Class 12

Odisha State Board CHSE Odisha Class 12 Approaches to English Book 2 Solutions Poem 3 Of a Questionable Conviction Textbook Exercise Questions and Answers.

CHSE Odisha 12th Class Alternative English Solutions Poem 3 Of a Questionable Conviction

Pre-Reading activity:
Have you ever written poems? If so, what was your purpose in writing them? Why does a poet write poems?To persuade the readers into his own beliefs? Is it always possible to know if the poet is sincerely expressing his conviction in his poems? The poem you are going to read has
the title of a questionable conviction. Can you guess what the poem is about? Now read the poem and find out.

Notes On The Poet:
Jayanta Mahapatra from(1928) is foremost among the Indian English poets. His poetry expresses an ‘unhealable rift” a profound sense of loss. A tireless experimenter, Mahapatra’s poetic idiom is subtle and suggestive. His well known poetry collection are “Close the Sky”, Ten by Ten (1971), ‘A Rain Rated (1976). ‘A Father’s Hours’ (1976) and Relationship (1980) and ‘Burden of waves and Fruit’ (1986). ‘Ofa Questionable Conviction’ addresses the issue of how important convictions are to the writing of a poem.

Questions For Discussion:

Question 1.
Why does the poet think the man who talks of pain invented it himself?
Answer:
The poet thinks that he man who talks of pain invented it himself because he has made virtue out of it.

Question 2.
What is the main idea in Stanza 1?
Answer:
The main idea in Stanza 1 is that it is perhaps he who has invented pain himself because he has made virtue out of it.

Question 3.
What, according to the poet ‘his excuse to live’?
Answer:
Waiting for hours in the night towards another night is, according to the poet, his excuse.

Question 4.
Can the walls be lonely? Why does the poet use this expression?
Answer:
Walls, in poetry, can be lonely. The poet makes use of this expression to bring about a kind of personification in the poem.

Question 5.
“For months together the window has been deceiving him”. What could the poet mean by these lines?
Answer:
The window has been deceiving him for months together, because light has been simply coming in and going out. He has not been able to polish and heart by means of the light.

Question 6.
The poet described in he poem ‘has been trying to polish the light on his heart? How?
Answer:
The light on his heart is poor and the poet is not able to polish his heart by means of the light that enters and exits through the window.

Question 7.
You can easily see that in the first three stanza only the present tense forms are used. In the final stanza, however, past tense forms are used. Why is this shift in the concluding stanza?
Answer:
The use of present perfect progressive in the third stanza makes it rather explicit that the action started some times in the past and it continues into the present. Hence, it has a past reference. The past tense from in the concluding stanza tell that the poet has been doing that from the past till the present.

Question 8.
Would you consider the ending of the poem satisfactory? Explain your point of view.
Answer:
The ending of the poem is satisfactory because writing poem beside a lonely wall does not harm anybody.

Question 9.
What can you say about the tone of the poem?
Answer:
The tone of the poem is pain and simple.

Question 10.
What do you mean by –
(i) “The empty window in his lonely wall” (stanza 2)
(ii) “The eyes saw the pain in the mirror” (stanza 4)
(iii) “They did not grudge him that.” (stanza 4)
Answer:
(i) The only well without window doors.
(ii) The poet’s eyes saw the pain in the mirror.
(in) The people did not give or allow anything unwillingly.

Of a Questionable Conviction Summary in English

This is a man who talks of pain as though it belonged to him alone. Perhaps he has invented it himself and made virtue of it. And this man may be a poet. He waits in the night for hours. He keeps waiting for another night because, that is his excuse to live. The empty window in his lonely wall belongs to him alone. The window has been deceiving him for months together. Light comes in and goes out of its own without any hindrance. He has been trying to polish the light on his heart. Everybody says that he is a poet. His eyes witnessed the pain in the mirror which reflected his image. They did not grudge him that such a harmless pastime never ruined anybody’s sleep.

