Class 11

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2

Question 1.
Without expanding the determinant, prove that
Solution:

Question 2.
Show that
Solution:

Question 3.
Prove that
Solution:
LHS
Taking a from C₁, b from C₂ and c from C₃ we get

Expanding along R₁ we get
(2c) (abc) (1) [ab + ab] = abc (2c) (2ab)
1 = (abc) (4abc) = 4a²b²c²
= RHS

Question 4.

Solution:

Question 5.
Prove that
Solution:

Question 6.
Show that
Solution:

Question 7.
Write the general form of a 3 × 3 skew-symmetric matrix and prove that its determinant is 0.
Solution:

Question 8.

Solution:

we get – (aα² + 2bα + c) [ac – b²]
So Δ = 0 ⇒ (aα² + 2bα + c) (ac -b²) = – 0 = 0
⇒ aα² + 2bα + c = 0 or ac – b² = 0
(i.e.) a is a root of ax² + 2bx + c = 0
or ac = b²
⇒ a, b, c are in G.P.

Question 9.
Prove that
Solution:

Question 10.
If a, b, c are p^th, q^th and r^th terms of an A.P., find the value of
Solution:
We are given a = t_p,b = t_q and c = t_r
Let a be the first term and d be the common difference

Question 11.
Show that is divisible by x⁴
Solution:

Multiplying R₁ by a, R₂ by b and R₃ by c and
taking out a from C₁ b from C₂ and c from C₃ we get
==

Question 12.
If a, b, c are all positive, and are p^th, q^th and r^th terms of a G.P., show that

Solution:

Question 13.
Find the value of if x, y, z ≠ 1.
Solution:
Expanding the determinant along R₁

Question 14.

Solution:

Question 15.
Without expanding, evaluate the following determinants:

Solution:

Question 16.
If A is a square matrix and |A| = 2, find the value of |AA^T|.
Solution:
Given |A| = 2
[Property 1: The determinant of a matrix remains unaltered if its rows are changed into columns and columns. That is, |A| = |A^T|]

|A^T| = |A| = 2
∴ |A A^T| = |A| |A^T|
= 2 × 2 = 4

Question 17.
If A and B are square matrices of order 3 such that |A| = -1 and |B| = 3, find the value of |3AB|.
Solution:
Given A and B are square matrices of order 3 such that |A| = -1 and |B| = 3
[It A is a square matrix of order n then det ( kA) = |kA| = kⁿ |A|.]
A and B are square matrices of order 3. Therefore,
AB is also a square matrix of order 3.
|3 AB| = 3³ |AB|
= 27 |A| |B|
= 27 × – 1 × 3
|3 AB| = – 81

Question 18.
If λ = -2, determine the value of
Solution:
Given λ = -2
∴ 2λ = -4; λ² = (-2)²; 3λ² + 1 = 3 (4) + 1 = 13
6λ – 1 = 6(-2) – 1 = -13

expanding along R₁
0(0) + 4 (0 + 13) + 1 (-52 + 0) = 52 – 52 = 0
Aliter: The determinant value of a skew-symmetric matrix is zero

Question 19.
Determine the roots of the equation
Solution:

Given the determinant value is 0
⇒ 30(1 + x) (2 – x) = 0
⇒ 1 + x = 0 or 2 – x = 0
⇒ x = -1 or x = 2
So, x = -1 or 2.

Question 20.
Verify that det (AB) = (det A) (det B) for
Solution:


{(-20)(52) (-19) + (10)(38)(—49) + (2)(64)(-17)} – {(-49)(52) (2) + (-17)(38)(-20) + (-19)(64)(10)}
= (19760 – 18620 – 2176) – (-5096 + 12920 – 12160)
= (19760 + 5096 + 12160) – (18620 + 2176 + 12920)
= 37016 – 33716 = 3300 ….(3)
Now (1) × (2) = (3)
(i.e.,) (-33) (-100) = 3300
⇒ det (AB) = (det A), (det B)

Question 21.
Using cofactors of elements of the second row, evaluate |A|, where
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 Additional Problems

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4

Question 1.
Find the area of the triangle whose vertices are (0, 0), (1, 2) and (4, 3)
Solution:
Area of triangle with vertices

∴ Area of A with vertices (0, 0), (1, 2) and (4, 3) is

(as the area cannot be negative).

Question 2.
If (k, 2), (2, 4) and (3, 2) are vertices of the triangle of area 4 square units then determine the value of k.
Solution:

Question 3.
Identify the singular and non-singular matrices:

Solution:
(i) For a given square matrix A,
1. If |A| = 0 then it is a singular matrix.
2. If |A| ≠ 0 then it is a non singular matrix.

⇒ A is a singular matrix.

Which is a skew symmetric matrix
∴ |A| = 0 ⇒ A is a singular matrix.

Question 4.
Determine the value of a and b so that the following matrices are singular:

Solution:

expanding along R₁
b(4 + 4) + 7 (-6 – 1) = 0 (given)
8b + 7 (-7) = 0
(i.e.,) 8b – 49 = 0 ⇒ 8b = 49 ⇒ b = 49/8

Question 5.
If cos 2θ = 0, determine \(\left[\begin{array}{ccc}{\theta} & {\cos \theta} & {\sin \theta} \\ {\cos \theta} & {\sin \theta} & {0} \\ {\sin \theta} & {0} & {\cos \theta}\end{array}\right]^{2}\)
Solution:

Question 6.
Find the value of the product;
Sol:

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 Additional Problems

Question 1.
Identify the singular and non-singular matrix.

Solution:
If the determinant value of a square matrix is zero it is called a singular matrix. Otherwise it is non-singular.

Question 2.
Show that
Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.2

Question 1.
Verify whether the following ratios are direction cosines of some vector or not.

