Class 12

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 11 Differential Equations Ex 11(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 11 Differential Equations Exercise 11(a)

Question 1.
Determine the order and degree of each of the following differential equations.
(i) y sec² x dx + tan x dy = 0
Solution:
Order: 1, Degree: 1

(ii) \(\left(\frac{d y}{d x}\right)^4\) + y⁵ = \(\frac{d^3 y}{d x^3}\)
Solution:
Order: 3, Degree: 1

(iii) a\(\frac{d^2 y}{d x^2}\) = \(\left\{1+\left(\frac{d y}{d x}\right)^2\right\}^{\frac{3}{2}}\)
Solution:
Order: 2, Degree: 2

(iv) tan^-1\(\sqrt{\frac{d y}{d x}}\) = x
Solution:
Order: 1, Degree: 1

(v) ln\(\left(\frac{d^2 y}{d x^2}\right)\) = y
Solution:
Order: 2, Degree: 1

(vi) \(\frac{\frac{d y}{d t}}{y+\frac{d y}{d t}}\) = \(\frac{y t}{d y}\)
Solution:
Order: 1, Degree: 2

(vii) \(\frac{d^2 y}{d u^2}\) = \(\frac{3 y+\frac{d y}{d u}}{\sqrt{\frac{d^2 y}{d u^2}}}\)
Solution:
Order: 2, Degree: 3

(viii) \(e^{\frac{d z}{d x}}\) = x²
Solution:
Order: 1, Degree: 1

Question 2.
Form the differential equation by eliminating the arbitrary constants in each of the following cases.
(i) y = A sec x
Solution:
y = A sec x
Then \(\frac{d y}{d x}\) = A sec x tan x = y tan x

(ii) y = C tan^-1 x
Solution:

(iii) y = Ae^t + Be^2t
Solution:

(iv) y = Ax² + Bx
Solution:

(v) y = -acos x + b sin x
Solution:

(vi) y = a sin^-1 x + b cos^-1 x
Solution:

(vii) y = at + be^t
Solution:

(viii) y = a sin t + be^t
Solution:

(ix) ax² + by = 1
Solution:

Question 3.
Find the general solution ofthe following differential equations.
(i) \(\frac{d y}{d x}\) = \(\frac{e^{2 x}+1}{e^x}\)
Solution:
\(\frac{d y}{d x}\) = \(\frac{e^{2 x}+1}{e^x}\)
⇒ y = ∫(e^x + e^-x) dx = e^x – e^-x + C

(ii) \(\frac{d y}{d x}\) = x cos x
Solution:
\(\frac{d y}{d x}\) = x cos x
⇒ y = ∫x cos x dx
= x . sin x – ∫sin x dx – x sin x + cos x + C

(iii) \(\frac{d y}{d x}\) = t⁵ log t
Solution:

(iv) \(\frac{d y}{d x}\) = 3t² + 4t + sec² t
Solution:
\(\frac{d y}{d x}\) = 3t² + 4t + sec² t
⇒ y = t³ + 2t² + tan t + C

(v) \(\frac{d y}{d x}\) = \(\frac{1}{x^2-7 x+12}\)
Solution:

(vi) \(\frac{d y}{d u}\) = \(\frac{u+1}{\sqrt{3 u^2+6 u+5}}\)
Solution:

(vii) (x² + 3x + 2) dy – dx = 0
Solution:

(viii) \(\frac{d y}{d t}\) = \(\frac{\sin ^{-1} t e^{\sin ^{-1} t}}{\sqrt{1-t^2}}\)
Solution:

Question 4.
Solve the following differential equations.
(i) \(\frac{d y}{d x}\) = y + 2
Solution:

(ii) \(\frac{d y}{d t}\) = \(\sqrt{1-y^2}\)
Solution:
\(\frac{d y}{d t}\) = \(\sqrt{1-y^2}\)
⇒ \(\frac{d y}{\sqrt{1-y^2}}\) = dt
⇒ sin^-1 y = t + C

(iii) \(\frac{d y}{d z}\) = sec y
Solution:
\(\frac{d y}{d z}\) = sec y
⇒ cos y dy = dz
⇒ sin y = z + C

(iv) \(\frac{d y}{d x}\) = e^y
Solution:

(v) \(\frac{d y}{d x}\) = y² + 2y
Solution:

(vi) dy + (y² + 1) dx = 0
Solution:

(vii) \(\frac{d y}{d x}\) + \(\frac{e^y}{y}\) = 0
Solution:

(viii) dx + cot x dt = 0
Solution:
dx + cot x dt = 0
⇒ tan x dx + dt = 0
⇒ ∫tan x dx + ∫dt = C₁
⇒ In sec x + t = C₁
⇒ In sec x = C₁ – t
⇒ sec x = \(e^{C_1}\) . e^-t
⇒ cos x = \(e^{-C_1}\) . e^t
⇒ cos x = Ce^t where C = \(e^{-C_1}\)

Question 5.
Obtain the general solution of the following differential equations.
(i) \(\frac{d y}{d x}\) = (x² + 1) (y² + 1)
Solution:

(ii) \(\frac{d y}{d t}\) = e^2t+3y
Solution:

⇒ 2e^-3y + 3e^2t + 6C₁ = 0
⇒ 2e^-3y + 3e^2t = C
where C = -6C₁

(iii) \(\frac{d y}{d z}\) = \(\frac{\sqrt{1-y^2}}{\sqrt{1-z^2}}\)
Solution:

(iv) \(\frac{d y}{d z}\) = \(\frac{x \log x}{3 y^2+4 y}\)
Solution:

(v) x²\(\sqrt{y^2+3}\) dx + y\(\sqrt{x^3+1}\) dy = 0
Solution:

(vi) tan y dx + cot x dy = 0
Solution:
tan y dx + cot x dy = 0
⇒ tan x . dx + cot y dy = 0
⇒ ∫tan x dx + ∫cot y dy = 0
⇒ -ln cos x + ln siny = ln C
⇒ ln\(\frac{\sin y}{\cos x}\) = ln C
⇒ \(\frac{\sin y}{\cos x}\) = C
⇒ sin y = C cos x

(vii) (x² + 7x + 12) dy + (y² – 6y + 5) dx = 0
Solution:

(viii) y dy + e^-y x sin x dx = 0
Solution:
y dy + e^-y x sin x dx = 0
⇒ ye^y dy + x sin x dx = 0
⇒ ∫ye^y dy + ∫x sin dx = C
[Integrating by parts.
⇒ ye^y – ∫e^y dy + x(-cos x) – ∫(-cos x) dx = C
⇒ ye^y – e^y – x cos x + sin x = C
⇒ (y – 1) e^y – x cos x + sin x = C

Question 6.
Solve the following second order equations.
(i) \(\frac{d^2 y}{d x^2}\) = 12x² + 2x
Solution:

(ii) \(\frac{d^2 y}{d t^2}\) =e^2t +e^-t
Solution:

(iii) \(\frac{d^2 y}{d \vartheta^2}\) = -sin υ + cos υ + sec² υ
Solution:
\(\frac{d^2 y}{d \vartheta^2}\) = -sin υ + cos υ + sec² υ
Integrating we get
\(\frac{d y}{d υ}\) = ∫sin υ dυ + ∫cos υ dυ + ∫sec² υ dυ
= cos υ + sin υ + tan υ + A
Again integratingwe get
y = ∫(cos υ + sin υ + tan υ + A)dυ + B
where A, B are arbritrary constants.
⇒ y = sin υ – cos υ + ln |sec υ| + A.υ. + B

(iv) cosec x \(\frac{d^2 y}{d x^2}\) = x
Solution:
cosec x \(\frac{d^2 y}{d x^2}\) = x
\(\frac{d^2 y}{d x^2}\) = x sin x
Integrating we get
\(\frac{d y}{d x}\) = ∫x sin x dx + A
= x . (-cos x) – ∫(-cos x) dx + A
= -x cos x + ∫cos x dx + A
= -x cos x + sin x + A
Again integrating we get
y = -∫x cos x dx + ∫sin x + ∫A dx + B
= -{x sin x -∫1 . sin x dx} – cos x + Ax + B
= -x sin x – 2cos x + Ax + B

(v) x²\(\frac{d^2 y}{d x^2}\) + 2 = 0
Solution:

(vi) sec x \(\frac{d^2 y}{d x^2}\) = sec 3x
Solution:

(vii) \(\frac{d^2 y}{d x^2}\) = sec² x + cos² x
Solution:

(viii) e^-x\(\frac{d^2 y}{d x^2}\) = x
Solution:
e^–x\(\frac{d^2 y}{d x^2}\) = x
⇒ \(\frac{d^2 y}{d x^2}\) = xe^x
Integrating we get
\(\frac{d y}{d x}\) = ∫xe^x dx = ∫e^x dx + Ax + B
= xe^x – e^x – e^x + Ax + B
= (x – 2)e^x + Ax + B