Analytical Outlines:

• The poet talks about a man.
• That man talks of pain.
• The pain belongs to him along.
• Perhaps, he has invented it himself.
• He makes virtue of it.
• That man may be a poet.
• He waits in the night for hours.
• He keeps waiting for another night.
• Because that is his excuse to live.
• It is the empty window in his lonely wall.
• It belongs to him alone.
• The window has been deceiving him.
• It has been deceiving him for months together.
• Light comes in.
• Light goes out of its own.
• It is without any hindrance.
• He has been trying to polish the light.
• He will polish it in his heart.
• Everybody says that he is a poet.
• His eyes witnessed the pain.
• It witnessed it in the mirror.
• The mirror reflected his image.
• They did not grudge him.
• It is a harmless pastime.
• It never ruined anybody’s sleep

Meaning Of Difficult Words:

conviction – a strong belief or opinion.
virtue – good quality
deceiving – misleading, cheating, betraying
hindrance- obstacle, impediment, obstruction
polish- to refine, to make or become smooth and shining.
witness- to see, to give evidence
image- a statue, an idol, an idea, a reflection
reflect- to express, to throw or come back
grudge- give or allow something very unwillingly
pastime- an enjoyable or interesting activity

Odisha State Board CHSE Odisha Class 12 Approaches to English Book 2 Solutions Poem 2 The Goat Paths Textbook Exercise Questions and Answers.

CHSE Odisha 12th Class Alternative English Solutions Poem 2 The Goat Paths

Pre-Reading activity:
The goat is a very common animal in our part of the country. What qualities do you associate with goats? Think up five words that come to your mind when you think of goats. The poem you are about to read has the title. “The Goat-Paths”. Can you guess what the poem is about? Write your guess here. Now read the poem is see what it is really about.

Notes On The Poet:
James Stephens (1882-1950), a British poet and novelist, is known for such work as Insurrections (1909). The Crack of gold (1912). The Hill of Vision (1912). Green Branches (1912) and Collected Poems (1936). His ‘delicates inspired’ poetry makes effective use of imagery, the familiar scene is often ‘translated into an image for a human meaning’. In this poem, the poet likes the goats, seeks harmony with Nature.

Questions For Discussion:

Question 1.
Where are the straying goats found?
Answer:
The goats prefer to go to the heather and not to a grassy field because there was nothing stir.

Question 2.
Why do the goats go to the heather and not to a grassy field?
Answer:
The goats prefer to go to the heather and not to a grassy field because there was nothing stir.

Question 3.
Are these goats different from the others of their kind?
Answer:
These goats are different from the others of their kind because they behave in different way from the others.

Question 4.
How many times para the words relating to ‘quiet’ used in this poem?
Answer:
Five times.

Question 5.
How does the poet related ‘quietly in quietness’ (line 13 to ‘the moving sky’?
(Line 16). If you were to make two sections of the poem how would you divided it?
Answer:
“Quietly in quietness’ means being alone in a lonely place. The moving sky implies the limitless sky.

Question 6.
How would you read of someone called the first section ‘observation’ and the second ‘reflection’?
Answer:
It would be right to say that the first section of the poem is based on ‘observation’ and the second section on ‘reflection’.

Question 7.
What do you notice in the voice of the poet? urgency, defiance, arrogance, envy, frustration, expectation or a sense of oneness with God’s creation?
Answer:
Urgency, defiance, expectation, a sense of oneness with God’s creation.

Question 8.
What does the poet say he would do when his ‘sunny solitude’ is distributed?
Answer:
He would stray apart and brood, he would beat a hidden way through the quiet heather spray in the sunny solitude.

Question 9.
Why does the poet call the goats ‘wise’?
Answer:
The poet calls the goats wise because they brood and beat the hidden way through the quiet heather spray in the sunny solitude.

Question 10.
The expression ‘IfI were’ conveys a wish. What is the poet’s wish?
Answer:
The poet’s wish is to be wise enough to brood and beat the hidden way through the quiet heather spray to a sunny solitude. He would run away and make an angry sound and would stare and turn and bound to the deeper silence where nothing disturbs.

Question 11.
I would think until I found/something I can never find’. How would you comment on the complexity of thought implicit in these lines?
Answer:
It is just like attaining the unattainable. The first lines is in simple past tense while the second line occurs in simple present form. The use of simple past in the present delineates are unfulfilled of he poet who would never find it. Hence, it is unfulfilled wish.