Solution:

Question 2.
Find the direction cosines of a vectors whose direction ratios are
(i) 1, 2, 3
(ii) 3, -1, 3
(iii) 0, 0, 7
Solution:

Question 3.
Find the direction cosines and direction ratios for the following vectors

Solution:

Question 4.
A triangle is formed by joining the points (1, 0, 0), (0, 1, 0) and (0, 0, 1). Find the direction cosines of the medians.
Solution:

Question 5.
If \(\frac{1}{2}, \frac{1}{\sqrt{2}}\), a are the direction cosines of some vector, then find a.
Solution:

Question 6.
If (a, a + b, a + b + c) is one set of direction ratios of the line joining (1, 0, 0) and (0, 1, 0), then find a set of values of a, b, c.
Solution:
Given (a, a + b, a + b + c) is one set of direction ratios of the line joining (1, 0, 0) and (0, 1, 0) then find a set of values of a, b, c.
Let the given points be A (1, 0, 0) and B (0, 1, 0)
[The direction ratios of the line joining the points A (x₁, y₁, z₁) and B (x₂, y₂, z₂) are x₂ – x₁, y₂ – y₁, z₂ – z₁ or x₁ – x₂, y₂ – y₂, z1,  z₁ – z₂ j
The direction ratios of the line joining the points A(1, 0, 0) and B(0, 1, 0) are
(0 – 1, 1 – 0, 0 – 0)
(- 1, 1, 0) ………. (1)
Also the direction ratios are
(1 – 0, 0 – 1, 0 – 0)
(1, -1, 0) ………. (2)
Given the direction ratios are
(a, a + b , a + b + c) ………. (3)
Comparing (1) and (3) we have
(-1, 1, 0) = (a, a + b, a + b + c)
a = -1, a + b = 1 ⇒ – 1 + b = 1 ⇒ b = 2
a + b + c = 0 ⇒ – 1 + 2 + c = 0 ⇒ c = – 1
Comparing (2) and (3) we get
a = 1 , a + b = – 1 ⇒ 1 + b = – 1 ⇒ b = – 2
a + b + c = 0 ⇒ 1 – 2 + c = 0 ⇒ c = 1
∴ The required set of values of a, b, c are
a = -1, b = 2, c = – 1 and
a = 1, b = – 2, c = 1

Question 7.
Show that the vectors \(2 \hat{i}-\hat{j}+\hat{k}, 3 \hat{i}-4 \hat{j}-4 \hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}\) form a right angled triangle.
Sol:

⇒ The given vectors form the sides of a right-angled triangle.

Question 8.
Find the value of k for which the vectors \(\vec{a}=3 \hat{i}+2 \hat{j}+9 \hat{k}\) and \(\vec{b}=\hat{i}+\lambda \hat{j}+3 \hat{k}\) are parallel.
Solution:

Question 9.
Show that the following vectors are coplanar.

Solution:
Let the given three vectors be \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\). When we are able to write one vector as a linear combination of the other two vectors, then the given vectors are called coplanar vectors.




We are able to write \(\vec{a}\) as a linear combination of \(\vec{b}\) and \(\vec{c}\)
∴ The vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar

Question 10.
Show that the points whose position vectors and are coplanar
Solution:
Let the given points be A, B, C and D. To prove that the points A, B, C, D are coplanar, we have to prove that the vectors \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}}\) and \(\overrightarrow{\mathrm{AC}}\) are coplanar



∴ we are able to write one vector as a linear combination of the other two vectors ⇒ the given vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are coplanar.
(i.e.,) the given points A, B, C, D are coplanar.

Question 11.
If \(\vec{a}=2 \hat{i}+3 \hat{j}-4 \hat{k}\), \(\vec{b}=3 \hat{i}-4 \hat{j}-5 \hat{k}\) and \(\vec{c}=-3 \hat{i}+2 \hat{j}+3 \hat{k}\), find the magnitude and direction cosines of
(i) \(\vec{a}+\vec{b}+\vec{c}\)
(ii) \(3 \vec{a}-2 \vec{b}+5 \vec{c}\)
Solution:

>

Question 12.
The position vectors of the vertices of a triangle are and . Find the perimeter of the triangle
Solution:
Let A, B, C be the vertices of the triangle ABC,

Question 13.
Find the unit vector parallel to and
Solution:

Question 14.
The position vector \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) three points satisfy the relation \(2 \vec{a}-7 \vec{b}+5 \vec{c}=\overrightarrow{0}\). Are these points collinear?
Solution:

Question 15.
The position vectors of the points P, Q, R, S are and respectively. Prove that the line PQ and RS are parallel.
Solution:

Question 16.
Find the value or values of m for which \(m(\hat{i}+\hat{j}+\hat{k})\) is a unit vector
Solution:

Question 17.
Show that points A(1, 1, 1), B(1, 2, 3), and C(2, -1, 1) are vertices of an isosceles triangle.
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.2 Additional Problems

Question 1.
Show that the points whose position vectors given by


Solution:

Question 2.
Find the unit vectors parallel to the sum of \(3 \hat{i}-5 \hat{j}+8 \hat{k}\) and \(-2 \hat{j}-2 \hat{k}\)
Solution:

Question 3.
The vertices of a triangle have position vectors Prove that the triangle is equilateral.
Solution:

Question 4.
Prove that the points form an equilateral triangle.
Solution:

Question 5.
Examine whether the vectors are coplanar
Solution:


⇒ We are not able to write one vector as a linear combination of the other two vectors
⇒ the given vectors are not coplanar.

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.4

Question 1.
For the curve y = x³ given in Figure, draw
(i) r = -x³
(ii) y = x³ + 1
(iii) y = x³ – 1
(iv) y = (x + 1)³ with the same scale.





