Class 11

Odisha State Board CHSE Odisha Class 11 Invitation to English 3 Solutions Writing Using Graphics Textbook Activity Questions and Answers.

CHSE Odisha 11th Class English Writing Using Graphics

Using Graphics

We think of composition as the writing of paragraphs or essays in words. But sometimes a description can be made clearer and easier to understand by the use of some kind of pictorial representation.
There are different kinds of graphic representations – they are :
(a) The table
(b) The Bar chart
(c) The Pie chart
(d) The Organisation chart
(e) Pictograms
An example of a graphic representation is given below :

Another example of a graphic representation is given below for reference.

Activity 1

Now, read the following passages and say if they can be represented graphically. Draw neat labelled diagrams to go with or to replace these passages. (More than one answer is possible.)

(a) Iron is not found pure in nature. It is found in the form of iron compounds, particularly compounds of iron and oxygen. Such compounds are called ores. To get iron from its ore, layers of iron ore are placed in a blast furnace with coke and limestone. Coke has a high proportion of carbon in it. The mixture of iron ore, coke and limestone is heated in the blast furnace. A blast of hot air enters the blast furnace from the bottom and passes upwards.

The carbon in the coke combines with the oxygen in the iron ore and forms a colourless gas called carbon monoxide. The oxygen in the iron oxide is thus removed. The other constituents of the ore combine with the limestone to form a slag which floats at the top. The slag is removed from the furnace through a tap. The metal which remains is called pig iron. It is 91 per cent pure.

(b) In Britain most schools are financed by the state and for the children attending these schools, they are free. However, about 5 per cent of the school population attend private schools, and these are financed from pupils’ fees. For state-run as well as private schools the general pattern of schooling in Britain is as follows. All children must start school at the age of 5. At the age of 11, children move to different schools, called Secondary Schools, where they are made to go through a set timetable of subjects for a period of 5 years.

After this, they take their first public examination at the age of 16. After the first public examination, most pupils leave school. Only about 30 per cent continue. Those that stay on spend the next two years concentrating on a small number of subjects. They take their second public examination at the age of 18. In Britain, there is also a provision for pre-primary schooling. However, pre-primary schooling is not compulsory. On the other hand, it is voluntary and is offered both by state-run schools and private schools.

(c) It was the season’s hottest day, in Orissa today. The temperature in many places went beyond 40° Celsius. Of all the places in Orissa, Angul recorded the highest temperature of 43.5°. The other places that experienced an equally warm day were Bolangir and Jharsuguda. In these two places, the mercury soared up to 42.7°. Bhubaneswar, the capital city, recorded a high temperature of 42.2°.

The other places that were as warm as Bhubaneswar were Sambalpur and Cuttack with a recording of 41.8° and 41.2° respectively. However, Puri, Gopalpur, Paradip and Balasore, places near the sea, were relatively cooler. They recorded maximum temperatures of around 32° Celsius.

(d) Ramesh walks to school at 10. It takes him hardly 20 minutes to get there. After prayers, he attends classes from 10.30 a.m. to 4 p.m. which includes games- break for half an hour, from 1 to 1.30. At 4, when the classes are over, he comes back home with his friends. He greatly enjoys the return trip home as his friends and he usually amuses themselves with innocent jokes, makes fun and indulge in different kinds of innocuous mischief. On reaching home he gobbles up something quickly and scampers away to join his friends either in volleyball or in some country games, which he plays for just one hour.

At 6 he has a wash, says his prayers and sits down to study, from 6.30 to 9.30 in the evening. At 9.30 he has dinner, does some light reading and goes to bed by 10.30. Next morning he is up by 6.30. Within one hour he brushes his teeth, has a wash, etc. and by 7.30 he is at his study table. He does his homework from 7.30 to 9.30. At 9.30 he has an early lunch, packs his books in the bag and is ready to go to school.

(e) The poor in Indian do not have a lot of money. Their annual family income, on an average, comes to Rs. 20,000. But they make a lot of purchases and constitute the mainstay for most products in the country. For example, families belonging to his category own 47 per cent of all bicycles. 17 per cent of mopeds, 9 per cent of scooters, 11 per cent of motor cycles and 10 per cent of colour televisions. They even own washing machines.

The rich, in contrast, have money, but their share of purchases of these products is comparatively much less. Despite the money they have, they buy only 3 per cent of bicycles, 7 per cent of mopeds, 17 per cent of scooters, 20 per cent of motor cycles and 19 per cent of colour TVs. The poor earn less, but spend more. And the rich earn more, but spend less.

(f) Go straight along the National Highway upto Jayadev Vihar. Then turn left and take the Nandan Kanan Road. Keep going for about two kilometres and then turn right at NALCO Chhak. Continue until you come to the Sainik School, on the left. Don’t stop. Go straight ahead. You will reach Vani Vihar after half a kilometre. You can see the gates of Vani Vihar on the left, just 400 metres after Acharya Vihar traffic post. Enter, and go straight to the Administrative Block of the university.

(g) Junagadh is an ancient city in Gujarat. It is situated among the shadows of Mount Gimar. The name “Junagadh” – Juna (old) and Gadh (fort) – literally means “old fort”. On the outskirts of the city, there is a dark basalt rock. It stands on the way to Mount Gimar. The rock hols the inscriptions of three mighty dynasties. They included the Maurya and Gupta dynasties. The incriptions are in Sanskrit. In a jungle nearby, there is a stupa and some Buddhist caves.

They were built between 100-700 A.D. At the foot of Mount Gimar there is the sacred Damodar Kund (pond). It is one of the most important places of pilgrimage for the Jains. Another place of interest is the 19th Century Rang Mahal Palace, which presently houses government offices.

(h) Two leaves are removed from a destarched plant. The upper side of one and the lower side of the other are greased with vaseline. The stalk of each leaf is dipped in water and the leaves are left in light for four hours so that photosynthesis takes place. Most of the vaseline is wiped off and the leaves are placed in a solution of potassium iodide. The leaf greased on the upper-side develops a blue colour, showing that starch has been formed by photosynthesis from carbon dioxide, which entered through the leaf pores which are mainly on the under side. No colour develops in the other leaf in which vaseline blocked the pores.

Answer:
(a)

(b)

(c) Look at the table below, showing the maximum temperature recorded at different places in Orissa :

(d)

(e) Here is a chart that deals with the purchasing habits of the poor in India.

The following chart presents a contrasting picture of the purchasing habits of the rich :

 

 

(f)

(g) Let us look at the table that deals with the beautiful spots of the nearby areas of Junagarh an ancient of Gujarat.

Places of interest	Description of these places
Rock	dark and basalt with the inscriptions of mighii dynasties — Maurya and Gupta
Stupa and some Buddhist caves	builLhetween 100-7(X) A.D.
Damodar Kund	most important places of pilgrimage for the jams, situateLat the foot of Mount Girnar
19th century Rang Mahal Palace	presently housing government offices

(h) The flow-chart throwing light on an experiment concerning photosynthesis

Additional Questions With Answers

Question 1.
Write about 150 words in each of the following:
(a) The camel
(b) The ostrich
Answer:

(a) The camel:
The camel is a big animal. There are two kinds. One has two humps on its back; it is called the Bactrian camel. The other has one hump; it is called the Arabian camel. The camel is called “the ship of the desert”. Why? Because, as men cross the sea in ships, so they cross the desert on camels. Camels can go a long time with very little food or drink.

How can they do this? They carry stores of water in their stomachs, and much fat in their “humps”. A fasting camel is really feeding on its hump. The camel is also very strong and can go on day after day without getting tired. So it can make long journeys over the hot, dry sandy deserts. It is a fine sight to see a line of camels marching in a caravan. They hold their heads up proudly and tread steadily on.

(b) The ostrich:
The ostrich is the largest of all birds. Some ostriches are eight feet high. The ostrich is so strong that it can carry a man, like a horse. It has a long neck and a small head. It carries its head and neck erect. Its wings are small, and it cannot fly. But it has very strong legs. It can run faster than a horse. It defends itself by kicking. A kick from an ostrich can lame a horse, or kill a man. The ostrich is found in Africa.

It lives in dry sandy wastes, or a country covered with low bushes. It feeds on grass, leaves, seeds and berries. Wild ostriches are very shy. They are hard to catch, for they run so fast. Ostriches are hunted for the sake of their feathers. Ostrich feathers are very costly. Ladies buy them for their hats. Ostriches are now bred in large farms in Africa. The sale of their feathers pays well.

Question 2.
Write a short paragraph on each of the following:
(a) Electronic calculator
(b) A type-writer
Answer:

(a) Electronic calculator:
Electronic calculators are useful gadgets to have in homes, offices, shops and schools. They come in various sizes. There are desk models which are the size of a book or bigger. For personal use, there are mini-calculators that one can hold comfortably in one’s hand. Other models are even smaller. Every calculator has a keyboard showing numbers 0 to 9 plus mathematical symbols i.e. +, -, x, %. Above the keyboard is a display window and above the window, at one comer, is the switch with which one can turn the calculator on or off.

The calculator works this way. The operator presses the keys to feed a problem, say 13 x 12 = into the machine. The answer, 156, appears almost instantly in the form of some lighted numerals in the display window. All calculators help to do the basic functions of arithmetic. Some calculators, those with memory, can store numbers for use in future calculations.

(b) A type-writer:
A typewriter is a familiar machine in offices and homes. It is worked by hand. It prints letters or figures on a sheet of paper, one at a time. The type-writer consists of a metal frame, movable rollers and a set of keys. The metal frame also contains a set of small hammers with tiny letters or figures on the ends. The typewriter has a ribbon soaked in special ink which is held in position by two spools on either side of the metal frame. A sheet of paper is clipped against the roller. When one of the keys is tapped, a hammer goes and strikes the ribbon and presses it against the blank paper, thus printing the required letter. Then the roller automatically moves along, another key is tapped and the next letter is printed after the first.

Question 3.
Describe the following simple processes:
(a) Kite-making
(b) Mending a puncture
Answer:

(a) Kite-making:
To make a kite, we take two sticks and tie them together in the shape of a cross. Then we tie a string, passing along the tips of the sticks to make a frame for the kite. We then cut a piece of thin paper to a size slightly larger than the frame and folding the edges over the frame, stick them with glue. We tie another string about IV2 times the length of the longer stick to the two tips of the longer stick; this is the bridle. We tie the flying cord to the bridle. We attach a tail to the kite at the narrower end.

(b) Mending a puncture:
The process of mending a puncture starts with the removal of the tube from the rim. Then we blow up the tube a little and dip it into a pan of water to find the puncture. We clean the tube around the puncture. Then we put on a sticky substance. We put on a patch and allow it to dry. We put the tube back on the rim after it gets dried. At last, we blow up the tyre. This is how the puncture is mended.

Question 4.
Prepare outlines and write an argument in support of the following proposition.
The Government should ban smoking in public places.

A. Stating the proposition :
The government should ban smoking in public places.

B. Reasons for the ban :
a. Cigarette smoking is responsible for a host of illnesses ranging from heart attacks, hypertension and strokes to cancers of the mouth, lung and kidneys. These illnesses cause thousands of deaths each year and a staggering number of hospital visits and bills.
b. Ban on cigarette advertising in several countries, and health warning required by law on every cigarette packet on sale.
c. Smoking injurious to the health of the non-smoker.
d. Unpleasant for non-smokers to sit in smoke-filled rooms and halls.
e. Smoking is a serious fire risk in crowded places.
f. Still people smoke; hence the need for a ban by the government.

C. The other side of the case :
a. There are already enough no-smoking areas such as cinema halls, for example, therefore total ban is unnecessary.
b. Ban will further curtail their freedom to smoke.

D. Smokers’ objections refuted
a. Smoking is harmful to smokers and non-smokers. Non-smokers are as much risk as smokers.
b. Smokers should not harm the health of non-smokers in the name of freedom to smoke.

Answer:
The paragraph in the outlines :
Medical researchers have established cigarette smoking as the cause of a host of potentially fatal illnesses ranging from heart attacks, hypertension and strokes, to cancers of the mouth, lungs and kidneys. These illnesses are responsible for thousands of deaths each year and a staggering number of hospital visits and bills. Several countries have as a result banned cigarette advertising and made it compulsory by law for cigarette manufacturers to print a health warning on every cigarette packet on sale.

What is at stake is not just the health of the smoker; the non-smoker is as much at risk as a result of the passive smoking he or she is exposed to. It is extremely unpleasant for non-smokers to sit in smoke-filled rooms and halls. Finally, smoking is a serious fire risk in crowded places. Despite all this, people still smoke. The government, therefore, should intervene and ban smoking in public places.

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

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

Question 1.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(ii) yz-plane
Solution:
x-axis

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(ii) cos x
Solution:

(iii) sin x
Solution:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 1.
The rows n = 6 and n = 7 in the pascal triangle have been kept vacant. Fill in the gaps
Solution:

Question 2.
Write down the expansion of (a + b)⁸ using Pascal’s triangle.
Solution:
The row n = 8 in Pascal’s triangle is 1, 8, 28, 56, 70, 56, 28, 8, 1.
∴ (a + b)⁸ = a⁸ + 8a^8-1b¹ + 28a^8-2b² + 56a^8-3b³ + 70a^8-4b⁴ + 56a^8-5b⁵ + 28a^8-6b⁶ + 8a^8-7b⁷ + b⁸
= a⁸ + 8a⁷b + 28a⁶b² + 56a⁵b³ + 70a⁴b⁴ + 56a³b⁵ + 28a²b⁶ + 8ab⁷ + b⁸

Question 3.
Find the 3rd term in the expansion of \(\left(2 x^3-\frac{1}{x^6}\right)^4\) using rules of Pascal triangle.
Solution:

Question 4.
Expand the following :
(a) (7a + 3b)⁶
Solution:
(7a + 3b)⁶ = (7a)⁶ + ⁶C₁(7a)^6-1(3b)¹ + ⁶C₂(7a)^6-2(3b)² + ….. + (3b)⁶7⁶a⁶ + 6(7a)⁵(3b) + 15(7a⁴) × 9b² + …. + 3⁶b⁶
= 7a⁶ + 18 × 7⁵a⁵b + 135 × 7⁴a⁴b² + ….. + 3⁶b⁶

(b) \(\left(\frac{-9}{2} a+b\right)^7\)
Solution:

(c) \(\left(a-\frac{7}{3} c\right)^4\)
Solution:

Question 5.
Apply Binominal Theorem to find the value of (1.01)⁵.
Solution:
= 1 + ⁵C₁(0.01)¹ + ⁵C₂(0.01)² + ⁵C₃(0.01)³ + ⁵C₄(0.01)⁴ + (0.01)⁵
= 1 + 5 × 0.01 + 10(0.0001) + 10(0.000001) + 5(0.00000001) + 0.0000000001
= 1 + 0.05 + 0.001 + 0.00001 + 0.00000005 + 0.000000001
= 1.0510100501

Question 6.
State true or false.
(a) The number of terms in the expansion of \(\left(x^2-2+\frac{1}{x^2}\right)^6\) is equal to 7.
Solution:
False

(b) There is a term independent of both x and y in the expansion of \(\left(x^2+\frac{1}{y^2}\right)^9\)
Solution:
False

(c) The highest power in the expansion of \(x^{40}\left(x^2+\frac{1}{x^2}\right)^{20}\) is equal to 40.
Solution:
False

(d) The product of K consecutive natural numbers is divisible by K!
Solution:
True

Question 7.
Answer the following :
(a) If the 6th term in the expansion of (x + *)ⁿ is equal to ⁿC₅x^n-10 find *
Solution:
Let the 6th term (x + y)ⁿ is ⁿC₅x^n-10
∴ ⁿC₅x^n-5y⁵ = ⁿC₅x^n-10 = ⁿC₅x^n-5.x^-5
⇒ y⁵ = x^-5 = \(\frac{1}{x^5}\)
∴ y = \(\frac{1}{x}\) . Hence * = \(\frac{1}{x}\)

(b) Find the number of terms in the expansion of (1 + x)ⁿ (1 – x)ⁿ.
Solution:
(1 + x)ⁿ (1 – x)ⁿ = (1 – x²)ⁿ
∴ The number of terms in this expansion is (n + 1)

(c) Find the value of \(\frac{{ }^n \mathrm{C}_{r-1}}{{ }^n \mathrm{C}_r}\)
Solution:

(d) How many terms in the expansion of \(\left(\frac{3}{a}+\frac{a}{3}\right)^{10}\) have positive powers of a? How many have negative powers of a?
Solution:

Question 8.
Find the middle term(s) in the expansion of the following.
(a) \(\left(\frac{a}{b}+\frac{b}{a}\right)^6\)
Solution:
Here there is only one middle term i.e. the 4th term.
∴ 4th term i.e. (3 + 1)th term in the expansion of

(b) \(\left(x+\frac{1}{x}\right)^9\)
Solution:
Here there are two middle terms i.e. 5th and 6th terms.
∴ 5th term in the above expansion is

(c) \(\left(x^{\frac{3}{2}}-y^{\frac{3}{2}}\right)^8\)
Solution:
Here there is only one middle term i.e. 5th term.
∴ 5th term i.e. (4 + 1)th term in the expansion of

Question 9.
Find the 6th term in the expansion of \(\left(x^2+\frac{a^4}{y^2}\right)^{10}\)
Solution:
6th term i.e. (5 + 1)th term in the expansion of
(x² + \(\frac{a^4}{y^2}\))¹⁰ is ¹⁰C₅(x²)^10-5 (\(\frac{a^4}{y^2}\))⁵

Question 10.
(a) Find the fifth term in the expansion of (6x – \(\frac{a^3}{x}\))¹⁰
Solution:

(b) Is there a term independent of x? If yes find it out.
Solution:
Let (r + 1)th term in the expansion of  (6x – \(\frac{a^3}{x}\))¹⁰ is independent of x.
∴ (r + 1)th term = ¹⁰C_r(6x)^10-r(\(\frac{-a^3}{x}\))^r
= ¹⁰C_r6^10-rx^10-r(-1)^ra^3rx^-r
= (-1)^{r 10}C_r6^10-ra^3rx^10-2r
∴ x^10-2r = 1 = x⁰
or, 10 – 2r = 0 or, r = 5
∴ 6th term is term independent of x in the above expansion and the term is (-1)^{5 10}C₅6^10-5a^3.5
= – ¹⁰C₅6⁵a¹⁵ = – 252 × 6⁵a¹⁵

Question 11.
(a) Find the coefficient of \(\frac{1}{y^{10}}\) in the expansion of \(\left(y^3+\frac{a^7}{y^5}\right)^{10}\)
Solution:

(b) Does there exist a term independent of y in the above expansion?
Solution:
Let (r + 1)th term is independent of y.
∴ y^30-8r = 1 = y⁰ or, 30 – 8r = 0
or, r = \(\frac{30}{8}\) = \(\frac{15}{4}\) which is not possible as r∈ N or zero.
∴ There is no term in the expansion of \(\left(y^3+\frac{a^7}{y^5}\right)^{10}\) which is independent of y.

