरसागर’ अमूल्य निधि है।

प्रश्न  3.
सूरदास के गुरु कौन थे?
उत्तर:
सूरदास के गुरु बल्लभायार्य थे।

प्रश्न 4.
यशोमति क्या देखकर हर्षित हो जाति है?
उत्तर:
यशोमति कृष्ण की बाललीला को देखकर हर्षित हो जाती है ।

प्रश्न  5.
‘जनमत ही को धूत’ का अर्थ क्या है?
उत्तर:
‘जनमत ही को धूत’ का अर्थ है जन्म से शरारती।

प्रश्न  6.
श्रीकृष्ण अपने छोटे-छोटे पैरों से क्या करते हैं?
उत्तर:
श्रीकृष्ण अपने छोटे-छोटे पैरों से नाचते हैं।

प्रश्न  7.
श्रीकृष्ण बाँहें उठाकर क्या करते हैं?
उत्तर:
श्रीकृष्ण बाँहें उठाकर कजरी और धौली गायों को बुलाते हैं।

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

प्रश्न  1.
कौन कान्हा को बहुत खिझाता है?
उत्तर:
दाऊ

प्रश्न  2.
दाऊ के खिझाने से कान्हा के नाराज होने पर माँ क्या कहकर समझाती है?
उत्तर:
मैं माता तू पुत्र

प्रश्न  3.
कान्हा क्यों खेलने नहीं जाते?
उत्तर:
रिस के मारे

प्रश्न  4.
दाऊ किसको खिझाते हैं?
उत्तर:
कृष्ण को

प्रश्न  5.
यशोदा कृष्ण मुँह से क्या सुनकर खुश हो जाती हैं?
उत्तर:
शिकायत

प्रश्न  6.
ग्वाले किस तरह हँसते हैं?
उत्तर:
चुटकी बजाकर

प्रश्न  7.
यशोमती क्या देखकर हर्षित हो जाती हैं?
उत्तर:
कृष्ण की बाललीला

प्रश्न  8.
माता यशोदा किसकी सौगंध खाती हैं?
उत्तर:
गोधन को

प्रश्न  9.
जो चुटकी बजाकर नाचते हैं, वे कौन हैं?
उत्तर:
ग्वाले

प्रश्न 10.
कौन जन्म से धूर्त हैं?
उत्तर:
बलभद्र

प्रश्न  11.
‘लौनी’ शब्द का अर्थ क्या है?
उत्तर:
मक्खन

C. रिक्त स्थानों की पूर्ति कीजिए।

प्रश्न 1.
…………………. दाऊ पर नहीं खीझती है।
उत्तर:
यशोदा

प्रश्न 2.
बालकृष्ण के हाथ में ………………… है।
उत्तर:
मक्खन

प्रश्न 3.
कृष्ण खंभे में ……………….. देखते हैं।
उत्तर:
प्रतिविम्ब

प्रश्न 4.
बाल कृष्ण ………………. हाथ उठाकर बुलाते हैं
उत्तर:
गायों को

प्रश्न 5.
………………. अपने आँगन में नाचते-गाते हैं।
उत्तर:
कृष्ण

प्रश्न 6.
कृष्ण ……………… माखन खिलाते हैं।
उत्तर:
प्रतिबिंब को

प्रश्न 7.
मैया मोहि दाउ बहुत दाउ बहुत खिझायो – इस पंक्ति में मोहि पद ……………….. के लिए प्रयोग हुआ है।
उत्तर:
कृष्ण

प्रश्न 8.
कृष्ण खलते समय अपने बदन पर ………………….. लगाते है।
उत्तर:
माखन

प्रश्न 9.
………………….. अपने आँगन में कछु गावत।
उत्तर:
हरी

प्रश्न 10.
ग्वाले बालक ………………… बजाकर हँसते हैं।
उत्तर:
चुटकी

प्रश्न 11.
दूरी देखति …………………. यह लीला।
उत्तर:
जसुमति

D. ठिक् या भूल लिखिए।

प्रश्न 1.
सूरदास के गुरु वल्लभाचार्य थे।
उत्तर:
ठिक्

प्रश्न 2.
बलराम कहते हैं कि ‘दाउहि कबहुँ न खीझै ‘?
उत्तर:
भूल

प्रश्न 3.
बलराम शरीर का रंग श्याम है
उत्तर:
भूल

प्रश्न 4.
यशोदा गोधन की कसम खाती हैं।
उत्तर:
ठिक्

प्रश्न 5.
माता यशोदा गोपियों को मारना सीखा है।
उत्तर:
भूल

प्रश्न 6.
मोहन की गुस्सैल वाणी सुनकर यशोमती प्रसन्न हो जाती है।
उत्तर:
ठिक्

प्रश्न 7.
‘गोरे नंद यशोदा गोरी, तू कत स्याम शरीर’ – इस पंक्ति के रचयिता कबीरदास हैं।
उत्तर:
भूल

प्रश्न 8.
कृष्ण बीमारी के कारण खेलने नहीं जाते।
उत्तर:
भूल

प्रश्न 9.
वलराम जन्म से चुगलखोर है।
उत्तर:
ठिक्

प्रश्न 10.
कृष्ण कजरी धौरी गायों को बुलाते हैं।
उत्तर:
ठिक्

प्रश्न 11.
कृष्ण बाँह उठाकर नंद बाबा को बुलाते हैं।
उत्तर:
भूल

Multiple Choice Questions (mcqs) with Answers

सही उत्तर चुनिए : (MCQs)

1. कौन कान्हा को बहुत खिझाता है?
(A) दाऊ
(B) नन्द
(C) यशोदा
(D) ग्वाल बाल
उत्तर:
(A) दाऊ

2. दाऊ के खिझाने से कान्ह के नाराज होने पर माँ क्या कहकर समझाती है?
(A) बलभद्र मेरा पुत्र नहीं है
(B) मैं माता तू पुत्र
(C) मैं बहुत खुश हूँ
(D) तू नाराज न हो
उत्तर:
(B) मैं माता तू पुत्र

3. कान्ह क्यों खेलने नहीं जाते?
(A) दुःख के मारे
(B) रिस के मारे
(C) खुसी के मारे
(D) डर के मारे
उत्तर:
(B) रिस के मारे

4. दाऊ किसको खिझातें हैं?
(A) कृष्ण को
(B) ग्वालों को
(C) गोपियों को
(D) माता यशोदा को
उत्तर:
(A) कृष्ण को

5. यशोदा कृष्ण के मुँह से क्या सुनकर खुश हो जाती हैं?
(A) शिकायत
(B)खेलने न जाने की बात
(C) क्रोधपूर्ण बातें
(D) बलराम के चिढ़ाने की बात
उत्तर:
(A) शिकायत

6. बाललीला (पद) के रचयिता कौन हैं?
(A) रामदास
(B) हरीदास
(C) सूरदास
(D) भक्तदास
उत्तर:
(C) सूरदास

7. वाले किस तरह हँसते हैं?
(A) चुटकी बजाकर
(B) गाना गाकर
(C) माखन खाकर
(D) तालियाँ बजाकर
उत्तर:
(A) चुटकी बजाकर

8. कौन दाऊ पर नहीं खीझती है?
(A) कृष्ण
(B) बलराम
(C) यशोदा
(D) बिजली कन्या
उत्तर:
(C) यशोदा

9. सूरदास के गुरु कौन थे?
(A) रामानन्द सागर
(B) वल्लभाचार्य
(C) रामकृष्ण परमहंस
(D) विवेकानन्द
उत्तर:
(B) वल्लभाचार्य

10. यशोमती क्या देखकर हर्षित हो जाती हैं?
(A) कृष्ण की बाललीला
(B) अर्जुन की विश्वरुप
(C) कृष्ण की गोपलीला
(D) बिजली कन्या के जन्म
उत्तर:
(A) कृष्ण की बाललीला

पद: (ପଦ)

1. मैया मोहि दाऊ बहुत खिझायो।
मोसों कहत मोल को लीन्हों, तू जसुमति कब जायो॥
कहा कहौं एही रिस के मारे, खेलन हौं नहीं जात।
पुनि-पुनि कहत कौन है माता, कौन है तुमरो तात॥
गोरे नन्द, जसोदा गोरी, तू कत स्याम सरीर।
चुटकी दै दै हँसत ग्वाल सब, सिखै देते बलबीर॥
तू मोही को मारन सीखी, दाउहि कबहूँ न खीझै।
मोहन मुख रिस की ये बातौं, जसुमति सुनि – सुनि रीझै॥
सुनहू कान्ह बलभद्र चबाई, जनमत ही को धूत।
सूर स्याम मोहि गोधन की सौं, हौं माता तू पूत॥

ମୈୟା ମୋହି ଦାଉ ବହୁତ ଖ୍ରୀୟୋ।
ମୋର୍ସ କହତ ମୋଲ୍ କୋ ଲୀକ୍ଷ୍ନୌ, ତୂ ଜସୁମତି କବ ଜାୟେ ॥
କହା କହାଁ ଏହୀ ରିସ କେ ମାତା, ମାରେ, ଖେଲନ ହେଁ ନହିଁ ଜାତ।
ପୁନି-ପୁନି କହତ କୌନ୍ ହୈ କୌନ ହି ତୁମରୋ ତାତ॥
ମାରେ, ଖେଲନ ହେଁ ନହିଁ ଜାତ। ମାତା, କୌନ ହି ତୁମରୋ ତାତ ॥
ଗୋରେ ନନ୍ଦ, ଜସୋଦା ଗୋରୀ, ତୂ କତ ସ୍ୟାମ ସରୀର।
ଚୁଟକୀ ଦୈ ଦୈ ହଁସତ ଗ୍ବାଲ ସବ, ସିଖି ଦେତେ ବଲବୀର॥
ତୂ ମୋହୀ କୋ ମାରନ ସୀଖୀ, ଦାଉହି କବହୁଁ ନ ଖୀର୍ଷେ।
ମୋହନ ମୁଖ ରିସ କୀ ୟେ ବାର୍ଡୋ, ଜସୁମିତ ସୁନି-ସୁନି ରୀର୍ଥେ ॥
ସୁନହ୍ନି କାହ୍ନ ବଲଭଦ୍ର ଚବାଈ, ଜନମତ ହୀ କୋ ଧୂତ।
ସୂର ସ୍ୟାମ ମୋହି ଗୋଧନ କୀ ସୌ, ହେଁ ମାତା ତୃ ପୂତ॥

हिन्दी व्याख्या: इस पद में कृष्ण की बाललीला का वर्णन है। बालक कृष्ण माँ यशोदा के पास शिकायत करते हैं, “माँ! मुझे बलराम भैया चिढ़ाते हैं। वे मुझे कहते हैं कि तुझे खरीद कर लिया गया है। जसुमति ने तुझे जन्म नहीं दिया है। इसलिए मैं उनके साथ खेलने नहीं जाता। वे बार-बार मुझे पूछते हैं कि कौन तेरी माता और कौन तेरे पिता हैं? और कहते हैं कि नंद गोरे हैं, यशोदा गोरी है, तू क्यों श्यामल/काला है। यह सुनकर मुझे चिढ़ाने के लिए ग्वाल बालक चुटकी बजाकर नाचते हैं।

बलराम भैया उन्हें सिखा देते हैं। तूने सिर्फ मुझे मारना सीखा है। बलराम भैया पर खीझती भी नहीं ।” मोहन के मुख पर गुस्सा देखकर और उनकी गुस्सैली वाणी सुनकर यशोमति प्रसन्न हो जाती हैं। माँ कहती है कि कान्हा सुन, यह बलराम चुगलखोर है, वह जन्म से शरारती है। मैं गोधन की कसम खाकर कहती हूँ कि मैं तेरी माता और तू मेरा पुत्र है।

ଓଡ଼ିଆ ଅନୁବାଦ:
ଏହି ପଦରେ କୃଷ୍ଣଙ୍କର ବାଲ୍ୟଲୀଳାର ବର୍ଣ୍ଣନା କରାଯାଇଅଛି। କୃଷ୍ଣ ମାତା ଯଶୋଦା ପାଖରେ ଆପରି କରୁଛନ୍ତି କି ହେ ମାତା! ମୋତେ ବଳରାମ ଭାଇ ଚିଡ଼ାଉଛନ୍ତି। ମୋତେ କହୁଛନ୍ତି ତୋତେ କିଣିକରି ଅଣାଯାଇଅଛି। ମାତା ଯଶୋଦା ତୋତେ ଜନ୍ମ ଦେଇନାହାଁନ୍ତି। ଏଣୁ ମୁଁ ତାଙ୍କ ସହିତ ଖେଳିବାକୁ ଯାଉନାହିଁ। ସେ ବାରମ୍ବାର ମୋତେ ପଚାରୁଛନ୍ତି ତୋ’ ମାତା ପିତା କିଏ?

