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Chapter – 9 Soil | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter - 9 Soil

Q1. Tick the most suitable answer in question 1 and 2.
In addition to the rock particles, the soil contains
(i) Air and water
(ii) Water and plants
(iii) Minerals, organic matter, air and water
(iv) Water, air and plants
Answer:
In addition to the rock particles, the soil contains
(iii) Minerals, organic matter, air and water.

Q2. The water holding capacity is the highest in
(i) Sandy soil
(ii) Clayey soil
(iii) Loamy soil
(iv) Mixture of sand and loam
Answer:
The water holding capacity is the highest in
(ii) Clayey soil

Q3. Match the items in Column I with those in Column II:

Column I	Column II
(i) A home for living organisms	(a) Large particles
(ii) Upper layer of the soil	(b) All kinds of soil
(iii) Sandy soil	(c) Dark in colour
(iv) Middle layer of the soil	(d) Small particles and packed tight
(v) Clayey soil	(e) Lesser amount of humus

Answer:

Column I	Column II
(i) A home for living organisms	(b) All kinds of soil
(ii) Upper layer of the soil	(c) Dark in colour
(iii) Sandy soil	(a) Large particles
(iv) Middle layer of the soil	(e) Lesser amount of humus
(v) Clayey soil	(d) Small particles and packed tight

Q4. Explain how soil is formed.
Answer:
Soil is formed through the process of weathering. Weathering is a process of physical breakdown and chemical decomposition of rocks and minerals near or at the surface of the earth. This physical and chemical decomposition is primarily done by wind, water, and climate. As a result of these processes, large rock pieces are converted into smaller pieces and eventually to soil.

Q5. How is clayey soil useful for crops?
Answer:
Following are the properties of clayey soil:

It has very good water holding capacity.
It is rich in organic matter.

For growing crops such as wheat, gram, and paddy, the soil that is good at retaining water and rich in organic matter is suitable. Therefore, clayey soils having these characteristics are useful for such kind of crops.

Q6. List the differences between clayey soil and sandy soil.
Answer:

Clayey Soil	Loamy Soil
(i) It has much smaller particles.	(i) It has much larger particles.
(ii) It can hold good amount of water.	(ii) It cannot hold water.
(iii) It is fertile.	(iii) It is not fertile.
(iv) Air content is low.	(iv) Air get trapped between the particles.
(iv) Particles are tightly packed	(iv) Particles are loosely packed
(iv) Good for growing various crops.	(iv) Not suitable for growing crops.

Q7. Sketch the cross section of soil and label the various layers.
Answer:

Q8. Razia conducted an experiment in the field related to the rate of percolation. She observed that it took 40 min for 200 mL of water to percolate through the soil sample. Calculate the rate of percolation.
Answer:

9. Explain how soil pollution and soil erosion could be prevented.
Answer:
Prevention of soil pollution:
The persistent build-up of toxic compounds in the soil is defined as soil pollution. To prevent soil pollution, its causes must be controlled.

Reduce the use of plastics: Plastics and polythene bags destroy the fertility of soil. Hence, these should be disposed off properly and if possible, their use should be avoided.
Industrial pollutants: Some waste products from industries and homes pollute soil. These pollutants should be treated chemically to make them harmless before they are disposed off.
Insecticides: Other pollutants of soil include pesticides and insecticides. Therefore, excessive use of these substances should be avoided.

Prevention of soil erosion:
Removal of the upper fertile layer of the soil (top soil) by strong wind and flowing water is known as soil erosion. Following steps can be taken to reduce soil erosion:

Mass awareness to reduce deforestation for industrial purposes.
Helping local people to regenerate degrading forest.
Planting trees.

10. Solve the following crossword puzzle with the clues given:

Across
2. Plantation prevents it.
5. Use should be banned to avoid soil pollution.
6. Type of soil used for making pottery.
7. Living organism in the soil.

Down
1. In desert soil erosion occurs through.
3. Clay and loam are suitable for cereals like.
4. This type of soil can hold very little water.
5. Collective name for layers of soil.

Answer:

Across
2. Plantation prevents it. → Erosion
5. Use should be banned to avoid soil pollution. → Polythene
6. Type of soil used for making pottery. → Clay
7. Living organism in the soil. → Earthworm

Down
1. In desert soil erosion occurs through. → Wind
3. Clay and loam are suitable for cereals like. → Wheat
4. This type of soil can hold very little water. → Sandy
5. Collective name for layers of soil. → Profile

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Chapter – 8 Winds | Storms and Climate | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter 8 Winds | Storms and Climate

Q.1.Fill the missing word in the blank spaces in the following statements:
(a) Wind is ___________air.
(b) Winds are generated due to _________ heating on the earth.
(c) Near the earth’s surface _________ air rises up whereas _________ air comes down.

(d) Air moves from a region _________of pressure to a region _________of pressure.
Ans.(a) moving (b) uneven
(c) warm, cooler (d) high, low

Q.2. Suggest two methods to find out wind direction at a given place.
Ans.(i) Take a piece of a paper in your hand. Allow it to fall from your hand. It will flow in the direction in which wind is blowing.
(ii) You can also use a wind-pane which helps us to know accurate wind direction.

Q.3.State two experiences that made you think that air exerts pressure.
Ans.(i) Balloons and balls can be used only when they are inflated with air. When balloon is overfilled with air it bursts due to excessive air pressure.
(ii) Compressed air is used in the brake system for stopping trains.

Q.4.You want to buy a house. Would you like to buy a house having windows but no ventilators? Explain your answer.
Ans.No, a house which has no ventilators is not a safe or healthy house to live in. The air circulation is not there in such a house. So, it has no fresh air. Because warm air rises up and goes out through ventilators and fresh air comes in through windows.

Q.5.Explain why holes are made in hanging banners and hoardings.
Ans.Air exerts pressure. Due to this pressure banners and hoarding flutter and torn when wind is blowing fast. Holes are made in banners and hoardings so that wind passes through the holes and they do not become loose and fall down.

Q.6.How will you help your neighbours in case cyclone approaches your village/town?
Ans. (i) I will make them aware of cyclone forecast and warning service.
(ii) Rapid communication of warning to the government agencies and all the important places.
(iii) Construction of cyclone shelters in the cyclone prone areas.
(iv) Helping them to shift essential goods, domestic animals etc. to safer places.

Q.7.What planning is required in advance to deal with the situation created by a cyclone?
Ans.To deal with cyclone, it is important to follow the following points :

(i) carefully listening the warnings transmitted on T.V. and radio.
(ii) moving to the safer places.
(iii) keeping an emergency kit ready.
(iv) store food in waterproof bags. .
(v) not venturing into sea.
(vi) keeping all the emergency numbers.

Q.8. Which one of the following places is unlikely to be affected by cyclone?
(i) Chennai (ii) Mangalore
(iii) Amritsar (iv) Puri
Ans.(iv) Amritsar

Q.9.Which of the statements given below is correct?
(i) In winter the winds flow from the land to the ocean.
(ii) In summer the winds flow from the land towards the ocean.
(iii) A cyclone is formed by a very high-pressure system with very high-speed winds revolving around it.
(iv) The coastline of India is not vulnerable to cyclones.
Ans.(i) In winter the winds flow from the land to the ocean.

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Chapter – 7 Weather, Climate and Adaptations of Animals to Climate | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter 7 Weather, Climate and Adaptations of Animals to Climate

Q.1.Name the elements that determine the weather of a place.
Ans. The temperature, humidity, rainfall, wind-speed, etc. are called the elements that determine the weather of a place.

Q.2. When are the maximum and minimum temperature likely to occur during the day?
Ans.The maximum temperature of the day occurs generally in the afternoon and the minimum temperature occurs in the early morning.

Q.3.Fill in the blanks
(i) The average weather taken over a long time is called __________
(ii) A place receives very little rainfall and the temperature is high throughout the year, the climate of that place will be ________ and _________
(iii) The two regions of the earth with extreme climatic conditions are __________ and ____________
Ans. climate of the place (ii) hot, dry (iii) polar, tropical regions

Q.4.Indicate the type of climate of the following areas:
(a) Jammu and Kashmir:
(b) Kerala:,
(c) Rajasthan:
(d) North-east India:
Ans.(a) Jammu and Kashmir—moderately hot and moderately wet climate.
(b) Kerala—very hot and wet climate.
(c) Rajasthan—hot and dry climate.
(d) North-east India—The north eastern India receives rain for a major part of the year, hence wet climate.

Q.5.Which of the two changes frequently, weather or climate?
Ans.Weather

Q.6.Followings are some of the characteristics of animals:
(i) Diets heavy on fruits , (ii) White fur (iii) Need to migrate (iv) Loud voice
(v) Sticky pads on feet (vi) Layer of fat under skin
(vii) Wide and large paws (viii) Bright colours
(ix) Strong tails (x) Long and large beak
For each characteristic indicate whether it is adaptation for tropical rainforests or polar regions. Do you think that some of these characteristics can be adapted for both regions?

Ans:
.-

Q.7. The tropical rainforests has a large population of animals. Explain why it is so.
Ans. Because of continuous warmth and rain, the tropical region supports an enormous number and a wide variety of animals

Q.8.Explain with examples, why we find animals of certain kind living in particular climatic conditions.
Ans. Animals are adapted to survive in the conditions in which they live. Features and habits which help them to adapt to their surroundings are the result of evolution. So, to survive in a particular type of climate the animals must have certain adapted features. This is the reason we find animals of certain kind living in particular climatic conditions. For example, animals in the polar region are adapted to the extremely cold climate. They have special characteristics, such as white fur, strong sense of smell, a layer of fat under the skin, wide and large paws for swimming and walking in snow etc

Q.9. How do elephants living in the tropical rainforests adapt themselves?
Ans. The elephant has adapted to the conditions of rainforest in many remarkable ways. It has a trunk that it uses as a nose because of this it has a strong sense of smell. The trunk is also used by it for picking up food. Its tusks are modified teeth. These can tear the bark of trees that an elephant loves to eat. So, the elephant is able to handle the competition for food very well. Large ears of the elephant help it to hear even very soft sounds. They also help the elephant to keep cool in the hot* humid climate of the rainforest.

Q.10.Choose the correct option which answers the following question:
A carnivore with stripes on its body moves very fast while catching its prey. It is likely to be found in:
(i) polar regions (ii) deserts
(iii) oceans (iv) tropical rainforests
Ans. (iv) tropical rainforests

Q.11.Which features adapt polar bears to live in extremely cold climate?
(i) A white fur, fat below skin, keen sense of smell.
(ii) Thin skin, large eyes, a white fur.
(iii) A long tail, strong claws, white large paws.
(iv) White body, paws for swimming, gills for respiration.
Ans.(iv) A white fur, fat below skin, keen sense of smell.

