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You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.2

Question 1.
Find the equation of the lines passing through the point (1, 1)
(i) With y-intercept (- 4)
(ii) With slope 3
(iii) And (-2, 3)
(iv) And the perpendicular from the origin makes an angle 60° with x-axis.
Solution:
(i) Given y intercept = – 4,
Let x intercept be a

(ii) with slope 3
The equation the line passing through the point (x₁, y₁) and having slope m is
y – y₁ = m(x – x₁)
Given (x₁, y₁) = (1, 1), m = 3
∴ The required equation of the line is
y – 1 = 3(x – 1)
y – 1 = 3x – 3
3x – y – 3 + 1 = 0
3x – y – 2 = 0

(iii) Passing through (1, 1) and (-2, 3)

Question 2.
If P (r, c) is mid point of a line segment between the axes, then show that \(\frac{x}{r}+\frac{y}{c}=\) 2.
Solution:
P (r, c) is the midpoint of AB.
⇒ A = (2r, 0) and B = (0, 2c)
(i.e) x intercept = 2r and
y intercept = 2c .

Question 3.
Find the equation of the line passing through the point (1, 5) and also divides the co-ordinate axes in the ratio 3 : 10.
Solution:
Let x-intercept be 3a and y-intercept be 10a

Question 4.
If p is length of perpendicular from origin to the line whose intercepts on the axes are a and b, then show that \(\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}\)
Solution:

Question 5.
The normal boiling point of water is 100°C or 212°F and the freezing point of water is 0°C or 32°F.
(i) Find the linear relationship between C and F
(ii) Find the value of C for 98.6° F and
(iii) The value of F for 38°C.
Solution:
Given when C = 100, F = 212 and when C = 0, F = 32

Question 6.
An object was launched from a place P in constant speed to hit a target. At the 15th second, it was 1400m away from the target and at the 18th second 800m away. Find
(i) The distance between the place and the target
(ii) The distance covered by it in 15 seconds,
(iii) Time is taken to hit the target.
Solution:
Taking time = x and distance = y
We are given at x = 15, y = 1400 and at x = 18, y = 800

Question 7.
The population of a city in the years 2005 and 2010 are 1,35,000 and 1,45,000 respectively. Find the approximate population in the year 2015. (assuming that the growth of population is constant).
Solution:
Taking the year as x and population as y
We are given when x = 2005,
y = 1,35,000 and
when x = 2010,
y = 1,45,000

y – 135000 = 2000 (x – 2005)
y = 2000(x – 2005) + 135000
At x = 2015, y = 2000 (2015 – 2005) + 135000
(i.e) y = 2000 (10) + 135000 = 20000 + 135000 = 1,55,000
The approximate population in the year 2015 is 1,55,000

Question 8.
Find the equation of the line, if the perpendicular drawn from the origin makes an angle 30° with x – axis and its length is 12.
Solution:
The equation of the line is x cos α + y sin α = p

Question 9.
Find the equation of the straight lines passing through (8, 3) and having intercepts whose sum is 1.
Solution:
Given sum of the intercepts = 1 ⇒ when x-intercept = a then y-intercept = 1 – a

8 (1 – a) + 3a = a (1 – a)
8 – 8a + 3a = a – a²

Question 10.

Solution:

⇒ The points A, B, C lie on a line
⇒ The points A, B, C are collinear

Question 11.
A straight line is passing through the point A (1, 2) with slope \(\frac{5}{12}\). Find points on the line which are 13 units away from A.
Solution:

Question 12.
A 150m long train is moving with a constant velocity of 12.5 m/s. Find
(i) The equation of the motion of the train,
(ii) Time taken to cross a pole,
(iii) The time taken to cross the bridge of length 850 m is?
Solution:
(i) Now m = \(\frac{y}{x}\) = 12.5m / second,
The equation of the line is y = mx + c ….(1)
Put c = -150, m = 12.5 m,
The equation of motion of the train is y = 12.5x – 150

(ii) To find the time taken to cross a pole we take y = 0 in (1)
⇒ 0 = 12.5x – 150 ⇒ 12.5x = 150

(iii) When y = 850 in (1)
850 = 12.5 x – 150 ⇒ 12.5x = 850 + 150 = 1000

Question 13.
A spring was hung from a hook in the ceiling. A number of different weights were attached to the spring to make it stretch, and the total length of the spring was measured each time shown in the following table.

(i) Draw a graph showing the results.
(ii) Find the equation relating the length of the spring to the weight on it.
(iii) What is the actual length of the spring.
(iv) If the spring has to stretch to 9 cm long, how much weight should be added?
(v) How long will the spring be when 6 kilograms of weight on it?
Solution:
Taking weight (kg) as x values and length (cm) as y values we get (x₁, y₁) = (2, 3), (x₂, y₂) = (4, 4)

The equation of the line passing through the above two points is

(iii) When x = 0, 2y = 4 ⇒ y = 2 cm

(iv) When y = 9 cm, x – 18 = – 4
x = -4 + 18 = 14 kg

(v) When x = 6 (kg), 6 – 2y = – 4, -2y = -4 – 6 = -10
⇒ 2y = 10 ⇒ y = 10/2 = 5 cm.

Question 14.
A family is using Liquefied petroleum gas (LPG) of weight 14.2 kg for consumption. (Full weight 29.5 kg includes the empty cylinders tare weight of 15.3 kg.). If it is used at a constant rate then it lasts for 24 days. Then the new cylinder is replaced
(i) Find the equation relating the quantity of gas in the cylinder to the days.
(ii) Draw the graph for the first 96 days.
Solution:

Question 15.
In a shopping mall, there is a hall of cuboid shape with dimension 800 × 800 × 720 units, which needs to be added the facility of an escalator in the path as shown by the dotted line in the figure. Find
(i) The minimum total length of the escalator,
(ii) The heights at which the escalator changes its direction,
(iii) The slopes of the escalator at the turning points.

Solution:
(i) the minimum total length of the escalator.
Shape of the hall in the shopping mall is cuboid. When you open out the cuboid, the not of the cuboid will be as shown in the following diagram.

The path of the escalator is from OA to AB to BC to CD

The minimum length = 3280 units

(ii) The height at which the escalator changes its direction.

(iii) Slope of the escalator at the turning points

Since ∆OAE = ∆ABB’ = ∆BCC’ = ∆CAD
Slope at the points B, C will be \(\frac{9}{40}\)

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.2 Additional Questions Solved

Question 1.
Find the equation of a straight line on which length of the perpendicular from the origin is four units and the line makes an angle of 120° with the positive direction of the x-axis.
Solution:

Question 2.
Find the equation of the line which passes through the point (- 4, 3), and the portion of the line intercepted between the axes is divided internally in the ratio 5 : 3 by this point.
Solution:
Let AB be a line passing through a point (-4, 3) and meets the x-axis at A (a, 0) and the y-axis at B (0, b).

Question 3.
If the intercept of a line between the coordinate axes is divided by the point (-5, 4) in the ratio 1 : 2, then find the equation of the line.
Solution:
Let a and b be the intercepts on the given line.


Hence, the required equation is 8x – 5y + 60 = 0

Question 4.
Find the equation of the straight line which passes through the point (1, -2) and cuts off equal intercepts from axes.
Solution:
Intercept form of a straight line is \(\frac{x}{a}+\frac{y}{b}\) = 1, where a and b are the intercepts on the axis

If equation (1) passes through the point (1, -2) we get

So, equation of the straight line is x v

Hence, the required equation x + y + 1 = 0

Question 5.
Find the distance of the line 4x – y = 0 from the point P(4, 1) measured along the line making an angle 135° with the positive x-axis
Solution:
The equation in distance form of the line passing through P(4, 1) and making an angle of 135° with the positive x-axis

Question 6.
The line 2x – y = 5 turns about the point on it, whose ordinate and abscissa are equal, through an angle of 45° in the anti-clockwise direction, find the equation of the line in the new position.
Solution:
If the line 2x – y = 5 makes an angle θ with the x-axis.
Then, tan θ = 2. Let P (α, α) be a point on the line 2x – y = 5.
Then, 2 α – α = 5 ⇒ α = 5

So, the coordinates of P are (5, 5). If the line 2x – y – 5 = 0 is rotated about point
P through 45° in an anti-clockwise direction, then the line in its new position makes angle θ + 45° with the x-axis.
Let m’ be the slope of the line in its new position. Then,

Thus, the line in its new pdsition passes through P (5, 5) and has slope m’ = -3
So, its equation y – 5 = m’ (x – 5) or, y – 5 = -3 (x – 5) or, 3x + y – 20 = 0

You can Download Samacheer Kalvi 10th Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 8 Periodic Classification of Elements

Samacheer Kalvi 10th Science Periodic Classification of Elements Textual Evaluation Solved

I. Choose the best answer.

Question 1.
The number of periods and groups in the periodic table are ____.
(a) 6, 16
(b) 7, 17
(c) 8, 18
(d) 7, 18.
Answer:
(d) 7, 48.

Question 2.
The basis of modern periodic law is:
(a) atomic number
(b) atomic mass
(c) isotopic mass
(d) number of neutrons
Answer:
(a) atomic number

Question 3.
_____ group contains the member of the halogen family.
(a) 17th
(b) 15th
(c) 18th
(d) 16th.
Answer:
(a) 17th

Question 4.
……….is a relative periodic property.
(a) atomic radii
(b) ionic radii
(c) electron affinity
(d) electronegativity
Answer:
(d) electronegativity

Question 5.
Chemical formula of rust is ____.
(a) FeO.xH₂O
(b) FeO₄.xH₂O
(c) Fe₂O₃.xH₂O
(d) FeO.
Answer:
(c) Fe₂O₃.xH₂O

Question 6.
In the aluminothermic process the role of Al is ____.
(a) oxidizing agent
(b) reducing agent
(c) hydrogenating agent
(d) sulphurising agent.
Answer:
(b) reducing agent

Question 7.
The process of coating the surface of metal with a thin layer of zinc is called:
(a) painting
(b) thinning
(c) galvanization
(d) electroplating
Answer:
(c) galvanization

Question 8.
Which of the following have inert gases 2 electrons in the outermost shell?
(a) He
(b) Ne
(c) Ar
(d) Kr.
Answer:
(a) He

Question 9.
Neon shows zero electron affinity due to:
(a) stable arrangement of neutrons
(b) stable configuration of electrons
(c) reduced size
(d) increased density
Answer:
(b) stable configuration of electrons

Question 10.
____ is an important metal to form an amalgam.
(a) Ag
(b) Hg
(c) Mg
(d) Al.
Answer:
(b) Hg

II. Fill in the blanks.

Question 1.
If the electronegativity difference between two bonded atoms in a molecule is greater than 1.7, the nature of bonding is ____.
Answer:
Ionic bond.

Question 2.
____ is the longest period in the periodical table.
Answer:
Sixth period.

Question 3.
______ forms the basis of modem periodic table.
Answer:
Atomic number.

Question 4.
If the distance between two Cl atoms in Cl₂ molecule is 1.98 Å, then the radius of the Cl atom is ____.
Answer:
0.99 Å.

Question 5.
Among the given species A^–, A⁺, and A, the smallest one in size is ______.
Answer:
A^+.

Question 6.
The scientist who propounded the modem periodic law is ______.
Answer:
Henry Moseley.

Question 7.
Across the period, ionic radii ______ (increases,decreases)
Answer:
Decreases.

Question 8.
_______ and ______ are called inner transition elements.
Answer:
Lanthanides and Actinides.

Question 9.
The chief ore of Aluminium is ______.
Answer:
Bauxite (Al₂O₃.2H₂O).

Question 10.
The chemical name of rust is ______.
Answer:
Hydrated ferric oxide.

III. Match the following.

Question 1.

1. Galvanisation	(a) Noble gas elements
2. Calcination	(b) Coating with Zn
3. Redox reaction	(c) Silver-tin amalgam
4. Dental filling	(d) Alumino thermic process
5. Group 18 elements	(e) Heating in the absence of air

Answer:
1 – (b), 2 – (e), 3 – (d), 4 – (c), 5 – (a).

IV. True or False: (If false give the correct statement)

Question 1.
Moseley’s periodic table is based on atomic mass.
Answer:
False.
Correct Statement: Moseley’s periodic table is based on atomic number.

Question 2.
Ionic radius increases across the period from left to right.
Answer:
False.
Correct Statement: Ionic radius decreases across the period from left to right.

Question 3.
All ores are minerals, but all minerals cannot be called as ores.
Answer:
True.

Question 4.
Al wires are used as electric cables due to their silvery – white colour.
Answer:
False.
Correct Statement: Al wires are used as electric cables due to their good conductor of electricity.

Question 5.
An alloy is a heterogeneous mixture of metals.
Answer:
False.
Correct Statement: An alloy is a homogenous mixture of metals.

V. Assertion and Reason:

Answer the following questions using the data given below.
(i) A and R are correct, R explains the A.
(ii) A is correct, R is wrong.
(iii) A is wrong, R is correct.
(iv) A and R are correct, R doesn’t explain A.

Question 1.
Assertion: The nature of bond in HF molecule is ionic
Reason: The electronegativity difference between H and F is 1.9.
Answer:
(i) A and R are correct, R explains the A.

Question 2.
Assertion: Magnesium is used to protect steel from rusting.
Reason: Magnesium is more reactive than iron.
Answer:
(i) A and R are correct, R explains the A.

Question 3.
Assertion: An uncleaned copper vessel is covered with a greenish layer.
Reason: copper is not attacked by alkali
Answer:
(iv) A and R are correct, R doesn’t explain A.

VI. Short Answer Questions.

Question 1.
A is a reddish-brown metal, which combines with O₂ at < 1370 K gives B, a black coloured compound. At a temperature > 1370 K, A gives C which is red in colour. Find A, B and C with reaction.
Answer:
(i) Reddish brown metal (A) is copper.

(ii) (A) reacts with O₂ at bleow 1370 K gives Copper (II) oxide (B), which is black in colour.
2Cu + O₂ \(\xrightarrow [ 1370k ]{ below }\) 2CuO (Copper (II) oxide) (B).

(iii) (A) reacts with O₂ at above 1370 K gives Copper (I) oxide (C), which is red in colour.

Question 2.
A is a silvery – white metal. A combines with O₂ to form B at 800°C, the alloy of A is used in making the aircraft. Find A and B.
Answer:
(i) Silver – white metal (A) is Aluminium.

(ii) (A) combines with O₂ to form aluminium oxide at 800°C.

(iii) Duralumin is the alloy of Al, which is used to make aircraft

Question 3.
What is rust? Give the equation for formation of rust.
Answer:
Rust is hydrated ferric oxide, Fe₂O₃.xH₂O. It is formed when iron is exposed to moist air.
4 Fe + 3O₂ + xH₂O > 2Fe₂O₃.xH₂O

Question 4.
State two conditions necessary for rusting of iron.
Answer:
Conditions for rusting of iron:

• The presence of water and oxygen is essential for the rusting of iron.
• Impurities in the iron, the presence of water vapour, acids, salts and carbon dioxide hasten to rust.
• Pure iron does not rust in dry and CO₂ free air. It also does not rust in pure water, free from dissolved salts.

VII. Long Answer Questions

Question 1.
(a) State the reason for the addition of caustic alkali to bauxite ore during purification of bauxite.
(b) Along with cryolite and alumina, another substance is added to the electrolyte mixture. Name the substance and give one reason for the addition.
Answer:
(a) Naturally, Bauxite is not soluble in normal solvents. Therefore the addition of caustic alkali to bauxite plays an important role while extraction of aluminium. Caustic alkali dissolves bauxite forming soluble sodium meta aluminate while impurities remain insoluble and precipitate as red mud.

(b) Along with cryolite and alumina, another substance is added to the electrolyte mixture is Fluorspar. Adding of fluorspar lowers the fusion temperature of the electrolyte.

Question 2.
The electronic configuration of metal A is 2, 8, 18, 1. The metal A when exposed to air and moisture forms B a green layered compound. A with conc. H₂SO₄ forms C and D along with water. D is a gaseous compound. Find A, B, C and D.
Answer:
(i) The electronic configuration of metal (A) is 2, 8, 18, 1. A is copper (Z = 29)

(ii) (A) Copper exposed to air and moisture forms green layered compound (B) that is a copper carbonate.

(iii) Copper (A) reacts with conc.H₂SO₄ to give copper sulphate (C) and Sulphur dioxide (D).

Question 3.
Explain the smelting process.
Answer:
Smelting (in a Blast Furnace): The charge consisting of roasted ore, coke and limestone in the ratio 8 : 4 : 1 is smelted in a blast furnace by introducing it through the cup and cone arrangement at the top. There are three important regions in the furnace.

(a) The Lower Region (Combustion Zone): The temperature is at 1500°C. In this region, coke bums with oxygen to form CO₂ when the charge comes in contact with a hot blast of air.

It is an exothermic reaction since heat is liberated.

(b) The Middle Region (Fusion Zone): The temperature prevails at 1000°C. In this region, CO₂ is reduced to CO.