Question 12.
(a) Find the coefficient of x⁴ in the expansion of (1 + 3x + 10x²)(x + \(\frac{1}{x}\))¹⁰
Solution:

(b) Find the term independent of x in the above expansion.
Solution:

Question 13.
Show that the coefficient of a^m and aⁿ in expansion of (1 + a)^m+n are equal.
Solution:
(m + 1)th and (n + 1)th terms in the expansion of (1 + a)^m+n are ^m+nC_ma^m and ^m+nC_naⁿ
∴ The coefficient of a^m and aⁿ are ^m+nC_m and ^m+nC_n which are equal.

Question 14.
An expression of the form (a + b + c + d + …. ) consisting of a sum of many distinct symbols called a multinomial. Show that (a + b + c)ⁿ is
the sum of all terms of the form \(\frac{\boldsymbol{n} !}{\boldsymbol{p} ! \boldsymbol{q} ! \boldsymbol{r} !}\) a^pb^qc^r where p, q and r range over all positive triples of non-negative integers such that p + q + r = n.
Solution:

Question 15.
State and prove a multinominal Theorem.
Solution:
Multinominal Theorem:
(P₁ + P₂ + ……. + P_m)ⁿ
\(=\sum \frac{n !}{n_{1} ! n_{2} ! \ldots n_{m} !} p_1^{n_1} p_2^{n_2} \ldots p_m^{n_m}\)
where n₁ + n₂ + ……. + n_m = n
The proof of this theorem is beyond the syllabus.

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

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

Question 1.
Find the number of ways in which 5 different books can be arranged on a shelf.
Solution:
The number of ways in which 5 different books can be arranged on a shelf is 5! = 5. 4. 3. 2. 1 = 120

Question 2.
Compute ⁿP_r for
(i) n = 8, r = 4
Solution:
∴ ⁿP_r = \(\frac{n !}{(n-r) !}=\frac{8 !}{(8-4) !}\)
\(=\frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 . !}{4 !}\) = 8.7.6.5 = 1680

(ii) n = 10, r = 3
Solution:
n = 10, r = 3
∴ ⁿP_r = \(\frac{n !}{(n-r) !}=\frac{10 !}{7 !}\)

(iii) n = 11, r = 0
Solution:
n = 11, r = 0
∴ ⁿP_r = ¹¹P₀ = 1

Question 3.
Compute the following :
(i) \(\frac{10 !}{5 !}\)
Solution:
\(\frac{10 !}{5 !}\) = 10. 9. 8. 7. 6 = 30240

(ii) 5! + 6!
Solution:
5 ! + 6! = 5 ! + 6.5 !
= 5 ! (1 + 6) = 120. 7 = 840

(iii) 3! × 4!
Solution:
3 ! × 4 ! = 6 × 24 = 144

(iv) \(\frac{1}{8 !}+\frac{1}{9 !}+\frac{1}{10 !}\)
Solution:

(v) 2!^3! = 2⁶ = 64
(vi) 2³! = 8! =40320

Question 4.
Show that 2.6.10 ……. to n factors = \(\frac{(2 n) !}{n !}\)
Solution:
2.6.10 ……. to n factors = \(\frac{(2 n) !}{n !}\)

Question 5.
Find r if P(20, r) = 13. P (20, r – 1).
Solution:

Question 6.
Find n if P(n, 4) = 12. P(n, 2).
Solution:
ⁿP₄ = 12 × ⁿP₂
or, \(\frac{n !}{(n-4) !}=12 \times \frac{n !}{(n-2) !}\)
or, (n – 2)! = 12 (n – 4)!
or, (n – 2) (n – 3) (n – 4)! = 12 (n – 4)!
or, (n – 2) (n – 3) = 12
or, n² – 5n – 6 = 60
or, (n – 6) (n + 1) = 0
or, n = 6 – 1
Hence n = 6 as n is a natural number.

Question 7.
If P (n – 1, 3) : P (n + 1, 3) = 5: 12, Find n.
Solution:

Question 8.
Find m and n  if P(m + n, 2) = 56, P(m – n, 2) = 12
Solution:
^m+nP₂ = 56, ^m-nP₂ = 12
or, \(\frac{(m+n) !}{(m+n-2) !}=56, \frac{(m-n) !}{(m-n-2) !}=12\)
or, (m + n) (m + n – 1) = 8 × 7
(m – n) (m – n – 1) = 4 × 3
∴ (m + n) = 8, m – n = 4
∴ m = 6, n = 2

Question 9.
Show that
(i) P(n, n) = P(n, n – 1) for all positive integers.
Solution:

(ii) P(m, 1) + P(n, 1) = P(m + n, 1) for all positive integers m, n.
Solution:
^mp₁ + ⁿp₁ = ^m+np₁ ∀ m, n ∈ Z
∴ L.H.S.
= ^mp₁ + ⁿp₁ = m + n = ^m+np₁ = R.H.S.
(∴ ⁿp₁ = n)

Question 10.
How many two-digit even numbers of distinct digits can be formed with the digits 1, 2, 3, 4, and 5?
Solution:
Two-digit even numbers of distinct digits are to be formed with the digits 1, 2, 3, 4, and 5. Here the even numbers must end with 2 or 4. When 2 is placed in the unit place, the tenth place can be filled up by the other 4 digits in 4 different ways. Similarly, when 4 is placed in the unit place, the tenth place can be filled up in 4 different ways.
∴ The total number of two-digit even numbers = 4 + 4 = 8.

Question 11.
How many 5-digit odd numbers with distinct digits can be formed with the digits 0, 1, 2, 3, and 4?
Solution:
5-digit odd numbers are to be formed with distinct digits from the digits 0, 1, 2, 3, and 4. The numbers are to end with 1 or 3 and must not begin with 0.
The 5th place can be filled up by any one of 1 or 3 by 2 ways.

1st

2nd

3rd

4th

5th

Places
The 1st place can be filled by the rest 3 digits except 0 and the digit in 5th place.
The 2nd, 3rd, and 4th places can be filled up by the rest 3 digits in 3! ways.
So total no. of ways = 2 × 3 × 3 ! = 2 × 3 × 2 = 36 ways.

Question 12.
How many numbers, each less than 400 can be formed with the digits 1, 2, 3, 4, 5, and 6 if repetition of digits is allowed?
Solution:
Different numbers each less than 400 are to be formed with the digits 1, 2, 3, 4, 5, and 6 with repetition. Here the numbers are 1-digit, 2-digit, and 3-digit.
∴ The number of 1-digit numbers = 6.
The number of 2-digit numbers = 6² = 36.
The 3-digit number each less than 400 must begin with 1, 2, or 3. So the hundred’s place can be filled by 3 digits and ten’s and unit place can be filled by 6 digits each. So the number of 3 digit numbers = 3 × 6 × 6 = 108.
∴ The total number of numbers = 6 + 36 + 108 = 150.

Question 13.
How many four-digit even numbers with distinct digits can be formed out of digits 0, 1, 2, 3, 4, and 5, 6?
Solution:
Four-digit even numbers mean, they must end with 0, 4, 2, 6. When 0 is placed in last, the 1st place is filled by 1, 2, 3, 4, 5, 6, and the remaining 2 places can be filled by ⁵P₂ ways.
The number of numbers ending with 0 = ⁵P₂ × 6 = 120
Similarly, the number of numbers ending. with 2, 4 and 6 = ⁵P₂  × 5 × 3 = 300
∴ The total number of numbers = 420.

Question 14.
How many integers between 100 and 1000 (both inclusive) consist of distinct odd digits?
Solution:
Integers are to be formed with distinct odd digits between 100 and 1000.
The numbers between 100 and 1000 are 3-digited.
The odd digits are 1, 3, 5, 1, and 9.
The number of distinct 3-digit odd numbers = ⁵P₃ = 5.4.3 = 60.

Question 15.
An unbiased die of six faces, marked with the integers 1, 2, 3, 4, 5, 6, one on each face, thrown thrice in succession. What is the total number of outcomes?
Solution:
An unbiased die of six faces, marked with the integers 1, 2, 3, 4, 5, 6, one on each face is thrown thrice in succession.
∴ The total number of outcomes = 6³ = 216.

Question 16.
What is the total number of integers with distinct digits that exceed 5500 and do not contain 0, 7, and 9?
Solution:
The integer exceeding 55000 must be 4-digited, 5-digited, 6-digited, and 7-digited, as there are seven digits i.e. 1, 2, 3, 4, 5, 6, and 8 to be used for the purpose.
In the 4-digit integers, when 1st place is filled by 5 and 2nd place by .6, the rest two places can be filled by the remaining 5 digits in ⁵P₂ ways. Similarly, when 5 is in 1st place and 8 in 2nd place, the remaining 5 digits be used in ⁵P₂ ways. So the number of 4-digit integers beginning with 5 is 2 × 5P = 40.
When 6 is placed in 1st place, the remaining 3 placed be filled by the remaining 6 digits in 6P3 ways. Similarly, when 8 is placed in 1 st place, the remaining 3 places be filled by the remaining 6 digits in ⁶P₃ ways.
So the number of 4-digit integers starting with 6 and 8 is 2 × ⁶P₃  = 240.
∴ The total number of 4-digit numbers is 40 + 240 = 280.
The number of 5-digit integers is ⁷P₅ = 2520.
The number of 6-digit numbers is ⁷P₆ = 5040 and the number of 7 -digit numbers is ⁷P₆  = 5040.
∴ The total number of integers exceeding 5500 and not containing 0, 7, and 9 is 280 + 2520 + 5040 + 5040 = 12880.

Question 17.
Find the total number of ways in which the letters of the word PRESENTATION can be arranged.
Solution:
The number of letters in the word “PRESENTATION” is 12, out of which there are 2N’s, 2E’s, and 2T’s. So the total number of arrangements.
\(=\frac{12 !}{2 ! 2 ! 2 !}=\frac{1}{8}(12) !\)

Question 18.
Find the numbers of all 4-lettered words (not necessarily having meaning) that can be formed using the letters of the word BOOKLET.
Solution:
We have to form 4-lettered words using the letters of the word BOOKLET. The word contains 7 letters out of which there are 20’s. So there are 6 letters.
∴ The number of 4-lettered words = ⁷P₄ – ⁶P₄ = 7.6.5.4 – 6.5.4.3 = 480

Question 19.
In how many ways can 2 boys and3 girls sit in a row so that no two girls sit side by side?
Solution:
Two boys and 3 girls sit in a row so that no two girls sit side by side.
So the only possibility is boys should be situated in between the two girls. In between 3 girls there are 2 gaps where 2 boys will be site.
The girls will be arranged in 3! and boys in 2! ways.
∴ The total number of ways = 2! × 3!=2 × 6= 12

Question 20.
Five red marbles, four white marbles, and three blue marbles the same shape and size are placed in a row. Find the total number of possible arrangements.
Solution:
5 red, 4 white, and 3 blue marbles of the same size and shape are placed in a row.
∴ The total number of marbles is 12 out of which 5 are of one kind, 4 are of 2nd kind and 3 are of 3rd kind
∴ The total number of possible arrangements
\(=\frac{12 !}{5 ! 4 ! 3 !}=\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 3 \cdot 2}\)
= 12.11.10.3.7 = 27720.

Question 21.
Solve example 2.
Solution:
We have |A| = n, |B| = m
∴ The number of one-one functions from A to B is ^mP_n = \(\frac{m !}{(m-n) !}\) when m> n.
If m = n, the number of one-one functions is \(\frac{m !}{(m-m) !}=\frac{m !}{0 !}\) = m! = n!
If m < n, then there is no possibility of one-one functions.

Question 22.
In how many ways can three men and three women sit at a round table so that no two men can occupy adjacent positions?
Solution:
Since no 2 men are to sit together, there are 4 places available for them corresponding to any one way of sitting of 2 men i.e., two places between the women and 2 places at two ends.

Let m₁ be fixed m₂ can sit in 2 places, w₁ can sit in 3 places, m₃ can sit in 1 place, w₂ can sit in 2 places w₃ can sit in 1 place.
∴ The total number of ways = 2 × 3 × 1 × 2 × 1 = 12

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 7 Linear Inequalities Ex 7(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 7 Linear Inequalities Exercise 7(c)

Solve the following systems of linear inequalities graphically.
Question 1.
2x – y ≥ 0, x – 2y ≤ 0, x ≤ 2, y ≤ 2 [Hint: You may consider the point (2, 2) to determine the SR of the first two inequalities.]
Solution:
2x – y ≥ 0
x – 2y ≤ 0
x ≤ 2
y ≤ 2
Step – 1: Let us draw the lines.
2x – y = 0, x – 2y = 0, x = 2, y = 2
2x – y = 0

X	0	1
y	0	2

x – 2y = 0

X	0	2
y	0	1

Step – 2: Let us consider point (1, 0) which does not line on any of these lines.
Putting x = 1, y = 0 in the inequations we get
2 ≥ 0 (True)
1 ≤ 0 (False)
1 ≤ 2 (True)
0 ≤ 2 (True)
Point (1, 0) satisfies all inequality except x – 2y < 0.
∴ Thus the shaded region is the solution region.

Question 2.
x – y < 1, y – x < 1
Solution:
x – y < 1
y – x < 1
Step – 1: Let us draw the dotted lines.
x – y = 1 and y – x = 1
x – y = 1 ⇒ y = x – 1

X	1	0
y	0	-1

y – x = 1 ⇒ y = x + 1

X	0	-1
y	1	0

Step – 2: Let us consider the point (0, 0) which does not lie on these lines.
Putting x = 0, y = 0 in the inequations
we get
0 < 1 (True)
0 < 1 (True)
∴ (0, 0) satisfies both the inequations.
∴ Thus the shaded region is the feasible region.

Question 3.
x – 2y + 2 < 0, x > 0
Solution:
x – 2y + 2 < 0, x > 0
Step – 1: Let us draw the dotted line x – 2y + 2 = 0
⇒ y = \(\frac{x+2}{2}\)

X	-2	0
y	0	1

Step – 2: Let us consider the point (1, 0) that does not lie on the lines  putting x = 0, y = 0 in the inequation, we get
2 < 0 (false)
1 > 0 (True)
⇒ (1, 0) satisfies x > 0 and does not satisfy x – 2y + 2 < 0.
∴ Thus the shaded region is the solution region.

Question 4.
x – y + 1 ≥ 0, 3x + 4y ≤ 12, x ≥ 0, y ≥ 0
Solution:
x – y + 1 ≥ 0
3x + 4y ≤ 12
x ≥ 0, y ≥ 0
Step – 1: Let us draw the lines.
x – y + 1 = 0
3x + 4y = 12
Now, x – y + 1 = 0 ⇒ y = x + 1

X	0	-1
y	1	0

3x + 4y = 12

X	4	0
y	0	3

Step – 2: Let us consider the point (0, 0) which does not lie on these lines.
Putting x = 0, y = 0 in the inequations
we get
1 ≥ 0 (True)
0 ≤ 12 (False)
∴ (0, 0) satisfies both the inequations and x > 0, y > 0 is the first quadrant.
∴ Thus the shaded region is the solution region.

Question 5.
x + y > 1, 3x – y < 3, x – 3y + 3 > 0
Solution:
x + y > 1
3x – y < 3
x – 3y + 3 > 0
Step – 1: Let us draw the lines.
x + y = 1
3x – y = 3
x – 3y + 3 = 0
Now x + y = 1
⇒ y = 1 –  x

X	1	0
y	0	1

3x – y = 3
⇒ y = 3x – 3

X	1	0
y	0	-3

x – 3y + 3 = 0
⇒ y = \(\frac{x+3}{3}\)

X	-3	0
y	0	1

Step – 2: Let us consider the point (0, 0) that does not lie on these lines. Putting x = 0, y = 0 in the inequations we get,
0 > 1 (False)
0 < 3 (True)
3 > 0 (True)
Thus (0, 0) satisfies 3x – y < 3 and x – 3y + 3 > 0 but does not satisfy x + y > 1
∴ The shaded region is the solution region.

Question 6.
x > y, x < 1, y > 0
Solution:
x > y, x < 1, y > 0
Step – 1: Let us draw the dotted lines.

Step – 2: Let us consider a point (2, 1) that does not lie on any of the lines.
Putting x = 2, y = 2 in the inequations
we get,
2 > 1 (True)
2 < 1 (False)
1 > 0 (True)
⇒ (2, 1) satisfies x > y and y > 0 but does not satisfy x < 1.
∴ Thus the shaded region is the solution region.

Question 7.
x < y, x > 0, y < 1
Solution:
x < y
x > 0
y < 1
Step – 1: Let us draw the dotted lines.
x = y
x = 0
and y = 1

Step – 2: Let us consider point (1, 0) that does not lie on these lines.
Putting x = 0, y = 0 in the inequations
we get
1 < 0 (False)
1 > 0 (True)
0 < 1 (True)
Clearly (1, 0) satisfies x > 0, y < 1 but does not satisfy x < y.
∴ The shaded region is the solution region.

