Category: Class 11

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

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

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 3.3 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

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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.

Students can download 11th Business Maths Chapter 2 Algebra Ex 2.7 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 2 Algebra Ex 2.7

Samacheer Kalvi 11th Business Maths Algebra Ex 2.7 Text Book Back Questions and Answers

Choose the correct answer.

Question 1.
If nC₃ = nC₂ then the value of nC₄ is:
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(d) 5
Hint:
Given that nC₃ = nC₂
We know that if nC_x = nC_y then x + y = n or x = y
Here 3 + 2 = n
∴ n = 5

Question 2.
The value of n, when np₂ = 20 is:
(a) 3
(b) 6
(c) 5
(d) 4
Answer:
(c) 5
Hint:
nP₂ = 20
n(n – 1) = 20
n(n – 1) = 5 × 4
∴ n = 5

Question 3.
The number of ways selecting 4 players out of 5 is:
(a) 4!
(b) 20
(c) 25
(d) 5
Answer:
(d) 5
Hint:
5C₄ = 5C₁ = 5

Question 4.
If nP_r = 720(nC_r), then r is equal to:
(a) 4
(b) 5
(c) 6
(d) 7
Answer:
(c) 6
Hint:
Given nP_r = 720(nC_r)
\(\frac{n !}{(n-r) !}=720 \frac{n !}{r !(n-r) !}\)
1 = \(\frac{720}{r !}\)
r! = 720
r! = 6 × 5 × 4 × 3 × 2 × 1
r! = 6!
r = 6

Question 5.
The possible outcomes when a coin is tossed five times:
(a) 25
(b) 52
(c) 10
(d) \(\frac{5}{2}\)
Answer:
(a) 25
Hint:
Number of possible outcomes When a coin is tossed is 2
∴ When five coins are tossed (same as a coin is tossed five times)
Possible outcomes = 2 × 2 × 2 × 2 × 2 = 2⁵

Question 6.
The number of diagonals in a polygon of n sides is equal to:
(a) nC₂
(b) nC₂ – 2
(c) nC₂ – n
(d) nC₂ – 1
Answer:
(c) nC₂ – n

Question 7.
The greatest positive integer which divide n(n + 1) (n + 2) (n + 3) for all n ∈ N is:
(a) 2
(b) 6
(c) 20
(d) 24
Answer:
(d) 24
Hint:
Put n = 1 in n(n + 1) (n + 2) (n + 3)
= 1 × 2 × 3 × 4
= 24

Question 8.
If n is a positive integer, then the number of terms in the expansion of (x + a)ⁿ is:
(a) n
(b) n + 1
(c) n – 1
(d) 2n
Answer:
(b) n + 1

Question 9.
For all n > 0, nC₁ + nC₂ + nC₃ + …… + nC_n is equal to:
(a) 2n
(b) 2ⁿ – 1
(c) n²
(d) n² – 1
Answer:
(b) 2ⁿ – 1
Hint:
Sum of binomial coefficients 2n
i.e., nC₀ + nC₁ + nC₂ + nC₃ + ……. + nC_n = 2ⁿ
nC₁ + nC₂ + nC₃ + ……. + nC_n = 2ⁿ – nC₀ = 2ⁿ – 1

Question 10.
The term containing x³ in the expansion of (x – 2y)⁷ is:
(a) 3rd
(b) 4th
(c) 5th
(d) 6th
Answer: (c) 5th
Hint:
First-term contains x⁷.
The second term contains x⁶.
The fifth term contains x³.

Question 11.
The middle term in the expansion of \(\left(x+\frac{1}{x}\right)^{10}\) is:
(a) 10C₄ \(\left(\frac{1}{x}\right)\)
(b) 10C₅
(c) 10C₆
(d) 10C₇ x²
Answer:
(b) 10C₅
Hint:
x is x, a = \(\frac{1}{x}\), n = 10 which is even.
So the middle term is

Question 12.
The constant term in the expansion of \(\left(x+\frac{2}{x}\right)^{6}\) is:
(a) 156
(b) 165
(c) 162
(d) 160
Answer:
(d) 160
Hint:
Here x is x, a is \(\frac{2}{x}\) (Note that each term x will vanish)
∴ Constant term occurs only in middle term
n = 6
∴ middle term = \(t_{\frac{6}{2}+1}\) = t₃₊₁

Question 13.
The last term in the expansion of (3 + √2 )⁸ is:
(a) 81
(b) 16
(c) 8
(d) 2
Answer:
(b) 16
Hint:
(√2)⁸ = \(\left(2^{\frac{1}{2}}\right)^{8}\) = 2⁴ = 16

Question 14.
If \(\frac{k x}{(x+4)(2 x-1)}=\frac{4}{x+4}+\frac{1}{2 x-1}\) then k is equal to:
(a) 9
(b) 11
(c) 5
(d) 7
Answer:
(a) 9
Hint:
\(\frac{k x}{(x+4)(x-1)}=\frac{4}{x+4}+\frac{1}{2 x-1}\)
kx = 8x – 4 + x + 4
kx = 9x
k = 9

Question 15.
The number of 3 letter words that can be formed from the letters of the word ‘NUMBER’ when the repetition is allowed are:
(a) 206
(b) 133
(c) 216
(d) 300
Answer:
(c) 216
Hint:
Number of letters in NUMBER is 5
From 5 letters we can form 3 letter ways = 6 × 6 × 6 = 216.