Solution:
(i) It is the reflection on y axis
(ii) The graph of y = x³ + 1 is shifted upward to 1 unit.
(iii) The graph of y = x³ – 1 is shifted downward to 1 unit.
(iv) The graph of y = (x + 1)³ is shifted to the left for 1 unit.

Question 2.

Solution:

Question 3.
Graph the functions f(x) = x³ and g(x) = \(\sqrt[3]{x}\) on the same coordinate plane. Find fog and graph it on the plane as well. Explain your results.
Solution:

Question 4.
Write the steps to obtain the graph of the function y = 3(x – 1)² + 5 from the graph y = x².
Solution:
Draw the graph of y = x²
To get y = (x – 1)² we have to shift the curve 1 unit to the right.
Then we have to draw the curve y = 3(x – 1)² and finally, we have to draw y = 3(x – 1)² + 5

Question 5.
From the curve y = sin x, graph the functions
(i) y = sin(-x)
(ii) y = -sin(-x)
(iii) y = sin (\(\frac{\pi}{2}\) + x) which is cos x
(iv) y = sin (\(\frac{\pi}{2}\) – x) which is also cos x (refer trigonometry)
Solution:
First we have to draw the curve y = sin x

(i) y = sin (-x) = – sin x = f(x)

(ii) y = -sin(-x) = -[-sin x] = sin x

Question 6.
From the curve y = x, draw
(i) y = -x
(ii) y = 2x
(iii) y = x + 1
(iv) y = \(\frac{1}{2}\)x + 1
(v) 2x + y + 3 = 0. 2
Solution:
y = x

(i) y = -x

(ii) y = 2x


y = 2x the graph moves away from the x-axis, as multiplying factor is 2 which is greater than one.

(iii) y = 2x + 1

(iv) y = 1/2x + 1


y = \(\frac{1}{2}\)x moves towards x – axis by a side factor \(\frac{1}{2}\) which is less than y = \(\frac{1}{2}\)x + 1 upwards by 1 unit.

(v) y = -2x – 3

Question 7.
From the curve y = |x|, draw
(i) y= |x – 1| + 1
(ii) y = |x + 1| – 1
(iii) y = |x + 2| – 3.
Solution:
Given, y = |x|
If y = x

If y = -x

(i) y = |x – 1| + 1
y = x – 1 + 1
y = -x + 1 + 1 = x

(ii) y = |x + 1| – 1
y = x + 1 – 1 = x
y = -x -1 – 1
y = -x – 2
y = – (x + 2)

(iii) y = |x + 2| – 3
y = x + 2 – 3 ⇒ x – 1
y = -x – 2 + 3 = 1 – x
y = -(x – 1)

Question 8.
From the curves = sin x, draw y = sin |x| (Hint: sin(-x) = -sin x.)
Solution:
y = sin |x|
∴ y = sin x
∴ y = sin (-x) = – sin x
y = – sin x

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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

Question 1.
Find the sum of the first 20-terms of the arithmetic progression having the sum of first 10 terms as 52 and the sum of the first 15 terms as 77.
Solution:
Let ‘a’ be the first term and d be the common difference of A.P.

Question 2.

Solution:

Question 3.
Compute the sum of first n terms of the following series:
(i) 8 + 88 + 888 + 8888 + ……
(i) 6 + 66 + 666 + 6666 + …..
Solution:

Question 4.
Compute the sum of first n terms of 1 + (1 + 4) + (1 + 4 + 4²) + (1 + 4 + 4² + 4³) + ……..
Solution:

Question 5.
Find the general term and sum to n terms of the sequence \(1, \frac{4}{3}, \frac{7}{9}, \frac{10}{27}, \ldots \ldots \ldots\)
Solution:

Denominator 1, 3, 9, 27, which is a G.P. a = 1, r = 3

It is an arithmetico Geometric series. Here the n^th term is t_n = [a + (n – 1)d]r^{n – 1} where a = 1, d = 3 and r = 1/3
Now the sum to n terms is

Question 6.

Solution:

Question 7.
Show that the sum of (m + n)^th and (m – n)^th term of an A.P. is equal to twice the m^th term.
Solution:
The n^th term of an A.P is T_n = a + (n – 1) d
T_m+n = a + (m + n – 1) d
T_m-n = a + (m – n – 1) d
T_m+n + T_m-n = a + (m + n – 1) d + a + (m – n – 1)d
= 2a + [m + n – 1 + m – n – 1) d
= 2a + [2m – 2] d
= 2a + 2 (m – 1) d
= 2 [a + (m – 1] d
T_m+n + T_m-n = 2T_m
Hence the sum of the ( m + n)^th term and ( m – n)^th term of A.P is equal to twice the m^th term.

Question 8.
A man repays an amount of ₹ 3250 by paying ₹ 20 in the first month and then increases the payment by ₹ 15 per month. How long will it take him to clear the amount?
Solution:
a = 20, d = 15, S_n = 3250 to find n.

3n² + 5n – 1300 = 0
3n² – 60n + 65n – 1300 = 0
3n (n – 20) + 65 (n – 20) = 0
(3n + 65) (n – 20) = 0
n = – 65/3 or 20
n = – 65/3 is not possible
∴ n = 20
So he will take 20 months to clear the amount.

Question 9.
In a race, 20 balls are placed in a line at intervals of 4 meters with the first ball 24 meters away from the starting point. A contestant is required to bring the balls back to the starting place one at a time. How far would the contestant run to bring back all balls?
Solution:
t₁ = 24 × 2 = 48, t₂ =48 + 8 = 56 or (24 + 4)2, t₃ =(28 + 4)2 = 64 which is an A.P.
Here a = 48,
d = 56 – 48 = 8

The contestant has to run 2480 m to bring all the balls.