ନନ୍ଦ-ଯଶୋଦା ତ ଗୋରା, ତୁ କାହିଁକି କଳା ହେଲୁ? ଏହା ଶୁଣି ଗୋପାଳ ପିଲାମାନେ ମଧ୍ଯ ଚୁଟୁକି ମାରି ନାଚୁଛନ୍ତି। ବଳରାମ ଭାଇ ତାଙ୍କୁ ଶିଖାଉଛନ୍ତି। ତୁମେ କେବଳ ମୋତେ ମାରିବା ଜାଣିଛ ମାତା କହୁଛନ୍ତି: କୃଷ୍ଣ ଶୁଣ, ବଳରାମ ମିଛୁଆଟା ଏବଂ ଜନ୍ମରୁ ଦୁଷ୍ଟ। ମୁଁ ଗାଈର ଶପଥ ଖାଇ କହୁଛି ଯେ ତୁ ମୋର ପୁଅ ଏବଂ ମୁଁ ତୋ’ ର ମାଆ

2. हरि अपने आँगन कछु गाबत।
तनक-तनक चरनन सौं नाचत, मनहिं – मनहिं रिझाबत ॥
बाँह उचाइ काजरी – धौरी, गौयनि टेरि बुलाबत।
कबहुँक बाबा नंद पुकारत, कबहुँक घर मैं आबत॥
माखन तनक आपने कर लै, तनक-बदन मैं नाबत।
कबहुँ चितौ प्रतिबिम्ब खंभ मैं, लौनी लिए खबावत॥
दुरि खति जसुमति यह लीला, हरष आनंद बढ़ावत।
सूर स्याम के बाल – चरित ये, तिन देखन मन भावत॥

ହରି ଅପନେ ଆଗନ କଛୁ ଗାବତ୍।
ତନକ-ତନକ ଚରନନ ସୌ ନାଚତ, ମନହିଁ-ମନହିଁ ରିଝାବତ ॥
ବାହ ଉଚାଇ କାଜରୀ-ଧୌରୀ, ଗୌୟନି ଟେରି ବୁଲାବତ।
କବର୍ଦ୍ଧକ ବାବା ନଂଦ ପୁକାରତ, କବଚ୍ଛୁକ ଘର ମେଁ ଆବତ॥
ମାଖନ ତନକ ଆପନେ କର ଲୈ, ତନକ-ବଦନ ମେଁ ନାବତ।
କବହୁଁ ଚିତୌ ପ୍ରିତବିମ୍ବ ଗଂଭ ମେଁ, ଲୌନୀ ଲିଏ ଖବାୱତ॥
ଦୁରି ଦେଖତି ଜସୁମତି ୟହ ଲୀଲା, ହରଷ ଆନଂଦ ବଢ଼ାୱତ।
ସୂର ସ୍ୟାମ କେ ବାଲ-ଚରିତ ଯେ, ନିତ ଦେଖତ ମନ ଭାବତ॥

हिन्दी व्याख्या: बालक कृष्ण घर के आंगन में अकेले खेल रहे हैं। उनका यह खेल सबके मन को मोह लेता है। यह वर्णन बहुत ही हृदयग्राही है। भगवान कृष्ण अपने आप कुछ गा रहे हैं। वे गाते-गाते नन्हें चरणों से नाचते भी हैं और मन-मन खुश हो रहे हैं। कभी वे हाथ उठाकर काली एवं सफेद गायों को बुलाते हैं, तो कभी नंद बाबा को पुकारते हैं। वे कभी घर के भीतर चले जाते हैं।

घर में जाकर थोड़ा मक्खन हाथ में लेकर खाते हैं, और थोड़ा-सा मुँह में लगा लेते हैं। माता यशोदा दूर से ही खड़ी होकर यह लीला देख रही हैं। और आनंदित हो रही हैं। सूरदास कह रहे हैं कि कन्हैया की यह बाललीला रोज-रोज देखने पर भी प्यारी लगती है। इससे मन तृप्त नहीं होता।

ଓଡ଼ିଆ ଅନୁବାଦ:
କୃଷ୍ଣ ଏକୁଟିଆ ଅଗଣାରେ ଖେଳୁଛନ୍ତି। ତାଙ୍କର ଏହି ଖେଳ ସମସ୍ତଙ୍କ ମନକୁ ମୋହିତ କରୁଅଛି। ଏହା ଅତ୍ୟନ୍ତ ହୃଦୟଗ୍ରାହୀ। କୃଷ୍ଣ ଆପଣା ଛାଏଁ କିଛି ଗୀତ ଗାଇ ଚାଲିଛନ୍ତି। ଗାଉ ଗାଉ ଛୋଟ ପାଦରେ ସେ ନାଚୁଛନ୍ତି ମଧ୍ୟ ଏବଂ ମଗ୍ନ ମଧ୍ୟ ହୋଇଯାଉଛନ୍ତି। କେତେବେଳେ ହାତଠାରି କାଳୀ ଓ ଧଳୀ ଗାଈଙ୍କୁ ଡାକୁଛନ୍ତି ତ କେତେବେଳେ ନନ୍ଦ ବାବାଙ୍କୁ ମଧ୍ୟ ଡାକୁଛନ୍ତି।

କେତେବେଳେ ଘର ଭିତରକୁ ଚାଲିଯାଇ ଲହୁଣି ଖାଉଛନ୍ତି ତ ଲହୁଣି ହାତରେ ଧରି ନିଜ ମୁହଁରେ ବୋଳି ମଧ୍ୟ ବହୁତ ସୁନ୍ଦର ଲାଗୁଛି। ଏତିକିରେ ମନର ତୃପ୍ତି ହେଉନାହିଁ। ଥାଇ ଏହା ଦେଖୁଛନ୍ତି ଏବଂ ଖୁସି ହେଉଛନ୍ତି। ସୁରଦାସ କହୁଛନ୍ତି ଯେ କୃଷ୍ଣଙ୍କର ଏହି ବାଲ୍ୟଲୀଳା ସବୁଦିନ ଦେଖୁଥିଲେ ମଧ୍ୟ ବହୁତ ସୁନ୍ଦର ଲାଗୁଛି। ଏତିକିରେ ମନର ତୃପ୍ତି ହେଉନାହିଁ

शबनार: (ଶରାର୍ଥି)

मैया – माँ (ମା’)

दाऊ – बड़ा भाई बलराम (ବଡ଼ଭାଇ ବଳରାମ)

खिझायो – चिढ़ाया (ବଡ଼ଭାଇ ବଳରାମ)

कहत – कहते है (ଚିଡ଼ାଇଲେ)।

लीन्हों – लाया गया (କୁହନ୍ତି)।

जायो – जन्मा (ଅଣାଯାଇଅଛି)।

एही – यही (ଜନ୍ମ ହେବା)।

खेलन – खेलने (ଏହି)।

पुनि-पुनि – बारबार (ଖେଳିବାକୁ)।

तात – पिता (ବାରମ୍ବାର)।

सरीर – शरीर (ପିତା)।

चुटकी – अंगूठे और पास की अंगुली से बजाना (ଶରୀର)।

दै- देकर (ଦେଇକରି)।

ग्वाल – गोपाल (ଗୋପାଳ)।

बलवीर – बलराम (ବଳରାମ)।

मारन – मारना (ମାରିବା)।

दाउहि – बडे भाई से ( କଥା)।

बातें – बातें (ବଡ଼ଭାଇ)।

सुनहू – सुनकर (ଶୁଣି ଶୁଣି)।

चबाई – चुगलखोर (ମିଛକୁହା)।

धूत – शरारती (ବଦମାସ୍)।

गोधन – गायरूपी धन (ଗୋଧନ)।

मोहि – मुझे (ମୋତେ)।

मोसों – मुझसे (ମୋତେ)।

मोल – खरीद कर (କିଣି କରି)।

जसुमति – यशोदा (ଯଶୋଦା)।

कहैं – कहें (କହିବ)

रिंस – गुस्सा (ରାଗ)।

जात – जाएँ (ଯାଏ)।

तुमरो – तुम्हारा (ତୁମର)।

कत – क्यों (କାହିଁକି)।

स्याम – श्यामल (ଶ୍ୟାମଳ ରଙ୍ଗ)।

सिखै – सिखाने (ଲିବା)।

मोहि – मुझे (ଶିଖାନ୍ତି)।

सीखी – सीखना (ଶିଖିବ)।

कबहूँ – कभी (କେତେ)।

रीझै – प्रसन्न (ଖୁଡି ହେବ)।

कान्ह – कान्हा (କୃଷ୍ଟ)।

जनमत – जन्म (କନ୍ନ)।

पूत – पुत्र (ପୁତ୍ର)।

सौं – सौगंध (ଶପଥ)।

हरि – कृष्ण (କୃଷ୍ଣ)।

कछु – कुछ (କିଛି)।

तनक-तनक – नन्हें-नन्हें (ଛୋଟ ଛୋଟ)।

मनहिं-मनहिं – मन में (ମନେମନେ)।

बाँह – हाथ (ହାଥ)

काजरी – काली (କାଳ)

कबहुँक – कभी (କେବେ)

आवत – आते हैं (ଅପନ୍ତି)

तनक – कुछ (କିଛି)

लै – लेकर (ନେଇକରି)

नावत – लगाते (ଲଗାଇବ)

दुरि – दूर (ହର)

बढ़ावत – बढ़ाना (ଗଢାଇବା)

आँगन – (ଅଗଣା)।

गावत – गाना (ଚାଇବା)।

चरनन – चरण (ପ।ଦ)।

रिझावत – मगन (ମଗ୍ନ)

उचाइ – उठाकर (ହାପଠାରି)

धौ. – सफेद (ଧଳା)

पुकारत – पुकारते हैं (ପାରନ୍ତି)

माखन – मक्खन (ଲତୁଣି)

कर – हाथ (घाथ)। (ତ୍ୱାଥ)

बदन – मुँह (ପୁହିଁ)

खबावत – खिलाते हैं (ଖୁଆନି)

लिए – लेकर (ନେଇ)

भावत – अच्छा लगता (ଭଲ ଲାଗ)

कवि परिचय

भक्तिकाल के श्रेष्ठ कवि सूरदास हिन्दी के सूर्य जैसे तेजस्वी कवि हैं। उनका जन्म सन् 1478 में दिल्ली के निकट सीही गाँव में एक गरीब ब्राह्मण परिवार में हुआ। वे अंधे थे, पर मालूम पड़ता है कि वे जन्मांध नहीं थे। मथुरा और आगरा के बीच यमुना नदी के तटपर स्थित गऊघाट पर उन्होंने संगीत, काव्य और शास्त्र का अभ्यास किया और विनय के भाव से पदों की रचना की। आगे चलकर वे वल्लभाचार्य के शिष्य बन गये और ब्रज जाकर गोवर्धन के पास पारसोली नामक जगह पर अपना स्थायी निवास बनाकर पद लिखते रहे।

सूरदास मानव: मन के बड़े पारखी थे। वात्सल्य भाव के तो वे मर्मज्ञ थे। श्रीमद् भागवत महापुराण के आधार पर रचित उनका विशाल ग्रंथ ‘सूर सागर’ हिंदी की अमूल्य निधि है। ये बच्चों कें, माताओं के, साथियों के, नारी और पुरुषों के मनोभावों के पारंगम कवि थे।

यह पद : (ଏହି ପଦ)
प्रस्तुत पाठ में महाकवि सूरदास द्वारा रचित कृष्ण की बाललीला के दो पदों को रखा गया है। ये पद ‘सूरसागर’ से संकलित हैं।
ପ୍ରସ୍ତୁତ ପାଠରେ ମହାକବି ସୁରଦାସଙ୍କଦ୍ଵାରା ରଚିତ କୃଷ୍ଣଙ୍କର ବାଲ୍ୟଲୀଳାର ଦୁଇଟି ପଦ ରଖାଯାଇଅଛି। ଏହି ପଦ ‘ସୁର ସାଗର’ରୁ ସଙ୍କଳିତ ହୋଇଅଛି।

Odisha State Board BSE Odisha 7th Class English Solutions Follow-Up Lesson 5 Loveliest of Trees Textbook Exercise Questions and Answers.

BSE Odisha Class 7 English Solutions Follow-Up Lesson 5 Loveliest of Trees

BSE Odisha 7th Class English Follow-Up Lesson 5 Loveliest of Trees Text Book Questions and Answers

Pre-Reading (ପ୍ରାକ୍-ପଠନ):

Your teacher will introduce the lesson in the following way.

Question 1.
Can you recognize these flowers? (Teacher show the flowers in picture)
( ଛବିରେ ଫୁଲମାନକୁ ଦେଖାଇ) ( ଏ ଫୁଲଗୁଡ଼ିକୁ ଚିହ୍ନି ପାରୁଛ ?)
Answer:
Yes, we can recognize these flowers. They are zinnia, hibiscus (china rose), lotus, and rose flowers respectively.