Limestone decomposes to calcium oxide and CO₂

These two reactions are endothermic due to absorption of heat. Calcium oxide combines with silica to form calcium silicate slag.
CaO + SiO₂ → CaSiO₃

(c) The Upper Region (Reduction Zone): The temperature prevails at 400°C . In this region carbon monoxide reduces ferric oxide to form a fairly pure spongy iron.
\(\mathrm{Fe}_{2} \mathrm{O}_{3}+3 \mathrm{CO} \stackrel{400^{\circ} \mathrm{C}}{\longrightarrow} 2 \mathrm{Fe}+3 \mathrm{CO}_{2}\)
The molten iron is collected at the bottom of the furnace after removing the slag.
The iron thus formed is called pig iron. It is remelted and cast into different moulds. This iron is called cast iron.

VIII. HOT Questions.

Question 1.
Metal A belongs to period 3 and group 13. A in red hot condition reacts with steam to form B. A with strong alkali forms C. Find A, B and C with reactions.
Answer:
(i) Metal (A) belongs to period 3 and group 13, is Aluminium (Al).

(ii) (A) Al in red hot condition reacts with steam to form Aluminium oxide (B).

(iii) Aluminium (A) reacts with strong alkali forms of sodium meta aluminate (C).

Question 2.
Name the acid that renders aluminium passive. Why?
Answer:
Dil. or cone. HNO₃ does not attack aluminium, but renders Al passive due to the formation of oxide film on its surface.

Question 3.
(a) Identify the bond between H and F in the HF molecule.
(b) What property forms the basis of identification?
(c) How does the property7 vary in periods and in groups?
Answer:
(a) The nature of the bond in the HF molecule is ionic.
(b) Electronegativity.
(c) Along the period, from left to right in the periodic table the electronegativity increases because of the increase in the nuclear charge which in turn attracts the electrons more strongly. On moving down a group, the electronegativity of the elements decreases because of the increased number of energy levels.

Samacheer Kalvi 10th Science Periodic Classification of Elements Additional Questions Solved

I. Choose the best answer.

Question 1.
Which period contains only two elements?
(a) Second
(b) First
(c) Third
(d) Fifth.
Answer:
(b) First

Question 2.
The sixth period contains inner transition elements.
(a) 18
(b) 14
(c) 10
(d) 8
Answer:
(b) 14

Question 3.
Lanthanides and Actinides are called as _____.
(a) Alkali metals
(b) Inner transition elements
(c) Transition elements
(d) Representative elements.
Answer:
(b) Inner transition elements

Question 4.
The number of valence electrons present in Halogens is:
(a) One
(b) Seven
(c) Zero
(d) Two
Answer:
(b) Seven

Question 5.
The distance between the two hydrogen nuclei of the molecule is 0.74 Å. So its covalent radius is _____.
(a) 0.74 Å
(b) 0.99 Å
(c) 0.37 Å
(d) 7.4 Å.
Answer:
(c) 0.37 Å

Question 6.
Along the groups, atomic radius _____.
(a) decreases
(b) increases
(c) decreases then increase
(d) no change.
Answer:
(b) increases.

Question 7.
Which one of the following elements will have the highest electronegativity?
(a) chlorine
(b) nitrogen
(c) caesium
(d) fluorine
Answer:
(d) fluorine

Question 8.
Electron affinity is measured in _____.
(a) kJ^-1
(b) mol^-1
(c) kJ/mol
(d) kJ/mol².
Answer:
(c) kJ/mol

Question 9.
Noble gases have ______ electron affinity.
(a) positive
(b) negative
(c) zero
(d) high.
Answer:
(c) zero

Question 10.
The element with positive electron gain enthalpy is:
(a) Hydrogen
(b) Sodium
(c) Argon
(d) Fluorine
Answer:
(b) Sodium

Question 11.
The mineral from which a metal can be readily and economically extracted on a large scale is said to be a / an _____.
(a) Ore
(b) Flux
(c) Slag
(d) Gangue.
Answer:
(a) Ore

Question 12.
Flux + Gangue → _____?
(a) Mineral
(b) Matrix
(c) Slag
(d) Smog.
Answer:
(c) Slag

Question 13.
………… is used for anodizing process.
(a) Zinc
(b) iron
(c) Aluminium
(d) Tin
Answer:
(c) Aluminium

Question 14.
The ore which can be purified by gravity separation method is _____.
(a) Haematite
(b) oxide ores
(c) sulphide ores
(d) both (a) and (b).
Answer:
(d) both (a) and (b).

Question 15.
The calcium silicate slag is formed in ………. zone.
(a) combustion zone
(b) fusion zone
(c) reduction
(d) molten
Answer:
(b) fusion zone

Question 16.
Zinc blende is purified by _____.
(a) Hydraulic method
(b) Magnetic separation method
(c) Froth floatation method
(d) Chemical method.
Answer:
(c) Froth floatation method

Question 17.
Bauxite ore is purified by _____.
(a) Leaching process
(b) Hydraulic method
(c) Froth floatation method
(d) Magnetic separation method.
Answer:
(a) Leaching process

Question 18.
When copper reacts with dil.HNO₃ it liberates …………. gas.
(a) SO₂
(b) NO₂
(C) CO₂
(d) NO
Answer:
(d) NO

Question 19.
Which metal process low melting point?
(a) Gallium
(b) Caesium
(c) Aluminium
(d) Copper.
Answer:
(a) Gallium

Question 20.
Which one of the following is not an ore of aluminium?
(a) Bauxite
(b) Haematite
(c) Cryolite
(d) Corundum.
Answer:
(b) Haematite

Question 21.
……….. is stored in Kerosene.
(a) Iron
(b) Silver
(c) Sodium
(d) Aluminium
Answer:
(c) Sodium

Question 22.
Electrolytic reduction of alumina into aluminium is ______.
(a) Hall’s process
(b) Alumino thermic process
(c) Baeyer’s process
(d) Bessemerisation process.
Answer:
(a) Hall’s process

Question 23.
In Hall’s process, cathode used is ______.
(a) Iron tank
(b) Graphite
(c) Pure alumina
(d) Iron tank linked with graphite.
Answer:
(d) Iron tank linked with graphite.

Question 24.
The metal oxides are usually……….. in nature.
(a) basic
(b) acidic
(c) amphoteric
(d) neither acidic nor basic
Answer:
(a) basic

Question 25.
A silvery-white metal is ______.
(a) Aluminium
(b) Copper
(c) Iron
(d) Zinc.
Answer:
(a) Aluminium

Question 26.
Aluminium reacts with NaOH to give ______.
(a) Al₂O₃
(b) AlCl₃
(c) NaAlO₂
(d) Al(OH)₃
Answer:
(c) NaAlO₂

Question 27.
Atoms of the elements belonging to the same group of periodic table will have:
(a) same number of protons
(b) same number of electrons in the valence shell
(c) same number of neutrons
(d) same number of electrons
Answer:
(b) same number of electrons in the valence shell

Question 28.
Chief ore of copper is ______.
(a) CuFeS₂
(b) Cu₂O
(c) Cu₂S
(d) CuSO₄
Answer:
(a) CuFeS₂

Question 29.
Molecular formula for copper pyrites ______.
(a) Cu₂O
(b) Cu₂S
(c) CuCO₃
(d) CuFeS₂
Answer:
(d) CuFeS₂

Question 30.
The electronegativity of fluorine is:
(a) 4.0
(b) 3
(c) 2.8
(d) 2.1
Answer:
(a) 4.0

Question 31.
The second most abundant metal available next to aluminium is ______.
(a) Cu
(b) Ag
(c) Au
(d) Fe.
Answer:
(d) Fe.

Question 32.
Most important ore of iron is ______.
(a) Haematite
(b) Magnetite
(c) Iron pyrite
(d) Cryolite.
Answer:
(a) Haematite

Question 33.
The volatile impurities present in haematite are:
(a) Carbon and Nitrogen
(b) Oxygen and Nitrogen
(c) Helium and Oxygen
(d) Sulphur and Phosphorus
Answer:
(d) Sulphur and Phosphorus

Question 34.
The solute present in brass is:
(a) copper
(b) zinc
(c) tin
(d) magnesium
Answer:
(b) zinc

Question 35.
Which one of the following is used for making pressure cookers?
(a) Brass
(b) Magnalium
(c) Duralumin
(d) Nickel steel
Answer:
(c) Duralumin

Question 36.
Which is used as propeller?
(a) Stainless steel
(b) Nickel steel
(c) Brass
(d) Magnalium.
Answer:
(b) Nickel steel

Question 37.
Gold does not occur in the combined form. It does not react with air or water. It is in the ______ state.
(a) native
(b) combined
(c) complex
(d) molten.
Answer:
(a) native

Question 38.
Which of the following metal is not found in a free state?
(a) Ag
(b) Au
(c) Pt
(d) Al.
Answer:
(d) Al.

Question 39.
Which one of the following does not react with copper?
(a) Oxygen
(b) Conc.H₂SO₄
(c) NaOH
(d) Conc.HNO₃
Answer:
(c) NaOH

Question 40.
An element which is an essential constituent of all organic compounds belongs to ______ group.
(a) 14th
(b) 15th
(c) 16th
(d) 17th.
Answer:
(a) 14th

Question 41.
The highest ionization energy is exhibited by ______.
(a) Halogens
(b) Alkaline earth metals
(c) Transition metals
(d) Nobel gases.
Answer:
(d) Nobel gases.

Question 42.
Which two elements of the following belongs to the same period? (Al, Si, Ba, O).
(a) Si, Ba
(b) Al, Ba
(c) Al, Si
(d) Al, O.
Answer:
(c) Al, Si

Question 43.
98% pure copper and 2% impurities is called ______.
(a) Matte
(b) Copper pyrites
(c) blister copper
(d) cuprite.
Answer:
(c) blister copper

Question 44.
_____ is used in making anchors and electromagnets.
(a) Steel
(b) Pig iron
(c) Cast iron
(d) Wrought iron.
Answer:
(d) Wrought iron.

Question 45.
Which reagent does not react with iron?
(a) Conc. HNO₃
(b) Conc.H₂SO₄
(c) Steam
(d) Dil. HNO₃
Answer:
(a) Conc. HNO₃

II. Fill in the blanks.

Question 1.
Matte is a mixture of ______.
Answer:
Cu₂S + FeS.

Question 2.
Second group elements are called ______.
Answer:
Alkaline earth metals.

Question 3.
Ionisation energy is measured in _____ unit.
Answer:
kJ/mol

Question 4.
The ionisation energy ______ along the period.
Answer:
Increases.

Question 5.
______ property, which predicts the nature of bonding between the atoms in a molecule.
Answer:
Electronegativity.

Question 6.
If the difference in electronegativity between two elements is 1.7, the bond has ____ and _____.
Answer:
50% ionic character, 50% covalent character.

Question 7.
If the difference in electronegativity between two elements is less than 1.7, the bond is considered to be ______.
Answer:
Covalent.

Question 8.
If the difference in electronegativity between two elements is greater than 1.7, the bond is considered to be ______.
Answer:
Ionic.

Question 9.
The process of extracting the ores from the earth’s crust is called ______.
Answer:
Mining.

Question 10.
______ is the main principle behind in Hydraulic method.
Answer:
Specific gravity.

Question 11.
Froth floatation process is preferable for ______ ores.
Answer:
Lighter
(or)
Sulphide.

Question 12.
On heating in air, iron forms ______.
Answer:
Fe₃O₄

Question 13.
Iron reacts with Chlorine to form _____ compound.
Answer:
Ferric chloride.

Question 14.
The corrosive action in the absence of moisture is called ______.
Answer:
Dry corrosion.

Question 15.
______ technique used to renovate the Pamban bridge.
Answer:
Protective coating.

Question 16.
The atomic number is the number of ____ in the nucleus or number of ______ revolving around the nucleus in an atom.
Answer:
Protons, electrons.

Question 17.
The long form of periodic table is based upon the _____ of elements.
Answer:
Electronic configuration.

Question 18.
In the periodic table, the horizontal rows are called _____ and vertical columns are called ______.
Answer:
Periods, groups.

Question 19.
The modem periodic table has been divided into ______ blocks known as _______ blocks.
Answer:
Four, s, p, d, f

Question 20.
The ______ of the elements in a period decreases from left to right and the atomic radii of the elements present in a group downwards.
Answer:
Atomic size, increases.

Question 21.
_______ period is the longest period and it contains ______ elements.
Answer:
Seventh, 32.

Question 22.
In the periodic table, there are ______ groups and _____ periods.
Answer:
18, 7.

Question 23.
Metals like Ti, Cr, Mn, Zr find their application in the manufacturing of defence equipment called ______.
Answer:
Strategic metals.

Question 24.
The metal ______ plays a vital role in nuclear reactions releasing nuclear energy and used in nuclear weapons.
Answer:
Uranium.

Question 25.
Copper, silver and gold are called _______ as they are used in making ____ and ______.
Answer:
Coinage metals, coins and jewellery.

Question 26.
Purity of gold is expressed in ______ and ______ is pure gold.
Answer:
Carats, 24 – carat gold.

Question 27.
_______ is an ore of aluminium and _____ is its mineral.
Answer:
Bauxite, clay.

Question 28.
All ______ cannot be called as ores but all ______ are minerals.
Answer:
Minerals, ores.

Question 29.
The process of extracting the ores from the earth’s crust is called ______.
Answer:
Mining.

Question 30.
The rocky impurity associated with the ore is called ____ or ______.
Answer:
Gangue, matrix.

Question 31.
_____ is a substance added to the ore to reduce the fusion temperature and to remove impurities.
Answer:
Flux.

Question 32.
____ is the process of reducing the roasted metallic oxide to metal.
Answer:
Smelting.

Question 33.
Slag is the fusible product formed when ____ reacts with ____ during the extraction of metals.
Answer:
Flux, gangue

Question 34.
The temperature applied in Hall’s process is _____ and the voltage used in ______.
Answer:
900 – 950°C, 5 – 6V

Question 35.
_____ is used in making manhole covers and drain pipes and _______ is used in making transmission cables and T.V. towers.
Answer:
Pig iron, steel.

Question 36.
______ is defined as the slow and steady destruction of a metal by the environment.
Answer:
Corrosion.

Question 37.
_______ is a process of coating zinc on iron sheets by using electric current.
Answer:
Galvanization.

III. Match the following.

Question 1.

i. Boron family	(a) Group 17
ii. Carbon family	(b) Group 16
iii. Nitrogen family	(c) Group 13
iv. Chalcogen family	(d) Group 15
v. Halogen family	(e) Group 14

Answer:
i – c, ii – e, iii – d, iv – b , v – a.

Question 2.

i. Alkali metals	(a) Lanthanides & Actinides
ii. Alkaline earth metals	(b) Groups 3 – 12
iii. Transition elements	(c) Group 2
iv. Inner transition elements	(d) Group 1

Answer:
i – d, ii – c, iii – b, iv – a.

Question 3.

i. Group 18	(a) Main group elements
ii. Group 3 – 12	(b) Noble gases
iii. Group 13 – 18	(c) Halogens
iv. Group 17	(d) Transition elements

Answer:
i – b, ii – d, iii – a, iv – c.

Question 4.

i. Copper	(a) Lustrous greyish white metal
ii. Iron	(b) Cn 112
iii. Aluminium	(c) Reddish-brown metal
iv. Copernicium	(d) Silvery white metal

Answer:
i – c, ii – a, iii – d, iv – b.

Question 5.

Alloys	Composition
i. Brass	(a) Al, Mg, Mn, Cu
ii. Duralumin	(b) Cu, Zn
iii. Bronze	(c) Al, Mg
iv. Magnalium	(d) Cu, Sn

Answer:
i – b, ii – a, iii – d, iv – c.

Question 6.

Elements	Electronegative value
i. F	(a) 2.5
ii. Cl	(b) 2.8
iii. Br	(c) 3.0
iv. I	(d) 4.0

Answer:
i – d, ii – c, iii – b, iv – a.

Question 7.

Process	Ores
i. Hydraulic process	(a) ZnS
ii. Magnetic separation	(b) Fe2O3
iii. Froth floatation process	(c) SnO2
iv. Leaching process	(d) Al2O3. 2H2O

Answer:
i – b, ii – c, iii – a, iv – d.

Question 8.

i. Cuprite	(a) Halide ore
ii. Marble	(b) Oxide ore
iii. Fluorspar	(c) Sulphide ore
iv. Galena	(d) Carbonate ore

Answer:
i – b, ii – d, iii – a, iv – c.

Question 9.

Alloy	Metals present	Uses
1. Brass	Fe, C, Ni	Statues, Coins
2. Bronze	Al, Mg, Mn, Cu	Aircraft, Pressure cookers
3. Duralumin	Fe, C, Ni, Cr	Cables, Propeller
4. Magnalium	Cu, Sn	Automobile parts, Utensils
5. Stainless steel	Cu, Zn	Scientific instruments, Air craft
6. Nickel steel	Al, Mg	Medals, decorative items

Answer:

Alloy	Metals present	Uses
1. Brass	Cu, Zn	Medals, decorative items
2. Bronze	Cu, Sn	Statues, Coins
3. Duralumin	Al, Mg, Mn, Cu	Aircraft, Pressure cookers
4. Magnalium	Al, Mg	Scientific instruments, Air craft
5. Stainless steel	Fe, C, Ni, Cr	Automobile parts, Utensils
6. Nickel steel	Fe, C, Ni	Cables, Propeller

Question 10.