Question 7.
Find the particular solutions of the following equations subject to the given conditions.
(i) \(\frac{d y}{d x}\) = cos x, given that y = 2 when x = 0.
Solution:
\(\frac{d y}{d x}\) = cos x
Integrating we get
y = ∫cos x dx = sin x + C
Given that when x = 0, y = 2
So 2 = C
∴ The particular solution is y = sin x + 2

(ii) \(\frac{d y}{d t}\) = cos² y subject to y = \(\frac{\pi}{4}\) when t = 0.
Solution:
\(\frac{d y}{d t}\) = cos² y
⇒ sec² y dy = dt
∫sec² dy = ∫dt
⇒ tan y = t + C
When t = 0, y = \(\frac{\pi}{4}\)
So tan \(\frac{\pi}{4}\) = C ⇒ C = 1
∴ The particular solution is tan y = t + 1

(iii) \(\frac{d y}{d x}\) = \(\frac{1+y^2}{1+x^2}\) given that y = √3 when x = 1.
Solution:

(iv) \(\frac{d^2 y}{d x^2}\) = 6x given that y = 1 and \(\frac{d y}{d x}\) = 2 when x = 0.
Solution:
\(\frac{d^2 y}{d x^2}\) = 6x ⇒ \(\frac{d y}{d x}\) = 3x² + 2
When x = 0, \(\frac{d y}{d x}\) = 2
So 2 = A
∴ \(\frac{d y}{d x}\) = 3x² + 2
Again integrating we get
y = x³ + 2x + B
When x = 0, y = 1
So B = 1.
∴ The particular solution is y = x³ + 2x + 1

Question 8.
(i) Solve : \(\frac{d y}{d x}\) = sec (x + y)
Solution:

(ii) Solve : \(\frac{d y}{d x}\) = sin(x + y) + cos(x + y)
Solution:

(iii) Solve : \(\frac{d y}{d x}\) = cos (x + y)
Solution:

(iv) Solve : \(\frac{d y}{d x}\) + 1 = e^x+y
Solution:

Odisha State Board CHSE Odisha Class 12 Approaches to English Book 1 Solutions Unit 1 Text B: Typing your own Blood Textbook Activity Questions and Answers.

CHSE Odisha 12th Class Alternative English Solutions Unit 1 Text B: Typing your own Blood

Activity – 8

Comprehension:
Question 1.
What does typing someone’s blood mean?
Answer:
Typing someone’s blood means determining the exact type of blood a person usually has. It was make one know one’s blood – group whether ‘A’ or ‘B’ or ‘O’.

Question 2.
What materials are necessary to type one’s blood? Which paragraphs tell you about these materials?
Answer:
Alcohol – soaked cotton balls, sterile lancet, a small test tube containing 1ml. of saline solution, anti-A anti- B and anti- Rhserum with individual eye droppers, two microscope slides, a grease pencil, a posture pipette, three applicator sticks and a warm fluorescent light or other low-heat sources are used on typing one’s blood.

Question 3.
What are the three stages of experimental process described in this text? Name them.
Answer:
First label one slide Rh with a grease pencil and place it under the low-heat source. Divide the cool slide into two equal portions labeling one side A and B and a drop of anti- Rh to warm the Rh slide. In the second stage, use an alcohol-soaked cotton ball to swab your middle or ring finger opening the sterile lancet prick the sterile finger once. Collect several drop of blood in the tube containing saline solution. In the third stage, using the porture pipette, transfer one drop of saline solution containing blood to each of the anti- A, anti- B and anti-Rh serums using a separate applicator stick. Two or three minutes after clumping should have appeared in one or three of the areas. This clumping determines what kind of blood a person has. The stages can be named as preparatory stage, experimental stage and conclusive stage.

Activity – 9

Remedial Grammar:
Like your Rh- slide experiment, you have only two tense forms of most of the English verbs, e.g. “go” and “went”. “Gone” is not a tense form. In association with the other auxiliary verbs, it gives a sense of completion of an activity“has gone”ora passive sense “is done”. Hence like Rh+ or Rh-. English verbs can be either in past tense or non-past tense.

Similarly like your blood grouping. A, B, AB or O, we can have the aspects of perfect (have + V + en), progressive (be + V + ing), perfect progressive (both combined or simple neither, perfect, nor progressive). These four aspects of either past or non-past give us the 8 types of verb groups. In addition to these two tenses and four aspects we can find do operations or model auxiliaries as elements ofa very group.

In the first sentences of the text, the verb is……used. You can see that it is be + v + en structure in simple non-past tense form. Hence, is a simple non-past passive structure. Similarly, find out the aspect, tense and voice of the following verb groups: Illustrates has finished is doing had been completed was being conducted.

Tense	Aspects	Voice
(i) Past	(a) simple	(i) Active
	(b)perfect
(ii) Non-past	(c) progressive	(ii) Passive
	(d)Perfect Progressive

Answer:

Verb groups	Tense	Aspects	Voice
Illustrate	Non-past	Simple	Active
has finished	Non-past	Perfective	Active
is doing	Non-past	Progressive	Active
had been completed	Past	Perfective	Passive
was being conducted	Past	Progressive	Passive

Activity -10

Composition:
In the passage you have step-by-step instructions on how to test and categorize your blood. Write instructions to carry out one of the following tasks.
(a) Teaching your friend how to make tea/cake/an omelette.
(b) Instructing a new friend how to reach your home.
(c) How to fix a fuse wire on your main switch.
Answer:
(a) How to make tea:
Ingredients: water, sugar, tea dust, boiled milk.
Instruments: stove, fry pan, a flat metal piece, spoon, a seive.

Preparation:
(i) Fire the stove.
(ii) Pour required cups of water.
(iii) Mix spoons of sugar as required.
(iv) Add one/two spoons of tea or as required.
(v) Serve the hot solution.
(vi) Add boiled milk to it.
(vii) Serve it in cups.

Extra Activity – 10(A)

B.(i) Derive adjectives from the following words in the text:
words – adjectives
thank – thankful
prepare – preparatory
talk – talkative
servility- servile
compel – compulsory
wisdom – wise
pleasure – pleasant
value- valuable
importance- important
success- successful
luck- lucky
proportion- proportional
enthusiasm- enthusiastic
completion- complete
reproach- reproachful
satisfy- satisfactory
reluctantly- reluctant
pleasure- pleasant
hastiness- haste
trouble- troublesome
persuade- persuasive
purpose- perposefiil
anger- angry
thought- thoughtful
child- children
despise- despicable
triviality- trivial
poverty- poor
necessity- necessary
deceive- deceptive/deceitful
sympathy- sympathetic
passion- passionate
clarity- clear
day- diurnal
night- nocturnal
truth- true
regularity- regular
respect- respectful
forget- forgetful
exhaust- exhaustive
fool- foolish
contempt- contemptuous
falsity- false
money- monetary
anxiety- anxious
continually- continual
pretend – pretentious
superiority- superior
misery- miserable

(ii) Derive adverbs from the following:

Words – Adverbs
thoughtful- thoughtfully
pleasant- pleasantly
reproachful- reproachfully
complete- completely
gradual- gradually
real- realty
excellent- excellently
passionate- passionately
filth- filthily
deep- deeply
full- folly
attract- attractively
possible- possibly
hunger- hungrily
exhaust- exhaustively
hesitate- hesitatingly
watch- watchfully
sharp- sharply
transitory- transitorily
rich- richly
strange- strangely
ordinary- ordinarily
desire- desirety
force- forcefully
strength- strongly
empty- emptity
foolish- foolishly
continual- continually
eternal- eternally
wonder- wonderfully
compel- compulsorily
respect- respectfully
necessary- necessarily
despicable- despicably
regular- regularly
contempt- contemptuously
anxiety- anxiously
misery- miserable

(iii) Say which words of the following in the text are nouns and which are adjectives:

happiness- Noun
good- Adjective
long- Adjective
flight- Noun
excess- Adjective
horrible- Adjective
ugly- Adjective
praise- Noun
bitter- Adjective
sleep- Noun
transitory- Adjective
happy- Adjective
mild- Noun
appearance- Noun
water- Noun
river- Noun
empty- Adjective
foolish- Adjective
years- Noun
folly- Noun
knowledge- Noun
Mortification- Noun
arrogance- Noun
intellectual- Adjective
penitence- Noun
voice- Noun
inward- Adjective
salvation- Adjective
power- Adjective
priest- Noun
madness- Noun
futile- Adjective
special- Adjective
crystal- Noun
depth- Noun
grateful- Adjective
new- Adjective
guest- Noun
hut- Noun
clothes – Noun
current- Noun
affection- Noun
secure- Adjective
bread- Noun
enjoyment- Noun
origin- Noun
despair- Noun
night- Noun
studies- Noun

(iv) Write antonyms of the following:
greatest- smallest
long- short
successful- unsuccessful
everywhere- nowhere
mature- immature
old- new
reality- appearance
real- unreal
presence- absence
true- false
agree- disagree
begin- end
wise- foolish
quickly- slowly
reach- depart
join- separate
remember- forget
small- big /great
back- front
conscious- unconscious
pallid- bright
obtain- lose
compare- contrast
fresh- stale
straighten- bend
hope- hopelessness/despair
win-lose /defeat
injustice- justice
take- give
sorrow- pleasure
much- less
stronger- weaker
completely- incompletely
remember- forget
compared- contrasted
difference- similarity
disappeared- appeared
secure- insecure
knowledge- ignorance
inward- outward
new-old
despair- hope

Section – C
Between men and women, who are the stronger? Who are more intelligent? Who are biologically superior? Why do you think so?
Discuss these questions in small groups and write down your most important arguments. Now read the following title on the differences between men and women.