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

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

Question 1.
If Z₁ and Z₂ are two complex numbers then show that
\(\begin{aligned}
& \left|1-z_1 \overline{z_2}\right|^2-\left|z_1-z_2\right|^2 \\
& =\left(1-\left|z_1\right|^2\right)\left(1-\left|z_2\right|^2\right)
\end{aligned}\)
Solution:
Let z₁ and z₂ be two complex numbers.
Let z₁ = a + ib, z₂ = c + id

Question 2.
If a, b, c are complex numbers satisfying a + b + c = 0 and a² + b² + c² = 0 then show that |a| = |b| = |c|
Solution:
Let a + b + c = 0 and a² + b² + c² = 0
Then (a + b + c)² = 0
⇒ a² + b² + c² + 2ab + 2bc + 2ca = 0
⇒ 2(ab + bc + ca) = 0
⇒ ab + bc = – ca
⇒ b(a + c) = – ca
⇒ b(- b) = – ca [a + b + c = 0]
⇒ b² = ca
⇒ b³ = abc
Similarly it can be shown that a³ = abc and c³ = abc
Thus a³ = b³ = c³
⇒ |a³|= |b³| = |c³|
⇒ |a|³ = |b|³ = |c|³
⇒ |a| = |b| = |c|

Question 3.
What do the following represent?
(i) { z : |z – a| + |z + a| = 2c } where |a| < c
Solution:
{ z : |z – a| + |z + a| = 2c } where |a| < c    …….(1)
Here z is a complex number.
Let z = x + iy.
∴ (x, y) is the point corresponding to the complex number z?
Let ‘a’ and ‘ – a’ be two fixed points.
∴ Eqn. (1) implies that the sum of the distances of the point (x, y) from two points la and a’ is constant i.e. 2c
∴ The locus is an ellipse.

(ii) {z : |z – a| – |z + a| = c }
Solution:
Here {z : |z – a| – |z + a| = c } implies that, the difference of the distances of the point (x, y) from two fixed points ‘ – a’ and ‘a’ is a constant i.e. c.
So the locus is a hyperbola.

(iii) What happens in (i) |a| > c?
Solution:
In(i), if |a| > c. then there is no locus. But if |a| = c, then the locus reduces to a straight line.

Question 4.
Given cos α + cos β + cos γ = sin α + sin β + sin γ = 0 Show that cos 3α + cos 3β + cos 3γ = 3 cos (α + β + γ)
Solution:
Let a = cos α + i sin α,
b = cos β + i sin β
c = cos γ + i sin γ
∴ a + b + c = ( cos α + cos β + cos γ) + i ( sin α + sin β + sin γ)
= 0 + i0 = 0
∴ a³ + b³ + c³ – 3 abc
= (a + b + c )( a² + b² + c² – ab – bc – ca) = 0
or, a³ + b³ + c³ = 3 abc
or, ( cos α + i sin α)³ + (cos β + i sin β)³ + ( cos γ + i sin γ)³
= 3( cos α + i sin α) (cos β + i sin β) ( cos γ + i sin γ)
or, cos 3α + i sin 3α + cos 3β + i sin 3β + cos 3γ + i sin 3γ
= 3[cos (α + β + γ) + i sin (α + β + γ)]
or, (cos 3α + cos 3β + cos 3γ) + i (sin 3α + sin 3β + sin 3γ)
= 3 cos (α + β + γ) + i 3 sin (a + β + γ)
∴ cos 3α + cos 3β + cos 3γ
= 3 cos (α + β + γ) and sin 3α + sin 3β + sin 3γ
= 3 sin (α + β + γ)

Question 5.
Binomial theorem for complex numbers. Show that (a+b)ⁿ = a^{n n}C₁a^n-1b +  …..+ ⁿc_ra^n-rb^r + …..+ bⁿ where a,b ∈ C and n, rule of multiplication of complex numbers and the relation ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r)
Solution:
Let a and b be two complex numbers
Let a = α₁ + iβ₁, b = α₂ + iβ₂
(a + b)¹ = (α₁ + iβ₁ + α₂ + iβ₂)¹
= α₁ + iβ₁ + α₂ + iβ₂
= (α₁ + iβ₁)¹ + ¹C₁(α₁ + iβ₁)^1-1(α₁ + iβ₁)¹
= a¹ + ¹C₁ a^1-1 b¹
∴ P₁ is true
Let P_k be true
i.e., (a + b)^k = a^k + ^kC₁ a^k-1 b¹ + … + b^k
where a.b ∈ C
Now ( a + b)^k+1 = (a + b)^k (a + b)¹
= (a^k + ^kC₁ a^k-1b¹ +…+ b^k) (a + b)
= a^k-1 + ba^k+ ^kC₁ a^kb + ^kC₁a^k-1b² + … + b^k+1
= a^k+1 + a^kb(^kC₁+ 1)+ …+ b^k+1
= a^k+1 + ^k+1C₁a^kb¹ + ^k+1 C₂a^k-1b² +… + b^k+1
∴ P_k+1 is true
∴ P_n is true for all values of n ∈ N

Question 6.
Use the Binomial theorem and De Moiver’s theorem to show
cos 3θ = 4 cos ³ θ – 3 cos θ,
sin 3θ = 3 sin θ – 4 sin ³ θ
Express cos nθ as a sum of the product of powers of sin θ and cos θ. Do the same thing for sin nθ.
Solution:
We have (cos θ + i sin θ)³
= cos 3θ + i sin 3θ       …..(1)
But by applying the Binomial theorem, we have
(cos θ + i sin θ)³
= cos ³ θ + ³C₁ (cos θ)^3-1 (i sin θ)¹ + ³C₂ (cos θ)^3-2  (i sin θ)² + (i sin θ)³
= cos3θ + 3i cos2θ sin θ + 3i² cos θ sin² θ + i³ sin³ θ
= (cos³ θ – 3 cos θ sin² θ) + i(3 cos² θ sin θ – sin³ θ)
∴ cos ³ θ =cos³ θ – 3 cos θ(1 – cos² θ)
= cos³ θ – 3 cos θ + 3 cos ³ θ
= 4 cos³ θ – 3 cos θ and
sin 3θ = 3 cos² θ sin³ θ – sin³ θ
= 3 (1 – sin² θ) sin θ – sin³ θ
= 3 sin θ – 3 sin³ θ – sin³ θ
= 3 sin θ – 4 sin³ θ (Proved)
Again, (cos θ + i sin θ)ⁿ
= cos nθ + i sin nθ         ….(3)
Also, (cos θ + i sin θ)ⁿ
= cosⁿ θ + ⁿC₁ cos^n-1 θ ( i sin θ) + ⁿc₂ cos ^n-2 θ (i sin θ)² + …+ (i sin θ)
= cosⁿ θ – ⁿC₂ cos^n-2 θ sin² θ + ⁿC₄ cos^n-4 θ sin⁴ θ – …) + i (ⁿC₁ cos^n-1 θ sin θ – ⁿC₃ cos^n-3 θ sin ³ θ) + ⁿC₅ cos^n-5 θ sin⁵ θ – …)    …..(4)
Equating real part and imaginary parts in (1) and (3), we have
cos nθ = cosⁿ θ – ⁿC₂ cos^n-2 θ × sin² θ + ⁿC₄ cos^n-4 θ sin⁴ θ …
and sin nθ = ⁿC₁ cos^n-1 θ sin θ – ⁿC₃ cos^n-3 θ sin³ θ + ⁿC₅ cos^n-5 θ sin⁵ θ…

Question 7.
Find the square root of
(i) – 5 + 12 √-1
Solution:

(ii) – 11 – 60 √-1
Solution:
Let \(\sqrt{-11-60 \sqrt{-1}}\) = x + iy
Squaring both sides we get
– 11 – 60i = (x + iy)²

(iii) – 47 + 8 √-1
Solution:

As 2ab = 8 > 0, a and b have the same sign.
∴ \(\sqrt{-47+8 i}\)
\(=\pm\left(\sqrt{\frac{\sqrt{2273}-47}{2}}+i \frac{\sqrt{2273}+47}{2}\right)\)

(iv) – 8 + √-1
Solution:

(v) a² – 1 +2a √-1
Solution:
a² – 1 + 2a √-1
= a² + i² + 2ai = (a + i)²
∴ \(\sqrt{a^2-1+2 a \sqrt{-1}}\) = ±(a + i)

(vi) 4ab – 2 (a² – b²) √-1
Solution:
4ab – 2 (a² – b²) √-1
= (a + b)² – (a – b)² – 2(a² – b²)i
= (a + b)² + (a – b)²i² – 2(a + b)(a – b)i
= (a + b) – i(a – b)²
∴ \(\sqrt{4 a b-2\left(a^2-b^2\right) \sqrt{-1}}\)
= ±[(a + b) – i (a – b)]

Question 8.
Find the values of cos72° ….
Solution:
Let 18° = θ then 5θ = 90°
⇒ 3θ = 9θ – 2θ
⇒ cos 3θ = cos (9θ – 2θ) = sin 2θ
⇒ 4 cos³ θ – 3 cos θ = 2 sin θ cos θ
⇒ 4 cos² 0 – 3 = 2 sin 0 (∴ cos θ = cos 18° ≠ 0)
⇒ 4(1 – sin² θ) – 3 = 2 sin θ
⇒ 4 sin² θ + 2 sin θ – 1 = 0
⇒ sin θ = \(\frac{-2 \pm \sqrt{4+16}}{2 \times 4}=\frac{-1 \pm \sqrt{5}}{4}\)
⇒ sin 18° = \(\frac{-1+\sqrt{5}}{4}=\frac{\sqrt{5}-1}{4}\)
(∴ 18° is a cut)
Now cos 72° = cos (90° – 18°)
= sin 18° = \(\frac{\sqrt{5}-1}{4}\)
For other methods refer 148 pages of the text book.

Question 9.
Find the value of cos 36°.
Solution:
We have 36° = \(\frac{\pi^0}{5}\)
∴ cos 36° = cos \(\frac{\pi}{5}\)
Let α = cos \(\frac{\pi}{5}\) + i sin \(\frac{\pi}{5}\)
be the root of the equation x⁵ + 1 = 0
Again, if x⁵ + 1 = 0
or, x⁵ = – 1 = cosπ + i sinπ
or, x = (cos π + i sin π )^1/5
= [cos (π + 2kπ)+ i sin (π + 2kπ)]^1/5
or, x = cos \(\frac{(2 k+1) \pi}{5}+i \sin \frac{(2 k+1) \pi}{5}\)
where 2kπ is the period of sine and cosine and k = 0, 1, 2, 3, 4
∴ The eqn x⁵ + 1 = 0 has 5 roots out of which -1 is one root which corresponds to k = 2
Again,
x⁵ + 1 = (x + 1)(x⁴ – x³ + x² – x + 1)
So their 4 roots will be obtained on solving the eqn.
x⁴ – x³ + x² – x + 1 = 0
we have, x⁴ – x³ + x² – x + 1 = 0
or, x² – x + 1 – \(\frac{1}{x}+\frac{1}{x^2}\) = 0
(Dividing both sides by x²)

Re α i.e. cos \(\frac{\pi}{5}=\frac{1+\sqrt{5}}{5}=\frac{\sqrt{5}+1}{4}\)
and cos.108° = \(\frac{1-\sqrt{5}}{4}\)

Question 10.
Evaluate cos \(\frac{2 \pi}{17}\) using the equation x¹⁷ – 1 = 0
Solution:
x¹⁷ – 1 = 0
or, x¹⁷ = 1 = cos 0° + i sin 0°
= cos (0 + 2kπ) + i sin (0 + 2kπ)
= cos 2kπ + i sin 2kπ
or x = (cos 2kπ + i sin 2kπ)^1/17
= cos \(\frac{2k \pi}{17}\) + i sin \(\frac{2k \pi}{17}\)
If k = 1, x = cos \(\frac{2 \pi}{17}\) + i sin \(\frac{2 \pi}{17}\)
If k = 0 , x = 1
As x¹⁷ – 1 = (x – 1) (x¹⁶ + x¹⁵ + …. +1)
So one root of the eqn. x¹⁷ – 1 = 0 is 1 and all other roots are the roots of the eqn.
x¹⁶ + x¹⁵ + ….+ 1 = 0
∴ The value of cos \(\frac{2 \pi}{17}\) can be found from the roots of the eqn.  (1)

Question 11.
Solve the equations.
(i) z⁷ = 1
Solution:
z⁷ = 1 = cos 0 + i sin 0
= cos (0 + 2kπ) + i sin (0 + 2kπ)
= cos 2kπ + i sin 2kπ
∴ z = (cos 2kπ + i sin 2kπ)^1/7
= cos \(\frac{2k \pi}{7}\) + i sin \(\frac{2k \pi}{7}\)
where k = 0, 1, 2, 3, 4, 5, 6

(ii) z³ = i
Solution:

(iii) z⁶ = – i
Solution:

(iv) z³ = 1 + i
Solution:

Question 12.
If sin α + sin β + cos γ = 0
= cos α + cos β + cos γ = 0
Show that
(i) sin 3α + sin 3β + sin 3γ = 3 sin (α + β + γ)
Solution:
Refer to Q. No. 4

(ii) sin² α + sin² β + sin² γ = cos² α + cos² β + cos² γ =3/2
Solution:
Let x = cos α + i sin α,
y = cos β + i sin β
z = cos β + sin β
∴ x + y + z = (cos α + cos β + cos γ) + i ( sin α + sin β + sin γ) = 0 +i0= 0
∴ xy + yz + zx = xyz (\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)) = 0
Since \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\) = cos α – i sin α + cos β – i sin β + cos γ – i sin γ
= (cos α + cos β + cos γ) – i (sin α + sin β + sin γ) = 0 – i0 =0
∴ (x + y + z)² = x ² + y² +z² + 2(xy + yz + zx)
= x² + y² + z² + 0 = x² + x² + z²
or 0 = x² + y² + z²
∴ x² +y² + z² =0
or, (cos α + i sin α)² + (cos β + i sin β)² + ( cos γ + i sin γ)² = 0
or, cos 2α + i sin 2α + cos 2β + i sin 2β + cos 2γ + i sin 2γ = 0
or, (cos 2α + cos 2β + cos 2γ) + i(sin 2α + sin 2β + sin 2γ) = 0
∴ cos 2α + cos 2β + cos 2γ = 0
or, cos² α – sin² α + cos² β – sin² β + cos² γ – sin² γ =0
or, (cos² α + cos² β + cos² γ) = (sin² α + sin² β + sin² γ)
But cos² α + sin² α + cos² β + sin² β + cos² γ + sin² γ = 1 + 1 + 1 = 3
∴ cos² α + cos² β + cos² γ = sin² α + sin² β + sin² γ = 3/2   (Proved)

Question 13.
If x + \(\frac{1}{x}\) = 2 cos θ
Show that \(x^n+\frac{1}{x^n}\) = 2 cos nθ

Question 14.
x_r = cos a^r + i sin a^r
r =1, 2, 3 and x₁ + x₂ + x₃ = 0 Show that \(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}\) = 0
Solution:

Question 15.
Show that \(\left(\frac{1+\sin \theta+i \cos \theta}{1+\sin \theta-i \cos \theta}\right)^n\) = \(\cos \left(\frac{n \pi}{2}-n \theta\right)+i \sin \left(\frac{n \pi}{2}-n \theta\right)\)
Solution:

Question 16.
If α and β are roots x² – 2x + 4 = 0 then show that \(\alpha^n+\beta^n=2^{n+1} \cos \frac{n \pi}{3}\)
Solution:
we have x² – 2x + 4 = 0

= \(2^n \times 2 \cos \frac{n \pi}{3}=2^{n+1} \cos \frac{n \pi}{3}\)

Question 17.
For a positive integer n show that
(i) (1 + i)ⁿ + (1 – i)ⁿ = \(2^{\frac{n+2}{2}} \cos \frac{n \pi}{4}\)
Solution:

(ii) (1 + i√3)ⁿ + (1 – i√3)ⁿ = \(2^{n+1} \cos \frac{n \pi}{3}\)
Solution:

Question 18.
Let x + \(\frac{1}{x}\) = 2 cos α, y + \(\frac{1}{y}\) = 2 cos β, z + \(\frac{1}{z}\) = 2 cos γ. Show that
(i) 2 cos (α + β + γ) = xyz + \(\frac{1}{xyz}\)
Solution:
We can take x = cos α + i sin α
y = cos β + i sin β, z = cos γ + i sin γ
∴ xyz = (cos α + i sin α ) (cos β + i sin β ) (cos γ + i sin γ)
= cos (α + β + γ) – i sin (α + β + γ)
∴ \(\frac{1}{xyz}\) = cos (α + β + γ) – i sin(α + β + γ)
∴ xyz + \(\frac{1}{xyz}\) = 2 cos(α + β + γ)

(ii) 2 cos (pα + qβ + rγ) = \(x^p y^q z^r+\frac{1}{x^p y^q z^r}\)
Solution:

Question 19.
Solve x⁹ + x⁵ – x⁴ = 1
Solution:
x⁹ + x⁵ – x⁴ = 1
or, x⁵(x⁴ + 1) – (x⁴ + 1) = 0
or, (x⁵ – 1) (x⁴ + 1) = 0
x⁴ + 1 = 0 and x⁵ – 1 = 0
x⁴ = – 1 = cos π + i sin π
= cos (π + 2nπ) + i sin (π + 2nπ)
∴ x = [cos (2n + 1) π + 1 sin (2n + 1) π]^1/4
= \(\cos \frac{2 n+1 \pi}{4}+i \sin \frac{2 n+1 \pi}{4}\)
for n = 0, 1, 2, 3
Again, x⁵ – 1 = 0 or, x⁵ = 1
or, x⁵ = cos 0 + i sin 0
= cos 2nπ + i sin 2nπ
or, x = (cos 2nπ + i sin 2nπ)^1/5
= \(\cos \frac{2 n \pi}{5}+i \sin \frac{2 n \pi}{5}\)
Where n = 0, 1, 2, 3, 4.