Q.12.Which option best describes a tropical region?
(i) hot and humid
(ii) moderate temperature, heavy rainfall (iii) cold and humid (iv) hot and dry
Ans.(i) Hot and humid

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RD SHARMA SOLUTION CHAPTER-19 Indefinite Integrals I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 19 Indefinite Integrals Ex. 19.1

Question 1

Integrate each of the following functions with respect to x:

(i) ∫ x⁴dx

(ii) ∫ x^5/4dx

(iii) begin mathsize 12px style integral 1 over straight x to the power of 5 dx end style

(iv) begin mathsize 12px style integral 1 over x to the power of 3 divided by 2 end exponent d x end style

(v) ∫ 3^xdx

(vi) $begin mathsize 12px style integral fraction numerator 1 over denominator square root of x squared end root end fraction d x end style$

(vii) ∫ 3^2log_3x dx

(viii) ∫ log_xx dx Solution 1

Question 2

Evaluate:

begin mathsize 12px style left parenthesis i right parenthesis space integral square root of fraction numerator 1 plus cos 2 x over denominator 2 end fraction end root d x
left parenthesis i i right parenthesis space integral square root of fraction numerator 1 minus cos 2 x over denominator 2 end fraction d x end root end style

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Chapter 19 Indefinite Integrals Ex. 19.2

Question 1

Evaluate begin mathsize 12px style integral open parentheses 3 cross times square root of x plus 4 square root of x plus 5 close parentheses d x end style Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

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Question 12

Solution 12

Question 13

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Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

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Question 18

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Question 19

Solution 19

Question 20

Solution 20

Question 21

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Question 22

Solution 22

Question 23

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Question 24

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Question 25

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Question 26

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Question 27

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Question 28

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Question 29

Solution 29

Question 30

Solution 30

Question 31

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Question 32

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Question 34

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Question 35

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Question 37

Solution 37

Question 38

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Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Evaluate : $integral fraction numerator cos x over denominator 1 plus cos x end fraction d x$ Solution 42

Question 43

Evalute :

integral fraction numerator 1 minus cos x over denominator 1 plus cos x end fraction d x

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Chapter 19 Indefinite Integrals Ex. 19.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Integrate begin mathsize 12px style integral sin x square root of 1 plus cos 2 x end root space d x end style Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Evaluate the following integrals:

begin mathsize 12px style integral fraction numerator 1 over denominator cos squared x left parenthesis 1 minus tan x right parenthesis squared end fraction end style

Solution 19

Chapter 19 – Indefinite Integrals Exercise Ex. 19.4

Question 1

Solution 1

Question 2

Integrate $begin mathsize 12px style integral fraction numerator x cubed over denominator x minus 2 end fraction d x end style$ Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Chapter 19 Indefinite Integrals Ex. 19.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Chapter 19 Indefinite Integral Ex. 19.6

Question 1

Solution 1

Question 2

Evalute the following integral :Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Integrate begin mathsize 12px style integral sin x square root of 1 minus cos 2 x end root space d x end style Solution 8

Chapter 19 Indefinite Integrals Ex. 19.7

Question 1

Integrate ∫ sin4x cos7x dxSolution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Chapter 19 Indefinite Integrals Ex. 19.8

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

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Question 15

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Question 19

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Solution 24

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Solution 25

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Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

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Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

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Question 37

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Question 46

Solution 46

Question 47

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Question 48

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Chapter 19 – Indefinite Integrals Exercise Ex. 19.9

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

begin mathsize 12px style Evaluate colon
integral fraction numerator 4 x plus 3 over denominator square root of 2 x squared plus 3 x plus 1 end root end fraction d x end style

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

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Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

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Question 39

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Question 40

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Question 41

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Question 42

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Question 43

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Question 44

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Question 45

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Question 46

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Question 47

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Question 48

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

begin mathsize 12px style Integrate space the space function
fraction numerator e to the power of m tan to the power of negative 1 end exponent x end exponent over denominator 1 plus x squared end fraction end style

Solution 54

L e t space tan to the power of minus 1 end exponent x equals t
D i f f e r e n t i a t i n g space t h e space a b o v e space f u n c t i o n space w i t h space r e s p e c t space t o comma space w comma space w e space h a v e comma
fraction numerator 1 over denominator 1 plus x squared end fraction d x equals d t
rightwards double arrow integral fraction numerator e to the power of m tan to the power of minus 1 end exponent x end exponent over denominator 1 plus x squared end fraction equals integral e to the power of m t end exponent cross times d t
rightwards double arrow integral fraction numerator e to the power of m tan to the power of minus 1 end exponent x end exponent over denominator 1 plus x squared end fraction equals e to the power of m t end exponent over m
R e s u b s t i t u t i n g space t h e space v a l u e space o f space t space i n space t h e space a b o v e space s o l u t i o n comma space w e space h a v e comma
space space space space space space rightwards double arrow integral fraction numerator e to the power of m tan to the power of minus 1 end exponent x end exponent over denominator 1 plus x squared end fraction equals e to the power of m tan to the power of minus 1 end exponent x end exponent over m plus C

Question 55

Solution 55

Question 56

Solution 56

Question 57

Solution 57

Question 58

Solution 58

Question 59

Solution 59

Question 60

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Question 62

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Question 63

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Question 67

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Question 70

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Question 71

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Question 72

Solution 72

Chapter 19 Indefinite Integrals Exercise Ex. 19.10

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Chapter 19 – Indefinite Integrals Exercise Ex. 19.11

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

begin mathsize 12px style Let space iota equals integral cos e c to the power of 4 3 x d x. space T h e n
iota equals integral cos e c squared 3 x space cos e c squared 3 d x
equals integral open parentheses 1 plus c o t squared 3 x close parentheses cos e c squared 3 x d x
equals integral open parentheses cos e c squared 3 x space plus space c o t squared 3 x space cos e c squared 3 x close parentheses d x
rightwards double arrow iota equals integral cos e c squared 3 x d x space plus space integral c o t squared 3 x space cos e c squared 3 x d x
Substituting space c o t space 3 x space equals t space a n d space cos e c t squared 3 x d x space equals negative d t space i n space 2 nd space integral comma space we space get
iota equals integral cos e c squared 3 x d x space minus space integral t squared fraction numerator d t over denominator 3 end fraction
equals fraction numerator negative 1 over denominator 3 end fraction c o t 3 x minus t cubed over 9 plus C equals fraction numerator negative 1 over denominator 3 end fraction c o t space 3 x minus 1 over 9 c o t cubed space 3 x plus c
therefore iota equals fraction numerator negative 1 over denominator 3 end fraction c o t space 3 x space minus space fraction numerator c o t cubed 3 x over denominator 9 end fraction plus c
end style

Question 9

Solution 9

begin mathsize 12px style L e t space iota equals integral c o t to the power of n x space cos e c squared x d x. n not equal to negative 1......... open parentheses i close parentheses
L e t space c o t space x equals t. space T h e n
d open parentheses c o t x close parentheses equals d t
rightwards double arrow negative cos e c squared x d x equals d t
rightwards double arrow cos e c squared x d x equals negative d t
P u t t i n g space c o t x space equals t space a n d space c os e c squared x d x equals negative d t space i n space e q u a t i o n space open parentheses i close parentheses comma space w e space g e t
iota equals integral t to the power of n cross times open parentheses negative d t close parentheses
equals negative fraction numerator t to the power of n plus 1 end exponent over denominator n plus 1 end fraction plus c
rightwards double arrow equals negative fraction numerator open parentheses c o t x close parentheses to the power of n plus 1 end exponent over denominator n plus 1 end fraction plus c end style

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Chapter 19 Indefinite Integrals Exercise Ex. 19.12

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Evaluate the following integral:

∫ sin⁵x cosx dxSolution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Chapter 19 Indefinite Integrals Exercise Ex. 19.14

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 19 Indefinite Integrals Exercise Ex. 19.15

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Chapter 19 – Indefinite Integrals Exercise Ex. 19.16

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Chapter 19 Indefinite Integrals Exercise Ex. 19.17

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Chapter 19 Indefinite Integrals Exercise Ex. 19.18

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Chapter 19 Indefinite Integrals Exercise Ex. 19.19

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Evaluate the following integrals:

Solution 10

Question 11

Solution 11

Question 12

Evaluate the following integrals:

Solution 12

Chapter 19 Indefinite Integrals Exercise Ex. 19.20

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 19 – Indefinite Integrals Exercise Ex. 19.32

Question 3

Solution 3

Question 1

Solution 1

Question 2

Solution 2

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Chapter 19 Indefinite Integrals Exercise Ex. 19.21

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Integrate the function

begin mathsize 12px style fraction numerator x minus 1 over denominator square root of x squared plus 1 end root end fraction end style

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Evaluate $begin mathsize 12px style integral square root of fraction numerator 1 minus x over denominator 1 plus x end fraction end root d x end style$ Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Evaluate the following integral

integral fraction numerator 5 x plus 3 over denominator square root of x squared plus 4 x plus 10 end root end fraction d x

Solution 17

Question 18

Evaluate the following integral:

Solution 18

Chapter 19 Indefinite Integrals Exercise Ex. 19.22

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Chapter 19 Indefinite Integrals Exercise Ex. 19.23

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Chapter 19 Indefinite Integrals Exercise Ex. 19.24

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Chapter 19 Indefinite Integrals Exercise Ex. 19.25

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Find ∫xe^xdxSolution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Evaluate the following integral:

Solution 27

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

begin mathsize 12px style Evaluate space the space integrals colon
integral sin to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction close parentheses dx end style

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 54

Solution 54

Question 55

Solution 55

Question 56

Solution 56

Question 57

Solution 57

Question 58

begin mathsize 12px style Evaluate space the space integrals colon
integral tan to the power of negative 1 end exponent square root of fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction end root dx end style

Solution 58

Question 59

Solution 59

Question 60

Solution 60

Question 61

Solution 61

Chapter 19 Indefinite Integrals Exercise Ex. 19.26

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

begin mathsize 12px style Evaluagte space the space following space integrals colon
integral straight e to the power of straight x open parentheses fraction numerator straight x minus 1 over denominator 2 straight x squared end fraction close parentheses dx end style

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Evaluate the following integral:

Solution 24

Chapter 19 Indefinite Integrals Exercise Ex. 19.27

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Chapter 19 Indefinite Integrals Exercise Ex. 19.28

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Chapter 19 Indefinite Integrals Exercise Ex. 19.29

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Evaluate the following integral:

Solution 11

Question 12

Evaluate the following integral:

Solution 12

Chapter 19 Indefinite Integrals Exercise Ex. 19.30

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 15

Solution 15

Question 16

Evaluate the following integral:

Solution 16

Question 17

Solution 17

Question 18

Evaluate the following integral:

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Evaluate the following integral:

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Evaluate the following integral:

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Question 50

Evaluate the following integral:

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

Solution 54

Question 55

Solution 55

Question 59

Solution 59

Question 60

Solution 60

Question 61

Solution 61

Question 62

Solution 62

Question 63

Solution 63

Question 64

Solution 64

Question 65

Solution 65

Chapter 19 – Indefinite Integrals Exercise Ex. 19.31

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Evaluate the following integral:

Solution 11

Chapter 19 – Indefinite Integrals Exercise MCQ

Question 1

Mark the correct alternative in each of the following:

begin mathsize 12px style integral fraction numerator straight x over denominator 4 plus straight x to the power of 4 end fraction dx space is space equal space to

open parentheses straight a close parentheses space 1 fourth tan to the power of negative 1 end exponent straight x squared space plus space straight C
open parentheses straight b close parentheses space 1 fourth tan to the power of negative 1 end exponent open parentheses straight x squared over 2 close parentheses
open parentheses straight c close parentheses space 1 half tan to the power of negative 1 end exponent open parentheses straight x squared over 2 close parentheses
open parentheses straight d close parentheses space none space of space these end style

Solution 1

begin mathsize 12px style Correct space option colon thin space left parenthesis straight b right parenthesis
straight I equals integral fraction numerator straight x over denominator 4 plus straight x to the power of 4 end fraction dx space
Put space straight x squared equals straight t
rightwards double arrow 2 xdx equals dt
rightwards double arrow xdx equals dt over 2
straight I equals integral fraction numerator begin display style dt over 2 end style over denominator 4 plus straight t squared end fraction
straight I equals 1 half tan to the power of negative 1 end exponent open parentheses straight t over 2 close parentheses plus straight C
straight I equals 1 half tan to the power of negative 1 end exponent open parentheses straight x squared over 2 close parentheses plus straight C
end style

Question 2

Mark the correct alternative in each of the following:

begin mathsize 12px style integral fraction numerator 1 over denominator cos space straight x space plus square root of 3 space sin space straight x end fraction dx space is space equal space to

open parentheses straight a close parentheses space log space tan open parentheses straight pi over 3 plus straight x over 2 close parentheses plus straight C
open parentheses straight b close parentheses space log space tan open parentheses straight x over 2 minus straight pi over 3 close parentheses plus straight C
open parentheses straight c close parentheses space 1 half space log space tan open parentheses straight x over 2 plus straight pi over 3 close parentheses plus straight C
open parentheses straight d close parentheses space none space of space these end style

Solution 2

begin mathsize 12px style Correct space option colon left parenthesis straight c right parenthesis
straight I equals integral fraction numerator 1 over denominator cosx plus square root of 3 sinx end fraction dx
straight I equals 1 half integral fraction numerator 1 over denominator begin display style cosx over 2 end style plus begin display style fraction numerator square root of 3 over denominator 2 end fraction end style sinx end fraction dx
straight I equals 1 half integral fraction numerator 1 over denominator cos open parentheses straight x minus begin display style straight pi over 6 end style close parentheses end fraction dx
straight I equals 1 half integral sec open parentheses straight x minus straight pi over 6 close parentheses dx
straight I equals 1 half ln open vertical bar tan open parentheses straight x over 2 plus straight pi over 3 close parentheses close vertical bar plus straight C end style

Question 3

Mark the correct alternative in each of the following:

begin mathsize 12px style integral straight x space sec space straight x squared space dx space is space equal space to

open parentheses straight a close parentheses space 1 half log open parentheses sec space straight x squared space plus space tan space straight x squared close parentheses plus straight C
open parentheses straight b close parentheses space straight x squared over 2 log open parentheses sec space straight x squared space plus space tan space straight x squared close parentheses plus straight C
open parentheses straight c close parentheses space 2 space log open parentheses sec space straight x squared space plus space tan space straight x squared close parentheses plus straight C

open parentheses straight d close parentheses space none space of space these
end style