Typing your own Blood Summary in English

Even if you do not wish to learn your blood type, the exercise is useful, because it familiarises you with some simple laboratory techniques, illustrates the use of basic equipment and prepare you to follow the stages of an orderly scientific procedure. In order to type your own blood, you need alcohol-soaked, soaked cotton balls, a sterile lancet, a small test tube containing 1 ml. of saline solution; anti – A, anti – B and anti – Rh serums with individual eye droppers, two microscope slides; a greased pencil; a pasture pipette, three applicator sticks and a warm fluorescent light or other low heat sources. One can make a typing of one’s own blood with these terms.

Analytical Outlines

• Of course, we don’t wish to learn about our blood type.
• Exercise is useful to us.
• It familiarises us with some sample laboratory techniques.
• It illustrates the use of basic equipment.
• It prepares us to follow the stages of an orderly scientific procedure.
• We can type our own blood.
• We need some equipment to do this.
• We require alcohol-soaked cotton balls.
• We require sterile lancet.
• We need a small test tube containing 1 ml of saline solution.
• We require anti-A
• We need anti-B
• We require anti-Rh serums with individual eye droppers.
• We require two microscope slides.
• We need a grease pencil.
• We require a Pasteur pipette.
• We need three applicator sticks.
• We also require a warm fluorescent light.
• We also need other low-heat sources.
• We require label one slide Rh with grease pencil.
• We have to place this slide under the low heat source.
• We have to divide the cool slide into two equal portion.
• We have to label one side as A and the other side B.
• We have to apply one drop of anti-A – A serum to slide – A.
• We have to apply one drop anti-B serum to slide – B.
• We have to apply one drop of anti – Rh serum to work Rh slide.
• We have to use an alcohol-soaked cotton ball to swab the middle or ring finger.
• We have to allow the excess alcohol to evaporate.
• After opening the sterile lancet, prick the sterile finger once.
• Now we have to collect several drops of blood in the test tube.
• The test tube also contains the saline solution.
• Now, we have to mix the solution.
• Again, we have to hold another sterile cotton ball over the cut.
• We have to allow the blood to clot.
• We have to transfer a drop of saline solution.
• It contains blood.
• It is transferred to anti-A.
• One drop is transferred to anti-B.
• Another drop is transferred to anti – Rh
• It is mixed using a separate applicator stick.
• It is allowed two or three minutes.
• Now, clumping should have appeared in A and B.
• Clumping denotes O blood.
• Rh – clumping means the blood is Rh- positive.
• The absence of Rh- clumping indicates it is Rh – negative

Meaning Of Difficult Words:

blood type – blood group
familiarise – intimate, make well-known
techniques – principles processes
illustrate – to explain, exemplify, show
I basic – fundamental, main, original
equipment – necessary instruments
procedure – principles, techniques
alcohol – pure spirit
soaked – absorbed
sterile – completely free from the seeds of disease
contain – comprise
saline – pertaining to salt
serum- liquid from of blood.
microscope – a magnifying instrument

Odisha State Board CHSE Odisha Class 12 Approaches to English Book 1 Solutions Unit 1 Text A: ‘Cures’ for the Common Cold Textbook Activity Questions and Answers.

CHSE Odisha 12th Class Alternative English Solutions Unit 1 Text A: ‘Cures’ for the Common Cold

Activity -1

Relation Between Parts of a Text:
If you are asked to divide the lesson into 5 sections in order to make notes, where possibly could you draw the lines separating the sections?
Write the paragraph number and the last word of the paragraph after which you will start a new section. Suggest a title for each section.
Answer:
Section- 1: Paragraph-1 …………… harmful
Title: Old Fashioned Remedies for Cold
Section- 2: Paragraphs – 2-4 …………avoided
Title: Morphine, Codeine and Papaverine as Remedies.
Section- 3: Paragraphs 5-6 ………… cold
Title: What The Scientists Studied.
Section- 4: Paragraphs 7-8 ………………..complications
Title: Opium Derivatives and Bed Rest.
Section- 5: Paragraph-9 ……………………..before
Title: Hot Baths and Cold Measurement

Activity – 2

Summary skill:
Of the following six statements only three are main points of the passage. Identify them:
(i) Many widely advertised cures and home remedies for cold are worthless or harmful
(ii) Students treated with sugar tablets showed little improvement.
(iii) Neither vaccines, nor vitamins and any other dietary measures prevent cold.
(iv) Nasal drops and sprays are found to be dangerous.
(v) Staying in bed for the duration of the cold was the only remedy that showed any result.

Activity-3

Comprehension:
Question 1.
The second paragraph possesses a question, what is it?
Answer:
The question is: Is there any remedy, then, of value in the treatment of colds?

Question 2.
What answer does the writer suggest?
Answer:
The writer says that there is scarcely any viable alternative for the treatment of common cold. However, there are a good many kinds of medicines which can be administered against cold.

Question 3.
How does the writer establish his answer?
Answer:
The writer picks up names like Morphine, Codeine, Papaverine combination, quinine hot water, air and stream baths were used as common therapies for cold but not as permanent cures.

Question 4.
What further recommendations did the writer make?
Answer:
The writer also brought out the names of different experts such as Dr. Russell Cecil, Dr. FitzHutter, De Quineeywhose findings were the best recommendations in the treatment of cold.

Activity – 4

Sequence In an Experiment:
What are the steps of the experiments mentioned in paragraph 4 and 5? Rearrange the steps given below in proper order:
(i) Record the health conditions of the patients at regular intervals.
(ii) Record initial health conditions of all the patients.
(iii) Compare the health conditions of the experiments group with that of the control group.
(iv) Prepare dummy to mixture.
(v) Draw inference after analysis of findings.
(vi) Divide the patients into experimental and control groups.
(vii) Select sample patients.
(viii) Prepare Codeine- Papaverine mixture.

Activity -5

Composition:
In this part of the country Tulsi leaves with honey are considered remedies for common cold. If you have to conduct an investigation to ascertain the truth of this belief, how will you organise the experiment? You can take clues from the reading passage and write down the steps of your proposed experiment.

Activity – 6

Remedial Grammar:
Morphine (which is) a derivative of opium, showed excellent results. (Paragraph – 3)
This preparation (which is) common called copavin, is not advertised to the public (Paragraph- 5)
In these sentences you have seen examples of non-defining relative clauses. Such clauses are separated from the main clauses with the help of commas. Secondly, the relative pronoun (like ‘which) and the ‘be’ verb can be omitted. The relative clauses without the relative pronoun and the ‘be’ verb are called the reduced relative clause. Similar reduction is possible in defining relative clauses also. Now reduce the relative clauses in the following sentences:
(a) They stood on the bridge which was connecting Cuttack with Jagatpur.
(b) The girl who is standing at the bus stop over there is my sister.
(c) The weapon that was used in the murder has been found.
(d) The boys who are being chosen for the college team are all under 18.
(e) The wooden beams which were holding up the roof have been damaged.

Activity – 7

Remedial Grammar:
1. Nasal congestion and stillness are reduced.
2. It was found that powered opium and Dover’s powder were beneficial.
3. The progress of the cold seemed to be arrested.
4. Commercial remedies are still sold.
In scientific tests were offer to see the examples of passive sentences. Whatever reduced nasal congestion, whoever found it out are unimportant in the first two sentences above. Similarly, we get examples of get-passive and have-passive scientific texts e.g.

When the boy gets chilled ___________.
I had my eyes tested.
Now rewrite the following sentences using passive structures like have/get + v + past participle.

The first one has been done for you.
1 . Our houses looked ugly. Its paint was pelling off.
So we got /had it painted.
2. Raman’s watch book. He could not afford to buy a new one.
So _____________
3. Lili split coffee on her favourite dress. She could not wash it by hand.
So _____________
4. In the super cyclone the roof was flown flourished and a wall fell down.
So _____________
5. Sharukh’s car was not starting well and seemed to be using too much petrol. But he did not want to sell his lucky car.
So _____________

Answer:
2. Raman’s watch broke. He could not afford to buy a new one.
So he had it repaired.
3. Lili split coffee on her favourite dress. She could not wash it by hand.
So she got it washed.
4. In the super cyclone the roof was flown of four shed and a wall fell down.
So we had it rebuilt.
5. Sharukh’s car was not starting well and seemed to be using too much petrol But he did not want to sell his lucky car.
So he got it repaired.