Question 20.
Find the general value of θ if (cos θ + i sin θ) (cos 2θ + i sin 2θ),…..(cos nθ + i sin nθ) =1
Solution:

Question 21.
If z = x + iy show that |x| + |y| ≤ √2 |z|
Solution:
z = x + iy
∴ |z| = \(\sqrt{x^2+y^2}\)
∴ |z| = x² + y²
We have (|x| – |y|)² ≥ 0
⇒ |x|² + |y|² -2|x||y|> 0
⇒ 2(|x|² + |y|²) – 2|x||y| ≥ |x|² + |y|²
⇒ 2(|x|² + |y|²) ≥ |x|² + |y|² + 2|x||y|
⇒ 2|z|² ≥ (|x| + |y|)²
⇒ √2|z| ≥ |x| + |y|
⇒ |x| + |y| ≤ √2|z|

Question 22.
Show that
Re (Z₁Z₂) = Re z₁, Re z₂ – Im z₁, Im z₂
Im (Z₁Z₂) = Re z₁, Im z₂ + Re z₂ Im z₁
Solution:
Let z₁ = a + ib, z₂ = c + id
∴ z₁, z₂ = (a + ib) (c + id)
= ac + iad + ibc + i²bc
= (ac – bd) + i (ad + be)
∴ Re (z₁, z₂) = ac – bd – Re z₁. Re z₂
– Im z₁,. Im z₂
Again, Im z₁, z₂ = ad + be
= Re z₁. Im z₂ + Im z₁,. Re z₂

Question 23.
What is the value of arg ω + arg ω²?
Solution:
arg ω = arg ω² = arg (ω • ω²)
= arg (ω³) = arg (1) = 2nπ
∴ The principal agrument = 0.

Question 24.
If |z₁| ≤ 1, |z₂| ≤ 1 show that \(\left|1-z_1 \overline{z_2}\right|^2-\left|z_1-z_2\right|^2=\left(1-\left|z_1\right|^2\right)\left(1-\left|z_2\right|^2\right)\) Hence and otherwise show that. \(\left|\frac{z_1-z_2}{1-z_1 z_2}\right|<1 \text { if }\left|z_1\right|<1,\left|z_2\right|<1\)
Solution:

Question 25.
If z₁² + z₂² + z₃² – z₁z₂ – z₂z₃ – z₃z₄ = 0 Show that |z₁ – z₂| = |z₂ – z₃| = |z₃ – z₁|
Solution:
Let z₁² + z₂² + z₃² – z₁z₂ – z₂z₃ – z₃z₄ = 0
⇒ 2z₁² + 2z₂² + 2z₃² – 2z₁z₂ – 2z₂z₃ – 2z₃z₄ = 0
⇒ (z₁ – z₂)² + (z₂ – z₃)² + (z₃ – z₁)² = 0
Put a = z₁ – z₂, b = z₂ – z₃, c = z₃ – z₁
Then a + b + c = 0 and a² + b² + c² = 0 As in Q2 we can show that |a| = |b| = |c|
⇒ |z₁ – z₂| = |z₂ – z₃| = |z₃ – z₁|

Question 26.
If |a| < |c| show that there are complex numbers z satisfying |z – a| = |z + a| = 2|c|
Solution:
Let z = x + iy
∴ |z – a| + |z + a| = 2c
or, |x + iy – a| + |x + iy + a| = 2c

Question 27.
Solve \(\frac{(1-i) x+3 i}{2+i}+\frac{(3+2 i) y+i}{2-i}=-i\) where x, y, ∈ R.
Solution:

Question 28.
If (1 + x + x²)ⁿ = p₀ + p₁x + p₂x² + …..+ p_2nx²ⁿ, then prove that p₀ + p₃ + p₆ + …..+ 3^n-1
Solution:
Given, (1 + x + x²)ⁿ = p₀ + p₁x + p₂x² + …..+ p_2nx²ⁿ
putting x = ω we get

Question 29.
Find the region on the Argand plane on which z satisfying
[Hint Arg (x + iy) =\(\frac{\pi}{2}\) = 0, y>0]
(i) 1 < |z – 2i| < 3
Solution:
Let z = x + iy
The given inequality is
1 < |x + i(y – 2)| < 3
⇒ \(1<\sqrt{x^2+(y-2)^2}<3\)

(ii) arg \(\left(\frac{z}{z+i}\right)=\frac{\pi}{2}\)
Solution:

As all are +ve we have
1 < x² + (y – 2)² < 9
x² + (y – 2)² < 9 is the region inside the circle with center (0, 2) and radius 1.
x² + (y – 2)² > 1 is the region outside the circle with center (0, 2) and radius.
∴ 1 < |z – 2i| < 3 is the region between two concentric circles with center (0, 2) and radius 1 and 3 which is shows below.

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

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

Question 1.

Fill in the blanks choosing the correct answer from the brackets.

(i) The number of solutions of  2 sin θ – 1 = 0 is__________. (one, two, infinite)
Solution:
Infinite

(ii) If cos α = cos β, then α + β = ____________. (0, π, 2π)
Solution:
2π

(iii) The number of solution(s) of 2 sin θ + 1 = 0 is__________. (zero, two, infinite)
Solution:
Zero

(iv) If tan θ = tan α and 90° < α < 180°, then θ can be in ____________quadrant. (1st, 3rd, 4th)
Solution:
4th

(v) If tan x. tan 2x. tan 7x = tan x + tan 2x + tan 7x, then x = _____________. (\(\frac{\pi}{4}, \frac{\pi}{5}, \frac{\pi}{10}\))
Solution:
\(\frac{\pi}{10}\)

(vi) For_____________value of θ, sin θ + cos θ = √2. (\(\frac{\pi}{4}, \frac{\pi}{2}, \frac{\pi}{3}\))
Solution:
\(\frac{\pi}{4}\)

(vii) The number of values of x for which cos² x = 1 and x² ≤ 4 is______________. (1, 2, 3)
Solution:
1

(viii) In the 1st quadrant the solution of tan² θ = 3 is_____________. (\(\frac{\pi}{2}, \frac{\pi}{3}, \frac{\pi}{4}\))
Solution:
\(\frac{\pi}{3}\)

(ix) The least positive value of θ for which 1 + tan θ = 0 and √2 cos θ + 1 = 0 is___________. (\(\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}\))
Solution:
\(\frac{3 \pi}{4}\)

(x) the least positive value of x for which tan 3x = tan x is______________. (\(\frac{\pi}{2}, \frac{\pi}{3}, \pi\))
Solution:
\(\frac{\pi}{2}\)

Question 2.
Find the principal solution of the following equations:
(i) sin θ = sin 2θ
Solution:
sin θ = sin 2θ
or, 2θ = nπ + (-1)ⁿ θ
or, 2π – (-1)ⁿ θ = nπ
or, θ = \(\frac{n \pi}{2-(-1)^n}\)
when n = 0, θ = 0
when n = 1, θ = \(\frac{\pi}{3}\)
when n = 2, θ = 2π
when n = 3, θ = π
when n = 4, θ = 4π
when n = 5, θ = \(\frac{5 \pi}{3}\)
∴ The principal solution are 0, \(\frac{\pi}{3}\), π, \(\frac{5 \pi}{3}\)

(ii) √3 sin θ – cos θ = 2
Solution:
√3 sin θ – cos θ = 2
or, \(\frac{\sqrt{3}}{2}\) sin θ – 1/2 cos θ = 1

which is the only principal solution.

(iii) cos² θ + sin θ + 1 = 0
Solution:
cos² θ + sin θ + 1 = 0
or, 1 – sin² θ + sin θ + 1 = 0
or,  sin² θ – sin θ + 2 = 0
or, sin² θ – 2 sin θ + sin θ – 2 = 0
or, sinθ(sinθ – 2) + (sinθ – 2) = 0
or, (sinθ – 2) (sinθ + 1) = 0
∴ sinθ = 2, sinθ = – 1
= sin \(\left(-\frac{3 \pi}{2}\right)\) or, θ = – \(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\)
∴ The principal solution is \(\frac{3 \pi}{2}\).

(iv) sin 4x + sin 2x = 0
Solution:

(v) sin x + cos x = \(\frac{1}{\sqrt{2}}\)
Solution:

Question 3.
Find the general solution of the following equations:
(i) cos 2x = θ
Solution:
cos 2x = θ
or, 2x = (2n + 1)\(\frac{\pi}{2}\)
or, x = (2n + 1)\(\frac{\pi}{4}\), n∈Z

(ii) sin(x° + 40°) = \(\frac{1}{\sqrt{2}}\)
Solution:

(iii) sin 5θ = sin 3θ
Solution:
sin 5θ = sin 3θ
or, 5θ = nπ + (-1)ⁿ 3θ
or, 5θ – (-1)ⁿ 3θ = nπ
or, θ[5 – (-1)ⁿ3] = nπ
or, θ = \(\frac{n \pi}{5-(-1)^n 3}\)

(iv) tan ax = cot bx
Solution:

(v) tan² 3θ = 3
Solution:

Question 4.
Solve the following:
(Hints : cos x ≠ 0 and sin² x- sin x + 1/2 = 0)
(i) tan² x + sec² x = 3
Solution:
tan² x + sec² x = 3
or, tan² x + 1 + tan2 x = 3
or, 2tan² x = 2
or, tan² x = 1
or, tan x = ± 1 = tan \(\left(\pm \frac{\pi}{4}\right)\)
∴ x = nπ ± \(\frac{\pi}{4}\)

(ii) 4 sin² x + 6 cos² x = 5
Solution:
4 sin² x + 6 cos² x = 5
or, 4 sin² x + 6(1 – sin² x) = 5
or, 4 sin² x + 6 – 6 sin² x = 5
or, 6- 2 sin² x = 5
or, 2 sin² x = 1
or, sin² x = 1/2
or, sin x = ± \(\frac{1}{\sqrt{2}}\) = sin \(\left(\pm \frac{\pi}{4}\right)\)
or, x = nπ + (-1)ⁿ \(\left(\pm \frac{\pi}{4}\right)\)

(iii) 3 sin x + 4 cos x = 5
Solution:

(iv) 3 tan x + cot x = 5 cosec x
Solution:

(v) cos x + √3 sin x = √2
Solution:

(vi) sin 3x – 2 cos² x = 0
Solution :
sin 2x – 2 cos² x = 0
or, 2 sin x cos x – 2 cos² x = 0
or, 2 cos x(sin x – cos x) = 0
∴ cos x = 0, sin x = cos x
∴ x = (2n + 1)\(\frac{\pi}{2}\), tan x = 1 = tan \(\frac{\pi}{4}\)
or, x = nπ + \(\frac{\pi}{4}\)

(vii) sec θ + tan θ = √3
Solution:

(viii) cos 2θ – cos θ = sin θ – sin 20
Solution:

(ix) sin θ + sin 2θ + sin 3θ + sin 4θ = 0
Solution:

(x) cos 2x° + cos x° – 2 = 0
Solution:
cos 2x° + cos x° – 2 = 0
or, 2 cos² x° – 1 + cos x° – 2 = 0
or, 2 cos² x° + cos x° – 3 = 0
or 2 cos² + 3cos x° – 2cos x°- 3 = 0
or, cos x°(2 cos x° + 3) – 1(2 cos x° + 3) = 0
or, (2 cos x° + 3)(cos x° – 1) = 0
∴ cos x° = 1 = cos 0°
∴ x° = 2nπ ± 0 = 2nπ
or, \(\frac{\pi x}{180}\) = 2nπ
or, x = 360 n
Again 2 cos x° + 3 = 0
⇒ cos x° = – 3/2 which has no solution.
Hence x = 360 n.

(xi) tan θ + tan 2θ = tan 3θ
Solution:

(xii) tan θ + tan (\(\theta+\frac{\pi}{3}\)) + tan (\(\theta+\frac{2\pi}{3}\)) = 3
Solution:

(xiii) cot² θ – tan² θ = 4 cot 2θ
Solution:

(xiv) cos 2θ = \((\sqrt{2}+1)\left(\cos \theta-\frac{1}{\sqrt{2}}\right)\)
Solution:

(xv) sec θ – 1 = \((\sqrt{2}-1)\) tan θ
Solution:

⇒ θ = 2nπ + \(\frac{\pi}{4}\)

(xvi) 3cot² θ – 2 sin θ = 0
Solution:

(xvii) 4 cos x. cos 2x . cos 3x = 1
Solution:
4 cos x cos 2x cos 3x = 1
⇒ 2 cos x cos 2x. 2 cos 3x = 1
⇒ (cos 3x + cos x) 2 cos 3x = 1
⇒ 2 cos² 3x + 2 cos 3x cos x = 1
⇒ 2 cos² 3x – 1 + cos 4x + cos 2x = 0
⇒ cos 6x + cos 4x + cos 2x = 0
⇒ cos 6x + cos 2x + cos 4x = 0
⇒ 2 cos 4x cos 2x + cos 4x = 0
⇒ cos 4x (2 cos 2x + 1) = 0
⇒ cos 4x = 0, cos 2x = – 1/2
cos 4x = 0 ⇒ 4x = (2n + 1) \(\frac{\pi}{2}\)

(xviii) cos 3x – cos 2x = sin 3x
Solution:

⇒ 1 – 2 sin x cos x = y²
∴ Equation (1) reduces to
1 – 2(1 – y²) + y = 0
⇒ 2y² + y – 1 = 0
⇒ (2y- 1) (y + 1) = 0

(xix) cos x + sin x = cos 2x + sin 2x
Solution:
cos x + sin x – cos 2x + sin 2x
[Refer (viii)]

(xx) tan x + tan 4x + tan 7x = tan x. tan 4x. tan 7x
Solution:
tan x + tan 4x + tan 7x = tan x tan 4x tan 7x
or, tan x + tan 4x
= – tan 7x + tan x tan 4x tan 7x
= – tan 7x (1 – tan x tan 4x)
or, \(\frac{\tan x+\tan 4 x}{1-\tan x \tan 4 x}\) = – tan 7x
or, tan (x + 4x) = tan (π – 7x)
or, tan 5x = tan (π – 7x)
or, 5x = nπ + π – 7x
or, 12x = π(n + 1)
or, x = \(\frac{\pi(n+1)}{12}\), n∈Z

(xxi) 2(sec² θ + sin² θ) = 5
Solution:

(xxii) \((\cos x)^{\sin ^2 x-\frac{3}{2} \sin x+\frac{1}{2}}=1\)
Solution:
\((\cos x)^{\sin ^2 x-\frac{3}{2} \sin x+\frac{1}{2}}=0\)
As cos x ≠ 0.
we have sin² – \(\frac{3}{2}\) sin x + \(\frac{1}{2}\) = 0
∴ 2 sin² x – 3 sin x + 1 = 0
or, 2 sin² x – 2 sin x – sin x + 1 = 0
or, (2sin x – 1)(sin x – 1) = 0
∴ sin x = \(\frac{1}{2}\) or, sin x = 1
But as cos x ≠ 0, we have sin x ≠ 1
∴ sin x = \(\frac{1}{2}\) = sin \(\frac{\pi}{6}\)
∴ x = nπ + (-1)ⁿ \(\frac{\pi}{6}\), n∈Z

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

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

Question 1.
In the following questions, write ‘T’ for true and ‘F’ for false statements.
(i) If tan x + tan y = 5 and tan x, tan y = 1/2  then cot (x + y) = 10
Solution:
False

(ii) √3 (1 + tan 15°) = 1 – tan 15°
Solution:
False

(iii) If θ lies in 3rd quadrant, then cos \(\frac{\theta}{2}\) + sin \(\frac{\theta}{2}\) is positive.
Solution:
True

(iv) 2 sin 105°. sin 15° = 1/2.
Solution:
True

(v) If cos A = cos B = 1 then tan\(\frac{A+B}{2}\). tan\(\frac{A+B}{2}\) = 1
Solution:
False

(vi) cos 15° cos\(7 \frac{1}{2}^{\circ}\). sin \(7 \frac{1}{2}^{\circ}\) = 1
Solution:
False

(vii) sin 20° (3 – 4 cos² 70°) = \(\frac{\sqrt{3}}{2}\)
Solution:
True

(viii) √3 (3 tan 10° – tan³ 10°) = 1 – 3tan² 10°
Solution:
True

(ix) \(\frac{2 \tan 7 \frac{1^{\circ}}{2}\left(1-\tan ^2 7 \frac{1^{\circ}}{2}\right)}{\left(1+\tan ^2 7 \frac{1^{\circ}}{2}\right)^2}\) = 1
Solution:
False

(x) The minimum value of sin θ. cos θ is (-1)².
Solution:
False

Question 2.
In the following questions, fill in the gaps with correct answers choice from the brackets.
(i) If α and β lie in 1st and 2nd quadrants respectively, and if sin α = 1/2, sin β = 1/3, then sin (α + β) = _______. \(\left(\frac{1}{2 \sqrt{3}}+\frac{\sqrt{2}}{3}, \frac{1}{2 \sqrt{3}}-\frac{\sqrt{2}}{3}, \frac{-1}{2 \sqrt{3}}+\frac{\sqrt{2}}{3}\right)\)
Solution:
\(\frac{1}{2 \sqrt{3}}+\frac{\sqrt{2}}{3}\)

(ii) if tan α = 1/2, tan β = 1/3, then α + β = ______      \(\left(\frac{\pi}{2}, \frac{\pi}{4}, \frac{\pi}{3}\right)\)
Solution:
\(\frac{\pi}{4}\)

(iii) The value of \(\frac{\cos 15^{\circ}+\sin 15^{\circ}}{\cos 15^{\circ}-\sin 15^{\circ}}\) = ______ \(\left(\frac{\sqrt{3}}{2}, \sqrt{3}, \frac{1}{\sqrt{3}}\right)\)
Solution:
√3