Solution 3

begin mathsize 12px style Correct space option colon left parenthesis straight a right parenthesis
straight I equals integral xsecx squared dx
Put space straight x squared equals straight t rightwards double arrow straight x equals square root of straight t
rightwards double arrow 2 xdx equals dt
rightwards double arrow xdx equals dt over 2
straight I equals integral sect dt over 2
straight I equals 1 half log open vertical bar sect plus tant close vertical bar plus straight C
straight I equals 1 half log open vertical bar sec straight x squared plus tan straight x squared close vertical bar plus straight C
end style

Question 4

begin mathsize 12px style If space integral fraction numerator 1 over denominator 5 plus 4 space sin space straight x end fraction dx equals straight A space tan to the power of negative 1 end exponent open parentheses straight B space tan straight x over 2 plus 4 over 3 close parentheses plus straight C comma space then

open parentheses straight a close parentheses space straight A equals 2 over 3 comma space straight B equals 5 over 3
open parentheses straight b close parentheses space straight A equals 1 third comma space straight B equals 2 over 3
open parentheses straight c close parentheses space straight A equals negative 2 over 3 comma space straight B equals 5 over 3
open parentheses straight d close parentheses space straight A equals 1 third comma space straight B equals negative 5 over 3 end style

Solution 4

begin mathsize 12px style Correct space option colon space left parenthesis straight a right parenthesis
integral fraction numerator 1 over denominator 5 plus 4 sinx end fraction dx equals Atan to the power of negative 1 end exponent open parentheses Btan straight x over 2 plus 4 over 3 close parentheses plus straight C
Put space tan straight x over 2 equals straight t rightwards double arrow straight x equals 2 tan to the power of negative 1 end exponent straight t
rightwards double arrow dx equals fraction numerator 2 dt over denominator 1 plus straight t squared end fraction
rightwards double arrow sinx equals fraction numerator 2 tan begin display style straight x over 2 end style over denominator 1 plus tan squared straight x over 2 end fraction equals fraction numerator 2 straight t over denominator 1 plus straight t squared end fraction
integral fraction numerator 1 over denominator 5 plus 4 sinx end fraction dx
equals integral fraction numerator begin display style fraction numerator 2 dt over denominator 1 plus straight t squared end fraction end style over denominator 5 plus 4 cross times fraction numerator 2 straight t over denominator 1 plus straight t squared end fraction end fraction
equals integral fraction numerator 2 dt over denominator 5 straight t squared plus 8 straight t plus 5 end fraction
equals 2 over 5 integral fraction numerator dt over denominator straight t squared plus begin display style 8 over 5 end style straight t plus 1 end fraction
Using space completing space square space method
we space get
straight I equals 2 over 3 tan to the power of negative 1 end exponent open parentheses 5 over 3 tan straight x over 2 plus 4 over 3 close parentheses plus straight C
straight A equals 2 over 3 space and space straight B equals 5 over 3
end style

Question 5

begin mathsize 12px style integral straight x to the power of sin space straight x end exponent open parentheses fraction numerator sin space straight x over denominator straight x end fraction plus cos space straight x. log space straight x close parentheses dx space is space equal space to

open parentheses straight a close parentheses space straight x to the power of sin space straight x end exponent plus straight C

open parentheses straight b close parentheses space straight x to the power of sin space straight x end exponent space cos space straight x plus straight C
open parentheses straight c close parentheses space open parentheses straight x to the power of sin space straight x end exponent close parentheses squared over 2 plus straight C
open parentheses straight d close parentheses space none space of space these end style

Solution 5

begin mathsize 12px style Correct space option colon thin space left parenthesis straight a right parenthesis
straight I equals integral straight x to the power of sinx open parentheses sinx over straight x plus cosxlogx close parentheses dx
Put space straight x to the power of sinx equals straight t
taking space log space on space both space sides comma
logt equals sinxlogx
1 over straight t dt equals sinx over straight x plus cosxlogx
rightwards double arrow straight I equals integral straight t cross times dt over straight t
straight I equals straight t plus straight C
straight I equals straight x to the power of sinx plus straight C
end style

Question 6

begin mathsize 12px style Integration space of space fraction numerator 1 over denominator 1 plus open parentheses log subscript straight e straight x close parentheses squared end fraction with space respect space to space log subscript straight e straight x space is

open parentheses straight a close parentheses space fraction numerator tan to the power of negative 1 end exponent open parentheses log subscript straight e straight x close parentheses over denominator straight x end fraction plus straight C
open parentheses straight b close parentheses space tan to the power of negative 1 end exponent open parentheses log subscript straight e straight x close parentheses space plus space straight C
open parentheses straight c close parentheses space fraction numerator tan to the power of negative 1 end exponent straight x over denominator straight x end fraction plus straight C
open parentheses straight d close parentheses space none space of space these end style

Solution 6

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
integral fraction numerator 1 over denominator 1 plus open parentheses log subscript straight e straight x close parentheses squared end fraction straight d open parentheses log subscript straight e straight x close parentheses
Put space log subscript straight e straight x equals straight t
integral fraction numerator dt over denominator 1 plus straight t squared end fraction equals tan to the power of negative 1 end exponent straight t plus straight C equals tan to the power of negative 1 end exponent open parentheses space log subscript straight e straight x close parentheses plus straight C
end style

Question 7

begin mathsize 12px style If space integral fraction numerator cos space 8 straight x plus 1 over denominator tan space 2 straight x space minus space cot space 2 straight x end fraction dx equals straight a space cos space 8 straight x space plus space straight C comma space then space straight a equals

open parentheses straight a close parentheses space minus 1 over 16
open parentheses straight b close parentheses space 1 over 8
open parentheses straight c close parentheses space 1 over 16
open parentheses straight d close parentheses space minus 1 over 8 end style

Solution 7

begin mathsize 12px style Correct space option colon thin space left parenthesis straight c right parenthesis
integral fraction numerator cos 8 straight x plus 1 over denominator tan 2 straight x minus cot 2 straight x end fraction dx
equals integral fraction numerator 2 cos squared 4 straight x over denominator begin display style fraction numerator sin 2 straight x over denominator cos 2 straight x end fraction end style minus begin display style fraction numerator cos 2 straight x over denominator sin 2 straight x end fraction end style end fraction dx
equals integral fraction numerator 2 cos squared 4 straight x over denominator sin squared 2 straight x minus cos squared 2 straight x end fraction cross times sin 2 xcos 2 xdx
equals integral negative fraction numerator cos squared 4 xsin 4 straight x over denominator cos 4 straight x end fraction dx
equals fraction numerator negative 1 over denominator 2 end fraction integral sin 8 xdx
equals fraction numerator cos 8 straight x over denominator 16 end fraction plus straight C
straight a equals 1 over 16 end style

Question 8

begin mathsize 12px style If space integral fraction numerator sin to the power of 8 space straight x space minus space cos to the power of 8 space straight x over denominator 1 minus 2 space sin squared space straight x space cos squared space straight x end fraction dx equals straight a space sin space 2 straight x space plus space straight C comma space then space straight a equals

open parentheses straight a close parentheses space minus 1 divided by 2

open parentheses straight b close parentheses space 1 divided by 2

open parentheses straight c close parentheses space minus 1

open parentheses straight d close parentheses space 1 end style

Solution 8

begin mathsize 12px style Correct space option colon left parenthesis straight a right parenthesis
integral fraction numerator sin to the power of 8 straight x minus cos to the power of 8 straight x over denominator 1 minus 2 sin squared xcos squared straight x end fraction dx
integral fraction numerator open parentheses sin to the power of 4 straight x plus cos to the power of 4 straight x close parentheses open parentheses sin to the power of 4 straight x minus cos to the power of 4 straight x close parentheses over denominator 1 minus 2 sin squared xcos squared straight x end fraction dx
integral fraction numerator open parentheses sin to the power of 4 straight x plus cos to the power of 4 straight x close parentheses open parentheses sin squared straight x plus cos squared straight x close parentheses open parentheses sin squared straight x minus cos squared straight x close parentheses over denominator open parentheses sin squared straight x plus cos squared straight x close parentheses squared minus 2 sin squared xcos squared straight x end fraction dx
integral fraction numerator open parentheses sin to the power of 4 straight x plus cos to the power of 4 straight x close parentheses open parentheses sin squared straight x minus cos squared straight x close parentheses over denominator open parentheses sin to the power of 4 straight x plus cos to the power of 4 straight x close parentheses end fraction dx
integral negative cos 2 xdx
fraction numerator negative sin 2 straight x over denominator 2 end fraction plus straight C
rightwards double arrow straight a equals fraction numerator negative 1 over denominator 2 end fraction end style

Question 9

begin mathsize 11px style integral open parentheses straight x minus 1 close parentheses straight e to the power of negative straight x end exponent space dx space is space equal space to

open parentheses straight a close parentheses space minus xe to the power of straight x plus straight C
open parentheses straight b close parentheses space xe to the power of straight x space plus space straight C
open parentheses straight c close parentheses space minus xe to the power of negative straight x end exponent plus straight C
open parentheses straight d close parentheses space xe to the power of negative straight x end exponent plus straight C
end style

Solution 9

begin mathsize 12px style Correct space option colon space left parenthesis straight a right parenthesis
integral open parentheses straight x minus 1 close parentheses straight e to the power of negative straight x end exponent dx
equals open parentheses straight x minus 1 close parentheses integral straight e to the power of negative straight x end exponent dx minus integral open parentheses open square brackets fraction numerator straight d open parentheses straight x minus 1 close parentheses over denominator dx end fraction close square brackets integral straight e to the power of negative straight x end exponent dx close parentheses dx
equals open parentheses straight x minus 1 close parentheses fraction numerator straight e to the power of negative straight x end exponent over denominator negative 1 end fraction minus integral fraction numerator straight e to the power of negative straight x end exponent over denominator negative 1 end fraction dx
equals negative open parentheses straight x minus 1 close parentheses straight e to the power of negative straight x end exponent plus fraction numerator straight e to the power of negative straight x end exponent over denominator negative 1 end fraction plus straight C
equals negative xe to the power of negative straight x end exponent plus straight e to the power of negative straight x end exponent minus straight e to the power of negative straight x end exponent plus straight C
equals negative xe to the power of negative straight x end exponent plus straight C
end style

Question 10

begin mathsize 11px style If space integral 2 to the power of 1 divided by straight x end exponent over straight x squared dx space equals space straight k space 2 to the power of 1 divided by straight x end exponent plus straight C comma space then space straight k space is space equal space to
open parentheses straight a close parentheses space minus fraction numerator 1 over denominator log subscript straight e 2 end fraction
open parentheses straight b close parentheses space minus log subscript straight e 2
open parentheses straight c close parentheses space minus 1
open parentheses straight d close parentheses space 1 half end style

Solution 10

begin mathsize 12px style Correct space option colon thin space left parenthesis straight a right parenthesis
straight I equals integral 2 to the power of begin display style 1 over straight x end style end exponent over straight x squared dx
Put space 1 over straight x equals straight t
rightwards double arrow fraction numerator negative 1 over denominator straight x squared end fraction dx equals dt rightwards double arrow 1 over straight x squared dx equals negative dt
straight I equals integral 2 to the power of straight t open parentheses negative dt close parentheses
straight I equals fraction numerator negative 2 to the power of straight t over denominator log subscript straight e 2 end fraction plus straight C
rightwards double arrow straight I equals fraction numerator negative 2 to the power of begin display style 1 over straight x end style end exponent over denominator log subscript straight e 2 end fraction plus straight C
straight k equals fraction numerator negative 1 over denominator log subscript straight e 2 end fraction
end style

Question 11

begin mathsize 11px style integral fraction numerator 1 over denominator 1 plus tan space straight x end fraction dx equals

open parentheses straight a close parentheses space log subscript straight e open parentheses straight x plus sin space straight x close parentheses plus straight C
open parentheses straight b close parentheses space log subscript straight e open parentheses sin space straight x plus cos space straight x close parentheses plus straight C
open parentheses straight c close parentheses space 2 space sec squared straight x over 2 plus straight C
open parentheses straight d close parentheses space 1 half open curly brackets straight x space plus space log open parentheses sin space straight x space plus space cos space straight x close parentheses close curly brackets space plus space straight C end style