Question 10.
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the 2^nd hour, 4^th hour, and n^th hour?
Solution:
Number of bacteria present initially = 30
Number of bacteria at the end of 1 hour = 2 × 30 = 60
Number of bacteria at the end of two hours = 2 × 60 = 120
Number of bacteria at the end of three hours = 2 × 120 = 240
Number of bacteria at the end of four hours = 2 × 240 = 480

∴ The sequence of a number of bacterias at the end of every hour is
30, 60, 120, 240, 480, …………………………..
30, 30 × 2, 30 × 4, 30 × 8, 30 × 16, …………………
30, 30 × 2, 30 × 2², 30 × 2³, 30 × 2⁴, …………………
This sequence is a Geometric sequence with first term a = 30, common ratio r = 2
Number of bacteria at the end of n^th hour t_n+1 = a. rⁿ
t_n+1 = 30(2ⁿ)
Number of bacteria at the end of 2^nd hour = 120
Number of bacteria at the end of 4^th hours = 480
Number of bacteria at the end of n^th hour = 30(2ⁿ)

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

Question 12.
In a certain town, a viral disease caused severe health hazards upon its people disturbing their normal life. It was found that on each day, the virus which caused the disease spread in Geometric Progression. The amount of infectious virus particle gets doubled each day, being 5 particles on the first day. Find the day when the infectious virus particles just grow over 1,50,000 units?
Solution:
a = 5, r = 2, t_n >150000, To find ‘n’

On the 15^th day it will grow over 150000 units.

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

Question 1.
If the ratio of the sums of m terms and n terms of an A.P. be m² : n², prove that the ratio of its m^th and n^th terms is (2m – 1) : (2n – 1).
Solution:
Let ‘a’ be the first term and d be the common difference of A.P.

⇒ 2an + (mn – n)d = 2am + (mn – m)d
⇒ 2an – 2am = (mn – m – mn + n)d
⇒ 2a(n – m) = (n – m)d ⇒ d = 2a, [n – m ≠ 0, as n ≠ m]

Hence, the required ratio is (2m – 1) : (2n – 1)

Question 2.
Determine the number of terms of geometric progression {a_n} if a₁ = 3, a_n = 96, S_n = 189.
Solution:

Question 3.
The sum of first three terms of a G.P. is to the sum of the first six terms as 125 : 152. Find the common ratio of the G.P.
Solution:

Hence, the common ratio of the G.P. is \(\frac{3}{5}\)

Question 4.
Find the sum to n terms the series: (x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + …
Solution:
(x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + … to n terms

Question 5.
Sum the series: (1 + x) + (1 + x + x²) + (1 + x + x² + x³) + … up to n terms
Solution:

Question 6.
Sum up to n terms the series: 7 + 77 + 777 + 7777 + …
Solution:
S = 1 + 77 + 777 + 7777 + … to n terms

Question 7.
If S₁, S₂, S₃ be respectively the sums of n, 2n, 3n, terms of a G.P. , then prove that S₁(S₃ – S₂) = (S₂ – S₁)₂.
Solution:
Let a be the first term and r be the common ratio of G.P.

Question 8.
If the sum of the n terms of a G.P be S, their product P and the sum of their reciprocals R, the prove that \(P^{2}=\left(\frac{S}{R}\right)^{n}\)
Solution:
Let a be the first term and r be the common ratio of the G.P.

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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

Question 1.
The sum of the digits at the 10th place of all numbers formed with the help of 2, 4, 5, 7 taken all at a time is ……..
(a) 432
(b) 108
(c) 36
(d) 18
Solution:
(b) 108
Hint.
The given digits are 2, 4, 5 ,7
The unit place can be filled with the digit 2. Then the remaining Ten’s , hundred’s, Thousand’s place can be filled with the remaining three digits 4, 5, 7 in 3P₃ ways.
∴ There will be 3 × 2 × 1 = 6 four digit numbers whose unit place is 2.
Similarly, there are 6 four digit numbers whose unit place is 4 , 6 four digit numbers whose unit place is 5 and 6 four digit numbers whose unit place is 7.
∴ Total in the unit place
= 6 × 2 + 6 × 4 + 6 × 5 + 6 × 7
= 6 × (2 + 4 + 5 + 7)
= 6 × 18 = 108
Sum of the digit at the unit place = 108
∴ Sum of the digit at the tens place = 108

Question 2.
In an examination, there are three multiple-choice questions and each question has 5 choices. A number of ways in which a student can fail to get all answer correct is ……..
(a) 125
(b) 124
(c) 64
(d) 63
Solution:
(b) 124
Hint.
Each question has 5 options in which 1 is correct.
So the number of ways of getting the correct answer for all three questions is 5³ = 125
So the number of ways in which a student can fail to get all answer correct is < 125 (i.e.) 125 – 1 = 124

Question 3.
The number of ways in which the following prize be given to a class of 30 boys first and second in mathematics, first and second in physics, first in chemistry and first in English is ……….
(a) 30⁴ × 29²
(b) 30³ × 29³
(c) 30² × 29⁴
(d) 30 × 29⁵
Solution:
(a) 30⁴ × 29²
Hint.
Number of students = 30
First prize can be given to any one of 30 students
= 30 × 30 × 30 × 30 = 30⁴ ways
Second prize can be given to anyone of the remaining 29 students = 29 × 29 = 29² ways
∴ Total number of ways prizes can be given = 30⁴ × 29² ways

Question 4.
The number of 5 digit numbers all digits of which are odd is ………
(a) 25
(b) 5⁵
(c) 5⁶
(d) 625
Solution:
(b) 5⁵
Hint. The odd number are 1, 3, 5, 7, 9
Number of odd numbers = 5
We need a five-digit number So the number of five-digit number = 5⁵

Question 5.
In 3 fingers, the number of ways four rings can be worn is …… ways.
(a) 4³ – 1
(b) 3⁴
(c) 68
(d) 64
Solution:
(b) 3⁴
Hint.
Number of rings = 4
Each finger can be worn rings in 4 ways.
∴ Number of ways of wearing four rings in three fingers
= 4 × 4 × 4
= 64

Question 6.