(iv) if \(\frac{1+\sin A}{\cos A}\) = √2 + 1, then the value of \(\frac{1-\sin A}{\cos A}\) is_________, \(\left(\frac{1}{\sqrt{2}-1}, \sqrt{2}-1, \sqrt{2}+1\right)\)
Solution:
√2 – 1

(v) sin 105°. cos 105° = \(\left(\frac{1}{2},-\frac{1}{4},-\frac{1}{2}\right)\)
Solution:
– 1/4

(vi) 2 sin\(67 \frac{1}{2}^{\circ}\) cos\(22 \frac{1}{2}^{\circ}\) = ___ \(\left(1-\frac{1}{\sqrt{3}}, 1+\frac{1}{\sqrt{2}},-1+\frac{1}{\sqrt{2}}\right)\)
Solution:
1 + \(\frac{1}{\sqrt{2}}\)

(vii) sin 35° + cos 5° =____ (2 cos 25°, √3 cos 25°, √3 sin 25°)
Solution:
√3 cos 25°

(viii) sin² 24° – sin² 26° =_____ \(\left(\frac{\sqrt{5}+1}{8}, \frac{\sqrt{5}-1}{8}, \frac{\sqrt{5}-1}{4}\right)\)
Solution:
\(\frac{\sqrt{5}-1}{8}\)

(ix) sin 70° (4 cos² 20° – 3) =_____ \(\left(\frac{\sqrt{3}}{2}, \frac{1}{2}, \sqrt{3}\right)\)
Solution:
1/2

(x) cos 3θ + sin 3θ is maximum if θ =_____ (60°, 15°, 45°)
Solution:
15°

(xi) sin 15° – cos 15° = _____ (1/2, 0, positive, negative)
Solution:
Negative

(xii) If θ lies in the third quadrant and tan θ = 2, then the value of sin θ is ____. \(\left(\frac{1}{\sqrt{5}},-\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}, \frac{-2}{\sqrt{5}}\right)\)
Solution:
\(\frac{-2}{\sqrt{5}}\)

(xiii) The correct expression is. (sin 1° > sin 1, sin 1° < sin 1, sin 1° = sin 1, sin 1° = \(\frac{\pi}{180^{\circ}}\) sin 1)
Solution:
sin 1° < sin 1

(xiv) The correct expression is —. (tan 1 > tan 2, tan 1 < tan 2, tan 1 =  1/2 tan 2, tan 1 < 0)
Solution:
tan 1 > tan 2

Question 3.
Prove the following
(i) sin A. sin (B – C) + sin B sin (C – A) + sin C. sin(A – B) = 0
Solution:
L. H. S
= sin A sin (B – C) + sin B sin (C – A) + sin C sin (A – B)
= sin A (sin B cos C – cos B sin C) + sin B (sin C cos A – cos C sin A) + sin C (sin A cos B – cos A sin B)
= sin A sin B cos C – sin A cos B sin C + cos A sin B sin C – sin A sin B cos C + sin A cos B sin C – cos A sin B sin C
= 0 = R. H. S

(ii) cos A. sin (B – C) + sin B sin (C – A) + cos C. sin(A – B) = 0
Solution:
L.H.S.
= cos A sin (B – C) + cos B sin (C – A) + cos C sin (A – B)
= cos A (sin B cos C – cos B sin C) + cos B (sin C cos A – cos C sin A) + cos C (sin A cos B – cos A sin B)
= cos A sin B cos C – cos A cos B sin C + cos A cos B sin C- sin A cos B cos C + sin A cos B cos C – cos A sin B cos C = 0 = R. H. S.

(iii) \(\frac{\sin (B-C)}{\sin B \cdot \sin C}\) + \(\frac{\sin (C-A)}{\sin C \cdot \sin A}\) + \(\frac{\sin (A-B)}{\sin A \cdot \sin B}\) = 0
Solution:
L. H. S.

(iv) tan² A – tan² B = \(\frac{\sin (\mathbf{A}+\mathbf{B}) \cdot \sin (\mathbf{A}-\mathbf{B})}{\cos ^2 \mathbf{A} \cdot \cos ^2 \mathbf{B}}\)
Solution:

= tan² A sec² B – tan² B sec² A
= tan² A (1 + tan² B) – tan² B (1 + tan² A)
= tan² A + tan² A tan² B – tan² B tan²A tan² B
= tan² A – tan² B = L. H. S

Question 4.
Prove the following :
Solution:
(i) tan 75° + cot 75° = 4
Solution:
L.H.S = tan 75° + cot 75° = tan 75° + \(\frac{1}{\tan 75^{\circ}}\)

(ii) sin² 18° + cos² 36° = 3/4
Solution:

(iii) sin 18°. cos 36° = 1/4
Solution:
sin 18° cos 36°
\(=\left(\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{5}+1}{4}\right)=\frac{5-1}{16}=\frac{1}{4}\)

(iv) sin 15° = \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)
Solution:
sin 15° = sin (45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°
\(=\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \cdot \frac{1}{2}=\frac{\sqrt{3}-1}{2 \sqrt{2}}\)

(v) cot \(\frac{\pi}{8}\) – tan \(\frac{\pi}{8}\) = 2
Solution:

(vi) \(\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}\) = tan 54°
Solution:

(vii) tan 10° + tan 35° + tan 10°. tan 35 = 1
Solution:

Question 5.
Prove the following:
(i) cot 2A = \(\frac{\cot ^2 A-1}{2 \cot A}\)
Solution:

(ii) \(\frac{\sin B}{\sin A}=\frac{\sin (2 A+B)}{\sin A}\) – 2cos(A + B)
Solution:

= \(\frac{\sin B}{\sin A}\) = L.H.S

(iii) \(\frac{\sin 2 A+\sin 2 B}{\sin 2 A-\sin 2 B}=\frac{\tan (A+B)}{\tan (A-B)}\)
Solution:

(iv) \(\frac{\cot A-\tan A}{\cot A+\tan A}\) = cos2A
Solution:

(v) \(\frac{\sin 2 A+\sin 5 A-\sin A}{\cos 2 A+\cos 5 A+\cos A}\) = tan 2A
Solution:

(vi) cot A – tan A = 2 cot 2A
Solution:

(vii) cot A – cosec 2A = cot 2A
Solution:

(viii) \(\frac{\cos A-\sin A}{\cos A+\sin A}\) = sec 2A – tan 2A
Solution:

(ix) tan θ (1 + sec 2θ) = tan 2θ
Solution:

(x) \(\frac{\sin A+\sin B}{\sin A-\sin B}\) = tan \(\frac{A+B}{2}\) . cot \(\frac{A-B}{2}\)
Solution:

(xi) sin 50° –  sin 70° + sin 10° = 0
Solution:
L. H. S. = sin 50° – sin 70° + sin 10°
= sin (60° – 10°) – sin (60° + 10°) + sin 10°
= – 2 cos 60° sin 10° + sin 10°
= – 2 × 1/2 sin 10° + sin 10°
= – sin 10° + sin 10° = 0 = R. H. S.

(xii) cos 80° + cos 40° – cos 20° = 0
Solution:
L. H. S. = cos 80° + cos 40° – cos 20°
= cos(60° + 20°) + cos (60° – 20°) – cos 20°
= 2 cos 60° cos 20° – cos 20° = 0

(xiii) 8 sin 10°. sin 50°. sin 70° = 1
Solution:
L. H. S. = 8 sin 10° sin 50° sin 70°
= 8 sin 10° sin (60° – 10°) sin (60° + 10°)
= 8 sin 10° (sin² 60° – sin² 10°)
= 8 sin 10°(3/4 – sin² 10°)
= 6 sin 10° – 8 sin³ 10°)
= 2 (3 sin 10° – 4 sin³ 10°)
= 2 sin (3 × 10°)
= 2 sin 30° = 2 × 1/2 = 1 = R.H.S

(xiv) 4 sin A sin (60° – A) sin (60° + A) – sin 3A = 0
Solution:
L. H. S. = 4 sin A sin (60° – A)
sin (60° + A) – sin 3A
= 4 sin A (sin² 60°- sin² A) – sin 3A
= 4 sin A. 3/4 – 4 sin³ A – sin 3A
= (3 sin A – 4 sin³A) – sin 3A
= sin 3A – sin 3A = 0

(xv) tan 3A – tan 2A – tan A = tan 3A tan 2A tan A
Solution:
We have tan 3A = tan (2A + A)
or, tan 3A = \(\frac{\tan 2 A+\tan A}{1-\tan 2 A \tan A}\)
or, tan 3A (1 – tan 2A tan A) = tan 2A + tan A
or, tan 3A – tan 3A tan 2A tan A = tan 2A + tan A
or, tan 3A – tan 2A – tan A = tan 3A tan 2A tan A (Proved)

Question 6.
Prove the following
(i) tan\(\frac{A}{2}\) = \(\sqrt{\frac{1-\cos A}{1+\cos A}}\)
Solution:

(ii) \(\sqrt{\frac{1+\sin A}{1-\sin A}}\) = tan\(\left(\frac{\pi}{4}+\frac{A}{2}\right)\)
Solution:

(iii) \(\frac{1+\tan \frac{A}{2}}{1-\tan \frac{A}{2}}\) = sec A + tan A
Solution:

(iv) sec θ + tan θ = tan\(\left(\frac{\pi}{4}+\frac{θ}{2}\right)\)
Solution:

(v) cot\(\frac{A}{2}\) = \(\frac{\sin A}{1-\cos A}\)
Solution:
R.H.S = \(\frac{\sin A}{1-\cos A}=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{2 \sin ^2 \frac{A}{2}}\)
= cot\(\frac{A}{2}\) R.H.S

Question 7.
Find the maximum value of the following.
(i) 5 sin x + 12 cos x
Solution:
5 sin x + 12 cos x
Let 5 = r cos θ, 12 = r sin θ 52
∴ 5² = r² cos² θ, 12² = r² sin² θ
∴ 5² + 12² = r² (cos² θ + sin² θ) = r²
∴ r = \(\sqrt{25+144}\) = 13
∴ The maximum value of 5 sin x + 12 cos x is 13.

(ii) 24 sin x – 7 cos x
Solution:
The maximum value of 24 sin x – 7 cos x is
\(\sqrt{(24)^2+(-7)^2}\)
= \(\sqrt{576+49}=\sqrt{625}\) = 25

(iii) 2 + 3 sin x + 4 cos x
Solution:
The maximum value of 3 sin x + 4 cos x is
\(\sqrt{3^2+4^2}\) = 5
∴ Maximum value of 2 + 3 sin x + 4 cos x is 2 + 5 = 7

(iv) 8 cos x – 15 sin x – 2
Solution:
Maximum value of 8 cos x- 15 sin x is \(\sqrt{(8)^2+(-15)^2}\)
= \(\sqrt{64+225}=\sqrt{289}\) = 17
∴ Maximum value of 8 cos x- 15 sin x – 2 is 17 – 2 = 15

Question 8.
Answer the following:
(i) If tan A = \(\frac{13}{27}\), tan B = \(\frac{7}{20}\) and A, B are acute, show that A + B = 45°.
Solution:

(ii) If tan θ = \(\frac{b}{a}\), find the value of a cos 2θ + b sin 2θ.
Solution:

(iii) If sec A – tan A = \(\frac{1}{2}\) and 0< A < 90° then show that sec A = \(\frac{5}{4}\)
Solution:
If sec A – tan A = \(\frac{1}{2}\),       ….(1)
0< A < 90°
⇒ sec A – tan A = \(\frac{\sec ^2 A-\tan ^2 A}{2}\)
= \(\frac{(\sec \mathrm{A}+\tan \mathrm{A})(\sec \mathrm{A}-\tan \mathrm{A})}{2}\)
or, sec A + tan A = 2    ……(2)
Now adding eqn. (1) and (2), We have
2 sec A = \(\frac{1}{2}\) + 2 = \(\frac{5}{2}\)
or, sec A = \(\frac{5}{4}\)

(iv) If sin θ + sin Φ = a and cos θ + cos Φ = b the show that tan \(\frac{1}{2}\) (θ + Φ) \(\frac{a}{b}\)
Solution:

(v) If tan θ = \(\frac{a \sin x+b \sin y}{a \cos x+b \cos y}\) then show that a sin (θ – x) + b sin (θ – y) = 0
Solution:
tan θ = \(\frac{a \sin x+b \sin y}{a \cos x+b \cos y}\)
or, \(\frac{\sin \theta}{\cos \theta}=\frac{a \sin x+b \sin y}{a \cos x+b \cos y}\)
or, a sin θ cos x + b sin θ cos y = a cos θ sin x + b cos θ sin y
or, a (sin θ cos x – cos θ sin x) + b (sin θ cos y – cos θ sin y) = 0
or, a sin (θ – x) + b sin (θ – y) = 0

(vi) If A + C = B, show that tan A. tan B. tan C = tan B – tan A – tan C.
Solution:
A + C = B
or, tan (A + C) = tan B
or, \(\frac{\tan A+\tan C}{1-\tan A \tan C}\)
or, tan A + tan C = tan B – tan A tan B tan C
or, tan A tan B tan C = tan B – tan A – tan C

(vii) If tan A = \(\frac{1}{5}\), tan B = \(\frac{2}{3}\) then show that cos 2A = sin 2B.
Solution:
tan A = \(\frac{1}{5}\), tan B = \(\frac{2}{3}\)
∴ cos 2A = \(\frac{1-\tan ^2 \mathrm{~A}}{1+\tan ^2 \mathrm{~A}}\)

(viii) If cos 2A = tan² B, then show that cos 2B = tan² A In Δ ABC, prove that.
Solution:

(ix) tan\(\frac{B+C}{2}\) = cot\(\frac{A}{2}\)
Solution:

(x) cos (A + B) + sin C = sin (A + B) – cos C
If A + B + C = π and cos A = cos B. cos C show that (x_i  and x_ii)
Solution:
We have A + B = π – C or, cos (A + B)
or, cos (A + B)
= cos (π – C) = – cos C
and sin (A + B) = sin (π – C) = sin C
∴ cos (A + B) + sin C
= – cos C + sin (A + B)
= sin (A + B) – cos C.

(xi) tan B + tan C = tan A
Solution:
[∴ A + B + C = π ⇒ B + C = π – A
⇒ sin (B + C) = sin (π – A) = sin A]
L. H. S. = tan B + tan C

(xii) 2 cot B. cot C = 1
Solution:
We have A + B + C = π
⇒ B + C = π – A
⇒ cos (B + C) = cos (π – A) = – cosA
⇒ cos B cos C – sin B sin C = – cos B cos C
(∵ cos B cos C = cos A)
⇒ 2 cos B cos C = sin B sin C
⇒ \(\frac{2 \cos B \cos C}{\sin B \sin C}\) = 1
⇒ 2 cot B cot C = 1

Question 9.
Prove the following:
(i) cos (A – D) sin (B – C) + cos (B – D) sin (C – A) + cos (C- D) sin (A – B) = 0
Solution:
L. H. S. = cos (A – D) sin (B – C) + cos (B – D) sin (C – A) + cos (C – D) sin (A- B)
= \(\frac{1}{2}\) [2 cos (A- D) sin (B – C) + 2 cos (B – D) sin (C – A) + 2 cos (C – D) sin (A – B)]
= \(\frac{1}{2}\) [ sin (A- D + B – C) – sin (A – D – B + C) + sin (B – D + C – A)- sin (B- D- C + A) + sin ( C – D + A – B)- sin (C – D- A + B)]
= \(\frac{1}{2}\) × 0 = 0 = R.H.S

(ii) sin 2A + sin 2B + sin 2 (A – B) = 4 sin A. cos B. cos (A – B)
Solution:
L. H. S. = sin 2A + sin 2B + sin 2 (A – B)
= 2 sin \(\frac{2 A+2 B}{2}\) cos \(\frac{2 A-2 B}{2}\) + 2 sin (A – B) cos (A – B)
= 2 sin (A + B) cos (A- B) + 2 sin (A – B) cos (A – B)
= 2 cos (A – B)[ sin (A-+ B) + sin(A – B)]
= 2 cos (A – B) × 2 sin A cos B
= 4 sin A cos B cos (A – B) = R. H. S.

(iii) cos 2A + cos 2B + cos 2 (A – B) + 1 = 4 cos A. cos B. cos (A – B)
Solution:
cos 2A + cos 2B + cos 2 (A- B) + 1
= 2 cos \(\frac{2 A+2 B}{2}\) cos \(\frac{2 A-2 B}{2}\) + 2 cos² (A – B)
= 2 cos (A + B) cos (A- B) + 2 cos2 (A – B)
= 2 cos (A – B) [cos (A + B) + cos (A – B)]
= 2 cos (A – B) x 2 cos A cos B
= 4 cos A cos B cos (A – B)
= R. H. S

(iv) sin 2A + sin 2B + sin 2C- sin 2 (A + B + C)
Solution:
L. H. S. = sin 2A + sin 2B + sin 2C- sin 2 (A + B + C)
= 2 sin \(\frac{2 A+2 B}{2}\) cos \(\frac{2 A-2 B}{2}\) = 2 cos \(\frac{2 \mathrm{C}+2(\mathrm{~A}+\mathrm{B}+\mathrm{C})}{2}\) × \(\frac{2 \mathrm{C}-2(\mathrm{~A}+\mathrm{B}+\mathrm{C})}{2}\)
= 2 sin (A + B) cos (A – B) + 2 cos (A + B + 2C) sin (A – B)
= 2 sin (A + B) [cos (A- B) – cos (A + B + 2C)]
= 2 sin (A + B) × 2 sin \(\frac{A-B+A+B+2 C}{2}\) sin \(\frac{A+B+2 C-A+B}{2}\)
= 4 sin (A + B) sin (C + A) sin (B + C) = R. H. S.

