Category: Class 11

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 6 Two Dimensional Analytical Geometry Ex 6.2 Text Book Back Questions and Answers, Notes.

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.
Answer:
(i) with y – intercept – 4
The equation the line with slope m and having y – intercept b is y = mx + b ——- (1)
Given b = – 4
∴ (1) ⇒ y = m x – 4
Given this line passes through the point (1, 1)
∴ 1 = m 1 – 4 ⇒ m = 5
∴ The required equation is y = 5x – 4

(ii) Slope m = 3, passing through (x₁, y₁) = (1, 1)
Equation of the line is y – y₁ = m(x – x₁)
(i.e) y- 1 = 3(x – 1) ⇒ y – 1 = 3x – 3 ⇒ 3x – y = 2

(iii) (-2, 3)
The equation of line joining the two points (x₁, y₁) and (x₂, y₂) is

Given (x₁, y₁) = (1, 1), (x₂, y₂) = (- 2, 3)
∴ The equation of the required line is

– 3 (y – 1) = 2 (x – 1)
– 3y + 3 = 2x – 2
2x + 3y – 2 – 3 = 0
2x + 3y – 5 = 0

(iv) The perpendicular from the origin makes an angle 60° with x – axis
The equation of the line in the normal form is x cos α + y sin α = p ——- (1)
Given α = 60°
∴ cos 60° = \(\frac{1}{2}\) , sin 60° = \(\frac{\sqrt{3}}{2}\)
(1) ⇒ x . \(\frac{1}{2}\) + y . \(\frac{\sqrt{3}}{2}\) = p
x + √3 = 2p —— (2)
This line passes through the point (1,1)
∴ 1 + √3 = 2p
Substituting for p in equation (2)
∴ The required equation is x + √3y = 1 + √3

Question 2.
If P (r, c) is mid point of a line segment between the axes then show that \(\frac{x}{\mathbf{r}}+\frac{\mathbf{y}}{\mathbf{c}}\) = 2
Answer:
Let AB be the line segment intercepted between the axes. Let A be (a, 0) and B be (0, b)

Given P (r, c) is the mid point of AB

The equation of the line AB having x-intercept a and y-intercept b is \(\frac{x}{a}+\frac{y}{b}\) = 1
Substituting for a, b in the above equation, we have

Question 3.
Find the equation of the line passing through the point (1, 5) and also divides the line segment between the coordinate axes in the ratio 3 : 10.
Answer:
Let the line divide the coordinate axis in the ratio 3 : 10.
∴ x-intercept = 3k
y-intercept = 10k
∴ The equation of the straight line is \(\frac{x}{3 k}+\frac{y}{10 k}\) = 1
This line passes through the point P (1, 5).

∴ The required equation is

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 \(\frac{1}{\mathbf{p}^{2}}=\frac{1}{\mathbf{a}^{2}}+\frac{1}{\mathbf{b}^{2}}\)
Answer:
The equation of the line with x – intercept a and y – intercept b is \(\frac{x}{a}+\frac{y}{b}\) = 1 ——- (1)
The length of the perpendicular from the origin (0, 0) to the line (1) is

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. Find
(ii) the value of C for 98.6° F and
(iii) the value of F for 38° C.
Answer:
(i) Choose Celsius degree along the x-axis and Fahrenheit degree along the y-axis.
Given a Freezing point in Celsius = 0°C
The freezing point in Fahrenheit degree = 32° F
∴ The Freezing point is (0°, 32°)
Also given Boiling point in Celsius = 100°C
The boiling point in Fahrenheit = 212° F
∴ The Boding point is (100°, 212°)
Let C denote the Celsius degree and F denote the Fahrenheit degree.
The equation of the path connecting the freezing point (0°, 32°) and the boiling point (100°, 212° ) is

which is the required relation connecting C and F.

(ii) To find the value of C for 98.6° F
(1) For 98.6° F To find C.

(iii) To find the value of F for 38° C,
For 38° C To find F

Question 6.
An object was launched from a place P in constant speed to hit a target. At the 15^th second, it was 1400 m from the target, and at the 18^th second 800 m away. Find
(i) the distance between the place and the target
(ii) The distance covered by it in 15 seconds
(iii) time taken to hit the target.
Answer:
Let us take the time T along the x-axis and the Distance D along the y-axis.
Given when time T = 15 s , the distance D = 1400 m
The corresponding point is (15, 1400)
Also when time T = 18 s , the distance D = 800 m.
The corresponding point is (18, 800)

(i) The distance between the place and the target:
∴ The relation connecting T and D is the equation of the straight line joining the points (15, 1400) and (18, 800)

To find the distance between the target and the place, Put T = 0 in equation (1)

4400 – D = 0 ⇒ D = 4400 m.
Required distance = 4400 m.

(ii) The distance covered by it in 15 seconds:
Put T = 15 in the above equation
15 = \(\frac{1400-\mathrm{D}}{200}\) + 15
∴ \(\frac{1400-\mathrm{D}}{200}\) = 0 ⇒ D = 1400 m.

(iii) Time taken to hit the Target:
When the target is reached D = 0
∴ (1) ⇒ T = \(\frac{1400-0}{200}\) + 15
T = \(\frac{1400}{200}\) + 15
T = 7 + 15 = 22 seconds
∴ The time taken to hit the target is 22 seconds

Question 7.
The population of the 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).
Answer:
Let us choose the year along the x-axis and the population of the city along the y-axis.
Given In the year 2005 population is 1,35,000
The corresponding point is (2005, 1,35,000)
In the year 2010, population is 1,45,000
The corresponding point is (2010, 1,45,000)
Let Y denote the year and P denote the population.
The relation connecting Y and P is the equation of the straight line joining the points (2005, 1,35,000) and (2010, 1,45,000)

When y = 2015

P = 1,35,000 + 10 × 2,000
P = 1,35,000 + 20,000 = 1,55,000
∴ The 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.
Answer:
Given length of the perpendicular p = 12
Angle made by the perpendicular α = 30°
The equation the straight line in the normal form is
x cos α + y sin α = p
∴ The required equation of the straight line is
x cos 30° + y sin 30° = 12
x\(\frac{\sqrt{3}}{2}\) + y × \(\frac{1}{2}\) = 12
√3x + y = 24

Question 9.
Find the equation of the straight lines passing through (8, 3) and having intercepts whose sum is 1.
Answer:
Let a and b be the x and y-intercepts of the line.
Given a + b = 1
b = 1 – a —— (1)
The equation of the straight line is

The line passes through the point (8, 3)

8(1 – a) + 3a = a(1 – a)
8 – 8a + 3a = a – a²
a² – 5a + 8 – a = 0
a² – 6a + 8 = 0
a² – 4a – 2a + 8 = 0
a(a – 4) – 2(a – 4) = 0
(a- 4) (a – 2) = 0
a = 4 or a = 2
When a = 2, b = 1 – 2 = – 1
When a = 4, b = 1 – 4 = – 3
∴ The equation of the straight lines are
x – 2y = 2 and 3x – 4y = 12

Question 10.
Show that the points (1, 3), (2, 1) and \(\left(\frac{1}{2}, 4\right)\) are collinear, by using
(i) concept of slope
(ii) using a straight line and
(iii) any other method.
Answer:
Let the given points be A (1 , 3), B (2 , 1) and \(\left(\frac{1}{2}, 4\right)\)

(i) Slope Method:
A (1 , 3 ), B (2 , 1 ), C\(\left(\frac{1}{2}, 4\right)\)

From equations (1) and (2)
Slope of AB = Slope of BC
∴ The given points A, B, C are collinear.

(ii) Using a straight line
A(1 , 3) , B(2 , 1) , C\(\left(\frac{1}{2}, 4\right)\)
The equation of the straight line joining the points
A( 1 , 3) , B(2 , 1) is

-2(x- 1) = y – 3
– 2x + 2 = y – 3
2x + y – 2 – 3 = 0
2x+ y – 5 = 0 ——- (2)
Substituting the third point C \(\left(\frac{1}{2}, 4\right)\) in equation (2)
we have 2\(\left(\frac{1}{2}\right)\) + 4 – 5 = 0
1 + 4 – 5 = 0
0 = 0
∴ The third point C\(\left(\frac{1}{2}, 4\right)\) lies on the straight line AB.
Hence the points A , B , C are collinear.

(iii) Distance method:
A(1 , 3) , B(2 , 1) , C\(\left(\frac{1}{2}, 4\right)\)


Thus BA + AC = BC
∴ 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 the points on the line which are 13 units away from A.
Answer:
Slope of the line m = tan θ = \(\frac{5}{12}\)
sin θ = \(\frac{5}{13}\), cos θ = \(\frac{12}{13}\)

The parametric equation of the line passing through the point (1, 2) making angle θ with x – axis is

Any point on this line is

(1 + r cos θ, 2 + r sin θ) ……… (1)
where r is the distance of any point from A (1, 2) on the line.
To find the point which is 13 units away from A (1, 2) on the line.

Substitute r = ± 13, cos θ = \(\frac{12}{13}\), sin θ = \(\frac{5}{13}\) in equation (1)
Required point = \(\left(1 \pm 13\left(\frac{12}{13}\right), 2 \pm 13\left(\frac{5}{13}\right)\right)\)
= (1 ± 12, 2 ± 5)
= (1 + 12, 2 + 5)
= (1 + 12, 2 + 5) or (1 – 12, 2 – 5)
= (13, 7) or (- 11, – 3)

Question 12.
A 150 m long train is moving with a constant velocity of 12.5 m/s.
(i) The equation of motion of the train.
(ii) Time taken to cross a pole.
(iii) The time to cross the bridge of length 850m is?
Answer:
Length of the train = 150 m
Constant velocity of the train = 12.5 m/s

(i) The equation of motion of the train:
Take time in seconds along the x-axis and distance in meters along the y-axis.
Let the train be at the origin.
∴ Length of the train = 150 m is the negative y-intercept
b = -150
The slope of the motion of the train m = 12.5 m/s
The equation of the line with slope-intercept form is
y = mx + b
∴ y = 12.5x – 150
which is the required equation of motion of the train.

(ii) Time taken to cross a pole:
The equation of motion is y = 12.5 x – 150
To find the time taken to cross the pole, Put y = 0
0 = 12.5 x – 150 ⇒ 12.5 x = 150
⇒ x = \(\frac{150}{12.5}\) = 12 sec

(iii) The time taken to cross the bridge of length 850 m
The equation of motion is y = 12.5 x – 150
To find the time taken to cross the bridge of length 850 m, put y = 850
850 = 12.5 x – 150
12.5 x = 850 + 150 = 1000
x = \(\frac{1000}{12.5}=\frac{10000}{125}\) = 80 sec

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.

(a) Draw a graph showing the results.
(b) Find the equation relating the length of the spring to the weight on it.
(c) What is the actual length of the spring?
(d) If the spring stretches to 9 cm long, how much weight should be added?
(e) How long will the spring be when 6 kilograms of weight on it?
Answer:
Choose the weight along the x-axis and Length along the y-axis.
(a)

(b) The points are(2, 3), (4, 4), (5, 4.5), (8, 6) The relation connecting weight and Length is the equation of the straight line joining the points (2, 3) and (4, 4)

x – 2 = 2(y – 3)
x – 2 = 2y – 6
x – 2y + 6 – 2 = 0
x – 2y + 4 = 0 —– (1)
which the required relation connecting weight and length.

(c) To find the actual length of the spring, put weight x = 0 in equation (1)
0 – 2y + 4 = 0 ⇒ 2y = 4 ⇒ y = 2
∴ The actual length of the spring is 2 cm.

(d) If the spring stretch to 9 cm long, To find the required weight, put y = 9, in equation (1)
(1) ⇒ x – 2 (9) + 4 = 0
x – 18 +4 = 0 ⇒ x = 14
Weight to be added is 14 kg.

(e) Next we find the length of the string when a weight of 6 kg is added.
Put x = 6 in equation (1)
6 – 2y + 4 = 0 ⇒ 2y = 10 ⇒ y = 5cm
∴ Required length is 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.
Answer:
(i) Find the equation relating the quantity of gas in the cylinder to the days.
Given Total weight of cylinder = 29.5 kg
Weight of the gas inside the cylinder = 14.2 kg
Let x denote the number of days of consumption of the gas, y denote the quantity of gas inside a cylinder.
Initially x = 0 then y = 14.2
The corresponding point is (0, 14.2)
The gas inside the cylinder lasts for 24 days
∴ When x = 24, we have y = 0
The corresponding point is (24, 0)
∴ The linear relation between the quantity of gas in the cylinder to the number of days of consumption is the equation of the line joining the points (0, 14.2 ) and (24, 0).

which is the required relation

(ii) Draw the graph for the first 96 days:
The relation connecting the quantity of gas to the number of days of consumption is
y = –\(\frac{71}{120}\)x + 14.2
Let f(x) = –\(\frac{71}{120}\) x + 14.2
Here f(x) is a periodic function of period 24
∴ f(x + 24) = f(x)
When x = 0
f(0) = –\(\frac{71}{120}\) × 0 + 14.2 ⇒ y = 14.2
The corresponding point is (0, 14.2)

When x = 24
f(24) = –\(\frac{71}{120}\) × 0 + 14.2 ⇒ y = 14.2
⇒ f(24) = –\(\frac{71}{5}\) + 14.2
= – 14.2 + 14.2 = 0 ⇒ y = 0
Corresponding point is (24 , 0)

When x = 48
f(48) = f(24 + 24 + 0) = f(24 + 0)
= f(o) = o
Corresponding point is (48, 0)

When x = 72
f(72) = f(24 + 24 + 24 + 0)
= f(24 + 24 + 0) = f(24 + 0)
= f(0) = 0
Corresponding point is (72, 0)
The required graph is

Question 15.
In the shopping mall, there is a hall of cuboid shape with dimension 800 x 800 x 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 height at which the escalator changes its direction
(iii) the slopes of the escalator at the turning points.
Answer:
Give the dimension of the cube = 800 × 800 × 720

(i) Minimum total length of the escalator:

The path of the escalator is
from OA to AB to BC to CD
OE = 800, EA = \(\frac{1}{4}\) × height of the building
EA = \(\frac{1}{4}\) × 720 = 180
Since there are four steps for the escalator
∴ OA² = OE² + EA²
= 800² + 180²
= (40 × 20)² + (9 × 20)²
= 40² × 20² + 92 × 20²
= 20² (40² + 9²)
= 20² ( 1600 + 81)
= 20² × 1681
OA² = 20² × 41²
OA = \(\sqrt{20^{2} \times 41^{2}}\) = 20 × 41 = 820
Since ∆OAE ≡ ∆ ABB’ ≡ ∆ BCC ≡ ∆ CDD’
We have OA = AB = BC = CD
Total length of the escalator
= OA + AB + BC + CD
= 820 + 820 + 820 + 820
= 4 × 820
= 3280
Minimum length of the escalator = 3280 units

(ii) Lengths at which the escalator changes its direction:
The heights at which the escalator changes its direction
First step EA = \(\frac{1}{4}\) × 720 = 180 unis
Second step FB = \(\frac{1}{2}\) × 720 = 360 units
Third step GC = \(\frac{1}{3}\) × 720 = 540 units

(iii) The slopes of the escalator at the turning points.
In the right angle ∆ OAE
OE = 800, EA = 180
Let ∠ AOE = θ
The slope of the escalator OE 180

∴ Slope at each turning points is \(\frac{9}{40}\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 6 Two Dimensional Analytical Geometry Ex 6.1 Text Book Back Questions and Answers, Notes.

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 coordinates of a moving point P is
(i) (9 cos α, 9 sin α)
(ii) (9 cos α, 6 sin α)
Answer:
(i) (9 cos α, 9 sin α)
Let P (h , k) be any point on the required path. Then by the given data we have h = 9 cos α, k = 9 sin α

The locus of P (h , k) is obtained by replacing h by x and k by y.
∴ The required locus becomes x² + y² = 81

(ii) ( 9 cos α, 6 sin α)
Let P (h, k) be any point on the required path. Then by the given data we have h = 9 cos α, k = 6 sin α

The locus of p(h , k) is obtained by replacing h by x and k by y
∴ The required locus is \(\frac{x^{2}}{81}+\frac{y^{2}}{36}\) = 1

Question 2.
Find the locus of a point P that moves a constant distant of
(i) two units from the x-axis
(ii) three units from the y-axis.
Answer:
(i) Two units from x-axis:

Let P (h, k) be any point on the required path. From the given data, we have k = 2
The locus of P (h, k) is obtained by replacing h by x and k by y.
∴ The required locus is y = 2

(ii) Three units from y-axis:

Let (h, k) be any point on the required path. From the given data, we have h = 3.
The locus of P (h, k) is obtained by replacing h by x and k by y.
∴ The required locus is x = 3

Question 3.
If 6 is a parameter, find the equation of the locus of a moving point, whose coordinates are x = a cos³θ, y = a sin³θ.
Answer:
The given moving points is (a cos³θ, a sin³θ)

Question 4.
Find the values of k and b. If the points P(-3, 1) and Q (2, b) lie on the locus of x² – 5x + ky = 0
Answer:
Given P (-3, 1) lie on x² – 5x + ky = 0
⇒ (-3)² – 5(-3) + k(1) = 0
9 + 15 + k = 0 ⇒ k = -24
Q (2, b) lie on x² – 5x + ky = 0
(2)² – 5(2) + k(b) = 0 ⇒ 4 – 5(2) – 24b = 0

Question 5.
A straight rod of length 8 units slides with its ends A and B always on the x and y axes respectively, then find the locus of the midpoint of the line segment AB.
Answer:
Given A and B are the ends of the straight rod of length 8 unit on the x and y-axes. Let A be (a, 0) and B (0, b).

Let M (h, k) be the midpoint of AB (h,k) =

In the right-angled ∆ OAB
AB² = OA² + OB²
8² = a² + b²
64 = (2h)² + (2k)
64 = 4h² + 4k²
h² + k² = \(\frac{64}{4}\) = 16
The locus of M (h, k) is obtained by replacing h by x and k by y
∴ The required locus is x² + y² = 16

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.
Answer:
Let P (h, k) be the moving point
Let the given point be A (3, 5) and B (1, -1)
We are given PA² + PB² = 20
⇒ (h – 3)² + (k – 5)² + (h – 1)² + (k + 1)² = 20
⇒ h² – 6h + 9 + k² – 10k + 25 + h² – 2h + 1 + k² + 2k + 1 = 20
(i.e.) 2h² + 2k² – 8h – 8k + 36 – 20 = 0
2h² + 2k² – 8h – 8k + 16 = 0
(÷ by 2 ) h² + k² – 4h – 4k + 8 = 0
So the locus of P 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.
Answer:

Given A (1, – 6) and B (4, – 2).
Let P (h, k) be a point such that the line segment AB subtends a right angle at P.
∴ ∆ APB is a right-angled triangle.
AB² = AP² + BP² ………… (1)
AB² = (4 – 1)² + (- 2 + 6)²
AB² = 3² + 4² = 9 + 16 = 25
AP² = (h – 1)² + (k + 6)²
BP² = (h – 4)² + (k + 2)²
(1) ⇒
25 = (h – 1)² + (k + 6)² + (h – 4)² + (k + 2)²
25 = h² – 2h + 1 + k² + 12k + 36 + h² – 8h + 16 + k² + 4k + 4
25 = 2h² + 2k² – 10h + 16k + 57
2h² + 2k² – 10h + 16k + 57 – 25 = 0
2h² + 2k² – 10h + 16k + 32 = 0
h² + k² – 5h + 8k + 16 =0
The locus of P (h , k) is obtained by replacing h by x and k by y.
∴ The required locus 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 segment OR.
Answer:

Let the variable point R be (x, y). Let M (h, k) be the midpoint of R.

But R(x , y) is a point on y² = 4x
∴ (2k)² = 4(2h)
4k² = 8h
k² = 2h
The locus of M (h , k) is obtained by replacing h by x and k by y.
∴ The required locus is y² = 2x

Question 9.
The coordinates of a moving point P are (\(\frac{\mathbf{a}}{2}\) (cosec θ + sin θ), \(\frac{\mathbf{b}}{2}\) (cosec θ – sin θ) where θ is a variable parameter. Show that the equation of the locus P is b² x² – a² y² = a² b²
Answer:
Let the moving point P be (h, k)
By the given data we have


The locus of P ( h , k ) is obtained by replacing h by x and k by y
∴ The required locus is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1
b² x² – a² y² = a² b²

Question 10.
If P (2, – 7) is given point and Q is a point on 2x² + 9y² = 18 then find the equations of the locus of the midpoint of PQ.
Answer:
Given P is (2, -7) and let Q be (x, y)
Given that Q is a point on 2x² + 9y² = 18
Let M (h, k) be the midpoint of PQ

2h = 2 + x ,                   2k = – 7 + y
x = 2h – 2,                     y = 2k + 7
But Q(x, y) is a point on 2x² + 9y² = 18
∴ 2 (2h – 2)² + 9 (2k + 7)² = 18
2 [4h² – 8h + 4] + 9 [4k² + 28k + 49] = 18
8h² – 16h + 8 + 36k² + 252k + 441 = 18
8h² + 36k² – 16h + 252k + 449 = 18
8h² + 36k² – 16h + 252k +431 =0
The locus of M ( h , k ) is obtained by replacing h by x and k by y.
∴ The required 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 P is a variable point on RQ with RP = b, PQ = a, then find the equation of locus of P.
Answer:
Given R is any point on the x-axis and Q is any point on the y-axis.
Let R be (x, 0) and Q be (0, y)
Let P (h, k ) be the variable point on RQ such that RP = b and PQ = a

The point P ( h, k ) divides the line joining the points R(x, 0) and Q (0, y) in the ratio b : a


Substituting in equation (1), we have

The locus of P (h, k) is obtained by replacing h by x and k by y.
∴ The required locus is \(\frac{x^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}\) = 1

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.
Answer:
Given P (6, 2), Q (-2 , 1), R (a, b) are the vertices of ∆ PQR where R (a, b) lies on y = x² – 3x + 4
∴ b = a² – 3a + 4 (1)
Let the centroid of ∆ PQR be G (h, k)

Substituting in equation (1) we have
(1) ⇒ 3k – 3 = (3h – 4)² – 3(3h – 4) + 4
3k – 3 = 9h² – 24h + 16 – 9h + 12 + 4
9h² – 33h + 32 – 3k + 3 = 0
9h² – 33h – 3k + 35 = 0
The locus of G (h, k) is obtained by replacing h by x and k by y.
∴ The required locus 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.
Answer:
Let Q be (a, b) lying on the locus
x² + y² + 4x – 3y + 7 = 0
∴ a² + b² + 4a – 3b + 7 = 0
Let the movable point P be (h , k)
Given P divides OQ externally in the ratio 3 : 4

Substituting in equation (1) we have

h² + k² – 12h + 9k + 63 = 0
The locus of P(h, k) is obtained by replacing h by x and k by y.
∴ The required locus is
x² + y² – 12x + 9y + 63 = 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)
Answer:
Given that the required pointis3unitsfrom x-axis and 5 units from the point P (5, 1). Let Q (h, 3) and K (h,- 3) be the required points.

∴ PQ = 5
\(\sqrt{(5-\mathrm{h})^{2}+(1-3)^{2}}\) = 5
(5 – h)² + (- 2)² = 25
25 – 10h + h² + 4 = 25
h² – 10h + 29 – 25 = 0
h² – 10h + 4 = 0

PR = 5

(5 – k)² + 4² = 25
25 – 10k + k² + 16 = 25
k² – 10k + 16 = 0


∴ R (8, – 3), (2, – 3)
∴ Required points are
(5 + √21, 3), (5 – √21, 3), (8, – 3), (2, – 3)

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 to the locus of the moving point.
Answer:
Let A be (4, 0) and B be (-4, 0). Let the moving point be p(h, k)
Given PA + PB = 10

16h² + 200h + 625 = 25[h² + 8h + 16 + k²]
16h² + 200h + 625 = 25h² + 200h + 400 + 25k²
9h² + 26k² = 225

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5

Choose the correct or the most suitable answer:

Question 1.
The value of 2 + 4 + 6 + …………… + 2n is
(1) \(\frac{n(n-1)}{2}\)
(2) \(\frac{n(n+1)}{2}\)
(3) \(\frac{2 n(2 n+1)}{2}\)
(4) n(n + 1)
Answer:
(4) n(n + 1)

Explaination:
2 + 4 + 6 + ……… + 2n = 2(1 + 2 + 3 + ………….. + n)
= 2 × \(\frac{n(n+1)}{2}\)
= n(n + 1)

Question 2.
The coefficient of x⁶ in (2 + 2x)¹⁰ is
(1) 10C₆
(2) 2₆
(3) 10C₆2⁶
(4) 10C₆2¹⁰
Answer:
(4) 10C₆2¹⁰

Explaination:

Question 3.
The coefficient of x⁸ y¹² in the expansion of (2x + 3y)²⁰ is
(1) 0
(2) 2⁸ 3¹²
(3) 2⁸ 3¹² + 2¹² 3⁸
(4) 20C₈ 2⁸ 3¹²
Answer:
(4) 20C₈ 2⁸ 3¹²

Explaination:

Question 4.
If nC₁₀ > nC_r for all possible r then a value of n is
(1) 10
(2) 21
(3) 19
(4) 20
Answer:
(4) 20

Explaination:
Out of ¹⁰C₁₀, ²¹C₁₀, ¹⁹C₁₀ and ²⁰C₁₀, ²⁰C₁₀ is larger.

Question 5.
If a is the Arithmetic mean and g is the Geometric mean of two numbers then
(1) a ≤ g
(2) a ≥ g
(3) a = g
(4) a > g
Answer:
(2) a ≥ g

Explaination:
Given Arithmetic mean = a,
Geometric mean = g
We have A. M ≥ G. M
∴ a ≥ g

Question 6.
If (1 + x²)² (1 + x)ⁿ = a₀ + a₁x + a₂x² + ………… + x^{n + 4} and if a₀, a₁, a₂ are in A. P then n is
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(3) 3

Explaination:

n² – 5n + 6 = 0
(n – 2) (n – 3) = 0
n = 2 or n = 3

Question 7.
If a, 8, b are in A .P , a, 4 , b are in G. P and if a, x ,b are in H . P then x is
(1) 2
(2) 1
(3) 4
(4) 16
Answer:
(1) 2

Explaination:
Given a, 8, b are in A. P ∴ 2 × 8 = a + b ⇒ a + b = 16 ——— (1)
Also a, 4, b are in G.P ∴ 4² = a . b ⇒ ab = 16 ——- (2)
Also a, x, b are in H.P. ∴ \(\frac{1}{a}, \frac{1}{x}, \frac{1}{b}\) are in A.P

Question 8.

(1) A. P
(2) G.P
(3) H.P
(4) AGP
Answer:
(3) H.P

Explaination:

Question 9.
The H.M of two positive numbers whose A.M and G.M are 16,8 respectively is
(1) 10
(2) 6
(3) 5
(4) 4
Answer:
(4) 4

Explaination:
Let a, b be the two numbers. Given A. M = \(\frac{a+b}{2}\) = 16
G.M = \(\sqrt{a b}\) = 8
a+b = 16 × 2 = 32
ab = 8² = 64
H.M = \(\frac{2 a b}{a+b}\)
= \(\frac{2 \times 64}{32}\) = 2 × 2 = 4

Question 10.
If S denote the sum of n terms of an A. P whose common difference is d, the value of S_n – 2S_{n- 1} + S_{n – 2} is
(1) d
(2) 2d
(3) 4d
(4) d²
Answer:
(1) d

Explaination:

= a + (n – 1) d – (a + (n – 2)d)
= a + (n – 1) d – a – (n – 2)d
= a + nd – d – a – nd + 2d = d

Question 11.
The remainder when 38¹⁵ is divided by 13 is
(1) 12
(2) 1
(3) 11
(4) 5
Answer:
(1) 12

Explaination:
38¹⁵ = (39 – 1)¹⁵ = 39¹⁵ – 15C₁ 39¹⁴(1) + 15C₂ (39)¹³(1)² – 15C₃ (39)¹²(1)³ ….. + 15C₁₄ (39)¹(1) – 15C₁₅(1)
Except -1 all other terms are divisible by 13.
∴ When 1 is added to it the number is divisible by 13. So the remainder is 13 – 1 = 12.

Question 12.
The nth term of the sequence 1, 2, 4, 7, 11, ………….. is
(1) n² + 3n² + 2n
(2) n³ – 3n² + 3n
(3) \(\frac{n(n+1)(n+2)}{3}\)
(4) \(\frac{n^{2}-n+2}{2}\)
Answer:
(4) \(\frac{n^{2}-n+2}{2}\)

Explaination:

Question 13.
The sum up to n terms of the series

(1) \(\sqrt{2 n+1}\)
(2) \(\frac{\sqrt{2 n+1}}{2}\)
(3) \(\sqrt{2 n+1}-1\)
(4) \(\frac{\sqrt{2 n+1}-1}{2}\)
Answer:
(4) \(\frac{\sqrt{2 n+1}-1}{2}\)

Explaination:

Question 14.
The n^th term of the sequence

(1) 2ⁿ – n – 1
(2) 1 – 2^-n
(3) 2^-n + n – 1
(4) 2^n-1
Answer:
(2) 1 – 2^-n

Explaination:

Question 15.
The sum up to n terms of the series

(1) \(\frac{\mathbf{n}(\mathbf{n}+1)}{2}\)
(2) 2n (n + 1)
(3) \(\frac{\mathbf{n}(\mathbf{n}+1)}{\sqrt{2}}\)
(4) 1
Answer:
(3) \(\frac{\mathbf{n}(\mathbf{n}+1)}{\sqrt{2}}\)

Explaination:

Question 16.
The value of the series

(1) 14
(2) 7
(3) 4
(4) 6
Answer:
(1) 14

Explaination:

Question 17.
The sum of an infinite G.P is 18. If the first term is 6 the common ratio is
(1) \(\frac{1}{3}\)
(2) \(\frac{2}{3}\)
(3) \(\frac{1}{6}\)
(4) \(\frac{3}{4}\)
Answer:
(2) \(\frac{2}{3}\)

Explaination:
Let the geometric series be a, ar, ar², …………… ar^n-1
Given a = 6, S_∞ = 18

Question 18.
The coefficient of x⁵ in the series e^-2x is
(1) \(\frac{2}{3}\)
(2) \(\frac{3}{2}\)
(3) \(-\frac{4}{15}\)
(4) \(\frac{4}{15}\)
Answer:
(3) \(-\frac{4}{15}\)

Explaination:

Question 19.
The value of

(1) \(\frac{e^{2}+1}{2 e}\)
(2) \(\frac{(e+1)^{2}}{2 e}\)
(3) \(\frac{(e-1)^{2}}{2 e}\)
(4) \(\frac{e^{2}+1}{2 e}\)
Answer:
(3) \(\frac{(e-1)^{2}}{2 e}\)

Explaination:

Question 20.
The value of

(1) log \(\left(\frac{5}{3}\right)\)
(2) \(\frac{3}{2}\) log \(\left(\frac{5}{3}\right)\)
(3) \(\frac{5}{3}\) log \(\left(\frac{5}{3}\right)\)
(4) \(\frac{2}{3}\) log \(\left(\frac{2}{3}\right)\)
Answer:
(2) \(\frac{3}{2}\) log \(\left(\frac{5}{3}\right)\)

Explaination:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.4

Question 1.
Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid.
(i) \(\frac{1}{5+x}\)
Answer:

(ii) \(\frac{2}{(3+4 x)^{2}}\)
Answer:

(iii) (5 + x²)^2/3
Answer:

(iv) \((x+2)^{-\frac{2}{3}}\)
Answer:

Question 2.
Find \(\sqrt[3]{1001}\) approximately. (two decimal places)
Answer:

Question 3.
Prove that \(\sqrt[3]{x^{3}+6}-\sqrt[3]{x^{3}+3}\) is approximately equal to \(\frac{1}{x^{2}}\) when x is sufficiently large.
Answer:

Question 4.
Prove that \(\sqrt{\frac{1-x}{1+x}}\) is approximately equal to 1 – x + \(\frac{x^{2}}{2}\) when x is very small.
Answer:

Question 5.
Write the first 6 terms of the exponential series
(i) e^5x
Answer:

(ii) e^-2x
Answer:

(iii) e^x/2
Answer:

Question 6.
Write the first 4 terms of the logarithmic series.
(i) log (1 + 4x)
(ii) log (1 – 2x)
(iii) log \(\left(\frac{1+3 x}{1-3 x}\right)\)
(iv) log \(\left(\frac{1-2 x}{1+2 x}\right)\)
Find the intervals on which the expansions are valid.
Answer:
(i) log ( 1 + 4x )

(ii) log (1 – 2x)

(iii) log \(\left(\frac{1+3 x}{1-3 x}\right)\)

(iv) log \(\left(\frac{1-2 x}{1+2 x}\right)\)

Question 7.

Answer:

Question 8.

Answer:

Question 9.
Find the coefficient of x⁴ in the expansion of \(\frac{3-4 x+x^{2}}{e^{2 x}}\)
Answer:

Question 10.
Find the value of
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3

Question 1.
Find the sum of the first 20 terms of the arithmetic progression having the sum of first 10 terms as 52 and the sum of the first 15 terms as 77 .
Answer:
Sum of the first n terms of an Arithmetic progression is S_n = \(\) [2a + (n – 1)d]
Given S₁₀ = 52
52 = \(\frac{10}{2}\) [ 2a + (10 – 1) d ]
52 = 5 [2a + 9d]
52 = 10a + 45d ……….. (1)
Also given S₁₅ = 77
77 = \(\frac{15}{2}\) [2a + (15 – 1)d]
77 = \(\frac{15}{2}\) [2a + 14d]
77 = 15 [a + 7d]
77 = 15a + 105d ………….. (2)

Substituting the value of d in (1)

Question 2.
Find the sum up to the 17^th term of the series
Answer:

Question 3.
Compute the sum of first n terms of the following series.
(i) 8 + 88 + 888 + 8888 + . . . . . . .
Answer:

(ii) 6 + 66 + 666 + 6666 + . . . . . . .
Answer:

Question 4.
Compute the sum of first n terms of 1 + (1 + 4) + (1 + 4 + 4²) + (1 + 4 + 4² + 4³) + …………..
Answer:
The given series is 1 + (1 + 4) + (1 + 4 + 4²) + (1 + 4 + 4² + 4³) + …………..
n^th term of the series is T_n = 1 + 4 + 4² + 4³ + ………………. + 4^n-1

Sum to n terms of the series
S_n = T₁ + T₂ + ………….. + T_n

Question 5.
Find the general terms and sum to n terms of the sequence 1, \(\frac{4}{3}\), \(\frac{7}{9}\), \(\frac{10}{27}\), ………….
Answer:
The given sequence as 1, \(\frac{4}{3}\), \(\frac{7}{9}\), \(\frac{10}{27}\), ………….
Consider the terms in the numerator 1, 4, 7, 10, ……………….. which is an Arithmetic progression
with first term a = 1, common difference d = 4 – 1 = 3
a_n = a + (n – 1)d = 1 + (n – 1)3 = 1 + 3n – 3 = 3n – 2
The given sequence can be written as 1, (1 + 3)\(\left(\frac{1}{3}\right)\), (1 + 2 × 3)\(\left(\frac{1}{3}\right)^{2}\) (1 + 3 × 3) \(\left(\frac{1}{3}\right)^{3}\), …………….
where \(1, \frac{1}{3}, \frac{1}{3^{2}}, \frac{1}{3^{3}}, \ldots \ldots\) is a G. P with first term a = 1 , common ratio r = \(\frac{1}{3}\)
∴ The given sequence is an Arithmetico – Geometric sequence.


which is an arithmetic – geometric sequence.
∴ The sum of first n terms of the arithmetico – geometric sequence is given by

Question 6.
Find the value of n if the sum to n terms of the series \(\sqrt{3}+\sqrt{75}+\sqrt{243}+\ldots \ldots . . . \text { is } 435 \sqrt{3}\)
Answer:

Question 7.
Show that the sum of ( m + n)^th and ( m – n)^th term of an A.P is equal to twice the m^th term.
Answer:
Let the A.P. be a, a + d, a + 2d, ……..
t_{m + n} = a + (m + n – 1)d
t_{m – n} = a + (m – n – 1)d
t_m = a + (m – 1)d
2t_m = 2[a + (m – 1)d]
To prove t_{m + n} + t_{m – n} = 2t_m
LHS t_{m + n} + t_{m – n} = a + (m + n – 1)d + a + (m – n – 1)d

= 2a + d [2m – 2]
= 2[a + (m – 1)d) = 2 t_m = RHS

Question 8.
A man repays an amount of Rs.3250 by paying Rs.20 in the first month and then increases the payment by Rs. 15 per month. How long will it take him to clear the amount?
Answer:
Amount of loan = 3250
Let n be the number of months taken to clear the loan
Amount paid in the first month a = 20
Increased payment in every month d = 15
∴ Amount paid in the second month = 20 + 15 = 35
Amount paid in the third month = 35 + 15 = 50
∴ The sequence of amount paid in every month is 20, 35, 50, …………. which is an A.P with first term a = 20 and common difference = 15
Given S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
S_n = 3250
3250 = \(\frac{n}{2}\) [2 × 20 + (n – 1) 15]
6500 = n[40 + 15n – 15]
6500 = n[25 + 15n]
6500 = 25n + 15n²
1300 = 5n + 3n²
3n² + 5n – 1300 = 0
3n² + 65n – 60n – 1300 = 0
n(3n + 65) – 20 (3n + 65) = 0
(n – 20) (3n + 65) = 0
n = 20 = 0 or 3n + 65 = 0
n = 20 or n = \(-\frac{65}{3}\)
n = \(-\frac{65}{3}\) is not possible ∴ n = 20
Thus, in 20 months the loan is cleared.

Question 9.
In a race, 20 balls are placed in a line at intervals of 4 meters with the first ball, 24 meters away from the starting point. A contestant is required to bring the balls back to the starting place one at a time. How far would the contestant run to bring back all balls?
Answer:
Number of balls placed in a line = 20
Let A be the starting point.
The distance of the first ball from A = 24 m
The distance of the second ball from the first ball = 4 m

A contestant starts from point A, travels a distance of 24 m and picks the first ball and brings it back to starting point A. This is continued for each ball.
Distance travelled by the contestant to bring the first ball = 24 + 24 = 2 × 24 = 48 m
Distance travelled to bring the second ball = 2 ( 24 + 4 ) = 2 × 28 = 56 m
Distance travelled to bring the third ball = 2 × (24 + 4 + 4) = 64 m

∴ The sequence of distances travelled are 48, 56, 64 ……………
This is an Arithmetic progression with first term a = 48 ,
common difference d = 56 – 48 = 8 ,
number of terms n = 20.
The sum of 20 terms of this A.P gives the total distance travelled by the contestant in bringing all balls to the starting place
S_n = \(\frac{\mathrm{n}}{2}\) [2a + (n – 1)d]
S₂₀ = \(\frac{\mathrm{20}}{2}\) [2 × 48 + (20 – 1)8] = 10 [96 + 19 × 8]
= 10 [ 96 + 152] = 10 × 248 = 2480
Total distance traveled = 2480 m

Question 10.
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally how many bacteria will be present at the end of the 2nd hour, 4th hour, and nth hour?
Answer:
Number of bacteria at the beginning = 30
Number of bacteria after 1 hour = 30 × 2 = 60
Number of bacteria after 2 hours = 30 × 2² = 120
Number of bacteria after 4 hours = 30 × 24 = 30 × 16 = 480
∴ Number of bacteria after n^th hour = 30 × 2ⁿ

Question 11.
What will Rs. 500 amounts to in 10 years after its deposit in a bank which pays an annual interest rate of 10 % compounded annually?
Answer:

Question 12.
In a certain town, a viral disease caused severe health hazards, upon its people disturbing their normal life. It was found that on each day, the virus which caused the disease spread in Geometric Progression. The amount of infected virus particle gets doubled each day, being 5 particles on the first day. Find the day when the infections virus particles just grow over 1,50,000 units?
Answer:
The number of viruses present at the beginning = 5, Given virus, gets doubled each day
∴ The sequence of a number of viruses in each day is 5, 10, 20, 40, 80, ………………. which is a G. P with First-term a = 5, Common ratio r = \(\frac{10}{5}\) = 2
n^th term t_n = ar^n-1

∴ On the 15^th day, the infectious virus grows over 1,50,000 units.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.2

Question 1.
Write the first 6 terms of the sequence whose nth terms are given below and classify them as Arithmetic progression Geometric progression, Arithmetic – geometric progression, Harmonic progression and none of them.
(i) \(\frac{1}{2^{n+1}}\)
Answer:

(ii)
Answer:

(iii) \(4\left(\frac{1}{2}\right)^{n}\)
Answer:

(iv) \(\frac{(-1)^{n}}{n}\)
Answer:

(v) \(\frac{2 n+3}{3 n+4}\)
Answer:

(vi) 2018
Answer:
The n^th term a_n = 2018
a₁ = 2018,
a₂ = 2018,
a₃ = 2018,
a₄ = 2018,
a₅ = 2018,
a₆ = 2018,
∴ The given sequence is 2018, 2018, 2018, 2018, 2018, 2018, ………….
This is a œnstant sequence which has same common ratio and common difference.
Hence this is an A. P, G . P and AGP.

(vii) \(\frac{3 n-2}{3^{n-1}}\)
Answer:

Question 2.
Write the first 6 terms of the sequences whose n^th term a_n is given below
(i)
n = 1, a_n = n + 1, a₁ = 1 + 1 = 2
n = 2, a_n = n, a₂ = 2
n = 3, a_n = n + 1, a₃ = 3 + 1 = 4
n = 4, a_n = n, a₄ = 4
n = 5, a_n = n + 1, a₅ = 5 + 1 = 6
n = 6, a_n = n, a₆ = 6
∴ The first six terms are 2, 2, 4, 4, 6, 6

(ii)
n = 1, a₁ = 1, n = 2, a₂ = 2
n = 3, a_n = a_n+1 + a_n-2, a₃ = a_{3 – 1} + a_{3 – 2} = a₂ + a₁ = 2 + 1 = 3

n = 4, a_n = a_n+1 + a_n-2, a₄ = a_{4 – 1} + a_{4 – 2} = a₃ + a₂ = 3 + 2 = 5

n = 5, a_n = a_n+1 + a_n-2, a₅ = a_{5 – 1} + a_{5 – 2} = a₄ + a₃ = 5 + 3 = 8

n = 6, a_n = a_n+1 + a_n-2, a₆ = a_{6 – 1} + a_{6 – 2} = a₅ + a₄ = 8 + 5 = 13
∴ The first six terms are 1, 2, 3, 5, 8, 13

(iii)
Answer:
n = 1, a_n = n, a₁ = 1
n = 2, a_n = n, a₂ = 1
n = 3, a_n = n, a₃ = 1

n = 4, a_n = a_n-1 + a_n-2 + a_n-3
a₄ = a_4-1 + a_4-2 + a_4-3
a₄ = a₃ + a₂ + a₁
a₄ = 3 + 2 + 1 = 6

n = 5, a_n = a_n-1 + a_n-2 + a_n-3
a₅ = a_5-1 + a_5-2 + a_5-3
a₅ = a₄ + a₃ + a₂
a₅ = 6 + 3 + 2 = 11

n = 6, a_n = a_n-1 + a_n-2 + a_n-3
a₆ = a_6-1 + a_6-2 + a_6-3
a₆ = a₅ + a₄ + a₃
a₆ = 11 + 6 + 3 = 20
∴ The first six terms are 1, 2, 3, 6, 11, 20

Question 3.
Write the n^th term of the following sequences.
(i) 2, 2, 4, 4, 6, 6, ……………..
Answer:
The odd terms are a₁ = 2, a₃ = 4, a₅ = 6
The even terms are a₂ = 2, a₄ = 4, a₆ = 6

(ii)
Answer:

The terms in the numerator are 1 , 2 , 3, 4
a = 1 , d = 2 – 1 = 1
a_n = a + (n – 1) d
a_n = 1 + (n – 1)(1) = 1 + n – 1 = n
a_n = n
The terms in the denominator are 2 , 3 , 4 , 5 , 6 .
a = 2, d = 3 – 2 = 1
a_n = a + (n – 1) d
a_n = 2 + (n – 1) (1) = 2 + n – 1 = n + 1
a_n = n + 1
∴ The n^th term of the given sequence is a_n = \(\frac{n}{n+1}\) for all n ∈ N

(iii)
Answer:

The terms in the numerator are 1, 3, 5, 7, 9, ………….
a = 1 , d = 3 – 1 = 2
a_n = a + (n – 1) d
a_n = 1 + (n – 1)2
a_n = 1 + 2n – 2 = 2n – 1
The terms in the denominator are 2, 4, 6, 8, 10, …………..
a = 2, d = 4 – 2 = 2
a_n = a + (n – 1) d
a_n = 2 + (n – 1)(2)
a_n = 2 + 2n – 2 = 2n
∴ The n^th term of the given sequence is a_n = \(\frac{2 n-1}{2 n}\) for all n ∈ N

(iv) 6, 10, 4, 12, 2, 14, 0, 16, – 2 …………………
Answer:
The given sequence is 6, 10, 4, 12, 2, 14, 0, 16, – 2 ……………..
The odd terms are a₁ = 6, a₃ = 4 , a₅ = 2 , a₇ = 0, a₉ = – 2
∴ a_n = n – 7, n is odd
The even terms are a₂ = 10, a₄ = 12 , a₆ = 14 , a₈ = 16
∴ a_n = 8 + n, n is even.

Question 4.
The product of three increasing numbers in a G.P is 5832 . If we add 6 to the second number and 9 to the third number, then resulting numbers form an A.P. Find the numbers in G.P.
Answer:
Let the increasing numbers in G.P be \(\), a, ar.
Given \(\frac{a}{r}\) × a × ar = 5832 ⇒ a³ = 5832 = 18³ ⇒ a = 18
Also given \(\frac{a}{r}\), a + 6, ar + 9 form an A.P.
∴ 2(a + b) = \(\frac{a}{r}\) + (ar + 9)
⇒ (a + 6) + (a + 6) = \(\frac{a}{r}\) + (ar + 9)
⇒ (a + 6) – \(\frac{a}{r}\) = (ar + 9) – (a + 6)
⇒ a + 6 – \(\frac{a}{r}\) = ar + 9 – a – 6
⇒ a + 6 – \(\frac{a}{r}\) = ar – a + 3
Substituting the value of a = 18, we get


39r = 18r² + 18
18r² – 39r + 18 = 0
(2r – 3)(3r -2) = 0
2r – 3 = 0 or 3r – 2 = 0
r = \(\frac{3}{2}\) or r = \(\frac{2}{3}\)

Case (i) When a = 18, r = \(\frac{3}{2}\) the numbers in G.P are

Case (ii) When a = 18, r = \(\frac{2}{3}\)

Question 5.
Write the n^th term of the sequence \(\frac{3}{1^{2} \cdot 2^{2}}, \frac{5}{2^{2} \cdot 3^{2}}, \frac{7}{3^{2} \cdot 4^{2}}\), …………….. as a difference of two terms.
Answer:

The terms in the numerator are 3,5, 7
which forms an A. P with first term a = 3 and common difference d = 5 – 3 = 2
n^th term t_n = a + (n – 1) d
= 3 + (n – 1)(2)
= 3 + 2n – 2 = 2n + 1
t_n = 2n + 1
The terms in the denominator are 1² . 2², 2² . 3², 3² . 4² ……………….
n^th term t_n = n² . (n + 1)²

Question 6.
If t_k is the k^th term of a G.P then show that t_{n – k}, t_k, t_{n + k} also form a G.F for any positive integer k.
Answer:
Given t^k is the k^th term of a G.P. We have n^th term of a G.P is t_n = ar^n-1

Question 7.
If a, b, c are in geometric progression and if a^1/x = b^1/y = c^1/z are in Arithmetic progression.
Answer:

Question 8.
The A.M of two numbers exceeds their G.M by 10 and H.M by 16. Find the numbers.
Answer:
Let the numbers be a and b

Question 9.
If the roots of the equation (q – r)x² + (r – p)x + (p – q) = 0 are equal then show that p , q and r are in A. P.
Answer:
The roots are equal ⇒ ∆ = 0
(i.e.) b² – 4ac = 0
Hence, a = q – r ; b = r – p ; c = p – q
b² – 4ac = 0
⇒ (r – p)² – 4(q – r)(p – q) = 0
r² + p² – 2pr – 4[qr – q² – pr + pq] = 0
r² + p² – 2pr – 4qr + 4q² + 4pr – 4pq = 0
(i.e.) p² + 4q² + r² – 4pq – 4qr + 2pr = 0
(i.e.) (p – 2q + r)² = 0
⇒ p – 2q + r = 0
⇒ p + r = 2q
⇒ p, q, r are in A.P.

Question 10.
If a , b , c are respectively the p^th, q^th and r^th terms of a G . P show that (q – r) log a + (r – p) log b + (p – q) log c = 0
Answer:
Let A be first term and R be the jmnon ratio of the G.P.
Given a = p^th term of the G.P
General term of a G. P with first term A and common ratio R is t_n = AR^{n – 1}
∴ a = t_p = AR^{P – 1}
log a = log AR^p-1 = log A + log R^p-1 = log A + (p – 1) log R

b = q^th term of the G.P
b = t_q = AR^q-1
log b = log AR^q-1 = log A + log R^q-1 = log A + (q – r)log R

c = r^th term of the G.P
c = t_r = AR^r-1
log c = log AR^r-1 = log A + log R^r-1 = log A + (r – 1) log R

(q – r) log a + (r – p) log b + (p – q) log c
= (q – r) [log A + (p – 1) log R] + (r – p) [ log A + (q – 1) log R ] + (P – q) [ log A + (r – 1) log R]
= (q – r) log A + (q – r) (p – 1 ) log R + (r – p) log A + (r – p) (q – 1) log R + (P – q) log A + (p – q) (r – 1 ) log R
= [ q – r + r – p + p – q ] log A + [ (q – r) (p – 1) + (r – p) (q – 1) + (p – q)(r – 1)] log R
= 0 × log A + [pq – q – rp + r + rq – r – pq + p + pr – p – rq + q] log R
= 0 × log R = 0
∴ (q – r) log a + (r – p) log b + (p – q) log c = 0

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1

Question 1.
Expand
(i) \(\left(2 x^{2}-\frac{3}{x}\right)^{3}\)
(ii) \(\left(2 x^{2}-3 \sqrt{1-x^{2}}\right)^{4}+\left(2 x^{2}+3 \sqrt{1-x^{2}}\right)^{4}\)
Answer:
(i) \(\left(2 x^{2}-\frac{3}{x}\right)^{3}\)

(ii) \(\left(2 x^{2}-3 \sqrt{1-x^{2}}\right)^{4}+\left(2 x^{2}+3 \sqrt{1-x^{2}}\right)^{4}\)

Question 2.
Compute
(i) 102⁴
(ii) 99⁴
(iii) 9⁷
Answer:
(i) 102⁴
102⁴ = (100 + 2)⁴

= 100000000 + 8 × 1000000 + 24 × 10000 + 3200 + 16
= 100000000 + 8000000 + 240000 + 3216
= 108243216

(ii) 99⁴
99⁴ = (100 – 1)⁴

= 100000000 – 4 × 1000000 + 6 × 10000 – 400 + 1
= 100000000 – 4000000 + 60000 -400+1
= 96059601

(iii) 9⁷
9⁷ = (10 – 1)⁷


= 10³ (10⁴ – 7 × 10³ + 21 × 10² – 35 × 10 + 35) – 21 × 100 + 70 – 1
= 10³ (10000 – 7000 + 2100 – 350 + 35 ) – 2100 + 70 – 1
= 10³ (12135 – 7350) – 2031
= 10³ × 4785 – 2031
= 4785000 – 2031
= 4782969

Question 3.
Using binomial theorem, indicate which of the following two number is larger. (l.Ol)1000000, 10000 Answer:

= 1000000C₀ + 1000000C₁( . 01) + …………………….
= 1 + 1000000 × 0 . 01 + ……………………..
= 1 + 10000 + other positive terms
= 10001 + other positive terms
(1.01)^1000000 – 10000 = 10001 + other positive terms – 10000 > 0
∴ (1.01)^1000000 > 10000 ⇒ (1.01)^1000000 is larger.

Question 4.
Find the coefficient of x¹⁵ in \(\left(x^{2}+\frac{1}{x^{3}}\right)^{10}\)
Answer:
General term T_r+1 = nC_r x^n-r . a^r
∴ In the expansion \(\left(x^{2}+\frac{1}{x^{3}}\right)^{10}\)

To find the coefficient of x¹⁵, Put 20 – 5r = 15
20 – 15 = 5r ⇒ 5r = 5 ⇒ r = 1
∴ T_{1 + 1} = 10C₁x^{20 – 5} ⇒ T₂ = 10 . x¹⁵
∴ The coefficient of x¹⁵ is 10

Question 5.
Find the coefficient of x6 and the coefficient of x² in \(\left(x^{2}-\frac{1}{x^{3}}\right)^{6}\)
Answer:
The general terms is T_r+1 = nC_ra^n-r . a^r
∴ The general term in the expansion of \(\left(x^{2}-\frac{1}{x^{3}}\right)^{6}\) is

To find the coefficient of x⁶ , Put 12 – 5r = 6
12 – 6 = 5r ⇒ 5r = 6 ⇒ r = \(\frac{6}{5}\) which is impossible.
∴ There is no x⁶ term in the expansion.
To find the coefficient of x²,
Put 12 – 5r = 2
⇒ 12 – 2 = 5r
⇒ 10 = 5r
⇒ r = \(\frac{10}{5}\) = 5
Substituting in (1) we have

T₃ = 15
∴ The coefficient of x² is 15

Question 6.
Find the coefficient of x⁴ in the expansion of (1 + x³)⁵⁰ \(\left(x^{2}+\frac{1}{x}\right)^{5}\)
Answer:


= 10x⁴ × 1 + 5 × 1225 x⁴ + 19600 x⁴ + 500 x⁴
= x⁴ (10 + 6125 + 19600 + 500) = 26235 . x⁴
∴ The coefficient of x⁴ is 26235.

Question 7.
Find the constant term of \(\left(x^{3}-\frac{1}{3 x^{2}}\right)^{5}\)
Answer:


To get the constant term,
Put 15 – 5r = 0
⇒ 5r = 15
⇒ r = 3

Question 8.
Find the last two digits of the number 3⁶⁰⁰.
Answer:
Consider 3⁶⁰⁰
3⁶⁰⁰ = (3²)³⁰⁰ = 9³⁰⁰ = (1o – 1)³⁰⁰

= 10³⁰⁰ – 300 (10)²⁹⁹ + ………………. + 300 C₁ × 10 × – 1 + 1 × 1 × 1
= 10³⁰⁰ – 300 (1o)²⁹⁹ + …………….. – 300 × 10 + 1
= 10³⁰⁰ – 300 × 10²⁹⁹ + ……………… – 3000 + 1
All the terms except the last are multiples of 100 and hence divisible by 100.
∴ The last two digits will be 01.

Question 9.
If n is a positive integer show that 9ⁿ⁺¹ – 8n – 9 is always divisible by 64.
Answer:

which is divisible by 64 for all positive integer n.
∴ 9ⁿ – 8n – 1 is divisible by 64 for all positive integer n.
Put n = n + 1 we get
9^{n + 1} – 8 (n + 1) – 1 is divisible by 64 for all possible integer n
(9^{n + 1} – 8n – 8 – 1) is divisible by 64
∴ 9^{n + 1} – 8n – 9 is always divisible by 64

Question 10.
If n is an odd positive integer, prove that the coefficients of the middle terms in the expansion of (x + y)ⁿ are equal.
Answer:
Given (x + y)ⁿ
If n is odd the middle term in the expansion of (x + y)ⁿ are \(\frac{T_{n-1}}{2}+1\) and \(\mathrm{T}_{\frac{\mathrm{n}+1}{2}+1}\)

Question 11.
If n is a positive integer and r is a non-negative integer, prove that the coefficients of x^r and x^n-r in the expansion of (1 + x)ⁿ are equal.
Answer:
Given (1 + x)ⁿ.
General term T_r+1 = nC_r x^{n – r} . a^r
∴ The general term in the expansion of (1 + x)ⁿ is T_r+1 = nC_r . (1)^{n – r} . x^r
T_r+1 = nC_r . x^r ……….. (1)
∴ Coefficient of x^r is nC_r,
Put r = n – r in (1)
T_{n – r + 1} = nC_{n – r} . x^n-r
∴ The coefficient of x^n-r is nC_n-r …………. (2)
we know nC_r = nC^n-r
∴ The coefficient of x^r and coefficient of x^n-r are equal, (by (1) & (2))

Question 12.
If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer.
Answer:
Let a = a + b – b
= b + (a – b)
aⁿ = [b + (a – b)]ⁿ
Using binomial expansion

Question 13.
In the binomial expansion of (a + b )n , the coefficients of the 4th and 13th terms are equal to each other , find n.
Answer:
In (a + b)ⁿ general term is t_{r + 1} = ⁿC_r a^{n – r} b^r
So, t₄ = t_{3 + 1} = ⁿC₃ = ⁿC₁₂
⇒ n = 12 + 3 = 15
We are given that their coefficients are equal ⇒ ⁿC₃ = ⁿC₁₂ ⇒ n = 12 + 3 = 15
[ⁿC_x = ⁿC_y ⇒ x = y (or) x + y = n]

Question 14.
If the binomial co – efficients of three consecutive terms in the expansion of (a + x)ⁿ are in the ratio 1 : 7 : 42, then find n.
Answer:
The general term in the expansion of (a + b)ⁿ is T_r+1 = nC_r . a^n-r . b^r
∴ The general term in the expansion of (a + x)ⁿ is T_r+1 = nC_r . a^n-r . x^r
Let the three consecutive terms be r^th term, (r + 1 )^th term, (r + 2 )^th term.

Question 15.
In the binomial coefficients of (1 + x)ⁿ, the coefficients of the 5^th, 6^th, and 7^th terms are in A.P. Find all values of n.
Answer:
Given (1 + x)ⁿ

12(n – 4) = 30 + n² – 5n – 4n + 20 ⇒ 12n – 48 = 30 + n² – 9n + 20
⇒ n² – 9n – 12n + 50 + 48 = 0 ⇒ n² – 21n + 98 = 0
⇒ n² – 14n – 7n + 98 = 0 ⇒ n(n – 14) – 7(n – 14) = 0
⇒ (n – 7) (n – 14) = 0 ⇒ n = 7 or n = 14

Question 16.
prove that

Answer:

This relation is true for all values of n. Equating coefficient of xⁿ on both sides, we have
The general term in the expansion of (1 + x)²ⁿ is T_{r + 1} = 2nC_r (1)^{2n – r} . x^r
Put r = n we get, T_n+1 = 2nC_n . xⁿ
∴ The coefficient of xⁿ in the expansion of (1 + x)²ⁿ is 2nC_n

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 4 Combinatorics and Mathematical Induction Ex 4.5 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.5

Choose the correct or the most suitable answer:

Question 1.
The sum if the digits at the 10^th place of all numbers formed with the help of 2, 4, 5, 7 taken all at a time is
(1) 432
(2) 108
(3) 36
(4) 18
Answer:
(2) 108

Explaination:

Number of digits given = 2, 4, 5, 7
Number of 4 digit numbers formed = 4! =24
So each digit occur \(\frac{24}{4}\) = 6 times
Sum of the digits = 2 + 4 + 5 + 7 = 18
So sum of the digits in each place = 18 × 6 = 108

Question 2.
In an examination, there are three multiple-choice questions and each question has 5 choices. The number of ways in which a student can fail to get all answer correct is
(1) 125
(2) 124
(3) 64
(4) 63
Answer:
(2) 124

Explanation:
Number of multiple-choice questions = 3
Number of choices in each question = 5
Number of ways of answering = 5 × 5 × 5 = 5³ = 125
Number of ways of getting all answer correct = 1
∴ Number of ways of getting incorrect answer = 125 – 1 = 124

Question 3.
The number of ways in which the following prize be given to a class of 30 boys first and second in mathematics, first and second in Physics, first in Chemistry and first in English is
(1) 30⁴ × 29²
(2) 30² × 29³
(3) 30² × 29⁴
(4) 30 × 29⁵
Answer:
(1) 30⁴ × 29²

Explanation:
I and II in maths can be given in 30 × 29 ways.
I and II in physics can be given in 30 × 29 ways.
I and chemistry can be given in 30 ways.
I in English can be given in 30 ways.
So total number of ways = 30 × 29 × 30 × 29 × 30 × 30 = 30⁴ × 29²

Question 4.
The number of 5 digit numbers all digits of which are odd is
(1) 25
(2) 5⁵
(3) 5⁶
(4) 625
Answer:
(2) 5⁵

Explanation:
The odd digit numbers = 1, 3, 5, 7, 9
Since repetition is allowed, each digit can be filled by 1, 3, 5, 7, 9
Unit’s place can be filled in 5 ways
Ten’s place can be filled in 5 ways etc.
∴ The number of 5 digit numbers with an odd digit in each place = 5 × 5 × 5 × 5 × 5 = 5⁵ ways

Question 5.
In 3 fingers, the number of ways four rings can be worn is ————- ways.
(1) 4³ – 1
(2) 3⁴
(3) 68
(4) 64
Answer:
(4) 64

Explanation:
Each letter can be ported in 3 ways
∴ 4 letter is 3⁴ ways

Question 6.

then the value of n are
(1) 7 and 11
(2) 6 and 7
(3) 2 and 11
(4) 2 and 6
Answer:
(2) 6 and 7

Explaination:

( n + 5 ) ( n + 4) = 2 × 11 × (n – 1)
n² + 5n + 4n + 20 = 22n – 22
n² + 9n + 20 – 22n + 22 = 0
n² – 13n + 42 = 0
n (n – 6) – 7 (n – 6) = 0
(n – 7) (n – 6) = 0
n – 7 = 0 or n – 6 = 0
n = 7 and n = 6

Question 7.
The product of r consecutive positive integers is divisible by
(1) r !
(2) (r – 1) !
(3) ( r + 1 ) !
(4) r^r
Answer:
(1) r !

Explanation:
1(2) (3) ….. (r) = r! which is ÷ by r!

Question 8.
The number of five-digit telephone numbers having at least one of their digits repeated is
(1) 90000
(2) 10000
(3) 30240
(4) 69760
Answer:
(4) 69760

Explaination:
Case (i) When zero is allowed in the first place.
The number of five-digit telephone numbers which can
be formed using the digits 0, 1, 2, …………,9 is 10⁵.
The number of five-digit telephone numbers which have none of their digits repeated is 10P⁵
= 10 × 9 × 8 × 7 × 6 = 30240
∴ The required number of telephone number
= 10⁵ – 30240
= 100000 – 30240
= 69,760

Case (ii) When zero is not allowed in the first place,
(a) Repetition allowed:

The number of ways of filling the first box, second box, …………….. , fifth box
= 9 × 10 × 10 × 10 × 10
= 90000

(b) Repetition riot allowed:
Number of ways of filling 1 box = 9
Number of ways of filling 2^nd box = 9
Number of ways of filling 3^rd box = 8
Number of ways of filling 4^th box = 7
Number of ways of filling 5^th box = 6
∴ Number of numbers with no digit is repeated = 9 × 9 × 8 × 7 × 6 = 37216
∴ Number of numbers having atleast one of the digits repeated = 90000 – 37216 = 52,784

Question 9.
If (a² – a ) C₂ = ( a² – a) C₄ then the value of a is
(1) 2
(2) 3
(3) 4
(4) 5
Answer:
(2) 3

Explaination:
(a² – a)C₂ = (a² – a) C₄
(a² – a)C₂ = (a² – a) C_{(a² – a) – 4} [nC_r = nC_n-r]
(a² – a)C₂ = (a² – a) C_{a² – a – 4}
∴ 2 = a² – a – 4
a² – a – 4 – 2 = 0 ⇒ a² – a – 6 = 0
a² – 3a + 2a – 6 = 0
a (a – 3) + 2 (a – 3) = 0
⇒ (a + 2)(a – 3) = 0 ⇒ a = – 2 or 3

Question 10.
There are 10 points in a plane and 4 of them are collinear. The number of straight lines joining any two points is
(1) 45
(2) 40
(3) 39
(4) 38
Answer:
(2) 40

Explaination:
Total number of points = 10
The number of straight lines joining any two points is same as the number ways of selecting 2 points from 10 points.
∴ The number of straight lines = 10C₂
Given 4 points are collinear.
∴ The number of straight lines that these 4 points contribute = 4C₂
Since 4 points are collinear they lie on a line.
∴ Required number of lines = 10C₂ – 4C₂ + 1

= 5 × 9 – 2 × 3 + 1
= 45 – 6 + 1 = 40

Question 11.
The number of ways in which a host lady invite for a party of 8 out of 12 people of whom two do not want to attend the party together is
(1) 2 × 11C₇ + 10C₈
(2) 11C₇ + 10C₈
(3) 12C₈ – 10C₆
(4) 10C₆ + 2!
Answer:
(3) 12C₈ – 10C₆

Explaination:
Number of people = 12
Number of people invited for the party = 8
Number of people not willing to attend the party = 2
Number of ways of selecting 8 people from 12 people is 12C₈
Two people do not attend party out of 10, remaining people 8 can attend in 10C₆ ways.
∴ Number of ways in which two of them do not attend together = 12C₈ – 10C₆

Question 12.
The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines.
(1) 6
(2) 9
(3) 12
(4) 18
Answer:
(4) 18

Explaination:
Number of first set of parallel lines = 4
Number of second set of parallel lines = 3
Number of parallelograms = 4C₂ × 3C₂
= \(\frac{4 \times 3}{1 \times 2} \times \frac{3 \times 2}{1 \times 2}\)
= 6 × 3
= 18

Question 13.
Everybody in a room shakes hands with everybody else. The total number of shake hands is 66. The number of persons in the room is
(1) 11
(2) 12
(3) 10
(4) 6
Answer:
(2) 12

Explaination:
Number of shake hands = 66
Let the number of persons = n
First person shakes hands with the remaining n – 1 persons.
Second person shakes hands with the remaining n – 2 persons.

∴ Total number of shake hands
= (n – 1) + (n – 2) + …………. + 2 + 1
= \(\frac{(\mathrm{n}-1)(\mathrm{n}-1+1)}{2}\)
Given 66 = \(\frac{(\mathrm{n}-1) \mathrm{n}}{2}\)
n² – n = 132
n² – n – 132 = 0
(n – 12) (n + 11) = 0
n = 12 or n = – 11
n = – 11 is not possible.
∴ n = 12

Question 14.
Number of sides of a polygon having 44 diagonals is
(1) 4
(2) 4!
(3) 11
(4) 22
Answer:
(3) 11

Explaination:
Number of diagonals of a polygon with n sides
= nC₂ – n
= \(\frac{n(n-1)}{2}\)
Given \(\frac{n(n-1)}{2}\) – n = 44
n² – n + 2n = 88
n² – 3n – 88 = 0
(n – 11 ) (n + 8) = 0
n = 11 and n = – 8
n = – 8 is not possible.
∴ n = 11

Question 15.
If 10 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, then the total number of point of intersection are
(1) 45
(2) 40
(3) 10!
(4) 210
Answer:
(1) 45

Explaination:
Number of non-parallel lines = 10
Two lines will interest.
∴ Number of points of intersection
= 10C₂ = \(\frac{10 \times 9}{1 \times 2}\) = 45

Question 16.
In a plane there are 10 points are there out of which 4 points are collinear, then the number of triangles formed is
(1) 110
(2) 10C₃
(3) 120
(4) 116
Answer:
(4) 116

Explaination:
Number of points = 10
Number of triangles formed by using 10 points is same as number of ways of choosing 3 points from 10 points = 10C₃
Also given 4 points are collinear.
∴ These 4 points do not contribute triangles.
Thus, total number of triangles = 10C₃ – 4C₃
= \(\frac{10 \times 9 \times 8}{1 \times 2 \times 3}\) – 4
= 5 × 3 × 8 – 4
= 120 – 4
= 116

Question 17.
In 2nC₃ : nC₃ = 11 : 1 then
(1) 5
(2) 6
(3) 11
(4) 7
Answer:
(2) 6

Explaination:
Given 2nC₃ : nC₃ = 11 :

4(2n – 1) = 11 (n – 2)
8n – 4 = 11 n – 22
-4 + 22 = 11 n – 8n
18 = 3n
⇒ n = 6

Question 18.
(n – 1) C_r + ( n – 1 ) C_{r – 1} is
(1) (n + 1) C_r
(2) (n – 1) C_r
(3) nC_r
(4) nC_{r – 1}
Answer:
(3) nC_r

Explaination:
[nC_r + nC_{r – 1} = (n + 1 ) C_r
(n – 1)C_r + (n – 1)C_{r – 1} = (n – 1 + 1) C_r
= nC_r

Question 19.
The number of ways of choosing 5 cards out of a deck of 52 cards which include atleast one king is
(1) 52C₅
(2) 48 C₅
(3) 52C₅ + 48C₅
(4) 52C₅ – 48C₅
Answer:
(4) 52C₅ – 48C₅

Explaination:
Number of cards = 52,
Number of kings = 4
Number of ways of choosing 5 cards out of 52 cards = 52 C₅
Number of cards without kings = 48
Number of ways of choosing 5 cards with no kings from 48 cards = 48 C₅
∴ Number of ways of choosing 5 cards out of 52 cards
= 52C₅ – 48C₅

Question 20.
The number of rectangles that a chessboard has
(1) 81
(2) 9⁹
(3) 1296
(4) 6561
Answer:
(3) 1296

Explaination:
Number of rectangles in a chessboard is

= 36 × 36
= 1296

Question 21.
The number of 10 digit number that can be written by using digits 2 and 3 is
(1) 10C₂ + 9C₂
(2) 2¹⁰
(3) 2¹⁰ – 2
(4) 10!
Answer:
(2) 2¹⁰

Explaination:

Each box can be filled in two ways using digits 2 and 3.
∴ A number of 10 digit number that can be written by using the digits 2 and 3 is 2 × 2 × 2 × …………. 10 times. = 210

Question 22.
If Pr stands for rPr, then the sum of the series 1 + P₁ + 2P₃ + 3P₃ + ……… + nP_n is
(1) P_{n + 1}
(2) P_{n + 1} – 1
(3) P_{n + 1} + 1
(4) (n + 1)P_{n – 1}
Answer:
(1) P_{n + 1}

Explaination:
Given P, = rP_r = r!
1 + 1 × P₁! + 2 × P₂ + 3 × P₃ + …………… + n × P_n!
= 1 + (1 × 1! + 2 × 2! + 3 × 3! + …………. + n × n!)
= (n + 1 )!
= P_{n + 1}
[(1 × 1 !) + (2 × 2!) + (3× 3 !) + ….. + (n × n !) = (n + 1)! – 1]

Question 23.
The product of first n odd natural numbers equals
(1) 2nC_n × nP_n
(2) \(\left(\frac{1}{2}\right)^{n}\) 2nC_n × nP_n
(3) \(\left(\frac{1}{4}\right)^{n}\) × 2nC_n × 2nP_n
(4) nC_n × nP_n
Answer:
(2) \(\left(\frac{1}{2}\right)^{n}\) 2nC_n × nP_n

Explaination:
1 . 3 . 5 …………. (2n – 1)

Question 24.
If nC₄, nC₅, nC₆ are in A. P. the value of n can be
(1) 14
(2) 11
(3) 9
(4) 5
Answer:
(1) 14

Explaination:
Given nC_n, nC₅, nC₆ are in A. P.
∴ 2 × nC₅ = nC₄ + nC₆

12(n – 4) = 30 + (n – 4) (n – 5)
12n – 48 = 30 + n² – 5n – 4n + 20
12n – 48 = 50 + n² – 9n
n² – 9n – 50 – 12n + 48 = 0
n² – 21n + 98 = 0
n² – 14n – 7n + 98 = 0
n(n – 14) – 7(n – 14) = 0
(n – 7) (n – 14) = 0
n – 7 = 0 or n – 14 = 0
n = 7 or n = 14

Question 25.
1 + 3 + 5 + 7 + …………. + 17 is equal to
(1) 101
(2) 81
(3) 71
(4) 61
Answer:
(2) 81

Explaination:
Given 1 + 3 + 5 + 7 + ………… + 17
This is an A.P. with first term a = 1
Common difference d = 2
Last term t_n = 17
a + (n – 1)d = 17
1 + (n – 1)2 = 17
(n – 1)2 = 17 – 1
(n – 1)2 = 16
n – 1 = \(\frac{16}{2}\) = 8
n = 9

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 Text Book Back Questions and Answers, Notes.

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
1² + 2² + 3² + ……….. + n³ = \(\left(\frac{\mathbf{n}(\mathbf{n}+\mathbf{1})}{2}\right)^{2}\)
Answer:
Let P(n) = 1² + 2² + 3² + ……….. + n³
p(n) = \(\left(\frac{\mathbf{n}(\mathbf{n}+\mathbf{1})}{2}\right)^{2}\)

Step 1:
Let us verify the statement for n = 1
p(1) = 1³ = \(\left(\frac{1(1+1)}{2}\right)^{2}\)

P(1) = 1
∴ The given statement is true for n = 1

Step 2 :
Let us assume that the given statement is true for n = k
P(k) = 1³ + 2³ + 3³ + …………… + k³

Step 3 :
Let us prove the statement is true for n = k + 1
P(k + 1 ) = 1³ + 2³ + 3³ + …………… + k³ + (k + 1)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.
1² + 2² + 3² + ……….. + n³ = \(\left(\frac{\mathbf{n}(\mathbf{n}+\mathbf{1})}{2}\right)^{2}\)
is true for all natural numbers n.

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

Answer:

Step 1 :
Let us verify the statement for n = 1

∴ The given statement is true for n = 1

Step 2 :
Let us assume that the given statement is true for n = k

Step 3:
Let us prove the statement is true for n = k + 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.

is true for all natural numbers n.

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

Step 1:
P( 1) : 2 = 1² + 1 = 2 which is true for P( 1)

Step 2:
P(k): 2 + 4 + 6+ …+ 2k = k² + k. Let it be true.

Step 3:
P(k + 1) : 2 + 4 + 6 + … + 2k + (2k + 2)
= k²+ k + (2k + 2) = k² + 3k + 2
= k² + 2k + k + 1 + 1
= (k+ 1)²+ (k + 1)
Which is true for P(k + 1)
So, P(k + 1) is true whenever P(k) is true.

Question 4.
By the principle of mathematical induction, prove that, for n ≥ 1, 1 . 2 + 2 . 3 + 3 . 4 + …………. + n . (n + 1) = \(\frac{\mathbf{n}(\mathbf{n}+1)(\mathbf{n}+2)}{3}\)
Answer:
Let P(n) = 1 . 2 + 2 . 3 + 3 . 4 + ………… + n . (n + 1) = \(\frac{\mathbf{n}(\mathbf{n}+1)(\mathbf{n}+2)}{3}\)

Step 1:
Let us verify the statement for n = 1
P(1) = 1 . 2 = \(\frac{1(1+1)(1+2)}{3}\)
= 1 . 2 = \(\frac{1 \cdot 2 \cdot 3}{3}\)
∴ The result is true for n = 1

Step 2:
Let us assume that the result is true for n = k
P (k) = 1-2 + 2-3 + 3- 4 + + k (k + 1 ) =

Step 3:
Let us prove the result for n = k + 1
P(k + 1) = 1 . 2 + 2 . 3 + 3 . 4 + …………… + k (k + 1) (k + 1) (k + 1 + 1)
P (k + 1) = 1 . 2 + 2 . 3 + 3 . 4 + ………….. + k (k + 1) + (k + 1 ) (k + 2)
P (k + 1 ) = P(k) + (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.
1 . 2 + 2 . 3 + 3 . 4 + ………. + n (n + 1) = \(\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}{3}\)
is true for all natural numbers n ≥ 1

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

Answer:
Let

The first stage is n = 2

Step 1:
Let us verify the result for n = 2

∴ The given result is true for n = 2.

Step 2:
Let us assume that the result is true for n = k.

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


∴ 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.

is true for all natural numbers n ≥ 2.

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

Answer:

Step 1:
Since n ≥ 2 the first stage is n = 2
Let us verify the result for n = 2

∴ P (2 ) is true.
The result is true for n = 2.

Step 2:
Let us assume that the result is true for n = k.

Step 3:
Let us prove the result for n = k + 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 ≥ 2.

is true for all natural numbers n ≥ 2

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

Answer:

Step 1:
First, let us verify the result for n = 1

∴ The given result is true for n = 1

Step 2:
Let us assume the result for n = k

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

Factorizing k³ + 6k² + 9k + 4
f (k) = k³ + 6k² + 9k + 4
f(- 1) = (- 1)³ + 6(- 1)² + 9(- 1) + 4
f(- 1) = – 1 + 6 – 9 + 4 = 0
∴ (k + 1) is a factor of f(k)

k³ + 6k² + 9k + 4 = (k + 1) (k² + 5k + 4).
= (k + 1) (k² + 4k + k + 4)
= (k + 1) [k(k + 4) + 1(k + 4)]
= (k + 1) (k + 4) (k + 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.

is true for all natural numbers n

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

Answer:

Step 1:
First, let us verify the result for n = 1


∴ The result is true for n = 1

Step 2:
Let us assume that the result is true for n = k

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

Question 9.
Prove by mathematical induction that
1! + (2 × 2 !) + (3 × 3!) +…. .+ (n × n!) = (n + 1)! – 1
Answer:
P(n) is the statement
1! + (2 × 2!) + (3 × 3!) + ….. + (n × n!) = (n + 1)! – 1
To prove for n = 1
LHS = 1! = 1
RHS = (1 + 1)! – 1 = 2! – 1 = 2 – 1 = 1
LHS = RHS ⇒ P(1) is true
Assume that the given statement is true for n = k
(i.e.) 1! + (2 × 2!) + (3 × 3!) + … + (k × k!) = (k + 1)! – 1 is true
To prove P(k + 1) is true
p(k + 1) = p(k) + t_{(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
= (k + 1 + 1)! – 1
∴ P(k + 1) is true
⇒ P(k) is true, So by the principle of mathematical induction P(n) is true.

Question 10.
Using the Mathematical induction, show that for any natural number n, x²ⁿ – y²ⁿ is divisible by x + y.
Answer:
Let P(n) = x²ⁿ – y²ⁿ is divisible by (x + y)
For n = 1
P(1) = x^{2 × 1} – y^{2 × 1} is divisible by (x + y)
⇒ (x + y) (x – y) is divisible by (x + y)
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k) = x^2k – y^2k is divisible by (x + y)
⇒ x^2k – y^2k = λ(x + y) …… (i)
For n = k + 1
⇒ P(k + 1) = x^{2(k + 1)} – y^{2(k + 1)} is divisible by (x + y)
Now x^{2(k + 2)} – y^{2(k + 2)}
= x^{2k + 2} – x^2ky² + x^2ky² – y^{2k + 2}
= x^2k.x² – x^2ky² + x^2ky² – y^2ky²
= x^2k (x² – y²) + y²λ (x + y) [Using (i)]
⇒ x^{2k + 2} – y^{2k + 2} is divisible by (x + y)
∴ 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 11.
By the principle of mathematical induction, prove that, for n ≥ 1, 1² + 2² + 3² + ………….. + n² > \(\frac{n^{3}}{3}\)
Answer:
Let P (n) = 1² + 2² + 3² + ………….. + n² > \(\frac{n^{3}}{3}\)

Step 1:
Let us first verify the result for n = 1

∴ The result is true for n = 1

Step 2:
Let us assume that the result is true for n = k.
P(k) = 1² + 2² + 3² + …………… + k² > \(\frac{\mathrm{k}^{3}}{3}\)

Step 3:
Let us prove the result for n = k + 1
P(k + 1) = 1² + 2² + 3² + …………… + k² + (k + 1)²
P(k + 1) = P(k) + ( k + 1)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.
1² + 2² + 3² + ………….. + n² > \(\frac{n^{3}}{3}\)

Question 12.
Use induction to prove that n³ – 7n + 3, is divisible by 3, for all natural numbers n.
Answer:
Let P(n) : n³ – 7n + 3
Step 1:
P(1) = (1)³ – 7(1) + 3
= 1 – 7 + 3 = -3 which is divisible by 3
So, it is true for P(1).

Step 2:
P(k) : k³ – 7k + 3 = 3λ. Let it be true
⇒ k³ = 3λ + 7k – 3

Step 3:
P(k + 1) = (k + 1)³ – 7(k + 1) + 3
= k³ + 1 + 3k² + 3k – 7k – 7 + 3
= k³ + 3k² – 4k – 3
= (3λ + 7k – 3) + 3k² – 4k – 3 (from Step 2)
= 3k² + 3k + 3λ – 6
= 3(k² + k + λ – 2) which is divisible by 3.
So it is true for P(k + 1).
Hence, P(k + 1) is true whenever it is true for P(k).

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.
Answer:
To prove 5^{n + 1} + 4 × 6ⁿ when divided by 20 leaves a remainder 9
That is to prove 5^{n + 1} + 4 × 6ⁿ – 9 is divisible by 20
Let P (n) be the statement 5^{n + 1} + 4 × 6ⁿ
which is divisible by 20
P (n) = 5^{n + 1} + 4 × 6ⁿ – 9 = 20λ

Step 1 :
Let us verify the statement for n = 1
P (1) = 5^{i + 1} + 4 × 6¹ – 9
= 5² + 4 × 6 – 9
= 25 + 24 – 9
= 49 – 9 = 40
which is divisible by 20
∴ The statement is true for n = 1

Step 2:
Let us assume that the statement is true for n = k
P(k) = 5^{k + 1} + 4 × 6^k – 9 is divisible by 20
P(k) = 5^{k + 1} + 4 × 6^k – 9 = 20λ …………… (1)

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

Hence, P(k + 1) is divisible by 20.
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.
Thus, 5^{n + 1} + 4 × 6ⁿ when divided by 20 leaves a remainder 9, for all natural numbers n.

Question 14.
Use induction to prove that 10ⁿ + 3 × 4^{n + 2} + 5 is divisible by 9, for all natural numbers n.
Answer:
Let P( n) = 10ⁿ + 3 × 4^{n + 2} + 5 is divisible by 9

Step 1:
Let us verify the result for n = 1
P ( 1 ) = 10¹ + 3 × 4^{1 + 2} + 5
= 15 + 3 × 4³
= 15 + 3 × 64
= 15 + 192 = 207
which is divisible by 9
Thus we have verified the result for n = 1

Step 2:
Let us assume the result is true for n = k
P(k) = 10^k + 3 × 4^{k + 2} + 5 is divisible by 9
∴ 10^k + 3 × 4^{k + 2} + 5 = 9m for some m
10^k = 9m – 5 – 3 × 4^{k + 2} for some m

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

which is divisible by 9
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.
10ⁿ + 3 × 4^{n + 2} + 5 is divisible by 9,
for all natural numbers n.

Question 15.
Prove that using mathematical induction

Answer:

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.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 Text Book Back Questions and Answers, Notes.

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

Question 1.
If nC₁₂ = nC₉ find 21C_n
Answer:
Given nC₁₂ = nC₉
We have nC_x = nC_y ⇒ x = y or x + y = n
nC₁₂ = nC₉
⇒ 12 + 9 = n ⇒ n = 21
⇒ 21C_n = 21C₂₁ = 1

Question 2.
If 15C_{2r – 1} = 15C_{2r + 4}, find F.
Answer:
Given 15C_{2r – 1} = 15C_{2r + 4}
We have nC_x = nC_y ⇒ x = y or x + y = n
15C_{2r – 1} = 15C_{2r + 4}
⇒ (2r – 1) + (2r + 4) = 15
4r + 3 = 15 ⇒ 4r = 12
r = \(\frac{12}{4}\) = 3

Question 3.
If nP_r = 720, If nC_r = 120, find n, r = ?
Answer:
Given nP_r = 720, If nC_r = 120

r ! = 6
r ! = 3 × 2 × 1 = 3!
r = 3

n (n – 1) (n – 2) = 720
n(n – 1) (n – 2) = 1o × 9 × 8
n = 10
∴ r = 3, n = 10

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

Question 5.
Prove that 35C₅ + \(\sum_{r=0}^{4}\)(39 – r) C₄ = 40C₅
Answer:

Question 6.
If (n + 1)C₈: (n – 3)P₄ = 57 : 16, find n
Answer:


(n + 1) n(n – 1)(n – 2) = 57 × 7 × 6 × 5 × 4 × 3
= 3 × 19 × 7 × 6 × 5 × 4 × 3
= (7 × 3) × (5 × 4)19 × (6 × 3)
= 21 × 20 × 19 × 18
= (20 + 1) × (20) × (20 – 1) × (20 – 2)
∴ n = 20

Question 7.

Answer:

Question 8.
prove that if 1 ≤ r ≤ n, then n × (n – 1)C_{r – 1} = (n – r + 1) . nC_{r – 1}
Answer:

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 ?
Answer:
Number of kabaddi players = 14
7 players must be selected from 14 players
Number of ways of selecting 7 players from 14 players is

(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?
Answer:
Total number of person in the party = 15
Given if each 2 of the 15 persons shakes bands with each other.
∴ The total number of hand shakes is same as the number of ways of selecting 2 persons among 15 persons. This can be done in 15C₂ ways.
Number of hand shakes = 15C₂

(iii) How many chords can be drawn through 20 points on a circle ?
Answer:
Number given points on the circle = 20
A chord is obtained by joining any two points on the circle.‘Number of chords drawn through 20 points is same as the number of ways of selecting 2 points out of 20 points. This can be done 20C₂ ways.
∴ The total number of chords = 20C₂

(iv) In a parking lot one hundred, one year old cars are parked. Out of them five are to be chosen at random for to cheek its pollution devices. How many different set of five cars can be chosen?
Answer:
Number of cars in the parking lot = 100
Number of cars to be selected to check pollution device = 5
Number of ways of selecting 5 cars out of 100 cars

(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 ?
Answer:
Number of boys = 5
Number of girls = 4
Number of transgender = 2
Number of ways of selecting 3 boys from 5 boys = 5c₃
Number of ways of selecting 2 girls from 4 girls = 4C₂
Number of ways of selecting one transgender from 2 transgenders = 2C₁
Total number of ways of selection

Question 10.
Find the total number of subsets of a set with
(i) 4 elements
(ii) 5 elements
(iii) n elements.
Answer:
(i) Subsets with 4 elements
Number of subsets with no element = 4C_o
Number of subsets with one element = 4C₁
Number of subsets with two elements = 4C₂
Number of subsets with three elements = 4C₃
Number of subsets with four elements = 4C₄
∴ Total number of subsets

(ii) Subsets with 5 elements:
Number of subsets with no element = 5C₀
Number of subsets with one element = 5C₁
Number of subsets with 2 elements = 5 C₂
Number of subsets with 3 elements = 5C₃
Number of subjects with 4 elements = 5C₄
Number of subsets with 5 elements = 5C₅
Total number of subjects

(iii) Subsets with n elements
Number of subsets with no element = nC₀
Number of subsets with 1 , 2 , 3, 4, …………. n elements are nC₁, nC₂, nC₃, nC₄ …………… nC_nrespectively.
∴ Total number of subjects
= nC₀ + nC₁ + nC₂ + nC₃ + ………… + nC_n
= Sum of the coefficients in the binomial expansion (x + a)ⁿ
= 2ⁿ

Question 11.
A Trust has 25 members.
(i) How many ways 3 officers can be selected?
(ii) In how many ways can a president, vice president and a secretary be selected?
Answer:
Number of members in the trust = 25
(i) How many ways 3 officers can be selected?
The number of ways of selecting 3 officers from 25 members is

(ii) In how many ways can a president, vice president, and secretary be selected?
Answer:
From the 25 members, a president can be selected in 25 ways
After the president is selected, 24 persons are left out.
So a Vice President can be selected in (from 24 persons) 24 ways.
After the selection of Vice President, 23 persons are left out
So a secretary can be selected (from the remaining 23 persons) in 23 ways
So a president, Vice president, and a secretary can be selected in ²⁵P₃ ways
²⁵P₃ = 25 × 24 × 23 = 13800 ways

Question 12.
How many ways a committee of six-person from 10 persons can be chosen along with a chairperson and a secretary?
Answer:
Number of persons = 10
Number of committee members to be selected = 6
A committee must have a chairperson and a secretary, the remaining 4 members.
The number of ways of selecting a chairperson and a secretary from 10 persons is

After the selection of chairperson and secretary remaining number of persons = 8
Number of ways of selecting remaining 4 committee members from the remaining 8 persons

= 2 × 7 × 5 = 70
∴ A number of ways of selecting the six committee members from 10 persons.
= 45 × 70 = 3150 ways

Question 13.
How many different selections of 5 books can be made from 12 different books if,
(i) Two particular books are always selected?
Answer:
Total number of books = 12
Number of books to be selected = 5
Given Two books are always selected.
Remaining number of books to be selected = 3
The number of ways of selecting the remaining 3 books from the remaining 10 books = 10C₃

(ii) Two particular books are never selected. Since two books are never selected, the total number of books is 10.
∴ The number of ways of selecting 5 books from 10 books

= 2 × 9 × 2 × 7 = 252 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?
Answer:
The number of teachers = 5
Number of students = 20
The number of ways of selecting 2 teachers from 5 teachers is = 5C₂ ways.

The number of ways of selecting 3 students from 20 students is = 20C₃

= 20 × 19 × 3 = 1140 ways
∴ The total number of selection of the committees with 2 teachers and 5 students is = 10 × 1140 = 11400

(i) A particular teacher is included.
Given a particular teacher is selected. Therefore, the remaining 1 teacher is selected from the remaining 4 teachers. Therefore, the number of ways of selecting 1 teacher from the remaining 4 teacher = 4C₁ ways = 4 ways
The number of ways of selecting 3 students from 20 students = 20C₃


Hence the required number of committees
= 1 × 4 × 1140 = 4560

(ii) A particular student is excluded The number of ways of selecting 2 teachers from 5 teachers is = 5C₂

A particular student is excluded
∴ The number of remaining students = 19
Number of ways of selecting 3 students from 19 students

= 19 × 3 × 17 = 969
∴ The required number of committees = 10 × 969 = 9690

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 a student can answer the questions?
Answer:
Total number of questions = 9
Number of questions to be answered = 5
Since 2 questions are compulsory, a student must select 3 questions from dying remaining 7 questions* The number of ways of selecting 3 questions from 7 questions is = 7C₃
∴ The number of ways of answering 5 questions = 7C₃

Question 16.
Determine the number of 5 card combinations out of a check of 52 cards if there is exactly three aces in each combination.
Answer:
Total number of cards in a pack 52
Number of aces = 4
Number of cards to be selected = 5
The number of ways of selecting 3 aces from 4 aces is 4C₃
The number of ways of selecting the remaining 2 cards
from the remaining 48 cards (52 – 4 aces cards) = 48C₂
∴ Required number of ways of selection
= 4C₃ × 48C₂
= 4C₁ × 48C₂

= 4 × 24 × 47 = 4512

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.
Answer:
Number of Indians = 7
Number of Americans = 5
Number of members in the committee = 5
Selection of 5 members committee with majority Indians

Case (i) 3 Indians and 2 Americans
The number of ways of selecting 3 Indians from 7 Indians is = 7C₃
The number of ways of selecting 2 Americans from 5 Americans is = 5C₂
Total number of ways in this case is = 7C₃ × 5C₂

Case (ii) 4 Indians and 1 American
The number of ways of selecting 4 Indians from 7 Indians is = 7C₄
The number of ways of selecting I American from 5 Americans is = 5C₁
The total number of ways, in this case, is = 7C₄ × 5C₁

Case (iii) 5 Indians no American
Number of ways of selecting 5 Indians from
7lndiansis = 7C₅
Total number of ways, in this case, = 7C₅ × 5C₀
∴ Total number of ways of forming the committee


= 7 × 5 × 5 × 24 – 7 × 5 × 5 + 7 × 3
= 350 + 175 + 21 = 546

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) Atleast 3 women?
(iii) Almost 3 women?
Answer:
Number of men = 8
Number of women = 4
Number of peoples in the committee = 7

(i) Exactly 3 women
In a 7 member committee, women must be 3. Therefore, the remaining 4 must be men.
The number of ways of selecting 3 women from 4 women = 4C₃
The number of ways of selecting 4 men from 8 men = 8C₄
∴ The total number of ways of selection is = 4C₃ × 8C₄

= 8 × 7 × 5 = 280

(ii) At least 3 women
The 7 members committee must contain atleast 3 women
∴ We have the following possibilities
(i) 4 women + 3 men
(ii) 3 women + 4 men

case (i) 4 women + 3 men
The number ways of selecting 4 women .from
4 women is = 4C₄ = 1 way
The number of ways of selecting 3 men from 8 men = 8C₃

∴ The total number of ways = 1 × 56 = 56

case (ii) 3 women + 4 men
The number of ways of selecting 3 women from 4 women is = 4C₃
The number of ways of selecting 4 men from 8 men is = 8C₄
∴ The total number of ways = 4C₃ × 8C₄

= 8 × 7 × 5 = 280
∴ The required number of ways of forming the committee = 56 + 280 = 336

(iii) atmost 3 women
The 7 members must contain at most 3 women,
∴ we have the following possibilities
(i) No women + 7 men
(ii) 1 women + 6 men
(iii) 2 women + men
(iv) 3 women + 4 men

Case (i) 0 women + 7 men
The number of ways of selecting O women from 4 women is = 4C₀
The number of ways of selecting 7 men from 8 men is = 8C₇
Total number of ways = 4C₀ × 8C₇

Case (ii) 1 women + 6 men
The number of ways of selecting 1 woman from 4 women is = 4C₁
The number of ways of selecting 6 men from 8 men is = 8C₆
Total number of ways = 4C₁ × 8C₆

Case (iii) 2 women + 5 men
The number of ways of selecting 2 women from 4 women is = 4C₃
The number of ways of selecting 4 men from 8 men is = 8C₄
∴ Total number of ways = 4C₃ × 8C₄
∴ The required number of ways of forming the committee

Question 19.
7 relatives of a man comprise 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 the men’s relatives and 3 of the wife’s relatives?
Answer:
Number of relatives of a man = 7
4 ladies and 3 gentlemen
Number of relatives of the man’s wife = 7
3 ladies and 4 gentlemen
The dinner party consists of 3 ladies and 3 gentlemen.
For the dinner party, 3 persons from the man’s relatives and 3 persons from the man’s wife’s relatives are invited.
Then we have the following possibilities for the different possible arrangements.

Required number of ways

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 atleast one black ball is to be included in the draw?
Answer:
Number of white balls = 2
Number of black balls = 3
Number of red balls = 4
Number of balls drawn = 3 balls with atleast 1 blackball
We have the following possibilities


Required number of ways of drawing 3 balls with atleast one black ball.

Question 21.
Find the number of strings of 4 letters that can be formed with the letters of the word EXAMINATION.
Answer:
There are 11 fetters in the word EXAMINATION,
They are AA, II, NN, E, X, M, T, O The four-letter strings may have
(i) 2 alike letters of one kind and 2 alike tetters of the second kind.
(ii) 2 alike fetters and 2 different letters.
(iii) All different fetters.

(i) 2 alike letter of one kind and 2 alike letters of the second kind:
There are three sets of 2 alike letters AA, II, NN. Out of these sets, two sets can be selected in 3C₂ ways. So there are 3C₂ groups each of which còntains 4 letter
strings out of which 2 are alike of one kind type and 2 are alike of the second type.
4 letters in each group can be arranged in \(\frac{4 !}{2 ! \times 2 !}\) ways.
Hence the total number of strings consisting of two alike letters of one kind and 2 alike letters of the second kind

(ii) 2 alike letter and 2 different letters:
Out of sets of two alike letters, one set can be chosen in 3C₁ ways. From the remaining 7 distinct letters, 2 letters can be chosen in 7C₂ ways. Thus 2 alike letters and 2 different letters can be selected in (3C₁ × 7C₂) ways. There are ( 3C₁ × 7C₂) groups of 4 letters each. Now letters of each group can be arranged among themselves in \(\frac{4 !}{2 !}\) ways.

Hence the total number of strings consisting of 2 alike and 2 distinct letters,

(iii) All different letters
There are 8 different letters
E, X, A, M, I, N, T, O

Out of which 4 can be selected in 8C₄ ways. So there are 8C₄ groups of 4 letters each The letter in each of 8C₄ groups’ can be arranged in 4! ways.
∴ The total number of 4. letter strings in which all letters are distinct = 8C4 × 4!

= 56 × 30 = 1680 strings
Hence the total number of 4 letter strings
= 18 +756 + 1680
= 2454 strings

Question 22.
How many triangles can be formed by joining 15 points on the plane, in which no line joining any three points?
Answer:
Number of points 15
To form a triangle we need 3 non – collinear points.
The number of ways of se1ecting 3 non-collinear points from 15 points is = 15C₃

Question 23.
How many triangles can be formed by 15 points of which 7 of them 11e on one line and the remaining 8 on another parallel ‘line?
Answer:
To form a triangle we need 3 non-collinear points. Take the 7 points lying on one line be group A and the remaining 8 points lying on another parallel line be group B.
We have the following possibilities

∴ Required number of ways of forming the triangle

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 obtained from the pairs of these points?
(ii) The number of triangles that can be formed which the vertices as their points.
Answer:
Number of points in a plane = 11
No three of these points lie in the same straight line except 4 points.

(i) The number of straight lines that can be obtained from the pairs of these points
Through any two points, we can draw a straight line.
∴The number straight lines through any two points of the given 11 points = 11C₂

Given that 4 points are collinear. The number of straight lines through any two points of these 4 points is

Since these 4 points are collinear, these 4 points contribute only one line instead of the 6 lines.
The total number of straight lines that can be drawn through 11 points on a plane with 4 of the points being collinear is = 55 – 6 + 1 = 50

(ii) The number of triangles that can be formed for which the points are their vertices.
To form a triangle we need 3 non – collinear points. We have the following possibilities.

(a) If we take one point from 4 collinear points and 2 from the remaining 7 points and join them.

The number of ways of selecting one point from the 4 collinear points is = 4C₁ ways
The number of ways of selecting 2 points from the remaining 7 points is = 7C₂
The total number of triangles obtained in this case is = 4C₁ × 7C₂
∴ The total number of triangles obtained in this case is

(b) If we select two points from the 4 collinear points and one point from the remaining 7 points then the number of triangles formed is

(c) If we select all the three points from the 7 points then the number of triangles formed is

∴ The total number of triangles formed are
= 84 + 42 + 35 = 161

Question 25.
A polygon has 90 diagonals. Find the number of sides.
Answer:
Let the number sides of the polygon be n.
The number of diagonals of a polygon with n sides = \(\frac{n(n-3)}{2}\)
Given \(\frac{n(n-3)}{2}\) = 90
n² – 3n = 180
n² – 3n = 180
n² – 3n – 180 = 0
n² – 15n + 12n – 180 = 0
n (n – 15 ) + 12 (n – 15) = 0
(n + 12) (n – 15) = 0
n + 12 = 0 or n – 15 = 0
n = – 12 or n = 15
But n = – 12 is not possible
∴ n = 15
∴ The number of sides of the polygon having 90 diagonals is 15.