Metal	Ore	Chemical formula
1. Copper	Bauxite	ZnCO3
2. Aluminium	Haematite	CuFeS2
3. Iron	Copper pyrite	Fe2O3
4. Zinc	Calamine	Al2O3.2H2O

Answer:

Metal	Ore	Chemical formula
1. Copper	Copper pyrite	CuFeS2
2. Aluminium	Bauxite	Al2O3.2H2O
3. Iron	Haematite	Fe2O3
4. Zinc	Calamine	ZnCO3

IV. State whether true or false. If false, give the correct statement.

Question 1.
First period contains only one element.
Answer:
False.
Correct Statement: First period contains two elements.(Hydrogen & Ftelium)

Question 2.
The valency of all alkali metals is one.
Answer:
True.

Question 3.
Noble gases are more reactive.
Answer:
False.
Correct Statement: Noble gases are less reactive.

Question 4.
The atomic radius decreases from Li to B?
Answer:
True.

Question 5.
As the positive charge increases, the size of the cation also increases.
Answer:
False.
Correct Statement: As the positive charge increases, the size of the cation also decreases.

Question 6.
Copper pyrite ore is concentrated by gravity separation method.
False
Correct Statement: Copper pyrite ore is concentrated by froth floatation process.

Question 7.
Aluminium alloyed with gold and silver for making coins and jewels.
Answer:
False.
Correct Statement: Copper alloyed with gold and silver for making coins and jewels.

Question 8.
The corrosive action in the presence of moisture is called wet corrosion.
Answer:
True

Question 9.
The physical and chemical properties of elements are the periodic function of their atomic numbers – modern periodic law.
Answer:
True.

Question 10.
The long form of periodic table consists of horizontal rows called groups and vertical columns called periods.
Answer:
False.
Correct statement: The long form of periodic table consists of horizontal rows called periods and vertical columns called groups.

Question 11.
The first period in the periodic table is the shortest period and contains 8 elements from Lithium to Neon.
Answer:
False.
Correct statement: The first period in the periodic table is the shortest period and contains 2 elements Hydrogen and Helium.
(OR)
The second period in the periodic table is the short period and contains 8 elements from Lithium to Neon.

Question 12.
The sixth period in the periodic table is the longest period and contains 32 elements.
Answer:
True.

Question 13.
Group 1, 2 and 13 – 18 are called normal elements.
(or)
Main group elements.
(or)
Representative elements.
Answer:
True.

Question 14.
The atomic size of the elements in a period increases from left to right.
Answer:
False.
Correct statement: The atomic size of the elements in a period decreases from left to right.

Question 15.
In a period, the metallic character of the element increases while their non-metallic character decreases.
Answer:
False.
Correct statement: In a period, the metallic character of the element decreases while their non-metallic character increases.

Question 16.
The last element authenticated by IUPAC is Cn 112 [Copemicium].
Answer:
True.

Question 17.
Silver was the first metal to be used in making utensils and weapons.
Answer:
False.
Correct statement: Copper was the first metal to be used in making utensils and weapons.

Question 18.
The strategic metals such as copper, silver and gold are used in the manufacturing of defence equipment.
Answer:
False.
Correct statement: The strategic metals such as titanium, chromium manganese, zirconium are used in the manufacturing of defence equipment.

Question 19.
Copper, silver and gold are called coinage metals.
Answer:
True.

Question 20.
For making ornaments, 24 – carat gold is used which is pure gold.
Answer:
False.
Correct statement: For making ornaments, 22 – carat gold is used which contains 22 parts of gold by weight and 2 parts of copper by weight.

Question 21.
The mineral from which a metal can be readily and economically extracted on a large scale is said to be ore.
Answer:
True.

Question 22.
The rocky impurity associated with the ore is called flux.
Answer:
False.
Correct statement: The rocky impurity associated with the ore is called gangue or matrix.

Question 23.
Slag is the fusible product formed when flux reacts with gangue during the extraction of metals.
Answer:
True.

Question 24.
Metals which have high chemical reactivity are found in a free state or in the native state.
Answer:
False.
Correct statement: Metals which have low chemical reactivity are found in a free state or in the native state.

Question 25.
Aluminium is the metal found most abundantly in the earth’s crust.
Answer:
True

Question 26.
Aluminium is a reddish-brown metal and it is a bad conductor of heat and electricity.
Answer:
False.
Correct statement: Aluminium is a silvery – white metal and it is a good conductor of heat and electricity.

Question 27.
Aluminium reacts with strong caustic alkalis forming aluminates.
Answer:
True

Question 28.
Conc. Nitric acid renders aluminium active due to the formation of nitride film on its surface.
Answer:
False.
Correct statement: Cone. Nitric acid renders aluminium passive due to the formation of oxide film on its surface.

Question 29.
Aluminium is a powerful reducing agent.
Answer:
True.

Question 30.
Duralumin alloy is light, having high tensile strength and corrosion-resistant.
Answer:
True.

Question 31.
Fe and Al₂O₃ are used in thermite welding.
Answer:
False.
Correct statement: Al powder and Fe₂O₃ is used in thermite welding.

Question 32.
The chief ore of copper is Ruby copper.
Answer:
False.
Correct statement: The chief ore of copper is copper pyrite.

Question 33.
Iron is a lustrous greyish white metal and can be magnetised.
Answer:
True.

Question 34.
The rust has the chemical formula as Fe₃O₄.
Answer:
False.
Correct statement: The rust has the chemical formula as Fe₂O₃. xH₂O.

V. Assertion and Reason.

Question 1.
Assertion (A): Nobel gas is unreactive.
Reason (R): They have unstable electronic configuration in their valence shells.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are not correct
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(c) (A) is correct but (R) is wrong

Question 2.
Assertion (A): The nature of bond in NaI molecule is covalent.
Reason (R): The electronegativity difference between Na and I is 1.5
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are not correct
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 3.
Assertion (A): Haematite ore was purified by Hydraulic method.
Reason (R): Haematite is oxide ore.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 4.
Assertion (A): Corundum is a chief ore of aluminium.
Reason (R): Molecular formula of Corundum is Al₂O₃
(a) (A) and (R) are correct, (R) explains the (A)
(b) (A) is correct, (R) is wrong
(c) (A) is wrong, (R) is correct
(d) (A) and (R) are correct, (R) doesn’t explain (A).
Answer:
(c) (A) is wrong, (R) is correct

Question 5.
Assertion (A): The chemical properties of the elements in the same period are not similar.
Reason (R): As the electronic configuration changes across the period, the chemical properties of the elements are not similar.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are not correct
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 6.
Assertion (A): Copper, Silver and Gold are used in making coins and jewellery. So they are called coinage metals.
Reason (R): These metals release an enormous amount of nuclear energy.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(c) (A) is correct but (R) is wrong

Question 7.
Assertion (A): Metals like Titanium, Chromium, Manganese and Zirconium are called strategic metals!
Reason (R): They find their applications in the manufacturing of defence equipment.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 8.
Assertion (A): Gold, Silver and Platinum are the metals that are found in a free state.
Reason (R): Those metals have low chemical reactivity and are found in a free state or in the native state.
(a) Both (A) and (R) are wrong
(b) Both (A) and (R) are correct
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct

Question 9.
Assertion (A): Aluminium occurs in the combined state.
Reason (R): It is a reactive metal and so it occurs in combined state.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 10.
Assertion (A): In the aluminothermic process, Iron oxide is reduced to iron by igniting with Aluminium powder.
Reason (R): Aluminium is a powerful reducing agent.
(a) Both (A) and (R) are wrong
(b) (A) is correct but (R) is wrong
(c) (A) is wrong but (R) is correct
(d) Both (A) and (R) are correct.
Answer:
(d) Both (A) and (R) are correct

Question 11.
Assertion (A): When iron is dipped in conc.HNO₃ it becomes chemically inert (or) passive.
Reason (R): Iron becomes passive when treated with nitric acid is due to the formation of a layer of iron oxide Fe₃O₄ on its surface.
(a) Both (A) and (R) are correct
(b) (A) is correct but (R) is wrong
(c) (A) is wrong but (R) is correct
(d) Both (A) and (R) are wrong.
Answer:
(d) Both (A) and (R) are wrong

Question 12.
Assertion (A): Duralumin is used in making aircraft, tools and pressure cookers.
Reason (R): Duralumin is an alloy that is light, strong, resistant to corrosion.
(a) Both (A) and (R) are wrong
(b) Both (A) and (R) are correct
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct

Question 13.
Assertion (A): Nickel steel is used in making cables, aircraft parts and propeller.
Reason (R): Nickel steel alloy is hard, brittle and polishable.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(c) (A) is correct but (R) is wrong

Question 14.
Assertion (A): Magnesium is used in a sacrificial protection method to prevent corrosion.
Reason (R): Magnesium is more reactive than iron. When it is coated on the articles made of steel, it sacrifices itself to protect steel.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 15.
Assertion (A): Electroplating method not only protects but also enhances the metallic appearance.
Reason (R): Electroplating is a method of coating one metal with another by passing current.
(a) (A) is right, (R) is wrong
(b) (A) is right, (R) is not relevant
(c) (A) is right, (R) is relevant
(d) Both (A) and (R) are wrong.
Answer:
(c) (A) is right, (R) is relevant

VI. Short Answer Questions.

Question 1.
State modern periodic law.
Answer:
Modem periodic law states that the physical and chemical properties of elements are the periodic function of their atomic numbers.

Question 2.
Write the flow chart of the long form of the periodic table.
Answer:

Question 3.
The distance between the adjacent copper atoms in solid copper is 2.56 Å. Calculate what is the metallic radius of Cu?
Answer:
Metallic radius of copper = \(\frac{2.56}{2}\) = 1.28 Å

Question 4.
Briefly write any four characteristics of a group in the periodic table.
Answer:

• The elements present in a group have the same valency.
• The elements present in a group have identical chemical properties.
• The physical properties of the elements in the group very gradually.
• The atomic radii of die elements present in a group increase downwards.

Question 5.
What is the principle behind Froth floatation?
Answer:
This process depends on the preferential wettability of the ore with oil (pine oil) and the gangue particles by water. Lighter ores, such as sulphide ores, are concentrated by this method. Eg: Zinc blende (ZnS).

Question 6.
What are minerals?
Answer:
A mineral may be a single compound or a complex mixture of various compounds of metals found in the earth, e.g. Clay Al₂O₃. 2SiO₂. 2H₂O is the mineral of Aluminium.

Question 7.
Write the increasing order of radii of the following species.
(a) Na, Na⁺, Cl, Cl^–
(b) Li, Na, K, Rb
Answer:
(a) Na⁺, Cl, Na, Cl^–
(b) Li, Na, K, Rb

Question 8.
Differentiate ore and mineral.
Answer:

Ore	Mineral
1. Ores contain a large percentage of metal.	1. Minerals contain a low percentage of metal.
2. Ores can be used for die extraction of metals on a large scale readily and economically.	2. Metals cannot be extracted easily from minerals.
3. Bauxite Al2O3. 2H2O is the ore of aluminium.	3. Clay Al2O3.2SiO.2H2O is the mineral of aluminium.

Question 9.
Define metallurgy.
Answer:
The various steps involved in the extraction of metals from their ores as well as refining of crude metals are collectively known as metallurgy.

Question 10.
Elements a, b, c and d have the following electronic configurations
(a) ls² 2s² 2p⁶
(b) Is², 2s², 2p⁶, 3s², 3p¹
(c) Is², 2s², 2p⁶, 3s², 3p⁶
(d) Is², 2s², Ip¹
Answer:
(a) and (c), (b) and (d). Because the number of valence electrons are the same.

Question 11.
What is slag? Give an example.
Answer:
Slag is a fusible product formed when flux reacts with gangue during the extraction of metals.
Flux + gangue → slag
CaO + SiO₂ → CaSiO₃

Question 12.
Why the electron affinities of noble gases are zero?
Answer:
Noble gases show no tendency to accept electrons because the outer 5 and p orbitals of noble gases are completely filled. No more electrons can be added to them and hence their electron affinities are zero.

Question 13.
Why does gold, silver and platinum occur in free state?
Answer:
Gold, silver and platinum have low chemical reactivity and so they are found in the free state or in a native state.

Question 14.
How does Aluminium reacts with air?
Answer:
Reaction with air: It is not affected by dry air. On heating at 800°C, aluminium bums very brightly forming it’s oxide and nitride.
4Al + 3O₂ → Al₂O₃(Aluminium oxide)
2Al + N₂ → 2 AlN (Aluminium nitride)

Question 15.
How does Aluminium react with caustic soda? Give an equation.
Answer:
Aluminium reacts with caustic soda to give sodium meta aluminate with the liberation of H₂ gas.

Question 16.
Prove that aluminium is a powerful reducing agent.
Answer:
Aluminium is a powerful reducing agent When a mixture of aluminium powder and iron oxide is ignited, iron oxide is reduced to iron. This process is known as the aluminothermic process.

Question 17.
Write a note on Aluminothermic process.
Answer:
As reducing agent: Aluminium is a powerful reducing agent. When a mixture of aluminium powder and iron oxide is ignited, the latter is reduced to metal. This process is known as aluminothermic process.
Fe₂O₃ + 2Al → 2Fe + Al₂O₃ + heat

Question 18.
What is the action of heat on copper?
Answer:
On heating at different temperatures in the presence of oxygen, copper forms two types of oxides CuO, Cu₂O.

Question 19.
Explain the action of dilute nitric acid with copper.
Answer:
Copper reacts with dil.HNO₃ with the liberation of Nitric oxide gas.

Question 20.
What happens when copper is treated with conc.HNO₃ and with conc.H₂SO₄?
Answer:

Question 21.
What do you mean by Ferrous and Non-Ferrous alloys? Given an example for each.
Answer:
Ferrous alloys : Contain iron as a major component.
Eg: Stainless steel, Nickel steel

Non-Ferrous alloys : These alloys do not contain iron as a major part.
Eg: Aluminium alloy, Copper alloy.

Question 22.
Explain the action of air with iron.
Answer:
3Fe + 2O₂ → Fe₃O₄ [Magnetic oxide (Black)].

Question 23.
What are amalgams? How are they prepared?
Answer:
An amalgam is an alloy of mercury with another metal.
Amalgams are formed by the metallic bonding with the electrostatic force of attraction between the electrons and the positively charged metal ions.

Question 24.
Explain the action of steam with iron.
Answer:

Question 25.
Mention the types of iron on the basis of carbon content.
Answer:

Pig iron	Iron with 2 – 4.5% carbon
Wrought iron	Iron with < 0.25% carbon
Steel	Iron with 0.25 – 2% carbon

Question 26.
What is dry corrosion or chemical corrosion?
Answer:
The corrosive action in the absence of moisture is called dry corrosion. It is the process of a chemical attack on a metal by corrosive liquids or gases such as O₂, N₂, SO₂, H₂S etc… It occurs at high temperature, of all the gases mentioned above O₂ is the most reactive gas to impart the chemical attack.

Question 27.
What is an amalgam? Give one example with its use.
Answer:

• An amalgam is an alloy of mercury with metals such as sodium, gold and silver.
• Dental amalgam is an alloy of mercury with silver and tin and it is used in the dental filling.

Question 28.
Explain sacrificial protection.
Answer:
Magnesium is more reactive than iron. When it is coated on the articles made of steel, it sacrifices itself to protect steel.

Question 29.
Define corrosion.
Answer:
Corrosion is defined as the slow and steady destruction of a metal by the environment. It results in the deterioration of the metal to form metal compounds by means of chemical reactions with the environment.

Question 30.
What is the action of Air and moisture on copper?
Answer:
Copper gets covered with a green layer of basic carbonate in the presence of CO₂ and moisture.
2 Cu + O₂ + CO₂ + H₂O → CUCO₃.CU(OH)₂

Question 31.
Give any two uses of aluminium.
Answer:

• Aluminium metal is a corrosion – resistant and a good conductor of heat. So it is used in making utensils.
• Aluminium is used in welding as thermite and a very good reducing agent.

Question 32.
What is the modern periodic table?
Answer:
The modem periodic table is a tabular arrangement of elements in rows and columns, highlighting the regular repetition of properties of the elements.

Question 33.
Explain the smelting process of iron in the Blast furnace.
Answer:
The charge consisting of roasted ore, coke and limestone in the ratio 8 : 4 : 1 is smelted in a blast furnace by introducing it through the cup and cone arrangement at the top.

Question 34.
What are the periodic properties?
Answer:
Properties such as atomic radius, ionic radius, ionisation energy, electronegativity, electron affinity, show a regular periodicity and hence they are called periodic properties.

Question 35.
Define Atomic radius.
Answer:
The atomic radius of an atom is defined as the distance between the centre of its nucleus and the outermost shell containing the valence electron.