Question 2.
How do you feel when you see a tree full of flowers?
(ବୃକ୍ଷଟିଏ ଫୁଲରେ ଭରିଥୁବା ଦେଖ‌ିଲେ ତୁମେ କିଭଳି ଅନୁଭବ କର ? )
Answer:
When we see a tree is full of flowers we think the tree is laughing with merriment. We feel pleasure from this sight.
Let’s see how a poet feels when he sees a tree full of flowers.
(ଚାଲ ଦେଖିବା ଗୋଟିଏ ଗଛ ଫୁଲରେ ଭରପୂର ଥ‌ିବା ଦେଖ୍ ଜଣେ କବି କ’ଣ ଅନୁଭବ କରନ୍ତି ?)

While-Reading (ପଠନକାଳୀନ ):

1. Your teacher will read the poem aloud, and you will listen to him/her without opening your books.
(ତୁମ ଶିକ୍ଷକ କବିତାଟି ଉଚ୍ଚସ୍ବରରେ ପଢ଼ିବେ । ବହି ନଖୋଲି ତୁମେ କେବ ତାଙ୍କୁ ଶୁଣିବ ।)
2. S/he will read the poem aloud for the second time and you will listen to him / her following the poem in your books.
(ତୁମ ଶିକ୍ଷକ ଦ୍ଵିତୀୟବାର ଉଚ୍ଚସ୍ବରରେ ପଢ଼ିବେ ଏବଂ ତୁମେ ବହିରୁ କବିତାଟିକୁ ଧାଡ଼ି ଧାଡ଼ି କରି ଦେଖ ।)
3. Read the poem silently and try to answer the questions asked by your teacher.
(କବିତାଟିକୁ ତୁମେ ନୀରବରେ ପାଠ କର ଏବଂ ଶିକ୍ଷକ ପଚାରିଥିବା ପ୍ରଶ୍ନସବୁର ଉତ୍ତର ଦେବାକୁ ଚେଷ୍ଟାକର ।)

Text (ପାଠ୍ୟବସ୍ତୁ):

Loveliest of trees, the cherry now
Is hung with bloom along the bough.
And stands about the woodland ride
Wearing white for Eastertide

Now my threescore years and ten,
Twenty will not come again,
And take from seventy springs a score,
It only leaves me fifty more.

And since to look at things in bloom
Fifty springs are little room,
About the woodlands, I go
To see the cherry hung with snow.

ଓଡ଼ିଆ ଅନୁବାଦ:
ସୁନ୍ଦରତମ ବୃକ୍ଷରାଜି, ଆଜି ସେହି ଚେରି
ଝୁଲୁଛି ଫୁଟା ଫୁଲଧରି ଶାଖା ସବୁ ଭରି,
ପର୍ବତିଆ କ୍ଷେତ ଧାରେ ଧାରେ ଠିଆ ହୋଇ
ଶୁଭ୍ର ଆଚ୍ଛାଦିତ ଆଭୂଷିତ ଇଷ୍ଟର ଢେଉରେ ।

ଅଧୁନା ମୋ ତିନିକୋଡ଼ି ବର୍ଷ ଏବଂ ଦଶ
କୋଡ଼ିଏ ତ ବାହୁଡ଼ି ନ ଆସେ,
ଏବଂ ଘେନିଯାଏ ସତୁରୀରୁ କୋଡ଼ିଏ ବସନ୍ତ
କେବଳ ସେ ମୋତେ ଛାଡ଼ିଦିଏ ଅଧୂକୁ ପଚାଶ ।

ଏବଂ ସେହିଠାରୁ ଚାହିଁରହେ ଫୁଲ ଜୁଆରକୁ
ପଚାଶ ବସନ୍ତ କିଛି ନୁହେଁ ପରା ତା’ର
ଯେବେ ଯାଏ ମୁଁ ତା’ପାଶେ ପାହାଡ଼ ପ୍ରାନ୍ତରେ
ଚେରିଗଛ ଦେଖ‌ିବାକୁ ବରଫ ଦୋଳିରେ ।

Notes And Glossary (ଶବ୍ଦାର୍ଥ):

loveliest (ଲଭ୍ଲିଏଷ୍ଟ) – ସୁନ୍ଦରତମ
cherry (ଚେରି) – ନାଲି ରଙ୍ଗର ଚେରିକୋଳି
hung (ହଙ୍ଗ୍) – ଝୁଲି ରହିବା
bloom (ବ୍ଲୁମ୍) – ଫୁଲ ଫୁଟିବା
bough (ବାଓ) – ଶାଖାପ୍ରଶାଖା
ride (ରାଇଡ଼୍) – ଆରୋହଣ କରିବା
woodland (ଉଡଲ୍ୟାଣ୍ଡ୍) – ଜଙ୍ଗଲିଆ ଅଞ୍ଚଳ
wearing (ଓୟରିଙ୍ଗ୍) – ପିନ୍ଧିଥ‌ିବା
tide (ଟାଇଡ୍) – ଡେଉ ବା ଜୁଆର
score (ସ୍କାର) – କୋଡ଼ିଏ
spring (ସ୍ପ୍ରିଙ୍ଗ୍) – ବସନ୍ତ
leaves (ଲିଭସ୍) – ଛାଡିଦିଏ
since (ସିନ୍ସ) – ପରଠାରୁ
snow (ତୁଷାର) – ବରଫ

Comprehension Questions (ବୋଧପରିମାପକ ପ୍ରଶ୍ନବଳୀ):

Question 1.
What is the poem about?
( କବିତାଟି କେଉଁ ସମ୍ବନ୍ଧରେ ?)
Answer:
The poem is about the beauty and life of a cherry tree.

Question 2.
Which word tells that ’cherry with flower’ is the nicest of all trees?
(କେଉଁ ଶବ୍ଦ କହୁଛି ଯେ, ଚେରି ଫୁଲରେ ସଜେଇ ହୋଇ ସବୁ ଗଛ ମଧ୍ୟରେ ସୁନ୍ଦରତମ ?)
Answer:
The phrase “hung with bloom” tells that cherry with flowers is the nicest of all trees.

Question 3.
Where does the tree stand?
((ଚେରି) ଗଛ କେଉଁଠି ଥାଏ ?)
Answer:
The tree stands in woodland.

Question 4.
What do the trees wear for Eastertide?
((ଚେରି) ଗଛସବୁ କ’ଣ ପିନ୍ଧିଥାଏ ଇଷ୍ଟର ଢେଉ ପାଇଁ ? )
Answer:
‘Easter’ is a Christian festival. The people do merrymaking. So the cherry enjoys the festival wearing white flowers.

Question 5.
What does ‘white’ refer to here?
(ଧଳା ବା ଶୁଭ୍ରତା ଏଠି କାହାକୁ ସୂଚାଏ ?)
Answer:
“White” refers to a white dress. As it is an Easter festival, the cherry wears a white dress.

Question 6.
Why is cherry dressed in white?
(ଚେରିଗଛ କାହିଁକି ଶୁଭ୍ର ବା ଧଳା ପୋଷାକ ପରିଧାନ କରିଥାଏ ?)
Answer:
As it is the Easter festival, the cherry wears white flowers to enjoy the festival.

Question 7.
The first stanza describes the cherry tree. How is the second stanza different from the first stanza?
(ପ୍ରଥମ ପଦ ଚେରି ଗଛ ବିଷୟରେ ବର୍ଣ୍ଣନା କରେ । ଦ୍ୱିତୀୟ ପଦ କିଭଳି ପ୍ରଥମ ପଦଠାରୁ ଭିନ୍ନ ?)
Answer:
The first stanza describes the cherry tree. But the second stanza describes the normal life of man.

Question 8.
How many years are there in a score?
(କୋଡ଼ିରେ କେତେ ବର୍ଷ ଥାଏ ?)
Answer:
There are twenty years in a score.

Question 9.
How many years are there in three scores?
(ତିନି କୋଡ଼ିରେ କେତେ ବର୍ଷ ଥାଏ ? )
Answer:
There are sixty years in three Scores.

Question 10.
What is the normal life span of humans according to the Bible?
(ବାଇବେଲ୍‌ ବା ଖ୍ରୀଷ୍ଟ ଧର୍ମଗ୍ରନ୍ଥ ଅନୁସାରେ ମଣିଷର ସାଧାରଣତ ଜୀବନକାଳ କେତେ ବର୍ଷ ?)
Answer:
According to Bible the normal life span of humans is seventy.

Question 11.
How old is the poet?
(କବିଙ୍କୁ କେତେ ବର୍ଷ ?)
Answer:
The poet is twenty years old.

Question 12.
Why does he deduct twenty years from seventy years?
(ସେ କାହିଁକି ୭୦ ବର୍ଷରୁ ୨୦ ବର୍ଷ ବାଦ୍ ଦିଅନ୍ତି ?)
Answer:
The poet has deducted twenty years from seventy years because he has already enjoyed twenty years from the normal life of seventy.

Question 13.
How many years does he suppose to live?
(ସେ କେତେ ବର୍ଷ ପର୍ଯ୍ୟନ୍ତ ବନ୍ଧୁ ରହିବାକୁ ଆଶା ରଖନ୍ତି ?)
Answer:
The poet is supposed to live another fifty years.

Question 14.
Is a fifty year life enough to see and enjoy the nature?
(ପଚାଶ ବର୍ଷର ଜୀବନ କ’ଣ ପ୍ରକୃତିକୁ ଦେଖ୍ ଉପଭୋଗ କରିବା ନିମିତ୍ତ ଯଥେଷ୍ଟ ? )
Answer:
No, fifty years is not enough to see and enjoy the nature.

Question 15.
Why is the poet in a hurry to go to the woodland?
(କବି କାହିଁକି ପାହାଡ଼ ପ୍ରାନ୍ତକୁ ଯିବାକୁ ବ୍ୟଗ୍ର ହୁଅନ୍ତି ?)
Answer:
The poet wants more and more to enjoy nature. But fifty years is not enough to enjoy the beauties of nature. So he is in a hurry to go to the woodland to enjoy beauties nature.

Question 16.
Does the third stanza start from the second stanza? How?
(ତୃତୀୟ ପଦ କ’ଣ ଦ୍ଵିତୀୟ ପଦଠାରୁ ଆରମ୍ଭ ହୋଇଛି ? କିପରି ?)
Answer:
Yes, the third stanza starts from the second stanza. The third stanza begins with the word ‘And’ and this means it is a continuation of the 2nd stanza.

Question 17.
Is the poet a lover of nature? How do you know?
(କବି କ’ଣ ପ୍ରକୃତିପ୍ରେମୀ ? ତୁମେ କିପରି ଜାଣୁଛ ?)
Answer:
Yes, the poet is a lover of nature. We know this from his hurries to go to the woodland to enjoy cherry hung with blossom.

Question 18.
What does ‘springs’ mean?
(ବସନ୍ତଗୁଡ଼ିକ କହିଲେ କ’ଣ ବୁଝୁଛ ?)
Answer:
Spring refers to one year. Because spring comes after one year. So here springs refer to years.

Question 19.
When cherry blooms, it does not have any leaf. Name a flower plant in our country similar to cherry.
(ଯେତେବେଳେ ଚେରି ଫୁଟେ, ତା’ର କୌଣସି ପତ୍ର ନଥାଏ । ଆମ ଦେଶରେ ଥ‌ିବା ଚେରି ଭଳି ଏକ ଫୁଲଗଛର ନାମ କୁହ ।)
Answer:
Lily is similar to cherry.

