Category: Class 11

Students can download 11th Business Maths Chapter 3 Analytical Geometry Ex 3.4 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 3 Analytical Geometry Ex 3.4

Samacheer Kalvi 11th Business Maths Analytical Geometry Ex 3.4 Text Book Back Questions and Answers

Question 1.
Find the equation of the following circles having
(i) the centre (3, 5) and radius 5 units.
(ii) the centre (0, 0) and radius 2 units.
Solution:
(i) Equation of the circle is (x – h)² + (y – k)² = r²
Centre (h, k) = (3, 5) and radius r = 5
∴ Equation of the circle is (x – 3)² + (y – 5)² = 5²
⇒ x² – 6x + 9 + y² – 10y + 25 = 25
⇒ x² + y² – 6x – 10y + 9 = 0

(ii) Equation of the circle when centre origin (0, 0) and radius r is x² + y² = r²
⇒ x² + y² = 2²
⇒ x² + y² = 4
⇒ x² + y² – 4 = 0

Question 2.
Find the centre and radius of the circle
(i) x² + y² = 16
(ii) x² + y² – 22x – 4y + 25 = 0
(iii) 5x² + 5y²+ 4x – 8y – 16 = 0
(iv) (x + 2) (x – 5) + (y – 2) (y – 1) = 0
Solution:
(i) x² + y² = 16
⇒ x² + y² = 4²
This is a circle whose centre is origin (0, 0), radius 4.

(ii) Comparing x² + y² – 22x – 4y + 25 = 0 with general equation of circle x² + y² + 2gx + 2fy + c = 0
We get 2g = -22, 2f = -4, c = 25
g = -11, f = -2, c = 25
Centre = (-g, -f) = (11, 2)

(iii) 5x² + 5y² + 4x – 8y – 16 = 0
To make coefficient of x² unity, divide the equation by 5 we get,
\(x^{2}+y^{2}+\frac{4}{5} x-\frac{8}{5} y-\frac{16}{5}=0\)
Comparing the above equation with x² + y² + 2gx + 2fy + c = 0 we get,

(iv) Equation of the circle is (x + 2) (x – 5) + (y – 2) (y – 1) = 0
x² – 3x – 10 + y² – 3y + 2 = 0
x² + y² – 3x – 3y – 8 = 0
Comparing this with x² + y² + 2gx + 2fy + c = 0
We get 2g = -3, 2f = -3, c = -8
g = \(\frac{-3}{2}\), f = \(\frac{-3}{2}\), c = -8
Centre (-g, -f) = \(\left(\frac{3}{2}, \frac{3}{2}\right)\)

Question 3.
Find the equation of the circle whose centre is (-3, -2) and having circumference 16π.
Solution:
Circumference, 2πr = 16π
⇒ 2r = 16
⇒ r = 8
Equation of the circle when centre and radius are known is (x – h)² + (y – k)² = r²
⇒ (x + 3)² + (y + 2)² = 8²
⇒ x² + 6x + 9 + y² + 4y + 4 = 64
⇒ x² + y² + 6x + 4y + 13 = 64
⇒ x² + y² + 6x + 4y – 51 = 0

Question 4.
Find the equation of the circle whose centre is (2, 3) and which passes through (1, 4).
Solution:

Centre (h, k) = (2, 3)

Equation of the circle with centre (h, k) and radius r is (x – h)² + (y – k)² = r²
⇒ (x – 2)² + (y – 3)² = (√2)²
⇒ x² – 4x + 4 + y² – 6y + 9 = 2
⇒ x² + y² – 4x – 6y + 11 = 0

Question 5.
Find the equation of the circle passing through the points (0, 1), (4, 3) and (1, -1).
Solution:
Let the required of the circle be x² + y² + 2gx + 2fy + c = 0 ……… (1)
It passes through (0, 1)
0 + 1 + 2g(0) + 2f(1) + c = 0
1 + 2f + c = 0
2f + c = -1 …….. (2)
Again the circle (1) passes through (4, 3)
4² + 3² + 2g(4) + 2f(3) + c = 0
16 + 9 + 8g + 6f + c = 0
8g + 6f + c = -25 …….. (3)
Again the circle (1) passes through (1, -1)
1² + (-1)² + 2g(1) + 2f(-1) + c = 0
1 + 1 + 2g – 2f + c = 0
2g – 2f + c = -2 ……… (4)
8g + 6f + c = -25
(4) × 4 subtracting we get, 8g – 8f + 4c = -8
14f – 3c = -17 ………. (5)
14f – 3c = -17
(2) × 3 ⇒ 6f + 3c = -3
Adding we get 20f = -20
f = -1
Using f = -1 in (2) we get, 2(-1) + c = -1
c = -1 + 2
c = 1
Using f = -1, c = 1 in (3) we get
8g + 6(-1)+1 = -25
8g – 6 + 1 = -25
8g – 5 = -25
8g = -20
g = \(\frac{-20}{8}=\frac{-5}{2}\)
using g = \(\frac{-5}{2}\), f = -1, c = 1 in (1) we get the equation of the circle.
x² + y² + 2(\(\frac{-5}{2}\))x + 2(-1)y + 1 = 0
x² + y² – 5x – 2y + 1 = 0

Question 6.
Find the equation of the circle on the line joining the points (1, 0), (0, 1), and having its centre on the line x + y = 1.
Solution:

Let the equation of the circle be
x² + y² + 2gx + 2fy + c = 0 ……… (1)
The circle passes through (1, 0)
1² + 0² + 2g(1) + 2f(0) + c = 0
1 + 2g + c = 0
2g + c = 1 …….. (2)
Again the circle (1) passes through (0, 1)
0² + 1² + 2g(0) + 2f(1) + c = 0
1 + 2f + c = 0
2f + c = -1 ……. (3)
(2) – (3) gives 2g – 2f = 0 (or) g – f = 0 ………. (4)
Given that the centre of the circle (-g, -f) lies on the line x + y = 1
-g – f = 1 …….. (5)
(4) + (5) gives -2f = 1 ⇒ f = \(-\frac{1}{2}\)
Using f = \(-\frac{1}{2}\) in (5) we get
-g – (\(-\frac{1}{2}\)) = 1
-g = 1 – \(-\frac{1}{2}\) = \(\frac{1}{2}\)
g = \(-\frac{1}{2}\)
Using g = \(-\frac{1}{2}\) in (2) we get
2(\(-\frac{1}{2}\)) + c = -1
-1 + c = -1
c = 0
using g = \(-\frac{1}{2}\), f = \(-\frac{1}{2}\), c = 0 in (1) we get the equation of the circle,
x² + y² + 2(\(-\frac{1}{2}\))x + 2(\(-\frac{1}{2}\))y + 0 = 0
x² + y² – x – y = 0

Question 7.
If the lines x + y = 6 and x + 2y = 4 are diameters of the circle, and the circle passes through the point (2, 6) then find its equation.
Solution:
To get coordinates of centre we should solve the equations of the diameters x + y = 6, x + 2y = 4.
x + y = 6 ……. (1)
x + 2y = 4 ………. (2)
(1) – (2) ⇒ -y = 2
y = -2
Using y = -2 in (1) we get x – 2 = 6
x = 8
Centre is (8, -2) the circle passes through the point (2, 6).

Equation of the circle with centre (h, k) and radius r is (x – h)² + (y – k)² = r²
⇒ (x – 8)² + (y + 2)² = 10²
⇒ x² + y² – 16x + 4y + 64 + 4 = 100
⇒ x² + y² – 16x + 4y – 32 = 0

Question 8.
Find the equation of the circle having (4, 7) and (-2, 5) as the extremities of a diameter.
Solution:
The equation of the circle when entremities (x₁, y₁) and (x₂, y₂) are given is (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
⇒ (x – 4) (x + 2) + (y – 7) (y – 5) = 0
⇒ x² – 2x – 8 + y² – 12y + 35 = 0
⇒ x² + y² – 2x – 12y + 27 = 0

Question 9.
Find the Cartesian equation of the circle whose parametric equations are x = 3 cos θ, y = 3 sin θ, 0 ≤ θ ≤ 2π.
Solution:
Given x = 3 cos θ, y = 3 sin θ
Now x² + y² = 9 cos²θ + 9 sin²θ
x² + y² = 9 (cos²θ + sin²θ)
x² + y² = 9 which is the Cartesian equation of the required circle.

Students can download 11th Business Maths Chapter 3 Analytical Geometry Ex 3.7 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 3 Analytical Geometry Ex 3.7

Samacheer Kalvi 11th Business Maths Analytical Geometry Ex 3.7 Text Book Back Questions and Answers

Question 1.
If m₁ and m₂ are the slopes of the pair of lines given by ax² + 2hxy + by² = 0, then the value of m₁ + m₂ is:
(a) \(\frac{2 h}{b}\)
(b) \(-\frac{2 h}{b}\)
(c) \(\frac{2 h}{a}\)
(d) \(-\frac{2 h}{a}\)
Answer:
(b) \(-\frac{2 h}{b}\)

Question 2.
The angle between the pair of straight lines x² – 7xy + 4y² = 0 is:
(a) \(\tan ^{-1}\left(\frac{1}{3}\right)\)
(b) \(\tan ^{-1}\left(\frac{1}{2}\right)\)
(c) \(\tan ^{-1}\left(\frac{\sqrt{33}}{5}\right)\)
(d) \(\tan ^{-1}\left(\frac{5}{\sqrt{33}}\right)\)
Answer:
(c) \(\tan ^{-1}\left(\frac{\sqrt{33}}{5}\right)\)
Hint:
x² – 7xy + 4y² = 0
Here 2h = -7, a = 1, b = 4

Question 3.
If the lines 2x – 3y – 5 = 0 and 3x – 4y – 7 = 0 are the diameters of a circle, then its centre is:
(a) (-1, 1)
(b) (1, 1)
(c) ( 1, -1)
(d) (-1, -1)
Answer:
(c) ( 1, -1)
Hint:
To get centre we must solve the given equations
2x – 3y – 5 = 0 …….(1)
3x – 4y – 7 = 0 ………(2)
(1) × 3 ⇒ 6x – 9y = 15
(2) × 2 ⇒ 6x – 8y = 14
Subtracting, -y = 1 ⇒ y = -1
Using y = -1 in (1) we get
2x + 3 – 5 = 0
⇒ 2x = 2
⇒ x = 1

Question 4.
The x-intercept of the straight line 3x + 2y – 1 = 0 is
(a) 3
(b) 2
(c) \(\frac{1}{3}\)
(d) \(\frac{1}{2}\)
Answer:
(c) \(\frac{1}{3}\)
Hint:
To get x-intercept put y = 0 in 3x + 2y – 1 = 0 we get
3x – 1 = 0
x = \(\frac{1}{3}\)

Question 5.
The slope of the line 7x + 5y – 8 = 0 is:
(a) \(\frac{7}{5}\)
(b) \(-\frac{7}{5}\)
(c) \(\frac{5}{7}\)
(d) \(-\frac{5}{7}\)
Answer:
(b) \(-\frac{7}{5}\)
Hint:
Slope of 7x + 5y – 8 = 0 is = \(\frac{-x \text { coefficient }}{y \text { coefficient }}\) = \(-\frac{7}{5}\)

Question 6.
The locus of the point P which moves such that P is at equidistance from their coordinate axes is:
(a) y = \(\frac{1}{x}\)
(b) y = -x
(c) y = x
(d) y = \(\frac{-1}{x}\)
Answer:
(c) y = x
Hint:

Given PA = PB
y₁ = x₁
∴ Locus is y = x

Question 7.
The locus of the point P which moves such that P is always at equidistance from the line x + 2y + 7 = 0:
(a) x + 2y + 2 = 0
(b) x – 2y + 1 = 0
(c) 2x – y + 2 = 0
(d) 3x + y + 1 = 0
Answer:
(a) x + 2y + 2 = 0
Hint:
Locus is line parallel to line x + 2y + 7 = 0 which is x + 2y + 2 = 0

Question 8.
If kx² + 3xy – 2y² = 0 represent a pair of lines which are perpendicular then k is equal to:
(a) \(\frac{1}{2}\)
(b) \(-\frac{1}{2}\)
(c) 2
(d) -2
Answer:
(c) 2
Hint:
Here a = k, b = -2
Condition for perpendicular is
a + b = 0
⇒ k – 2 = 0
⇒ k = 2

Question 9.
(1, -2) is the centre of the circle x² + y² + ax + by – 4 = 0, then its radius:
(a) 3
(b) 2
(c) 4
(d) 1
Answer:
(a) 3
Hint:
Given centre (-g, -f) = (1, -2)
From the given equation c = -4
Radius = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{1+4-(-4)}=\sqrt{9}\) = 3

Question 10.
The length of the tangent from (4, 5) to the circle x² + y² = 16 is:
(a) 4
(b) 5
(c) 16
(d) 25
Answer:
(b) 5
Hint:
Length of the tangent from (x₁, y₁) to the circle x² + y² = 16 is \(\sqrt{x_{1}^{2}+y_{1}^{2}-16}=5\)

Question 11.
The focus of the parabola x² = 16y is:
(a) (4 , 0)
(b) (-4, 0)
(c) (0, 4)
(d) (0, -4)
Answer:
(c) (0, 4)
Hint:
x² = 16y
Here 4a = 16 ⇒ a = 4
Focus is (0, a) = (0, 4)

Question 12.
Length of the latus rectum of the parabola y² = -25x:
(a) 25
(b) -5
(c) 5
(d) -25
Answer:
(a) 25
Hint:
y² = -25a
Here 4a = 25 which is the length of the latus rectum.

Question 13.
The centre of the circle x² + y² – 2x + 2y – 9 = 0 is:
(a) (1, 1)
(b) (-1, 1)
(c) (-1, 1)
(d) (1, -1)
Answer:
(d) (1, -1)
Hint:
2g = -2, 2f = 2
g = -1, f = 1
Centre = (-g, -f) = (1, -1)

Question 14.
The equation of the circle with centre on the x axis and passing through the origin is:
(a) x² – 2ax + y² = 0
(b) y² – 2ay + x² = 0
(c) x² + y² = a²
(d) x² – 2ay + y² = 0
Answer:
(a) x² – 2ax + y² = 0
Hint:
Let the centre on the x-axis as (a, 0).
This circle passing through the origin so the radius

Now centre (h, k) = (a, 0)
Radius = a
Equation of the circle is (x – a)² + (y – 0)² = a²
⇒ x² – 2ax + a² + y² = a²
⇒ x² – 2ax + y² = 0

Question 15.
If the centre of the circle is (-a, -b) and radius is \(\sqrt{a^{2}-b^{2}}\) then the equation of circle is:
(a) x² + y² + 2ax + 2by + 2b² = 0
(b) x² + y² + 2ax + 2by – 2b² = 0
(c) x² + y² – 2ax – 2by – 2b² = 0
(d) x² + y² – 2ax – 2by + 2b² = 0
Answer:
(a) x² + y² + 2ax + 2by + 2b² = 0
Hint:
Equation of the circle is (x – h)² + (y – k)² = r²
⇒ (x + a)² + (y + b)² = a² – b²
⇒ x² + y² + 2ax + 2by + a² + b² = a² – b²
⇒ x² + y² + 2ax + 2by + 2b² = 0

Question 16.
Combined equation of co-ordinate axes is:
(a) x² – y² = 0
(b) x² + y² = 0
(c) xy = c
(d) xy = 0
Answer:
(d) xy = 0
Hint:
Equation of x-axis is y = 0
Equation of y-axis is x = 0
Combine equation is xy = 0

Question 17.
ax² + 4xy + 2y² = 0 represents a pair of parallel lines then ‘a’ is:
(a) 2
(b) -2
(c) 4
(d) -4
Answer:
(a) 2
Hint:
Here a = 0, h = 2, b = 2
Condition for pair of parallel lines is b² – ab = 0
4 – a(2) = 0
⇒ -2a = -4
⇒ a = 2

Question 18.
In the equation of the circle x² + y² = 16 then v intercept is (are):
(a) 4
(b) 16
(c) ±4
(d) ±16
Answer:
(c) ±4
Hint:
To get y-intercept put x = 0 in the circle equation we get
0 + y² = 16
∴ y = ±4

Question 19.
If the perimeter of the circle is 8π units and centre is (2, 2) then the equation of the circle is:
(a) (x – 2)² + (y – 2)² = 4
(b) (x – 2)² + (y – 2)² = 16
(c) (x – 4)² + (y – 4)² = 16
(d) x² + y² = 4
Answer:
(c) (x – 2)² + (y – 2)² = 16
Hint:
Perimeter, 2πr = 8π
r = 4
Centre is (2, 2)
Equation of the circle is (x – 2)² + (y – 2)² = 4² = 16

Question 20.
The equation of the circle with centre (3, -4) and touches the x-axis is:
(a) (x – 3)² + (y – 4)² = 4
(b) (x – 3)² + (y + 4)² = 16
(c) (x – 3)² + (y – 4)² = 16
(d) x² + y² = 16
Answer:
(b) (x – 3)² + (y + 4)² = 16
Hint:
Centre (3, -4).
It touches the x-axis.
The absolute value of y-coordinate is the radius, i.e., radius = 4.
Equation is (x – 3)² + (y + 4)² = 16

Question 21.
If the circle touches the x-axis, y-axis, and the line x = 6 then the length of the diameter of the circle is:
(a) 6
(b) 3
(c) 12
(d) 4
Answer:
(a) 6
Hint:

Question 22.
The eccentricity of the parabola is:
(a) 3
(b) 2
(c) 0
(d) 1
Answer:
(d) 1

Question 23.
The double ordinate passing through the focus is:
(a) focal chord
(b) latus rectum
(c) directrix
(d) axis
Answer:
(b) latus rectum
Hint:

Question 24.
The distance between directrix and focus of a parabola y² = 4ax is:
(a) a
(b) 2a
(c) 4a
(d) 3a
Answer:
(b) 2a

Question 25.
The equation of directrix of the parabola y² = -x is:
(a) 4x + 1 = 0
(b) 4x – 1 = 0
(c) x – 1 = 0
(d) x + 4 = 0
Answer:
(b) 4x – 1 = 0
Hint:
y² = -x.
It is a parabola open leftwards.
Here 4a = 1 ⇒ a = \(\frac{1}{4}\)
Equation of directrix is x = a.
i.e., x = \(\frac{1}{4}\) (or) 4x – 1 = 0

Students can download 11th Business Maths Chapter 3 Analytical Geometry Ex 3.2 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 3 Analytical Geometry Ex 3.2

Samacheer Kalvi 11th Business Maths Analytical Geometry Ex 3.2 Text Book Back Questions and Answers

Question 1.
Find the angle between the lines whose slopes are \(\frac{1}{2}\) and 3.
Solution:
Given that m₁ = \(\frac{1}{2}\) and m₂ = 3.
Let θ be the angle between the lines then

tan θ = 1
tan θ = tan 45°
∴ θ = 45°

Question 2.
Find the distance of the point (4, 1) from the line 3x – 4y + 12 = 0.
Solution:
The length of perpendicular from a point (x₁, y₁) to the line ax + by + c = 0 is d = \(\left|\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\right|\)
∴ The distance of the point (4, 1) to the line 3x – 4y + 12 = 0 is
[Here (x₁, y₁) = (4, 1), a = 3, b = -4, c = 12]

Question 3.
Show that the straight lines x + y – 4 = 0, 3x + 2 = 0 and 3x – 3y + 16 = 0 are concurrent.
Solution:
The lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0, a₃x + b₃y + c₃ = 0 are concurrent if
\(\left|\begin{array}{lll}
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2} \\
a_{3} & b_{3} & c_{3}
\end{array}\right|=0\)
The given lines x + y – 4 = 0, 3x + 0y + 2 = 0, 3x – 3y + 16 = 0
\(\left|\begin{array}{lll}
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2} \\
a_{3} & b_{3} & c_{3}
\end{array}\right|=\left|\begin{array}{rrr}
1 & 1 & -4 \\
3 & 0 & 2 \\
3 & -3 & 16
\end{array}\right|\)
= 1(0 + 6) – 1(48 – 6) – 4(-9 – 0)
= 6 – (42) + 36
= 42 – 42
= 0
The given lines are concurrent.

Question 4.
Find the value of ‘a’ for which the straight lines 3x + 4y = 13; 2x – 7y = -1 and ax – y – 14 = 0 are concurrent.
Solution:
The lines 3x + 4y = 13, 2x – 7y = -1 and ax – y – 14 = 0 are concurrent.
\(\left|\begin{array}{rrr}
3 & 4 & -13 \\
2 & -7 & 1 \\
a & -1 & -14
\end{array}\right|=0\)
⇒ 3(98 + 1) – 4(-28 – a) – 13(-2 + 7a) = 0
⇒ 3(99) + 112 + 4a + 26 – 91a = 0
⇒ 297 + 112 + 26 + 4a – 91a = 0
⇒ 435 – 87a = 0
⇒ -87a = -435
⇒ a = \(\frac{-435}{-87}\) = 5

Question 5.
A manufacturer produces 80 TV sets at a cost of ₹ 2,20,000 and 125 TV sets at a cost of ₹ 2,87,500. Assuming the cost curve to be linear, find the linear expression of the given information. Also, estimate the cost of 95 TV sets.
Solution:
Let x represent the TV sets, andy represent the cost.

The equation of straight line expressing the given information as a linear equation in x and y is

1(y – 2,20,000) = (x – 80)1500
y – 2,20,000 = 1500x – 80 × 1500
y = 1500x – 1,20,000 + 2,20,000
y = 1500x + 1,00,000 which is the required linear expression.
When x = 95,
y = 1,500 × 95 + 1,00,000
= 1,42,500 + 1,00,000
= 2,42,500
∴ The cost of 95 TV sets is ₹ 2,42,500.

Students can download 11th Business Maths Chapter 3 Analytical Geometry Ex 3.1 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 3 Analytical Geometry Ex 3.1

Samacheer Kalvi 11th Business Maths Analytical Geometry Ex 3.1 Text Book Back Questions and Answers

Question 1.
Find the locus of a point which is equidistant from (1, 3) and x-axis.
Solution:

Let P(x₁, y₁) be any point on the locus.
Let A be the point (1, 3)
The distance from the x-axis on the moving pint P(x₁, y₁) is y₁.
Given that AP = y₁
AP² = \(y_{1}^{2}\)

∴ The locus of the point (x₁, y₁) is x² – 2x – 6y + 10 = 0

Question 2.
A point moves so that it is always at a distance of 4 units from the point (3, -2).
Solution:
Let P(x₁, y₁) be any point on the locus.
Let A be the point (3, -2)
Given that PA = 4
PA² = 16

∴ The locus of the point (x₁, y₁) is x² + y² – 6x + 4y – 3 = 0

Question 3.
If the distance of a point from the points (2, 1) and (1, 2) are in the ratio 2 : 1, then find the locus of the point.
Solution:
Let P(x₁, y₁) be any point on the locus.
Let A(2, 1) and B(1, 2) be the given point.
Given that PA : PB = 2 : 1
i.e., \(\frac{P A}{P B}=\frac{2}{1}\)
PA = 2PB
PA² = 4PB²

∴ The locus of the point (x₁, y₁) is 3x² + 3y² – 4x – 14y + 15 = 0

Question 4.
Find a point on the x-axis which is equidistant from the points (7, -6) and (3, 4).
Solution:
Let P(x₁, 0) be any point on the x-axis.
Let A(7, -6) and B(3, 4) be the given points.
Given that PA = PB
PA² = PB²

∴ The required point is \(\left(\frac{15}{2}, 0\right)\)

Question 5.
If A(-1, 1) and B(2, 3) are two fixed points, then find the locus of a point P so that the area of triangle APB = 8 sq.units.
Solution:
Let the point P(x₁, y₁).
Fixed points are A(-1, 1) and B(2, 3).
Given area (formed by these points) of the triangle APB = 8
⇒ \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 8
⇒ \(\frac{1}{2}\) [x₁(1 – 3) + (-1) (3 – y₁) + 2(y₁ – 1)] = 8
⇒ \(\frac{1}{2}\) [-2x₁ – 3 + y₁ + 2y₁ – 2] = 8
⇒ \(\frac{1}{2}\) [-2x₁ + 3y₁ – 5] = 8
⇒ -2x₁ + 3y₁ – 5 = 16
⇒ -2x₁ + 3y₁ – 21 = 0
⇒ 2x₁ – 3y₁ + 21 = 0
∴ The locus of the point P(x₁, y₁) is 2x – 3y + 21 = 0.

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 3 Analytical Geometry Ex 3.5

Samacheer Kalvi 11th Business Maths Analytical Geometry Ex 3.5 Text Book Back Questions and Answers

Question 1.
Find the equation of the tangent to the circle x² + y² – 4x + 4y – 8 = 0 at (-2, -2).
Solution:
The equation of the tangent to the circle x² + y² – 4x + 4y – 8 = 0 at (x₁, y₁) is
xx₁ + yy₁ – 4\(\frac{\left(x+x_{1}\right)}{2}\) + 4\(\frac{\left(y+y_{1}\right)}{2}\) – 8 = 0
Here (x₁, y₁) = (-2, -2)
⇒ x(-2) + y(-2) – 2(x – 2) + 2(y – 2) – 8 = 0
⇒ -2x – 2y – 2x + 4 + 2y – 4 – 8 = 0
⇒ -4x – 8 = 0
⇒ x + 2 = 0

Question 2.
Determine whether the points P(1, 0), Q(2, 1) and R(2, 3) lie outside the circle, on the circle or inside the circle x² + y² – 4x – 6y + 9 = 0.
Solution:
The equation of the circle is x² + y² – 4x – 6y + 9 = 0
PT² = \(x_{1}^{2}+y_{1}^{2}\) – 4x₁ – 6y₁ + 9
At P(1, 0), PT² = 1 + 0 – 4 – 0 + 9 = 6 > 0
At Q(2, 1), PT² = 4 + 1 – 8 – 6 + 9 = 0
At R(2, 3), PT² = 4 + 9 – 8 – 18 + 9 = -4 < 0
The point P lies outside the circle.
The point Q lies on the circle.
The point R lies inside the circle.

Question 3.
Find the length of the tangent from (1, 2) to the circle x² + y² – 2x + 4y + 9 = 0.
Solution:
The length of the tangent from (x₁, y₁) to the circle x² + y² – 2x + 4y + 9 = 0 is \(\sqrt{x_{1}^{2}+y_{1}^{2}-2 x_{1}+4 y_{1}+9}\)
Length of the tangent from (1, 2) = \(\sqrt{1^{2}+2^{2}-2(1)+4(2)+9}\)
= \(\sqrt{1+4-2+8+9}\)
= \(\sqrt{20}\)
= \(\sqrt{4 \times 5}\)
= 2√5 units

Question 4.
Find the value of P if the line 3x + 4y – P = 0 is a tangent to the circle x² + y² = 16.
Solution:
The condition for a line y = mx + c to be a tangent to the circle x² + y² = a² is c² = a² (1 + m²)
Equation of the line is 3x + 4y – P = 0
Equation of the circle is x² + y² = 16
4y = -3x + P
y = \(\frac{-3}{4} x+\frac{P}{4}\)
∴ m = \(\frac{-3}{4}\), c = \(\frac{P}{4}\)
Equation of the circle is x² + y² = 16
∴ a² = 16
Condition for tangency we have c² = a²(1 + m²)
⇒ \(\left(\frac{P}{4}\right)^{2}=16\left(1+\frac{9}{16}\right)\)
⇒ \(\frac{P^{2}}{16}=16\left(\frac{25}{16}\right)\)
⇒ P² = 16 × 25
⇒ P = ±√16√25
⇒ P = ±4 × 5
⇒ P = ±20

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 2 Algebra Ex 2.5

Samacheer Kalvi 11th Business Maths Algebra Ex 2.5 Text Book Back Questions and Answers

By the principle of mathematical induction, prove the following:

Question 1.
1³ + 2³ + 3³ + ….. + n³ = \(\frac{n^{2}(n+1)^{2}}{4}\) for all x ∈ N.
Solution:
Let P(n) be the statement 1³ + 2³ + …… + n³ = \(\frac{n^{2}(n+1)^{2}}{4}\) for all n ∈ N.
i.e., p(n) = 1³ + 2³ + …… + n³ = \(\frac{n^{2}(n+1)^{2}}{4}\), for all n ∈ N
Put n = 1
LHS = 1³ = 1
RHS = \(\frac{1^{2}(1+1)^{2}}{4}\)
= \(\frac{1 \times 2^{2}}{4}\)
= \(\frac{4}{4}\)
= 1
∴ P(1) is true.
Assume that P(n) is true n = k
P(k): 1³ + 2³ + …… + k³ = \(\frac{k^{2}(k+1)^{2}}{4}\)
To prove P(k + 1) is true.
i.e., to prove 1³ + 2³ + ……. + k³ + (k + 1)³ = \(\frac{(k+1)^{2}((k+1)+1)^{2}}{4}=\frac{(k+1)^{2}(k+2)^{2}}{4}\)
Consider 1³ + 2³ + …… + k³ + (k + 1)³ = \(\frac{k^{2}(k+1)^{2}}{4}\) + (k + 1)³
= (k + 1)² [\(\frac{k^{2}}{4}\) + (k + 1)]
= (k + 1)² \(\left[\frac{k^{2}+4(k+1)}{4}\right]\)
= \(\frac{(k+1)^{2}(k+2)^{2}}{4}\)
⇒ P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction P(n) is true for all n ∈ N.

Question 2.
1.2 + 2.3 + 3.4 + …… + n(n + 1) = \(\frac{n(n+1)(n+2)}{3}\), for all n ∈ N.
Solution:
Let P(n) denote the statement
1.2 + 2.3 + 3.4 + …… + n(n + 1) = \(\frac{n(n+1)(n+2)}{3}\)
Put n = 1
LHS = 1(1 + 1) = 2
RHS = \(\frac{1(1+1)(1+2)}{3}=\frac{1(2)(3)}{3}\) = 2
∴ P(1) is true.
Now assume that the statement be true for n = k
(i.e.,) assume P(k) be true
(i.e.,) assume 1.2 + 2.3 + 3.4 + …… + k(k + 1) = \(\frac{k(k+1)(k+2)}{3}\) be true
To prove: P(k + 1) is true
(i.e.,) to prove: 1.2 + 2.3 + 3.4 + …… + k(k + 1) + (k + 1) (k + 2) = \(\frac{(k+1)(k+2)(k+3)}{3}\)
Consider 1.2 + 2.3 + 3.4 + ……. + k(k + 1) + (k + 1) (k + 2)
= [1.2 + 23 + …… + k(k + 1)] + (k + 1) (k + 2)
= \(\frac{k(k+1)(k+2)}{3}\) + (k + 1) (k + 2)
= \(\frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}\)
= \(\frac{(k+1)(k+2)(k+3)}{3}\)
∴ P(k + 1) is true.
Thus if P(k) is true, P(k + 1) is true.
By the principle of Mathematical ‘induction, P(n) is true for all n ∈ N.
1.2 + 2.3 + 3.4 + …… + n(n + 1) = \(\frac{n(n+1)(n+2)}{3}\)

Question 3.
4 + 8 + 12 + ……. + 4n = 2n(n + 1), for all n ∈ N.
Solution:
Let P(n) denote the statement 4 + 8 + …….. + 4n = 2n(n + 1)
i.e., P(n) : 4 + 8 + 12 + … + 4n = 2n(n + 1)
Put n = 1,
P(1): LHS = 4
RHS = 2 (1)(1 + 1) = 4
P(1) is true.
Assume that P(n) is true for n = k
P(k): 4 + 8 + 12 + ……. + 4k = 2k(k + 1)
To prove P(k + 1)
i.e., to prove 4 + 8 + 12 + ……. + 4k + 4(k + 1) = 2(k + 1) (k + 1 + 1)
4 + 8 + 12 + …… + 4k + (4k + 4) = 2(k + 1) (k + 2)
Consider, 4 + 8 + 12 + …….. + 4k + (4k + 4) = 2k(k + 1) + (4k + 4)
= 2k(k + 1) + 4(k + 1)
= 2k² + 2k + 4k + 4
= 2k² + 6k + 4
= 2(k + 1)(k + 2)
P(k + 1) is also true.
∴ By Mathematical Induction, P(n) for all value n ∈ N.

Question 4.
1 + 4 + 7 + ……. + (3n – 2) = \(\frac{n(3 n-1)}{2}\) for all n ∈ N.
Solution:
Let P(n) : 1 + 4 + 7 + ……. + (3n – 2) = \(\frac{n(3 n-1)}{2}\)
Put n = 1,
LHS = 1
RHS = \(\frac{1(3-1)}{2}\) = 1
∴ P(1) is true.
Assume P(k) is true for n = k
P(k): 1 + 4 + 7 + ……. + (3k – 2) = \(\frac{k(3 k-1)}{2}\)
To prove P(k + 1) is true, i.e., to prove
1 + 4 + 7 + ……. + (3k – 2) + (3(k + 1) – 2) = \(\frac{(k+1)(3(k+1)-1)}{2}\)
1 + 4 + 7 + ……. + (3k – 2) + (3k + 3 – 2) = \(\frac{(k+1)(3 k+2)}{2}\)
1 + 4 + 7 + …… + (3k + 1) = \(\frac{(k+1)(3 k+2)}{2}\)
1 + 4 + 7 + …… + (3k – 2) + (3k + 1) = \(\frac{k(3 k-1)}{2}\) + (3k + 1)
= \(\frac{k(3 k-1)+2(3 k+1)}{2}\)
= \(\frac{3 k^{2}-k+6 k+2}{2}\)
= \(\frac{3 k^{2}+5 k+2}{2}\)
= \(\frac{(k+1)(3 k+2)}{2}\)
∴ P(k + 1) is true whenever P(k) is true.
∴ By the Principle of Mathematical Induction, P(n) is true for all n ∈ N.

Question 5
3²ⁿ – 1 is divisible by 8, for all n ∈ N.
Solution:
Let P(n) denote the statement 3²ⁿ – 1 is divisible by 8 for all n ∈ N
Put n = 1
P(1) is the statement 3²⁽¹⁾ – 1 = 3² – 1 = 9 – 1 = 8, which is divisible by 8
∴ P(1) is true.
Assume that P(k) is true for n = k.
i.e., 3^2k – 1 is divisible by 8 to be true.
Let 3^2k – 1 = 8m
To prove P(k + 1) is true.
i.e., to prove 3^2(k+1) – 1 is divisible by 8
Consider 3^2(k+1) – 1 = 3^2k+2 – 1
= 3^2k . 3² – 1
= 3^2k (9) – 1
= 3^2k (8 + 1) – 1
= 3^2k × 8 + 3^2k × 1 – 1
= 3^2k (8) + 3^2k – 1
= 3^2k (8) + 8m (∵ 3^2k – 1 = 8m)
= 8(3^2k + m), which is divisible by 8.
∴ P(k + 1) is true wherever P(k) is true.
∴ By principle of Mathematical Induction, P(n) is true for all n ∈ N.

Question 6.
aⁿ – bⁿ is divisible by a – b, for all n ∈ N.
Solution:
Let P(n) denote the statement aⁿ – bⁿ is divisible by a – b.
Put n = 1. Then P(1) is the statement: a¹ – b¹ = a – b is divisible by a – b
∴ P(1) is true. Now assume that the statement be true for n = k
(i.e.,) assume P(k) be true, (i.e.,) a^k – b^k is divisible by (a – b) be true.
⇒ \(\frac{a^{k}-b^{k}}{a-b}\) = m (say) where m ∈ N
⇒ a^k – b^k = m(a – b)
⇒ a^k = b^k + m(a – b) ……. (1)
Now to prove P(k + 1) is true, (i.e.,) to prove: a^k+1 – b^k+1 is divisible by a – b
Consider a^k+1 – b^k+1 = a^k . a – b^k . b
= [b^k + m(a – b)] a – b^k . b [∵ a^k = bm + k(a – b)]
= b^k . a + am(a – b) – b^k . b
= b^k . a – b^k . b + am(a – b)
= b^k(a – b) + am(a – b)
= (a – b) (b^k + am) is divisible by (a – b)
∴ P(k + 1) is true.
By the principle of Mathematical induction. P(n) is true for all n ∈ N.
∴ aⁿ – bⁿ is divisible by a – b for n ∈ N.

Question 7.
5²ⁿ – 1 is divisible by 24, for all n ∈ N.
Solution:
Let P(n) be the proposition that 5²ⁿ – 1 is divisible by 24.
For n = 1, P(1) is: 5² – 1 = 25 – 1 = 24, 24 is divisible by 24.
Assume that P(k) is true.
i.e., 5^2k – 1 is divisible by 24
Let 5^2k – 1 = 24m
To prove P(k + 1) is true.
i.e., to prove 5^2(k+1) – 1 is divisible by 24.
P(k): 5^2k – 1 is divisible by 24.
P(k + 1) = 5^2(k+1) – 1
= 5^2k . 5² – 1
= 5^2k (25) – 1
= 5^2k (24 + 1) – 1
= 24 . 5^2k + 5^2k – 1
= 24 . 5^2k + 24m
= 24 [5^2k + 24]
which is divisible by 24 ⇒ P(k + 1) is also true.
Hence by mathematical induction, P(n) is true for all values n ∈ N.

Question 8.
n(n + 1) (n + 2) is divisible by 6, for all n ∈ N.
Solution:
P(n): n(n + 1) (n + 2) is divisible by 6.
P(1): 1 (2) (3) = 6 is divisible by 6
∴ P(1) is true.
Let us assume that P(k) is true for n = k
That is, k (k + 1) (k + 2) = 6m for some m
To prove P(k + 1) is true i.e. to prove (k + 1) (k + 2)(k + 3) is divisible by 6.
P(k + 1) = (k + 1) (k + 2) (k + 3)
= (k + 1)(k + 2)k + 3(k + 1)(k + 2)
= 6m + 3(k + 1)(k + 2)
In the second term either k + 1 or k + 2 will be even, whatever be the value of k.
Hence second term is also divisible by 6.
∴ P (k + 1) is also true whenever P(k) is true.
By Mathematical Induction P (n) is true for all values of n.

Question 9.
2ⁿ > n, for all n ∈ N.
Solution:
Let P(n) denote the statement 2ⁿ > n for all n ∈ N
i.e., P(n): 2ⁿ > n for n ≥ 1
Put n = 1, P(1): 2¹ > 1 which is true.
Assume that P(k) is true for n = k
i.e., 2^k > k for k ≥ 1
To prove P(k + 1) is true.
i.e., to prove 2^k+1 > k + 1 for k ≥ 1
Since 2^k > k
Multiply both sides by 2
2 . 2^k > 2k
2^k+1 > k + k
i.e., 2^k+1 > k + 1 (∵ k ≥ 1)
∴ P(k + 1) is true whenever P(k) is true.
∴ By principal of mathematical induction P(n) is true for all n ∈ N.

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 2 Algebra Ex 2.6

Samacheer Kalvi 11th Business Maths Algebra Ex 2.6 Text Book Back Questions and Answers

Question 1.
Expand the following by using binomial theorem:
(i) (2a – 3b)⁴
(ii) \(\left(x+\frac{1}{y}\right)^{7}\)
(iii) \(\left(x+\frac{1}{x^{2}}\right)^{6}\)
Solution:

Question 2.
Evaluate the following using binomial theorem:
(i) (101)⁴
(ii) (999)⁵
Solution:
(i) (x + a)ⁿ = nC₀ xⁿ a⁰ + nC₁ x^n-1 a¹ + nC₂ x^n-2 a² + ……… + nC_r x^n-r a^r + …… + nC_n aⁿ
(101)⁴ = (100 + 1)⁴ = 4C₀ (100)⁴ + 4C₁ (100)³ (1)¹ + 4C₂ (100)² (1)² + 4C₃ (100)¹ (1)³ + 4C₄ (1)⁴
= 1 × (100000000) + 4 × (1000000) + 6 × (10000) + 4 × 100 + 1 × 1
= 100000000 + 4000000 + 60000 + 400 + 1
= 10,40,60,401

(ii) (x + a)ⁿ = nC₀ xⁿ a⁰ + nC₁ x^n-1 a¹ + nC₂ x^n-2 a² + ……… + nC_r x^n-r a^r + …… + nC_n aⁿ
(999)⁵ = (1000 – 1)⁵ = 5C₀ (1000)⁵ – 5C₁ (1000)⁴ (1)¹ + 5C₂ (1000)³ (1)² – 5C₃ (1000)² (1)³ + 5C₄ (1000)⁵ (1)⁴ – 5C₅ (1)⁵
= 1(1000)⁵ – 5(1000)⁴ – 10(1000)³ – 10(1000)² + 5(1000) – 1
= 1000000000000000 – 5000000000000 + 10000000000 – 10000000 + 5000 – 1
= 995009990004999

Question 3.
Find the 5th term in the expansion of (x – 2y)¹³.
Solution:
General term is t_r+1 = nC_r x^n-r a^r
(x – 2y)¹³ = (x + (-2y))¹³
Here x is x, a is (-2y) and n = 13
5th term = t₅ = t₄₊₁ = 13C₄ x^13-4 (-2y)⁴
= 13C₄ x⁹ 2⁴ y⁴
= \(\frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1}\) × 2 × 2 × 2 × 2× x⁹y⁴
= 13 × 11 × 10 × 8x⁹y⁴
= 13 × 880x⁹y⁴
= 11440x⁹y⁴

Question 4.
Find the middle terms in the expansion of
(i) \(\left(x+\frac{1}{x}\right)^{11}\)
(ii) \(\left(3 x+\frac{x^{2}}{2}\right)^{8}\)
(iii) \(\left(2 x^{2}-\frac{3}{x^{3}}\right)^{10}\)
Solution:
(i) General term is t_r+1 = nC_r x^n-r a^r
Here x is x, a is \(\frac{1}{x}\) and n = 11, which is odd.
So the middle terms are \(\frac{t_{n+1}}{2}=\frac{t_{11+1}}{2}, \frac{t_{n+3}}{2}=\frac{t_{11+3}}{2}\)
i.e. the middle terms are t₆, t₇

(ii) Here x is 3x, a is \(\frac{x^{2}}{2}\), n = 8, which is even.
∴ The only one middle term = \(\frac{t_{n+1}}{2}=\frac{t_{8+1}}{2}\) = t₅
General term t_r+1 = nC_r x^n-r a^r

(iii) \(\left(2 x^{2}-\frac{3}{x^{3}}\right)^{10}=\left(2 x^{2}+\frac{-3}{x^{3}}\right)^{10}\) compare with the (x + a)ⁿ
Here x is 2x², a is \(\frac{-3}{x^{3}}\), n = 10, which is even.
So the only middle term is \(\frac{t_{n+1}}{2}=\frac{t_{10}}{2}+1\) = t₆
General term t_r+1 = nC_r x^n-r a^r
t₆ = t₅₊₁ = t_r+1

Question 5.
Find the term in dependent of x in the expansion of
(i) \(\left(x^{2}-\frac{2}{3 x}\right)^{9}\)
(ii) \(\left(x-\frac{2}{x^{2}}\right)^{15}\)
(iii) \(\left(2 x^{2}+\frac{1}{x}\right)^{12}\)
Solution:
(i) Let the independent form of x occurs in the general term, t_r+1 = nC_r x^n-r a^r
Here x is x², a is \(\frac{-2}{3 x}\) and n = 9

Independent term occurs only when x power is zero.
18 – 3r = 0
⇒ 18 = 3r
⇒ r = 6
Put r = 6 in (1) we get the independent term as 9C₆ x⁰ \(\frac{(-2)^{6}}{3^{6}}\) = 9C₃ \(\left(\frac{2}{3}\right)^{6}\)
[∵ 9C₆ = 9C_9-6 = 9C₃]

(ii) \(\left(x-\frac{2}{x^{2}}\right)^{15}=\left(x+\frac{-2}{x^{2}}\right)^{15}\) compare with the (x + a)ⁿ
Here x is x, a is \(\frac{-2}{x^{2}}\), n = 15.
Let the independent term of x occurs in the general term

Independent term occurs only when x power is zero.
15 – 3r = 0
15 = 3r
r = 5
Using r = 5 in (1) we get the independent term
= 15C₅ x⁰ (-2)⁵ [∵ (-2)⁵ = (-1)⁵ 2⁵ = -2⁵]
= -32(15C₅)

(iii) \(\left(2 x^{2}+\frac{1}{x}\right)^{12}\) Compare with the (x + a)ⁿ.
Here x is 2x², a is \(\frac{1}{x}\), n = 12.
Let the independent term of x occurs in the general term.


Independent term occurs only when x power is zero
24 – 3r = 0
24 = 3r
r = 8
Put r = 8 in (1) we get the independent term as
= 12C₈ 2^12-8 x⁰
= 12C₄ × 2⁴ × 1
= 7920

Question 6.
Prove that the term independent of x in the expansion of \(\left(x+\frac{1}{x}\right)^{2 n}\) is \(\frac{1 \cdot 3 \cdot 5 \cdot \cdot \cdot(2 n-1) 2^{n}}{n !}\)
Solution:
There are (2n + 1) terms in expansion.
∴ t_n+1 is the middle term.

Question 7.
Show that the middle term in the expansion of is (1 + x)²ⁿ is \(\frac{1 \cdot 3 \cdot 5 \ldots(2 n-1) 2^{n} x^{n}}{n !}\)
Solution:
There are 2n + 1 terms in expansion of (1 + x)²ⁿ.
∴ The middle term is t_n+1.

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Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 2 Algebra Ex 2.4

Samacheer Kalvi 11th Business Maths Algebra Ex 2.4 Text Book Back Questions and Answers

Question 1.
If ⁿP_r = 1680 and ⁿC_r = 70, find n and r.
Solution:
Given that ⁿP_r = 1680, ⁿC_r = 70
We know that ⁿC_r = \(\frac{n \mathrm{P}_{r}}{r !}\)
70 = \(\frac{1680}{r !}\)
r! = \(\frac{1680}{70}\) = 24
r! = 4 × 3 × 2 × 1 = 4!
∴ r = 4

Question 2.
Verify that ⁸C₄ + ⁸C₃ = ⁹C₄.
Solution:
LHS = ⁸C₄ + ⁸C₃
= \(\frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1}+\frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1}\)
= 7 × 2 × 5 + 8 × 7
= 70 + 56
= 126
RHS = ⁹C₄
= \(\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}\)
= 9 × 7 × 2
= 126
∴ LHS = RHS
Hence verified.

Question 3.
How many chords can be drawn through 21 points on a circle?
Solution:
To draw a chord we need two points on a circle.
∴ Number chords through 21 points on a circle = ²¹C₂ = \(\frac{21 \times 20}{2 \times 1}\) = 210.

Question 4.
How many triangles can be formed by joining the vertices of a hexagon?
Solution:
A hexagon has six vertices. By joining any three vertices of a hexagon we get a triangle.
∴ Number of triangles formed by joining the vertices of a hexagon = ⁶C₃ = \(\frac{6 \times 5 \times 4}{3 \times 2 \times 1}\) = 20.

Question 5.
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
Solution:
In this problem first, we have to select consonants and vowels.
Then we arrange a five-letter word using 3 consonants and 2 vowels.
Therefore here both combination and permutation involved.
The number of ways of selecting 3 consonants from 7 is ⁷C₃.
The number of ways of selecting 2 vowels from 4 is ⁴C₃.
The number of ways selecting 3 consonants from 7 and 2 vowels from 4 is ⁷C₃ × ⁴C₂.
Now with every selection number of ways of arranging 5 letter word
= 5! × ⁷C₃ × ⁴C₂
= 120 × \(\frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times \frac{4 \times 3}{2 \times 1}\)
= 25200

Question 6.
If four dice are rolled, find the number of possible outcomes in which atleast one die shows 2.
Solution:
When a die is rolled number of possible outcomes is selecting an event from 6 events = ⁶C₁
When four dice are rolled number of possible outcomes = ⁶C₁ × ⁶C₁ × ⁶C₁ × ⁶C₁
When a die is rolled number of possible outcomes in which ‘2’ does not appear is selecting an event from 5 events = ⁵C₁
When four dice are rolled number of possible outcomes in which 2 does not appear = ⁵C₁ × ⁵C₁ × ⁵C₁ × ⁵C₁
Therefore the number of possible outcomes in which atleast one die shows 2
= ⁶C₁ × ⁶C₁ × ⁶C₁ × ⁶C₁ – ⁵C₁ × ⁵C₁ × ⁵C₁ × ⁵C₁
= 6 × 6 × 6 × 6 – 5 × 5 × 5 × 5
= 1296 – 625
= 671
Note: when two dice are rolled number of possible outcomes is 36 and the number of possible outcomes in which 2 doesn’t appear = 25. When two dice are rolled the number of possible outcomes in which atleast one die shows 2 = 36 – 25 = 11. Use the sample space, S = {(1, 1), (1, 2),… (6, 6)}.

Question 7.
There are 18 guests at a dinner party. They have to sit 9 guests on either side of a long table, three particular persons decide to sit on one side and two others on the other side. In how many ways can the guests to be seated?
Solution:
Let A and B be two sides of the table 9 guests sit on either side of the table in 9! × 9! ways.
Out of 18 guests, three particular persons decide to sit namely inside A and two on the other side B. remaining guest = 18 – 3 – 2 = 13.
From 13 guests we can select 6 more guests for side A and 7 for the side.
Selecting 6 guests from 13 can be done in ¹³C₆ ways.
Therefore total number of ways the guest to be seated = ¹³C₆ × 9! × 9!
= \(\frac{13 !}{6 !(13-6) !} \times 9 ! \times 9 !\)
= \(\frac{13 !}{6 ! \times 7 !} \times 9 ! \times 9 !\)

Question 8.
If a polygon has 44 diagonals, find the number of its sides.
Solution:
A polygon of n sides has n vertices. By joining any two vertices of a polygon, we obtain either a side or a diagonal of the polygon.
A number of line segments obtained by joining the vertices of a n sided polygon taken two at a time = Number of ways of selecting 2 out of n.
= ⁿC₂
= \(\frac{n(n-1)}{2}\)
Out of these lines, n lines are the sides of the polygon, Sides can’t be diagonals.
∴ Number of diagonals of the polygon = \(\frac{n(n-1)}{2}\) – n = \(\frac{n(n-3)}{2}\)
Given that a polygon has 44 diagonals.
Let n be the number of sides of the polygon.
\(\frac{n(n-3)}{2}\) = 44
⇒ n(n – 3) = 88
⇒ n² – 3n – 88 = 0
⇒ (n + 8) (n – 11)
⇒ n = -8 (or) n = 11
n cannot be negative.
∴ n = 11 is number of sides of polygon is 11.

Question 9.
In how many ways can a cricket team of 11 players be chosen out of a batch of 15 players?
(i) There is no restriction on the selection.
(ii) A particular player is always chosen.
(iii) A particular player is never chosen.
Solution:
(i) Number of ways choosing 11 players from 15 is ¹⁵C₁₁ = ¹⁵C₄
= \(\frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1}\)
= 15 × 7 × 13
= 1365.

(ii) If a particular is always chosen there will be only 14 players left put, in which 10 are to selected in ¹⁴C₁₀ ways.
¹⁴C₁₀ = ¹⁴C₄
= \(\frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1}\)
= \(\frac{14 \times 13 \times 11}{2}\)
= 91 × 11
= 1001 ways

(iii) If a particular player is never chosen we have to select 11 players out of remaining 14 players in ¹⁴C₁₁ ways.
i.e., ¹⁴C₃ ways = \(\frac{14 \times 13 \times 12}{3 \times 2 \times 1}\) = 364 ways.

Question 10.
A committee of 5 is to be formed out of 6 gents and 4 ladies. In how many ways this can be done when
(i) atleast two ladies are included.
(ii) atmost two ladies are included.
Solution:
(i) A committee of 5 is to be formed.

(ii) Almost two ladies are included means maximum of two ladies are included.

Students can download 11th Business Maths Chapter 2 Algebra Ex 2.3 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 2 Algebra Ex 2.3

Samacheer Kalvi 11th Business Maths Algebra Ex 2.3 Text Book Back Questions and Answers

Question 1.
If ⁿP₄ = 12(ⁿP₂), find n.
Solution:
Given that ⁿP₄ = 12(ⁿP₂)
n(n – 1) (n – 2) (n – 3) = 12n(n – 1)
Cancelling n(n – 1) on both sides we get
(n – 2) (n – 3) = 4 × 3
We have product of consecutive number on both sides with decreasing order.
n – 2 = 4
∴ n = 6

Question 2.
In how many ways 5 boys and 3 girls can be seated in a row so that no two girls are together?
Solution:
5 boys can be seated among themselves in ⁵P₅ = 5! Ways. After this arrangement, we have to arrange the three girls in such a way that in between two girls there atleast one boy. So the possible places girls can be placed with the × symbol given below.
× B × B × B × B × B ×
∴ There are 6 places to seated by 3 girls which can be done 6P3 ways.
∴ Total number of ways = 5! × ⁶P₃
= 120 × (6 × 5 × 4)
= 120 × 120
= 14400

Question 3.
How many 6-digit telephone numbers can be constructed with the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 if each number starts with 35 and no digit appears more than once?
Solution:
Given that each number starts with 35. We need a 6 digit number. So we have to fill only one’s place, 10’s place, 100th place, and 1000th places. We have to use 10 digits.

In these digits, 3 and 5 should not be used as a repetition of digits is not allowed. Except for these two digits, we have to use 8 digits. One’s place can be filled by any of the 8 digits in different ways, 10’s place can be filled by the remaining 7 digits in 7 different ways.

100th place can be filled by the remaining 6 different ways and 1000th place can be filled by the remaining 5 digits in 5 different ways.

∴ Number of 6 digit telephone numbers = 8 × 7 × 6 × 5 = 1680

Question 4.
Find the number of arrangements that can be made out of the letters of the word “ASSASSINATION”.
Solution:
The number of letters of the word “ASSASSINATION” is 13.
The letter A occurs 3 times
The letter S occurs 4 times
The letter I occur 2 times
The letter N occurs 2 times
The letter T occurs 1 time
The letter O occurs 1 time
∴ Number of arrangements = \(\frac{13 !}{3 ! 4 ! 2 ! 2 ! 1 ! 1 !}=\frac{13 !}{3 ! 4 ! 2 ! 2 !}\)

Question 5.
(a) In how many ways can 8 identical beads be strung on a necklace?
(b) In how many ways can 8 boys form a ring?
Solution:
(a) Number of ways 8 identical beads can be stringed by \(\frac{(8-1) !}{2}=\frac{7 !}{2}\)
(b) Number of ways 8 boys form a ring = (8 – 1)! = 7!

Question 6.
Find the rank of the word ‘CHAT’ in the dictionary.
Solution:
The letters of the word CHAT in alphabetical order are A, C, H, T. To arrive the word CHAT, first, we have to go through the word that begins with A. If A is fixed as the first letter remaining three letters C, H, T can be arranged among themselves in 3! ways. Next, we select C as the first letter and start arranging the remaining letters in alphabetical order. Now C and A is fixed remaining two letters can be arranged in 2! ways. Next, we move on H with C, A, and H is fixed the letter T can be arranged in 1! ways.
∴ Rank of the word CHAT = 3! + 2! + 1! = 6 + 2 + 1 = 9

Note: The rank of a given word is basically finding out the position of the word when possible words have been formed using all the letters of the given word exactly once and arranged in alphabetical order as in the case of dictionary. The possible arrangement of the word CHAT are (i) ACHT, (ii) ACTH, (iii) AHCT, (iv) AHTC, (v) ATCH, (vi) ATHC, (vii) CAHT, (viii) CATH, (ix) CHAT. So the rank of the word occurs in the ninth position.
∴ The rank of the word CHAT is 9.

Students can download 11th Business Maths Chapter 3 Analytical Geometry Ex 3.3 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 3 Analytical Geometry Ex 3.3

Samacheer Kalvi 11th Business Maths Analytical Geometry Ex 3.3 Text Book Back Questions and Answers

Question 1.
If the equation ax² + 5xy – 6y² + 12x + 5y + c = 0 represents a pair of perpendicular straight lines, find a and c.
Solution:
Comparing ax² + 5xy – 6y² + 12x + 5y + c = 0 with ax² + 2hxy + by² + 2gx + 2fy + c = 0
We get a = a, 2h = 5, (or) h = \(\frac{5}{2}\), b = -6, 2g = 12 (or) g = 6, 2f = 5 (or) f = \(\frac{5}{2}\), c = c
Condition for pair of straight lines to be perpendicular is a + b = 0
a + (-6) = 0
a = 6
Next to find c. Condition for the given equation to represent a pair of straight lines is

R₁ → R₁ – R₃
Expanding along first row we get 0 – 0 + (6 – c) [\(\frac{25}{4}\) + 36] = 0
(6-c) [\(\frac{25}{4}\) + 36] = 0
6 – c = 0
6 = c (or) c = 6

Question 2.
Show that the equation 12x² – 10xy + 2y² + 14x – 5y + 2 = 0 represents a pair of straight lines and also find the separate equations of the straight lines.
Solution:
Comparing 12x² – 10xy + 2y² + 14x – 5y + 2 = 0 with ax² + 2hxy + by² + 2gh + 2fy + c = 0
We get a = 12, 2h = -10, (or) h = -5, b = 2, 2g = 14 (or) g = 7, 2f = -5 (or) f = \(-\frac{5}{2}\), c = 2
Condition for the given equation to represent a pair of straight lines is \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=0\)
\(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{rrr}
12 & -5 & 7 \\
-5 & 2 & \frac{-5}{2} \\
7 & \frac{-5}{2} & 2
\end{array}\right|\)

= \(\frac{1}{4}\) [12(16 – 25) + 5(-40 + 70) + 7(50 – 56)]
= \(\frac{1}{4}\) [12(-9) + 5(30) + 7(-6)]
= \(\frac{1}{4}\) [-108 + 150 – 42]
= \(\frac{1}{4}\) [0]
= 0
∴ The given equation represents a pair of straight lines.
Consider 12x² – 10xy + 2y² = 2[6x² – 5xy + y²] = 2[(3x – y)(2x – y)] = (6x – 2y)(2x – y)
Let the separate equations be 6x – 2y + l = 0, 2x – y + m = 0
To find l, m
Let 12x² – 10xy + 2y² + 14x – 5y + 2 = (6x – 2y + l) (2x – y + m) ……. (1)
Equating coefficient of y on both sides of (1) we get
2l + 6m = 14 (or) l + 3m = 7 ………… (2)
Equating coefficient of x on both sides of (1) we get
-l – 2m = -5 ……… (3)
(2) + (3) ⇒ m = 2
Using m = 2 in (2) we get
l + 3(2) = 7
l = 7 – 6
l = 1
∴ The separate equations are 6x – 2y + 1 = 0, 2x – y + 2 = 0.

Question 3.
Show that the pair of straight lines 4x² + 12xy + 9y² – 6x – 9y + 2 = 0 represents two parallel straight lines and also find the separate equations of the straight lines.
Solution:
The given equation is 4x² + 12xy + 9y² – 6x – 9y + 2 = 0
Here a = 4, 2h = 12, (or) h = 6 and b = 9
h² – ab = 6² – 4 × 9 = 36 – 36 = 0
∴ The given equation represents a pair of parallel straight lines
Consider 4x² + 12xy + 9y² = (2x)² + 12xy + (3y)²
= (2x)² + 2(2x)(3y) + (3y)²
= (2x + 3y)²
Here we have repeated factors.
Now consider, 4x² + 12xy + 9y² – 6x – 9y + 2 = 0
(2x + 3y)² – 3(2x + 3y) + 2 = 0
t² – 3t + 2 = 0 where t = 2x + 3y
(t – 1)(t – 2) = 0
(2x + 3y – 1) (2x + 3y – 2) = 0
∴ Separate equations are 2x + 3y – 1 = 0, 2x + 3y – 2 = 0

Question 4.
Find the angle between the pair of straight lines 3x² – 5xy – 2y² + 17x + y + 10 = 0.
Solution:
The given equation is 3x² – 5xy – 2y² + 17x + y + 10 = 0
Here a = 3, 2h = -5, b = -2
If θ is the angle between the given straight lines then