Section – A
New look at the little of the first passage. “Cures for The Common Cold.” What possible cures can you think of? Do you know that science has not yet brought us a cure for this disease? However, the quest continues to find a possible remedy, can you guess any home remedy that may cure common cold?
Now go through the text quickly and see if you guess right. You have only two minutes to do so. Read the text again and identify the cures that have been short-listed.

Section – B
In section A we read about a sequence of experiments to find a cure for the common cold. In Section B we shall read about a different kind of experiment whose purpose is to find out the types of human blood. What’s more interesting, you can learn how to determine your blood type as well as that of others.

‘Cures’ for the Common Cold Summary in English

Cures for the common cold comprise general skepticism. Millions of dollars is being spent for this every year. Obsolete cures like asafetida and camphor are not longer in vogue and popular remedies like vitamins, vaccines, nasal medications and other drugs have substituted them. Advertised remedies now available in the market sometimes prove worthless and harmful. There is absolutely to effective prevention of the common cold. Morphine which is a derivative of opium showed excellent results, but was rejected on account of its danger. But some other derivatives of opium which are less toxic and carry no practical danger of habituation proved to be definitely valuable. Codline and papaverine both proved valuable in the treatment of acute colds.

The codlin-papaverine combination proved to be, after Morphine, the most valuable of all cold medications. A preparation, consisting of one quarter grain of codeine and one quarter of grain of papaverine was finally selected as the most effective dosage. The main efficacy was a marked decrease or complete disappearance of nasal congestion and discharge. Most of the students were up and doing while taking this medication. Had they remained in bed while using it is probable that even better result have been obtained. This preparation commonly called copavin, is not advertised to the public. But it is available through physicians who should decide when and in what dosage it should be used. Dr. Russel CecilofNew York and Dr. Fritz Hutter of Vienna, both found that the codeine, paparine mixture was particularly beneficial if used by their patients at the very beginning of the affection.

Dr. Quincy, in his “Confessions” wrote that during the years in which he had taken opium he “never once caught cold, one the phrase in nor even the slightest cough. But after discontinuing the use of opium, a violent cold attacked me and a cough soon after.” Less effective, but still of moderate value were several other opium derivatives. In addition to codeine and papaverine it was found that powered opium and the old fashioned Dover’s powder were beneficial. Quinine also came to be included in this group of moderately valuable medications. In the end, certain general hygienic measures are helpful in the treatment of colds. Going to bed and remaining there until recovery is good advice.

The value of bed rest lies in protecting others from exposure, in necessary general resistance and in keeping the body warm. Hot baths for the treatment of colds may consist of hot water, hot air stream. The effect of these baths is to dialate the blood vessels of the skin and to increase blood flow through them. As a result, nasal congestion and stiffness are reduced. Other effects may be obtained with message of or other forms of physiotherapy, with hot or cold compresses, mustard plasters and certain, medicated ointments. If such treatments are followed by rest in bed with sufficient covers to prevent cooling, the effect is prolonged and the possibility of their being more than temporary benefit is increased.

Analytical Outlines:

• Cures for the common cold comprise of general skepticism.
• Millions of dollars is being spent for this every year.
• Asafetida and camphor are considered as obsole cures.
• These are no longer in vogue.
• These have been so far substituted.
• The substitutions are popular remedies.
• These are vitamins, vaccines, nasal medications etc.
• Now advertised remedies are available in the market.
• These are proved worthless and harmful.
• There is absolutely no effective prevention of common cold.
• Morphine is a derivative of opium.
• Morphine should excellent results.
• But it was rejected on account of its danger.
• However, some other derivatives of opium are taken.
• These are less toxic.
• They also carry no practical danger of habituation.
• Hence, it proved to be definitely valuable.
• Codeine and papaverine both proved valuable in the treatment of acute cold.
• The codeine-papaverine combination proved to be the most valuable of all cold medication after morphine.
• The preparation is made.
• One-quarter grain of codeine and one-quarter grain of papaverine are prepared together.
• It is finally selected as the most effective dosage.
• The main result was the marked decrease or complete disappearance of nasal congestion and discharge.
• Most of the students were up and doing while taking this medication.
• They had to remain in bed.
• So that they would have obtained better results.
• This preparation is commonly called copavin.
• It is not advertised to the public.
• But it is available through physicians.
• He is to decide about the dosage.
• Dr. Russell of New York and Dr. Fritz Hutter of Vienna found something about it.
• They found something beneficial about the mixture of codeine and papaverine.
• It is particularly beneficial for the patients at the very beginning of affection.
• Dr. Quincy in his “Confessions” wrote something.
• He wrote that the had taken opium for something.
• He marked that the had never caught by cold once.
• There was not even the slighest cough.
• But he discontinued the use of opium.
• He was attacked by a variant cold then.
• It was followed by a cough soon after.
• Several other opium derivative were less effective.
• They were still having with moderate value.
• It was found that powered opium and the old fashioned Dover’s powder were beneficial.
• Quinine belongs to this group.
• It is also accepted as the moderately valuable medication
• Certain general hygienic measures are considered.
• They are found helpful in the treatment of cold.
• Complete bed rest up to full recovery is a good advice.
• It lies in protecting others from exposure.
• It is necessary for general resistance.
• Again it keeps the body warm.
• We can accept hot baths for the treatment of cold.
• It may consist of hot water, hot air or stream
• Its effect is very important.
• It can dialate the blood vessels of the skin.
• Again, it can increase blood flow through them
• As a result of this, nasal congestion and stiffness are cured.
• Other effects may be obtained with message.
• We can also adopt other forms of physiotherapy.
• This can be done with hot or cold compresses.
• This can be done with hot or cold compresses.
• It can also be done by other medicated ointments.
• Such treatment should be followed with complete bed rest.
• It should be with sufficient covers to prevent cooling.
• Its effect to some extent prolonged.
• Again, the possibility of temporary benefit is increased.

Meaning Of Difficult Words:

remedies – panaceas: ways and means of cure.
in vogue – in prevalence, in operation
investigation – searching or examining carefully, enquiry into a matter.
uniformly – identically, almost the same, equally
distinct – distinguished, different, separate
discarded – cast off rejected, thrown away, not accepted
one- quarter- one-fourth something
De Quincey – Thomas De Quincey (1 785 – 59), English essayist and critic famous for “Confession of an English Opium- eater”, fascinating memories of distinguished by great imaginative power and splendid prose.
Octean – Jean Octean (1 889 – 1963) French poet, novelist, dramatist, film writer and director who was in the vanguard of almost every experimental artistic movement of the 20th century.

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 11 Differential Equations Ex 11(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 11 Differential Equations Exercise 11(c)

Find the solutions of the following differential equations:
Question 1.
(x + y) dy + (x – y) dx = 0
Solution:
Given equation can be written as

Question 2.
\(\frac{d y}{d x}\) = \(\frac{1}{2}\left(\frac{y}{x}+\frac{y^2}{x^2}\right)\)
Solution:

Question 3.
(x² – y²) dx + 2xy dy = 0
Solution:
Given equation can be written as

Question 4.
x\(\frac{d y}{d x}\) + \(\sqrt{x^2+y^2}\) = y
Solution:
Given equation can be written as

Question 5.
x (x + y) dy = (x² + y²) dx
Solution:


This is the required solution.

Question 6.
y² + x² \(\frac{d y}{d x}\) = xy \(\frac{d y}{d x}\)
Solution:
Given equation can be written as

This is the required solution.

Question 7.
x sin\(\frac{y}{x}\) dy = \(\left(y \sin \frac{y}{x}-x\right)\)dx
Solution:
Given equation can be written as

Question 8.
x dy – y dx= \(\sqrt{x^2+y^2}\) dx
Solution:

This is the required solution.

Question 9.
\(\frac{d y}{d x}\) = \(\frac{y-x+1}{y+x+5}\)
Solution:
Given equation is


This is the required solution.

Question 10.
(x – y) dy = (x + y + 1) dx
Solution:
Given equation can be written as


This is the required solution.

Question 11.
(x – y – 2) dx + (x – 2y – 3) dy = 0
Solution:
Given equation can be written as


This is the required solution.

Question 12.
\(\frac{d y}{d x}\) = \(\frac{3 x-7 y+7}{3 y-7 x-3}\)
Solution:
Given equation can be written as


This is the required solution.

Question 13.
(2x + y + 1) dx + (4x + 2y – 1) dy = 0
Solution:
Given equation can be written as

⇒ 2z + ln (z – 1) = 3x + C
⇒ 2 (2x + y) + ln (2x + y – 1 ) = 3x + C
⇒ (x + 2y) + ln (2x + y – 1 ) = C
This is the required solution.

Question 14.
(2x + 3y – 5)\(\frac{d y}{d x}\) + 3x + 2y – 5 = 0
Solution:
Given equation can be written as

Question 15.
(4x + 6y + 5) dx – (2x + 3y + 4) dy = 0
Solution:
Given equation can be written as


This is the required solution.