(v) sin A + sin 3A + sin 5A = sin 3A (1 + 2 cos 2A)
Solution:
L. H. S. = sin A + sin 3A + sin 5A
= sin 3A + sin 5A + sin A
= sin 3A + 2 sin \(\frac{5 \mathrm{~A}+\mathrm{A}}{2} cos \frac{5 \mathrm{~A}-\mathrm{A}}{2}\)
= sin 3A + 2 sin 3A cos 2A
= sin 3A (1 + 2 cos 2A) = R. H. S

(vi) sin A – sin 3A + sin 5A = sin 3A (2 cos 2A – 1)
Solution:
L. H. S. = sin A – sin 3A + sin 5A
= sin A + sin 5A – sin 3A
= sin 5A + sin A – sin 3A
= 2 sin \(\frac{5 \mathrm{~A}+\mathrm{A}}{2} cos \frac{5 \mathrm{~A}-\mathrm{A}}{2}\) – sin 3A
= 2 sin 3A cos 2A – sin 3A
= sin 3A (2 cos 2A – 1) = R. H. S.

(vii) cos (A + B) + sin (A – B) = 2 sin (45° + A) cos (45° + B)
Solution:
R. H. S. = 2 sin (45° + A) cos (45° + B)
= sin (45° + A + 45° + B) + sin (45° + A – 45° – B)
= sin (90° + A + B) sin (A – B)
= cos (A + B) sin (A – B) = L. H. S.

(viii) cos (120° + A) cos (120° – A) + cos (120° + A) cos A + cos A cos (120° – A) + \(\frac{3}{4}\) = 0
Solution:
L. H. S.
= cos (120° + A) cos (120° – A) + cos (120° + A) cos A + cos A cos (120° – A) + \(\frac{3}{4}\)
= cos² A – sin² 120° + cos A [ cos (120° + A) + cos (120° – A)] + \(\frac{3}{4}\)
= cos² A – \(\frac{3}{4}\) + cos A ( 2 cos 120°. cos A) + \(\frac{3}{4}\)
= cos² A + 2 cos 120°. cos² A
= cos² A + 2 \(\left(-\frac{1}{2}\right)\) . cos² A
= cos² A – cos² A = 0

(ix) cos 4A – cos 4B = 8 (cos A – cos B) (cos A + cos B) (cos A – sin B) (cos A + sin B)
Solution:
R. H. S. = 8 (cos A- cos B) (cos A + cos B)
(cos A – sin B) (cos A + sin B)
=  8 (cos² A – cos² B) (cos² A – sin² B)
= – 8(cos² B – cos² A)(cos² A – sin² B)
= – 8 sin (A +B) sin (A – B) cos (A + B) cos (A – B)
= – 2 x [ 2 sin (A + B) cos (A + B)] [2 sin (A – B) cos (A – B)]
= – 2 sin (2A + 2B) sin (2A – 2B)
= – [cos (2A + 2B – 2A + 2B) – cos (2A + 2B + 2A – 2B)]
= – cos 4B + cos 4A
= cos 4A – cos 4B = L. H. S.

Question 10.
Prove the following:
(i) \(\frac{1-\tan ^2\left(45^{\circ}-A\right)}{1+\tan ^2\left(45^{\circ}-A\right)}\) = sin 2A
Solution:
L.H.S \(\frac{1-\tan ^2\left(45^{\circ}-A\right)}{1+\tan ^2\left(45^{\circ}-A\right)}\) = sin 2A = cos 2(45° – A)
= cos (90° – 2A) = sin 2A = R. H. S.

(ii) \(\frac{\cos A+\sin A}{\cos A-\sin A}-\frac{\cos A-\sin A}{\cos A+\sin A}\) = 2 tan 2A
Solution:

(iii) \(\frac{1-\cos 2 A+\sin 2 A}{1+\cos 2 A+\sin 2 A}\) = tan A
Solution:

(iv) \(\frac{\sin (\mathbf{A}+\mathbf{B})+\cos (\mathbf{A}-\mathbf{B})}{\sin (\mathbf{A}-\mathbf{B})+\cos (\mathbf{A}+\mathbf{B})}\) = sec 2B + tan 2B
Solution:

(v) \(\frac{\cos 7 \alpha+\cos 3 \alpha-\cos 5 \alpha-\cos \alpha}{\sin 7 \alpha-\sin 3 \alpha-\sin 5 \alpha+\sin \alpha}\) = cot 2α
Solution:
L.H.S = \(\begin{array}{r}
\cos 7 \alpha+\cos 3 \alpha \\
-\cos 5 \alpha-\cos 3 \alpha \\
\hline \sin 7 \alpha-\sin 3 \alpha \\
-\sin 5 \alpha+\sin \alpha
\end{array}\)

(vi) \(\frac{\sin \theta+\sin 3 \theta+\sin 5 \theta+\sin 7 \theta}{\cos \theta+\cos 3 \theta+\cos 5 \theta+\cos 7 \theta}\) = tan 4θ
Solution:

Question 11.
Prove the following:
(i) Express 4 cos A. cos B. cos C as the sum of four cosines.
Solution:
4 cos A cos B cos C
= 2 (2cos A cos B) cos C
= 2 [ cos (A + B) + cos (A – B)] cos C
= 2 cos (A + B) cos C + 2 cos (A – B) cos C
= cos (A + B + C) + cos (A + B – C) + cos (A – B + C) + cos (A – B – C)

(ii) Express cos 2A + cos 2B + cos 2C + cos 2 (A + B + C) as the product of three cosines.
Solution:
cos 2A + cos 2B + cos 2C + cos 2(A + B + C)
= 2 cos \(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\) cos \(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\) = 2 cos \(\frac{2 C+2(A+B+C)}{2}\) × cos \(\frac{2 C-2(A+B+C)}{2}\)
= 2 cos (A + B) cos (A- B) + 2 cos (A + B + 2C) cos (A + B)
= 2 cos (A + B) [cos (A- B) + cos (A + B + 2C)]
= 2 cos (A + B) × 2 cos \(\frac{(A-B+A+B+2 C)}{2}\) cos \(\frac{(A-B-A-B-2 C)}{2}\)
= 4 cos (A + B) cos (C + A) cos (B + C)

Question 12.
Prove the following:
(i) cos⁶ A – sin⁶ A = cos 2A(1 – \(\frac{1}{4}\) sin² 2A)
Solution:
L.H.S = cos⁶ A – sin⁶ A
= (cos² A)³– (sin² A)³ = (cos² A – sin² A)³ + 3 cos² A sin² A (cos² A – sin² A)
= cos³ 2A + \(\frac{3}{4}\) sin² 2A cos 2A
= cos² A (cos² 2A + \(\frac{3}{4}\) sin² 2A)
= cos 2A (1 – sin² 2A + \(\frac{3}{4}\) sin² 2A)
= cos 2A (1 – \(\frac{1}{4}\) sin² 2A) = R.H.S

(ii) cos⁶A + sin⁶ A = \(\frac{1}{4}\) (1 + 3 cos²2A)
Solution:
L.H.S = cos⁶ A + sin⁶ A
= (cos² A)³ + (sin² A)³
= (cos² A)³ + (sin² A)³ – 3 cos² A. sin² A (cos² A + sin² A)
= 1 – \(\frac{3}{4}\) sin² 2A = 1 – \(\frac{3}{4}\) (1 – cos² 2A)
= 1 – \(\frac{3}{4}\) + \(\frac{3}{4}\) cos² 2A
= \(\frac{1}{4}\) + \(\frac{3}{4}\) cos² 2A = \(\frac{1}{4}\) (1 + 3cos² 2A) = R.H.S.

(iii) cos³ A. cos 3A + sin³ A sin³ A = cos³ 2A
Solution:
L.H.S = cos³ A cos 3A + sin³ A sin 3A
= cos³ A (4 cos³ A- 3 cos A) + sin³ A (3 sin A- 4 sin³ A)
= 4 cos⁶A- 3 cos⁴A + 3 sin⁴A- 4 sin⁶A
= 4 (cos⁶A- sin⁶A) – 3(cos⁴A- sin⁴A)
= 4 {(cos²A)³– (sin²A)³} – 3 {(cos²A)²– (sin²A)²}
= 4 (cos²A- sin²A) {(cos²A)² + cos²A sin²A + (sin²A)²} – 3 (cos²A- sin²A) (cos²A + sin²A)
= (cos²A- sin²A) [4 {(cos²A + sin²A)²-2 cos²A sin²A + cos²A sin²A} -3×1]
= cos 2A ( 4- 4 sin²A cos²A- 3)
= cos 2A ( 1 – 4 sin²A cos²A)
= cos 2A (1 – sin²2A)
= cos 2A cos²2A = cos³2A = R. H. S.

(iv) sin⁴ θ = \(\frac{3}{8}\) – \(\frac{1}{2}\) cos 2θ + \(\frac{1}{8}\) cos 4θ
Solution:
R.H.S = \(\frac{3}{8}\) – \(\frac{1}{2}\) cos 2θ + \(\frac{1}{8}\) cos 4θ
= \(\frac{3}{8}\) – \(\frac{1}{2}\) (1 – 2 sin² θ) + \(\frac{1}{8}\) (1 – 2 sin² θ)
= \(\frac{3}{8}\) – \(\frac{1}{2}\) + sin² θ + \(\frac{1}{8}\) – \(\frac{1}{4}\)  sin² 2θ
= sin² θ – \(\frac{1}{4}\) × 4 sin² θ cos² θ
= sin² θ (1 – cos² θ)
= sin² θ sin² θ = sin⁴ = L.H.S

(v) cot 3A = \(\frac{\cot ^3 A-3 \cot A}{3 \cot ^2 A-1}\)
Solution:

(vi) tan 4θ = \(\frac{4 \tan \theta-4 \tan ^3 \theta}{1-6 \tan ^2 \theta+\tan ^4 \theta}\)
Solution:

(vii) \(\frac{1}{\tan 3 A-\tan A}-\frac{1}{\cot 3 A-\cot A}\) = cot 2A
Solution:

(viii) \(\frac{\cot A}{\cot A-\cot 3 A}-\frac{\tan A}{\tan 3 A-\tan A}\) = 1
Solution:

Question 13.
Find the value of
sin 3°,cos 3°, 2 sin \(\frac{\pi}{32}\)
Solution:
sin 3° = sin (18° – 15°)
= sin 18° cos 15° – cos 18° sin 15°

Question 14.
If sin A + sin B = a and cos A + cos B = b, then show that
(i) tan(A + B) = \(\frac{2 a b}{b^2-a^2}\)
Solution:

(ii) sin (A + B) = \(\frac{2 a b}{b^2-a^2}\)

(iii) cos (A + B) = \(\frac{b^2-a^2}{b^2+a^2}\)

Question 15.
Prove the following:
(i) \(\frac{1+\sin A-\cos A}{1+\sin A+\cos A}\) = tan \(\frac{A}{2}\)
Solution:

(ii) \(8 \sin ^4 \frac{1}{2} \theta-8 \sin ^2 \frac{1}{2} \theta\) + 1 = cos 2θ
Solution:

(iii) \(\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}\) \(+\cos ^4 \frac{7 \pi}{8}=\frac{3}{2}\)
Solution:

(iv) cos² \(\frac{\alpha}{2}\) (1- 2cos α)² + sin² α(1+ 2 cos α)² =1
Solution:

Question 16.
Prove the following:
(i) sin 20°. sin 40°. sin 60°. sin 80° = \(\frac{3}{16}\)
Solution:

(ii) cos 36°. cos 72°. cos 108°. cos144° = \(\frac{7}{16}\)
Solution:

(iii) cos 10°. cos 30°. cos 50°. cos 70° = \(\frac{3}{16}\)
Solution:

(iv) cos 20°. cos 40°. cos 60°. cos 80° = \(\frac{1}{16}\)
Solution:

(v) tan 6°. tan 42°.tan 66°. tan 78° = 1 [Hints: Use the identity tan 3A = tan A tan (60° – A) tan (60° + A)]
Solution:
We have tan 3A
= tan A tan (60° – A) tan (60° + A)
Now putting A = 6° and 18°.
in (1) we have
tan 18° = tan 6° tan 54° tan 66°     …(2)
and tan 54°
= tan 18° tan 42° tan 78°            ……(3)
Multiplying (2) and (3) we have
tan 18° tan 54°
= tan 6° tan 54° tan 66° tan 18° . tan 42°. tan 78°
or, 1 = tan 6° tan 42° tan 66° tan 78°.

Question 17.
Prove the following:
(i) cos 7 \(\frac{1}{2}\)° = √6 + √3 + √2 + 2
Solution:

(ii) cot 22 \(\frac{1}{2}\)° = √2 + 1
Solution:

(iii) cot 37 \(\frac{1}{2}\)° = √6 – √3 – √2 + 2
Solution:

(iv) tan 37 \(\frac{1}{2}\)° = √6 + √3 – √2 + 2
Solution:

(v) cos \(\frac{\pi}{16}\) = 1/2 \(\sqrt{2+\sqrt{2+\sqrt{2}}}\)
∴ 2 cos \(\frac{\pi}{16}\)

Question 18.
(i) If sin A = K sin B, prove that tan 1/2(A – B) = \(\frac{K-1}{K+1}\) tan 1/2(A – B)
Solution:

(ii) Ifa cos (x + α) = b cos (x – α) show that (a + b) tan x = (a – b) cot α.
Solution:

or, cot x cot α = \(\frac{a+b}{a-b}\)
or, (a – b) cot α = \(\frac{a+b}{\cot x}\) = (a + b) tan x
or, (a + b) tan x =(a-b) cot α

(iii) An angle 0 is divided into two parts α, β such that tan α: tan β = x : y.
Prove that sin (α – β) = \(\frac{x-y}{x+y}\) sin θ.
Solution:

(iv) If sin θ + sin Φ = a, cos θ + cos Φ = b, show that
\(\frac{\sin \frac{\theta+\phi}{2}}{a}=\frac{\cos \frac{\theta+\phi}{2}}{b}=2 \frac{\cos \frac{\theta-\phi}{2}}{a^2-b^2}\)
Solution:
sin θ + sin Φ = a
cos θ + cos Φ = b
We have
a2 + b2 = (sin θ + sin Φ)² + (sin θ + sin Φ)²
= sin² θ + sin² Φ + 2 sin θ. sin Φ + cos² θ +cos² Φ + 2 cos θ. cos Φ
= 2 + 2 (cos θ cos Φ + sin θ sin Φ)
= 2 + 2 cos (θ – Φ)
= 2 [1+ cos (θ – Φ)]

(v) If a cos α + b sin α = c = a cos β + b sin α then prove that
\(\frac{a}{\cos \frac{1}{2}(\alpha+\beta)}=\frac{b}{\sin \frac{1}{2}(\alpha+\beta)}\) \(=\frac{c}{\cos \frac{1}{2}(\alpha-\beta)}\)
Solution:

(vi) Prove that \(\left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)^n+\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)^n\) = 2 cosⁿ \(\frac{A-B}{2}\) or zero according as n is even or odd.
Solution:

Question 19.
(i) If (1 – e) tan² \(\frac{\boldsymbol{\beta}}{2}\) = (1 + e) tan² \(\frac{\boldsymbol{\alpha}}{2}\), prove that cos β = \(\frac{\cos \alpha-e}{1-e \cos \alpha}\)
Solution:

(ii) If cos θ = \(\frac{\cos \mathbf{A}-\cos \mathbf{B}}{1-\cos \mathbf{A} \cdot \cos \mathbf{B}}\) prove that one of the values of
tan \(\frac{θ}{2}\) is tan \(\frac{A}{2}\) . tan \(\frac{B}{2}\).
Solution:

(iii) If tan θ = \(\frac{\sin x \cdot \sin y}{\cos x+\cos y}\) then prove that one of the values of tan \(\frac{1}{2}\) θ tan \(\frac{1}{2}\) x and tan \(\frac{1}{2}\) y.
Solution:

(iv) If sec (Φ + a) + sec (Φ – α) = 2 sec Φ. show that cos Φ = ± √2 cos \(\frac{α}{2}\)
Solution:

(v) If tan A + tan B = a and cot A + cot B = b. then show that cot (A + B) = \(\frac{1}{a}\) – \(\frac{1}{b}\).
Solution:

(vi) If cot θ = cos (x + y) and cot Φ = cos (x – y) show that tan (θ = Φ) = \(\frac{2 \sin x \cdot \sin y}{\cos ^2 x+\cos ^2 y}\)
Solution:

(vii) If tan β = \(\frac{n^2 \sin \alpha \cdot \cos \alpha}{1-n^2 \sin ^2 \alpha}\), then show that \(\frac{\tan (\alpha-\beta)}{\tan \alpha}\) = 1 – n²
Solution:

(viii)If 2 tan α = 3 tan β, then prove that tan(α – β) = \(\frac{\sin 2 \beta}{5-\cos 2 \beta}\)
Solution:

(ix) If α, β are acute angles and cos 2α = \(\frac{3 \cos 2 \beta-1}{3-\cos 2 \beta}\) then prove that tan α = √2 tan β.
Solution:

Question 20.
If A + B + C = π, then prove the following.
(i) cos 2A + cos 2B + cos 2C + 1 + 4 cos A. cos B. cos C = 0
Solution:
A + B + C = π  or, A + B = π – C
or, cos (A + B) = cos (π – C) = – cos C
∴ cos 2A + cos 2B + cos 2C
= 2 cos \(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\) cos \(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\)
= 2 cos (A + B) cos (A- B) + 2 cos² C – 1
= – 2 cos C cos (A – B) + 2 cos² C –  1
= – 1 – 2 cos C cos (A – B) – cos C
= – 1 – 2 cos C [cos (A- B) + cos (A + B)]
= – 1 – 2 cos C × 2 cos A cos B
= – 1 – 4 cos A cos B cos C
∴ L. H. S. = cos 2A + cos 2B + cos 2C + 1 + 4 cos A cos B cos C
= – 1 – 4 cos A cos B cos C + 1 + 4 cos A cos B cos C = 0 = R. H. S.