Word Note:
(The words/phrases have been defined mostly on their contextual meanings)

beams (ବିମ୍ସ) – sunlight (ସୂର୍ଯ୍ୟକିରଣ)
boughs (ବାଓଜ୍) – branches (ଗଛର ଶାଖା ବା ଡାଳ)
cherry (ଚେରୀ) – a small sweet fruit (ଚେରିକୋଳି)
drowsy (ଡ୍ରୋଜି) – sleepy (ନିଦ୍ରାଳୁ ବା ନିଦ୍ରା ବିଜଡ଼ିତ ଭାବ)
Eastertide (ଇଷ୍ଟରଟାଇଡ୍) – Eastertime, Easter festival time of the Christians (ଖ୍ରୀଷ୍ଟଧର୍ମୀମାନଙ୍କର ଇଷ୍ଟର ପର୍ବ ସମୟ)
floats (ଫ୍ଲୋସ୍) – moves in the sky (ଆକାଶରେ ଭାସେ ବା ଦେଖାଯାଏ )
gather (ଗ୍ୟାଦର) – collect (ଏକତ୍ରିତ କରେ-ପକ୍ଷୀମାନେ ବସା ବାନ୍ଧି ରୁହନ୍ତି)
harm (ହାର୍ମ) – hurt, damage (କ୍ଷତି କରିବା)
hum (ହମ୍) – sing in very low voice (ଗୁଣୁଗୁଣୁ ହୋଇ ଗାଇବା)
hung with bloom (ହଙ୍ଗ୍ ଉଇଥ୍ ବ୍ଲମ୍) – ripe cherry hanging from the tree (ବୃକ୍ଷର ଶାଖାପ୍ରଶାଖାରେ ପାଚି ଓହଳିଥିବା ଚେରିକୋଳି)
lullaby (ଲୁଆବେ) – song to make children to sleep (ଶିଶୁକୁ ଶୁଆଇବା ପାଇଁ ଗୀତ-ଧୋରେ ବାଇଆ ଧୋ ପରି)
score (ସ୍କୋର) – twenty (କୋଡିଏ)
three scores years and ten (ଥ୍ରି ସ୍କୋର୍‌ସ୍ ଇୟର୍‌ସ ଆଣ୍ଡ୍ ଟେନ୍) – 70 years (the normal life span (time) of man).
woodland (ଉଲ୍ୟାଣ୍ଡ୍ ) – forest area (ବନଭୂମି)

BSE Odisha 7th Class English Solutions Follow-Up Lesson 5 Loveliest of Trees Important Questions and Answers

(A) Choose the right answer from the options.

Question 1.
The poem is about the
(a) nature
(b) beauties of nature
(c) cherry trees
(d) human life
Answer:
(c) cherry trees

Question 2.
The poet of this poem is _____________.
(a) A. E. Housman
(b) W. Wordsworth
(c) Lord Tennyson
(d) none of these
Answer:
(a) A.E. Housman

Question 3.
The cherry hung with bloom
(a) under the bough
(b) along the bough
(c) on the top branches of the tree
(d) on the lower branches of the tree.
Answer:
(b) along the bough

Question 4.
What is the age of the poet now?
(a) 20
(b) 50
(c) 70
(d) 90
Answer:
(a) 20

Question 5.
In the woodland cherry hung _____________.
(a) with snow
(b) with dust
(c) in snow
(d) none of these
Answer:
(a) with snow

(B) Answer the following questions.

Question 1.
Which line tells that the poet has already enjoyed twenty years of age?
Answer:
The line “Twenty will not come again”. This line shows that the poet has already enjoyed twenty years of age.

Question 2.
“Fifty springs are little room”. Explain.
Answer:
The man nowadays lives for seventy years on average. The poet expects to live another fifty years. But it is not much for him to enjoy the beauties of nature.

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Exercise 6(a)

Question 1.
Multiply (2√-3 + 3√-2) by (4√-3 – 5√-2)
Solution:
(2√-3 + 3√-2) by (4√-3 – 5√-2)
= (2√3i + 3√2i) (4√3i – 5√2i)
= i² (2√3 + 3√2) (4√3 – 5√2)
= – 1(24 – 10√6 + 12√6 – 30)
= – 1(- 6 + 2√6) = 6 – 2√6

Question 2.
Multiply (3√-7 – 5√-2) (3√-2 + 5√-2)
Solution:
(3√-7 – 5√-2) (3√-2 + 5√-2)
= (3√7i – 5√-2i) (3√2i + 5√2i)
= i² (3√7- 5√2) (3√2 + 5√2)
= (- 1)8√2(3√7 – 5√2 )
= – 24√14 + 80

Question 3.
Multiply (√-1 +√-1) (a – b√-1)
Solution:
(√-1 +√-1) (a – b√-1)
= (i + i) (a – bi) = 2i(a – bi)
= 2ai – 2bi² = 2ai + 2b

Question 4.
Multiply (x – \(\frac{1+\sqrt{-3}}{2}\)) (x – \(\frac{1-\sqrt{3}}{2}\))
Solution:
(x – \(\frac{1+\sqrt{-3}}{2}\)) (x – \(\frac{1-\sqrt{3}}{2}\))
= (x + \(\frac{-1-i \sqrt{3}}{2}\)) (x + \(\frac{-1+i \sqrt{3}}{2}\))
= (x + ω) (x + ω²) = x² + ω²x + ωx + ω³
= x² + x (ω ² + ω) + 1 = x² –  x + 1

Question 5.
Express with rational denominator. \(\frac{1}{3-\sqrt{-2}}\)
Solution:
\(\begin{aligned}
& \frac{1}{3-\sqrt{-2}}=\frac{1}{3-\sqrt{2} \mathrm{i}}=\frac{3+\sqrt{2} \mathrm{i}}{(3-\sqrt{2} \mathrm{i})(3+\sqrt{2} \mathrm{i})} \\
& =\frac{3+\sqrt{2} \mathrm{i}}{9-2 \mathrm{i}^2}=\frac{3+\sqrt{2} \mathrm{i}}{9+2}=\frac{3+\mathrm{i} \sqrt{2}}{11}
\end{aligned}\)

Question 6.
\(\frac{3 \sqrt{-2}+2(-5)}{3 \sqrt{-2}-2 \sqrt{-2}}\)
Solution:

Question 7.
\(\frac{3+2 \sqrt{-1}}{2-5 \sqrt{-1}}+\frac{3-2 \sqrt{-1}}{2+5 \sqrt{-1}}\)
Solution:

Question 8.
\(\frac{a+x \sqrt{-1}}{a-x \sqrt{-1}}-\frac{a-x \sqrt{-1}}{a+x \sqrt{-1}}\)
Solution:

Question 9.
\(\frac{(x+\sqrt{-1})^2}{x-\sqrt{-1}}-\frac{\left(x-\sqrt{-1^2}\right)}{x+\sqrt{-1}}\)
Solution:

Question 10.
\(\frac{(a+\sqrt{-1})^3-(a-\sqrt{-1})^3}{(a+\sqrt{-1})^2-(a-\sqrt{-1})^2}\)
Solution:

Question 11.
Find the value of (- i)⁴ⁿ⁺³; when n is positive.
Solution:
(- i)⁴ⁿ⁺³
= (-i⁴ⁿ) (-i)³ = 1(- i³) = – (-i) = i

Question 12.
Find the square root of (a + 40i) + \(\sqrt{9-40 \sqrt{-i}}\)
Solution:

Question 13.
Express in the form of a + ib:
(i) \(\frac{3+5 i}{2-3 i}\)
Solution:

(ii) \(\frac{\sqrt{3}-i \sqrt{2}}{2 \sqrt{3}-i \sqrt{3}}\)
Solution

(iii) \(\frac{(\mathrm{I}+\mathrm{i})^2}{3-\mathrm{i}}\)
Solution:

(iv) \(\frac{(a+i b)^3}{a-i b}-\frac{(a-i b)^3}{a+i b}\)
Solution:

(v) \(\frac{1+i}{1-i}\)
Solution:
\(\frac{1+i}{1-i}\) = \(\frac{(1+i)^2}{2}=\frac{1-1+2 i}{2}\)
= i = 0 + i

Question 14.
Express the following points geometrically in the Argand plane.
(i) 1
Solution:
1 = 1 + i0 = (1, 0)

(ii) 3i
Solution:
3i = 0 + 3i = (0, 3)

(iii) – 2
Solution:
– 2 = – 2 + i0 = (- 2, 0)

(iv) 3 + 2i
Solution:
3 + 2i = (3, 2)

(v) – 3 + i
Solution:
– 3 + i = (- 3, 1)

(vi) 1-i
Solution:
1 – i = (1, – 1)

Question 15.
Show that the following numbers are equidistant from the origin:
√2 +i, 1 + i√2, i√3
Solution:
|√2 + i| = \(\sqrt{(\sqrt{2})^2+1^2}\) = √3
|1 + i√2| = \(\sqrt{1^2+(\sqrt{2})^2}\) = √3
and |i√3| = √3
∴ The points are equidistant from the origin.

Question 16.
Express each of the above complex numbers in the polar form.
Solution:

Question 17.
If 1, ω, ω² are the three cube roots of unity, prove that (1 + ω²)⁴ = ω
Solution:
L.H.S. = (1 + ω²)⁴ = (- ω)⁴ = ω⁴
= ω³ .ω = 1. ω = ω
= R.H.S. (Proved)

Question 18.
(1 – ω+ ω² ) (1 + ω – ω² ) = 4
Solution:
L.H.S. = (1 – ω+ ω² ) (1 + ω – ω² )
= (- ω – ω )(- ω² – ω² ) (∴ 1 + ω + ω² = 0)
= (- 2ω² – 2ω²) = 4ω³ = 4 = R.H.S.

Question 19.
(1 – ω) (1 – ω)² (1 – ω⁴) (1 – ω⁵) = 9
Solution:
L.H.S. =
(1 – ω) (1 – ω)² (1 – ω⁴) (1 – ω⁵)
= (1 – ω) (1 – ω²) (1 – ω) (1 – ω²)
= (1 – ω)² (1 – ω²)²
= {(1 – ω) (1 – ω²)}²
= (1 – ω² – ω + ω³)²
= {3 – (ω² + ω + 1)}²
= (3)² = 9 = R.H.S.

Question 20.
(2 + 5ω + 2ω² )⁶ = (2 + 2ω + 5ω² )⁶ =729
Solution:
L.H.S. = (2 + 5ω + 2ω² )⁶
(2 + 2ω² + 5ω)⁶ = {2(1 + ω² ) + 5ω}⁶
(- 2ω + 5ω)⁶ = (3ω)⁶ = 729ω⁶ = 729
Again, (2 + 2ω + 5ω² )⁶
= {2(1 + ω) + 5ω² )⁶
= (- 2ω² + 5ω² )⁶ = (3ω²)⁶
= 729ω¹² =729
∴ (2 + 5ω + 2ω² )⁶ = (2 + 2ω + 5ω² )⁶ =729

Question 21.
(1 – ω + ω² ) (1 – ω² + ω⁴) (1 – ω⁴ +ω²)  ….to 2n factors = 2²ⁿ
Solution:
L.H.S. = (1 – ω + ω² ) (1 – ω² + ω⁴) (1 – ω⁴ +ω²)  ….to 2n factors = 2²ⁿ
= (- ω – ω) (1 – ω² + ω) (1 – ω + ω² )  ….to 2n factors = 2²ⁿ
= (- 2ω) (- ω² – ω² ) (- ω – ω)  ….to 2n factors = 2²ⁿ
= [(- 2ω)(- 2ω) … to n factors] × [(- 2ω²)(- 2ω²) …. to n factors]
= (- 2ω)ⁿ × (- 2ω²)ⁿ = (4ω³)ⁿ = 4ⁿ = 2²ⁿ
R.H.S. (Proved)

Question 22.
Prove that x³ + y³ + z³– 3xyz
= (x + y + z) (x + ωy + ω²z) (x + yω² + zω)
Solution:
R.H.S. = (x + y + z) (x + ωy + ω²z) (x + yω² + zω)
= (x + y + z) (x + xyω² + zxω + xyω + y²ω³+ yz ω2 + zxω² + yzω⁴ + z²ω³)
= (x + y + z) [x² + y² + z² + xy (ω² + ω) +yz (ω² + ω) + zx(ω² + ω)]
= (x + y + z) (x² + y² + z² – xy – yz – zx)
= x³ + y³ + z³ – 3xyz = L.H.S. (Proved)

Question 23.
If x = a + b, y = aω + b ω², z = aω² + bω show that
(1) xyz = a³ + b³
Solution:
L.H.S. = xyz
= (a + b) (a ω + b ω²) (a ω² + b ω)
= (a + b) (a² ω³ + ab ω² + ab ω⁴ + b² ω³)
= (a + b) {a² + b² + ab(ω² + ω)}
= (a + b) (a² – ab + b²) = a³ + b³ = R.H.S. (Proved)

(2) x² + y² + z² = 6ab
Solution:
L.H.S. = x² + y² + z²
= (a + b)² + (a ω + b ω²)² (a ω² + b ω)²
= a² + b² + 2ab + a² ω² + b² ω⁴ + 2ab ω³ + a² ω⁴ + b² ω² + 2ab ω³
= a² + a² ω² + a² ω + b² + b² ω + b² ω² + 2ab + 2ab + 2ab
= a²(1 + ω² + ω) + b² (1 + ω + ω²) + 6ab
= 0 + 0 + 6ab = 6ab = R.H.S. (Proved)

(3) x³ + y³ + z³ = 3(a³ + b³)
Solution:
L.H.S. = x³ + y³ + z³
= (a + b)³ + (a ω + b ω²)³ + (aω² + b²)³
= a³ + 3a²b + 3ab² + b³ + a³ ω³ + 3a²b² bω² + 3a ωb² ω⁴ + b³ ω⁶ + a³ ω⁶ + 3a² ω⁴bω + 3a ω²b² ω² + b³ ω³
= a³ + a³ + a³ + b³ + b³ + b³ + 3a²b(1 + ω⁴ + ω⁵) + 3ab² (1 + ω⁵ + ω⁴)
= 3a³ + 3b³ + 0 + 0 = 3 (a³ + b³) = R.H.S. (Proved)

Question 24.
If ax + by + cz = X, cx + by + az = Y, bx + ay + cz = Z
show that (a² + b² + c² – ab – bc- ca) (x² +y² + z² – xy – yz – zx) = X² + Y² + Z² – YZ – ZX – XY
Solution:
L.H.S.
= (a² + b² + c² – ab – bc – ca) (x² + y² + z² – xy – yz – zx)
= (a + b ω + cω²) (a + bω² + cω) (x + yω + zω)² (x + yω² + z ω) (Refer Q.No.22)
= {(a + bω + cω²) (x + yω + zω²)} {(a + bω² + cω) (x + yω² + zω)}
= (ax + ayω + azω² + bx ω + byω² + bzω³ + cxω² + cyω³ + czω⁴) × (ax + ay ω² + azω + bxω² + byω⁴ + bzω³ + cxω + cyω³ + czω²)
= {(ax + cy + bz) + (cx + by + az) ω + (bx + ay + cz) ω²} x (ax + cy + bz) + (cx + by + az) ω + (bx + ay + cz) ω²}
= (X + Yω² + Zω) (X + Yω + Zω²)
= X² + Y² + Z² – XY – YZ – ZX (Refer Q. No. 22)
(Where X = ax + cy + bz.
Y = cx + by + az.
Z = bx + ay + cz).