Question 16.
The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is:
(a) 18
(b) 12
(c) 9
(d) 6
Answer:
(a) 18
Hint:

To form a parallelogram we need 2 parallel lines from 4 and 2 intersecting lines from 3.
Number of parallelograms = 4C₂ × 3C₂
= \(\frac{4 \times 3}{2 \times 1} \times 3\)
= 18

Question 17.
There are 10 true or false questions in an examination. Then these questions can be answered in
(a) 240 ways
(b) 120 ways
(c) 1024 ways
(d) 100 ways
Answer:
(c) 1024 ways
Hint:
For each question, there are two ways of answering it.
for 10 questions the numbers of ways to answer = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2¹⁰
= 1024 ways

Question 18.
The value of (5C₀ + 5C₁) + (5C₁ + 5C₂) + (5C₂ + 5C₃) + (5C₃ + 5C₄) + (5C₄ + 5C₅) is:
(a) 2⁶ – 2
(b) 2⁵ – 1
(c) 2⁸
(d) 2⁷
Answer:
(a) 2⁶ – 2
Hint:
(5C₀ + 5C₁ + 5C₂ + 5C₃ + 5C₄ + 5C₅) + (5C₁ + 5C₂ + 5C₃ + 5C₄)
= 2⁵ + (5C₀ + 5C₁ + 5C₂ + 5C₃ + 5C₄ + 5C₅) – (5C₀ + 5C₅)
= 2⁵ + 2⁵ – (1 + 1) (∵ Adding and subtracting of 5C₀ and 5C₅)
= 2(2⁵) – 2 (∵ 5C₀ = 5C₅ = 1)
= 2⁶ – 2

Question 19.
The total number of 9 digit number which has all different digit is:
(a) 10!
(b) 9!
(c) 9 × 9!
(d) 10 × 10!
Answer:
(c) 9 × 9!
Hint:
Here we can use the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
They are in 10 in total. We have to form a nine-digit number.

The first place from the left can be filled up by anyone of the digits other than zero in 9 ways. The second place can be filled up by anyone of the remaining (10 – 1) digits (including zero) in 9 ways, the third place in 8 ways, fourth place in 7 ways, fifth place in 6 ways, sixth place in 5 ways, seventh place in 4 ways, eighth place in 3 ways and ninth place in 2 ways.
∴ The number of ways of making 9 digit numbers = 9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 = 9 × 9!

Question 20.
The number of ways to arrange the letters of the word “CHEESE”:
(a) 120
(b) 240
(c) 720
(d) 6
Answer:
(a) 120
Hint: Here there are 6 letters.
The letter C occurs one time
The letter H occurs one time
The letter E occurs three times
The letter S occurs one time
Number of arrangements = \(\frac{6 !}{1 ! 1 ! 3 ! 1 !}=\frac{6 !}{3 !}=\frac{6 \times 5 \times 4 \times 3 !}{3 !}\) = 120

Question 21.
Thirteen guests have participated in a dinner. The number of handshakes that happened in the dinner is:
(a) 715
(b) 78
(c) 286
(d) 13
(b) 78
Hint:
To handshakes, we need two guests.
Number of selecting 2 guests from 13 is 13C₂ = \(\frac{13 \times 12}{2 \times 1}\) = 78

Question 22.
The number of words with or without meaning that can be formed using letters of the word “EQUATION”, with no repetition of letters is:
(a) 7!
(b) 3!
(c) 8!
(d) 5!
Answer:
(c) 8!
Hint:
There are 8 letters.
From 8 letters number of words is formed = 8P₈ = 8!

Question 23.
Sum of binomial coefficient in a particular expansion is 256, then number of terms in the expansion is:
(a) 8
(b) 7
(c) 6
(d) 9
Answer:
(a) 8
Hint:
Sum of binomial coefficient = 256
i.e., 2ⁿ = 256
2ⁿ = 2⁸
n = 8

Question 24..
The number of permutation of n different things taken r at a time, when the repetition is allowed is:
(a) rⁿ
(b) n^r
(c) \(\frac{n !}{(n-r) !}\)
(d) \(\frac{n !}{(n+r) !}\)
Answer:
(b) n^r

Question 25.
The sum of the binomial coefficients is:
(a) 2ⁿ
(b) n²
(c) 2n
(d) n + 17
Answer:
(a) 2ⁿ

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:

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.