Question 36.
Write the action of dil.HCl and dil.H₂SO₄ with iron.
Answer:
(i) Fe + 2HCl → FeCl₂ + H₂↑
(ii) Fe + H₂SO₄ → FeSO₄ + H₂↑

Question 37.
What is a covalent radius?
Answer:
It is defined as half the distance between the nuclei of two covalently bonded atoms of the same element in a molecule.

Question 38.
Define Ionisation energy.
Answer:
Ionisation energy is the minimum energy required to remove an electron from a gaseous atom in its ground state to form a cation.

Question 39.
What is an alloy?
Answer:
An alloy is a homogeneous mixture of two or more metals or of one or more metals with certain non-metallic elements. The properties of alloys are often different from the component metals.

Question 40.
What is Electronegativity?
Answer:
Electronegativity of an element is the measure of the tendency of its atom to attract the shared pair of electrons towards itself in a covalent bond.

Question 41.
Write the steps involved in metallurgical process.
Answer:
A metallurgical process involves three main steps as follows:

• Concentration or Separation of the ore: It is the process of removal of impurities from the ore.
• Production of the metal: It is the conversion of the ore into metal.
• Refining of the metal: It is the process of purification of the metal.

Question 42.
Give a single term for each of the following:
(i) The process of extracting the ores from the Earth’s crust is called:
Answer:
Mining

(ii) The rocky impurities associated with an ore is called:
Answer:
Gangue or matrix

(iii) The substance added to the ore to reduce fusion temperature:
Answer:
Flux

(iv) Noble metals occur in this state:
Answer:
Native

Question 43.
How will you convert copper into copper carbonate?
Answer:
Copper reacts with oxygen in the presence of CO₂ and moisture to give copper carbonate.
2Cu + O₂ + CO₂ + H₂O → CuCO₃. Cu(OH)₂ (Copper Carbonate).

Question 44.
Mention the uses of iron.
Answer:
Uses of iron:

1. Pig iron (Iron with 2 – 4.5 % of carbon): It is used in making pipes, stoves, radiators, railings, manhole covers and drain pipes.
2. Steel (Iron with < 0.25 % of carbon): It is used in the construction of buildings, machinery, transmission cables and T.V towers and in making alloys.
3. Wrought iron (Iron with 0.25 – 2 % of wrought carbon): It is used in making springs, anchors and electromagnets.

Question 45.
What are the chemical properties of metals in terms of

1. valence electrons
2. Atomicity.

Answer:

1. Valence electrons: Metals usually have 1, 2 or 3 electrons in their outermost shell.
2. Atomicity: Metals are usually monoatomic in their vapour state.

Question 46.
Why alloys are said to solid solutions?
Answer:
Alloys can be considered solid solutions in which the metal with high concentration is solvent and other metals are solute.
Example: brass is a solid solution of zinc (solute) in copper (solvent).

Question 47.
Write a note on Dry corrosion.
Answer:

• The corrosive action in the absence of moisture is called dry corrosion.
• It is the process of a chemical attack on a metal by corrosive liquids or gases such as O₂, N₂, SO₂, H₂S etc. in which O₂ is more reactive.
• It occurs at high temperature

Question 48.
Explain Wet Corrosion.
Answer:

• The corrosive action in the presence of moisture is called wet corrosion.
• It occurs as a result of the electrochemical reaction of metal with water or an aqueous solution of salt or acids or bases.

Question 49.
What is electroplating?
Answer:
Electroplating is a method of coating one metal over another metal by passing an electric current.

VII. Long Answer Questions.

Question 1.
Explain the variation of ionisation energy along the group and period.
Answer:

• As the atomic size decreases from left to right in a period, more energy is required to remove the electrons. So, the ionisation energy increases throughout the period.
• Down the group, the atomic size increases and hence the valence electrons are loosely bound. They require relatively less energy for the removal. Thus, ionisation energy decreases down the group in the periodic table.

Question 2.
Explain the electrolytic refining of copper.
Answer:
(i) Blister copper contains 98% of pure Cu and 2 % of impurities and is purified by electrolytic refining.
For electrolytic refining of Cu we use:
Cathode: A thin plate of pure Cu.
Anode: A block of impure Cu
Electrolyte: CuSO₄ + dil H₂SO₄

(ii) When electric current is passed through the electrolytic solution, Pure Cu gets deposited at the cathode and the impurities settle at the bottom of the anode in the form of anode mud.

Question 3.
Explain Gravity separation method.
Answer:
Gravity Separation (or) Hydraulic method:
1. Principle: The difference in the densities or specific gravities of the ore and the gangue is the main principle behind this method. Oxide ores are purified by this method,
e.g., Haematite Fe₂O₃ the ore of iron.

2. Method: The ore is poured over a sloping, vibrating corrugated table with grooves and a jet of water is allowed to flow over it. The denser ore particles settle down in the grooves and lighter gangue particles are washed down by the water.

Question 4.
Discuss the magnetic separation methods.
Answer:
Magnetic separation method:
Principle: The magnetic properties of the ores from the basis of separation. When either the ore or the gangue is magnetic, this method is employed, e.g., Tinstone SnO₂, the ore of tin.

Method: The crushed ore is placed over a conveyer belt which rotates around two metal wheels, one of which is magnetic. The magnetic particles are attracted to the magnetic wheel and fall separately apart from the nonmagnetic particles.

Question 5.
Explain the froth floatation process.
Answer:
Froth floatation Process:
Principle: This process depends on the preferential wettability of the ore with oil (pine oil) and the gangue particles by water. Lighter ores, such as sulphide ores, are concentrated by this method, e.g., Zinc blende (ZnS).

Method: The crushed ore is taken in a large tank containing oil and water and agitated with a current of compressed air. The ore is wetted by the oil and gets separated from the gangue in the form of froth. Since the ore is lighter, it comes on the surface with the froth and the impurities are left behind, e.g., Zinc blende (ZnS).

Question 6.
How will you extract aluminium from its ore?
Answer:
The extraction of aluminium from bauxite involves two steps:
(i) Conversion of bauxite into alumina – Baeyer’s Process
The conversion of Bauxite into Alumina involves the following steps:
Bauxite ore is finely ground and heated under pressure with a solution of concentrated caustic soda solution at 150° C to obtain sodium metal aluminate.
On diluting sodium meta aluminate with water, a precipitate of aluminium hydroxide is formed.
The precipitate is filtered, washed, dried and ignited at 1000°C to get alumina.
\(2 \mathrm{Al}(\mathrm{OH})_{3} \stackrel{1000^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{Al}_{2} \mathrm{O}_{3}+3 \mathrm{H}_{2} \mathrm{O}\)

(ii) Electrolytic reduction of alumina – Hall’s Process
Aluminium is produced by the electrolytic reduction of fused alumina (Al₂O₃) in the electrolytic cell.

Cathode: Iron tank linked with graphite
Anode: A bunch of graphite rods suspended in a molten electrolyte.
Electrolyte: Pure alumina + molten cryolite + fluorspar (fluorspar lowers the fusion temperature of electrolyte)
Temperature: 900 – 950°C
The voltage used: 5 – 6 V
Aluminium is deposited at the cathode and oxygen gas is liberated at the anode. Oxygen combines with graphite to form CO₂.

Question 7.
Explain the extraction of copper from copper pyrites.
Answer:
Extraction of copper from copper pyrites involves the following steps:
(i) The concentration of ore: The ore is crushed and then concentrated by froth floatation process.

(ii) Roasting: The concentrated ore is roasted in excess of air. During the process of roasting, the moisture and volatile impurities are removed. Sulphur, phosphorus, arsenic and antimony are removed as oxides. Copper pyrite is partly converted into sulphides of copper and iron.
\(2 \mathrm{CuFeS}_{2}+\mathrm{O}_{2} \rightarrow \mathrm{Cu}_{2} \mathrm{S}+2 \mathrm{FeS}+\mathrm{SO}_{2} \uparrow\)

(iii) Smelting: The roasted ore is mixed with powdered coke and sand and is heated in a blast furnace to obtain matte (Cu₂S + FeS) and slag. The slag is removed as waste.

(iv) Bessemerisation: The molten matte is transferred to the Bessemer converter in order to obtain blister copper. Ferrous sulphide from matte is oxidized to ferrous oxide, which is removed as slag using silica.

(v) Refining: Blister copper contains 98% of pure copper and 2% of impurities and is purified A by electrolytic refining. This method is used to get metal of a high degree of purity. For electrolytic refining of copper, we use:
Cathode: A thin plate of pure copper metal.
Anode: A block of impure copper metal.
Electrolyte: Copper sulphate solution acidified with sulphuric acid.
When an electric current is passed through the electrolytic solution, pure copper gets deposited at the cathode and the impurities settle at the bottom of the anode in the form of sludge called anode mud.

Question 8.
Explain the metallurgy of iron.
Answer:
Iron is chiefly extracted from haematite ore (Fe₂O₃):
(i) Concentration by Gravity Separation: The powdered ore is washed with steam of water. As a result, the lighter sand particles and other impurities are washed away and the heavier ore particles settle down.

(ii) Roasting and Calcination: The concentrated ore is strongly heated in a limited supply of air in a reverberatory furnace. As a result, moisture is driven out and sulphur, arsenic and phosphorus impurities are oxidized off.

(iii) Smelting (in a Blast Furnace): The charge consisting of roasted ore, coke and limestone in the ratio 8 : 4 : 1 is smelted in a blast furnace by introducing it through the cup and cone arrangement at the top. There are three important regions in the furnace.

a. The Lower Region (Combustion Zone):
The temperature is at 1500°C. In this region, coke bums with oxygen to form CO₂ when the charge comes in contact with a hot blast of air.

It is an exothermic reaction since heat is liberated.

b. The Middle Region (Fusion Zone):
The temperature prevails at 1000°C. In this region, CO₂ is reduced to CO.

Limestone decomposes to calcium oxide and CO₂

These two reactions are endothermic due to absorption of heat. Calcium oxide combines with silica to form calcium silicate slag.
CaO + SiO₂ → CaSiO₃

c. The Upper Region (Reduction Zone): The temperature prevails at 400°C. In this region carbon monoxide reduces ferric oxide to form a fairly pure spongy iron.
\(\mathrm{Fe}_{2} \mathrm{O}_{3}+3 \mathrm{CO} \stackrel{400^{\circ} \mathrm{C}}{\longrightarrow} 2 \mathrm{Fe}+3 \mathrm{CO}_{2}\)
The molten iron is collected at the bottom of the furnace after removing the slag.
The iron thus formed is called pig iron. It is remelted and cast into different moulds. This iron is called cast iron.

Question 9.
Explain the types of alloys.
Answer:
Based on the presence or absence of Iron, alloys can be classified into:
1. Ferrous alloys: Contain Iron as a major component.
A few examples of ferrous alloys are Stainless Steel, Nickel Steel etc.

2. Non – ferrous alloys: These alloys do not contain Iron as a major component.
For example, Aluminium alloy, Copper alloy etc.
Copper Alloys (Non – ferrous):

Alloys	Uses
Brass (Cu, Zn)	Electrical fittings, medal, decorative items, hardware
Bronze (Cu, Sn)	Statues, coins, bells, gongs

Aluminium Alloys (Non – ferrous):

Alloys	Uses
Duralumin (Al, Mg, Mn, Cu)	Aircraft tools, pressure cookers
Magnalium (Al, Mg)	Aircraft, scientific instruments

Iron Alloys(Ferrous):

Alloys	Uses
Stainless steel (Fe, C, Ni, Cr)	Utensils, cutlery, automobile parts
Nickel steel (Fe, C, Ni)	Cables, aircraft parts, propeller

VIII. HOT Questions.

Question 1.
What would be the atomic number of the next

1. alkali metal
2. alkaline earth metal
3. Halogens
4. inert gas, if discovered in future.

Answer:

1. Alkali metal: 118 + 1 = 119
2. Alkaline earth metal: 120
3. Halogens: 117
4. Inert gas: 118

Question 2.
Explain the mechanism of rusting?
Answer:
Rust is chemically known as hydrated ferric oxide (it is formulated as Fe₂O₃.xH₂O).
Rusting results in the formation of scaling reddish – brown hydrated ferric oxide on the surface of iron and iron-containing materials.

Question 3.
All ores are minerals, but ait minerals are not ores. Why?
Answer:

• The elements or compounds of the metals which occur in nature in the earth’s crust are called minerals whereas ore is a mineral from which the metal is profitably extracted.
• For example, aluminium exists in the two mineral forms, that is clay and bauxite. But aluminium is mainly extracted from bauxite which contains 70 % aluminium oxide. So, bauxite is an ore of aluminium whereas clay is not ore.
• So, all ores are minerals but all minerals need not be ores.

Question 4.
Why the graphite rods acting as anode in the Hall’s process should be replaced periodically?
Answer:
During Hall’s Process, the carbon electrodes get consumed, so they have to be replaced periodically.

Question 5.
Anionic radius Is higher than the corresponding neutral atom. Give reason.
Answer:
When an atom gains one or more electrons it forms an anion. During the formation of anion, the number of orbital electrons becomes greater than the nuclear charge. Hence, the electrons are not strongly attracted by the lesser number of nuclear charges. Hence anionic radius is higher than the corresponding neutral atom.

Question 6.
A reddish – brown metal A, when exposed to moist air, forms a green layer B. When A is heated at different temperatures in the presence of O₂, it forms two types of oxides – C (black) and D (red). Identify A, B, C, D and write the balanced equation.
Answer:
(i) A reddish – brown metal A is a copper (Cu).

(ii) When copper (A) is exposed to moist air it forms a green layer (B) is copper carbonate.

(iii) When copper is heated at different temperature in the presence of oxygen, it forms two types of oxides CuO and Cu₂O. (C and D)

Question 7.
A silvery – white metal on treatment with NaOH and HCl liberates H₂ gas to form B and C respectively. The metal A will not react with acid D due to the formation of a passive film on the surface. Hence it is used for transporting acid D. Identify A, B, C, D and support your answer with balanced equations.
Answer:
(i) A silvery – white metal (A) is Aluminium (Al).
(ii) Aluminium reacts with NaOH to form B which is known as sodium meta aluminate with the liberation of H₂ gas.

(iii) Aluminium reacts with HCl to form Aluminium chloride which is known as C with the liberation of H₂ gas.

(iv) Aluminium does not react with conc. nitric acid (HNO₃) which is known as D, due to the formation of a passive film on the surface.

A	Aluminium	Al
B	Sodium meta aluminate	NaAlO2
C	Aluminium chloride	AlCl3
D	Nitric acid	HNO3

Question 8.
Metal A belongs to period 4 and group 8. A in red hot condition reacts with steam to form B. A reacts with dilute HNO₃ to give C. A again reacts with conc. H₂SO₄ to give D. Find A, B, C and D with suitable reaction.
Answer:
(i) Metal (A) belongs to period 4 and group 8 is iron (Fe).

(ii) Iron (A) reacts with steam to form magnetic oxide (B)

(iii) Iron (A) reacts with dilute HNO₃ in cold condition to give ferrous nitrate (C).

(iv) Iron (A) reacts with conc.H₂SO₄ to form Ferric Sulphate (D).

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.1

Question 1.
Find the locus of P, if for all values of a, the co-ordinates of a moving point P is
(i) (9 cos α, 9 sin α)
(ii) (9 cos α, 6 sin α)
Solution:
(i) Let P(h, k) be the moving point.
We are given h = 9 cos α and k = 9 sin α and

∴ locus of the point is x² + y² = 81

(ii) Let P(h , k) be a moving point.
We are given h = 9 cos α and k = 6 sin α

Question 2.
Find the locus of a point P that moves at a constant distance of
(i) Two units from the x-axis
(ii) Three units from the y-axis.
Solution:
(i) Let the point (x, y) be the moving point.
The equation of a line at a distance of 2 units from the x-axis is k = 2
So the locus is y = 2 (i.e.) y – 2 = 0

(ii) Equation of a line at a distance of 3 units from y-axis is h = 3
So the locus is x = 3 (i.e.) x – 3 = 0

Question 3.
If θ is a parameter, find the equation of the locus of a moving point, whose coordinates are x = a cos³ θ, y = a sin³ θ
Solution:

Question 4.
Find the value of k and b, if the points P (-3, 1) and Q (2, b) lie on the locus of x² – 5x + ky = 0.
Solution:
Given P (-3, 1) lies on the locus of
x² – 5x + ky = 0
∴ (- 3)² – 5 (-3) + k(1) = 0
9 + 15 + k = 0
⇒ k = -24
Also given Q(2 , b) lies on the locus of
x² – 5x + ky = 0
x² – 5x – 24y = 0
∴ (2)² – 5(2) – 24(b) = 0
4 – 10 – 24b = 0 ⇒ – 6 – 24b = 0
⇒ 24b = -6 ⇒ b = \(-\frac{6}{24}\) = \(-\frac{1}{4}\)
Thus k = -24, b = \(-\frac{1}{4}\)

Question 5.
A straight rod of length 8 units slides with its ends A and B always on the x and y-axis respectively. Find the locus of the midpoint of the line segment AB.
Solution:
Let P (h, k) be the moving point A (a, 0) and B (0, b) P is the midpoint of AB.

Question 6.
Find the equation of the locus of a point such that the sum of the squares of the distance from the points (3, 5), (1, -1) is equal to 20.
Solution:
Let the given points be A (3, 5) and (1, -1).
Let P (h, k ) be the point such that
PA² + PB² = 20 ………….. (1)
PA² = (3 – h)² + (5 – k)²
PB² = (1 – h)² + (- 1 – k)²
(1) ⇒ (3 – h)² + (5 – k)² + (1 – h)² + (1 + k)² = 20
9 – 6h + h² + 25 – 10k + k² + 1 – 2h + h² + 1 + 2k + k² = 20
2h² + 2k² – 8h – 8k + 36 = 20
2h² + 2k² – 8h – 8k + 16 = 0
h² + k² – 4h – 4k + 8 = 0
The locus of P ( h , k ) is obtained by replacing h by x and k by y
∴ The required locus is x² + y² – 4x – 4y + 8 = 0

Question 7.
Find the equation of the locus of the point P such that the line segment AB, joining the points A (1, -6) and B (4, -2), subtends a right angle at P.
Solution:
Let P (h, k) be the moving point

Given A (1, – 6) and B (4, – 2),
Since ∆APB = 90°, PA² + PB² = AB²
(i.e.) (h – 1)² + (k + 6)² + (h – 4)² + (k + 2)² = (4 – 1)² + (-2 + 6)²
(i.e) h² + 1 – 2h + k² + 36 + 12k + h² + 16 – 8h + k² + 4 + 4k = 3² + 4² = 25
2h² + 2k² -10h + 16k + 57 – 25 = 0
2h² + 2k² – 10h + 16k + 32 = 0
(÷ by 2)h² + k² – 5h + 8k + 16 = 0
So the locus of P is x² + y² – 5x + 8y + 16 = 0

Question 8.
If O is origin and R is a variable point on y² = 4x, then find the equation of the locus of the mid-point of the line segment OR.
Solution:
Let P(h, k) be the moving point
We are given O (0, 0). Let R = (a, b)

Substituting a, b values is y² = 4x
we get (2k)² = 4 (2h)
(i.e) 4k² = 8h
(÷ by 4) k² = 2h
So the locus of P is y² = 2x

Question 9.

Solution:

Question 10.
If P (2, -7) is a given point and Q is a point on 2x² + 9y² = 18, then find the equations of the locus of the mid-point of PQ.
Solution:
P = (2, -7); Let (h, k) be the moving point Q = (a, b)

⇒ a = 2h – 2,
b = 2k + l
Q is a point on 2x² + 9y² = 18 (i.e) (a, b) is on 2x² + 9y² = 18
⇒ 2(2h – 2)² + 9 (2k + 7)² = 18
(i.e) 2 [4h² + 4 – 8h] + 9 [4k² + 49 + 28k] – 18 = 0
(i.e) 8h² + 8 – 16h + 36k² + 441 + 252k – 18 = 0
8h² + 36k² – 16h + 252k + 431 = 0
The locus is 8x² + 36y² – 16x + 252y + 431 = 0

Question 11.
If R is any point on the x-axis and Q is any point on the y-axis and Pis a variable point on RQ with RP = b, PQ = a. then find the equation of locus of P.
Solution:
P = (x, 0), Q = (0, y), R (h, k) be a point on RQ such that PR : RQ = b : a

From the right-angled triangle OQR, OR² + OQ² = QR²

Question 12.
If the points P (6, 2) and Q (-2, 1) and R are the vertices of a ∆PQR and R is the point on the locus y = x² – 3x + 4, then find the equation of the locus of the centroid of ∆PQR.
Solution:
P (6, 2), Q (-2, 1). Let R = (a, b) be a point on y = x² – 3x + 4.

But (a, b) is a point on y = x² – 3x + 4
b = a² – 3a + 4
(i.e) 3k – 3 = (3h – 4)² – 3(3h – 4) + 4
(i.e) 3k – 3 = 9h² + 16 – 24h – 9h + 12 + 4
⇒ 9h² – 24h – 9h + 32 – 3k + 3 = 0
(i.e) 9h² – 33h – 3k + 35 = 0,
Locus of (h, k) is 9x² – 33x – 3y + 35 = 0

Question 13.
If Q is a point on the locus of x² + y² + 4x – 3y + 7 = 0 then find the equation of locus of P which divides segment OQ externally in the ratio 3 : 4, where O is origin.
Solution:
Let (h, k) be the moving point O = (0, 0);
Let PQ = (a, b) on x² + y² + 4x – 3y + 7 = 0

Question 14.
Find the points on the locus of points that are 3 units from the x-axis and 5 units from the point (5, 1).
Solution:
A line parallel to the x-axis is of the form y = k.
Here k = 3 ⇒ y = 3
A point on this line is taken as P (a, 3).
The distance of P (a, 3) from (5, 1) is given as 5 units
⇒ (a – 5)² + (3 – 1)² = 5²
a² + 25 – 10a + 9 + 1 – 6 = 25
a² – 10a + 25 + 4 – 25 = 0
a² – 10a + 4 = 0

Question 15.
The sum of the distance of a moving point from the points (4, 0) and (-4, 0) is always 10 units. Find the equation of the locus of the moving point.
Solution:
Let P (h, k) be a moving point
Here A = (4, 0) and B = (-4, 0)
Given PA + PB = 10

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.1 Additional Questions

Question 1.
If the sum of the distance of a moving point in a plane from the axis is 1, then find the locus of the point.
Solution:
Let coordinates of a moving point P be (x, y).
Given that the sum of the distances from the axis to the point is always 1.

∴ |x| + |y| = 1 ⇒ x + y = 1
⇒ -x – y = 1 ⇒ x – y = 1
Hence, these equations give us the locus of the point P which is a square.

Question 2.
A point moves so that square of its distance from the point (3, -2) is numerically equal to its distance from the line 5x -12y = 3. The equation of its locus is ……..
Solution:
The given equation of line is 5x – 12y = 3 and the given point is (3, -2).
Let (a, b) be any moving point.


⇒ 13a² + 13b² – 78a + 52b + 169 = 5a – 12b – 3
⇒ 13a² + 13b² – 83a + 64b + 172 = 0
So, the locus of the point is 13x² + 13y² – 83x + 64y + 172 = 0

Question 3.
Find the Locus of the midpoints of the portion of the line x cos θ + y sin θ = p intercepted between the axis.
Solution:
Given the equation of the line is x cos θ + y sin θ = p … (i)
Let C (h, k) be the midpoint of the given line AB where it meets the two-axis at A (a, 0) and B (0, b).
Since (a, 0) lies on eq (i) then “a cos θ + θ = p”


B (0, b) also lies on the eq (i) then 0 + b sin θ = p

Since C (h, k) is the midpoint of AB

Putting the values of a and b is eq (ii) and (iii) we get P

Squaring and adding eq (iv) and (v) we get

Question 4.

Solution:

Here, α is a variable. To find the locus of P (h, k), we have to eliminate α.
From (i), we obtain

Question 5.

Solution:

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.3

Solve the following problems by using factor theorem

Question 1.
Show that < = (x – a)² (x + 2a)
Solution:

⇒ (x + 2d) is a factor of A.
Now degree of Δ is 3 (x × x × x = x³) and we have 3 factors for A
∴ There can be a constant as a factor for A.
(i.e.,) Δ = k(x – a)² (x + 2d)
equating coefficient of x³ on either sides we get k = 1

∴ Δ = (x – a)² (x + 2a)

Question 2.
Show that
Solution:

Similarly b and c are factors of Δ.
The product of the leading diagonal elements is (b + c) (c + a) (a + b)
The degree is 3. And we got 3 factors for Δ ∴ m = 3 – 3 = 0
∴ there can be a constant k as a factor for Δ.
>

Question 3.

Solution:

⇒ x = 0, 0 are roots.
Now the degree of the leading diagonal elements is 3.
∴ the equation is of degree 3, so the roots are 0, 0, – (a + b + c)

Question 4.
Show that = (a + b + c) (a – b) (b – c) (c – a)
Solution:

⇒ (a – b) is a factor of Δ.
Similarly, (b – c) and (c – a) are factors of Δ.
The degree of the product of elements along the leading diagonal is 1 + 1 + 2 = 4 and we got 3 factors for Δ.
m = 4 – 3 = 1
∴ There can be one more factor symmetric with a, b, c which is of the form k (a + b + c).

Question 5.
Solve
Solution:

Question 6.
Show that = (x – y) (y – z) (z – x)
Solution:

⇒ (x – y) is a factor of Δ.
Similarly (y – z) and (z – x) are factors of Δ.
Now degree of Δ = 0 + 1 + 2 = 3 and we have 3 factors of Δ.
and so there can be a constant k as a factor of Δ.

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.3 Additional Problems

Question 1.
Prove that = (a – b) (b – c) (c – a) (a + b + c).
Solution:

∴ (a – b) is a factor of Δ.
Similarly, we observe that Δ is symmetric in a, b, c by putting b = c, c = a, we get Δ = 0.
Hence (b – c) and (c – a) are also factors of Δ.
∴ The product (a – b) (b – c) (c – a) is a factor of Δ. The degree of this product is 3.
The product of leading diagonal elements is 1. bc³. The degree of this product is 4.
∴ By cyclic and symmetric properties, the remaining symmetric factor of the first degree must be k (a + b + c), where k is any non-zero constant.

Question 2.
Using factor method show that = (a – b) (b – c) (c – a)
Solution:

⇒ (a – b) is a factor of Δ.
similarly, (b – c) and (c – a) are factors of Δ.
The product of leading diagonal elements is bc². The degree of the product is 1 + 2 = 3.
∴ there will be three factors for Δ.
We got 3 factors for Δ as (a – b), (b – c) and (c – a). Its degree = 3.
∴ m = 3 – 3 = 0
∴ there can be a constant k as a factor of Δ.

Question 3.

Solution:

⇒ (a – b) is a factor of A.
Similarly, (b – c) and (c – a) are factors of Δ.
The degree of Δ = 5 and degree of product of factors = 3.

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3

Question 1.
If ⁿC₁₂ = ⁿC₉ find ²¹C_n.
Solution:
ⁿC_x = ⁿC_y ⇒ x = y or x + y = n
Here ⁿC₁₂ = ⁿC₉ ⇒ 12 ≠ 9 so 12 + 9 = n (i.e) n = 21

Question 2.
If ¹⁵C_{2r – 1} = ¹⁵C_{2r + 4}, find r.
Solution:

Question 3.
If ⁿP_r = 720 and ⁿC_r = 120, find n, r.
Solution:

Question 4.
Prove that ¹⁵C₃ + 2 × ¹⁵C₄ + ¹⁵C₅ = ¹⁷C₅
Solution:

Question 5.

Solution:

Question 6.
If ^{(n + 1)}C₈ : ^{(n – 3)}P₄ = 57 : 16, find the value of n.
Solution:

Question 7.

Solution:

Question 8.
Prove that if 1 ≤ r ≤ n then n × ^{(n – 1)}C_{r – 1} = (n – r + 1)C_{r – 1}.
Solution:


(1) = (2) ⇒ LHS = RHS

Question 9.
(i) A Kabaddi coach has 14 players ready to play. How many different teams of 7 players could the coach put on the court?
Solution:
No. of players in the team = 14
We need 7 players
So selecting 7 from 14 players can be done is ¹⁴C₇ = 3432 ways

(ii) There are 15 persons in a party and if, each 2 of them shakes hands with each other, how many handshakes happen in the party?
Solution:
Total No. of persons = 15
Every two persons shake hands

(iii) How many chords can be drawn through 20 points on a circle?
Solution:
A chord is a line join of 2 points
No. of points given = 20
Selecting 2 from 20 can be done in ²⁰C₂ ways

(iv) In a parking lot one hundred, one-year-old cars are parked. Out of the five are to be chosen at random to check its pollution devices. How many different sets of five cars are possible?
Solution:
Number of cars =100
Select 5 from 100 cars can be done in ¹⁰⁰C₅ ways

(v) How many ways can a team of 3 boys, 2 girls and 1 transgender be selected from 5 boys, 4 girls and 2 transgenders?
Solution:
We have 5 boys, 4 girls, and 2 transgenders. We need 3 boys, 2 girls and 1 transgender The selection can be done as follows Selecting 3 boys from 5 boys can be done in ⁵C₃ ways

Selecting 2 girls from 4 girls can be done in ⁴C₂ ways

Selecting 1 transgender from 2 can be done in ²C₁ = 2 ways
∴ Selecting 3 boys, 2 girls and 1 transgender can be done in 10 × 6 × 2 = 120 ways

Question 10.
Find the total number of subsets of a set with
(i) 4 elements
(ii) 5 elements
(iii) n elements

Solution:
If a set has n elements then the number of its subsets = 2ⁿ

(i) Here n = 4
So number of subsets = 2⁴ = 16

(ii) n = 5
So number of subsets = 2⁵ = 32

(iii) n = n
So number of subsets = 2

Question 11.
A trust has 25 members.

(i) How many ways 3 officers can be selected?
Solution:
Selecting 3 from 25 can be done in ²⁵C₃ ways

(ii) In how many ways can a President, Vice President, and secretary be selected?
Solution:
The number of ways of selecting a president from 25 members = 25C₁ = 25
After the selection of the president, the remaining number of members in the trust is 24
The number of ways of selecting a vice president
from the remaining 24 members of the trust is = 24C₁ 24
After the selection of the president and vice president, the number of remaining members in the trust = 23
The number of ways of selecting a secretary from the remaining 23 members of the trust is = 23 C₁ = 23
∴ Total number of ways of selection = 25 × 24 × 23 = 13800

Question 12.
How many ways a committee of six persons from 10 persons can be chosen along with a chairperson and a secretary?
Solution:
Selecting a chairperson from the 10 persons can be done in 10 ways
After the selection of chairperson, only 9 persons are left out so selecting a secretary (from the remaining persons) can be done in 9 ways.
The remaining persons = 8
Totally we need to select 6 persons
We have selected 2 persons.
So we have to select 4 persons
Selecting 4 from 8 can be done in ⁸C₄ ways

Question 13.
How many different selections of 5 books can be made from 12 different books if,
Solution:
No. of books given = 12
No. of books to be selected = 5

(i) Two particular books are always selected?
Solution:
So we need to select 3 more books from (12 – 2) 10 books which can be done in ¹⁰C₃ ways

(ii) Two particular books are never selected?
Solution:
Two particular books never to be selected.
So only 10 books are there and we have to select 5 books which can be done in ¹⁰C₅ ways

Question 14.
There are 5 teachers and 20 students. Out of them, a committee of 2 teachers and 3 students is to be formed. Find the number of ways in which this can be done. Further, find in how many of these committees
(i) a particular teacher is included?
(ii) a particular student is excluded?
Solution:
No. of teachers = 5
No of students = 20
We need to select 2 teachers and 3 students

(i) A particular teacher should be included. So from the remaining 4 teachers, one teacher is to be selected which can be done in ⁴C₁ = 4 ways

So selecting 2 teachers and 3 students can be done in 4 × 1140 = 4560 ways

(ii) the particular student should be excluded.
So we have to select 3 students from 19 students which can be done in ¹⁹C₃ ways

∴ 2 teachers and 3 students can be selected in 969 × 10 = 9690 ways

Question 15.
In an examination, a student has to answer 5 questions, out of 9 questions in which 2 are compulsory. In how many ways students can answer the questions?
Solution:
No. of questions given = 9
No. of questions to be answered = 5
But 2 questions are compulsory
So the student has to answer the remaining 3 questions (5 – 2 = 3) from the remaining 7 (9 – 2 = 7) questions which can be done in ⁷C₃ ways

Question 16.
Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly three aces in each combination.
Solution:
No. of cards = 52
In that number of aces = 4
No. of cards needed = 5
In that 5 cards number of aces needed = 3
So the 3 aces can be selected from 4 aces in ⁴C₃ = ⁴C₁ = 4 ways
So the remaining = 5 – 3 = 2
These 2 cards can be selected in ⁴⁸C₂ ways

Question 17.
Find the number of ways of forming a committee of 5 members out of 7 Indians and 5 Americans, so that always Indians will be the majority’ in the committee.
Solution:
We need a majority of Indian’s which is obtained as follows.

The possible ways are (5I) or (4I and 1A) or (3I and 2A)

Question 18.
A committee of 7 peoples has to be formed from 8 men and 4 women. In how many ways can this be done when the committee consists of
(i) exactly 3 women?
(ii) at least 3 women?
(iii) at most 3 women?
Solution:

We need a committee of 7 people with 3 women and 4 men.
This can be done in (⁴C₃) (⁸C₄) ways

The number of ways = (70) (4) = 280

(ii) Atleast 3 women

So the possible ways are (3W and 4M) or (4W and 3M)

The number of ways (4) (70) + (1) (56) = 280 + 56 = 336

(iii) Atmost 3 women

Question 19.
7 relatives of a man comprises 4 ladies and 3 gentlemen, his wife also has 7 relatives; 3 of them are ladies and 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of man’s relative and 3 of the wife’s relatives?
Solution:

We need 3 ladies and 3 gentlemen for the party which consist of 3 Husbands relative and 3 wife’s relative.
This can be done as follows

Question 20.
A box contains two white balls, three black balls, and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?
Solution:
The box contains 2 white, 3 black, and 4 red balls
We have to draw 3 balls in which there should be at least 1 black ball
The possible draws are as follows
Black balls = 3
Red and White = 2 + 4 = 6

Question 21.
Find the number of strings of 4 letters that can be formed with the letters of the word EXAMINATION.
Solution:
EXAMINATION
(i.e.) A, I, N are repeated twice. So the number of distinct letters = 8
From the 8 letters, we have to select and arrange 4 letters to form a 4 letter word which can
be done in ⁸P₄ = 8 × 7 × 6 × 5 = 1680
From the letters A, A, I, I, N, N when any 2 letters are taken as AA, II or AA, NN or II, NN

From AA, II, NN we select one of them and from the remaining we select and arrange 3 which can be done in ways

Total number of 4 letter word = 1680 + 18 + 756 = 2454

Question 22.
How many triangles can be formed by joining 15 points on the plane, in which no line joining any three points?
Solution:
No. of non-collinear points = 15
To draw a Triangle we need 3 points
∴ Selecting 3 from 15 points can be done in ¹⁵C₃ ways.

Question 23.
How many triangles can be formed by 15 points, in which 7 of them lie on one line and the remaining 8 on another parallel line?
Solution:

7 points lie on one line and the other 8 points parallel on another paraller line.
A triangle is obtained by taking one point from one line and second points from the other parallel line which can be done as follows.

∴ Number of triangles = (7) (28) + (21) (8) = 196 + 168 = 364

Question 24.
There are 11 points in a plane. No three of these lie in the same straight line except 4 points, which are collinear. Find,
(i) The number of straight lines that can be obtained from the pairs of these points?
Solution:
4 points are collinear

Total number of points 11.
To get a line we need 2 points

But in that 4 points are collinear

From (1) Joining the 4 points we get 1 line
∴ Number of lines = ¹¹C₂ – ⁴C₂ + 1 = 55 – 6 + 1 = 50

(ii) The number of triangles that can be formed for which the points as their vertices?
A triangle is obtained by joining 3 points.
So selecting 3 from 11 points can be

But of the 11 points, 4 points are collinear. So we have to subtract ⁴C₃ = ⁴C₁ = 4
∴ Number of triangles = 165 – 4 = 161

Question 25.
A polygon has 90 diagonals. Find the number of its sides?
Solution:

∴ n = 15

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 Additional Questions Solved

Question 1.
A group consists of 4 girls and 7 boys. In bow many ways can a team of 5 members be selected, if the team has
(i) no girls
(ii) atleast one boy and one girl.
(iii) at least three girls
Solution:
We have 4 girls and 7 boys and a team of 5 members is to be selected.

(i) If no girl is selected, then all the 5 members are to be selected out of 7 boys

(ii) When at least one boy and one girl are to be selected, then


Hence the required number of ways are (i) 21 ways (ii) 441 ways (iii) 91 ways

Question 2.
A committee of 6 is to be chosen from 10 men and 7 women so as to contain atleast 3 men and 2 women. In how many different ways can this be done it two particular women refuse to serve on the same committee?
Solution:
We have 10 men and 7 women out of which a committee of 6 is to be formed which contain atleast 3 men and 2 women

∴ Total number of committee = 8610 – 810 = 7800
Hence, the value of the filler is 7800

Question 3.
Using the digits 1, 2, 3,4, 5, 6, 7 a number of 4 different digits is formed. Find

Solution:
(a) Total of 4 digit number formed with 1, 2, 3, 4, 5, 6, 7

(b) When a number is divisible by 2 = 4 × 5 × 6 × 3 = 360
(c) Total numbers which are divisible by 25 = 40
(d) Total numbers which are divisible by 4 (last two digits is divisble by 4) = 200
Hence, the required matching is (a) ⟷ (z), (b) ⟷ (iii), (c) ⟷ (iv), (d) ⟷ (ii)

Question 4.
If ²²P_{r + 1} : ²⁰P_{r + 2} = 11 : 52, find r.
Solution:

Question 5.

Solution:

Question 6.
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:
(i) exactly 3 girls?
(ii) at least 3 girls?
(iii) almost 3 girls?
Solution:

(ii) We have to select at least 3 girls. So the committee consists of 3 girls and 4 boys or 4 girls and 3 boys.
∴ Number of ways of selection = ⁴C₃ × ⁹C₄ + ⁴C₄ × ⁹C₃

(iii) We have to select at most 3 girls. So the committee consists of no girl and 7 boys or 1 girl and 6 boys or 2 girls and 5 boys or 3 girls and 4 boys.

Question 7.
Determine n if
(i) ²ⁿC₃ : ⁿC₂ = 12 : 1
(ii) ²ⁿC₃ : ⁿC₃ = 11 : 1
Solution:

Question 8.
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
Solution:
There are 6 red balls, 5 white balls and 5 blue balls.
We have to select 3 balls of each colour.

Question 9.

Solution:

Question 10.
If ⁿC₄, ⁿC₅ and ⁿC₄ are in A.P. then find n.
[Hint: ²ⁿC₅ = ⁿC₆ + ⁿC₄]
Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.4

Question 1.
By the principle of mathematical induction, prove that, for n ≥ 1

Solution:

∴ P(k+ 1) is true.
Thus P(K) is true ⇒ (k + 1) is true.
Hence by principle of mathematical induction, P(n) is true for all n ∈ N.

Question 2.
By the principle of mathematical induction, prove that, for n > 1

Solution:

∴ P(1) is true
Let P(n) be true for n = k

∴ P(k + 1) is true
Thus P(k) is true ⇒ P(k + 1) is true. Hence by the principle of mathematical induction, P(k) is true for all n ∈ N.

Question 3.
Prove that the sum of the first n non-zero even numbers is n² + n.
Solution:
Let P(n) = 2 + 4 + 6 + ………….. + 2n = n² + n

Step 1:
Let us verify the statement for n = 1
P (1 ) = 2 = 1² + 1 = 1 + 1 = 2.
∴ The given result is true for n = 1.

Step 2:
Let us assume that the given result is true for n = k
P ( k) = 2 + 4 + 6 + ………… + 2k = k² + k

Step 3:
Let us prove the result for n = k + 1
P (k+ 1 ) = 2 + 4 + 6+ + 2k + (2k + 2 )
P(k+ 1 ) = P(k) + (2k + 2)
= k² + k + 2k + 2
= k² + 3k + 2
= k² + 2k + k + 2
= k(k +2) + 1(k + 2)
P(k+ 1 ) = (k+ 1) (k + 2) ……….. (1)
P (k) = k² + k
= k (k + 1 )
P(k+ 1) = (k + 1) (k + 1 + 1)
= (k + 1) (k + 2)
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
2 + 4 + 6 + ………….. + 2n = n² + n
is true for all natural numbers n.

Question 4.
By the principle of Mathematical induction, prove that, for n ≥ 1.

Solution:


∴ P(k + 1) is true
Thus P(k) is true ⇒ P(k + 1) is true
Hence by the principle of mathematical induction, P(n) is true for all n ∈ N

Question 5.
Using the Mathematical induction, show that for any natural number n ≥ 2,

Solution:


⇒ P(k + 1) is true when P(k) is true so by the principle of mathematical induction P(n) is true.

Question 6.
Using the Mathematical induction, show that for any natural number n ≥ 2,

Solution:


⇒ P(k + 1) is true when P(k) is true so by the principle of mathematical induction P(n) is true for n ≥ 2.

Question 7.
Using the Mathematical induction, show that for any natural number n

Solution:


∴ P(k + 1) is true
Thus p(k) is true ⇒ P(k + 1) is true
Hence by the principle of mathematical induction,
p(n) is true for all n ∈ z

Question 8.
Using the Mathematical induction, show that for any natural number n,

Solution:


∴ P(k + 1) is true
Thus P(k) is true ⇒ P(k + 1) is true. Hence by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 9.
Prove by Mathematical Induction that
1! + (2 × 2!) + (3 × 3!) + … + (n × n!) = (n + 1)! – 1
Solution:
Let p(n) = 1! + (2 × 2 !) + (3 × 3!) +…. .+ (n × n!) = (n + 1)! – 1

Step 1:
First let us verify the result for n = 1
P(1) = 1! = (1 + 1)! – 1
P(1) = 1! = 2! – 1
P(1) = 1 = 2 – 1 = 1
∴ We have verified the result for n = 1.

Step 2:
Let us assume that the result is true for n = k
P(k) = (1 × 1 !) + (2 × 2!) + (3 × 3!) + …………. + (k × k!) = (k + 1)! – 1

Step 3:
Let us prove the result for n = k + 1
P(k + 1)=(1 × 1!) + (2 × 2!) + (3 × 3!) + ………….. + (k × k!) + ((k + 1) × (k + 1)!)
P(k + 1) = P(k) + ((k + 1) × (k + 1)!)
P(k + 1) = (k + 1)! – 1 + (k + 1) × (k + 1)!
= (k + 1)! + (k + 1) (k + 1)! – 1
= (k + 1)! (1 + k + 1) – 1
= (k + 1)! (k + 2) – 1
= (k + 2)! – 1
P(k + 1) = ((k + 1) + 1)! – 1
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
(1 × 1!) + (2 × 2!) + (3 × 3!) + …………… + (n × n!) = (n + 1)! – 1
is true for all natural numbers n.

Question 10.
Using the Mathematical induction, show that for any natural number n, x²ⁿ – y²ⁿ is divisible by x +y.
Solution:
Let P(n) = x²ⁿ – y²ⁿ is divisible by x + y
Step 1:
First, let us verify the result for n = 1.
P ( 1 ) = x²⁽¹⁾ – y²⁽¹⁾ = x² – y²
P(1) = (x + y) (x – y) which is divisible by x + y
∴ The result is true for n = 1

Step 2:
Let us assume that the result is true for n = k
P(k) = x^2k – y^2k which is divisible by x + y
∴ P (k) = x^2k – y^2k = λ (x + y) where λ ∈ N ——— (1)

Step 3:
Let us prove the result for n = k + 1

∴ P ( k + 1) is divisible by x + y
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
x²ⁿ – y²ⁿ is divisible by x + y
for all natural numbers n.

Question 11.
By the principle of mathematical induction, prove that, for n ≥ 1,

Solution:

Question 12.
Use induction to prove that n³ – 7n + 3, is divisible by 3, for all natural numbers n.
Solution:
Let P ( n) = n³ – 7n + 3 is divisible by 3

Step 1:
First let us verify the results for n = 1
P(I) = 1³ – 7 × 1 + 3
= 1 – 7 + 3
P (1) = – 3
which is divisible by 3
∴ The result is true for n = 1

Step 2:
Let us assume that the result is true for n = k
P(k) = k³ – 7k + 3 is divisible by 3
P(k) = k³ – 7k + 3 = 3λ where λ ∈ N

Step 3:
Let us prove the result for n = k + 1
P(k + 1) = (k + 1)³ – 7(k + 1 ) + 3
= k³ + 3k² + 3k + 1 – 7k – 7 + 3
= k³ + 3k² – 4k – 3
= k³ – 4k – 3k + 3k – 3 + 6 – 6 + 3k²
= k³ – 7k + 3 + 3k – 6 + 3k²
= (k³ – 7k + 3) + 3(k² + k – 2)
= 3λ + 3 (k² + k – 2)
P(k + 1) = 3 (λ + k² + k – 2 )
which is a multiple of 3, hence divisible by 3
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
n³ – 7n + 3 is divisible by 3 for all natural numbers n.

Question 13.
Use induction to prove that 5^{n + 1} + 4 × 6ⁿ when divided by 20 leaves a remainder 9, for all natural numbers n.
Solution:
P(n) is the statement 5^{n + 1} + 4 × 6ⁿ – 9 is ÷ by 20
P(1) = 5^{1 + 1} + 4 × 6¹ – 9 = 5² + 24 – 9
= 25 + 24 – 9 = 40 ÷ by 20
So P(1) is true
Assume that the given statement is true for n = k
(i.e) 5^{k + 1} + 4 × 6ⁿ – 9 is ÷ by 20
P(1) = 5^{1 + 1} + 4 × 6¹ – 9
= 25 + 24 – 9
So P(1) is true
To prove P(k + 1) is true
P(k + 1) = 5^{k + 1 + 1} + 4 × 6^{k + 1 + 1} – 9
= 5 × 5 ^{k + 1} + 4 × 6 × 6^k – 9
= 5[20C + 9 – 4 × 6^k] + 24 × 6^k – 9 [from(1)]
= 100C + 45 – 206^k + 246^k – 9
= 100C + 46^k + 36
= 100C + 4(9 + 6^k)
Now for k = 1 ⇒ 4(9 + 6^k) = 4(9 + 6)
= 4 × 15 = 60 ÷ by 20 .
for k = 2 = 4(9 + 6²) = 4 × 45 = 180 ÷ 20
So by the principle of mathematical induction 4(9 + 6^k) is ÷ by 20
Now 100C is ÷ by 20.
So 100C + 4(9 + 6^k) is ÷ by 20
⇒ P(k + 1) is true whenever P(k) is true. So by the principle of mathematical induction P(n) is true.

Question 14.
Use induction to prove that 10ⁿ + 3 × 4^{n + 2} + 5, is divisible by 9, for all natural numbers n.
Solution:
P(n) is the statement 10ⁿ + 3 × 4^{n + 2} + 5 is ÷ by 9
P(1) = 10¹ + 3 × 4² + 5 = 10 + 3 × 16 + 5
= 10 + 48 + 5 = 63 ÷ by 9
So P(1) is true. Assume that P(k) is true
(i.e.) 10^k + 3 × 4^{k + 2} + 5 is ÷ by 9
(i.e.) 10^k + 3 × 4^{k + 2} + 5 = 9C (where C is an integer)
⇒ 10^k = 9C – 5 – 3 × 4^{k + 2} ……(1)
To prove P(k + 1) is true.
Now P(k + 1) = 10^{k + 1} + 3 × 4^{k + 3} + 5
= 10 × 10^k + 3 × 4^{k + 2} × 4 + 5
= 10[9C – 5 – 3 × 4^{k + 2}] + 3 × 4^{k + 2} × 4 + 5
= 10[9C – 5 – 3 × 4^{k + 2}] + 12 × 4^{k + 2} + 5
= 90C – 50 – 30 × 4^{k + 2} + 12 × 4^{k + 2} + 5
= 90C – 45 – 18 × 4^{k + 2}
= 9[10C – 5 – 2 × 4^{k + 2}] which is ÷ by 9
So P(k + 1) is true whenever P(K) is true. So by the principle of mathematical induction P(n) is true.

Question 15.
Prove that using the Mathematical induction

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 Additional Questions

Question 1.
Prove by induction the inequality (1 + x)ⁿ ≥ 1 + nx, whenever x is positive and n is a positive integer.
Solution:
P(n) : (1 +x)ⁿ ≥ 1 +nx
P(1): (1 + x)¹ ≥ 1 + x
⇒ 1 + x ≥ 1 + x, which is true.
Hence, P(1) is true.
Let P(k) be true
(i.e.) (1 + x)^k ≥ 1 + kx
We have to prove that P(k + 1) is true.
(i.e.) (1 + x)^{k + 1} ≥ 1 + (k + 1)x
Now, (1 + x)^{k + 1} ≥ 1 + kx [∵ p(k) is true]
Multiplying both sides by (1 + x), we get
(1 + x)^k(1 + x) ≥ (1 + kx)(1 + x)
⇒ (1 + x)^{k + 1} ≥ 1 + kx + x + kx²
⇒ (1 + x)^{k + 1} ≥ 1 + (k + 1)x + kx² ….. (1)
Now, 1 + (k + 1) x + kx² ≥ 1 + (k + 1)x …… (2)
[∵ kx² > 0]
From (1) and (2), we get
(1 + x)^{k + 1} ≥ 1 + (k + 1)x
∴ P(k + 1) is true if P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all values, of n.

Question 2.
3²ⁿ – 1 is divisible by 8.
Solution:
P(n) = 3²ⁿ – 1 is divisible by 8
For n = 1, we get
P(1) = 3^2.1 – 1 = 9 – 1 = 8
P(1) = 8, which is divisible by 8.
Let P(n) be true for n = k
P(k) = 3^2k – 1 is divisible by 8 ….. (1)
Now, P(k + 1) = 3^{(2k + 2)} – 1 = 3^2k.3² – 1
= 3²(3^2k – 1) + 8
Now, 3^2k – 1 is divisible by 9. [Using (1)]
∴ 3² (3^2k – 1) + 8 is also divisible by 8.
Hence, 3²ⁿ – 1 is divisible by 8 ∀ n E N

Question 3.
Prove by the principle of mathematical induction if x and y are any two distinct integers, then xⁿ – yⁿ is divisible by x – y. [OR]
xⁿ – yⁿ is divisible by x – y, where x – y ≠ 0.
Solution:
Let the given statement be P(n).
(i.e.) P(n): xⁿ – yⁿ = M(x – y), x – y ≠ 0

Step I.
When n = 1,
xⁿ – yⁿ = x – y = M(x – y) ….(1)
⇒ P(1) is true.

Step II.
Assume that P(k) is true.
(i.e.) x^k – y^k = M(x – y), x – y ≠ 0
We shall now show that P(k + 1) is true
Now, x^{k + 1} – y^{k + 1} = x^{k + 1} – x^ky + x^{k + 1}y – y^{k + 1}
= x^k(x – y) + y(x^k – y^k)
= x^k(x – y) + yM(x – y) [Usng ….. (1)]
= (x – y)(x^k – yM)
∴ By the principle of mathematical induction, P(n) is true for all n ∈ N

Question 4.
Prove by the principle of mathematical induction that for every natural number n, 3^{2n + 2} – 8n – 9 is divisible by 8.
Solution:
Let P(n): 3^{2n + 2} – 8n – 9 is divisible by 8.
Then, P(1): 3^{2.1 + 2} – 8.1 – 9 is divisible by 8.
(i.e.) 3⁴ – 8 – 9 is divisible by 8 or 81 – 8 – 9 is divisible by 8
(or) 64 is divisible by 8, which is true.
Suppose P(k) is true, then
P(k) : 3^{2k + 2} – 8k – 9 is divisible by 8
(i.e.) 3^{2k + 2} – 8k – 9 = 8m, where m ∈ N (or)
3^{2k + 2} = 8m + 8k + 9
P(k + 1) is the statement given by, …(1)
P(k + 1) : 3^{2(k + 1) + 2} – 8(k + 1) – 9

∴ P(k + 1) is true
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N

Question 5.
Use the principle of mathematical induction to prove that for every natural number n.

Solution:
Let P(n) be the given statement, i.e.

⇒ P(1) is true.
We note that P(n) is true for n = 1.
Assume that P(k) is true

Now, we shall prove that P(k + 1) is true whenever P(k) is true. We have,

∴ P(k + 1) is also true whenever P(k) is true
Hence, by the principle of mathematical induction, P(n) is also true for all n ∈ N.

Question 6.
n³ – n is divisible by 6, for each natural number n ≥ 2.
Solution:
Let P(n) : n³ – n

Step 1 :
P(2): 2³ – 2 = 6 which is divisible by 6. So it is true for P(2).

Step 2 :
P(A): k³ – k = 6λ. Let it is be true for k ≥ 2
⇒ k³ = 6λ + k …(i)

Step 3 :
P(k + 1) = (k + 1)³ – (k + 1)
= k³ + 1 + 3k² + 3k – k – 1 = k³ – k + 3(k² + k)
= k³ – k + 3(k² + k) = 6λ + k – k + 3(k² + k)
= 6λ + 3(k² + k) [from (i)]
We know that 3(k² + k) is divisible by 6 for every value of k ∈ N.
Hence P(k + 1) is true whenever P(k) is true.

Question 7.
For any natural number n, 7ⁿ – 2ⁿ is divisible by 5.
Solution:
Let P(n) : 7ⁿ – 2ⁿ

Step 1:
P(1) : 7¹ – 2¹ = 5λ which is divisible by 5. So it is true for P(1).

Step 2:
P(k): 7^k – 2^k = 5λ. Let it be true for P(k)

Step 3:
P(k + 1) = 7^{k + 1} – 2^{k + 1}

So, it is true for P(k + 1)
Hence, P(k + 1) is true whenever P(k) is true.

Question 8.
n² < 2ⁿ, for all natural numbers n ≥ 5.
Solution:
Let P(n) : n² < 2ⁿ for all natural numbers, n ≥ 5

Step 1 :
P(5) : 1⁵ < 2⁵ ⇒ 1 < 32 which is true for P(5)

Step 2 :
P(k): k² < 2k. Let it be true for k ∈ N

Step 3 :
P(k + 1): (k + 1)² < 2^{k + 1}
From Step 2, we get k² < 2^k
⇒ k² < 2^{k + 1} < 2k + 2k + 1


From eqn. (i) and (ii), we get (k + 1)² < 2^{k + 1}
Hence, P(k + 1) is true whenever P(k) is true for k ∈ N, n ≥ 5.

Question 9.
In 2n < (n + 2)! for all natural number n.
Solution:
Let P(n) : 2n < (n + 2)! for all k ∈ N.

Hence, P(k + 1) is true whenever P(k) is true.

Question 10.
1 + 5 + 9 + … + (4n – 3) = n(2n – 1), ∀ n ∈ N.
Solution:
Let P(n) : 1 + 5 + 9 + … + (4n – 3) = n(2n – 1), ∀ n ∈ N

Step 1:
P(1) : 1 = 1(2.1 – 1) = 1 which is true for P(1)

Step 2:
P(k) : 1 + 5 + 9 + … + (4k – 3) = k(2k – 1). Let it be true.

Step 3:
P(k + 1) : 1 + 5 + 9 + … + (4k – 3) = k(4k + 1)
= k(2k – 1) + (4k + 1) = 2k² – k + 4k + 1
= 2k² + 3k + 1 = 2k² + 2k + k + 1
= 2k(k + 1) + 1 (k + 1) = (2k + 1)(k + 1)
= (k+ 1) (2k + 2 – 1) = (k + 1) [2(k + 1) – 1]
Which is true for P(k + 1).
Hence, P(k + 1) is true whenever P(k) is true.

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.3

Question 1.
Suppose that 120 students are studying in 4 sections of eleventh standard in a school. Let A denote the set of students and B denote the set of the sections. Define a relation from A to B as “x related toy if the student x belongs to the section y”. Is this relation a function? What can you say about the inverse relation? Explain your answer.
Solution:
(i) A = {set of students in 11^th standard}
B = {set of sections in 11sup>th standard}
R : A ➝ B ⇒ x related to y
⇒ Every students in eleventh Standard must in one section of the eleventh standard.
⇒ It is a function.
Inverse relation cannot be a function since every section of eleventh standard cannot be related to one student in eleventh standard.

Question 2.
Write the values of f at – 4, 1, -2, 7, 0 if

Solution:
f(-4) = -(-4) + 4 = 8
f(1) = 1 – 1² = 0
f(-2) = (-2)² – (-2) = 4 + 2 = 6
f(7) = 0
f(0) = 0

Question 3.
Write the values of f at -3, 5, 2, -1, 0 if

Solution:
f(-3) = (-3)² – 3 – 5 = 9 – 8 = 1
f(5) = (5)² + 3(5) – 2 = 25 + 15 – 2 = 38
f(2) = 4 – 3 = 1
f(-1) = (-1)² + (-1) – 5 = 1 – 6 = -5
f(0) = 0 – 3 = -3

Question 4.
State whether the following relations are functions or not. If it is a function check for one-to-oneness and ontoness. If it is not a function, state why?
(i) If A = {a, b, c] and/= {(a, c), (b, c), (c, b)};(f: A ➝ A).
(ii) If X = {x, y, z} and/= {(x, y), (x, z), (z, x)}; (f: X ➝ X).
Solution:
(i) f : A ➝ A

It is a function but it is not 1 – 1 and not onto function.

(ii) f : X ➝ X

x ∈ X (Domain) has two images in the co-domain x. It is not a function.

Question 5.
Let A = {1, 2, 3, 4} and B = {a, b, c, d}. Give a function from A ➝ B for each of the following:
(i) neither one-to-one nor onto.
(ii) not one-to-one but onto.
(iii) one-to-one but not onto.
(iv) one-to-one and onto.
Solution:
A = {1, 2, 3, 4}
B = {a, b, c, d}.

R = {(1, b) (2, b) (3, c) (4, d)} is not 1-1 and not onto

(iii) Not possible

(iv)

Question 6.
Find the domain of \(\frac{1}{1-2 \sin x}\)
Solution:

Question 7.
Find the largest possible domain of the real valued function f(x) = \(\frac{\sqrt{4-x^{2}}}{\sqrt{x^{2}-9}}\)
Solution:

∴ No largest possible domain
The domain is null set

Question 8.
Find the range of the function \(\frac{1}{2 \cos x-1}\)
Solution:
The range of cos x is – 1 to 1

Question 9.
Show that the relation xy = -2 is a function for a suitable domain. Find the domain and the range of the function.
Solution:
(i) Let f: R → R defined as f: x → \(-\frac{2}{x}\) then
f(x) = \(-\frac{2}{x}\) or y = \(-\frac{2}{x}\)
⇒ xy = – 2
f (x) is not a function since f(x) is not defined for x = 0

(ii) Let f: R – {0} → R defined as f(x) = \(-\frac{2}{x}\)
⇒ y = \(-\frac{2}{x}\) = xy = – 2
f is one – one but not onto because 0 has no preimage.
f : R – {0} → R {0} is a function which is one- one and onto
Domain = R – {0}
Range = R – {0}

Question 10.
If f, g : R ➝ R are defined by f(x) = |x| + x and g(x) = |x| – x, find gof and fog.
Solution:

Question 11.
If f, g, h are real-valued functions defined on R, then prove that
(f + g)oh = foh + goh. What can you say about fo(g + h)? Justify your answer.
Solution:
Let f + g = k

= (f + g((h(x))
= f[h(x)] + g [h(x)]
= foh + goh
(i.e.,)(f + g)(o)h = foh + goh
fo(g + h) is also a function

Question 12.
If f: R ➝ R is defined by f(x) = 3x – 5, prove that f is a bijection and find its inverse.
Solution:
P(x) = 3x – 5
Let y = 3x – 5 ⇒ 3x = y + 5

Question 13.
The weight of the muscles of a man is a function of his bodyweight x and can be expressed as W(x) = 0.35x. Determine the domain of this function.
Solution:
Given W(x) = 0.35x
W(0) = W(1) = 0.35, W(2) = 0.7 ………….. W ( ∞ ) = ∞
Since x. denotes the bodyweight of a man, it will take only positive integers. That is x > 0.
W(x) : (0, ∞) → (0, ∞)
Domain = (0, ∞) , Range = (0, ∞)

Question 14.
The distance of an object falling is a function of time t and can be expressed as s(t) = -16t². Graph the function and determine if it is one-to-one.
Solution:
s(t) = -16t²
Suppose S(t₁) = S(t₂)

since time cannot be negative, we to take t₁ = t₂
Hence it is one-one.

t	0	1	2	3
s	0	-16	-64	-144

Question 15.
The total cost of airfare on a given route is comprised of the base cost C and the fuel surcharge S in rupee. Both C and S are functions of the mileage m; C(m) = 0.4m + 50 and S(m) = 0.03m. Determine a function for the total cost of a ticket in terms of the mileage and find the airfare for flying 1600 miles.
Solution:
Given the cost of airfare function and fuel surcharge functions are as follows.
C(m) = 0.4 m+ 50 ———- (1)
S (m) = 0.03 m ———- (2)
Total cost of a ticket = C(m) + S(m)
f(x) = 0.4 m + 50 + 0.03 m
f(x) = 0.43 m + 50
Given m = 1600 miles
The cost of Airfare for flying 1600 miles
f( 1600 ) = 0.43 × 1600 + 50
= 688 + 50
= 738
∴ Airfare for flying 1600 miles is Rs. 738.

Question 16.
A salesperson whose annual earnings can be represented by the function A(x) = 30, 000 + 0.04x, where x is the rupee value of the merchandise he sells. His son is also in sales and his earnings are represented by the function S(x) = 25, 000 + 0.05x. Find (A + S)(x) and determine the total family income if they each sell Rupees 1,50,00,000 worth of merchandise.
Solution:
A(x) = 30, 000 + 0.04x, where x is merchandise rupee value
S(x) = 25000 + 0.05 x
(A + S) (x) = A(x) + S(x)
= 30000 + 0.04x + 25000 + 0.05 x
= 55000 + 0.09x
(A + S) (x) = 55000+ 0.09x
They each sell x = 1,50,00,000 worth of merchandise
(A + S) x = 55000 + 0.09 (1,50,00,000)
= 55000 + 13,50,000
∴ Total income of family = ₹ 14,05,000

Question 17.
The function for exchanging American dollars for Singapore Dollar on a given day is f(x) = 1.23x, where x represents the number of American dollars. On the same day, the function for exchanging Singapore Dollar to Indian Rupee is g(y) = 50.50y, where y represents the number of Singapore dollars. Write a function which will give the exchange rate of American dollars in terms of the Indian rupee.
Solution:
f(x) = 1. 23x where x is number of American dollars.
g(y) = 50.50y where y is number of Singapore dollars.
gof(x) = g(f(x))
= g(1. 23x)
= 50.50 (1.23x)
= 62.115 x

Question 18.
The owner of a small restaurant can prepare a particular meal at a cost of Rupees 100. He estimates that if the menu price of the meal is x rupees, then the number of customers who will order that meal at that price in an evening is given by the function D(x) = 200 – x. Express his day revenue, total cost and profit on this meal as functions of x.
Solution:
Number of customers = 200 – x
Cost of one meal = Rs. 100
Cost of (200 – x) meals = (200 – x) × 100
Menu price of the meal = Rs. x
∴ Total menu price of (200 – x) meals = (200 – x) x
Profit = Menu price – Cost
= (200 – x) x – (200 – x) 100
Profit = (200 – x) (x – 100)

Question 19.
The formula for converting from Fahrenheit to Celsius temperatures is \(y=\frac{5 x}{9}-\frac{160}{9}\)
Find the inverse of this function and determine whether the inverse is also a function.
Solution:

Question 20.
A simple cipher takes a number and codes it, using the function f(x) = 3x – 4. Find the inverse of this function, determine whether the inverse is also a function and verify the symmetrical property about the line y = x (by drawing the lines).
Solution:
f(x) = 3x – 4
Let y = 3x – 4

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.3 Additional Questions

Question 1.
Find the domain and range of the function \(f(x)=\frac{1}{\sqrt{x-5}}\)
Solution:
Given that : f(x) \(f(x)=\frac{1}{\sqrt{x-5}}\)
Here, it is clear that / (x) is real when x – 5 > 0 ⇒ x > 5
Hence, the domain = (5, ∞)
Now to find the range put

For x ∈ (5, ∞), y ∈ R⁺.
Hence, the range of f = R⁺.

Question 2.
If \(f(x)=\frac{x-1}{x+1}\), then show that

Solution:

Question 3.
Find the domain of each of the following functions given by:
\(f(x)=\frac{x^{3}-x+3}{x^{2}-1}\)
Solution:

Here, f(x) is not defined if x² – 1 ≠ 0
(x – 1) (x + 1) ≠ 0
x ≠ 1, x ≠ -1
Hence, the domain of f = R – {-1, 1}

Question 4.
Find the range of the following functions given by
f(x) = 1 + 3 cos 2x
Solution:
Given that: f(x) = 1 + 3 cos 2x
We know that -1 ≤ cos 2x ≤ 1
⇒ -3 ≤ 3 cos 2x ≤ 3 ⇒ -3 + 1 ≤ 1 + 3 cos 2x ≤ 3 + 1
⇒ -2 ≤ 1 + 3 cos 2x ≤ 4 ⇒ -2 ≤ f(x) ≤ 4
Hence the range of f = [-2, 4]

Question 5.
Find the domain and range of the function \(f(x)=\frac{x^{2}-9}{x-3}\)
Solution:

Domain off: Clearly f(x) is not defined for x – 3 = 0 i.e. x = 3.
Therefore, Domain (f) = R – {3}
Range off: Let f(x) = y. Then,

It follows from the above relation that y takes all real values except 6 when x takes values in the set R – {3}. Therefore, Range (f) = R {6}.

Question 6.
Find the range of the following functions given by f(x) = \(\frac{1}{2-\sin 3 x}\)
Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2

Question 1.
Without expanding the determinant, prove that
Solution:

Question 2.
Show that
Solution:

Question 3.
Prove that
Solution:
LHS
Taking a from C₁, b from C₂ and c from C₃ we get

Expanding along R₁ we get
(2c) (abc) (1) [ab + ab] = abc (2c) (2ab)
1 = (abc) (4abc) = 4a²b²c²
= RHS

Question 4.

Solution:

Question 5.
Prove that
Solution:

Question 6.
Show that
Solution:

Question 7.
Write the general form of a 3 × 3 skew-symmetric matrix and prove that its determinant is 0.
Solution:

Question 8.

Solution:

we get – (aα² + 2bα + c) [ac – b²]
So Δ = 0 ⇒ (aα² + 2bα + c) (ac -b²) = – 0 = 0
⇒ aα² + 2bα + c = 0 or ac – b² = 0
(i.e.) a is a root of ax² + 2bx + c = 0
or ac = b²
⇒ a, b, c are in G.P.

Question 9.
Prove that
Solution:

Question 10.
If a, b, c are p^th, q^th and r^th terms of an A.P., find the value of
Solution:
We are given a = t_p,b = t_q and c = t_r
Let a be the first term and d be the common difference

Question 11.
Show that is divisible by x⁴
Solution:

Multiplying R₁ by a, R₂ by b and R₃ by c and
taking out a from C₁ b from C₂ and c from C₃ we get
==

Question 12.
If a, b, c are all positive, and are p^th, q^th and r^th terms of a G.P., show that

Solution:

Question 13.
Find the value of if x, y, z ≠ 1.
Solution:
Expanding the determinant along R₁

Question 14.

Solution:

Question 15.
Without expanding, evaluate the following determinants:

Solution:

Question 16.
If A is a square matrix and |A| = 2, find the value of |AA^T|.
Solution:
Given |A| = 2
[Property 1: The determinant of a matrix remains unaltered if its rows are changed into columns and columns. That is, |A| = |A^T|]

|A^T| = |A| = 2
∴ |A A^T| = |A| |A^T|
= 2 × 2 = 4

Question 17.
If A and B are square matrices of order 3 such that |A| = -1 and |B| = 3, find the value of |3AB|.
Solution:
Given A and B are square matrices of order 3 such that |A| = -1 and |B| = 3
[It A is a square matrix of order n then det ( kA) = |kA| = kⁿ |A|.]
A and B are square matrices of order 3. Therefore,
AB is also a square matrix of order 3.
|3 AB| = 3³ |AB|
= 27 |A| |B|
= 27 × – 1 × 3
|3 AB| = – 81

Question 18.
If λ = -2, determine the value of
Solution:
Given λ = -2
∴ 2λ = -4; λ² = (-2)²; 3λ² + 1 = 3 (4) + 1 = 13
6λ – 1 = 6(-2) – 1 = -13

expanding along R₁
0(0) + 4 (0 + 13) + 1 (-52 + 0) = 52 – 52 = 0
Aliter: The determinant value of a skew-symmetric matrix is zero

Question 19.
Determine the roots of the equation
Solution:

Given the determinant value is 0
⇒ 30(1 + x) (2 – x) = 0
⇒ 1 + x = 0 or 2 – x = 0
⇒ x = -1 or x = 2
So, x = -1 or 2.

Question 20.
Verify that det (AB) = (det A) (det B) for
Solution:


{(-20)(52) (-19) + (10)(38)(—49) + (2)(64)(-17)} – {(-49)(52) (2) + (-17)(38)(-20) + (-19)(64)(10)}
= (19760 – 18620 – 2176) – (-5096 + 12920 – 12160)
= (19760 + 5096 + 12160) – (18620 + 2176 + 12920)
= 37016 – 33716 = 3300 ….(3)
Now (1) × (2) = (3)
(i.e.,) (-33) (-100) = 3300
⇒ det (AB) = (det A), (det B)

Question 21.
Using cofactors of elements of the second row, evaluate |A|, where
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 Additional Problems

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.1

Question 1.
Represent graphically the displacement of
(i) 45 cm 30 ° north of east
(ii) 80 km, 60° south of west
Solution:
(i) 45 cm 30 0 north of east

(ii) 80 km 60° south of west

Question 2.
Prove that the relation R defined on the set V of all vectors by \(\vec{a}\) R \(\vec{b}\) if \(\vec{a}=\vec{b}\) is an equivalence relation on V.
Solution:
\(\vec{a}\) R \(\vec{b}\) is given as \(\vec{a}=\vec{b}\).
(i) \(\vec{a}\) = \(\vec{a}\) ⇒ \(\vec{a}\) R \(\vec{a}\)
(i.e.,) the relation is reflexive.

(ii) \(\vec{a}=\vec{b}\) ⇒ \(\vec{b}\) = \(\vec{a}\)
(i.e.,) \(\vec{a}\) R \(\vec{b}\) – \(\vec{b}\) R \(\vec{a}\)
So, the relation is symmetric.

(iii) \(\vec{a}=\vec{b} ; \vec{b}=\vec{c} \Rightarrow \vec{a}=\vec{c}\)
(i.e) \(\vec{a}\) R \(\vec{b}\) ; \(\vec{b}\) R \(\vec{c}\) ⇒ \(\vec{a}\) R \(\vec{c}\)
So the given relation is transitive
So, it is an equivalence relation.

Question 3.
Let \(\vec{a}\) and \(\vec{a}\) be the position vectors of points A and B. Prove that the position vectors of the points which trisect the line segment AB are
Solution:

Question 4.
If D and E are the midpoints of the sides AB and AC of a triangle ABC, prove that

Solution:

Question 5.
Prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side whose length is half of the length of the third side.
Solution:

Question 6.
Prove that the line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram.
Solution:

In a quadrilateral when opposite sides are equal and parallel it is a parallelogram So, PQRS is a parallelogram, from (1) and (2).

Question 7.
If \(\vec{a}\) and \(\vec{b}\) represent a side and a diagonal of a parallelogram, find the other sides and the other diagonal.
Solution:
OABC is a parallelogram where

Question 8.
If \(\overrightarrow{\mathrm{PO}}+\overrightarrow{\mathrm{OQ}}=\overrightarrow{\mathrm{QO}}+\overrightarrow{\mathrm{OR}}\), prove that the points P, Q, R are collinear.
Solution:

But Q is a common point.
⇒ P, Q, R are collinear.

Question 9.
If D is the midpoint of the side BC of a triangle ABC, prove that \(\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{A C}}=2 \overrightarrow{\mathbf{A D}}\)
Solution:
D is the midpoint of ∆ ABC.

Question 10.
If G is the centroid of a triangle ABC, prove that \(\overrightarrow{\mathrm{GA}}+\overrightarrow{\mathrm{GB}}+\overrightarrow{\mathrm{GC}}=0\)
Solution:
For any triangle ABC,
\(\overrightarrow{\mathrm{GA}}+\overrightarrow{\mathrm{GB}}+\overrightarrow{\mathrm{GC}}=0\)
Now G is the centroid of ∆ABC, which divides the medians (AD, BE and CF) in the ratio 2 : 1.

Question 11.
Let A, B, and C be the vertices of a triangle. Let D, E, and F be the midpoints of the sides BC, CA, and AB respectively. Show that \(\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{BE}}+\overrightarrow{\mathrm{CF}}=\overrightarrow{0}\)
Solution:
In ∆ABC, D, E, F are the midpoints of BC, CA, and AB respectively.

Question 12.
If ABCD is a quadrilateral and E and F are the midpoints of AC and BD respectively, then prove that \(\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{A D}}+\overrightarrow{\mathbf{C B}}+\overrightarrow{\mathbf{C D}}=4 \overrightarrow{\mathrm{EF}}\)
Solution:
ABCD is a quadrilateral in which E and F are the midpoints of AC and BD respectively.

Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.1 Additional Problems

Question 1.
Shown that the points with position vectors are collinear.
Solution:
To prove the points P, Q, R are collinear we have to prove that \(\overrightarrow{\mathrm{PQ}}\) = t \(\overrightarrow{\mathrm{PR}}\) where t is a scalar.
Let the given points be P, Q, R.

So, the points P, Q, R are collinear (i.e,) the given points are collinear.

Question 2.
If ABC and A’B’C’ are two triangles and G, G’ be their corresponding centroids, prove that \(\overrightarrow{\mathrm{AA}^{\prime}}+\overrightarrow{\mathrm{BB}^{\prime}}+\overrightarrow{\mathrm{CC}^{\prime}}=3 \overrightarrow{\mathrm{GG}}\)
Solution:
Let O be the origin.
We know when G is the centroid of ∆ ABC,

Question 3.
Prove using vectors the mid-points of two opposite sides of a quadrilateral and the mid-points of the diagonals are the vertices of a parallelogram.
Solution:
ABCD is a quadrilateral with position vectors
OA = \(\vec{a}\), OB = \(\vec{b}\), OC = \(\vec{c}\) and OD = \(\vec{d}\)
P is the midpoint of BC and R is the midpoint of AD.
Q is the midpoint of AC and S is the midpoint of BD.
To prove PQRS is a parallelogram. We have to prove that \(\overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{SR}}\)

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.2

Question 1.
Discuss the following relations for reflexivity, symmetricity and transitivity:

(i) The relation R defined on the set of all positive integers by “mRn if m divides n”.
Solution:
Let Z = {1, 2, 3, ……….}
R is a relation defined on the set of all positive integers by m R n if m divides n
R = { (m, n) : \(\frac{\mathrm{m}}{\mathrm{n}}\) for all m, n ∈ Z } n

(a) Reflexive:
m divides m for all m ∈ Z
∴ (m, m) ∈ R for all m ∈ Z
Hence R is reflexive

(b) Symmetric:
Let (m, n) ∈ R ⇒ m divides n
⇒ n = km for some integers k
But km need not divide m, ie. n need not divide m
∴ (n, m) ∉ R
Hence R is not symmetric.

(c) Transitive:
Let (m, n), (n, r) ∈ R
Then m divides n ⇒ n = km and
n divides r ⇒ r = k₁n
r = k₁(km) = (k₁k) m
m divides r
∴ (m, r) ∈ R
Hence R is transitive.

(ii) Let P denote the set of all straight lines in a plane. The relation R defined by “lRm if l is perpendicular to m”.
Solution:
P = {set of all straight lines in a plane}
lRm ⇒ l is perpendicular to m

(a) lRl ⇒ l is not perpendicular to l
⇒ It is not reflexive

(b) lRm ⇒ l is perpendicular to m
mRl ⇒ m is perpendicular to l
It is symmetric

(c) l perpendicular to m ⇒ m perpendicular to n ⇒ l is parallel to n It is not transitive

(iii) Let A be the set consisting of all the members of a family. The relation R defined by “aRb if a is not a sister of b”.
Solution:
Let F = Father,
M = Mother
G = Male child
H = Female child
A = { F, M, G, H }
The relation R is defined by
a R b if a is not a sister of b.
R = {(F, F), (F, M), (F, G), (F, H), (M, F), (M, M), (M, G), (M, H), (G, F), (G, M), (G, G), (G, H), (H, F), (H, M), (H, H)}

(a) Reflexive:
(F, F) , (M , M), (G, G), ( H, H ) ∈ R
∴ R is reflexive.

(b) Symmetric:
For (G, H) ∈ R, we have (H, G) ∉ R
∴ R is not symmetric.

(c) Transitive:
Suppose in a family if we take mother M , male child-G and female child-H.
H is not a sister of M ⇒ HRM, (H, M) ∈ R
M is not a sister of G ⇒ MRG, (M, G) ∈ R
But H is a sister of G ⇒ HRG, (H, G) ∉ R
Thus, for (H, M), (M, G) ∈ R
we have (H, G) ∉ R
∴ R is transitive.

(iv) Let A be the set consisting of all the female members of a family. The relation R
defined by “aRb if a is not a sister of b”.
Solution:
A = {set of all female members of a family}

(a) aRa ⇒ a is a sister of a
It is reflexive

(b) aRb ⇒ a is a sister of b
bRa ⇒ b is the sister of a
⇒ It is symmetric

(c) aRb ⇒ a is a sister of b bRc ⇒ b is a sister of c aRc ⇒ a can be the sister of c It is not transitive.

(v) On the set of natural numbers the relation R defined by “xRy if x + 2y = 1”.
Solution:
x + 2y = 1 for x, y ∈ N
There is no x , y ∈ N satisfying x + 2y = 1
∴ The relation R is an empty relation.
An empty relation is symmetric and transitive.
∴ R is symmetric and transitive.
R is not reflexive

Question 2.
Let X = {a, b, c, d} and R = {(a, a), {b, b), (a, c)}. Write down the minimum number
of ordered pairs to be included to R to make it
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Solution:
X = {a, b, c, d}
R = {(a, a), (b, b), (a, c)}
(i) To make R reflexive we need to include (c, c) and (d, d)
(ii) To make R symmetric we need to include (c, a)
(iii) R is transitive
(iv) To make R reflexive we need to include (c, c)
To make R symmetric we need to include (c, c) and (c, a) for transitive
∴ The relation now becomes
R = {(a, a), (b, b), (a, c), (c, c), (c, a)}
∴ R is equivalence relation.

Question 3.
Let A = {a, b, c} and R = {(a, a), (b, b), (a, c)}. Write down the minimum number of ordered pairs to be included to R to make it
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Solution:
Given A = {a, b, c }
R = { (a, a), (b, b),(a, c) }
(i) The minimum ordered pair to be included to R in order to make it reflexive is (c, c).
(ii) The minimum ordered pair to be included to R in order to make it symmetrical is (c, a).
(iii) R is transitive. We need not add any pair.
(iv) After including the ordered pairs (c, c),(c, a) to R the new relation becomes
R₁ = { (a, a), (b, b), (c, c) , (a, c) , (c, a) }
R₁ is reflexive symmetric and transitive.
∴ R₁ is an equivalence relation.

Question 4.
Let P be the set of all triangles in a plane and R be the relation defined on P as aRb if a is
similar to b. Prove that R is an equivalence relation.
Solution:
P = {set of all triangles in a plane}
aRb ⇒ a similar to b

(a) aRa ⇒ every triangle is similar to itself
∴ aRa is reflexive

(b) aRb ⇒ if a is similar to b ⇒ b is also similar to a.
⇒ It is symmetric

(c) aRb ⇒ bRc ⇒ aRc
a is similar to b and b is similar to c
⇒ a is similar to a
⇒ It is transitive
∴ R is an equivalence relation

Question 5.
On the set of natural numbers let R be the relation defined by aRb if 2a + 3b = 30. Write down the relation by listing all the pairs. Check whether it is
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Solution:
N = {set of natural numbers}
R ={(3, 8), (6, 6), (9, 4), (12, 2)}

(a) (3, 3) ∉ R ⇒ R is not reflexive
2a + 3b = 30
3b = 30 – 2a
b = \(\frac{30-2 a}{3}\)

(b) (3, 8) ∈ R(8, 3) ∉ R
⇒ R is not symmetric

(c) (a, b) (b, c) ∉ R ⇒ R is transitive
∴ It is not an equivalence relation.

Question 6.
Prove that the relation “friendship” is not an equivalence relation on the set of all people in Chennai.
Solution:
If a is a friend of b and b is a friend of c, then a need not be a friend of c.
a R b and b R c does not imply a R c.
∴ R is not transitive.
∴ The relation is not an equivalence relation.

Question 7.
On the set of natural numbers let R be the relation defined by aRb if a + b ≤ 6. Write down the relation by listing all the pairs. Check whether it is
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Solution:
Set of all natural numbers aRb if a + b ≤ 6
R= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)}
(i) (5, 1) ∈ R but(5, 5) ∉ R
It is not reflexive

(ii) aRb ⇒ bRa ⇒ It is symmetric

(iii) (4, 2), (2, 3) ∈ R ⇒ (4, 3) ∉ R
∴ It is not transitive

(iv) ∴ It is not an equivalence relation

Question 8.
Let A = {a, b, c}. What is the equivalence relation of smallest cardinality on A? What is the equivalence relation of largest cardinality on A?
Solution:
A = { a, b, c }
Let R₁ = { (a, a),(b, b),(c, c) }
Clearly, R₁ is reflexive, symmetric, and transitive.
Thus R₁ is the equivalence relation on A of smallest cardinality, n (R₁) = 3
Let R₂ = { (a, a), (b, b), (c, c), (a, b), (b, a), (b, c), (c, b), (c, a), (a, c)}

(i) Reflexive:
(a, a) , (b, b) , (c, c) ∈ R
∴ R₂ is reflexive.

(ii) Symmetric:
(a , b) ∈ R₂ we have (b, a) ∈ R₂
(b , c) ∈ R₂ we have (c, b) ∈ R₂
(c , a) ∈ R₂ we have (a, c) ∈ R₂
∴ R₂ is symmetric.

(iii) Transitive:
(a, b), (b, c) ∈ R₂ ⇒ (a, c) ∈ R₂
(b, c), (c, a) ∈ R₂ ⇒ (b, a) ∈ R₂
(c, a) , (a, b) ∈ R₂ ⇒ (c, b) ∈ R₂
∴ R₂ is transitive and R₂ is an equivalence relation of largest cardinality.
n (R₂) = 9

Question 9.
In the set Z of integers, define mRn if m – n is divisible by 7. Prove that R is an equivalence relation.
Solution:
mRn if m – n is divisible by 7
(a) mRm = m – m = 0
0 is divisible by 7
∴ It is reflexive

(b) mRn = {m – n) is divisible by 7
nRm = (n – m) = – {m – n) is also divisible by 7
It is symmetric

It is transitive
mRn if m – n is divisible by 7
∴ R is an equivalence relation.

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.2 Additional Questions

Question 1.
Find the range of the function.
f = {(1, x), (1, y), (2, x), (2, y), (3, z)}
Solution:
The range of the function is {x, y, z}.

Question 2.
For n, m ∈ N, It means that it is a factor of n & m. Then find whether the given relation is an equivalence relation.
Solution:
Since n is a factor of n. So the relation is reflexive.
When n is a factor of m (where m ≠ n) then m cannot be a factor of n.
So the relation is not symmetric when n is a factor of m and m is a factor of p then n will be a factor of p. So the given relation is transitive. So it is not an equivalence relation.

Question 3.
Verify whether the relation “is greater than” is an equivalence relation.
Solution:
You can do it yourself.