Solution 11

begin mathsize 12px style Correct space option colon space left parenthesis straight d right parenthesis
straight I equals integral fraction numerator 1 over denominator 1 plus tanx end fraction dx
equals integral fraction numerator cosx over denominator sinx plus cosx end fraction dx
Numerator space can space be space written space as colon
cosx equals straight A open parentheses sinx plus cosx close parentheses plus straight B fraction numerator straight d open parentheses sinx plus cosx close parentheses over denominator dx end fraction
cosx equals open parentheses straight A minus straight B close parentheses sinx plus open parentheses straight A plus straight B close parentheses cosx
rightwards double arrow straight A minus straight B equals 0 space and space straight A plus straight B equals 1
rightwards double arrow straight A equals 1 half equals straight B
straight I equals integral fraction numerator open square brackets begin display style 1 half end style open parentheses sinx plus cosx close parentheses plus 1 half open parentheses cosx minus sinx close parentheses close square brackets dx over denominator sinx plus cosx end fraction
straight I equals 1 half integral open parentheses 1 plus fraction numerator cosx minus sinx over denominator sinx plus cosx end fraction close parentheses dx
straight I equals 1 half open square brackets 1 plus ln open parentheses sinx plus cosx close parentheses close square brackets plus straight C end style

Question 12

begin mathsize 12px style integral open vertical bar straight x close vertical bar cubed space dx space is space equal space to

open parentheses straight a close parentheses space fraction numerator negative straight x to the power of 4 over denominator 4 end fraction plus straight C
open parentheses straight b close parentheses space open vertical bar straight x close vertical bar to the power of 4 over 4 plus straight C
open parentheses straight c close parentheses space straight x to the power of 4 over 4 plus straight C
open parentheses straight d close parentheses space none space of space these end style

Solution 12

begin mathsize 12px style Correct space option colon thin space left parenthesis straight d right parenthesis
integral open vertical bar straight x close vertical bar cubed dx
If space straight x greater than 0
rightwards double arrow integral straight x cubed dx
equals straight x to the power of 4 over 4 plus straight C
If space straight x less than 0
rightwards double arrow integral negative straight x cubed dx
equals negative straight x to the power of 4 over 4 plus straight C end style

Question 13

begin mathsize 12px style The space value space of space integral fraction numerator cos space square root of straight x over denominator square root of straight x end fraction dx space is
open parentheses straight a close parentheses space 2 space cos square root of straight x plus straight C
open parentheses straight b close parentheses space square root of fraction numerator cos space straight x over denominator straight x end fraction end root plus straight C
open parentheses straight c close parentheses space sin space square root of straight x plus straight C
open parentheses straight d close parentheses space 2 space sin space square root of straight x plus space straight C end style

Solution 13

begin mathsize 12px style Correct space option colon thin space left parenthesis straight d right parenthesis
straight I equals integral fraction numerator cos square root of straight x over denominator square root of straight x end fraction dx
Put comma space
square root of straight x space equals straight t
fraction numerator 1 over denominator 2 square root of straight x end fraction dx equals dt
rightwards double arrow fraction numerator 1 over denominator square root of straight x end fraction dx equals 2 dt
straight I equals integral cost space 2 dt
straight I equals 2 sint plus straight C equals 2 sin square root of straight x plus straight C end style

Question 14

begin mathsize 12px style integral straight e to the power of straight x open parentheses 1 minus cot space straight x space plus space cot squared straight x close parentheses space dx equals

open parentheses straight a close parentheses space straight e to the power of straight x space cot space straight x space plus space straight C
open parentheses straight b close parentheses space minus straight e to the power of straight x space cot space straight x space plus space straight C
open parentheses straight c close parentheses space straight e to the power of straight x space cosec space straight x space plus space straight C
open parentheses straight d close parentheses space minus straight e to the power of straight x space cosec space straight x space plus space straight C end style

Solution 14

begin mathsize 12px style Correct space option colon space left parenthesis straight b right parenthesis
straight I equals integral straight e to the power of straight x open parentheses 1 minus cotx plus cot squared straight x close parentheses dx
straight I equals integral straight e to the power of straight x open parentheses 1 plus cot squared straight x minus cotx close parentheses dx
straight I equals integral straight e to the power of straight x open parentheses cosec squared straight x minus cotx close parentheses dx
Here comma space straight f left parenthesis straight x right parenthesis equals negative cotx rightwards double arrow straight f apostrophe left parenthesis straight x right parenthesis equals cosec squared straight x
straight I equals negative straight e to the power of straight x cotx plus straight C end style

Question 15

Error converting from MathML to accessible text.

Solution 15

Question 16

begin mathsize 12px style integral fraction numerator 1 over denominator 7 plus 5 cos space straight x end fraction dx equals

open parentheses straight a close parentheses space fraction numerator 1 over denominator square root of 6 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator square root of 6 end fraction tan straight x over 2 close parentheses plus straight C
open parentheses straight b close parentheses space fraction numerator begin display style 1 end style over denominator begin display style square root of 3 end style end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 1 end style over denominator begin display style square root of 3 end style end fraction tan fraction numerator begin display style straight x end style over denominator begin display style 2 end style end fraction close parentheses plus straight C
open parentheses straight c close parentheses space fraction numerator begin display style 1 end style over denominator begin display style 4 end style end fraction tan to the power of negative 1 end exponent open parentheses tan fraction numerator begin display style straight x end style over denominator begin display style 2 end style end fraction close parentheses plus straight C
open parentheses straight d close parentheses space fraction numerator begin display style 1 end style over denominator begin display style 7 end style end fraction tan to the power of negative 1 end exponent open parentheses tan fraction numerator begin display style straight x end style over denominator begin display style 2 end style end fraction close parentheses plus straight C end style

Solution 16

begin mathsize 12px style Correct space option colon space left parenthesis straight a right parenthesis
straight I equals integral fraction numerator dx over denominator 7 plus 5 cosx end fraction
put space tan straight x over 2 equals straight t rightwards double arrow dx equals fraction numerator 2 dt over denominator 1 plus straight t squared end fraction
rightwards double arrow cosx equals fraction numerator 1 minus tan squared begin display style straight x over 2 end style over denominator 1 plus tan squared straight x over 2 end fraction equals fraction numerator 1 minus straight t squared over denominator 1 plus straight t squared end fraction
straight I equals integral fraction numerator fraction numerator 2 dt over denominator 1 plus straight t squared end fraction over denominator 7 plus 5 cross times fraction numerator 1 minus straight t squared over denominator 1 plus straight t squared end fraction end fraction
rightwards double arrow straight I equals integral fraction numerator 1 over denominator straight t squared plus 6 end fraction dt
straight I equals fraction numerator 1 over denominator square root of 6 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator straight t over denominator square root of 6 end fraction close parentheses plus straight C
rightwards double arrow fraction numerator 1 over denominator square root of 6 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator tan begin display style straight x over 2 end style over denominator square root of 6 end fraction close parentheses plus straight C
end style

Question 17

begin mathsize 12px style integral fraction numerator 1 over denominator 1 minus cosx minus sinx end fraction dx equals
open parentheses straight a close parentheses space log space open vertical bar 1 plus cot straight x over 2 close vertical bar plus straight C
open parentheses straight b close parentheses space log space open vertical bar 1 minus tan fraction numerator begin display style straight x end style over denominator begin display style 2 end style end fraction close vertical bar plus straight C
open parentheses straight c close parentheses space log space open vertical bar 1 minus cot fraction numerator begin display style straight x end style over denominator begin display style 2 end style end fraction close vertical bar plus straight C
open parentheses straight d close parentheses space log space open vertical bar 1 plus tan fraction numerator begin display style straight x end style over denominator begin display style 2 end style end fraction close vertical bar plus straight C end style

Solution 17

begin mathsize 12px style Correct space option colon left parenthesis straight c right parenthesis
integral fraction numerator dx over denominator 1 minus cosx minus sinx end fraction
Put space straight t equals space tan straight x over 2 rightwards double arrow dx equals fraction numerator 2 dt over denominator 1 plus straight t squared end fraction
cosx equals fraction numerator 1 minus tan squared straight x over 2 over denominator 1 plus tan squared straight x over 2 end fraction equals fraction numerator 1 minus straight t squared over denominator 1 plus straight t squared end fraction
sinx equals fraction numerator 2 tan straight x over 2 over denominator 1 plus tan squared straight x over 2 end fraction equals fraction numerator 2 straight t over denominator 1 plus straight t squared end fraction
put space in space the space question
straight I equals integral fraction numerator begin display style fraction numerator 2 dt over denominator 1 plus straight t squared end fraction end style over denominator 1 minus fraction numerator 1 minus straight t squared over denominator 1 plus straight t squared end fraction minus fraction numerator 2 straight t over denominator 1 plus straight t squared end fraction end fraction
straight I equals integral fraction numerator dt over denominator straight t squared minus straight t end fraction
straight I equals integral fraction numerator dt over denominator straight t squared minus straight t plus begin display style 1 fourth end style minus begin display style 1 fourth end style end fraction
rightwards double arrow straight I equals ln open vertical bar 1 minus cot straight x over 2 close vertical bar plus straight C

end style

Question 18

begin mathsize 12px style integral fraction numerator straight x plus 3 over denominator open parentheses straight x plus 4 close parentheses squared end fraction straight e to the power of straight x dx equals
open parentheses straight a close parentheses space fraction numerator straight e to the power of straight x over denominator straight x plus 4 end fraction plus straight C
open parentheses straight b close parentheses space fraction numerator begin display style straight e to the power of straight x end style over denominator begin display style straight x plus 3 end style end fraction plus straight C
open parentheses straight c close parentheses space fraction numerator begin display style 1 end style over denominator begin display style open parentheses straight x plus 4 close parentheses squared end style end fraction plus straight C
open parentheses straight d close parentheses space fraction numerator begin display style straight e to the power of straight x end style over denominator begin display style open parentheses straight x plus 4 close parentheses squared end style end fraction plus straight C end style

Solution 18

begin mathsize 12px style Correct space option colon thin space left parenthesis straight a right parenthesis
straight I equals integral fraction numerator straight x plus 3 over denominator open parentheses straight x plus 4 close parentheses squared end fraction straight e to the power of straight x dx
straight I equals integral open parentheses fraction numerator straight x plus 4 minus 1 over denominator open parentheses straight x plus 4 close parentheses squared end fraction close parentheses straight e to the power of straight x dx
straight I equals integral open parentheses fraction numerator 1 over denominator straight x plus 4 end fraction minus 1 over open parentheses straight x plus 4 close parentheses squared close parentheses straight e to the power of straight x dx
straight f left parenthesis straight x right parenthesis equals fraction numerator 1 over denominator straight x plus 4 end fraction rightwards double arrow straight f apostrophe left parenthesis straight x right parenthesis equals negative 1 over open parentheses straight x plus 4 close parentheses squared
rightwards double arrow straight I equals fraction numerator straight e to the power of straight x over denominator straight x plus 4 end fraction plus straight C
end style

Question 19

begin mathsize 12px style integral fraction numerator sin space straight x over denominator 3 space plus space 4 space cos squared straight x end fraction dx
open parentheses straight a close parentheses space log space open parentheses 3 plus 4 cos squared straight x close parentheses plus straight C
open parentheses straight b close parentheses space fraction numerator 1 over denominator 2 square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator cos begin display style space end style begin display style straight x end style over denominator square root of 3 end fraction close parentheses plus straight C
open parentheses straight c close parentheses space minus fraction numerator begin display style 1 end style over denominator begin display style 2 square root of 3 end style end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 2 space cos space straight x end style over denominator square root of 3 end fraction close parentheses plus straight C
open parentheses straight d close parentheses space fraction numerator begin display style 1 end style over denominator begin display style 2 square root of 3 end style end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 2 space cos space straight x end style over denominator square root of 3 end fraction close parentheses plus straight C end style

Solution 19

begin mathsize 12px style Correct space option colon thin space left parenthesis straight c right parenthesis
straight I equals integral fraction numerator sinx over denominator 3 plus 4 cos squared straight x end fraction dx
Put space cosx equals straight t
rightwards double arrow negative sinxdx equals dt
rightwards double arrow sinxdx equals negative dt
straight I equals integral fraction numerator negative dt over denominator 3 plus 4 straight t squared end fraction
straight I equals fraction numerator negative 1 over denominator 2 square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight t over denominator square root of 3 end fraction close parentheses plus straight C
straight I equals fraction numerator negative 1 over denominator 2 square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator 2 cosx over denominator square root of 3 end fraction close parentheses plus straight C end style

