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

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)
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.4 Q2

(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,
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.4 Q2.1
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.4 Q2.2

(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) = \(\left(\frac{3}{2}, \frac{3}{2}\right)\)
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.4 Q2.3

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

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:
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.4 Q4
Centre (h, k) = (2, 3)
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.4 Q4.1
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

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

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:
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.4 Q6
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

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

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).
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.4 Q7
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

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

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.

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

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
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.7 Q2

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

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 = \(\frac{1}{3}\)

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

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:
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.7 Q6
Given PA = PB
y1 = x1
∴ Locus is y = x

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

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

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

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 \(\sqrt{x_{1}^{2}+y_{1}^{2}-16}=5\)

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

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.

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

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
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.7 Q14
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

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

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

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

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

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

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

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

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:
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.7 Q21

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

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

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:
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.7 Q23

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

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

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

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

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
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.2 Q1
tan θ = 1
tan θ = tan 45°
∴ θ = 45°

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

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]
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.2 Q2

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.

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

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.
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.2 Q5
The equation of straight line expressing the given information as a linear equation in x and y is
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.2 Q5.1
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.

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

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:
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.1 Q1
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 = \(y_{1}^{2}\)
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.1 Q1.1
∴ 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
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.1 Q2
∴ 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
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.1 Q3
∴ 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
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.1 Q4
∴ 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.

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

Students can download 11th Business Maths Chapter 3 Analytical Geometry Ex 3.5 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 3.5

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

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

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

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

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

Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.5

Students can download 11th Business Maths Chapter 2 Algebra Ex 2.5 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 2.5

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) = \(\frac{n(n+1)(n+2)}{3}\)

Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.5

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.

Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.5

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., 32k – 1 is divisible by 8 to be true.
Let 32k – 1 = 8m
To prove P(k + 1) is true.
i.e., to prove 32(k+1) – 1 is divisible by 8
Consider 32(k+1) – 1 = 32k+2 – 1
= 32k . 32 – 1
= 32k (9) – 1
= 32k (8 + 1) – 1
= 32k × 8 + 32k × 1 – 1
= 32k (8) + 32k – 1
= 32k (8) + 8m (∵ 32k – 1 = 8m)
= 8(32k + 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.,) ak – bk is divisible by (a – b) be true.
⇒ \(\frac{a^{k}-b^{k}}{a-b}\) = m (say) where m ∈ N
⇒ ak – bk = m(a – b)
⇒ ak = bk + m(a – b) ……. (1)
Now to prove P(k + 1) is true, (i.e.,) to prove: ak+1 – bk+1 is divisible by a – b
Consider ak+1 – bk+1 = ak . a – bk . b
= [bk + m(a – b)] a – bk . b [∵ ak = bm + k(a – b)]
= bk . a + am(a – b) – bk . b
= bk . a – bk . b + am(a – b)
= bk(a – b) + am(a – b)
= (a – b) (bk + 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.

Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.5

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., 52k – 1 is divisible by 24
Let 52k – 1 = 24m
To prove P(k + 1) is true.
i.e., to prove 52(k+1) – 1 is divisible by 24.
P(k): 52k – 1 is divisible by 24.
P(k + 1) = 52(k+1) – 1
= 52k . 52 – 1
= 52k (25) – 1
= 52k (24 + 1) – 1
= 24 . 52k + 52k – 1
= 24 . 52k + 24m
= 24 [52k + 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.

Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.5

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., 2k > k for k ≥ 1
To prove P(k + 1) is true.
i.e., to prove 2k+1 > k + 1 for k ≥ 1
Since 2k > k
Multiply both sides by 2
2 . 2k > 2k
2k+1 > k + k
i.e., 2k+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.

Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6

Students can download 11th Business Maths Chapter 2 Algebra Ex 2.6 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 2.6

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:
Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6 Q1
Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6 Q1.1
Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6 Q1.2
Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6 Q1.3
Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6 Q1.4Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6 Q1.5
Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6 Q1.6
Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6 Q1.7

Question 2.
Evaluate the following using binomial theorem:
(i) (101)4
(ii) (999)5
Solution:
(i) (x + a)n = nC0 xn a0 + nC1 xn-1 a1 + nC2 xn-2 a2 + ……… + nCr xn-r ar + …… + nCn 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 xn-1 a1 + nC2 xn-2 a2 + ……… + nCr xn-r ar + …… + nCn 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

Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6

Question 3.
Find the 5th term in the expansion of (x – 2y)13.
Solution:
General term is tr+1 = nCr xn-r ar
(x – 2y)13 = (x + (-2y))13
Here x is x, a is (-2y) and n = 13
5th term = t5 = t4+1 = 13C4 x13-4 (-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 tr+1 = nCr xn-r ar
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
Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6 Q4

(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 tr+1 = nCr xn-r ar
Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6 Q4.1

(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 tr+1 = nCr xn-r ar
t6 = t5+1 = tr+1
Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6 Q4.2

Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6

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, tr+1 = nCr xn-r ar
Here x is x2, a is \(\frac{-2}{3 x}\) and n = 9
Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6 Q5
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 = 9C9-6 = 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
Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6 Q5.1
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.
Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6 Q5.2
Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6 Q5.3
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 212-8 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.
∴ tn+1 is the middle term.
Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6 Q6

Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6

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 tn+1.
Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.6 Q7

Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.4

Students can download 11th Business Maths Chapter 2 Algebra Ex 2.4 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 2 Algebra Ex 2.4

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

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

Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.4

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.

Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.4

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

Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.4

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 × 6C15C1 × 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!
= \(\frac{13 !}{6 !(13-6) !} \times 9 ! \times 9 !\)
= \(\frac{13 !}{6 ! \times 7 !} \times 9 ! \times 9 !\)

Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.4

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.

Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.4

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.
Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.4 Q10.1

(ii) Almost two ladies are included means maximum of two ladies are included.
Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.4 Q10

Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.3

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

Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.3

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

Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.3

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!

Samacheer Kalvi 11th Business Maths Guide Chapter 2 Algebra Ex 2.3

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.

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

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
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.3 Q1
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

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

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 \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=0\)
\(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{rrr}
12 & -5 & 7 \\
-5 & 2 & \frac{-5}{2} \\
7 & \frac{-5}{2} & 2
\end{array}\right|\)
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.3 Q2
= \(\frac{1}{4}\) [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.

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

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

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

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
Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.3 Q4