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 8 Permutations and Combinations Ex 8(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Exercise 8(a)

Question 1.
What is the total number of functions that can be defined from the set {1, 2} to the set {1, 2, 3}?
Solution:
The total number of functions that can be defined from the set {1, 2} to the set {1, 2, 3} is 3² = 9.

Question 2.
A die of six faces marked with the integers 1, 2, 3, 4, 5, and 6 one on each face is thrown twice in succession, what is the total number of outcomes thus obtained?
Solution:
A die of six faces marked with the integers 1, 2, 3, 4, 5, and 6 one on each face, is thrown twice in succession.
∴ The total number of outcomes is 6² = 36.

Question 3.
Five cities A, B, C, D, and E are connected to each other by straight roads. What is the total number of such roads?
Solution:
Five cities A, B, C, D, and E are connected to each other by straight roads.
∴ The number of such roads is \(\frac{{ }^5 \mathrm{P}_2}{2 !}\) = 10

Question 4.
What is the total number of different diagonals of a given pentagon?
Solution:
The total number of different diagonals of a given pentagon is
\(\frac{{ }^5 \mathrm{P}_2}{2 !}\) – 5 = 10 – 5 = 5

Question 5.
There are two routes joining city A to city B and three routes joining B to city C. In how many ways can a person perform a journey from A to C?
Solution:
There are two routes joining city A to city B and three routes joining B to city C.
∴ By the fundamental principle of counting, the total number of journeys in which a person can perform from A to C is 2 × 3 = 6.

Question 6.
How many different four-lettered words can be formed by using the four letters a, b, c, and d, while the letters can be repeated?
Solution:
The number of different words that can be formed by using the four letters a, b, c, and d, while the letters can be repeated is 4⁴ = 256.

Question 7.
What is the sum of all three-digit numbers formed by using the digits 1, 2, and 3?
Solution:
The 3-digit numbers that can be formed by using the digits 1, 2, and 3 are 123, 132, 213, 231, 312, and 321.
∴ Their sum = 1332.

Question 8.
How many different words with two letters can be formed by using the letters of the word JUNGLE, each containing one vowel and one consonant?
Solution:
The word ‘Jungle’ contains 2 vowels and 4 consonants. Each word contains one vowel and one consonant.
The number of different words formed.
2 × ⁴P₁ × ²P₁ = 2 × 4 × 2 = 16

Question 9.
There are four doors leading to the inside of a cinema hall. In how many ways can a person enter into it and come out?
Solution:
There are four doors leading to the inside of a cinema hall. A person can enter into it and come out in 4² = 16 different ways. (By the principle of counting.)

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 5 Principles Of Mathematical Induction Ex 5 Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Exercise 5

Prove the following by induction.

Question 1.
1 + 2 + 3 + …… + n = \(\frac{n(n+1)}{2}\)
Solution:
Let p_n be the given statement

Question 2.
1² + 2² + …… + n² = \(\frac{n(n+1)(2 n+1)}{6}\)
Solution:
Let p_n be the given statement
when n = 1
1² =1 = \(\frac{1(1+1)(2 \times 1+1)}{6}\)
P₁ is true
Let P_k be true.
i.e. 1² + 2² + … + k² = \(\frac{k(k+1)(2 k+1)}{6}\)
we shall prove P_k + 1 is true i.e., 1² + 2² + … + k² + (k + 1)²
\(=\frac{(k+1)(k+2)(2 k+3)}{6}\)

∴ P_k+1 is true
∴ P_n is true of all values of n∈N.

Question 3.
1 + r + r²+ …. + rⁿ = \(\frac{r^{n+1}-1}{r-1}\)
Solution:
Let P_n be the given statement

Question 4.
5ⁿ – 1 is divisible by 4.
Solution:
Let P_n = 5ⁿ – 1
When n = 1,
5¹ – 1 = 4 is divisible by 4.
∴ P₁, is true.
Let P_k be true i.e.,
5^k – 1 is divisible by 4.
Let 5^k+1 – 1 = 4m, me Z
Now 5k + 1 – 1 = 5^k. 5 – 5 + 4
= 5 (5^k – 1) + 4
= 5 × 4m + 4 = 4 (5m + 1)
which is divisible by 4.
∴ P_k+1 is true.
∴ P_n is true for all values of n ∈ N

Question 5.
7²ⁿ + 2^3n-3  3^n-1 is divisible by 25 for any natural number n > 1.
Solution:
Let 7²ⁿ + 2^3n-3 . 3^n-1
when n = 1, 7¹ + 2⁰ . 3⁰ ⇒ 49 + 1 = 50
which is divisible by 25.
∴ P₁ is true. Let P_k be true.
i.e., 72^k + 2^3k-3 3^k-1 is divisible by 35
Now \(7^{2 \overline{k+1}}+2^{3 \overline{k+1}-3} 3^{\overline{k+1}-1}\)
=7^2k+2 + 2^3k. 3^k
= 7^2k. 7² + 2^3k-3 2³. 3^k-1. 3¹
= 7^2k. 49 + 2^3k-3. 3^k-1. 24
= 7^2k (25 + 24) + 24. 2^3k-3. 3^k-1
= 7^2k. 25 + 24 (7^2k + 2^3k-3. 3^k-1)
= 7^2k. 25 + 24 × 25m
Which is divisible by 25 (∵ P_k is true)
∴ P_k+1 is true
∴ P_n is true for all values of n > 1.

Question 6.
7. 5^2n-1 + 23ⁿ⁺¹ is divisible by 17 for every natural number n ≥ 1.
Solution:
Let P_n = 7. 5²ⁿ – 1 + 2³ⁿ⁺¹.
When n = 1, 7.5 + 2⁴ = 35 + 16 = 51
Which is divisible by 17.
P₁, is true.
Let P_k be true i.e., 7.5^2k-1 + 2^3k+1  is divisible by 17.
Let 7.5^2k-1 + 2^3k+1 = 17 m, m ∈ Z
Now, \(7.5^{2 \overline{k+1}-1}+2^{3 \overline{k+1}+1}\)
= 7.5^2k-1 + 2^3k+4
= 7.5^2k-1 . 5² + 2^3k+1 . 2³
= 25. 7. 5^2k-1 + 8. 2^3k+1
= (17 + 8) 7.5^2k-1 + 8. 2^3k+1
= 17. 7. 5^2k-1 + 8 (7. 5^2k-1 + 2^3k+1)
= 17 × 7 × 5^2k-1 + 8 × 17m
Which is divisible by 17.
Hence P_k+1. is true.
∴ P_n is true for all values of n ≥ 1.

Question 7.
4ⁿ⁺¹ + 15n + 14 is divisible by 9 for every natural number n ≥ 0.
Solution:
Let P_n = 4ⁿ⁺¹+ 15 n + 14
when n = 1, 4² + 15 + 14 = 45 is divisible by 9.
∴ P₁ is true. Let P_k be true.
i.e., 4^k+1 + 15k + 14 is divisible by 9.
Now, 4^k+1+1 + 15 (k + 1) + 14
= 4^k+2 + 15k + 29
= 4^k+1. 4 + 60k + 56 – 45k – 27
= 4 (4^k+1 + 15k + 14) – 9 (5k + 3)
Which is divisible by 9.
∴ P_k+1, is true.
∴ P_n is true for all values of n ≥ 0.

Question 8.
3^(2n-1) + 7 is divisible by 9 for every natural number n ≥ 2.
Solution:
Let P_n = 3^2(n-1) + 7
When n = 2. 3² + 7 = 16 is divisible by 8.
∴ P₂ is true.
Let P_k be true.
i.e., 5^2(k-1) + 7 is divisible by 8.
Let 3^2k-2 + 7 = 8m. m ∈ Z.
Now \(3^{2(\overline{k+1}-1)}\) + 7 = 3^2k + 7
= 3^2k-2. 3² + 63 – 56
= 9(3^2k-2 + 7) – 56
= 9 × 8m – 56 = 8 (9m – 7)
Which is divisible by 8.
P_k+1 is true.
P_n is true for all values on n ≥ 2.

Question 9.
5^(2n-4)  – 6n + 32 is divisible by 9 for every natural number n ≥ 5.
Solution:
Let P = 5^2(n-4) – 6n + 32
For n = 5, P₅ = 5² – 6. 5 + 32
= 25 – 30 + 32 = 27
Which is divisible by 9.
Hence P₅ is true.
Let P_k is true.
Let P_k is divisible by 9.
Let P_k = 5^2(k-4) – 6k + 32 = 9., m ≥ Z
5^2k+2 = 576 m + 24k  + 25 … (1)
we shall prove that P_k+1 is true.
Now 5^2(K+1)+2 – 24(k+1) – 25
= 5² (5^2k+2) – 24k – 24 – 25
= 5²[576m + 24k + 25] – 24k – 24 – 25
= 25 × 576 m + 25 × 24k + 25 × 25 – 24k – 24 – 25
= 25 × 576 m + 576 k + 576
= 576 [25 m + k + 1]
which is divisible by 576
∴ P_k+1 is true.
So by the method of induction P_n is true for all n.
i.e., 5²ⁿ⁺²  – 24n – 25 is divisible by 576 for all n ∈ N.
Hence P_k+1 is true.
So by methods of induction P_n is true.
i.e., 5²ⁿ⁺² – 24n – 25 is divisible by 576 for all n.

Question 10.
\(\frac{1}{1.2}+\frac{1}{2.3}+\ldots+\frac{1}{n(n+1)}=\frac{n}{n+1}\)
Solution:
when n = 1,

Question 11.
1.3 + 2.4 + 3.5 + …….. + n(n + 2) = \(\frac{n(n+1)(2 n+7)}{6}\)
Solution:
when n = 1,
we have 1.3 = 3 = \(\frac{3 \times 6}{6}\)
\(=\frac{1 \times 2 \times 9}{6}=\frac{1(1+1)(2 \times 1+7)}{6}\)

Question 12.
xⁿ – yⁿ = (x – y)(x^n-1 + x^n-2 y + … + xy^n-2 + y^n-1); x, y ∈ R [Hint : Write xⁿ⁺¹ – yⁿ⁺¹ = x(xⁿ – yⁿ) + yⁿ(x – y)]
Solution:
Let p(n) is
xⁿ – yⁿ = (x – y)(x^n-1 + x^n-2 y + … + xy^n-2 + y^n-1); x, y ∈ R
Step – 1:
For n = 2
x² – y² = (x – y) (x + y) (True)
∴ P(1) is true.
Step – 2:
Let P(k) is true.
i.e., x^k – y^k = (x – y)(x^k-1 + x^k-2y + … +xy^k-2 + y^k-1)
Step – 3:
Let us prove P_k+1 is true.
i.e., x^k+1 – y^k+1 = (x – y) (x^k + x^k-1y + … (xy^k-1 + y^k)
L.H.S. = x^k+1 – y^k+1
= x^k+1 – xy^k + xy^k – y^k+1
= x(x^k – y^k) + y^k (x – y)
= x(x – y)(x^k-1 + x^k-2 y + … + xy^k-2 + y^k-1) + y^k(x – y) [by (1)]
= (x – y) [x^k + x^k-1 y + … + xy^k-2 + xy^k-1 + y^k]
= R.H.S.
∴ P_(k+1) is true.
Step – 4:
By Principle Of Mathematical Induction P(n) is true for all n ∈ N.