Question 20

begin mathsize 12px style integral straight e to the power of straight x open parentheses fraction numerator 1 minus sin space straight x over denominator 1 minus cos space straight x end fraction close parentheses dx
open parentheses straight a close parentheses space minus straight e to the power of straight x space tan space straight x over 2 plus straight C
open parentheses straight b close parentheses space minus straight e to the power of straight x space cot space straight x over 2 plus straight C
open parentheses straight c close parentheses space minus 1 half space straight e to the power of straight x space tan space straight x over 2 plus straight C
open parentheses straight a close parentheses space minus 1 half space straight e to the power of straight x space cot space straight x over 2 plus straight C end style

Solution 20

Question 21

begin mathsize 12px style integral 2 over open parentheses straight e to the power of straight x space plus space straight e to the power of negative straight x end exponent close parentheses squared dx
open parentheses straight a close parentheses space fraction numerator negative straight e to the power of negative straight x end exponent over denominator straight e to the power of straight x plus straight e to the power of negative straight x end exponent end fraction plus straight C
open parentheses straight b close parentheses space minus fraction numerator begin display style 1 end style over denominator begin display style straight e to the power of straight x plus straight e to the power of negative straight x end exponent end style end fraction plus straight C
open parentheses straight c close parentheses space fraction numerator begin display style negative 1 end style over denominator begin display style open parentheses straight e to the power of straight x plus space 1 close parentheses squared end style end fraction plus straight C
open parentheses straight d close parentheses space fraction numerator begin display style 1 end style over denominator begin display style straight e to the power of straight x plus straight e to the power of negative straight x end exponent end style end fraction plus straight C end style

Solution 21

Question 22

begin mathsize 12px style integral fraction numerator straight e to the power of straight x open parentheses 1 plus straight x close parentheses over denominator cos squared open parentheses xe to the power of straight x close parentheses end fraction dx equals

open parentheses straight a close parentheses space 2 space log subscript straight e cos open parentheses xe to the power of straight x close parentheses plus straight C
open parentheses straight b close parentheses space sec open parentheses xe to the power of straight x close parentheses plus straight C
open parentheses straight c close parentheses space tan open parentheses xe to the power of straight x close parentheses plus straight C
open parentheses straight d close parentheses space tan open parentheses straight x plus straight e to the power of straight x close parentheses plus straight C end style

Solution 22

Question 23

begin mathsize 12px style integral fraction numerator sin squared straight x over denominator cos to the power of 4 straight x end fraction dx equals

open parentheses straight a close parentheses space 1 third space tan squared straight x space plus space straight C
open parentheses straight b close parentheses space 1 half space tan squared straight x space plus space straight C
open parentheses straight c close parentheses space 1 third space tan cubed straight x space plus space straight C
open parentheses straight d close parentheses space none space of space these end style

Solution 23

Question 24

begin mathsize 12px style The space primitive space of space the space function space straight f left parenthesis straight x right parenthesis space equals space open parentheses 1 minus 1 over straight x squared close parentheses straight a to the power of straight x plus 1 over straight x end exponent comma space straight a space greater than space 0 space is
open parentheses straight a close parentheses space fraction numerator straight a to the power of straight x plus begin display style 1 over straight x end style end exponent over denominator log subscript straight e straight a end fraction
open parentheses straight b close parentheses space log subscript straight e straight a times straight a to the power of straight x plus 1 over straight x end exponent
open parentheses straight c close parentheses space straight a to the power of straight x plus begin display style 1 over straight x end style end exponent over straight x log subscript straight e straight a
open parentheses straight d close parentheses space straight x fraction numerator straight a to the power of straight x plus begin display style 1 over straight x end style end exponent over denominator log subscript straight e straight a end fraction end style

Solution 24

begin mathsize 12px style Correct space option colon left parenthesis straight a right parenthesis
straight f left parenthesis straight x right parenthesis equals open parentheses 1 minus 1 over straight x squared close parentheses straight a to the power of straight x plus 1 over straight x end exponent
rightwards double arrow integral straight f left parenthesis straight x right parenthesis dx equals integral open parentheses 1 minus 1 over straight x squared close parentheses straight a to the power of straight x plus 1 over straight x end exponent dx
Put space straight x plus 1 over straight x equals straight t
rightwards double arrow open parentheses 1 minus 1 over straight x squared close parentheses dx equals dt
straight I equals integral straight a to the power of straight t dt
straight I equals fraction numerator straight a to the power of straight t over denominator log subscript straight e straight a end fraction plus straight C
straight I equals fraction numerator straight a to the power of straight x plus 1 over straight x end exponent over denominator log subscript straight e straight a end fraction plus straight C end style

Question 25

begin mathsize 12px style The space value space of space integral fraction numerator 1 over denominator straight x plus straight x space log space straight x end fraction dx space is
open parentheses straight a close parentheses space 1 space plus space log space straight x
open parentheses straight b close parentheses space straight x space plus space log space straight x
open parentheses straight c close parentheses space straight x space log space open parentheses 1 space plus space log space straight x close parentheses
open parentheses straight d close parentheses space log open parentheses 1 space plus space log space straight x close parentheses end style

Solution 25

Question 26

begin mathsize 12px style integral square root of fraction numerator straight x over denominator 1 minus straight x end fraction end root dx space is space equal space to
open parentheses straight a close parentheses space sin to the power of negative 1 end exponent square root of straight x plus straight C
open parentheses straight b close parentheses space sin to the power of negative 1 end exponent open curly brackets square root of straight x minus square root of straight x left parenthesis 1 minus straight x right parenthesis end root close curly brackets plus straight C
open parentheses straight c close parentheses space sin to the power of negative 1 end exponent open curly brackets square root of straight x left parenthesis 1 minus straight x right parenthesis end root close curly brackets plus straight C
open parentheses straight d close parentheses space sin to the power of negative 1 end exponent square root of straight x minus square root of straight x left parenthesis 1 minus straight x right parenthesis end root plus straight C end style

Solution 26

begin mathsize 12px style Correct space option colon space left parenthesis straight d right parenthesis
straight I equals integral square root of fraction numerator straight x over denominator 1 minus straight x end fraction end root dx
straight I equals integral square root of fraction numerator straight x over denominator 1 minus straight x end fraction cross times straight x over straight x end root dx
straight I equals integral fraction numerator xdx over denominator square root of straight x minus straight x squared end root end fraction
Consider comma
straight x equals straight A fraction numerator straight d open parentheses straight x minus straight x squared close parentheses over denominator dx end fraction plus straight B
straight x equals straight A open parentheses 1 minus 2 straight x close parentheses plus straight B
straight x equals negative 2 Ax plus straight A plus straight B
minus 2 straight A equals 1 rightwards double arrow straight A equals fraction numerator negative 1 over denominator 2 end fraction
rightwards double arrow straight A plus straight B equals 0 rightwards double arrow straight B equals 1 half
straight I equals integral fraction numerator fraction numerator negative 1 over denominator 2 end fraction open parentheses 1 minus 2 straight x close parentheses plus 1 half over denominator square root of straight x minus straight x squared end root end fraction dx
straight I equals integral open parentheses fraction numerator negative 1 over denominator 2 end fraction fraction numerator 1 minus 2 straight x over denominator square root of straight x minus straight x squared end root end fraction plus fraction numerator 1 over denominator 2 square root of straight x minus straight x squared end root end fraction close parentheses dx
straight I equals fraction numerator negative 1 over denominator 2 end fraction cross times 2 square root of straight x minus straight x squared end root plus 1 half integral fraction numerator 1 over denominator square root of straight x minus straight x squared end root end fraction dx
Second space term space after space completing space square space method space you space will space get space as
straight I equals negative square root of straight x minus straight x squared end root plus sin to the power of negative 1 end exponent square root of straight x plus straight C end style

Question 27

begin mathsize 12px style integral straight e to the power of straight x open curly brackets straight f open parentheses straight x close parentheses space plus space straight f apostrophe open parentheses straight x close parentheses close curly brackets dx space equals

open parentheses straight a close parentheses space straight e to the power of straight x space straight f open parentheses straight x close parentheses space plus space straight C
open parentheses straight b close parentheses space straight e to the power of straight x space plus space straight f open parentheses straight x close parentheses space plus space straight C
open parentheses straight c close parentheses space 2 straight e to the power of straight x space straight f open parentheses straight x close parentheses space plus space straight C
open parentheses straight d close parentheses space straight e to the power of straight x space minus space straight f open parentheses straight x close parentheses space plus space straight C end style

Solution 27

begin mathsize 12px style Correct space option colon left parenthesis straight a right parenthesis
integral straight e to the power of straight x open curly brackets straight f open parentheses straight x close parentheses plus straight f apostrophe open parentheses straight x close parentheses close curly brackets dx equals straight e to the power of straight x straight f open parentheses straight x close parentheses plus straight C
end style

Question 28

begin mathsize 12px style The space value space of space integral fraction numerator sin space straight x space plus space cos space straight x over denominator square root of 1 minus sin space 2 straight x end root end fraction dx space is space equal space to
open parentheses straight a close parentheses space square root of sin space 2 straight x space end root plus space straight C
open parentheses straight b close parentheses space square root of cos 2 straight x end root space plus space straight C
open parentheses straight c close parentheses space plus-or-minus open parentheses sin space straight x space minus space cos space straight x close parentheses space plus space straight C
open parentheses straight d close parentheses space plus-or-minus space log space open parentheses sin space straight x space minus space cos space straight x close parentheses space plus space straight C end style

Solution 28

Question 29

begin mathsize 12px style If space integral straight x space sin space straight x space dx space equals space minus straight x space cos space straight x space plus space straight alpha comma space then space straight alpha space is space equal space to

open parentheses straight a close parentheses space sin space straight x space plus space straight C
open parentheses straight b close parentheses space cos space straight x space plus space straight C
open parentheses straight c close parentheses space straight C
open parentheses straight d close parentheses space none space fo space these end style

Solution 29

begin mathsize 12px style Correct space option colon space left parenthesis straight a right parenthesis
integral xsinxdx equals negative xcosx plus straight alpha
straight I equals integral xsinxdx
straight I equals straight x integral sinxdx minus integral open parentheses fraction numerator d straight x over denominator d straight x end fraction integral sinxdx close parentheses dx
straight I equals negative xcosx plus integral cosxdx
straight I equals xcosx plus sinx plus straight C
rightwards double arrow straight alpha equals sinx plus straight C
end style

Question 30

begin mathsize 12px style integral fraction numerator cos space 2 straight x space minus space 1 over denominator cos space 2 straight x space plus space 1 end fraction dx equals

open parentheses straight a close parentheses space tan space straight x space minus space straight x space plus space straight C
open parentheses straight b close parentheses space straight x space plus space tan space straight x space plus space straight C
open parentheses straight c close parentheses space straight x space minus space tan space straight x space plus space straight C
open parentheses straight d close parentheses space minus straight x space minus space cot space straight x space plus space straight C end style

Solution 30

begin mathsize 12px style Correct space option colon space left parenthesis straight c right parenthesis
straight I equals integral fraction numerator cos 2 straight x minus 1 over denominator cos 2 straight x plus 1 end fraction dx
straight I equals negative integral fraction numerator 1 minus cos 2 straight x over denominator 1 plus cos 2 straight x end fraction dx
straight I equals negative integral fraction numerator 2 sin squared straight x over denominator 2 cos squared straight x over 2 end fraction dx
straight I equals negative integral tan squared xdx
straight I equals negative integral open parentheses sec squared straight x minus 1 close parentheses dx
straight I equals negative open parentheses tanx minus straight x close parentheses plus straight C
straight I equals straight x minus tanx plus straight C end style

Question 31

begin mathsize 12px style integral fraction numerator cos space 2 straight x space minus space cos space 2 straight theta over denominator cos space straight x space minus space cos space straight theta end fraction dx space is space equal space to

open parentheses straight a close parentheses space 2 open parentheses sin space straight x space plus space straight x space cos space straight theta close parentheses space plus space straight C
open parentheses straight b close parentheses space 2 open parentheses sin space straight x space minus space straight x space cos space straight theta close parentheses space plus space straight C
open parentheses straight c close parentheses space 2 open parentheses sin space straight x space plus space 2 straight x space cos space straight theta close parentheses space plus space straight C
open parentheses straight d close parentheses space 2 open parentheses sin space straight x space minus space 2 straight x space cos space straight theta close parentheses space plus space straight C end style

Solution 31

begin mathsize 12px style Correct space option colon left parenthesis straight a right parenthesis
straight I equals integral fraction numerator cos 2 straight x minus cos 2 straight theta over denominator cosx minus cosθ end fraction dx
straight I equals integral fraction numerator 2 cos squared straight x minus 1 minus open parentheses 2 cos squared straight theta minus 1 close parentheses over denominator cosx minus cosθ end fraction dx
straight I equals integral fraction numerator 2 open parentheses cos squared straight x minus cos squared straight theta close parentheses over denominator cosx minus cosθ end fraction dx
straight I equals 2 integral open parentheses cosx plus cosθ close parentheses dx
straight I equals 2 open parentheses sinx plus xcosθ close parentheses plus straight C end style