(a) 7 and 11
(b) 6 and 7
(c) 2 and 11
(d) 2 and 6
Solution:
(b) 6 and 7

Question 7.
The product of r consecutive positive integers is divisible by ………
(a) r!
(b) (r – 1)!
(c) (r + 1)!
(d) r!
Solution:
(a) r!
Hint.
The product of r consecutive positive integers is
1 × 2 × 3 × …………… × r = r!
which is divisible by r!
Also, 1 × 2 × 3 × …………… × r = r!
= ( r – 1) ! × r
which is divisible by (r – 1) !

Question 8.
The number of 5 digit telephone numbers which have none of their digits repeated is
(a) 90000
(b) 10000
(c) 30240
(d) 69760
Solution:
(d) 69760
Hint.
The number of 5 digit telephone numbers which have none of their digits repeated is ¹⁰P₅ = 30240
Thus the required telephone number is 10⁵ – 30240 = 69760

Question 9.
If a² – ^aC₂ = a² – ^aC₄ then the value of ‘a’ is ….
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
(b) 3
Hint.
a₂ – a = 2 + 4 = 6
a₂ – a – 6 = 0
(a – 3) (a + 2) = 0 ⇒ a = 3

Question 10.
There are 10 points in a plane and 4 of them are collinear. The number of straight lines joining any two points is ……..
(a) 45
(b) 40
(c) 39
(d) 38
Solution:
(b) 40
Hint.

Question 11.
The number of ways in which a host lady invite 8 people for a party of 8 out of 12 people of whom two do not want to attend the party together is
(a) 2 × ¹¹C₇ + ¹⁰C₈
(b) ¹¹C₇ + ¹⁰C₈
(c) ¹²C₈ – ¹⁰C₆
(d) ¹⁰C₆ + 2!
Solution:
(c) ¹²C₈ – ¹⁰C₆
Hint.
Number of the way of selecting 8 people from 12 in ¹²C₈
∴ out of the remaining people, 8 can attend in ¹⁰C₈
The number of ways in which two of them do not attend together = ¹²C₈ – ¹⁰C₆

Question 12.
The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines …….
(a) 6
(b) 9
(c) 12
(d) 18
Solution:
(d) 18
Hint.
Number of parallelograms = ⁴C₂ × ³C₂ = 6 × 3 = 18

Question 13.
Everybody in a room shakes hands with everybody else. The total number of shake hands is 66. The number of persons in the room is …….
(a) 11
(b) 12
(c) 10
(d) 6
Solution:
(b) 12
Hint.

Question 14.
The number of sides of a polygon having 44 diagonals is ……….
(a) 4
(b) 4!
(c) 11
(d) 22
Solution:
(c) 11
Hint:

Question 15.
If 10 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, then the total number of points of intersection are ………
(a) 45
(b) 40
(c) 10!
(d) 2¹⁰
Solution:
(a) 45
Hint:

Question 16.
In a plane there are 10 points are there out of which 4 points are collinear, then the number of triangles formed is …….
(a) 110
(b) ¹⁰C₃
(c) 120
(d) 116
Solution:
(d) 116
Hint:

Question 17.
In ²ⁿC₃ : ⁿC₃ = 11 : 1 then n is ………
(a) 5
(b) 6
(c) 11
(d) 1
Solution:
(b) 6
Hint.

Question 18.
^{(n – 1)}C_r + ^{(n – 1)}C_{(r – 1)} is ………
(a) ^{(n + 1)}C_r
(b) ^{(n – 1)}C_r
(c) ⁿC_r
(d) ⁿC_{r – 1}
Solution:
(c) ⁿC_r

Question 19.
The number of ways of choosing 5 cards out of a deck of 52 cards which include at least one king is …….
(a) ⁵²C₅
(b) ⁴⁸C₅
(c) ⁵²C₅ + ⁴⁸C₅
(d) ⁵²C₅ – ⁴⁸C₅
Solution:
(d) ⁵²C₅ – ⁴⁸C₅
Hint.
Selecting 5 from 52 cards = ⁵²C₅
selecting 5 from the (non-king cards 48) = ⁴⁸C₅
∴ Number of ways is ⁵²C₅ – ⁴⁸C₅

Question 20.
The number of rectangles that a chessboard has ……
(a) 81
(b) 99
(c) 1296
(d) 6561
Solution:
(c) 1296
Hint. Number of horizontal times = 9
Number of vertical times = 9
Selecting 2 from 9 horizontal lines = ⁹C₂
Selecting 2 from 9 vertical lines = ⁹C₂

Question 21.
The number of 10 digit number that can be written by using the digits 2 and 3 is ……..
(a) ¹⁰C₂ + ⁹C₂
(b) 2¹⁰
(c) 2¹⁰ – 2
(d) 10!
Solution:
(b) 2¹⁰
Hint.
Selecting the number from (2 and 3)
For till the first digit can be done in 2 ways
For till the second digit can be done in 2 ways ….
For till the tenth digit can be done in 2 ways
So, total number of ways in 10 digit number = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2¹⁰

Question 22.
If P_r stands for ^rP_r then the sum of the series 1 + P₁ + 2P₂ + 3P₃ + … + nP_n is ……..
(a) P_{n + 1}
(b) P_{n + 1} – 1
(c) P_{n – 1} + 1
(d) ^{(n + 1)}P_{(n – 1)}
Solution:
(b) P_{n + 1} – 1
Hint:
1 + 1! + 2! + 3! + … + n!
Now 1 + 1 (1!) = 2 = (1 + 1)!
1 + 1 (1!) + 2(2!) = 1 + 1 + 4 = 6 = 3!
1 + 1(1!) + 2(2!)+ 3(3!) = 1 + 1 + 4 + 18 = 24 = 4!
1 + 1(1!) + 2(2!) + 3(3!) ….+ n(n!) = (n + 1) ! – 1
= ^{n + 1}P_{n + 1} – 1 = P_{n + 1} – 1

Question 23.
The product of first n odd natural numbers equals …….