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 9 Integration Additional Exercise Textbook Exercise questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Additional Exercise

Question 1.
∫\(\sqrt{1-\sin 2 x}\) dx
Solution:
I = ∫\(\sqrt{1-\sin 2 x}\) dx
= ∫\(\sqrt{(\cos x-\sin x)^2}\) dx
= ∫(cos x – sin x) dx
= sin x + cos x + c

Question 2.
∫\(\frac{d x}{1+\sin x}\)
Solution:
I = ∫\(\frac{d x}{1+\sin x}\)
= ∫\(\frac{1-\sin x}{\cos ^2 x}\)
= ∫sec² x – sec x tan x dx
= tan x – sec x + c

Question 3.
∫\(\frac{\sin x}{1+\sin x}\) dx
Solution:

Question 4.
∫\(\frac{\sec x}{\sec x+\tan x}\) dx
Solution:
I = ∫\(\frac{\sec x}{\sec x+\tan x}\) dx
= ∫\(\frac{\sec x(\sec x-\tan x)}{\sec ^2 x-\tan ^2 x}\) dx
= ∫sec² x – sec x tan x dx
= tan x – sec x + c

Question 5.
∫\(\frac{1+\sin x}{1-\sin x}\) dx
Solution:
I = ∫\(\frac{1+\sin x}{1-\sin x}\) dx
= ∫\(\frac{(1+\sin x)^2}{\cos ^2 x}\) dx
= ∫[sec² x+ tan² x+ 2sec x tan x) dx
= ∫[2sec² x – 1 + 2sec x tan x) dx
= 2tan x – x + 2sec x + c

Question 6.
∫tan^-1 (sec x + tan x) dx
Solution:

Question 7.
∫\(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\) dx
Solution:
I = ∫\(\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}\) dx
= ∫\(\frac{\left(2 \cos ^2 x-1\right)-\left(2 \cos ^2 \alpha-1\right)}{\cos x-\cos \alpha}\) dx
= 2 ∫(cos x + cos α) dx
= 2 sin x + 2x cos α + c

Question 8.
∫tan-1\(\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\) dx
Solution:

Question 9.
∫\(\frac{d x}{\sqrt{x+1+} \sqrt{x+2}}\)
Solution:

Question 10.
∫\(\frac{2+3 x}{3-2 x}\) dx
Solution:

Question 11.
∫\(\frac{d x}{\sqrt{x}+x}\)
Solution:

Question 12.
∫\(\frac{d x}{1+\tan x}\)
Solution:

Question 13.
∫\(\frac{x+\sqrt{x+1}}{x+2}\) dx (Hints put : \(\sqrt{x+1}\) = t)
Solution:

Question 14.
∫sin^-1\(\sqrt{\frac{x}{a+x}}\) dx (Hints put : x = a tan² t)
Solution:

Question 15.
∫e^x\(\left(\frac{2+\sin 2 x}{1+\cos 2 x}\right)\) dx
Solution:

Question 16.
∫\(\frac{\left(x^2+1\right) e^x}{(x+1)^2}\) dx
Solution:

Question 17.
∫\(\frac{x^2-1}{x^4+x^2+1}\) dx
Solution:

Question 18.
∫\(\frac{x^2 d x}{x^4+x^2+1}\)
Solution:

Question 19.
∫\(\sqrt{\cot x}\) dx
Solution:

Question 20.
∫\((\sqrt{\tan x}+\sqrt{\cot x})\) dx
Solution:

Question 21.
∫\(\frac{\mathrm{dx}}{x\left(x^4+1\right)}\)
Solution:

Question 22.
∫\(\frac{\mathrm{dx}}{e^x-1}\)
Solution:

Question 23.
∫\(\frac{(x-1)(x-2)(x-3)}{(x+4)(x-5)(x-6)}\) dx
Solution:

Question 24.
∫\(\frac{d x}{\left(e^x-1\right)^2}\)
Solution:

Question 25.
∫\(\frac{d x}{\sin x \cos ^2 x}\)
Solution:

Question 26.
\(\int_2^4 \frac{\left(x^2+x\right) d x}{\sqrt{2 x+1}}\)
Solution:

Question 27.
\(\int_{-a}^a \sqrt{\frac{a-x}{a+x}}\) dx
Solution:

Let a² – x² = t²
-2x dx = 2t dt
x = -a ⇒ 0 t = 0
x = a ⇒ t = 0
= 0
I = aI₁ – I₂ = aπ

Question 28.
\(\int_0^{\pi / 2}(\sqrt{\tan x}+\sqrt{\cot x})\) dx
Solution:

Question 29.
\(\int_0^{\pi / 2} \frac{\cos x d x}{1+\cos x+\sin x}\)
Solution:

Question 30.
\(\int_0^1\)x (1 – x)ⁿ dx
Solution:

Question 31.
\(\int_0^{\pi / 2}\)sin 2x log (tan x) dx
Solution:

Question 32.
\(\int_0^{\pi / 2} \frac{\sin ^2 x d x}{\sin x+\cos x}\)
Solution:

Question 33.
\(\int_0^{\pi / 2} \frac{\sin ^2 x d x}{1+\sin x \cos x}\)
Solution:

Question 34.
\(\int_0^{\pi / 2} \frac{x d x}{\sin x+\cos x}\)
Solution:

Question 35.
Prove that \(\int_0^\pi\) x sin³ x dx = \(\frac{2 \pi}{3}\)
Solution:

Question 36.
\(\int_{\pi / 5}^{3 \pi / 10} \frac{\sin x d x}{\sin x+\cos x}\)
Solution:

Question 37.
\(\int_0^\pi\)|cos x| dx
Solution:

Question 38.
\(\int_1^4\)(|x – 1| + |x – 2| + |x – 3|) dx
Solution:

Question 39.
\(\int_{-\pi / 2}^{\pi / 2}\)(sin |x| + cos |x|) dx
Solution:

Question 40.
\(\int_0^\pi\)log (1 + cos x) dx
Solution:

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 9 Integration Ex 9(l) Textbook Exercise questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Exercise 9(l)

Question 1.
\(\int_0^{\frac{\pi}{2}}\)sin¹⁰ θ dθ
Solution:
\(\int_0^{\frac{\pi}{2}}\)sin¹⁰ θ dθ = \(\frac{9}{10} \cdot \frac{7}{8} \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}\) = \(\frac{405 \pi}{7680}\)

Question 2.
\(\int_0^{\frac{\pi}{2}}\)cos¹² θ dθ
Solution:
\(\int_0^{\frac{\pi}{2}}\)cos¹² θ dθ = \(\frac{11}{12} \cdot \frac{9}{10} \cdot \frac{7}{8} \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}\) = \(\frac{4455 \pi}{92160}\)

Question 3.
\(\int_0^{\frac{\pi}{2}}\)sin¹¹ θ dθ
Solution:
\(\int_0^{\frac{\pi}{2}}\)sin¹¹ θ dθ = \(\frac{10}{11} \cdot \frac{8}{9} \cdot \frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3}\) = \(\frac{3840}{4455}\)

Question 4.
\(\int_0^{\frac{\pi}{2}}\)cos⁹ θ dθ
Solution:
\(\int_0^{\frac{\pi}{2}}\)cos⁹ θ dθ = \(\frac{8}{9} \cdot \frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3}\) = \(\frac{384}{405}\)

Question 5.
\(\int_0^1 \frac{x^7}{\sqrt{1-x^2}}\) dx
Solution:

Question 6.
\(\int_0^1 \frac{x^5\left(4-x^2\right)}{\sqrt{1-x^2}}\) dx
Solution:

Question 7.
\(\int_0^a x^3\left(a^2-x^2\right)^{\frac{5}{2}}\) dx
Solution:

Question 8.
\(\int_0^1 x^5 \sqrt{\frac{1+x^2}{1-x^2}}\) dx
Solution:

Question 9.
\(\int_0^{\infty} \frac{x^2}{\left(1+x^6\right)^n}\) dx
Solution:

Question 10.
\(\int_0^\pi\)sin⁸ θ dθ
Solution:

Odisha State Board Elements of Mathematics Class 12 Solutions CHSE Odisha Chapter 9 Integration Ex 9(k) Textbook Exercise questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 9 Integration Exercise 9(k)

Evaluate the following Integrals:
Question 1.
(i) \(\int_0^{\frac{\pi}{2}} \frac{d x}{1+\tan x}\)dx
Solution:

(ii) \(\int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\)dx
Solution:

(iii) \(\int_0^1 \frac{\ln (1+x)}{2+x^2}\)dx (x = tan θ)
Solution:

(iv) \(\int_0^\pi \frac{x d x}{1+\sin x}\)
Solution:

Question 2.
(i) \(\int_{-a}^a\)x⁴ dx
Solution:

(ii) \(\int_{-a}^a\)(x⁵ + 2x² + x) dx
Solution:

(iii) \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\)cos² x dx
Solution:

(iv) \(\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\)sin⁵ x dx
Solution:
Let f(x) = sin⁵ x
Then f(-x) = sin⁵ (-x)
= -sin⁵ x = -f(x)
So f(x) is an odd function.
Thus \(\int_{-a}^a\)f(x) dx = 0
\(\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\)sin⁵ x dx = 0