Question 13.
1 + 3 + 5 + ……. +(2n – 1) = n²
Solution:
Let P(n) is : 1 + 3 + 5 + ……. +(2n – 1) = n².
Step – 1:
For n = 2
L.H.S. = 1 + 3 = 4 = 2² (R.H.S)
∴ P(1) is true.
Step – 2:
Let P(k) is true.
i.e., 1 + 3 + 5 … + (2k – 1) = k² …(1)
Step – 3:
We will prove that P(k + 1) is true
i.e., we want to prove.
1+ 3 + 5 + … + (2k – 1) + (2k + 1) = (k + 1)²
L.H.S. = 1 + 3 + 5 + … + (2k – 1) + (2k + 1)
= k² + 2k + 1          [By – (1)]
= (k + 1)² = R.H.S.
Step – 4:
By the Principle of Mathematical Induction P(n) is true for all n.
i.e., 1 + 3 + 5 ….+ (2n – 1) = n²

Question 14.
n > n; n is a natural number.
Solution:
Let P(n) is 2ⁿ > n
Step – 1:
2¹ > 1 (True)
∴ P(1) is true.
Step – 2:
Let P(k) is true.
⇒ 2^k > k
Step – 3:
We shall prove that P(k + 1) is true
i.e., 2^k+1 > k + 1
Now 2^k+1 = 2.2k > 2k ≥ k + 1 for k ∈ N.
∴ 2^k+1 > k + 1
⇒ P(k + 1) is true.
Step – 4:
By the Principle of Mathematical Induction P(n) is true for all n.
i.e., 2ⁿ > n for n ∈ N

Question 15.
(1, 2, 3 … n)³  > 8 (1³ + 2³ + 3³ + … + n³), for n > 3.
Solution:
Let P(n) is
(1, 2, 3 … n)³  > 8 (1³ + 2³ + 3³ + … + n³), for n > 3.
Step – 1:
For n = 4
(1. 2. 3. 4)³ = 24³ = 13824
8(1³+ 2³ + 3³ + 4³) = 808
∴ (1. 2 . 3 . 4)³ > 8(1³ + 2³ + 3³ + 4³)
∴ P(4) is true.
Step- 2:
Let P(k) is true.
(1. 2. 3…….k)³ > 8(1³ + 2³ + 3³ + …… + k³)
Step – 3:
We shall prove that P_(k+1) is true.
i.e., (1. 2. 3. …….. k(k+1))³ > 8(1³ + 2³ + … + k³ + (k + 1)³)
Now (1. 2. 3. …….. k(k + 1)³)
= (1. 2. 3 … k)³ (k + 1)³
> 8 (1³ + 2³ + … k³) (k + 1)³
> 8 (1³ + 2³ + … k³) + 8(k + 1)³
= 8 (1³ + 2³ + … + k³ + (k + 1)³)
P_(k+1) is true.
Step – 4:
By the Principle of Mathematical Induction P(n) is true for all n ∈ N and n > 3.

Question 16.
\(\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{3 n+1}>\) for every positive integer n.
Solution:
Let P(n) is
\(\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{3 n+1}>\)



Step-4:
By the Principle of Mathematical Induction P(n) is true for all n ∈ N.

Odisha State Board BSE Odisha 6th Class English Solutions Follow-Up Lesson 1 I Like Bats Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Follow-Up Lesson 1 I Like Bats

BSE Odisha 6th Class English Follow-Up Lesson 1 I Like Bats Text Book Questions and Answers

Session – 1 (ପ୍ରଥମ ପର୍ଯ୍ୟାୟ):
I. Pre-Reading (ପ୍ରାକ୍-ପଠନ):

→ Socialisation (ସାମାଜିକୀକରଣ):
→ The teacher thinks of a pre-reading activity.
(ଶିକ୍ଷକ ଏକ ପଢ଼ିବା ପୂର୍ବବର୍ତ୍ତୀ କାର୍ଯ୍ୟ ବିଷୟରେ ଚିନ୍ତା କରନ୍ତୁ ।)

→ You may use pictures. You may also link the poem with the poem of the main lesson.
(ତୁମେ ଛବିଗୁଡ଼ିକୁ ବ୍ୟବହାର କରିପାର । ତୁମେ କବିତାଟିକୁ ମୁଖ୍ୟପାଠର କବିତା ସହ ଯୋଡ଼ିପାର ।)

→ What are there in the picture? What do they look like? That’s how bats sleep and rest hanging upside down. What an interesting way of resting and relaxing! Do you like to rest like bats hanging upside down?
(ଛବିରେ କ’ଣ ଅଛି ? ସେମାନେ କିପରି ଦେଖାଯାଉଛନ୍ତି ? ସେହିପରି ଭାବରେ ବାଦୁଡ଼ିମାନେ ଉପରୁ ତଳକୁ ଝୁଲିରହି ଶୁଅନ୍ତି ଏବଂ ବିଶ୍ରାମ ନିଅନ୍ତି । ବିଶ୍ରାମ ନେବା ଓ ନିଦ୍ରାଯିବାର କି କୌତୂହଳଜନକ ଉପାୟ ! ତୁମେ ବାଦୁଡ଼ିମାନଙ୍କ ପରି ଉପରୁ ତଳକୁ ଝୁଲିରହି ବିଶ୍ରାମ କରିବାକୁ ଭଲ ପାଅ କି ?)

→ In the poem ‘Mice’ the poet likes mice. Let’s read this poem to see if the poet likes bats.
(‘Mice” କବିତାରେ, କବି ମୂଷାମାନଙ୍କୁ ଭଲ ପାଇଛନ୍ତି । ଆସ ଆମେ ଏହି କବିତାଟି ପଢ଼ିବା ଏବଂ କବି ବାଦୁଡ଼ିମାନଙ୍କୁ ଭଲ ପାଆନ୍ତି କି ନାହିଁ ଦେଖୁବା ।)

II. While-Reading (ପଢ଼ିବା ସମୟରେ ):
Follow the steps of the main lesson.
(ମୁଖ୍ୟ ବିଷୟର ସୋପାନଗୁଡ଼ିକୁ ଅନୁସରଣ କର ।)
TEXT (ବିଷୟବସ୍ତୁ):
Read the poem silently and answer the questions that follow.
(କବିତାଟିକୁ ନୀରବରେ ପଢ଼ ଏବଂ ପରବର୍ତ୍ତୀ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

I like bats
Hanging upside down
Like rats. Like silk-cotton fruits
Swinging in wind
What a way to relax and rewind.
I wish I could
Do that
Like a bat
A way to find
After a day’s work
To relax and rewind
Upside down
Hang like bats.
I like bats
Hanging upside down
Like rats.

ଓଡ଼ିଆ ଉଚ୍ଚାରଣ :
ଆଇ ଲାଇକ୍ ବ୍ୟାଟ୍‌ସ୍
ହ୍ୟାଙ୍ଗିଙ୍ଗ୍ ଅପସାଇଡ୍ ଡାଉନ୍
ଲାଇକ୍ ମ୍ୟାଟ୍‌ସ୍ । ଲାଇକ୍ ସିଲ୍‌କ୍-କଟନ୍ ଫୁସ୍
ସୁଇଙ୍ଗିଙ୍ଗ୍ ଇନ୍ ଉଇଣ୍ଡ୍
ଦ୍ଵାଟ୍ ଏ ୱେ ଟୁ ରିଲାକ୍ସ ଆଣ୍ଡ ରିୱାଇଣ୍ଡ୍ ।
ଆଇ ଉଇସ୍ ଆଇ କୁଡ଼୍
ଡୁ ଦ୍ଯାଟ୍
ଲାଇକ୍ ଏ ବ୍ୟାଟ୍
ଏ ୱେ ଟୁ ଫାଇଣ୍ଡ୍
ଆଫ୍‌ଟର୍ ଏ ଡେ’ଜ୍ ୱାର୍କ
ଟୁ ରିଲାକ୍ସ ଆଣ୍ଡ ରିୱାଇଣ୍ଡ୍
ଅପ୍‌ସାଇଡ୍ ଡାଉନ୍
ଆଇ ଲାଇକ୍ ବ୍ୟାଟ୍‌ସ୍ ।
ହ୍ୟାଙ୍ଗିଙ୍ଗ୍ ଅଫ୍‌ସାଇଡ୍ ଡାଉନ୍
ଲାଇକ୍ ଗ୍ୟାସ୍ ।

Knowing The Key Words (ମୁଖ୍ୟ ଶବ୍ଦଗୁଡ଼ିକୁ ଜାଣିବା):

like – ସଦୃଶ
bats – ବାଦୁଡି
hanging – ଫାଶୀ
upside – ଓଲଟା |
down – ତଳକୁ
like – ପରି
rats – ମୂଷା
silk – cotton – ଶିମିଳି-ତୁଳା
fruits – ଫଳ
swinging – ସୁଇଙ୍ଗ୍
in wind — ପବନରେ
could — କରିପାରନ୍ତି |
What a way — କି ଉପାୟ |
find — ଖୋଜିବା
relax — ବିଶ୍ରାମ କରିବା
day’s work — ଦିନର କାମ
rewind – ରିଭାଇଣ୍ଡ୍ |
wish — ଇଚ୍ଛା

ସାରକଥା | ଓଡ଼ିଆ ଅନୁବାଦ:
ମୁଁ ବାଦୁଡ଼ିମାନଙ୍କୁ ଭଲପାଏ
ସେମାନେ ଉପର ପାଖ ତଳକୁ କରି ଝୁଲୁଥା’ନ୍ତି ମୂଷାମାନଙ୍କ ସଦୃଶ । | ସେମାନେ ପବନରେ ଦୋଳି ଖେଳୁଥା’ନ୍ତି ଶିମିଳି ତୁଳା ଫଳଗୁଡ଼ିକ ପରି । | କି ସୁନ୍ଦର ଉପାୟ ବିଶ୍ରାମ କରିବାର ଏବଂ ପବନରେ ଦୋହଲିବାର । | ମୁଁ ଭାବୁଛି ମୁଁ ସେହିପରି କରି ପାରିଥା’ନ୍ତି ଏକ ବାଦୁଡ଼ି ପରି । | ଗୋଟିଏ ଦିନକର କାମ ପରେ ଏକ ଉପାୟ ଖୋଜି ପାଇବାକୁ ବିଶ୍ରାମ କରିବାକୁ ଏବଂ ପବନରେ ଦୋହଲିବାକୁ । | ଉପର ପାଖ ତଳକୁ କରି ଝୁଲୁଥା’ନ୍ତି ବାଦୁଡ଼ିମାନଙ୍କ ସଦୃଶ । ମୁଁ ଭଲପାଏ ବାଦୁଡ଼ିମାନଙ୍କୁ ଯେଉଁମାନେ ଉପର ପାଖ ତଳକୁ କରି ଝୁଲୁଥା’ନ୍ତି ମୂଷାମାନଙ୍କ ଭଳି ।

Comprehension Questions : (ବୋଧମୂଳକ ପ୍ରଶ୍ନବଳୀ)

The teacher is to try to frame his/her own questions. Here are some for him/her.
(ଶିକ୍ଷକ ତାଙ୍କର ନିଜର ପ୍ରଶ୍ନଗୁଡ଼ିକ ତିଆରି କରିବାକୁ ଚେଷ୍ଟା କରିବେ । ଏଠାରେ କେତେକ ତାଙ୍କ (ପୁ/ସ୍ତ୍ରୀ) ପାଇଁ ଅଛି ।)

Question 1.
How do bats hang?
(ବାଦୁଡ଼ିମାନେ କିପରି ଝୁଲୁଥା’ନ୍ତି ?)
Answer:
Bats hang upside down.

Question 2.
What are bats compared to ?
(ବାଦୁଡ଼ିମାନଙ୍କୁ କାହା ସହିତ ତୁଳନା କରାଯାଇଛି ?)
Answer:
Bats are compared to rats.

Question 3.
Have you seen bats hanging upside down on trees in great numbers?
(ଗଛଗୁଡ଼ିକରେ ବାଦୁଡ଼ିମାନେ ବହୁ ସଂଖ୍ୟାରେ ଝୁଲି ରହିଥ‌ିବାର ତୁମେ ଦେଖୁଛ କି ?)
Answer:
Yes, we have seen bats hanging upside down on trees in great numbers.