Question 12.
Can you write a simple paraphrase of the last stanza?
Answer:
The last stanza speaks of attaining of the unattainable. The poet makes use of the simple past tense form of verb, that is ‘found’ in the present context. The grammatical usage of such expression explains the unfulfilled wish of the poet. The poet is of the view that he will go on thinking until he has found what he can never find.

Question 13.
Point out same of the striking images in the poem?
Answer:
There are some striking images in the poem. They are-crooked paths, sunny quietness, crouching, beaten way etc.

Question 14.
What purpose do the words ‘quiet sunniness’. Sunny quietness’ ‘deeper sunniness’, ‘Quietly in quietness’. ‘Sunny quietude’. ‘Sunny solitude’ serve?
(i) They cast a spell on the reader.
(ii) They add to the musically of the poem
(iii) They are meant to create ambiguity for the reader.
Answer:
(iii) They are meant to create ambiguity for the reader.

Question 15.
Can you think of possible revisions of the following lines of the poem?
Line 17: “If you approach they run away.”
Line 30: “And should you come I’d run away”
Answer:
“If you approach they run away”
“If you come I’d run away”

Composition

Question 1.
Would you agree that the poem suggest the possibility of a harmony within the self and within the natural process of life and world”? Justify your answer.
Answer:
The poem‘ The Goat – Paths” is undoubtedly the master piece of James Stephens, an eminent and out-standing British poet and novelist. In this discussing poem, he expresses the possibility of a harmony within the selfand with the natural process of life and the world. The goats and goat paths are natural elements. The process of life is best reflected by means of these things. Life and life’s process are a part of nature. The outer reflections cost a deep spell on human life. It is because the external nature has a tremendous influence on the internal nature of man. Man internalises the outer nature which is at times didactic and delightful. The nature says words worth, ‘is the friend, philosopher and guide’, for man. In this discussing poem ‘The Goat – Paths,’ the poet James Stephens reflects the human self and natural processes. The poem ultimately leads the poet to identify himself with the goats! He says, “IfI were as wise as they I would stray apart and brood ….”. If a man appears on the scene, he would run away, he would like goats make an angry sound. He would is to a place where nothing stirs in the silence ofthe furze. He would also like to think in the airy quietness through the quiet sunniness, he would stray away to brood by hidden, beaten way in the sunny solitude. He would think until he found something he can never and something lying on the ground. As a matter offact, Stephens has tried his best to establish a harmony within the self and with the natural process of life and the world. The way Stephens explains it is superb and fantastic. On the whole. Stephens’s expression is most during, evaluating, inspiring and heart-touching.

Question 2.
Comment on the appropriateness of the title “The Goat-Paths”.
Answer:
The poem “The Goat-Paths” is, indeed, the best typical master-piece of James Stephens, a prominent and outstanding British poet and novelist of the twentieth century. His ‘delicately inspired’ poetry tinged with apt and suggestive titles is undoubtedly superband excellent. However, the title of any piece of art must be thoroughly exact, appropriate and suggestive. It should be like a colourful and attractive signboard of a shop. An attractive signboard automatically arrests the attention of the customers and makes them spell bound to have a visit to the shop. It also explains what a shop contains similarly, and apt and suggestive title arrests the attention of the reader instantly and makes them spell bound to go through it completely. It is like ‘the face is the index of heart.’ The discussing poem ‘The Goat-Paths”| bears the same ideas to prove the authenticity and appropriateness of the title of the poem. Moreover, the goat-paths are usually crooked and they lead uphill. They also wind about through the heather in and out of the quiet sunniness. The goats keep cropping here and there, they pause and turn and pass. When approaches they run away, leap and stare and they go away to the sunny quietude with a sudden angry sound. The poet brings oneness between the goats and his life. In feet, appropriateness of the title applies to the poem. The poem bears all qualities of being an apt and suggestive one. The external natural things like the goats, goat-paths have been identified with human life. The poet wishes to internalise these values and learn a lot from the goats. As a matter of feet, the title of the poem‘ The Goat-Paths” is apt and suggestive. The way the poet has expressed the central idea of the poem through the appropriate title is superb and outstanding. One the whole, the title of the poem is thought-provoking, inspiring and heart enduring.