Question 3.
(i) \(\int_0^\pi\)cos³ x dx
Solution:

(ii) \(\int_0^\pi\)cos² x dx
Solution:

(iii) \(\int_0^\pi\)sin³ x cos x dx
Solution:
\(\int_0^\pi\)sin³ x cos x dx
[Put sin x = t, then cos x dx = dt
When x = 0, t = 0, when x = π, t = 0
\(\int_0^\pi\)t³ dt = 0

(iv) \(\int_0^\pi\)sin x cos² x dx
Solution:

Question 4.
Show that
(i) \(\int_0^1 \frac{\ln x}{\sqrt{1-x^2}}\) dx = \(\frac{\pi}{2} \ln \frac{1}{2}\)
Solution:

(ii) \(\int_0^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1+\sin x \cos x}\) dx = 0
Solution:

(iii) \(\int_0^\pi\)x ln sin x dx = \(\frac{\pi^2}{2} \ln \frac{1}{2}\)
Solution:

Question 5.
(i) \(\int_0^{\pi / 2}\)ln (tan x + cot x) dx
Solution:

(ii) \(\int_0^\pi \frac{x \tan x-\sin x}{1+\sin x \cos x}\) dx
Solution:

(iii) \(\int_1^3 \frac{\sqrt{x} d x}{\sqrt{4-x}+\sqrt{x}}\)
Solution:

(iv) \(\int_0^\pi \frac{x \sin x d x}{1+\cos ^2 x}\)
Solution:

(v) \(\int_0^1\)x (1 – x)¹⁰⁰ dx
Solution:

(vi) \(\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\cot x}}\)
Solution:

(vii) \(\int_0^{50}\)e^x-[x] dx
Solution:

Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 10 Area Under Plane Curves Ex 10 Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 10 Area Under Plane Curves Exercise 10

Question 1.
Find the area bounded by
(i) y = e^x, y = 0, x = 4, x = 2
Solution:
Area = \(\int_2^4\)e^x dx
= \(\left[e^x\right]_2^4\)
= e⁴ – e²

(ii) y = x², y = 0, x = 1
Solution:
Area = \(\int_0^1\)x² dx
= \(\left[\frac{x^3}{3}\right]_0^1\)
= \(\frac{1}{3}\)

(iii) xy = a², y = 0, x = α, x = β (β > α > 0)
Solution:
Area = \(\int_\alpha^\beta y\)y dx
= \(\int_\alpha^\beta \frac{a^2}{x}\) dx
= a²\([\ln x]_\alpha^\beta\)
= a² ln (β/α)

(iv) y = sin x, y = 0, x = \(\frac{\pi}{2}\)
Solution:
Area = \(\int_0^{\frac{\pi}{2}}\)sin x dx
= \([-\cos x]_0^{\frac{\pi}{2}}\)
= -cos\(\frac{\pi}{2}\) + cos θ = 1

Question 2.
Find the area enclosed by
(i) y = e^x, x = 0, y = 2, y = 3
Solution:

(ii) y² = x, x = 0, y = 1
Solution:
Area = \(\int_0^1\)x dy
= \(\int_0^1\)y² dy
= \(\left[\frac{y^3}{3}\right]_0^1\)
= \(\frac{1}{3}\)

(iii) xy = a², x = 0, y = α, y = β (β > α > 0)
Solution:
Area = \(\int_\alpha^\beta\)x dy
= \(\int_\alpha^\beta \frac{a^2}{y}\)dy
= a²\([\ln y]_\alpha^\beta\)
= a² ln (β/α)

(iv) y² = x³, x = 0, y = 1
Solution:
Given curve is y² = x³
⇒ x = y^2/3
It passes through the origin. So the required area
= \(\int_0^1\)x dy
= \(\int_0^1 y^{\frac{2}{3}}\) dy
= \(\left[\frac{y^{\frac{5}{3}}}{5 / 3}\right]_0^1\)
= \(\frac{3}{5}\)

Question 3.
(i) Determineellipse the area with in the ellipse
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1
Solution:

The ellipse is symmetrical about x-axis and y-axis.It is divided into 4 equal parts by the coordinate axes. So required area

(ii) Find the area of the circle x² + y² = 2ax.
Solution:
Given circle is x² + y² = 2ax
⇒ x² – 2ax + a² + y² = a²
⇒ (x – a)² + y² = a² … (1)
The centre of the circle is (a, 0) and radius is a.

(iii) Find the area of the portion of the parabola y² = 4x bounded by the double ordinate through (3, 0).
Solution:
Given parabola is y² = 4x

(iv) Determine the area of the region bounded by y² = x³ and the double ordinate through (2, 0)
Solution:
Given curve is y² = x³
⇒ y = ±x^3/2 … (1)
The curve passes through the origin and symmetrical about x-axis because the power of y is even.

Question 4.
(i) Find the area of the regions into which the circle x² + y² = 4 is divided by the line x + √3y = 2.
Solution:
Given circle and the straight line are x²+ y² = 4 and x+ √3y = 2
The circle has the centre at (0, 0) and radius ‘2’.
The eqn. (2) can be written as
y = –\(\frac{1}{\sqrt{3}}\)x + \(\frac{2}{\sqrt{3}}\)
Slope of the strainght line = –\(\frac{1}{\sqrt{3}}\)
The line makes and angle of 150° with x-axis making intercept \(\frac{2}{\sqrt{3}}\) from y-axis.
It intersects x-axis at (2, 0).

Solveing (1) and (2),
ge wet (2 – 3√y)² + y² = 4
4 + 3√y² – 4√3y + y² = 4
4y² – 4√3y = 0
y(y – √3) = 0
y = 0 or y = √3
When y = 0, x = 2
When y = √3, x = -1
Thus the straight line intersects the circle at (2, 0) and (-1, √3).
Area of the portion ACBA.

(ii) Determine the area of the region between the curves y = cos x and y = sin x, bounded by x = 0.
Solution:

The curves y = cos x and y = sin x are shown in the above figure. The region included between these two curves in [0, \(\frac{\pi}{4}\)] is OABO.

(iii) Find the area enclosed by the two parabolas y² = 4 ax and x² = 4ay.
Solution:
The given parabolas are y² = 4ax and x² = 4ay.
The graphs of the two parabolas are shown in the figure.

⇒ x⁴ = 64 a⁴
⇒ x⁴ – 64 a³x = 0
⇒ x (x³ – (4a)³) = 0
⇒ x (x – 4a) (x² + 4ax + 16a²) = 0
⇒ x = 0, 4a
When x = 0, y = 0 and
when x = 4a, y = 4a
Thus the two parabolas intersect at (0, 0) and (4a, 4a).
Area between two parabolas

(iv) Determine the area common to the parabola y² = x and the circle x²+ y² = 2x.
Solution:
Gien parabola is y² = x
Given circle is
x² + y² = 2x ⇒ (x – 1)² + y² = 1
The centre is at (1, 0) and radius is 1.

Solving (1) and (2) we get
x² + x = 2x ⇒ x² – x = 0 ⇒ x(x – 1) = 0
⇒ x = 0, x = 1
When x = 0, y = 0 and when x = 1, y = 1.
Thus both the parabola and circle intersect at (0, 0) and (1, 1).
Required Area

CHSE Odisha 12th Class Political Science Book Solutions (+ 2 2nd Year)

CHSE Odisha Class 12 Political Science Book Solutions in English Medium

Unit 1 Democracy in India

• Democracy in India Objective Questions
• Democracy in India Short Answer Questions
• Democracy in India Long Answer Questions

Unit 2 Democratic Process in India-I

• Democratic Process in India-I Objective Questions
• Democratic Process in India-I Short Answer Questions
• Democratic Process in India-I Long Answer Questions

Unit 3 Democratic Process in India-II

• Democratic Process in India-II Objective Questions
• Democratic Process in India-II Short Answer Questions
• Democratic Process in India-II Long Answer Questions

Unit 4 India in World Politics

• India in World Politics Objective Questions
• India in World Politics Short Answer Questions
• India in World Politics Long Answer Questions

Unit 5 Issues in International Politics

• Issues in International Politics Objective Questions
• Issues in International Politics Short Answer Questions
• Issues in International Politics Long Answer Questions

CHSE Odisha Class 12 Political Science Book Solutions in Odia Medium

Chapter 1 ଗଣତନ୍ତ୍ର

• Objective Questions
• Short Answer Questions
• Long Answer Questions

Chapter 2 ଭାରତରେ ରାଜନୈତିକ ଦଳୀୟ ବ୍ୟବସ୍ଥା

• Objective Questions
• Short Answer Questions
• Long Answer Questions

Chapter 3 ଭାରତରେ ସଂଘୀୟବାଦ

• Objective Questions
• Short Answer Questions
• Long Answer Questions

Chapter 4 ଭାରତରେ ଗ୍ରାମାଞ୍ଚଳ ଓ ସହରାଞ୍ଚଳ ସ୍ୱାୟତ୍ତ ଶାସନ ବ୍ୟବସ୍ଥା

• Objective Questions
• Short Answer Questions
• Long Answer Questions Part-1
• Long Answer Questions Part-2