(ii) sin 2A + sin 2B – sin 2C = 4 cos A. cos B. sin C
Solution:
L. H. S. = sin 2A + sin 2B – sin 2C
= 2 sin \(\frac{2 A+2 B}{2}\) cos \(\frac{2 A-2 B}{2}\) – sin 2C
= 2 sin(A + B) cos (A – B) -2 sinC cos C
= 2 sin C cos (A- B) – 2 sin C cos C [ ∵ A + B= π – C]
or, sin (A + B) = sin (n- C) = sin C]
= 2 sin C [cos (A – B)- cos C]
= 2 sin C [cos (A – B) – cos (A + B)]
= 2 sin C. 2 cos A. cos B
= 4 cos A cos B sin C = R. H. S.

(iii) cos A + cos B + cos C = 1 + 4 sin \(\frac{1}{2}\) A. sin \(\frac{1}{2}\) B. sin \(\frac{1}{2}\) C.
Solution:

(iv) sin A + cos B- sin C = 4 sin \(\frac{1}{2}\) A. sin \(\frac{1}{2}\) B. cos \(\frac{1}{2}\) C.
Solution:

(v) cos² A + cos² B + 2cos A. cos B. cos C = sin² C
Solution:

(vi) sin² \(\frac{A}{2}\) + sin² \(\frac{B}{2}\) + sin² \(\frac{C}{2}\) = 1 – 2 sin \(\frac{A}{2}\). sin \(\frac{B}{2}\). sin \(\frac{C}{2}\)
Solution:

(vii) sin \(\frac{A}{2}\) + sin \(\frac{B}{2}\) + sin \(\frac{c}{2}\) = 4 sin \(\frac{\pi-A}{4}\) sin \(\frac{\pi-B}{4}\). sin \(\frac{\pi-C}{4}\) + 1
Solution:

(viii) cos² \(\frac{A}{2}\) + cos² \(\frac{B}{2}\) – cos² \(\frac{C}{2}\) = 2 cos \(\frac{A}{2}\). cos \(\frac{B}{2}\). sin \(\frac{C}{2}\)
Solution:

(ix) sin (B + 2C) + sin (C + 2A) + sin (A + 2B) = 4 sin \(\frac{B-C}{2}\). sin \(\frac{C-A}{2}\). sin \(\frac{A-B}{2}\)
Solution:

Question 21.
(i) Show that (2 cos θ – 1) (2 cos 2θ – 1) (2 cos² 2θ – 1) ….. (2 cos 2^n-1 θ – 1) \(=\frac{2 \cos 2^n \theta+1}{2 \cos \theta+1}\)
Solution:
We have
(2 cos θ + 1) (2 cos θ – 1)
= 4 cos² θ – 1=4 cos² θ – 2+1
= 1 + 2(2 cos² θ – 1)
= 1+2 cos 2θ = 2 cos 2θ+1
And, (2 cos 2θ + 1) (2 cos 2θ – 1)
= 4 cos² 2θ – 1 = 2 ( 2 cos² 2θ – 1) + 1
= 2 cos 4 θ + 1 = 2 cos 2² θ + 1
Similarly, we can prove,
( 2 cos 2² θ + 1) ( 2 cos 2² θ- 1)
= 2 cos 2³ θ + 1
Proceeding in this way, we can prove,
(2 cos 2θ + 1) (2 cos 2θ – 1)
(2 cos 2θ- 1) (2 cos 2² θ – 1) ……  (2 cos 2^n-1 θ-1)
= 2 cos 2ⁿ θ + 1
or, (2 cos θ-1) (2 cos 2θ-1) (2 cos 2² θ- 1) (2 cos 2^n-1 θ-1)
\(=\frac{2 \cos 2^n \theta+1}{2 \cos \theta+1}\)

(ii) Show that 2ⁿ cos θ. cos 2θ. cos2² θ ……… 2^n-1 θ = 1 If θ = \(\frac{\pi}{2^n+1}\)
Solution:
we have, θ = \(\frac{\pi}{2^n+1}\)
or, 2ⁿ θ + θ = π
or, 2ⁿ θ  = π – θ
or, 2ⁿ θ = sin (π – θ) = sin θ
or, \(\frac{\sin 2^n \theta}{\sin \theta}\) = 1   …….(1)
Again, sin 2ⁿ θ = 2 sin 2^n-1 θ cos 2^n-1 θ
= 2 × 2 sin 2^n-2 θ cos 2^n-2 θ cos 2^n-1 θ
= 2² × 2 sin 2^n-3 θ cos 2^n-3 θ  cos 2^n-2 θ cos 2^n-v θ
= 2³ sin 2^n-3 θ cos 2^n-3 θ cos 2^n-2 θ cos 2^n-1 θ

………………………..

………………………..

………………………..

= 2ⁿ sin θ cos θ cos 2θ cos 2² θ …….. cos 2^n-2 θ cos 2^n-1 θ

or,  2ⁿ cos θ cos 2θ cos 2² θ …… cos 2^n-1 θ = 1

(iii) Prove that \(\frac{\tan 2^n \theta}{\tan \theta}\) = = (1 + sec 2θ) (1 + sec2² θ) …. (1 + sec2ⁿ θ)
Solution:

Question 22.
If x + y + z = xyz, prove that
(i) \(\frac{x}{1-x^2}+\frac{y}{1-y^2}+\frac{z}{1-z^2}\) = =\(\frac{4 x y z}{\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)}\)
Solution:
x + y + z = xyz   (Given)
Let x = tan α, y = tan β, z = tan γ
∴ tan α + tan β + tan γ = tan α .tan β .tan γ or; tan α + tan β
= – tan γ + tan α tan β tan γ
= – tan γ(1 – tan α tan β)
or, \(\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\) = – tan γ
or, tan (α + β) = tan (π- γ)
or, α + β = π-γ
or, 2α + 2β = 2π- 2γ
or, tan (2α + 2β = tan (2π- 2γ)
or, tan (2α + 2β) = tan (2π- 2γ)

(ii) \(\frac{3 x-x^3}{1-3 x^2}+\frac{3 y-y^3}{1-3 y^2}+\frac{3 z-z^3}{1-3 z^2}\) = \(\frac{3 x-x^3}{1-3 x^2} \cdot \frac{3 y-y^3}{1-3 y^2} \cdot \frac{3 z-z^3}{1-3 z^2}\)
Solution:

Question 23.
If \(\text { If } \frac{\sin ^4 \alpha}{a}+\frac{\cos ^4 \alpha}{b}=\frac{1}{a+b}\) show that \(\frac{\sin ^8 \alpha}{a^3}+\frac{\cos ^8 \alpha}{b^3}=\frac{1}{(a+b)^3}\)
Solution:

Odisha State Board CHSE Odisha Class 11 Political Science Solutions Unit 3 Indian Constitution Short Answer Questions.

CHSE Odisha 11th Class Political Science Unit 3 Understanding Political Theory Short Answer Questions

Short Question With Answers

Question 1.
What were the functions of the constituent assembly of India?
Answer:
The constitution assembly was formed to prepare the new constitution of India and To act as the union legislature till the new parliament was formed.

Question 2.
What was the importance of the Drafting committee?
Answer:
The drafting committee played an important role in drafting the new constitution of India. It was the future ruling pattern of India.

Question 3.
What do you mean by objective resolution?
Answer:
Objective resolution refers to the aims and objectives of the constitutional frames expressed in the form of a resolution. It was approved by the constituent assembly on 22nd January 1947.

Question 4.
Why Indian constitution has been so large?
Answer:
The constitution of India is so large because it contains provisions both for the union government and states. The length of the constitution increased in view of the detailed analysis of every provision to make it clear to the people.

Question 5.
What is an ideal constitution?
Answer:
An ideal constitution is a document that has the ability to develop and V change in accordance with changes in time and circumstances. Its language must be simple clear andunamigous

Question 6.
What is the preamble?
Answer:
A preamble is a brief introduction about the constitution. It contains ideals and objectives of the constitution in a nutshell.

Question 7.
What does the term we the people of India imply?
Answer:
The term we the people of India implies the source of the constitution and its democratic character. It was framed by a group of representatives of Indian people.

Question 8.
What are the main sources of the constitution?
Answer:
The Government of India Act. 1935. The parliamentary statutes and debates. The constitution of UK, Canada, Australia, Ireland, Germany, Spain and South Africa etc.

Question 9.
Why India is called a secular state?
Answer:
India is called a secular state because
There is no state religion in India and All religious communities are treated equally by the state. People enjoy the freedom of religion and worship.

Question 10
Why India is called a sovereign state?
Answer:
India is called a sovereign state because it is independent and free from the influence of any foreign country. The decision of the government of India is found delating internal administration and foreign policy.

Question 11.
Why India is called a socialist state?
Answer:
India is called a socialistic state because, it seeks to provide social, economic, and political justice to the people. The government gives priority to the public sector and seeks to abolish economic exploitation and provide basic minimum needs to all.

Question 12.
Why India is called a democracy?
Answer:
India is called a democracy because the constitution has set up parliamentary representative democracy in India. The people enjoy fundamental rights and the media and judiciary remain free from Govt, control.

Question 13.
Why India is called a Republic?
Answer:
India is called a republic because the head of state, the president is elected by the people indirectly Again any ordinary citizen can contest and assume the office of the president of India.

Question 14.
Why Indian constitution is called an enacted constitution?
Answer:
Indian constitution is called an enacted constitution because it was ‘ framed by a constituent assembly.’ It was enacted on 26th November 1949 by the constituent assembly.

Question 15.
India is a union of states?
Answer:
Art. 1 declares India as a union of states. That means the federal structure of ” India can’t be changed by the decision of states.

Question 16.
Why India is called a quasi-federal state?
Answer:
Prof. K.C. where describes India as a quasi-federal state. It means that India is organized on federal lines but it works as a unitary state.

Question 17.
Is the Indian constitution rigid?
Answer:
Indian constitution can’t be considered exclusively either as a rigid or flexible constitution. The procedure of amendment is a mixture of both rigidity and flexibility.

Question 18.
What is single citizenship?
Answer:
Single citizenship means only the union government has the power to grant citizenship. It is found in a unitary state, but in India single citizenship is preferred to maintain national unity and integrity.

Question 19.
What is judicial review?
Answer:
Judicial review refers to the power of the supreme court to examine the constitutional validity of the law. If it finds any law or decision violating the constitution the court declares such law unconstitutional.

Question 20.
What do you mean by fundamental rights?
Answer:
The fundamental rights are the most valuable rights which are mentioned in part – III of the constitution. These rights are justiciable in nature, therefore in case of violation, the court takes immediate redressal measures.

Question 21.
What is Right to Equality?
Answer:
The right to equality is a fundamental right which is guaranteed under Art 14 to 18 of the constitution. This right is available both for citizens.

Question 22.
What is equality before law?
Answer:
Equality before law means all are equal in the eyes of law. The law courts the judges and the legal system gives equal protection to all citizens.

Question 23.
Importance of Art. 19?
Answer:
Art. 19 guarantees six fundamental freedoms to Indian citizens. These include freedom of speech and expression freedom to assemble peacefully without arms, freedom of association, freedom of movement, freedom of settlement & freedom of profession trade or business.

Question 24.
Traffic in human beings?
Answer:
Traffic in human beings means exploitation and unlawful treatment of weaker sections of society. It prohibits men and upper classes from exploiting women, children, and backward classes.

Question 25.
How the fundamental rights differ from legal rights?
Answer:
The fundamental rights are guaranteed and protected by the institution, but the legal rights are protected by law. The legal rights can be amended by ordinary legislation, but fundamental rights can be amended only by constitutional amendment.

Question 26.
Which fundamental rights are available both for citizens and aliens?
Answer:
Right to equality under Art. 14 and Right to life liberty and property under Art. 21 are available both for citizens and aliens

Question 27.
Right to education?
Answer:
Right to education is now a fundamental right under Art. 21 (A) It is inserted into the fundamental rights by the 86th amendment act. 2002. It provides for free and compulsory education to all children below 14 years of age.

Question 28.
Right against exploitation?
Answer:
This right is guaranteed under Art. 23 and 24 of the constitution. It seeks to give protection to the weaker sections of society, including women, children and backward classes from the exploitation of rich and male-dominated society.

Question 29.
What is protective discrimination?
Answer:
The constitution of India under Art. 15 (3) and provide for protective discrimination. It enables the scheduled castes, scheduled tribes, women- and backward classes to avail special reservations in admission to educational institutions and in public service.

Question 30.
Habeas Corpus?
Answer:
Habeas Corpus is a writ issued by a superior court against any public authority. It is issued against unlawful arrest and detention.

Question 31.
Mandamus?
Answer:
The writ of Mandamus is issued by a superior court to any person or ‘ public authority to perform a legal duty. It is a direction to perform ministerial acts.

Question 32.
What is Prohibition?
Answer:
The writ of prohibition is issued either by Supreme Court or High Court to an inferior court against excessive jurisdiction. It prevents a lower court from handling cases beyond the jurisdiction

Question 33.
Certiorari?
Answer:
A certiorari is a writ issued by the superior court to any lower court against excessive jurisdiction. It is issued to quash the order and decision of the court or tribunal.

Question 34.
Quo-Warranto?
Answer:
The writ of quo warm to is issued by a superior court to any public authority against the illegal assumption of public office. It prohibits illegal appointments to public service.

Question 35.
Importance of Art. 32?
Answer:
The Indian Constitution Under Art. 32 grants right to constitutional remedies. As per the article, in the case of. violation of the fundamental rights of a citizen, he can move to the Supreme Court for immediate redressal.

Question 36.
Objectives of Directive Principles?
Answer:
The objectives of directive principles of state policy are To make India a welfare state and To create conditions for socio-economic democracy in India.

Question 37.
Fundamental-Duties?
Answer:
The fundamental duties are the moral responsibilities of the citizens towards the state. These are non-justiciable in nature.

Question 38.
Right to the property?
Answer:
The right to property was a fundamental right up to 1978, but it was deleted in 1978 by 44th Amendment Act. Now it is a legal duty under Art. 300 A.

Question 39.
Right to Education?
Answer:
Right to Education is a fundamental right under Art. 21 (A). It provides for free and compulsory education to all children belonging to the age group of 6 to 14 years.

Question 40.
Right to life?
Answer:
Right to life is a fundamental right under Art. 21. It states that no one shall be deprived of his life and liberty except, according to procedure established by law.

Question 41.
Right to Information Act. 2005?
Answer:
The Right to Information Act, 2005 seeks to promote transparency and accountability in the working of the government. It helps in containing corruption and makes the citizens aware about the activities of the government.

Question 42.
What is preventive detention?
Answer:
The constitution of India under Art 22 provides for preventive detention of criminal and anti-socials beyond 24 hours. Under this provision any person whose presence is apprehensive of creating social disorder and violence, he may be kept under legal custody for three months even without violating law.

Question 43.
What constitutes the basic structure of the constitution?
Answer:
Basic structure comprises of some values ideals and principles of the constitution which lay its foundation, According to eminent justice, it includes parliamentary democracy, fundamental rights, preamble, secularism, republican model, independent judiciary and centralized federation, etc.

Question 44.
What is free legal aid?
Answer:
The constitution of India under Art. 39A under the Directive Principle makes provision for free legal aid to the poor and underprivileged sections of society; Under the provision, the state government bears the cost of fighting legal battles against those exploiting women, poor and distressed people.

Question 45.
Constituent Assembly of India?
Answer:
The new constitution of India was framed by die constituent Assembly, (b) It comprises of all the eminent lawyers. Intellectuals, social activists, and political leaders of India of that time. It was a representative body that prepared the constitution within 2 years 11 months and 18 days. It played the role of parliament till the new parliament was elected. Its members were elected indirectly by the provincial legislatures on the basis of population.

Question 46.
Cabinet mission plan?
Answer:
Cabinet mission. came to India in the year 1946 to discuss about the grant of independence to India. Prime minister clement allee sent three of his cabinet minister, Sir Stafford Cripps, A’.V. Alexander and Lord Pathik Lawrence as members of the mission. The mission held discussions with Indian leaders, leaders of the Muslim League, and administrators about the ways and means to concede independence. The mission proposed for a ‘Constituent assembly to be formed to prepare the new constitution of India. It recommended for a federal setup in India with Indian provinces and princely states.

Question 47.
Drafting committee?
Answer:
The drafting committee was instrumental for framing the new. constitution of India. It comprises of Dr. B .R.Ambedkar as the chairman and N. Gopalswamy Ayengarm, Alladi Krishnaswamy, Nyer, K.M. Munsi, Mahammad Saddik, N. Madhab Rao and T.T. Krishiiamachari as members. This committee prepared the draft of the whole constitution and got it approved. It had 55 sessions continuing for 114 days and on 4th November 1948 submitted the draft constitution for approved. Dr. Ambedkar played a leading role as its chairman for which he is regarded as the father of Indian constitution.

Question 48.
Objective Resolution?
Answer:
Objective Resolution of the constituent Assembly was initiated by Pandit Nehru on 13th December 1946. It was approved on. 22nd January 1947 by the constituent assembly. This resolution contained the aims and aspirations of the founding fathers relating to the future of India. It proposed to make India, a sovereign Independent Republic. Besides, it also resolved to ensure social political and economic justice to the people of India along with liberty and equality.

Question 49.
Write at least five important features of Indian Constitution?
Answer:

• Longest written constitution.
• A balance between rigidity and flexibility
• Federal in form but unitary in spirit
• Parliamentary democracy; and
• Secular State

Question 50.
What does the preamble to the constitution imply?
Answer:
The preamble implies an introduction to the Constitution. It embodies the objective, ideals, sentiments and aspirations of the nation. It is a key to understanding the nature and spirit of the Indian Policy. It speaks about the source of the Constitution and the date of its adoption. The preamble indicates the general purpose for which people have ordained and established Constitution.

Question 51.
What are the source of the Indian Constitution?
Answer:
The contents of the Indian Constitution have been derived from different sources out of which the important ones are given below.
The Government of India Act, of 1935 has been the most important source of the Constitution. The British Constitution has influenced it the most. The fundamental rights have come from the American constitution. The Irish constitution, the Weimar Constitution of Germany the Canadian Constitution and the Constitution of the South African Republic have also influenced the Indian Constitution.