Solution:

Hint:

Question 24.
If ⁿC₄, ⁿC₅, ⁿC₆ are in AP the value of n can be ………..
(a) 14
(b) 11
(c) 9
(d) 5
Solution:
(a) 14
Hint:

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

Question 25.
1 + 3 + 5 + 7 + + 17 is equal to ………
(a) 101
(b) 81
(c) 71
(d) 61
Solution:
(b) 81
Hint:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.2

Question 1.
If ^{(n – 1)} P₃ : ⁿP₄, find n:
Solution:

Question 2.
If ¹⁰P_{r – 1} = 2 × ⁶P_r, find r.
Solution:

Question 3.
(i) Suppose 8 people enter an event in a swimming meet. In how many ways could the gold, silver, and bronze prizes be awarded?
Solution:
Number people enter in a swimming meet = 8 Prizes awarded = Gold, silver, bronze.
The gold medal can be awarded to any one of the 8 people in 8 ways. The silver medal can be awarded to any one of the remaining 7 people in 7 ways. The bronze medal can be awarded to any one of the remaining 6 people in 6 ways,
∴ Total number of ways of awarding the prizes = 8 × 7 × 6 = 336

(ii) Three men have 4 coats, 5 waistcoats, and 6 caps. In how many ways can they wear them?
Solution:
Selecting and arranging 3 coats from 4 can be done in ⁴P₃ ways
Selecting and arranging 3 waistcoats from 5 can be done in ⁵P₃ ways Selecting and arranging 3 caps from 6 can be done in ⁶P₃ ways
∴ Total number of ways = ⁴P₃ × ⁵P₃ × ⁶P₃ = 172800 ways

Question 4.
Determine the number of permutations of the letters of the word SIMPLE if all are taken at a time?
Solution:
Number of letters in the word SIMPLE = 6
The total number of the word is equal to the number of arrangements of these letters, taken all at a time
∴ Total number of words = 6P₆ = 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720

Question 5.
A test consists of 10 multiple-choice questions. In how many ways can the test be answered if

(i) Each question has four choices?
Solution:
Total number of questions = 10
Each question has four choices.
Each question can be answered in 4 ways.
∴ The total number of ways of answering 10 questions
= 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4
= 4¹⁰

(ii) The first four questions have three choices and the remaining have five choices?
Solution:
The first four questions have three choices and the remaining have five choices
First, four questions have three choices.
∴ The number of ways of answering the first four questions is = 3 × 3 × 3 × 3 = 3⁴
The remaining six questions have 5 choices
∴ The number of ways of answering the remaining 6 questions is = 5 × 5 × 5 × 5 × 5 × 5 = 5⁶
∴ The total number of ways of answering the questions = 3⁴ × 5⁶

(iii) Question number n has n + 1 choices?
Solution:
Question number n has n + 1 choices.
The first question has a 1 + 1 choice.
∴ Number of ways of answering the first question = 2
The second question has 2 + 1 choices
∴ Number ways of answering second question = 3

Tenth question has (10 + 1) choices
∴ Number of ways of answering tenth question = 11
∴ A total number of ways of answering the given 10 questions = 2 × 3 × 4 × ……. × 11 = 11!

Question 6.
A student appears in an objective test which contain 5 multiple-choice questions. Each question has four choices out of which one correct answer.

(i) What is the maximum number of different answers can the students give?
Solution:
Number multiple-choice questions = 5
Number of ways of answering each question = 4
∴ The total number of ways of answering the five questions = 4 × 4 × 4 × 4 × 4 = 45
Hence, the maximum number of different answers = 4⁵

(ii) How will the answer change if each question may have more than one correct answers?
Solution:
When each question has more than 1 correct answer.
Selecting the correct choice from the 4 choice can be done is ⁴C₁ or ⁴C₂ or ⁴C₃ or ⁴C₄ ways.

Each question can be answered in 15 ways.
Number of questions = 5
∴ Total number of ways = 15⁵

Question 7.
How many strings can be formed from the letters of the word ARTICLE, so that vowels occupy the even places?
Solution:
The given word is ARTICLE
Number of letters in the word = 7
Vowels in the given word = A, I, E
Number of vowels in the given word = 3

∴ Number of even places = 3
3 Vowels can occupy the 3 even places in 3P3 = 3! ways
The remaining 4 letters can occupy the remaining places in
4P₄ = 4! ways
Hence a total number of ways of arrangement = 4! × 3! = 4 × 3 × 2 × 3 × 2 = 144

Question 8.
8 women and 6 men are standing in a line.
(i) How many arrangements are possible if any individual can stand in any position?
Solution:
Total number of persons in a line = 8 + 6 = 14
The number of ways of standing 14 persons in a line in any position = 14P₁₄ = 14!

(ii) In how many arrangements will all 6 men be standing next to one another?
Solution:
Consider 6 men as one unit.
8 women + 6 men as one unit = 9 can be arranged in 9! ways.
6 men can among themselves be arranged in 6! ways.
∴ A total number of ways of arrangement = 9! × 6!

(iii) In how many arrangements will no two men be standing next to one another?
Solution:
Since no two men be together they have to be placed between 8 women and before and after the women.
w | w | w | w | w | w | w | w
There are 9 places so the 6 men can be arranged in the 9 places in ⁹P₆ ways.
After this arrangement, the 8 women can be arranged in 8! ways.
∴ Total number of arrangements = (⁹P₆) × 8!