Question 4.
Have you seen silk-cotton fruit hanging in great numbers? Do they look alike?
(ଶିମିଳି-ତୁଳା ଫଳ ବହୁ ସଂଖ୍ୟାରେ ଝୁଲୁଥ‌ିବା ତୁମେ ଦେଖୁଛ କି ? ସେଗୁଡ଼ିକ ଏକାପରି ଦେଖାଯାଆନ୍ତି କି ?)
Answer:
Yes, we have seen silk-cotton fruit hanging in great numbers. Really, they look alike.

Question 5.
What is the meaning of the word ‘rewind’? See the dictionary at the end of this lesson.
(‘ରିୱାଇଣ୍ଡ୍’ ଶବ୍ଦର ଅର୍ଥ କ’ଣ ? ଏହି ଅଧ୍ୟାୟର ଶେଷରେ ଥ‌ିବା ଅଭିଧାନ ବା ଶବ୍ଦାର୍ଥ ଦେଖ ।)
Answer:
The meaning of the word ‘rewind’ is taking a rest with occasional backward movement.
Teacher is to frame some questions from the second stanza.
(ଶିକ୍ଷକ ଦ୍ଵିତୀୟ ପଦରୁ କିଛି ପ୍ରଶ୍ନ ତିଆରି କରିବେ ।)

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

(a) Answer the following questions.
(ନିମ୍ନଲିଖ ପ୍ରଶ୍ନଗୁଡ଼ିକର ଉତ୍ତର ଦିଅ ।)

(i) How do bats hang ?
(ବାଦୁଡ଼ିମାନେ କିପରି ଝୁଲନ୍ତି ?)
Bats hang ______________________________________.
Answer:
Bats hang upside down.

(ii) What are bats compared to ?
(ବାଦୁଡ଼ିମାନଙ୍କୁ କାହା ସହ ତୁଳନା କରାଯାଇଛି ?)
Bats are ______________________________________.
Answer:
Bats are compared to rats.

(iii) What swings in the wind?
(ପବନରେ କ’ଣ ଝୁଲିଥାଏ ?)
The silk – ______________________________________.
Answer:
The silk-cotton fruits swing in the wind.

(iv) Why does the poet like to hang like bats upside down?
(କବି କାହିଁକି ବାଦୁଡ଼ିମାନଙ୍କ ପରି ଉପରପାଖ ତଳକୁ କରି ଝୁଲିବାକୁ ଭଲପାଆନ୍ତି ?)
______________________________________to relax and ____________________________.
Answer:
The poet likes to hang like a bats upside down because it is a nice way to relax and rewind.

(b) Let us summarise the poem. Fill in the gaps.
(ଆସ ଆମେ କବିତାର ସାରାଂଶ ବାହାର କରିବା । ଶୂନ୍ୟସ୍ଥାନଗୁଡ଼ିକୁ ପୂରଣ କର ।) (Question with Answer)

The poet _____________________to see bats. _____________________. Bats hang like _____________. This is a good way to relax and ____________. The poet wants to_____________ _____________bats to relax ________________ ________________after the day’s ________________.
Answer:
The poet likes to see bats. They are hanging upside down. Bats hang like rats and like silk-cotton fruits swinging in wind. This is a good way to relax and rewind. The poet wants to hang upside down like a bat to relax and rewind after the day’s work.

(c) Think of writing a poem. Start with replacing ‘bats’ with some fruits and make minimum changes in the poem. Change the title accordingly.
(ଗୋଟିଏ କବିତା ଲେଖିବା କଥା ଭାବ । ‘ବାଦୁଡ଼ିମାନଙ୍କ’’ ବଦଳରେ କେତେକ ଫଳକୁ ନେଇ ଏବଂ କବିତାରେ ସ୍ଵଳ୍ପ ପରିବର୍ତ୍ତନ କରି ଆରମ୍ଭ କର । ସେହି ଅନୁସାରେ କବିତାର ଶିରୋନାମା ପରିବର୍ତ୍ତନ କର ।)

APPLES
Answer:
I like apples
Hanging upside down
Like rats. Like silk-cotton fruits
Swinging in wind
What a way to relax and rewind.
I wish I could
Do that
Like an apple
A way to find
After a day’s work
To relax and rewind
Upside down Hang like apples.
I like apples
Hanging upside down
Like rats.

Knowing The Key Words (ମୁଖ୍ୟ ଶବ୍ଦଗୁଡ଼ିକୁ ଜାଣିବା):
(The words/phrases have been defined mostly on contextual meanings.)

Nibble – gentle and playful bite of a mouse. ମୂଷାର କୁଟ୍ କୁଟ୍ କରି କାଟି ଖାଇବା
Pink – (colour) pale red, ଫିକା ନାଲି |
Swinging – hanging and moving (bats have)
relax and rewind – taking rest (with occassional backward movement) ଆରାମରେ ବିଶ୍ରାମ ନେବା
upside down – legs upward and head downward. ଗୋଡ଼ ଉପରକୁ ଓ ମୁଣ୍ଡ ତଳକୁ କରି ଓଲଟା ରହିବା ।

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

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Exercise 11(a)

Question 1.
Find the distance between the following pairs of points.
(i) (3, 4), (-2, 1);
Solution:
Distance between points (3, 4) and (-2, 1) is
\(\sqrt{(3+2)^2+(4-1)^2}=\sqrt{25+9}=\sqrt{34}\)

(ii) (-1, 0), (5, 3)
Solution:
The distance between the points (-1, 0) and (5, 3) is
\(\sqrt{(-1-5)^2+(0-3)^2}\)
= \(\sqrt{36+9}=\sqrt{45}=3 \sqrt{5}\)

Question 2.
If the distance between points (3, a) and (6, 1) is 5, find the value of a.
Solution:
Distance between the points. (3, a) and (6, 1) is
\(\sqrt{(3-6)^2+(a-1)^2}=\sqrt{9+(a-1)^2}\)
∴ \(\sqrt{9+(a-1)^2}=5\)
or, 9 + (a – 1)² = 25
or, (a – 1)² = 16
or, a – 1 = ± 4
a = 1 ± 4 = 5 or, – 3

Question 3.
Find the coordinate of the points which divides the line segment joining the points A (4, 6), B (-3, 1) in the ratio 2: 3 internally. Find also the coordinates of the point which divides \(\overline{\mathbf{A B}}\) in the same ratio externally.
Solution:

Question 4.
Find the coordinates of the mid-point of the following pairs of points.
(i) (-7, 3), (8, -4);
Solution:
Mid-point of the line segment joining the points (-7, 3) and (8, -4) are \(\left(\frac{-7+8}{2}, \frac{3-4}{2}\right)=\left(\frac{1}{2},-\frac{1}{2}\right)\)

(ii) (\(\frac{3}{4}\), -2), (\(\frac{-5}{2}\), 1)
Solution:
Mid-point of the line segment joining the points. (\(\frac{3}{4}\), -2) and (\(\frac{-5}{2}\), 1) is,
\(\left(\frac{\frac{3}{4}-\frac{5}{2}}{2}, \frac{-2+1}{2}\right)=\left(\frac{-7}{8}, \frac{-1}{2}\right)\)

Question 5.
Find the area of the triangle whose vertices are (1, 2), (3, 4) (\(\frac{1}{2}\), \(\frac{1}{4}\))
Solution:
Area of the triangle whose vertices are (1, 2), (3, 4) and (\(\frac{1}{2}\), \(\frac{1}{4}\)) is

Question 6.
If the area of the triangle with vertices (0, 0), (1, 0), (0, a) is 10 units, find the value of a.
Solution:
Area of the triangle with vertices (0, 0),(1,0), (0, a), is \(\frac{1}{2}\) × 1 × a = \(\frac{a}{2}\)
∴ \(\frac{a}{2}\) = 10 or a = 20

Question 7.
Find the value of a so that the points (1, 4), (2, 7), (3, a) are collinear.
Solution:
As points (1, 4), (2, 7), (3, a) are collinear, we have the area of the triangle with vertices (1, 4), (2, 7), and (3, a) is zero.
∴ \(\frac{1}{2}\) {1(7 – a) + 2(a – 4) + 3 (4 – 7)} = 0
or, 7 – a + 2a – 8 + 12 – 21 =0
⇒ a = 10

Question 8.
Find the slope of the lines whose inclinations are given.
(i) 30°
Solution:
The slope of the line whose inclination is 30°.
tan 30° = \(\frac{1}{\sqrt{3}}\)

(ii) 45°
Solution:
Slope = tan 45° = +1

(iii) 60°
Solution:
Slope = tan 60° = √3

(iv) 135°
Solution:
Slope = tan 135° = – 1

Question 9.
Find the inclination of the lines whose slopes are given below.
(i) \(\frac{1}{\sqrt{3}}\)
Solution:
The slope of the line is \(\frac{1}{\sqrt{3}}\)
∴ tan θ = \(\frac{1}{\sqrt{3}}\) or, θ = 30°
∴ The inclination of the line is 30°

(ii) 1
Solution:
Slope = 1 = tan 45°
∴ The inclination of the line is 45°.

(iii) √3
Solution:
Slope = √3 = tan 60°  ∴ θ = 60°
∴ Inclination = 60°

(iv) – 1
Solution:
Slope = – 1 = tan 135°
∴ Inclination = 135°

Question 10.
Find the angle between the pair of lines whose slopes are ;
(i) \(\frac{1}{\sqrt{3}}\), 1
Solution:

(ii) √3, -1
Solution:

Question 11.
(a) Show that the points (0, -1), (-2, 3), (6, 7), and (8, 3) are vertices of a rectangle.
Solution:


∴ The opposite sides are equal and two consecutive sides are perpendicular. So it is a rectangle.

(b) Show that the points (1, 1), (-1, -1), and (-√3, √3) are the vertices of an equilateral triangle.
Solution:

Question 12.
Find the coordinates of the point P(x, y) which is equidistant from (0, 0), (32, 10), and (42, 0).
Solution:

Question 13.
If the points (x, y) are equidistant from the points (a + b, b – a) and (a – b, a + b), prove that bx = ay.

Question 14.
The coordinate of the vertices of a triangle are (α₁, β₁), (α₂, β₂), and (α₃, β₃). Prove that the coordinates of its centroid is \(\left(\frac{\alpha_1+\alpha_2+\alpha_3}{3}, \frac{\beta_1+\beta_2+\beta_3}{3}\right)\)
Solution:

Question 15.
Two vertices of a triangle are (0, -4) and  (6, 0). If the medians meet at the point (2, 0), find the coordinates of the third vertex.
Solution:

∴ \(\frac{6+x}{3}\) = 2, \(\frac{-4+y}{3}\) = 0
⇒ x = 0, y = 4
∴ The coordinates of the 3rd vertex are (0, 4).

Question 16.
If the point (0, 4) divides the line segment joining(-4, 10) and (2, 1) internally, find the point which divides it externally in these same ratios.
Solution:

Question 17.
Find the ratios in which the line segment joining (-2, -3) arid (5, 4) is divided by the coordinate axes and hence find the coordinates of these points.
Solution:

Question 18.
In a triangle, one of the vertices is at (2, 5) and the centroid of the triangle is at (-1, 1). Find the coordinates of the midpoint of the side opposite to the given angular point.
Solution:

Question 19.
Find the coordinates of the vertices of a triangle whose sides have midpoints at (2, 1), (-1, 3), and (-2, 5).
Solution:

∴ x₂ + x₃ – 4 or, x₂ – 4 – 3 = 1
∴ x₁ = – 4 – x₂ = -4 – 1 = -5
Similarly y₁ + y₂ + y₃ = 5 + 1 + 3 = 9
As y₁ + y₂ = 10
we have y₃ = 9 – 10 = – 1
Again y₁ + y₃ = 6
or, y₁ = 6 – y₃ = 6 + 1 = 7
and y₂ = 10 – y₁ = 10 – 7 = 3
∴ The coordinates of A, B, and C are (-5, 7), (1, 3), and (3, -1).

Question 20.
If the vertices of a triangle have their coordinates given by rational numbers, prove that the triangle cannot be equilateral.
Solution:
Let us choose the contradiction method. Let the triangle is equilateral if the co¬ ordinate of the vertices is rational numbers.
Let ABC be an equilateral triangle with vertices A (a, 0), B (a, 0), and C (b, c) where a, b, c are rational.

⇒ a² = b² + c² = \(\frac{a^2}{4}\) + c²
⇒ c² =  a² – \(\frac{a^2}{4}\) = \(\frac{3a^2}{4}\) ⇒ c = \(\frac{\sqrt{3}}{2}\) a     ….(2)
Now b = \(\frac{a}{2}\), c = \(\frac{\sqrt{3}}{2}\) a
If a is rational then b is rational but c is irrational, i.e., the coordinates of the vertices are not rational, which contradicts the assumption.
Hence assumption is wrong.
So the triangle cannot be equilateral if the coordinate of the vertices is rational numbers.