Chapter 5 ଜାତି ଗଠନରେ ପ୍ରତିବନ୍ଧକ

• Objective & Short Answer Questions
• Long Answer Questions

Chapter 6 ଭାରତୀୟ ରାଜନୀତିରେ ସାଂପ୍ରତିକ ପ୍ରସଙ୍ଗ-ଜନପ୍ରିୟ ଆନ୍ଦୋଳନ

• Objective Questions
• Short Answer Questions
• Long Answer Questions

Chapter 7 ଭାରତର ବୈଦେଶିକ ନୀତି

• Objective Questions
• Short & Long Answer Questions

Chapter 8 ଆନ୍ତର୍ଜାତୀୟ ସଙ୍ଗଠନ

• Objective & Short Answer Questions
• Long Answer Questions

Chapter 9 ସାଂପ୍ରତିକ ବିଶ୍ଵରେ ନିରାପତ୍ତା ପ୍ରସଙ୍ଗର ପରିବର୍ତ୍ତିତ ପୃଷ୍ଠଭୂମି

• Objective Questions
• Short Answer Questions
• Long Answer Questions

Chapter 10 ପରିବେଶ ଓ ପ୍ରାକୃତିକ ସମ୍ପଦ

• Objective Questions
• Short Answer Questions
• Long Answer Questions

CHSE Odisha Class 12 Political Science Syllabus (+2 2nd Year)

There shall be two papers in Political Science modeled on the Syllabi of CBSE. Paper-I: TitleFoundation of Political Theory and Indian Government at work (For First Year). Paper-II: Title-Democracy and Nation Building in India and International affairs (For Second Year).

The subject of Political Science modeled on the Syllabi of CBSE consists of two papers as mentioned above. Paper-I is to be covered in the +2 First Year class and Paper-ll is to be covered in the +2 Second Year class. Each paper is divided into two sections and each section is further subdivided into two/three units. Thus there are five units in both Paper-I and Paper II. Periods have been allocated for the respective units approximately. Teachers are advised to take at least those numbers of periods to cover the particular unit. The major concepts and principles should be taught in such a manner as to stimulate higher mental abilities among students like application, logical thinking, analysis, etc., and not factual information. Paper-setters and Examiners are requested to keep the above in mind while setting questions and examining, respectively. Questions should of short (one word/ multiple-choice/ one sentence), medium (50/100 words/ five sentences), and long (500 words or thereabout). Also, Questions of the final/AHS Examination shall cover all five units of Paper -II.

Objectives of the course/syllabus are, as briefly mentioned above are:

• To enable the students to acquire knowledge about the important concepts, theories, principles, provisions, processes and Institutions of the Indian constitution, and some rudimentary knowledge about International affairs;
• To acquaint the students with the changing dimension of politics and political theory both in the national and international knowledge domain;
• To develop an interest among the students regarding problems of the political domain and to find out the possible solution to those problems.

Suggested Reading:

1. Political Theory- For Class-XI (Published by NCERT, New Delhi)
2. Indian Constitution at Work- For Class-XI (Published by NCERT, New Delhi)
3. Contemporary World Politics, For Class-XII (Published by NCERT, New Delhi)
4. Politics in India, For Class-XII (Published by NCERT, New Delhi)

Second Year CHSE (2022-2023)
Political Science Paper-II
(DEMOCRACY IN INDIA AND INTERNATIONAL POLITICS)

Part A: Politics in India

Unit I Democracy in India

1. Democracy: Meaning, Types, and Features; Challenge to Democratic Process in India – Inequality, Illiteracy, Regionalism, Naxalite Problem, Gender Inequality. (8 Periods)
2. Party System in India: Meaning, Types; One Party Dominance, Coalition Politics; Regional Parties. (8 Periods)

Unit II Democratic Process in India-I

1. Federalism in India: Features, Centre-State relation, Recent Trends in Indian Federalism. (8 Periods)
2. Local Government in India – Rural and Urban Local Bodies, Composition and Functions. (8 Periods)

Unit III Democratic Process in India-II

1. Challenges to Nation-Building: Meaning, Communalism, Casteism, Regionalism, Terrorism; Remedies. (8 Periods)
2. Contemporary issues in Indian Politics: Popular Movements – Women Movement, Environment protection Movement, Development – Displacement Movements. (8 Periods)

Part B: Contemporary World Politics

UNIT-IV (India in World Politics)

1. Indian Foreign Policy: Basic Features; India and its neighbours-China, Pakistan. (8 Periods)
2. International Organizations: UN: Major Organs-General Assembly; Security Council; International Court of Justice; Reforms of the UN. India’s position in the UN. International Economic Organizations- World Bank and the IMF. (8 Periods)

UNIT-IV (Issues in International Politics)

1. Changing Dimension of Security in Contemporary World: Traditional Security Concerns: Arms Race and Disarmament. Non-Traditional Security Concerns: Human security: Global Poverty, Inequality, Health, and Education. (8 Periods)
2. Environment and Natural Resources: Global Environmental Concerns; Development and Environment; Global Warming and Climate Change. (8 Periods)

BOOK PRESCRIBED:
Bureau’s Higher Secondary (+2) Political Science, Paper-II (English & Odia) Published by Odisha State Bureau of Textbook Preparation & Production, Bhubaneswar.

CHSE Odisha Class 12 Text Book Solutions

CHSE Odisha 12th Class Psychology Book Solutions (+ 2 2nd Year)

CHSE Odisha 12th Class Psychology Book Solutions in English Medium

Unit 1 Life Span Development & Self and Personality

• Unit 1 Objective & Short Answer Type Questions
• Unit 1 Long Answer Questions Part-1
• Unit 1 Long Answer Questions Part-2
• Unit 1 Long Answer Questions Part-3
• Unit 1 Long Answer Questions Part-4

Unit 2 Stress: Meeting Life Challenges & Physical Environment and Behaviour

• Unit 2 Objective & Short Answer Type Questions
• Unit 2 Long Answer Questions Part-1
• Unit 2 Long Answer Questions Part-2

Unit 3 Group Processess and Leadership & Counselling Process

• Unit 3 Objective & Short Answer Type Questions
• Unit 3 Long Answer Questions Part-1
• Unit 3 Long Answer Questions Part-2

Unit 4 Psychological Disorder & Therapeutic Approaches

• Unit 4 Objective & Short Answer Type Questions
• Unit 4 Long Answer Questions Part-1
• Unit 4 Long Answer Questions Part-2
• Unit 4 Long Answer Questions Part-3
• Unit 4 Long Answer Questions Part-4

Unit 5 Statistics in Psychology

• Unit 5 Objective Questions and Answers

CHSE Odisha 12th Class Psychology Book Solutions in Odia Medium

Unit 1 ଜୀବନବ୍ୟାପୀ ବିକାଶ & ସ୍ଵୟଂ ତଥା ବ୍ୟକ୍ତିତ୍ଵ

• Objective Questions
• Short Answer Questions
• Long Answer Questions

Unit 2 ମନୋ-ସାମାଜିକ ଚାପ & ଭୌତିକ ପରିବେଶ ଓ ବ୍ୟବହାର

Unit 3 ସମୂହ ପ୍ରକ୍ରିୟା ଓ ନେତୃତ୍ଵ & ପରାମର୍ଶ ପ୍ରକ୍ରିୟା

• Objective Questions
• Short & Long Answer Questions

Unit 4 ମାନସିକ ବିକାର & ମାନସିକ ବିକାର ଚିକିତ୍ସା

• Objective Questions
• Short & Long Answer Questions

Unit 5 ମନୋବିଜ୍ଞାନରେ ପରିସଂଖ୍ୟାନର ବ୍ୟବହାର

• Objective Questions
• Short & Long Answer Questions

CHSE Odisha Class 12 Psychology Syllabus (+2 2nd Year)

Psychology is introduced as an elective subject at the higher secondary stage of education. The course deals with Psychological knowledge and practices which are contextually rooted, It emphasizes the complexity of cognitive and behavioural processes of human beings within a socio-cultural context. It encourages critical reasoning, allowing students to appreciate the role of social factors in behaviour and illustrates how biology and experience shape behaviour.

In the second year, there will be a council examination and the theory paper carries 70 marks and the practical paper carries 30 marks. The question paper pattern for both papers is same.