Question 9.
Find the distinct permutations of the letters of the word MISSISSIPPI?
Solution:
MISSISSIPPI
Number of letters = 11
Here M – 1 time
I – 4 times
S – 4 times
P – 2 times

Question 10.
How many ways can the product a2b3c4 be expressed without exponents?
Solution:
a²b³c⁴ = aabbbcccc
Number of letters = 9
a = 2 times,
b = 3 times,
c = 4 times,

Question 11.
In how many ways 4 mathematics books, 3 physics books, 2 chemistry books and 1 biology book can be arranged on a shelf so that all books of the same subjects are together.
Solution:
Number of subjects = 4
4 subjects can be arranged in the shelf in 4! ways Number of mathematics books = 4
4 Mathematics books keeping together can be arranged in 4! ways
Number of physics books = 3
3 Physics books keeping together can be arranged in 3! ways.
Number of chemistry books =2
2 Chemistry books keeping together can be arranged in 2! ways.
Number of biology books = 1
1 biology book can be arranged in 1! way
Hence, the total numbers of ways of arranging the books
= 4! × 4! × 3! × 2! × 1!
= (4 × 3 × 2 × 1) (4 × 3 × 2 × 1) × (3 × 2) (2 × 1)
= 24 × 24 × 6 × 2
= 6912

Question 12.
In how many ways can the letters of the word SUCCESS be arranged so that all Ss are together?
Solution:
The given word is SUCCESS
Number of letters other than S = 4
Treating all S’s together as one letter
Total number of letters in the word = 5
Number of U’s = 1
Number of C’s = 2
Number of E’s = 1
Number of S’s (treated as one letter ) = 1
Number of ways of arranging = \(\frac{5 !}{2 ! \times 1 ! \times 1 ! \times 1 !}\)
= \(\frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}\)
= 5 × 4 × 3
= 60 ways

Question 13.
A coin is tossed 8 times,

(i) How many different sequences of heads and tails are possible?
Solution:
Number of coins tossed = 8
Number of outcome for each toss = 2
Total number of outcomes = 28

(ii) How many different sequences containing six heads and two tails are possible?
Solution:
Getting 6 heads and 2 tails can be done in ⁸P₆ or ⁸P₂ ways

Question 14.
How many strings are there using the letters of the word INTERMEDIATE, if

(i) The vowels and consonants are alternative
Solution:
INTERMEDIATE

The number of ways in which vowels and consonants are alternative = \(\frac{6 ! 6 !}{3 ! 2 !}=\) 43200

(ii) All the vowels are together
Solution:
The number of arrangements:
Keeping all the vowels as a single unit. Now we have 6 + 1 = 7 units which can be arranged in 7! ways.
Now the 6 consonants can be arranged in \(\frac{6 !}{2 !}\) (T occurs twice) ways in vowels, I – repeats thrice
and E – repeats twice

(iii) Vowels are never together (and) (iv) No two vowels are together.
Solution:
Vowels should not be together = No. of all arrangements – No. of all vowels together

So number of ways in which No two vowels are together = 19958400 – Number of ways in which vowels are together = 19958400 – 151200 = 19807200

Question 15.
Each of the digits 1,1, 2, 3, 3 and 4 is written on a separate card. The seven cards are then laid out in a row to form a 6-digit number.

(i) How many distinct 6-digit numbers are there?
Solution:
The given digits are 1, 1, 2, 3, 3, 4
The 6 digits can be arranged in 6! ways
In which 1 and 3 are repeated twice.

(ii) How many of these 6-digit numbers are even?
Solution:
To find the number even numbers
The digit in unit place is 2 or 4 which can be filled in 2 ways

(iii) How many of these 6-digit numbers are divisible by 4?
Solution:
To get a number -f- by 4 the last 2 digits should be -r- by 4 So the last two digits will be 12 or 24 or 32.
When the last 2 digits are 1 and 2.

When the last 2 digits are 3 and the number of 6 digit numbers (remaining number 1, 1, 3, 4)
So there of 6 digit numbers ÷ by 4 = 12 + 6 + 12 = 30

Question 16.
If the letters of the word GARDEN are permuted in all possible ways and the strings thus formed are arranged in the dictionary order, then find the ranks of the words
(i) GARDEN
(ii) DANGER.
Solution:
The given letters are GARDEN.
To find the rank of GARDEN:
The given letters in alphabetical order are A D E G N R

The rank of GARDEN is 379
To find the rank of DANGER

(ii) The No. of words starting with A = 5! =120

Question 17.
Find the number of strings that can be made using all letters of the word THING. If these words are written as in a dictionary, what will be the 85^th string?
Solution:
(i) Number of words formed = 5! = 120
(ii) The given word is THING
Taking the letters in alphabetical order G H I N T
To find the 85^th word

Question 18.
If the letters of the word FUNNY are permuted in all possible ways and the strings thus formed are arranged in the dictionary order, find the rank of the word FUNNY.
Solution:
The given word is FUNNY

Question 19.
Find the sum of all 4-digit numbers that can be formed using digits 1, 2, 3, 4, and 5 repetitions not allowed?
Solution:
The given digits are 1, 2, 3, 4, 5
The no. of 4 digit numbers

Sum of the digits = 1 + 2 + 3 + 4 + 5 = 15
Sum of number’s in each place = 24 × 15 = 360

Question 20.
Find the sum of all 4-digit numbers that can be formed using digits 0, 2, 5, 7, 8 without repetition?
Solution:
The given digits are 0, 2, 5, 7, 8
To get the number of 4 digit numbers
1000’s place can be filled in 4 ways (excluding 0)
100’s place can be filled in 4 ways (excluding one number and including 0)
10’s place can be filled in (4 – 1) = 3 ways
and unit place can be filled in (3 – 1) = 2 ways
So the number of 4 digit numbers = 4 × 4 × 3 × 2 = 96

To find the sum of 96 numbers:

In 1000’s place we have the digits 2, 5, 7, 8. So each number occurs \(\frac{96}{4}\) = 24 times.
Now in 100’s place 0 come 24 times. So the remaining digits 2, 5, 7, 8 occurs 96 – 24 = \(\frac{72}{4}\) = 18 times
Similarly in 10’s place and in-unit place 0 occurs 24 times and the remaining digits 2, 5, 7, 8 occurs 18 times.
Now sum of the digits = 2 + 5 + 7 + 8 = 22
Sum in 1000’s place = 22 × 24 = 528
Sum in 100’s, 10’s and in unit place = 22 × 18 = 396
∴ Sum of the 4 digit numbers is

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.2 Additional Questions

Question 1.