Question 21.
Prove that the area of any triangle is equal to four times the area of the triangle formed by joining the midpoints of its sides.
Solution:

∴ The area of triangle ABC is four times the area of triangle DEF. (Proved)

Question 22.
Find the condition that the point (x, y) may lie on the line joining (1, 2) and (5, -3).
Solution:

∴ As points A, B, and C are collinear, we have the area of the triangle ABC as 0.
∴ \(\frac{1}{2}\) {1(-3 – y) + 5(y – 2) + x(2 + 3)} = 0
or, – 3 – y + 5y – 10 + 5x = 0
or, 5x + 4y = 13

Question 23.
Show that the three distinct points (a², a), (b², b), and (c², c) can never be collinear.
Solution:
Area of the triangle with vertices (a², a), (b², b) , and (c², c) is
\(\frac{1}{2}\) {a²(b – c) + b²(c – a) c²(a – b)}
= (a – b)(b – c)(a – c)
which is never equal to zero except when a = b = c, hence the points are not collinear.

Question 24.
If A, B, and C are points (-1, 2), (3, 1), and (-2, -3) respectively, then show that the points which divide BC, CA, and AB in the ratios (1: 3), (4: 3) and (-9: 4) respectively are collinear.
Solution:
Let the points P, Q, and R divides \(\overline{\mathrm{BC}}\), \(\overline{\mathrm{CA}}\), in \(\overline{\mathrm{AB}}\) the ratio 1: 3, 4: 3 and -9: 4

Question 25.
Prove analytically :
(a) The line segment joining the midpoints of two sides of a triangle is parallel to the third and half of its length.

Solution:
Let the coordinates of the triangle ABC be (x₁, y₁), (x₂, y₂) and (x₃, y₃)
The points D and E are the midpoints of the sides \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{AC}}\)

(b) The altitudes of a triangle are concurrent.
Solution:

(c) The perpendicular bisectors of the sides of a triangle are concurrent.
Solution:

(d) An angle in a semicircle is a right angle.
Solution:

∴ The angle in a semicircle is a right angle. (Proved)

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

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

Question 1.
Which of the following in a sequence?
(i) f(x) = [x], x ∈ R
(ii) f(x) = |x|, x ∈ R
(iii) f(x) = \(\sqrt[n]{\pi}\) ,n∈ N
Solution:
(iii) f(x) = \(\sqrt[n]{\pi}\) ,n∈ N is a sequence f(n) : N → X, X ⊂ R.

Question 2.
Determine if (t_n) is an arithmetic sequence if :
(i) t_n = an² + bn
Solution:
t_n = an² + bn
⇒ t_n+1 = a(n + 1)² + b(n – 1)
⇒ t_n+1 – t_n = a{(n + 1)² – n²} + b{n + 1 – n}
= a(2n + 1) + b
Which is not independent of n.
∴ (t_n) is not an A.P.

(ii) t_n = an + b
Solution:
t_n = an + b
⇒ t_n+1 = a(n + 1) + b
Now t_n+1 – t_n
= {a(n + 1) + b} – {an + b}
= a (constant)
∴ (t_n) is an arithmetic sequence.

(iii) t_n = an² + b
Solution:
t_n = an² + b
⇒ t_n+1 = a(n + 1)² + b
∴ t_n+1 – t_n = a[(n + 1)² – n²] + b – b
= a(2n + 1)
(does not independent of n)
∴ (t_n) is not an arithmetic sequence.

Question 3.
If a geometric series converges which of the following is true about its common ratio r?
(i) r > 1
(ii) -1 < r < 1
(iii) r > 0
Solution:
(ii) -1 < r < 1

Question 4.
If an arithmetic series ∑tn converges, which of the following is true about t_n?
(i) t_n < 1
(ii) |t_n| < 1
(iii) t_n = 0
(iv) t_n → 0
Solution:
(iii) t_n = 0

Question 5.
Which of the following is an arithmetic-geometric series?
(i) 1 + 3x + 7x² + 15x³+ ….
(ii) x + \(\frac{1}{2}\)x + \(\frac{1}{3}\)x² + ….
(iii) x + (1 + 2)x² + (1 + 2 + 3)x³ +…
(iv) x + 3x² + 5x³ + 7x⁴ + …
Solution:
(iv) x + 3x² + 5x³ + 7x⁴ + … is an arithmetic geometric series with a = 1, d = 2, r = x.

Question 6.
For an arithmetic sequence (t_n) t_p = q, t_q = p, (p ≠ q), find t_n.
Solution:
t_p = q ⇒ a + (p – 1)d = q    ……(1)
t_q = p ⇒ a + (q – 1)d = p    ……(2)
From (1) and (2) we have (p – q)d = q – p
⇒ d = (-1)
Putting d = (-1) in (1)
we have a = p + q – 1
∴ t_n = a + (n – 1)d
= (P + q – 1) + (n – 1) (-1)
= p + q – n

Question 7.
For an arithmetic series, ∑a_n S_p = q and S_q = p (p ≠ q) find S_p+q
Solution:
S_p = q and S_q = p

Question 8.
The sum of a geometric series is 3. The series of squares of its terms have a sum of 18. Find the series.
Solution:

Question 9.
The sum of a geometric series is 14, and the series of cubes of its terms have a sum of 392 find the series.
Solution:


∴ The series is \(\sum_{n=1}^{\infty} \frac{7}{2^{n-1}}\)

Question 10.
Find the sum as directed
(i) 1 + 2a + 3a² + 4a³ + …..(first n terms(a ≠ 1))
Solution:

(ii) 1 + (1 + x)y + (1 + x + x²)y² + …..(to infinity)
Solution:
Let S_∞ = 1 + (1 + x)y + (1x + x²)y² + …
⇒ S_n = 1 + (1 + 1 + x)y + (1 + x + x²)y² + ……+(1 + x + …. x^n-1)y^n-1

(iii) 1 + \(\frac{3}{5}\) + \(\frac{7}{25}\) + \(\frac{15}{125}\) + \(\frac{31}{625}\) + …..(to infinity)
Solution:

(iv) 1 + 4x + 8x² + 13x³ + 19x⁴ + …..(to infinity). Assuming that the series has a sum for |x| < 1.
Solution:

(v) 3.2 + 5.2² + 7.2³ + …..(first n terms)
Solution:
Sn = 3.2 + 5.22 + 7.3 + ….n terms = 2[3 + 5.2 + 7.22 + ….n terms]

Question 11.
Find the sum of the infinite series.
(i) \(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots\)
Solution:

= 1 – \(\frac{1}{n+1}\)
∴ \(S_{\infty}=\lim _{n \rightarrow \infty} S_n=1\)

(ii) \(\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\ldots\)
Solution:
\(\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\ldots\)

(iii) \(\frac{1}{2 \cdot 5 \cdot 8}+\frac{1}{5 \cdot 8 \cdot 11}+\frac{1}{8 \cdot 11 \cdot 14}+\ldots\)
Solution:
Here t_n = \(\frac{1}{(3 n-1)(3 n+2)(3 n+5)}\)
The denominator of t_n is the product of 3 consecutive terms of A.P. Now multiplying and dividing by (3n + 5) – (3n – 1) we have
\(t_n=\frac{(3 n+5)-(3 n-1)}{6(3 n-1)(3 n+2)(3 n+5)}\)

(iv) \(\frac{3}{1^2 \cdot 2^2}+\frac{5}{2^2 \cdot 3^2}+\frac{7}{3^2 \cdot 4^2}+\ldots\) [Hint : take t_n = \(\frac{2 n+1}{n^2(n+1)^2}=\frac{(n+1)^2-n^2}{n^2(n+1)^2}\)]
Solution:

(v) \(\frac{1}{1 \cdot 5}+\frac{1}{3 \cdot 7}+\frac{1}{5 \cdot 9}+\ldots .\)
Solution:

Question 12.
Find S_n for the series.
(i) 1.2 + 2.3 + 3.4 + ….
Solution:

(ii) 1.2.3 + 2.3.4 + 3.4.5 + …
Solution:

(iii) 2.5.8 + 5.8.11 + 8.11.14 +…
Solution:

(iv) 1.2.3.4 + 2.3.4.5 + 3.4.5.6 + …
[Hint : t_n = (3n – 1) (3n + 2)(3n + 5)]
Solution:

(v) 1.5 + 2.6 + 3.7 + …
[Hint: t_n = n(n + 4) is not a product of two successive terms of an A.P. for the term following n should be n+1, not n+4. So the method of previous exercises is not applicable. Instead, write t_n = n² + 4n and find S_n = \(\sum_{k=1}^n k^2+4 \sum_{k=1}^n k\) applying formulae]
Solution:

(vi) 2.3 + 3.6 + 4.11 + …
[Hint : Take t_n = (n + 1) (n² + 2)]
Solution:

(vii) 1.3² + 2.5² + 3.7² + ….
Solution:

Question 13.
Find the sum of the first n terms of the series:
(i) 5 + 6 + 8 + 12 + 20 + …
Solution:

(ii) 4 + 5 + 8 + 13 + 20 + …
Solution:

Question 14.
(i) Find the sum of the product of 1,2,3….20 taken two at a time. [Hint: Required sum = \(\frac{1}{2}\left\{\left(\sum_{k=1}^{20} k\right)^2-\sum_{k=1}^{20} k^2\right\}\)]
Solution:
We know that
(x₁ + x₂ + x₃ + …. + x_n)²
= (x²₁ + xⁿ₂ + … + x²_n) + 2 (Sum of all possible Products taken two at a time)
∴ 2 (Sum of products of 1. 2, 3,…… 20 taken two at a time)
= (1 + 2 + 3 + … 20)² – (1² + 2² + … + 20²)
\(=\left(\frac{20 \times 21}{2}\right)^2-\frac{20(20+1)(40+1)}{6}\)
= (210)² – 70 × 41
= 44100 – 2870 = 41230
∴ The required sum = \(\frac{41230}{2}\) = 20615

(ii) Do the same for 1, 3, 5, 7,….19.
Solution:
Here the required sum

Question 15.
If a = 1 + x + x² + ….. and b = 1 + y + y² + ….|x| <  1 and |y| <  1, then prove that 1 – xy + x²y² + x³y³ + ….. =  \(\frac{a b}{a+b-1}\)
Solution:

Question 16.
If a, b, c are respectively the pth, qth, rth terms of an A.P., then prove that a(q – r) + b(r – p) + c(p – q) = 0
Solution:
Let the first term of an A.P. = A and the common difference = D.
According to the question
t_P = a, t_q = b, t_r = c
⇒ A + (p – q)D = a      …..(1)
A + (q – 1)D = b           …..(2)
A + (r – 1)D = c            …..(3)
L.H.S = a(q – r) + b (r – p) + c (p – q)
= (A + (p – 1)D) (q – r) + (A + (q – 1)D)
(r- p) + (A + (r – 1)D) (p – q)
= A (q – r + r – p + p – q) + D [(p – 1)
(q – r) + (q – 1) (r – p) + (r – 1) (p – q)]
= D (pq – pr + qr – pq + pr – qr) – D (q – r + r – p + p – q) = 0

Question 17.
If \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in A.P. and a + b + c ≠ 0, prove that \(\frac{\mathbf{b}+\mathbf{c}}{\mathbf{a}}, \frac{\mathbf{c}+\mathbf{a}}{\mathbf{b}}, \frac{\mathbf{a}+\mathbf{b}}{\mathbf{c}}\) are in A.P.
Solution:

Question 18.
If a², b², c² are in A.P. Prove that \(\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}\) are in A.P.
Solution:

Question 19.
If \(\frac{b+c}{a}, \frac{c+a}{b}, \frac{\mathbf{a}+\mathbf{b}}{c}\) are in A.P.,prove that \(\frac{\mathbf{1}}{\mathbf{a}}, \frac{\mathbf{1}}{\mathbf{b}}, \frac{\mathbf{1}}{\mathbf{c}}\) are inA.P.given a + b + c ≠ 0
Solution:

Question 20.
If (b – c)², (c – a)², (a – b)² are in A.P., prove that \(\frac{1}{b-c}, \frac{1}{c-a}, \frac{1}{a-b}\) are in A.P.
Solution:

Question 21.
If a, b, c are respectively the sum of p, q, r terms of an A.P., prove that \(\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)\) = 0
Solution:

Question 22.
If a, b, c,d are in G.P., prove that (a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)².
Solution:
Let a, b, c, and d are in G.P.
Let the common ratio = r
⇒ b = ar, c = ar², d = ar³
LHS = (a² + b² + c²) (b² + c² + d²)
= (a² + a²r² + a²r⁴) (a²r² + a²r⁴ + a²r⁶)
= a⁴r² (1 + r² + r⁴)²
= (a²r + a²r³ + a²r⁵)²
= (a.ar + ar.ar² + ar².ar³)²
= (ab + bc + cd)² = R.H.S. (proved)