The Goat Paths Summary in English

The crooked paths go every way upon the hill. They wind about though the heather in and out of the quiet sunniness. The goats there, day after day, stray in sunny quietness, cropping here and there as they halt and turn and pass. There is a bit of Heather spray, a mouthful of the grass in the deep sunniness in the place where nothing stirs quietly in the quietness. In quiet of the furze, they come and lie staring on the roving sky. When somebody approaches they run away. They leap and stare, they bound away with a sudden angry sound of the sunny quietude. Crouching down where nothing stirs in the silence of these furze, crouching down again to brood in the sunny solitude. If we were as wide as they were, he would stray apart and brood, he would beat a hidden way through the quiet heather spray to a sunny solitude. If someone came he would run away, he would make an angry sound and would stare and turn and bound to the deeper quietude to the place where nothing stirs in the silence of the furze. He would think as long as they in that airy quietness through the quiet sunniness he would stray away to brood by a hidden beaten way in the sunny solitude.

Analytical Outlines:

• The crooked paths go every way.
• It goes every way upon the hill.
• They wind about through the heather.
• They wind about in and out of the quiet sunniness.
• There, the goats stray in sunny quietness.
• The goats stray day after day.
• They halt here and there.
• They turn here and there
• They pass here and there
• They crop here and there
• There is a bit of heather spray.
• It is mouthful of the grass.
• It is in the deep sunniness in the place.
• There nothing stirs quietly.
• Actually, there is an entire quietness.
• They come in the quiet of the furze.
• They lie staring on the roving sky.
• They run away when somebody approaches.
• They leap.
• They stare
• They bound away with a sudden angry sound.
• They do it with the sound of the sunny quietude.
• They crouch down.
• Nothing stirs in the silence of the furze.
• They crouch down again.
• They brood in the sunny solitude.
• If he were as wise as they.
• He would stray apart.
• He would brood.
• He would beat a hidden way through the quiet heather.
• He would spray to a sunny solitude
• If someone came.
• He would run away.
• He would make an angry sound.
• He would stare.
• He would turn.
• He would bound to the deeper quietude.
• There nothing stirs in the silence of the furze.
• He would think as long as they were in that airy quietness.
• He would think this through the quiet sunniness.
• He would stray away to brood.
• He would brood by a hidden beaten way.
• He would do this in the sunny solitude.

Meaning Of difficult Words:

croocked- bent, not straight
heather- wasteland covered with strubs
cropping- grazing
furze- shrub with spikes
roving (sky) – moving (here, appearing to move)
quietude – calmness, serenity, solitude
crouching – lying close to the ground
beat- make a path
stare- to look fixedly, to glance
solitude- loneliness, calmness, pin drop silence
brood- act ofbreeding
stray- to wander
spray- to sprinkle

Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 12 Vectors Ex 12(d) Textbook Exercise Questions and Answers.

CHSE Odisha Class 12 Math Solutions Chapter 12 Vectors Exercise 12(d)

Question 1.
Each question given below has four possible answers out of which only one is correct. Choose the correct one.
(i) \(\vec{a} \cdot \vec{b} \times \vec{a}\) = _______.
(a) \(\overrightarrow{0}\)
(b) 0
(c) 1
(d) \(\vec{a}^2 \vec{b}\)
Solution:
\(\vec{a} \cdot(\vec{b} \times \vec{a})\) = \((\vec{b} \times \vec{a}) \cdot \vec{a}\)
= \(\vec{b} \cdot(\vec{a} \times \vec{a})\) = \(\vec{b} \cdot \overrightarrow{0}\)
= 0 [∴ Dot product is commutative and in the scalar triple product the dot and cross can be interchanged.]