Question 52.
How does the Preamble depict the characteristics of Indian Constitution?
Answer:
The preamble outlines the following characteristics of Indian Constitution. India is a sovereign independent community. India is a republic. India has a democratic form of government. The constitution ensures justice, liberty, equality and fraternity. It seeks to harmonize dignity of the individual with unity of the nation. It establishes popular sovereignty and seeks to follow socialistic and secular principles.

Question 53.
India is a Sovereign Socialist Secular Democratic Republic Discuss?
Answer:
The Preamble reads India as a Sovereign, Socialist Secular Democratic Republic. India is sovereign both internally and externally and it believes in socialistic principles. India has no state religion and the Constitution grants equal status to all religions. Indian policy is organized on democratic principles and the President being the head of state is elected hence it is a republic.

Question 54.
Why is Indian Constitution so long?
Answer:
Indian constitution is the longest written Constitution in the world having 395 Articles and 12 Schedules. The factors responsible for the extra length of the Constitution are as follows. The constitution is meant both for the Union and the States. Everything in the Constitution has been discussed in detail.

The diversity of the Indian culture necessitated special provisions for which the Constitution has been so long. Besides the above factors, detailed chapters on fundamental rights, Directive Principles, emergency provisions, center-state relations and fundamental duties have added to the size of the constitution.

Question 55.
Why is Indian Constitution supposed to be a bag of borrowings?
Answer:
The Indian Constitution was framed following the major democratic constitution of the world. The British Constitution America, Irish, Australian and Canadian Constitutions, etc. have significantly influenced the framers, but that does not mean that the principles and ideals of the Constitution are borrowed: The framers have borrowed most of the provisions from Western countries and modified them to suit to Indian conditions and the people. It will be wrong to brand as a borrowed constitution.

Question 56.
Indian Constitution is federal in form but unitary in spirit. Discuss?
Answer:
Indian constitution has designed the country on the principles of a federation but it works more or less as a unitary state. The framers adopted a federal form of government on account of its vastness but declared it as a union of states. Despite all federal features, the union government is given an upper hand in every sphere. The union Govt enjoys more powers and it can at times change the federal structure into a unitary one. The framers have made a happy blend between unitarism and federalism for which it is supposed to be federal in form but unitary in spirit.

Question 57.
Why is India called a Quasi Federal State?
Answer:
A quasi-federal state is one which is neither completely unitary nor federal in character. Prof. K.C. where has criticized India’s federal character as was quasi-federal. India is organized on the basis of the federation but works on unitary lines. The union government enjoys more powers than the states and the framer can suspend the state government on various grounds. The Union Govt, through the office of the Governor, All India Services, Planning Commission, National Development Council and Finance Commission, etc. can influence the Govt, at the state. Therefore, India is called a quasi-federal state.

Question 58.
Is Indian Constitution a rigid one?
Answer:
No, Indian Constitution does not come along the lines of a rigid Constitution as that of the U.S.A. Most parts of the Constitution can be amended by simple procedure of legislation. Only the federal features of the Constitution are rigid. Those matters can be amended if a majority of both houses of Parliament coupled with the 2/3rd majority of those present and voting support the motion.

Therefore, the Constitutional amendment is sent to the state legislatures for ratification. If at least half of the states approve the proposal the amendment is effected. This rigid procedure applies only to a few items.

Question 59.
Indian Constitution is flexible. Discuss?
Answer:
A flexible Constitution is one which can be amended easily, by the simple procedure of legislation. Most parts of the Constitution can be amended Iggy this process but there are still important matters which need to be amended by the rigid process. As the constitutional framers have made a compromise between rigidity and flexibility it can not be described as i flexible Constitution.

Question 60.
Briefly state the Preamble of the Indian Constitution?
Answer:
The Preamble of the Indian Constitution reads as follows: “WE ‘THE PEOPLE OF INDIA” laving solemnly resolved to constitute India intoa SOVEREIGN SOCIALIST SECULAR DEMOCRATIC REPUBLIC AND TO SECURE TO ALL ITS CITIZENS. JUSTICE, Social, Economic and Political LIBERTY of thought, expression, faith, belief and worship.

EQUALITY of status and opportunity, land to promote among them all FRATERNITY assuring the dignity of the individual and the unity and integrity of the nation. IN OUR CONSTITUENT ASSEMBLY, this twenty-sixth day of November 1949, do hereby ADOPT ENACT AND GIVE TO OURSELVES THIS CONSTITUTION.

Question 61.
What is the importance of the term ‘We the people of India’?
Answer:
The Preamble starts with the phrase ‘We the people of India which illustrates that the people of India are the maker of the Constitution. It speaks about the representative character of the Constitution and accepts people as the ultimate sovereign. The constitution is a popular document which is not imposed on the people but deliberately prepared by a representative Constituent Assembly.

Question 62.
Is India is a sovereign state?
Answer:
Yes, India is a sovereign state and the preamble of the constitution declares about its sovereign character. Internally and externally. India is supreme. Externally, the country is free from any outside authority and it has accepted the membership of international organizations voluntarily internally, it is competent to adopt its domestic policies and regulate the behavior of its nationals and organizations.

Question 63.
Is India is a Secular State?
Answer:
Yes, India is a secular state. A secular state is one which is neither anti-religious nor irreligious. It grants equal status and freedom to, all religions. There is no state religion in India and no religious discrimination is practiced. The Constitution has guaranteed right to freedom of religion to the citizens which symbolizes the secular character of the constitution.

Question 64.
Why is lndia called a democratic republic?
Answer:
The Preamble declares India as a democratic republic. The Indian political system is organized on a comprehensive democratic basis. The parliament and the state legislature are elective bodies and the franchise is extended to all adult citizens universally. The Govt, of India, is elected by the people periodically and remains. responsible.to them. It is a republic because the President being the head of State is elected indirectly. Thus India is called a democratic republic.

Question 65.
Is India a Welfare State?
Answer:
Yes, India is a secular state ordained under Art. 38 of the Constitution. The Directive Principles also embody the principles of a welfare state. The constitution states that the ownership and control of material resources shall be distributed equally to subserve common good. The Govt provides special assistance to the educationally, and economically backward sections of the community. The five-year plans and the socialistic principles are implemented in the country to make India a welfare state.

Question 66.
What is single citizenship?
Answer:
Single citizenship means that in India the power to grant citizenship has been conferred upon the union government only. The state has no authority in this regard. All citizens residing anywhere within India irrespective of their residence or place of birth, are the citizens of India. This idea is supposed to promote a sense of unity among Indians.

Question 67.
What are the five pillars of the Constitution?
Answer:
The five pillars of the constitution are

• the election commission,
• the Finance Commission
• the Union Public Service Commission
• the Attorney General and
• the Auditor-General of India.

These constitute the heart of the democratic structure.

Question 68.
What are the basic philosophy of the Constitution?
Answer:
The basic principles embodying the philosophy of the constitution are as follows:

• Popular sovereignty,
• Federal system
• Fundamental rights of the citizens,
• Directive principles of state policy,
• Judicial independence
• Parliamentary system
• Supremacy of the Union Government.

Question 69.
How the Indian Constitution can be amended?
Answer:
The Indian constitution can be amended by the Parliament under Art. 368. The parliament by required majority can make addition, variation or repeal any provision of the constitution. Such a proposal can be initiated in either House of Parliament and after approval, it is submitted to the President for assent. The President, can’t reject such a proposal and after President’s assent, necessary amendments may be effected.

Question 70.
Why India is called a socialist state?
Answer:
India is called a socialist state, because of the following reasons. The Govt, of India, acts on socialistic lines. The Govt is committed to securing socio-economic and political justice to the people by ending all forms of exploitation. It undertakes economic planning to ensure equitable distribution of wealth and income.

Question 71.
What is the importance of the Preamble to the constitution?
Answer:
A preamble is an introduction to the constitution. It underlines the philosophical foundations of the constitution and its objectives. It outlines in a nutshell the ideals and objectives of our constitution. It lays down the pattern of our political setup, that is to make India a sovereign, socialist, secular, democratic republic. It speaks about the source of constitution And the date of its, adoption.

Question 72.
Is, the Preamble a part of the constitution?
Answer:
This is a controversial question. As per the decision of the Supreme Court on the transfer of the Berubari Union and exchange of Enclaves, the Preamble is not a part of the constitution. But the apex court modified its decision: in the Keshavananda Bharati case and regarded the preamble as a part of the constitution. In fact, it forms a part of the basic structure as it defines the. structure and philosophy of the constitution.

Question 73.
How the Constitution of India has safeguarded the interests of S.Cs and STs?
Answer:
The Indian constitution under Part XVI has specified special provisions for safeguarding the interests of Scheduled Castes Under Art – 330 seats have been reserved for them in the Lok Sabha and under Art. 331 and 332 in the State Legislatures. They are given special facilities to join public service, to get promotions, and to get admission into colleges, universities, and- professional institutions.

Question 74.
Which four new languages have been added to the VIIIth Schedule by the 100th Amendment Act?
Answer:
The four languages which have got an entry into the VIIIth Schedule by the 100th Amendment Act are, Bodo, Dogri, Maithili and Santhali.

Question 75.
Name five major sources of Indian Constitution?
Answer:
The five major sources of Indian Constitution are:

• The Govt, of India Act, 1935
• Provisions of foreign constitutions.
• Records of debates and discussions in the constituent Assembly.
• Parliamentary statutes and judicial decisions.
• The ideals and values of freedom struggle.

Question 76.
86th Constitutional Amendment, 2002?
Answer:
The Parliament passed the 86th Amendment Act, in 2002. It provides for Inclusion of Art. 21 (A) Whereby all children under the age group of 6 to 14 years of age are given the fundamental right of free and compulsory education. It added a new duty in Art. 51(A) increasing the number to 11. By this amendment, it has become the duty of the parents to send their children from 6 to 14 years of age to school.

Question 77.
What do you mean by fundamental Rights?
Answer:

• Fundamental Rights are those rights guaranteed in the constitution.
• It ensures the development of individual personality.
• These rights are guaranteed and protected by the state.
• The fundamental rights are superior to ordinary laws.
• These rights are the bedrock of Indian democracy.
• The Government cannot make laws violating any of these rights.

Question 78.
Name the six Fundamental Rights?
Answer:

• Right to Equality
• Right to Freedom
• Right against Exploitation
• Right of Freedom of Religion
• Educational and Cultural Rights
• Right to Constitutional Remedies

Question 79.
What do you mean by Right to equality?
Answer:
Right to equality is the first fundamental right guaranteed to the citizens. It refers to equality before law and equal protection of laws. If eliminates the possibility of all discrimination against a citizen on grounds of religion, race, sex, caste or place of birth. It guarantees equality of opportunity to all in matters of public employment. It seeks to abolish untouchability to ensure social equality. The constitution prohibits all titles and honors for the Indian except that of military or academic character to maintain proper equality.

Question 80.
What are the six freedom guaranteed under Art19?
Answer:
The six freedoms guaranteed under Art. 19 are:

• Freedom of speech and expression.
• Freedom to assemble peacefully without arms.
• Freedom to form associations or unions.
• Freedom to move freely throughout the territory of India.
• Freedom to reside and settle in any part of India.
• Freedom to practice any profession, occupation, trade, or business.

Question 81.
What does Right against Exploitation imply?
Answer
The right against exploitation is a fundamental right which seeks to prohibit all forms of exploitation of the children and women. It prohibits forced labor and traffic in human beings. It prevents child below 14 years to be employed in any factory or mine where there is danger to life. It also opposes the illegal use of women for immoral purposes. This right is meant for the weaker sections of the community.

Question 82.
What does the Right to Freedom of Religion imply?
Answer:
The Right to Freedom of Religion implies secular character of the state. It has been explained under four articles:

• Art. 25 states that all persons are equally entitled to freedom of conscience and the right to profess, propagate or practice any religion.
• Art. 26 grants the freedom to establish and maintain religious institutions for religious or charitable purposes.
• Art. 27 provides that, no person shall be ” compelled to pay any tax for the promotion of any particular religion.
• Art. 28 states that no religious instruction shall be provided in any educational institution wholly maintained by the state funds.

Question 83.
What is the meaning of Right to constitutional remedies?
Answer:
This is the sixth fundamental right which provides remedial measures for the enforcement of fundamental rights. This right is conferred under Art. 32 of the Constitution and it empowers the Supreme Court to uphold the fundamental rights of the citizens. If the fundamental rights of a citizen are violated he can move to the Supreme Court under Art. 32 or High Court under Art. 226 for redressal. The court can issue appropriate writs to provide- relief to the affected persons.

Question 84.
What are the different kinds of writs issued by the courts for the redressal of fundamental rights?
Answer:
There are five kinds of writs issued by the courts for the redressal of fundamental rights. These writs are

• Habeas Corpus
• Mandamus
• Prohibition
• Certiorari, and
• Quo-Warranto

Question 85.
What is the purpose of the writ of Habeas Corpus?
Answer:
The writ of Habeas Corpus is issued to maintain and enforce the fundamental rights of citizens. It is a Latin word which means to have one’s own body. The purpose of Habeas Corpus is to protect a citizen from unlawful arrest and detention. If a person is arrested or detained without any valid reason the court orders for his release by issuing this writ. It preserves the liberty of a person who is confined without legal justification.

Question 86.
What for the writ of Mandamus is issued?
Answer:
The writ of mandamus is issued by the court commanding a person or authority to do his duty. This writ is used for public purposes to enforce the performance of public duties. It also enforces some privacy rights when they are withheld by the public.

Question 87.
What is prohibition?
Answer:
It is a judicial writ issued by a superior court to an inferior court to prevent the lower court from usurping jurisdiction. The writ of prohibition is meant to check the possibility of abuse in a jurisdiction. If any case pending before the court is beyond its jurisdiction, the higher court by iásue of the writ prohibits the lower cÓurt not to hear such case.

Question 88.
What is Certiorari?
Answer:
A certiorari is a writ issued by a higher court to a lower count for the withdrawal of a case from the latter to the former. Such a writ may be issued even before the trial of the court comes to know that anything beyond the jurisdiction of the court is pending for decision before it. Both the writs of prohibition and certiorari are intended to deal with complaints of excess of jurisdiction.

Question 89.
What is the purpose of Quo warranto?
Answer:
Quo-warranto is a writ which is issued to prevent the illegal assumption of public office. Quo-warrantà’means by what authority. The court issuing the writ has to consider whether there has b&n usurpation of public office or not. It has been there even before the framing of the constitution.

Question 90.
Can the Parliament of India amend the fundamental rights?
Answer:
Yes, the Parliament of India can amend any portion of the fundamental rights excluding the basis structure of the Keshavananda Bharati and others Vs the State of Kerì1a held that the f parliament has right to an end all párts of the Constitution including Part Ill without destroying the basic structure of the constitution.

Question 91.
What do you mean by Directive Principles of State Policy?
Answer:
Directive Principles of State Policy is an important feature of the Indian constitution. These principles are discussed under part IV of the constitution and they are in the nature of constitutional directions o the state and central government to implement in administration. These directives are nonjustifiable but they are considered fundamental in the governance of the country These principles intend to play a positive role in the establishment of a socialist welfare state.

Question 92.
What Is the importance of Directive Principles?
Answer:
The Directive Principles of State Policy have immense utility and importance in making every modem state welfare state. These principles set the foundation of socio-economic democracy in India. These are not obligatory, but they are positive guidelines for the states to observe and implement them in the day-to-day administration of the state. They are not enforceable in any court but they are considered fundamental in the governance of the country.

Question 93.
Name of the socialistic principles of the Directives?
Answer:
The socialistic principles of the Directives contain the following.

• The ownership and control of the material resources of the community must be distributed to subserve common good.
• There must be equal pay for equal work for both men and women.
• The operation of the economic system should not lead to concentration of wealth and means of production to common detriment.
• The health and strength of the workers and tender age of children should not be abused.

Question 94.
Mention at least three Gandhian Principles of the Directives?
Answer:

• Organization of village Panchayats and to provide them with the power of self-government.
• Promotion of educational and economic interests of the S.C. and S.T.
• Promotion of cottage industry on individual and cooperative lines in the rural sector.

Question 95.
What constitutes the liberal principles of the directives?
Answer:
The liberal principles of the directive principles include:

• Separation of executive from the judiciary in public services.
• Provision for uniform civil code throughout the country.
• Protection of monuments and objects of historical and artistic importance.
• Promotion of international peace and security and maintaining adjustable and honorable relations with nations.
• To foster respect for international law and treaty obligations and settlement of international disputes through arbitration.

Question 96.
Fundamental Duties?
Answer:
The fundamental duties are mentioned in Part – IV A, under Art – 51 (A) of the constitution. These are inserted into the text of the constitution by the 42nd Amendment Act of 1976 on the recommendations of Dr. Swaran Singh’s Committee.These are moral in character. These duties are ten in number and in 2002, the 86th Amendment has inserted another duty to the list.

Question 97.
Features of Indian Fundamental Rights?
Answer:
The basic features of Indian fundamental rights can be discussed below:

• These rights are elaborate and comprehensive.
• Both negative and positive rights.
• These rights are not absolute.
• There are certain rights for minorities.
• They are binding upon the union, states, and other state authorities.

Question 98.
Right to Education?
Answer:
The Indian Constitution by the 86 Amendment Act of 2002 has added a new fundamental right under Art.- 21 A, called right to education. Under this article, the govt provides free and compulsory education to all children from the age group of 6 to 14 years in such as manner as the state may by law determine.

Question 99.
The Fundamental Rights in India are defective in many respects.
Answer:
The constitution under part – III has prescribed a comprehensive list of fundamental rights for Indian citizens. But there are certain weaknesses inherent in these rights as are given below:

• The constitution has imposed several limitations on fundamental rights.
• These rights are coded in difficult languages.
• There are no socioeconomic rights.
• The Parliament can amend fundamental rights.
• No mention about natural rights.