Question 32

begin mathsize 12px style integral straight x to the power of 9 over open parentheses 4 straight x squared plus 1 close parentheses to the power of 6 dx space is space equal space to

open parentheses straight a close parentheses space fraction numerator 1 over denominator 5 straight x end fraction open parentheses 4 plus 1 over straight x squared close parentheses to the power of negative 5 end exponent plus straight C
open parentheses straight b close parentheses space fraction numerator begin display style 1 end style over denominator begin display style 5 end style end fraction open parentheses 4 plus fraction numerator begin display style 1 end style over denominator begin display style straight x squared end style end fraction close parentheses to the power of negative 5 end exponent plus straight C
open parentheses straight c close parentheses space fraction numerator begin display style 1 end style over denominator begin display style 10 straight x end style end fraction open parentheses fraction numerator begin display style 1 end style over denominator begin display style straight x squared end style end fraction plus 4 close parentheses to the power of negative 5 end exponent plus straight C
open parentheses straight d close parentheses space fraction numerator begin display style 1 end style over denominator begin display style 10 end style end fraction open parentheses fraction numerator begin display style 1 end style over denominator begin display style straight x squared end style end fraction plus 4 close parentheses to the power of negative 5 end exponent plus straight C
end style

Solution 32

begin mathsize 12px style Correct space option colon left parenthesis straight d right parenthesis
straight I equals integral straight x to the power of 9 over open parentheses 4 straight x squared plus 1 close parentheses to the power of 6 dx
straight I equals integral fraction numerator straight x to the power of 9 over denominator straight x to the power of 12 open parentheses 4 plus begin display style 1 over straight x squared end style close parentheses to the power of 6 end fraction dx
straight I equals integral fraction numerator dx over denominator straight x cubed open parentheses 4 plus 1 over straight x squared close parentheses to the power of 6 end fraction
Put space open parentheses 4 plus 1 over straight x squared close parentheses equals straight t rightwards double arrow fraction numerator negative 2 over denominator straight x cubed end fraction dx equals dt
straight I equals 1 half integral negative straight t to the power of negative 6 end exponent dt
straight I equals 1 half open parentheses fraction numerator negative straight t to the power of negative 5 end exponent over denominator negative 5 end fraction close parentheses plus straight C
straight I equals 1 over 10 1 over straight t to the power of 5 plus straight C
straight I equals 1 over 10 open parentheses 4 plus 1 over straight x squared close parentheses to the power of negative 5 end exponent plus straight C end style

Question 33

begin mathsize 12px style integral fraction numerator straight x cubed over denominator square root of 1 plus straight x squared end root end fraction dx equals straight a open parentheses 1 plus straight x squared close parentheses to the power of 3 divided by 2 end exponent space plus space straight b square root of 1 plus straight x squared end root plus straight C comma space then

open parentheses straight a close parentheses space straight a space equals space 1 third comma space straight b space equals space 1
open parentheses straight b close parentheses space straight a space equals space minus fraction numerator begin display style 1 end style over denominator begin display style 3 end style end fraction comma space straight b space equals space 1
open parentheses straight c close parentheses space straight a space equals negative fraction numerator begin display style 1 end style over denominator begin display style 3 end style end fraction comma space straight b space equals space minus 1
open parentheses straight d close parentheses space straight a space equals space fraction numerator begin display style 1 end style over denominator begin display style 3 end style end fraction comma space straight b space equals space minus 1 end style

Solution 33

Question 34

begin mathsize 12px style integral fraction numerator straight x cubed over denominator straight x plus 1 end fraction dx space is space equal space to

open parentheses straight a close parentheses space straight x space plus space straight x squared over 2 plus straight x cubed over 3 minus space log space open vertical bar 1 minus straight x close vertical bar plus straight C
open parentheses straight b close parentheses space straight x space plus space fraction numerator begin display style straight x squared end style over denominator begin display style 2 end style end fraction minus fraction numerator begin display style straight x cubed end style over denominator begin display style 3 end style end fraction minus space log space open vertical bar 1 minus straight x close vertical bar plus straight C
open parentheses straight c close parentheses space straight x space minus space fraction numerator begin display style straight x squared end style over denominator begin display style 2 end style end fraction minus fraction numerator begin display style straight x cubed end style over denominator begin display style 3 end style end fraction minus space log space open vertical bar 1 plus straight x close vertical bar plus straight C
open parentheses straight d close parentheses space straight x space minus space fraction numerator begin display style straight x squared end style over denominator begin display style 2 end style end fraction plus fraction numerator begin display style straight x cubed end style over denominator begin display style 3 end style end fraction minus space log space open vertical bar 1 plus straight x close vertical bar plus straight C
end style

Solution 34

begin mathsize 12px style Correct space option colon left parenthesis straight d right parenthesis
straight I equals integral fraction numerator straight x cubed over denominator straight x plus 1 end fraction dx
straight I equals integral fraction numerator straight x cubed plus 1 minus 1 over denominator straight x plus 1 end fraction dx
straight I equals integral fraction numerator open parentheses straight x plus 1 close parentheses open parentheses straight x squared minus straight x plus 1 close parentheses minus 1 over denominator straight x plus 1 end fraction dx
straight I equals integral open parentheses straight x squared minus straight x plus 1 minus fraction numerator 1 over denominator straight x plus 1 end fraction close parentheses dx
straight I equals straight x cubed over 3 minus straight x squared over 2 plus straight x minus log open vertical bar straight x plus 1 close vertical bar plus straight C end style

Question 35

begin mathsize 12px style If space integral fraction numerator 1 over denominator open parentheses straight x plus 2 close parentheses open parentheses straight x squared plus 1 close parentheses end fraction dx space equals space straight a space log space open vertical bar 1 plus straight x squared close vertical bar space plus space straight b space tan to the power of negative 1 end exponent straight x space plus space 1 fifth log open vertical bar straight x space plus space 2 close vertical bar space plus space straight C comma space then

open parentheses straight a close parentheses space straight a space equals space minus 1 over 10 comma space straight b space equals space minus 2 over 5
open parentheses straight b close parentheses space straight a space equals space 1 over 10 comma space straight b space equals space minus 2 over 5
open parentheses straight c close parentheses space straight a space equals space minus 1 over 10 comma space straight b space equals space 2 over 5
open parentheses straight d close parentheses space straight a space equals space 1 over 10 comma space straight b space equals space 2 over 5 end style

Solution 35

Chapter 19 – Indefinite Integrals Exercise Ex. 19VSAQ

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Write the value of ∫ e²x² + In x dxSolution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

Solution 54

Question 55

Solution 55

Question 56

Solution 56

Question 57

Solution 57

Question 58

Solution 58

Question 59

Solution 59

Question 60

Solution 60

Question 61

Solution 61

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Chapter – 6 Physical and Chemical Changes and Salts | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter 6 Physical and Chemical Changes

Q.1.Classify the changes involved in the following processes as physical or chemical changes:
(a) Photosynthesis
(b) Dissolving sugar in water
(c) Burning of coal
(d) Melting of wax
(e) Beating aluminium to make aluminium foil
(f) Digestion of food

Ans.(a) Chemical change (b) Physical change
(c) Chemical change (d) Physical change
(e) Physical change (/) Chemical change

Q.2.(a) Cutting a log of wood into pieces is a chemical change. (True/ False)
(b) Formation of manure from leaves is a physical change. (True/ False)

(c)Iron pipes coated with zinc do not get rusted easily. (True/ False)
(d)Iron and rust are the same substances. (True/ False)
(e)Condensation of steam is not a chemical change. (True/ False)
Ans. (a)False
Correct statement: Cutting a log of wood into pieces is an irreversible physical change.
(b)False
Correct statement: Formation of manure from leaves is a chemical change.
(c) True
(d)False
Correct statement: Iron and rust are two different chemical substances.
(e)True

Q.3.Fill in the blanks in the following statements:
(a) When carbon dioxide is passed through lime water, it turns milky due to the formation of .
(b) The chemical name of baking soda is .
(c) Two methods by which rusting of iron can be prevented are __________ and __________
(d) Changes in which only ____________ properties of a substance change are called physical changes.
(e) Changes in which new substances are formed are called _____________ changes..
Ans. (a)calcium carbonate
(b) sodium hydrogen carbonate
(c) painting or greasing, galvanisation
(d) physical
(e) chemical

Q.4. When baking soda is mixed with lemon juice, bubbles are formed with the evolution of a gas. What type of change is it? Explain.
Ans. The reaction between baking soda and lemon juice can be given as below:
Lemon juice + Baking soda ————-> C02 (bubbles) + Other substances
(Citric acid) (Sodium hydrogen carbonate) (Carbon dioxide)

It is a chemical change

Q.5. When a candle burns, both physical and chemical changes take place. Identify these changes. Give another example of a familiar process in which both the chemical and physical changes take place.
Ans. When a candle burns, both physical and chemical changes occur:

(i) Physical change: melting of wax, vapourisation of melted wax.
(ii) Chemical change: Burning of vapours of wax to give carbon dioxide, heat and light.
LPG is another example in which physical change occurs when LPG comes out of cylinder and is converted from liquid to gaseous state and a chemical change occurs when gas burns in air.

Q.6. How would you show that setting of a curd is a chemical change?
Ans. We can say that setting of curd is a chemical change because we can not get the original substance, i.e., milk back and a new substance is formed with different taste, smell and other chemical properties

Q.7. Explain why burning of wood and cutting it into small pieces are considered as two different types of changes. ~
Ans. Burning of wood is a chemical change because in burning new substances are formed as
Wood + Oxygen ———–> Charcoal + Carbon dioxide + Heat + Light
But cutting it into small pieces is physical change because no new substance is formed. We can only reduce the size of wood.

Q.8. Describe how crystals of copper sulphate are prepared.
Ans. Take a cupful of water in a beaker and add a few drops of dilute sulphuric acid. Heat the water. When it starts boiling, add copper sulphate powder slowly. Continue to add copper sulphate powder till no more powder can be dissolved. .During this process continuously stir the solution. Filter the solution. Leave it for cooling. Look it after some time, you can see the crystals of copper sulphate

Q.9. Explain how painting of an iron gate prevents it from rusting?
Ans. It is known that for rusting the presence of oxygen and moisture is essential. Painting prevents the iron gate from coming in contact with oxygen and moisture.

Q.10. Explain why rusting of iron objects is faster in coastal areas than in deserts.
Ans. As content of moisture in the air in coastal areas is higher than in the air in deserts. So, the process of rusting is faster in coastal areas.

Q.11. The gas we use in the kitchen is called liquified petroleum gas (LPG). In the cylinder it exists as a liquid. When it comes out from the cylinder it becomes a gas (Change- A) then it bums (Change-B). The following statements pertain to these changes. Choose the correct one.
(i) Process-A is a chemical change.
(ii) Process-B is a chemical change.
(iii) Both processes A and B are chemical changes.
(iv) None of these processes is a chemical change.
Ans. (ii) Process-B is a chemical change.

Q.12.Anaerobic bacteria digest animal waste and produce biogas (Change-A). The biogas is then burnt as fuel (Change-B). The following statements pertain to these changes. Choose the correct one.
(i) Process-A is a chemical change.
(ii) Process-B is a chemical change.
(iii) Both processes A and B are chemical changes.
(iv) None of these processes is a chemical change.
Ans.(iii) Both processes A and B are chemical change

Rohit0 comments

Chapter – 5 Acids Bases and Salts | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter 5 Acids Bases and Salts

Q.2.Ammonia is found in many household products, such as window cleaners. It turns red litmus blue. What is its nature?
Ans.Ammonia has basic nature.

Q.3.Name the source from which litmus solution is obtained. What is the use of this solution?
Ans.Litmus solution is extracted from lichens. It is used to determine whether the given solution is acidic or basic.

Q.4.Is the distilled water acidic/basic/neutral? How would you verify it?
Ans.Distilled water will be neutral. We can verify it by showing that neither blue nor red litmus paper changes its colour when dipped in it.

Q.5.Describe the process of neutralisation with the help of an example.
Ans.The reaction between an acid and a base is known as neutralisation. Salt and water are produced in this process with the evolution of heat.
Antacids like milk of magnesia (magnesium hydroxide), baking soda, etc. which contain a base are used for reducing acidity in stomach when excessive acid released by glands.