(ii) \((-\vec{a}) \cdot \vec{b} \times(-\vec{c}))\) = _______.
(a) \(\vec{a} \times \vec{b} \cdot \vec{c}\)
(b) \(-\vec{a} \cdot(\vec{b} \times \vec{c})\)
(c) \(\vec{a} \times \vec{c} \cdot \vec{b}\)
(d) \(\vec{a} \cdot(\vec{c} \times \vec{b})\)
Solution:
\((-\vec{a}) \cdot \vec{b} \times(-\vec{c})\) = \(\vec{a} \cdot(\vec{b} \times \vec{c})\)

(iii) For the non-zero vectors \(\vec{a}, \vec{b}\) and \(\vec{c}, \vec{a} \cdot(\vec{b} \times \vec{c})\) = 0 if
(a) \(\vec{b} \perp \vec{c}\)
(b) \(\vec{a} \perp \vec{b}\)
(c) \(\vec{a} \| \vec{c}\)
(d) \(\vec{a} \perp \vec{c}\)
Solution:
\(\vec{a} \cdot(\vec{b} \times \vec{c})\) = \((\vec{a} \times \vec{b}) \cdot \vec{c}\)
\(\vec{c} \perp(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})\)
but \(\vec{a} \times \vec{b}\) is perpendicular to \(\vec{a}\) and \(\vec{b}\)
∴ \(\vec{a} \| \vec{b}\)

Question 2.
Find the scalar triple product \(\vec{b} \cdot(\vec{c} \times \vec{a})\) where \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are respectively.
(i) î + ĵ, î – ĵ, 5î + 2ĵ + 3k̂
Solution:

= 1 (0 – 3) + 1 (0 – 3) + 0 (5 – 2)
= 3 – 3 = -6

(ii) 5î – ĵ + 4k̂, 2î + 3ĵ + 5k̂, 5î – 2ĵ
Solution:

= 5 (18 + 10) + 1 (12 – 25) + 4 (- 4 – 15)
= 140 – 13 – 76 = 140 – 89 = 51

Question 3.
Find the volume of the parallelopiped whose sides are given by the vectors.
(i) î + ĵ + k̂, k̂, 3î – ĵ + 2k̂
Solution:

= 1 (0 + 1) – 1 (0 – 3) + 1 (0 – 0)
= 1 + 3 = 4 cube units.

(ii) (1, 0, 0), (0, 1, 0), (0, 0, 1).
Solution:

Question 4.
Show that the following vector are co-planar
(i) î – 2ĵ + 2k̂, 3î + 4ĵ + 5k̂, -2î + 4ĵ – 4k̂
Solution:

(ii) î + 2ĵ + 3k̂, -2î – 4ĵ + 5k̂, 3î + 6
Solution:

Question 5.
Find the value of λ so that the three vectors are co-planar.
(i) î + 2ĵ + 3k̂, 4î + ĵ + λk̂ and λî – 4ĵ + k̂
Solution:

(ii) (2, -1, 1), (1, 2, -3) and (3, λ, 5)
Solution:

⇒ 2 (10 + 3λ) + 1 (5 + 9) + 1 (λ – 6) = 0
⇒ 20 + 6λ +14 + λ – 6 = 0
⇒ 7λ + 28 = 0 ⇒ λ = -4

Question 6.
If \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) mutually perpendiculars, show that \([\vec{a} .(\vec{b} \times \vec{c})]^2\) = a²b²c²
Solution:

Question 7.
Show that \([\vec{a}+\vec{b} \vec{b}+\vec{c} \vec{c}+\vec{a}]\) = 2\([\vec{a} \vec{b} \vec{c}]\)
Solution:

Question 8.
Prove that \([\vec{a} \times \vec{b} \vec{b} \times \vec{c} \vec{c} \times \vec{a}]\) = \([\vec{a} \vec{b} \vec{c}]^2\)
Solution:

Question 9.
For \(\vec{a}\) = î + ĵ, \(\vec{b}\) = -î + 2k̂, \(\vec{c}\) = ĵ + k̂ obtain \(\vec{a} \times(\vec{b} \times \vec{c})\) and also verify the formula \(\vec{a} \times(\vec{b} \times \vec{c})\) = \((\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}\).
Solution:

Question 10.
Prove that \(\vec{a} \times(\vec{b} \times \vec{c})+\vec{b} \times(\vec{c} \times \vec{a})+\vec{c} \times(\vec{a} \times \vec{b})\) and hence prove that \(\vec{a} \times(\vec{b} \times \vec{c}), \vec{b} \times(\vec{c} \times \vec{a}), \vec{c} \times(\vec{a} \times \vec{b})\) are coplanar.
Solution:

Question 11.
If \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) unit vectors and \(\hat{a} \times(\hat{b} \times \hat{c})=\frac{1}{2} \hat{b}\) find the angles that â makes with b̂ and ĉ, where b̂, ĉ are not parallel.
Solution:

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

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

Solve the following differential equations.
Question 1.
\(\frac{d y}{d x}\) + y = e^-x
Solution:
Given equation is \(\frac{d y}{d x}\) + y = e^-x … (1)
This is a linear differential equation.
Here P = 1, Q = e^-x
So the integrating factor
I.F. = e^{∫P dx} = e^∫dx = e^x
The solution of (1) is given by
ye^x = ∫e^-x . e^x dx = ∫dx = x + C
⇒ y – xe^-x + Ce^-x

Question 2.
(x² – 1)\(\frac{d y}{d x}\) + 2xy = 1
Solution:
Given equation is (x² – 1)\(\frac{d y}{d x}\) + 2xy = 1

Question 3.
(1 – x²)\(\frac{d y}{d x}\) + 2xy = x \(\sqrt{1-x^2}\)
Solution:
Given equation is

Question 4.
x log x \(\frac{d y}{d x}\) + y = 2 log x
Solution:
Given equation is

Question 5.
(1 + x²)\(\frac{d y}{d x}\) + 2xy = cos x
Solution:

Question 6.
\(\frac{d y}{d x}\) + y sec x = tan x
Solution:
Given equation is
\(\frac{d y}{d x}\) + y sec x = tan x
This is a linear equation where
P = sec x, Q = tan x
I.F. = e^{∫sec dx}
= e^{(sec x + tan x)} = sec x + tan x
The solution is y . (sec x + tan x)
= ∫(sec x + tan x) tan x dx
= ∫(sec x tan x + tan² x) dx
= ∫(sec x . tan x + sec² x – 1) dx
= ∫(sec x + tan x) – x + C
⇒ (y – 1) (sec x + tan x) + x = C

Question 7.
(x + tan y) dy = sin 2y dx
Given equation can be written as
Solution:

Question 8.
(x + 2y³)\(\frac{d y}{d x}\) = y
Solution:
Given equation can be written as

Question 9.
sin x\(\frac{d y}{d x}\)+ 3y = cos x
Solution:

Question 10.
(x + y + 1)\(\frac{d y}{d x}\) = 1
Solution:

Question 11.
(1 + y²) dx + (x – \(e^{-\tan ^{-1} y}\)) dy = 0
Solution:
Given equation can be written as

Question 12.
x\(\frac{d y}{d x}\) + y = xy²
Solution:
Given equation can be written as

⇒ z = -x ln x + Cx
⇒ \(\frac{1}{y}\) = -x ln x + Cx
⇒ 1 = -xy ln x + Cxy
∴ The solution is (C – ln x) xy = 1

Question 13.
\(\frac{d y}{d x}\) + y = y² log x
Solution:

Question 14.
(1 + x²)\(\frac{d y}{d x}\) = xy – y²
Solution:
The given equation can be written as

Question 15.
\(\frac{d y}{d x}\) + \(\frac{y}{x-1}\) = \(x y^{\frac{1}{2}}\)
Solution:
The given equation can be written as

Question 16.
\(\frac{d y}{d x}\) + \(\frac{y}{x}\) = x², y(1) = 1
Solution:
The given equation can be written as
\(\frac{d y}{d x}\) + \(\frac{y}{x}\) = x², y(1) = 1 … (1)
This is a linear equation.

Question 17.
\(\frac{d y}{d x}\) + 2y tan x = sin x, y\(\left(\frac{\pi}{3}\right)\) = 0.
Solution:

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: