Class 11

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 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 6.3

Question 1.
Show that the lines are 3x + 2y + 9 = 0 and 12x + 8y – 15 = 0 are parallel lines.
Solution:

Here m₁ = m₂ ⇒ the two lines are parallel.

Question 2.
Find the equation of the straight line parallel to 5x – 4y + 3 = 0 and having x – intercept 3.
Solution:
Equation of a line parallel to ax + by + c = 0 will be of the form ax + by + k = 0
So equation of a line parallel to 5x – 4y + 3 = 0 will be of the form 5x – 4y = k

Question 3.
Find the distance between the line 4x + 3y + 4 = 0 and a point
(i) (-2, 4)
(ii) (7, -3)
Solution:
The distance between the line ax + by + c = 0 and the point(x₁, y₁) is given by

(i) Now the distance between the line 4x + 3y + 4 = 0 and (-2, 4) is

(ii) The distance between the line 4x + 3y + 4 = 0 and (7, -3) is

Question 4.
Write the equation of the lines through the point (1, -1)
(i) Parallel to x + 3y – 4 = 0
(ii) Perpendicular to 3x + 4y = 6
Solution:
(i) Parallel to x + 3y – 4 = 0
The equation of any line parallel to the line
x + 3y – 4 = 0 is x + 3y + k = 0 ………… (1)
This line passes through the point (1, – 1)
∴ (1) ⇒ 1 + 3 (-1 ) + k = 0
1 – 3 + k = 0 ⇒ k = 2
∴ The equation of the required line is
x + 3y + 2 = 0

(ii) Perpendicular to 3x + 4y = 6
The equation of any line perpendicular to 3x + 4y = 6 is
4x – 3y + k = 0 …………. (2)
This line passes through the point (1,-1)
(2) ⇒ (4) 1 – 3 (-1) + k = 0
4 + 3 + k = 0 ⇒ k = – 7
∴ The required equation is 4x – 3y – 7 = 0

Question 5.
If (- 4, 7) is one vertex of a rhombus and if the equation of one diagonal is 5x – y + 7 = 0, then find the equation of another diagonal.
Solution:
Let the equation of the diagonal AC be 5x – y + 7 = 0 ……….. (1)
Since (-4, 7) does not satisfy equation (1), (- 4, 7 ) represents neither A nor C.
Let (-4, 7) represent the vertex D.
The diagonal BD is perpendicular to AC
The equation of any line perpendicular to line (1) is – x – 5y + k = 0 ……….. (2)
This line passes through the point D (-4, 7)
∴ (2) ⇒ -(-4) – 5(7) + k = 0
4 – 35 + k = 0 ⇒ k = 31
∴ The equation of the other diagonal is
-x – 5y + 31 = 0
x + 5y – 31 = 0

Question 6.
Find the equation of the lines passing through the point of intersection lines 4x – y + 3 = 0 and 5x + 2y +7 = 0, and
(i) Through the point (-1, 2)
(ii) Parallel to x – y + 5 = 0
(iii) Perpendicular to x – 2y + 1 = 0.
Solution:
To find the point of intersection of the lines we have to solve them

Substituting x = -1 in equation (2) we get
-5 + 2y = -7
⇒ 2y = – 7 + 5 = -2
⇒ y = -1
So the point of intersection is (-1, -1)

(ii) Parallel to x – y + 5 = 0
Given that the line (1) is parallel to the line
x – y + 5 = 0 ………. (2)
∴ Slope of line (1) = Slope of line (2)
(4x – y + 3) + λ (5x + 2y + 7) = 0
4x – y + 3 + 5λx + 2λy + 7λ = 0
(4 + 5λ)x + (2λ – 1)y + 3 + 7λ = 0
Slope of this line = \(-\frac{4+5 \lambda}{2 \lambda-1}\)
Slope of line (2) = –\(\frac{1}{-1}\) = 1
These two slopes are equal
\(-\frac{4+5 \lambda}{2 \lambda-1}\) = 1
– (4 + 5λ) = 2λ – 1
– 4 – 5λ = 2λ – 1
2λ + 5λ – 1 + 4 = 0
7λ + 3 = 0 ⇒ λ = –\(-\frac{3}{7}\)
Substituting the value of λ in equation (1), we have
(4x – y + 3) –\(-\frac{3}{7}\) (5x + 2y + 7) = 0
7 (4x – y + 3) – 3 (5x + 2y + 7) = 0
28x – 7y + 21 – 15x – 6y – 21 =0
13x – 13y = 0
x – y = 0

(iii) Equation of a line perpendicular to x – 2y+ 1 =0 will be of the form 2x + y + k = 0. It passes through (-1, -1) ⇒ -2 – 1 + k = 0 ⇒ k = 3.
So the required line is 2x + y + 3 = 0

Question 7.
Find the equations of two straight lines which are parallel to the line 12x + 5y + 2 = 0 and at a unit distance from the point (1, -1).
Solution:
Equation of a line parallel to 12x + 5y + 2 = 0 will be of the form 12x + 5y + k = 0.
We are given that the perpendicular distance form (1, -1) to the line 12x + 5y + k = 0 is 1 unit.

So the required line will be 12x + 5y + 6 = 0 or 12x + 5y – 20 = 0

Question 8.
Find the equations of straight lines which are perpendicular to the line 3x + 4y – 6 = 0 and are at a distance of 4 units from (2, 1).
Solution:
Given equation of line is 3x + 4y – 6 = 0.
Any line perpendicular to 3x + 4y – 6 = 0 will be of the form 4x – 3y + k = 0
Given perpendicular distance is 4 units from (2, 1) to line (1)

∴ 20 = + (5 + k) or 20 = – (5 + k)
⇒ k = 20 – 5 or k = -(20 + 5)
k = 15 or k : = -25
∴ Required equation of the lines are 4x – 3y + 15 = 0 and 4x – 3y – 25 = 0

Question 9.
Find the equation of a straight line parallel to 2x + 3y = 10 and which is such that the sum of its intercepts on the axes is 15.
Solution:
The equation of the line parallel to 2x + 3y = 10 will be of the form 2x + 3y = k .

Question 10.
Find the length of the perpendicular and the co-ordinates of the foot of the perpendicular from (-10, -2) to the line x + y – 2 = 0.
Solution:
Length of the perpendicular from (-10, -2) to x + y – 2 = 0 is

Question 11.
If p₁ and p₂ are the lengths of the perpendiculars from the origin to the straight lines. sec θ +y cosec θ = 2a and x cos θ – y sin θ = a cos 2θ, then prove that _{p_{1}^{2}+p_{2}^{2}=a^{2}}
Solution:
p₁ = length of perpendicular from (0, 0) to x sec θ + y cosec θ = 2a

Question 12.
Find the distance between the parallel lines
(i) 12x + 5y = 7 and 12x + 5y + 7 = 0
(ii) 3x – 4y + 5 = 0 and 6x – 8y – 15 = 0.
Solution:

Question 13.
Find the family of straight lines
(i) Perpendicular
(ii) Parallel to 3x + 4y – 12 = 0.
Solution:
(i) The equation of the family of straight lines perpendicular to 3x + 4y – 12 = 0 is 4x – 3y + k = 0 where k ∈ R
(ii) The equation of the family of straight lines parallel to the straight line 3x + 4y – 12 = 0 is 3x + 4y + λ = 0 , λ ∈ R

Question 14.
If the line joining two points A (2, 0) and B (3, 1) is rotated about A in an anti-clockwise direction through an angle of 15°, then find the equation of the line in the new position.
Solution:

This line is rotated about 15° in an anti-clockwise direction
⇒ New slope = tan (45° + 15°) = tan 60° = \(\sqrt{3}\) (i.e) m = \(\sqrt{3}\).
Point A = (2, 0)

Question 15.
A ray of light coming from the point (1, 2) is reflected at a point A on the x-axis and it passes through the point (5, 3). Find the coordinates of point A.
Solution:
The image of the point P (1, 2) will be P’ (1, -2).
Since ∠OAP = ∠XAQ (angle of inches = angle of reflection) So ∠OAP’ = ∠XAQ = a (Vertically opposite angles)
⇒ P’, A, Q lie on the same line.

Now equation of the line P’, Q is [where P’ = (1, -2), Q = (5, 3)]

Since we find a point of intersection with the x-axis we put y = 0.

Question 16.
A line is drawn perpendicular to 5x = y + 7. Find the equation of the line if the area of the triangle formed by this line with co-ordinate axes is 10 sq. units.
Solution:
Equation of the given lines 5x = y + 7 ⇒ 5x – y = 7.
So its perpendicular will be of the form x + 5y = 7

Question 17.
Find the image of the point (-2, 3) about the line x + 2y – 9 = 0.
Solution:
The coordinates of the image of the point (x₁, y₁) with respect to the line ax + by + c = 0 can be

Question 18.
A photocopy store charges Rs. 1.50 per copy for the first 10 copies and Rs. 1.00 per copy after the 10th copy. Let x be the number of copies, and let y be the total cost of photocopying.
(i) Draw a graph of the cost as x goes from 0 to 50 copies.
(ii) Find the cost of making 40 copies
Solution:

Question 19.
Find atleast two equations of the straight lines in the family of the lines y = 5x + b, for which b and the x-coordinate of the point of intersection of the lines with 3x – 4y = 6 are integers.
Solution:
y = 5x + b …….. (1)
3x-4y = 6 …….. (2)
Solving (1) and (2)
Substituting y value from (1) in (2) we get


Since x coordinate and 6 are integers 6 + 46 must be a multiple of 17

Question 20.
Find all the equations of the straight lines in the family of the lines y = mx – 3, for which m and the x-coordinate of the point of intersection of the lines with x – y = 6 are integers.
Solution:
Equation of the given lines are
y= mx – 3 …….. (1)
and x – y = 6 ……. (2)
Solving (1) and (2)
x – (mx – 3) = 6

Since m and x coordinates are integers
1 – m is the divisor of 3 (i.e) ± 1, ± 3

So equation of lines are (y = mx – 3) ,y = mx – 3
(i) When m = 0, y = -3
(ii) When m = 2, y = 2x – 3
(iii) When m = -2, y = -2x – 3 or 2x + y + 3 = 0
(iv) When m = 4, y = 4x – 3

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.3 Additional Questions

Question 1.
Find the equation of the line passing through the point (5, 2) and perpendicular to the line joining the points (2, 3) and (3, -1).
Solution:
Slope of the line joining the points (2, 3) and (3, -1) is

Slope of the required line which is perpendicular to it

Equation of the line passing through the point (5, 2) is

Hence, the required equation is x – 4y + 3 = 0.

Question 2.
Find the points on the line x + y = 4 which lie at a unit distance from the line 4x + 3y = 10. Solution:

So, the required point is (3, 1)
Now taking(-) sign, we have

So, the required point is (- 7, 11)
Hence, the required points on the given line are (3, 1) and (-7, 11).

Question 3.
Find the equation of the line passing through the point of intersection 2x + y = 5 and x + 3y + 8 = 0 and parallel to the line 3x +4y = 7.
Solution:

On putting the value of λ in equation (iv) we get
(2x + y – 5) + 1 (x + 3y + 8) = 0
⇒ 2x + y – 5 + x + 3y + 8 = 0
⇒ 3x + 4y + 3 = 0
Hence, the required equation is 3x + 4y + 3 = 0

Question 4.
A line passing through the points (a, 2a) and (-2, 3) is perpendicular to the line 4 x + 3y + 5 = 0, find the value of a.
Solution:

Question 5.
Find the equation of the straight line which passes through the intersection of the straight lines 2x + y = 8 and 3x – 2y + 7 = 0 and is parallel to the straight line 4x + y – 11 = 0.
Solution:
Equation of line through the intersection of straight lines 2x + y = 8 and 3x – 2y + 7 = 0 is 2x + y – 8 + k (3x – 2y + 7) = 0
x(2 + 3k) + y (1 – 2k) +(-8 + 7k) = 0

⇒ 28x + 7y – 74 = 0

Question 6.
Find the equation of the straight line passing through intersection of the straight lines 5x – 6y = 1 and 3x + 2y + 5 = 0 and perpendicular to the straight line 3x – 5y + 11 = 0.
Solution:
Equation of line through the intersection of straight lines 5x – 6y = 1 and 3x + 2y + 5 = 0 is
5x – 6y – 1 + k (3x + 2y + 5) = 0
x (5 + 3k) + y (-6 + 2k) + (-1 + 5k) = 0
This is perpendicular to 3x – 5y + 11 = 0
That is, the product of their slopes is -1

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 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.

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 7.1

Question 1.
Construct an m × n matrix A = [a_ij], where a_ij is given by

Solution:
(i) a_ij = \(\frac{(i-2 j)^{2}}{2}\)
Here m = 2, n = 3
So we have to construct a matrix of order 2 × 3

(ii) Here m = 3 and n = 4
So we have to construct a matrix order 3 × 4
The general form of a matrix of order 3 × 4 will be

Question 2.
Find the values of p, q, r and s if

Solution:
When two matrices (of the same order) are equal then their corresponding entries are equal.

p² – 1 = 1 ………… (1)
– 31 – q³ = – 4 ……….. (2)
r + 1 = \(\frac{3}{2}\) ……… (3)
s – 1 = – π ……….. (4)
(1) ⇒ p² – 1 ⇒ p² = 1 + 1 = 2
p = ±√2
(2) ⇒ – 31 – q³ – 4
31 + q³ = 4
q³ = 4 – 31 = – 27
q³ = (-3)³
q = – 3
(3) ⇒ r + 1 = \(\frac{3}{2}\)
r = \(\frac{3}{2}\) – 1 = \(\frac{1}{2}\)
(4) ⇒ s – 1 = – π
s = 1 – π
∴ The required values are
p = ±√2,
q = 3,
r = \(\frac{1}{2}\),
s = 1 – π

Question 3.
Determine the value of x + y if
Solution:
\(\left[ \begin{matrix} 2x\quad +\quad y & 4x \\ 5x\quad -\quad 7 & 4x \end{matrix} \right] \) = \(\left[ \begin{matrix} 7 & 7y\quad -\quad 13 \\ y & x\quad +\quad 6 \end{matrix} \right]\)
Equating the corresponding entries
2x + y = 7 ………. (1)
4x = 7y – 13
4x – 7y = – 13 ………. (2)
5x – 7 = y
5x – y = 7 ………. (3)
4x = x + 6
4x – x = 6
3x = 6
x = \(\frac{6}{3}\) = 2 ………. (4)
Substituting for x in equation (1)
(1) ⇒ 2 × 2 + y = 7
y = 7 – 4 = 3
The required values are x = 2 and y = 3
x + y = 2 + 3 = 5
x + y = 5

Question 4.
Determine the matrices A and B if they satisfy

Solution:

Question 5.
If A = \(\left[\begin{array}{ll}{\mathbf{1}} & {\boldsymbol{a}} \\ {\mathbf{0}} & {\mathbf{1}}\end{array}\right]\), then compute A⁴
Solution:

Question 6.
Consider the matrix A_α = \(\left[\begin{array}{cc}{\cos \alpha} & {-\sin \alpha} \\ {\sin \alpha} & {\cos \alpha}\end{array}\right]\)
(i) Show that A_αA_β = A_{α + β}.
(ii) Find all possible real values of satisfying the condition A_α + A^T_α = 1.
Solution:

General solution is α = 2nπ + \(\frac{\pi}{3}\), n ∈ Z

Question 7.
If A = \(\left[\begin{array}{rr}{4} & {2} \\ {-1} & {x}\end{array}\right]\) such that (A – 2I) (A – 3I) = 0, find the value of x.
Solution:

Question 8.
If A = \(\left[\begin{array}{ccc}{\mathbf{1}} & {\mathbf{0}} & {\mathbf{0}} \\ {\mathbf{0}} & {\mathbf{1}} & {\mathbf{0}} \\ {\boldsymbol{a}} & {\boldsymbol{b}} & {-\mathbf{1}}\end{array}\right]\), show that A² is a unit matrix.
Solution:

Question 9.
If A = and A³ – 6A² + 7A + KI = 0, find the value of k.
Solution:

Question 10.
Give your own examples of matrices satisfying the following conditions in each case:
(i) A and B such that AB ≠ BA.
(ii) A and B such that AB = 0 = BA, A ≠ 0 and B ≠ 0.
(iii) A and B such that AB = 0 and BA ≠ 0.
Solution:

Question 11.
Show that f(x) f(y) = f(x + y), where f(x) =
Solution:

Question 12.
If A is a square matrix such that A² = A, find the value of 7A – (I + A)³.
Solution:
Given A is a square matrix such that A² = A.
(I + A)³ = (I + A) (I + A) (I + A)
= (I . I + I . A + A . I + A . A) (I + A)
= (I + A + A + A²) (I + A)
= (I + 2A + A ) (I + A)
[Given A² = A]
= (I + 3A) (I + A)
= I . I + I . A + 3A . I + 3A . A
= I + A + 3A + 3A²
= I + 4A + 3A
[Given A² = A]
(I + A)³ = I + 7A
∴ 7A – (I + A)² = 7A – (I + 7A)
= 7A – I – 7A
7A – (I + A)² = – I

Question 13.
Verify the property A (B + C) = AB + AC, when the matrices A, B, and C are given by

Solution:

Question 14.
Find the matrix A which satisfies the matrix relation
Solution:

Question 15.

(i) (A + B)^T = A^T + B^T = B^T + A^T
(ii) (A – B)^T = A^T – B^T
(iii) (B^T)^T = B.
Solution:

Question 16.
If A is a 3 × 4 matrix and B is a matrix such that both ATB and BAT are defined, what is the order of the matrix B?
Solution:
Given Order of A = 3 × 4
∴ Order of A^T = 4 × 3
Given that A^TB is defined.
∴ Number of columns of A^T = Number of rows of B
Number of rows of B = 3
Also given BA^T is defined.
∴ Number of columns of B = Number of rows of A^T
Number of columns of B = 4
∴ Order of B = 3 × 4

Question 17.
Express the following matrices is the sum of a symmetric matrix and a skew-symmetric matrix:

Solution:

Question 18.
Find the matrix A such that
Solution:

Question 19.
If A = is a matrix such that AA^T = 9I, find the values of x and y.
Solution:

Question 20.

Solution:

Question 21.
Construct the matrix A = [a_ij]_3×3, where a_ij = i- j. State whether A is symmetric or skew- symmetric.
Solution:
Given A is a matrix of order 3 × 3

Here A^T = -A
⇒ A is skew-symmetric

Question 22.
Let A and B be two symmetric matrices. Prove that AB = BA if and only if AB is a symmetric matrix.
Solution:
Let A and B be two symmetric matrices
⇒ A^T = A and B^T = B …………….. (1)
Given that AB = BA (2)
To prove AB is symmetric:
Now (AB)^T = B^TA^T = BA
(from(1)) But (AB)^T = AB by ………….. (2)
⇒ AB is symmetric.
Conversely, let AB be a symmetric matrix.
⇒ (AB)^T = AB
i.e. B^TA^T = AB
i.e. BA = AB (from (1))
⇒ AB is symmetric

Question 23.
If A and B are symmetric matrices of the same order, prove that
(i) AB + BA is a symmetric matrix.
(li) AB – BA is a skew-symmetric matrix.
Solution:
Given A and B are symmetric matrices of the same order.
∴ A^T = A , B^T = B

(i) AB + BA is a symmetric matrix
(AB + BA)^T = (AB )^T + (BA)^T
= B^TA^T + A^TB^T
= BA + AB
= AB + BA
∴ AB + BA is a symmetric matrix.

(ii) AB – BA is a skew – symmetric matrix
(AB – BA)^T = (AB )^T (BA )^T
= B^TA^T – A^TB^T
= BA – AB
(AB – BA)^T = – (AB – BA)
∴ AB – BA is a skew symmetric matrix.

Question 24.
A shopkeeper in a Nuts and Spices shop makes gift packs of cashew nuts, raisins, and almonds.
The pack-I contains 100 gm of cashew nuts, 100 gm of raisins, and 50 gm of almonds. Pack-II contains 200 gm of cashew nuts, 100 gm of raisins, and 100 gm of almonds. Pack-III contains 250 gm of cashew nuts, 250 gm of raisins, and 150 gm of almonds. The cost of 50 gm of cashew nuts is ₹ 50, 50 gm of raisins is ₹ 10, and 50 gm of almonds is₹ 60. What is the cost of each gift pack?
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.1 Additional Problems

Question 1.

Prove that (i) AB ≠ BA
(ii) A(BC) = (AB) C
(iii) A(B + C) = AB + AC
(iv) AI = IA = A
Solution:

Question 2.
If A = \(\left[\begin{array}{ll}{2} & {3} \\ {4} & {5}\end{array}\right]\) find A² – 7A – 2I.
Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.
If A = \(\left[\begin{array}{rr}{3} & {-5} \\ {-4} & {2}\end{array}\right]\), show that A² – 5A – 14I = 0 where I is the unit matrix of order 2.
Solution:

Question 6.
If A = \(\left[\begin{array}{rr}{3} & {-2} \\ {4} & {-2}\end{array}\right]\), find k so that A² = kA – 2I
Solution:

Question 7.
If A = \(\left[\begin{array}{lll}{1} & {2} & {2} \\ {2} & {1} & {2} \\ {2} & {2} & {1}\end{array}\right]\), show that A² – 4A – 5I = 0
Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

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 5 Binomial Theorem, Sequences and Series Ex 5.4

Expanding Binomials Calculator online with solution and steps.

Question 1.
Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid.
Solution:

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

Question 3.

Solution:

Question 4.

Solution:

Question 5.
Write the first 6 terms of the exponential series
(i) e^5x
(ii) e^-2x
(iii) \(e^{\frac{x}{2}}\)
Solution:

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.
Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

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

Question 1.

Solution:

Question 2.
Write the four terms in the expansions of the following:

Solution:

Question 3.
Evaluate the following:

Solution:

Question 4.

Solution:

Question 5.

Solution:

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Samacheer Kalvi 11th Commerce Solutions Chapter 32 Direct Taxes

Samacheer Kalvi 11th Commerce Direct Taxes Textbook Exercise Questions and Answers

I. Choose the Correct Answer

Question 1.
Income Tax is ……………
(a) a business tax
(b) a direct tax
(c) an indirect tax
(d) none of these
Answer:
(b) a direct tax

Question 2.
Period of assessment year is ……………
(a) 1^st April to 31^st March
(b) 1^st March to 28^st Feb
(c) 1^st July to 30^st June
(d) 1^st Jan to 31^st Dec
Answer:
(a) 1^st April to 31^st March

Question 3.
The year in which income is earned is known as ……………
(a) Assessment Year
(b) Previous Year
(c) Light Year
(d) Calendar Year
Answer:
(b) Previous Year

Question 4.
The aggregate income under five heads is termed as ……………
(a) Gross Total Income
(b) Total Income
(c) Salary Income
(d) Business Income
Answer:
(b) Total Income

Question 5.
Agricultural income earned in India is ……………
(a) Fully Taxable
(b) Fully Exempted
(c) Not Considered for Income
(d) None of the above
Answer:
(b) Fully Exempted

II. Very Short Answer Questions

Question 1.
What is Income tax?
Answer:
Income tax is a direct tax under which tax is calculated on the income, gains or profits earned by a person such as individuals and other artificial entities.

Question 2.
What is meant by the previous year?
Answer:
The year in which income is earned is called the previous year. It is also normally consisting of a period of 12 months commencing on 1^st April every year and ending on 31^st March of the following year. It is also called a financial year immediately following the assessment year.

Question 3.
Define the term person?
Answer:
According to the Income Tax 1961, the term Person includes, “ individual, Hindu Undivided Family, Fìrrn, Company, local authority, Association of person or Body of Individual or any other artificial juridical persons.

Question 4.
Define the term assessee?
Answer:
Assessee means a person by whom any tax or any other sum of money is payable under this Act. It includes every person in respect of whom any proceeding has been taken for the assessment of his income or assessment of fringe benefits.

Question 5.
What is an assessment year?
Answer:
The term has been defined under section 2(9). The year in which tax is paid iš called the assessment year. It normally consisting of a period óf 12 months còrnmencing on 1st April every year and ending on 31 st March of the following year.

III. Short Answer Questions

Question 1.
What is Gross Total Income?
Answer:
Income from the five heads, namely – Salaries, House Property, Profits and Gains of Business or Profession, Capital Gains, and Other Sources – is computed separately according to the provisions given in the Act. Income computed under these heads shall be aggregated after adjusting past and present losses and the total so arrived at is known as ‘Gross Total Income’.

Question 2.
List out the five heads of income.
Answer:

1. Income from Salaries
2. Income from House Property
3. Income from Business or Profession
4. Income from Capital Gain
5. Income from Other Sources

Question 3.
Write a note on Agricultural Income.
Answer:
Any rent or revenue derived from land which is situated in India and is used for agriculture purposes. Agricultural income is fully exempted from tax u/s 10(1) and as such does not form part of total income.

Question 4.
What do you mean by Total Income?
Answer:
According to Section 2 (45) of the Income Tax Act 1961, the income which is earned out of the Gross Total Income, after allowing the deductions under section 80 is known as total income.

Question 5.
Write short notes on:

1. Direct Tax
2. Indirect Tax

Answer:
1. Direct Tax:
If a tax levied on the income or wealth of a person and is paid by that person (or his office) directly to the Government, , it is called a direct tax, for example, Income – Tax, Wealth Tax, Capital Gains Tax, Securities Transaction Tax, Fringe Benefits Tax (from 2005), Banking Cash Transaction Tax (for Rs,50,000 and above – from 2005), etc. In India, all direct taxes are levied and administered by the Central Board of Direct Taxes.

2. Indirect Tax:
If tax is levied on the goods or services of a person (seller). It is collected from, the buyers and is paid by the seller to the Government. It is called indirect tax example GST.

IV. Long Answer Questions

Question 1.
Elucidate any five features of Income Tax.
Answer:
Meaning: Income tax is a direct tax under which tax is calculated on the income, gains or profits earned by a person such as individuals and other artificial entitled.
Levied as Per the Constitution
Income tax is levied in India by virtue of entry No. 82 of list 1 (Union List) of Seventh Schedule to Article 246 of the Constitution of India.

Levied by Central Government
Income tax is charged by the Central Government on all incomes other than agricultural income. However, the power to charge income tax on agricultural income has been vested with the State Government as per entry 46 of List II, i.e., State List.

Direct Tax
Income tax is a direct tax. It is because the liability to deposit and ultimate burden are on the same person. The person earning income is liable to pay income tax out of his own pocket and cannot pass on the burden of tax to another person.

Annual Tax
Income tax is an annual tax because it is the income of a particular year which is chargeable to tax.

Tax on Person
It is a tax on income earned by a person. The term ‘person’ has been defined under the Income-tax Act. It includes individual, Hindu Undivided Family, Firm, Company, local authority, Association of person or Body of Individual or any other artificial juridical persons.

Question 2.
Define Tax. Explain the term direct tax and indirect tax with an example.
Answer:
Tax is a compulsory contribution to state revenue by the Government. It is levied on the income or profits from business of individuals and institutions. It may be added to the price of goods, services or transactions. Tax is the basic source of revenue to the Government. This revenue is utilised for the expenses of civil administration, internal and external security, building infrastructure, etc.

There are two types of taxes – direct taxes and indirect, taxes.

1. Direct Tax:
If a tax levied on the income or wealth of a person and is paid by that person (or his office) directly to the Government, it is called direct tax, e.g., Income – Tax, Wealth Tax, Capital Gains Tax, Securities Transaction Tax, Fringe Benefits Tax (from 2005), Banking Cash Transaction Tax (for Rs.50,000 and above – from 2005), etc. In India all direct taxes are levied and administered by Central Board of Direct Taxes.

2. Indirect Tax:
If tax is levied on the goods or services of a person (seller). It is collected from the buyers and is paid by seller to the Government. It is called indirect tax. example GST.

Question 3.
List out any ten kinds of incomes chargeable under the head income tax.
Answer:
According to section 2(24) of the Income Tax Act, Income includes the following:

1. Profits and gains of business or profession.
2. Dividend
3. Export incentives, like duty drawback, cash compensatory support, sale of licenses, etc.,
4. Interest, salary, bonus, commission or remuneration earned by a partner of a firm from such firm.
5. Capital gains chargeable u/s 45. viii. Profits and gains from the business of banking carried on by a co-operative society with its members.
6. Winnings from lotteries, crossword puzzles, races including horse races, card games, and other games of any sort or from gambling or betting of any form or nature whatsoever.
7. Deemed income u/s 41 or 59.
8. Amount received under key man. insurance policy including bonus thereon.
9. Benefit or perquisite received from a company, by a director or a person holding substantial interest or a relative of the director or such person.
10. Gift as defined u/s 56 (2) (vi) and others.

Question 4.
Discuss the various kinds of assessments.
Answer:
Assessee means a person by whom any tax or any other sum of money is payable under this Act. It includes every person in respect of whom any proceeding has been taken for the assessment of his income or assessment of fringe benefits. The term ‘person’ includes the following:

1. an individual
2. a Hindu Undivided Family (HUE)
3. a company
4. a firm
5. an Association Of Persons or a Body Of Individual, whether incorporated or not
6. a local authority, and
7. every artificial juridical person example an idol or deity.

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Tamilnadu Samacheer Kalvi 11th English Speaking Group Discussion

(i) Intelligent Quotient (IQ) Versus Emotional Quotient (EQ)

RAJ:
According to my opinion, those who have a good IQ level will always have the ability to reason out things and find an amicable solution to any problem that he/she faces in life. Even in today’s world, when a young aspirant is made to face an interview with a board of panelist, one of the key methods to finalise the situation is a problem solving abilities a candidate possesses. Nowadays, your mental appearance is more important than your physical appearance. Though eloquence in communication was given weight age, the skill and ability to convince others is much more important than flamboyant language. Presence of mind, common sense and practicality is more important than bookish knowledge. Therefore people with more IQ is needed than people with better EQ.

Chandran:
According to my understanding, I strongly contradict the views of Raj. It is the character that brings you huge success and I think good character, good personalities and good leaders are mostly driven by emotions.

Emotional quotient and intelligence quotient are undoubtedly the most important aspects of a person’s life and can make a huge difference between an average life and a successful life. To be successful, one may want to have a great IQ but, that success will not last long if your EQ is not appropriate. But on the contrary, if you have a good EQ and a little less IQ, then the chances of you becoming successful to carry on a successful career increases.

LEELA:
Everyone is equal. All should be recognised. One needs to be balanced with all qualities just as a balanced diet is needed for healthy living. It is illogical for learned people to say that IQ is much more needed than EQ or vice versa. I feel everyone is unique and is important. There is always something that one can contribute to. It needs to be balanced. Team play is vital at such a situation. A person with better IQ/EQ has his own workspace and methodology he contributes to the work differently as a person with different IQ/EQ. One who has a high IQ level can work smartly to do even the toughest tasks easily and hence procure popularity and wealth. Likewise, one who has better EQ can develop the same popularity and also leadership qualities. He/She can withstand adverse condition and overcome them.

Let me detail you with examples. To manage a team, a person of high EQ is needed whereas someone who needs to plan tasks and perform it in the best possible way, a person with high IQ should be there.

A person with higher IQ may work on any area and be extremely successful but may lack good social circle. This will surely lead to depression or emotional setbacks.

A person with higher EQ may win a large social circle of friends and supporters who are a big pillar of support at difficult times but due to their lack of technical abilities that needs better IQ, will turn out to be failure.

Hence, I want to reiterate that a person with an average IQ level and an equal amount of EQ level will be the best to survive in the present society. He can work anywhere happily with a fair wealth and fair emotional stability. He would be the most satisfied person.

Therefore a balanced IQ & EQ is the most desirable to be successful.

RAJ:
But I strongly feel that IQ is more important than EQ. Emotions are the biggest barrier to decision-making abilities. If we think about what others say and starts taking into consideration of others opinion, we get messed up and hence lack in making the right decision. Aren’t we living in a world where everything changes in an instant. So, we need someone who is confident of themselves and not get carried away by emotions. Hence I would prefer IQ over EQ,

Chandran:
I strongly opine that you consider the weight age of EQ more than IQ because if you . are perfect in EQ then you can manage with less IQ. For instance, if you have the qualities of EQ such as good personality skills, a better way of communication, and good attitude you can face any competitive situation. On the contrary, although you have strong IQ but poor EQ, then no one will take you seriously or you will be less focused leading to a few people joining you. So overall conclusion is that you don’t need to have a balanced or same level of IQ or EQ. If you are the best in EQ, then you can easily survive in this competitive era with less IQ.

LEELA:
There is no point in Mr. Raj arguing for IQ and Mr.Chandran taking sides with those who have better EQ, IQ and EQ have their own importance in our life and both should be in balanced proportion. I feel IQ makes us ambitious but EQ put limits on these ambitions. High IQ and Low EQ are the traits that can be observed in terrorists. So today, our topic is IQ vs EQ. I stay strong in both needed for a successful person. EQ is important for our social . life. AS we live in a society, we need good EQ levels for our personal and social behaviour and character to be well accepted in the society. Whereas, better IQ is surely the need for enhancing our knowledge to take the right decisions in this society and be a successful person. A balance between both is the need of the hour for both act as the two wheels in our journey of a successful life cycle.

(ii) To Live or Not

Kiruba:
We live in a world where most of us have a pet at home. When they fall sick, we rush them to the hospital. When our heart melts to see an animal or a bird suffer because of sickness, how will our conscience allow us to kill a fellow human being just because he is incurably ill? Moreover, he has done no harm to the society and his illness is not because of his fault.Therefore, we must provide him proper treatment and allow him to live as long as nature and God has willed his life span to be in this world.

Suganthi:
Exactly. Well said. God has gifted us life. So, he alone has the right to take it back. No human being has the right to interfere in His decisions or will for us in this world. Every man/woman has the right to live as long as God intends him/her to lead a life in this world.

Therefore, the illogical reasoning that a man suffering from an incurable disease, should be put to death is inexcusable beyond reason and a thought to be condemned.

Babu:
This world is governed by Darwin’s survival of the fittest principle. Darwin drew parallels between his own economic theories and his biological ones: “This survival of the fittest”, which he had sought to express in mechanical terms, is that which he has also called ‘natural selection’, or the preservation of favoured races in the struggle for life. An incurably diseased person is weak and has no value whatsoever to the society. More so, he has no means to live. Therefore, it would be in the fitness of things to kill him even against his wish. Therefore I bring to your kind notice that it is the natural selection to kill a person who cannot live befitting the needs of the society.

Kiruba:
Now a days, new vistas of progress have been opened in medical sciences and alternative ‘ medicine like Acupuncture, Acupressure, Reiki Pranik healing, Touch therapy, Herbal therapy, Diet therapy, etc. hold a ray of hope for the so called incurably diseased persons. So, why snatch life from them? When the kith and kin are ready to try all the means for survival, why should one interfere and speak things agaainst their survival. Miracles do happen and ‘ prayers are surely answered. Nothing wrong in giving a chance for someone fighting and struggling to live. We have no right to speak about putting another person to death.

Babu:
An incurably diseased person is the cause of constant worry to his family. His demands are unending and not withstanding the best possible attention, care and treatment given to him, he always remains dissatisfied and disgruntled. This adversely affects the peace of mind and comfort of the family members and all those who are trying to help them. Many a times, the physical worry and mental agony faced by the caretakers fall a prey to many other difficult situations and even fall sick being in an unhealthy environment for weeks and months. Therefore, the best way out of such a situation is to put an end to his life.

Suganthi:
It is not always the case that incurably diseased persons spread contagious diseases as some might argue. Even in those rare cases where it may be true, these persons are not real health hazards because it is medically established now that all incurable diseases are not contagious. However, as a precautionary measure, we should open separate hospitals or isolation wards for persons suffering from incurable contagious diseases and thus quarantine them. Isn’t that a better solution than thinking of putting an end to their life? To me, it’s a murder under legal permits.

Babu:
But these days we are burdened with the responsibility of reducing our rapidly increasing population. The many diseased persons constitute a good part of it. Even otherwise their contribution to society is a burden. It would be justified and reasonable not to allow them to drag on their agonizing life. Killing an incurably diseased person will put an end to research work in medical science. As I have already stated, there are many hospitals and many doctors who start experimenting with diseased persons. Even otherwise, suffering people have been the subject of research work for those who are into higher education. Thereby, its appropriate to free them from pain and suffering!

Kiruba:
Such an act only points out to the degenerated values in this society. It is almost compared to that of an individual throwing an appliance when it is old and doesn’t yield good results. Isn’t there a difference between living and non-living? How can a living human being be treated like a material object? What has happened to the values that students were taught or imbibed in their life? Such a thinking is only an assertion of degenerated moral values.

Conclusion:
India is a land of values. We are a secular society. We believe in respecting each other. Let us not be carried away by degenerated moral values and learn to respect one another and help them to live in peace without a feeling of guilt.