Q.6.Mark ‘T’ if the statement is true and ‘F’ if it is false:
(i) Nitric acid turns red litmus blue. (T/F)
(ii) Sodium hydroxide turns blue litmus red. {T/F)
(iii) Sodium hydroxide and hydrochloric acid neutralise each other and form salt and water. (T/F)
(id) Indicator is a substance which shows different colours in acidic and basic solutions. . (T/F)
(v) Tooth decay is caused by the presence of a base. (T/F)
Ans.(1) F (ii) F (iii) T (iv) T (V) F

Q.7. Dorji has a few bottles of soft drink in his restaurant. But, unfortunately, these are
not labelled. He has to serve the drinks on the demand of customers. One customer
wants acidic drink, another wants basic and third one wants neutral drink. How
will Dorji decide which drink is to be served to whom?
Ans.Dorji can decide with the help of litmus paper:
(i) The drink which would turn a red litmus blue would be basic.
(ii) If the drink turns a blue litmus to red would be acidic.
(iii) The drink which would not affect both red and blue litmus would be neutral.

Q.8.Explain why:
(a) An antacid tablet is taken when you suffer from acidity.
(b) Calamine solution is applied on the skin when an ant bites.
(c) Factory waste is neutralised before disposing it into the water bodies.
Ans.(a) We take an antacid such as milk of magnesia to neutralises the excessive acid released in stomach.
(b) Ant injects an acidic liquid (Formic acid) into the skin on biting which causes inflammation, to the skin. The effect of the acid can be neutralised by rubbing. Calamine solution which contains zinc carbonate which is very weak base and causes no harm to the skin.
(c) The wastes of factories contain acids. If acids are disposed off in the water body, the acids will harm the organisms. So factory wastes are neutralised by adding basic substances.

Q.9. Three liquids are given to you. One is hydrochloric acid, another is sodium hydroxide and third is a sugar solution. How will you identify them? You have only turmeric indicator.
Ans.Name of the substances Effect on turmeric indicator
1. Hydrochloric acid Yellow to blue
2. Sodium hydroxide Yellow to red
3. Sugar solution No change

Q.10. Blue litmus paper is dipped in a solution. It remains blue. What is the nature of the solution? Explain.
Ans. (i) It can be identified on the basis of the following observations : Bases change the colour of litmus paper to blue. As the colour of blue litmus paper is not affected, the solution must be basic.

(ii) If the solution is neutral, even then colour of litmus will not change.

Q. 11. Consider the following statements:
(a) Both acids and bases change colour of all indicators.
(b) If an indicator gives a colour change with an acid, it does not give a change with a base.
(c) If an indicator changes colour with a base, it does not change colour with an acid.
(d) Change of colour in an acid and a base depends on the type of the indicator. Which of these statements are correct?
(i) All four (ii) (a) and (d) (iii) (b) and (c) (iv) only (d)
Ans. (ii) (a) and (d)

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RD SHARMA SOLUTION CHAPTER- 33 Binomial Distribution I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 33 Binomial Distribution Exercise Ex. 33.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Required Probability =Question 9

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Question 10

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Question 11

Solution 11

Question 12

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Question 13

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Question 14

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Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Also, find the mean and variance of this distribution.Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

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Question 32

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Question 33

Solution 33

= 0.0256Question 34

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Question 35

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Question 48

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Question 50

Find the probability that in 10 throws of a fair die a score which is a multiple of 3 will be obtained in at least 8 of the throws.Solution 50

Question 51

A die is thrown 5 times. Find the probability that an odd number will come up exactly three times.Solution 51

Question 52

The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?Solution 52

Question 53

A factory produces bulbs. The probability that one bulb is defective is and they are packed in boxes of 10. From a single box, find the probability that

i. none of the bulbs is defective.

ii. exactly two bulls are defective.

iii. more than 8 bulbs work properly.Solution 53

Note: Answer given in the book is incorrect.

Chapter 33 Binomial Distribution Exercise Ex. 33.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

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Question 9

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Question 11

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Question 15

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Question 18

Solution 18

Question 19

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Question 20

Solution 20

Question 21

Solution 21

Question 22

From a lot of 15 bulbs which include 5 defective, sample of 4 bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence, find the mean of the distribution.Solution 22

Out of 15 bulbs 5 are defective.

begin mathsize 12px style table attributes columnalign left end attributes row cell text Hence , the probability that the drawn bulb is defective is end text end cell row cell text P end text left parenthesis text Defective end text right parenthesis equals 5 over 15 equals 1 third end cell row cell text P end text left parenthesis text Not defective end text right parenthesis equals 10 over 15 equals 2 over 3 end cell row cell text Let X denote the number of defective bulbs out of 4. end text end cell row cell text Then , X follows binomial distribution with end text end cell row cell straight n equals 4 comma text end text straight p equals 1 third text and end text straight q equals 2 over 3 text such that end text end cell row cell straight P left parenthesis straight X equals straight r right parenthesis equals straight C presuperscript 4 subscript straight r open parentheses 1 third close parentheses to the power of straight r open parentheses 2 over 3 close parentheses to the power of 4 minus straight r end exponent semicolon straight r equals 0 comma 1 comma 2 comma 3 comma 4 end cell row cell text Mean end text equals sum from straight r equals 0 to 4 of rP left parenthesis straight r right parenthesis equals 1 cross times straight C presuperscript 4 subscript 1 open parentheses 1 third close parentheses open parentheses 2 over 3 close parentheses cubed plus 2 cross times straight C presuperscript 4 subscript 2 open parentheses 1 third close parentheses squared open parentheses 2 over 3 close parentheses squared end cell row cell plus 3 cross times straight C presuperscript 4 subscript 3 open parentheses 1 third close parentheses cubed open parentheses 2 over 3 close parentheses plus 4 cross times straight C presuperscript 4 subscript 4 open parentheses 1 third close parentheses to the power of 4 open parentheses 2 over 3 close parentheses to the power of 0 end cell row cell equals 32 over 81 plus 48 over 81 plus 24 over 81 plus 4 over 81 equals 108 over 81 equals 4 over 3 end cell end table end style

Question 23

A die is thrown three times. Let X be’ the number of twos seen’. Find the expectation of X.Solution 23

Question 24

A die is thrown twice. A ‘success’ is getting an even number on a toss. Find the variance of number of successes.Solution 24

Question 25

Three cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability of the number spades. Hence, find the mean of the distribution.Solution 25

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Chapter – 4 Heat | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter 4 Heat

Q.1. State similarities and differences between the laboratory thermometer and the clinical thermometer.
Ans. Similarities:
(i) Both thermometers consist of long narrow uniform glass tubes.
(ii) Both have a bulb at one end.
(iii) Both contain mercury in bulb.
(iv) Both use Celsius scale on the glass tube.
Differences:
(i) A clinical thermometer reads temperature 35°C to 45°C while the range of laboratory thermometer is -10°C to 110°C.
(ii) Clinical thermometer has a kink near the bulb while there is no kink in the laboratory thermometer.
Due to kink mercury does not fall down on its own in clinical thermometer.

Q.2. Give two examples each of conductors and insulators of heat.
Ans. Conductors—aluminium, iron Insulators—plastic, wood.

Q.3.Fill in the blanks
The hotness of an object is detetmined by its ____________ .
(b) Temperature of boiling water cannot be measured by a ____________ thermometer.
(c) Temperature is measured in degree ____________ .

(d) No medium is required for transfer of heat by the process of ____________.
(e) A cold steel spoon is dipped in a cup of hot milk. It transfers heat to its other end by the process of ____________
(f) Clothes of ___________ colours absorb heat better than clothes of light colours.

Ans. (a) temperature (b) clinical (c) Celsius (d) radiation (e) conduction (f) dark

Q.5. Discuss why wearing more layers of clothing during winter keeps us warmer than wearing just one thick piece of clothing?
Ans.More layers of clothing keep us warm in winters as they have a lot of space between them. This space gets filled up with air. Air is a bad conductor, it does not allow the body heat to escape out.

Q.6. Look at figure 4.6. Mark where the heat is being transferred by conduction, by convection and by radiation
Ans.

Q.7. In places of hot climate it is advised that the outer walls of houses be painted white. Explain.
Ans.In places of hot climate it is advised that the outer wall of houses be painted white because white colour reflects heat and the houses do not heat up too much

Q.8. One litre of water at 30°C is mixed with one litre of water at 50°C. The temperature of the mixture will be:
(a) 80°C (b) More than 50°C but less than 80°C
(c) 20°C (d) Between 30°C and 50°C
Ans.(d) Between 30°C and 50°C.

Q.9. An iron ball at 40°C is dropped in a mug containing water at 40°C. The heat will:
(a) flow from iron ball to water.
(b) not flow from iron ball to water or from water to iron ball.
(c) flow from water to iron ball.
(d) increase the temperature of both.
Ans. (b) not flow from iron ball to water or from water to iron ball

Q.10. A wooden spoon is dipped in a cup of ice-cream. Its other end:
(a) becomes cold by the process of conduction
(b) becomes cold by the process of convection
(c) becomes cold by the process of radiation
(d) does not become cold
Ans.(d) does not become cold.

Q.11.Stainless steel pans are usually provided with copper bottoms. The reason for this could be that:
(a) copper bottom makes the pan more durable
(b) such pans appear colourful
(c) copper is a better conductor of heat than the stainless steel
(d) copper is easier to clean than the stainless steel
Ans.(c) copper is better conductor of heat than the stainless steel

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RD SHARMA SOLUTION CHAPTER- 32 Mean and Variance of a Random Variable I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 32 Mean and variance of a random variable Exercise Ex. 32.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

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Question 11

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Question 15

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Question 23

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Question 26

Solution 26

Question 27

Solution 27

Question 28

Four balls are to be drawn without replacement from a box containing 8 red and 4 white balls. If X denotes the number of red balls drawn, find the probability distribution of X.Solution 28

Question 29

The probability distribution of a random variable X is given below:

(i) Determine the value of k

(ii) Determine P (X 2) and P b(X > 2)

(iii) Find P (X 2) + P(X > 2)Solution 29

Chapter 32 Mean and variance of a random variable Exercise Ex. 32.2

Question 1(i)

Find the mean and standard deviation of each of the following probability distributions:

x_i : 2 3 4

p_i : 2.2 0.5 0.3Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 1(ix)

Find the mean and standard deviation of each of the following probability distributions:

Solution 1(ix)

Question 2

A discrete random variable X has the probability distribution given below:

X : 0.5 1 1.5 2

P(X) : k k² 2k² k

(i) Find the value of k.

(ii) Determine the mean of the distribution.Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

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Question 10

Solution 10

Question 11

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Question 12

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Question 13

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Question 14

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Question 15

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Question 16

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Question 17

Three cards are drawn at random (without replacement) from a well shuffled pack of 52 cards. Find the probability distribution of number of red cards. Hence find the mean of the distribution.Solution 17

begin mathsize 12px style table attributes columnalign left end attributes row cell Let text end text apostrophe straight X apostrophe text end text be text end text the text end text random text end text variable text end text which text end text can text end text assume end cell row cell values text end text from text end text 0 text end text to text end text 3. end cell row cell straight P left parenthesis straight X equals 0 right parenthesis equals fraction numerator blank to the power of 26 straight C subscript 3 over denominator blank to the power of 52 straight C subscript 3 end fraction equals 2600 over 22100 equals 2 over 17 end cell row cell straight P left parenthesis straight X equals 1 right parenthesis equals fraction numerator blank to the power of 26 straight C subscript 1 cross times to the power of 26 straight C subscript 2 over denominator blank to the power of 52 straight C subscript 3 end fraction equals 8450 over 22100 equals 13 over 34 end cell row cell straight P left parenthesis straight X equals 2 right parenthesis equals fraction numerator blank to the power of 26 straight C subscript 2 cross times to the power of 26 straight C subscript 1 over denominator blank to the power of 52 straight C subscript 3 end fraction equals 8450 over 22100 equals 13 over 34 end cell row cell straight P left parenthesis straight X equals 3 right parenthesis equals fraction numerator blank to the power of 26 straight C subscript 3 over denominator blank to the power of 52 straight C subscript 3 end fraction equals 2600 over 22100 equals 2 over 17 end cell row cell Probability text end text distribution text end text of text end text straight X colon end cell row cell table row cell straight X equals straight x subscript straight i end cell 0 1 2 3 row cell straight p left parenthesis straight X equals straight x subscript straight i right parenthesis end cell cell 2 over 17 end cell cell 13 over 34 end cell cell 13 over 34 end cell cell 2 over 17 end cell end table end cell row blank row cell Mean equals sum from straight i equals 0 to 3 of left parenthesis straight x subscript straight i cross times straight p subscript straight i right parenthesis end cell row cell equals straight x subscript 0 straight p subscript 0 plus straight x subscript 1 straight p subscript 1 plus straight x subscript 2 straight p subscript 2 plus straight x subscript 3 straight p subscript 3 end cell row cell equals 0 cross times 2 over 17 plus 1 cross times 13 over 34 plus 2 cross times 13 over 34 plus 3 cross times 2 over 17 end cell row cell equals fraction numerator 13 plus 26 plus 12 over denominator 34 end fraction end cell row cell equals 51 over 34 end cell row cell equals 3 over 2 end cell row cell equals 1.5 end cell end table end style

Question 18

An urn contains 5 are 2 black balls. Two balls are randomly drawn, without replacement. Let X represent the number of black balls drawn. What are the possible values of X? Is X a random variable? If yes, find the mean and variance of X.Solution 18

Question 19

Two numbers are selected at random (without replacement) from positive integers 2,3,4,5, 6 and 7. Let X denote the larger of the two number obtained. Find the mean and variance of the probability distribution of X.Solution 19

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RD SHARMA SOLUTION CHAPTER- 31 Probability I CLASS 12TH MATHEMATICS-EDUGROWN

Chapter 31 Probability Exercise Ex. 31.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Chapter 31 Probability Exercise Ex. 31.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is probability that both drawn balls are black?Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Chapter 31 Probability Exercise Ex. 31.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 5(iv)

If A and B are two events such that

Solution 5(iv)

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

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Question 15

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Question 26

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Question 27

Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls? Given that

(i) the youngest is a girl

(ii) at least one is girl.Solution 27

(i) Let ‘A’ be the event that both the children born are girls.