QUESTION PAPER PATTERN

Theory Paper:
Group-A: Objective type (Compulsory)
Q.No.1: Multiple choice (Fill up the blanks from all units) 1 mark each × 10 = 10 marks
Q.No.2: Statements “True” or “False” 1 mark each × 10 = 10 marks

Group-B: Short Type
Q.No.3: Short type answer (Answer within two /three sentences) and one has to answer 10 bits out of 12 bits.
2 marks each × 10 = 20 marks
Q.No.4: Short type answer (Answer within six sentences) and one has to answer 3 bits out of 5 bits.
3 mark each × 3 = 09 marks

Group-C: Long Type
Q.No.5: Question No-5 to Q.No-10 (Questions will be from all the units and one has to answer any three questions).
7 marks each × 3 = 21 marks

PRACTICAL
There will be 4 number of questions and the examinee is to choose/draw 2 number of questions through the lottery and is to conduct any one question out of the two questions.
Distribution of marks :
Record – 03
Viva -Voce – 07
Conduction & Report writing – 20
Total marks – 30

PSYCHOLOGY IN APPLICATION
SECOND YEAR
Total Marks – 100
Theory – 70 marks
Practical – 30 marks

Theory
UNIT-I
1. Life span development [10 Periods]
This chapter deals with variations in development and developmental tasks across the life span.
a) Meaning of development – Life span perspective
b) Principles of development
c) Stages of development: Pre-natal stage, Infancy, childhood stage, Adolescence, Adulthood and old age.

2. Self and Personality [9 Periods]
This chapter focuses on the study of self and personality in the context of different approaches in an effort to appraise the person. The assessment of personality will also be discussed.
a) Concept of self and personality
b) Personality types and traits
c) Assessment of Personality
This chapter deals with the nature of stress and strategies to cope with stress.
a) Meaning, Nature and causes of stress.
b) Coping strategies to deal with stress.

UNIT-II
3. Stress: Meeting life challenges [6 Periods]
This chapter deals with the nature of stress and strategies to cope with stress.
a) Meaning, Nature and causes of stress.
b) Coping strategies to deal with stress.

4. Physical environment and behaviour. [6 Periods]
This chapter focuses on the application of psychological understanding of human-environment relationship.
a) Human impact on the environment: Noise Pollution, Crowding, Natural disasters.
b) Impact of environment on human behaviour.

UNIT-III
5. Group Processes and leadership. [7 Periods]
This chapter deals with the concept of a group and the role of the leader in a group.
a) Groups: Nature, types and formation.
b) Leadership: Nature, functions and styles of leadership.

6. Counselling Processes [6 Periods]
This chapter focuses on helping the client in living a meaningful and fulfilling life.
a) Meaning and concept of counselling; Goals of counselling.
b) Characteristics of an effective counsellor.

UNIT-IV
7. Psychological disorder [10 Periods]
This chapter discusses the concept of normality and abnormality and the major psychological disorders.
a) Concept of normality and abnormality, criteria of studying abnormal behaviour
b) Causal factors associated with abnormal behaviour.
c) Major Psychological disorders: Anxiety disorders, somatoform disorders and mood disorders.

8. Therapeutic Approaches [6 Periods]
This chapter discusses the purpose and processes to treat Psychological disorders:
a) Nature and Processes of therapy
b) Types of therapy: Psychotherapy, Behaviour therapy, Cognitive therapy and Biomedical therapy.

UNIT-V
9. Statistics in Psychology [10 Periods]
This chapter deals with some basic statistical methods to be used in psychological studies.
a) Frequency distribution
b) Measures of Central Tendency: Computation and uses of mean, median and mode.

PRACTICALS

1. RCPM (Children) / RPM (Adults)
2. Case History Method (Preparation of at least one case profile)
3. Personality Test (Type A/B)
4. Piagetian Task (Conservation of Liquid Quantity)

Books Recommended:
1. Bureau’s Higher Secondary +2 Psychology Part-II Published by Odisha State Bureau of Test Book Preparation and Production, Bhubaneswar.
2. Psychology Part-II, NCERT.

CHSE Odisha Class 12 Text Book Solutions

CHSE Odisha 12th Class Education Book Solutions (+ 2 2nd Year)

CHSE Odisha Class 12 Education Book Solutions in English Medium

Unit 1 Contribution of Educators

• Contribution of Educators Objective Questions
• Contribution of Educators Short Answer Questions
• Contribution of Educators Long Answer Questions

Unit 2 Learning and Motivation

• Learning and Motivation Objective Questions
• Learning and Motivation Short Answer Questions
• Learning and Motivation Long Answer Questions

Unit 3 Current Issues in Education

• Current Issues in Education Objective Questions
• Current Issues in Education Short Answer Questions
• Current Issues in Education Long Answer Questions

Unit 4 Educational Statistics

• Educational Statistics Questions and Answers

CHSE Odisha Class 12 Education Book Solutions in Odia Medium

Unit 1 ଶିକ୍ଷାବିତମାନଙ୍କର ଦାନ

• Chapter 1 ମହାତ୍ମା ଗାନ୍ଧି
• Chapter 2 ପଣ୍ଡିତ ଗୋପବନ୍ଧୁ ଦାସ
• Chapter 3 ଶ୍ରୀ ଅରବିନ୍ଦ
• Chapter 4 ଜିନ୍ ଜାକୁଏସ୍ ଋଷୋ
• Chapter 5 ଜନ୍ ନ୍ୟୁଇ

Unit 2 ଶିକ୍ଷଣ ଏବଂ ଅଭିପ୍ରେରଣା

• Chapter 6 ଶିକ୍ଷଣ ପ୍ରକ୍ରିୟା ଓ ଶିକ୍ଷଣ ତତ୍ତ୍ବ
• Chapter 7 ପର୍ଯ୍ୟବେକ୍ଷଣାତ୍ମକ ଶିକ୍ଷଣ
• Chapter 8 ଶିକ୍ଷଣରେ ଅଭିପ୍ରେରଣା

Unit 3 ବିଦ୍ୟାଳୟ ଶିକ୍ଷା ସମସ୍ୟା

• Chapter 9 ପ୍ରାଥମିକ ଶିକ୍ଷାର ସାର୍ବଜନୀନୀକରଣ ଓ ଶିକ୍ଷାଧିକାର
• Chapter 10 ଜାତୀୟ ସଂହତି ଓ ଆନ୍ତର୍ଜାତିକ ବୁଝାମଣା ପାଇଁ ଶିକ୍ଷା
• Chapter 11 ପରିବେଶ ଶିକ୍ଷା
• Chapter 12 ମୂଲ୍ୟବୋଧ ଶିକ୍ଷା ଓ ମାନବାଧିକାର ଶିକ୍ଷା
• Chapter 13 ଶିକ୍ଷାରେ ସୂଚନା ଓ ସଞ୍ଚାର ପ୍ରଯୁକ୍ତି ବିଦ୍ୟା
• Chapter 14 ଜୀବନ ନୈପୁଣ୍ୟ (ଦକ୍ଷତା) ଶିକ୍ଷା

Unit 4 ଶୈକ୍ଷିକ ପରିସଂଖ୍ୟାନ

• Chapter 15 ଶୈକ୍ଷିକ ପରିସଂଖ୍ୟାନ (ପରିସଂଖ୍ୟାନ, ପୌନଃପୁନଃ ବିତରଣ, ତଥ୍ୟର ରେଖଚିତ୍ର ବା ଗ୍ରାଫୀୟ ପ୍ରଦର୍ଶନ, କେନ୍ଦ୍ରୀୟ ପ୍ରବଣତା ମାପ)

CHSE Odisha Class 12 Education Syllabus (+2 2nd Year)

EDUCATION ELECTIVE (Second Year)
Theory – 70 marks & Practical – 30 marks.
Theory Paper – II
FOUNDATIONS OF EDUCATION – II

Unit-I Fundamentals of Education (20 periods)
Contribution of Educators: Mahatma Gandhi, Pandit Gopabandhu Das, Sri Aurobindo, Jena Jacques Rousseau, John Dewey.

Unit-II Learning and Motivation (20 Periods)
Meaning, Nature and Factors of Learning, Theories of Learning: Trial and Error Theory and Laws of Learning, Classical Conditioning Theory, Insightful Learning, Learning and Construction of knowledge, Motivation in Learning: Meaning, Types, and Techniques of motivation.

Unit-III Current Issues in Education (20 Periods)
Universalisation of Elementary Education (UEE) and RTE, Education for National Integration and International Understanding, Environmental Education, Value Education, Human Rights Education, Information and Communication Technology (ICT) in education, and Life-skills Education.

Unit-IV Educational Statistics (20 Periods)
Statistics: Meaning, Nature and uses, Frequency Distribution, Graphical Representation of Data: Histogram, Polygon, and Pie-Chart, Measures of Central Tendency: Mean, Median, and mode – meaning, calculation, and uses.

Practical (60 Periods)
(To be examined by both external and Internal Examiners)
A. Practice Teaching Five Lessons in Classroom in the selected subject (30 Periods)
B. Preparation of Five Improvised Teaching Aids relating to the Five lessor plane along with their improvised teaching aids records (30 Periods)
For the Final Practical examination, the students shall deliver one lesson in his/her method subject.
Practice teaching records and improvised teaching aids records will be submitted during the final examination.

BOOKS RECOMMENDED:
1. Bureau Uchcha Madhyamik Siksha 2 (in Odia)
2. Bureau’s Higher Secondary Education II. Published by Odisha State Bureau of Textbook Preparation & Production, Bhubaneswar.

CHSE Odisha Class 12 Text Book Solutions