Let ‘B’ be the event that the youngest is a girl.

We have to find conditional probability P(A/B).

begin mathsize 12px style table attributes columnalign left end attributes row cell straight P left parenthesis straight A divided by straight B right parenthesis equals fraction numerator straight P left parenthesis straight A intersection straight B right parenthesis over denominator straight P left parenthesis straight B right parenthesis end fraction end cell row cell straight A subset of straight B rightwards double arrow straight A intersection straight B equals straight A end cell row cell rightwards double arrow straight P left parenthesis straight A intersection straight B right parenthesis equals straight P left parenthesis straight A right parenthesis equals straight P left parenthesis GG right parenthesis equals 1 half cross times 1 half equals 1 fourth end cell row cell straight P left parenthesis straight B right parenthesis equals straight P left parenthesis BG right parenthesis plus straight P left parenthesis GG right parenthesis equals 1 half cross times 1 half plus 1 half cross times 1 half equals 1 fourth plus 1 fourth equals 1 half end cell row cell text Hence , end text straight P left parenthesis straight A divided by straight B right parenthesis equals fraction numerator 1 divided by 4 over denominator 1 divided by 2 end fraction equals 1 half end cell end table end style

(ii) Let ‘A’ be the event that both the children born are girls.

Let ‘B’ be the event that at least one is a girl.

We have to find the conditional probability P(A/B).

Chapter 31 Probability Exercise Ex. 31.4

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

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Question 15

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Question 17

Solution 17

Question 18

Solution 18

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Question 21

Solution 21

Question 22

Solution 22

Question 23

begin mathsize 12px style table attributes columnalign left end attributes row cell text The probabilities of two students end text straight A text and end text straight B text coming end text end cell row cell text to the school in time are end text fraction numerator text 3 end text over denominator text 7 end text end fraction text and end text fraction numerator text 5 end text over denominator text 7 end text end fraction text respectively . end text end cell row cell text Assuming that the events , ' end text straight A text coming in time ' and ' end text straight B end cell row cell text coming in time ' are independent , find the probability of end text end cell row cell text only one of them coming to the school in time . end text end cell row cell text Write atleast one advantage of coming to school in time. end text end cell end table end style

Solution 23

Given that the events ‘A coming in time’ and ‘B coming in time’ are independent.

begin mathsize 12px style table attributes columnalign left end attributes row cell text Let ' A ' denote the event of ' A coming in time '. end text end cell row cell text Then , ' end text stack text A end text with bar on top apostrophe text denotes the complementary event of A. end text end cell row cell text Similarly we define B and end text stack text B end text with bar on top. end cell end table end style

begin mathsize 12px style table attributes columnalign left end attributes row cell straight P left parenthesis only text one coming in time end text right parenthesis equals straight P left parenthesis straight A intersection straight B with bar on top right parenthesis plus straight P left parenthesis straight A with bar on top intersection straight B right parenthesis end cell row cell equals straight P left parenthesis straight A right parenthesis cross times straight P left parenthesis straight B with bar on top right parenthesis plus straight P left parenthesis straight A with bar on top right parenthesis cross times straight P left parenthesis straight B right parenthesis... left parenthesis text since A and B are independent events end text right parenthesis end cell row cell equals 3 over 7 cross times 2 over 7 plus 4 over 7 cross times 5 over 7 equals 6 over 49 plus 20 over 49 equals 26 over 49 end cell end table end style

The advantage of coming to school in time is that you will not miss any part of the lecture and will be able to learn more.Question 24

Two dice are thrown together and the total score is noted. The event E, F and G are “a total 4”, “a total of 9 or more”, and “a total divisible by 5”, respectively. Calculate P (E), P(F) and P(G) and decide which pairs of events, if any, are independent.Solution 24

Question 25

Let A and B be two independent events such that P (A) = p₁ and P (B) = p₂. Describe in words the events whose probabilities are:

(i) p₁p₂ (ii) (1 – p₁)p₂ (iii) 1-(1- p₁) (1 – p₂) (iv) p₁ + p₂ = 2p₁p₂Solution 25

Chapter 31 Probability Exercise Ex. 31.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

In a hockey match, both teams A and B scored same number of goals upto the end of the game, so to decide the winner, the refree asked both the captains to throw a die alternately and decide that the team, whose captain gets a first six, will be declared the winner. If the captain of team A was asked to start, find their respective probabilities of winning the match and state whether the decision of the refree was fair or not.Solution 35

begin mathsize 12px style table attributes columnalign left end attributes row cell text Probability of getting six in any toss of a dice end text equals 1 over 6 end cell row cell text Probability of not getting six in any toss of a dice end text equals 5 over 6 end cell row cell text A and B toss the die alternatively . end text end cell row cell text Hence probability of A ' s win end text end cell row cell equals straight P left parenthesis straight A right parenthesis plus straight P left parenthesis straight A with bar on top straight B with bar on top straight A right parenthesis plus straight P left parenthesis straight A with bar on top straight B with bar on top straight A with bar on top straight B with bar on top straight A right parenthesis plus straight P left parenthesis straight A with bar on top straight B with bar on top straight A with bar on top straight B with bar on top straight A with bar on top straight B with bar on top straight A right parenthesis plus....... end cell row cell equals 1 over 6 plus 5 over 6 cross times 5 over 6 cross times 1 over 6 plus 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 1 over 6 plus 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 1 over 6 plus..... end cell row cell equals 1 over 6 plus open parentheses 5 over 6 close parentheses squared 1 over 6 plus open parentheses 5 over 6 close parentheses to the power of 4 1 over 6 plus open parentheses 5 over 6 close parentheses to the power of 6 1 over 6 plus..... end cell row cell equals fraction numerator 1 divided by 6 over denominator 1 minus open parentheses 5 divided by 6 close parentheses squared end fraction equals 1 over 6 cross times 36 over 11 equals 6 over 11 end cell row cell Similarly comma text probability of B ' s win end text end cell row cell equals straight P left parenthesis straight A with bar on top straight B right parenthesis plus straight P left parenthesis straight A with bar on top straight B with bar on top straight A with bar on top straight B right parenthesis plus straight P left parenthesis straight A with bar on top straight B with bar on top straight A with bar on top straight B with bar on top straight A with bar on top straight B right parenthesis plus...... end cell row cell equals 5 over 6 cross times 1 over 6 plus 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 1 over 6 plus 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 1 over 6 plus..... end cell row cell equals 5 over 6 cross times 1 over 6 plus open parentheses 5 over 6 close parentheses squared 5 over 6 cross times 1 over 6 plus open parentheses 5 over 6 close parentheses to the power of 4 5 over 6 cross times 1 over 6 plus open parentheses 5 over 6 close parentheses to the power of 6 5 over 6 cross times 1 over 6 plus..... end cell row cell equals fraction numerator 5 over 6 cross times 1 over 6 over denominator 1 minus open parentheses 5 over 6 close parentheses squared end fraction equals 5 over 36 cross times 36 over 11 equals 5 over 11 end cell row cell text Since the probabilities are not equal , end text end cell row cell text the decision of the refree was not a fair one. end text end cell end table end style

Chapter 31 Probability Exercise Ex. 31.6

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

There machines E₁, E₂, E₃ in a certain factory produce 50%, 25% and 25% respectively, of the total daily output of electric bulbs. It is known that 4% of the tubes produced one each of machines E₁ and E₂ are defective, and that 5% of those produced on E₃ are defective. If one tube is picked up at random from a day’s production, calculate the probability that it is defective.Solution 13

Chapter 31 Probability Exercise Ex. 31.7

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Suppose a girl throws a die. If she gets 1 or 2, she tosses a coin three times and notes the number of tails. If she gets 3,4, 5 or 6, she tosses a coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained exactly one ‘tail’, what is the probability that she threw 3, 4, 5 or 6 with the die?Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

An item is manufactured by three machine A, B and C. out of the total number of items manufactured during a specified period, 50% are manufacture on machine A 30% on B and 20% on C. 2% of the items produced on A and 2% of items produced on B are defective and 3% of these produced on C are defective. All the items stored at one godown. One items is drawn at random and is found to be defective. What is the probability that it was manufactured on machine A?Solution 14

Question 15

There are three coins. One is two-headed coin (having head on both faces), another is biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tail 40% of the times. One of the three coins is chosen at random and tossed, and it shows heads. What is the probability that it was the two-headed coin?Solution 15

begin mathsize 12px style table attributes columnalign left end attributes row cell text Let end text straight E subscript 1 comma straight E subscript 2 comma straight E subscript 3 text be the events that we choose the first coin , end text end cell row cell text second coin , and third coin respectively in a random toss. end text end cell row cell straight P left parenthesis straight E subscript 1 right parenthesis equals 1 third comma straight P left parenthesis straight E subscript 2 right parenthesis equals 1 third comma straight P left parenthesis straight E subscript 3 right parenthesis equals 1 third end cell row cell text Let A denote the event when the toss shows heads. end text end cell row cell text It is given that end text end cell row cell straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis equals 1 comma straight P left parenthesis straight A divided by straight E subscript 2 right parenthesis equals 0.75 comma straight P left parenthesis straight A divided by straight E subscript 3 right parenthesis equals.60 end cell row cell text We have to find end text straight P left parenthesis straight E subscript 1 divided by straight A right parenthesis. end cell row cell By text Baye ' s theorem end text end cell row cell straight P left parenthesis straight E subscript 1 divided by straight A right parenthesis equals fraction numerator straight P left parenthesis straight E subscript 1 right parenthesis straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis over denominator straight P left parenthesis straight E subscript 1 right parenthesis straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis plus straight P left parenthesis straight E subscript 2 right parenthesis straight P left parenthesis straight A divided by straight E subscript 2 right parenthesis plus straight P left parenthesis straight E subscript 3 right parenthesis straight P left parenthesis straight A divided by straight E subscript 3 right parenthesis end fraction equals end cell row cell equals fraction numerator 1 third left parenthesis 1 right parenthesis over denominator 1 third left parenthesis 1 right parenthesis plus 1 third left parenthesis 0.75 right parenthesis plus 1 third left parenthesis 0.60 right parenthesis end fraction equals fraction numerator 1 divided by 3 over denominator left parenthesis 1 divided by 3 right parenthesis plus left parenthesis 1 divided by 4 right parenthesis plus left parenthesis 1 divided by 5 right parenthesis end fraction end cell row cell equals fraction numerator 1 divided by 3 over denominator 47 divided by 60 end fraction equals 20 over 47 end cell end table end style

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

In a group of 400 people, 160 are smokers and non-vegetarian, 100 are smokers and vegetarian and the remaining are non-smokers and vegetarian. The probabilities of getting a special chest disease are 35%, 20% and 10% respectively. A person is chosen from the group at random and is found to be suffering from the disease. What is the probability that the selected person is a smoker and non-vegetarian?Solution 19