Solution:

Question 2.
How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7 if no digit is repeated?
Solution:
Here total number of digits = 6
The unit place can be filled with any one of the digits 2, 4, 6.

Question 3.
Find n if ^{n – 1}P₃ : ⁿP₄ = 1 : 9
Solution:

Question 4.
How many words can be formed by using the letters of the word ORIENTAL so that A and E always occupy the odd places?
Solution:
[Hint: There are 4 odd places in the word]

Now the remaining 6 places filled with remaining 6 letters

Hence the total number of permutations = 12 × 720 = 8640

Question 5.
Out of 18 points in a plane, no three are in the same line except five points which are collinear. Find the number of lines that can be formed joining the points.
Solution:
Total number of points = 18
Out of 18 numbers, 5 are collinear and we get a straight line by joining any two points.
∴ Total number of straight-line formed by joining 2 points out of 18 points = ¹⁸C₂
Number of straight lines formed by joining 2 points out of 5 points = ⁵C₂
But 5 points are collinear and we get only one line when they are joined pairwise.
So, the required number of straight lines are

Hence, the total number of straight lines = 144

Question 6.
We wish to select 6 person from 8 but, if the person A is chosen, then B must be chosen. In how many ways can selections be made?
Solution:
Total number of persons = 8
Number of persons to be selected = 6
Condition is that if A is chosen, B must be chosen

Case I: When A is chosen, B must be chosen
Number of ways = ⁶C₄
[∴ A and B are set to be choosen]

Case II: When A is not chosen, then B may be chosen
∴ Number of ways = ⁷C₆

Hence, the required number of ways = 22

Question 7.
How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
Solution:
For 3-digit even numbers, the unit’s place can be occupied by one of the 3 digits 2, 4 or 6. The remaining 5 digits can be arranged in the remaining 2 places in ⁵P₂ ways.

∴ By the multiplication rule, the required number of 3-digit even numbers is 3 × ⁵P₂ = 3 × 5 × 4 = 60.

Question 8.
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?
Solution:
For 4 digit numbers, we have to arrange the given 5 digits in 4 vacant places. This can be done in ⁵P₄ = 5 × 4 × 3 × 2 = 120 ways.

For 4-digit even numbers, the unit’s place can be occupied by one of the 2 digits 2 or 4. The remaining 4 digits can be arranged in the remaining 3 places in ⁴P₃ ways.
∴ By the multiplication rule, the required number of 4-digit even numbers is 2 × ⁴P₃ = 2 × 4 × 3 × 2 = 48.

Question 9.
Find r if

(i) ⁵P_r = 2 ⁶P_{r – 1} 4
Solution:

(ii) ⁵P_r = ⁶P_{r – 1}

⇒ 42 – 13r + r² = 6 ⇒ r² – 13r + 36 = 0
⇒ (r – 4)(r – 9) = 0 ⇒ r = 4, 9
Now, we know that ⁿP_r is meaningful only when r ≤ n.
∴ ⁵P_r and ⁶P_{r – 1} are meaningless when r ≤ 9.
∴ Rejecting r = 9, we have r = 4

Question 10.
How many words, with or without meaning, can be made from the letters of the word MONDAY, assuming that no letter is repeated, if

(i) 4 letters are used at a time
(ii) all letters are used at a time
(iii) all letters are used but first letter is a vowel?
Solution:
The word MONDAY has 6 distinct letters.

(i) 4 letters out of 6 can be arranged in ⁶P₄ ways.
∴ The required number of words = ⁶P₄ = 6 × 5 × 4 × 3 = 360

(ii) 6 letters can be arranged among themselves in 6P6 ways.
∴ The required number of words = ⁶P₆ = 6! = 1 × 2 × 3 × 4 × 5 × 6 = 720.

(iii) The first place can be filled by anyone Of the two vowels O or A in 2 ways. The remaining 5 letters can be arranged in the remaining 5 places II to VI in ⁵P₅ = 5! ways.
∴ By the multiplication rule, the required number of words = 2 × 5! = 2 × 1 × 2 × 3 × 4 × 5 = 240

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.2

Question 1.
Find the equation of the lines passing through the point (1, 1)
(i) With y-intercept (- 4)
(ii) With slope 3
(iii) And (-2, 3)
(iv) And the perpendicular from the origin makes an angle 60° with x-axis.
Solution:
(i) Given y intercept = – 4,
Let x intercept be a

(ii) with slope 3
The equation the line passing through the point (x₁, y₁) and having slope m is
y – y₁ = m(x – x₁)
Given (x₁, y₁) = (1, 1), m = 3
∴ The required equation of the line is
y – 1 = 3(x – 1)
y – 1 = 3x – 3
3x – y – 3 + 1 = 0
3x – y – 2 = 0

(iii) Passing through (1, 1) and (-2, 3)

Question 2.
If P (r, c) is mid point of a line segment between the axes, then show that \(\frac{x}{r}+\frac{y}{c}=\) 2.
Solution:
P (r, c) is the midpoint of AB.
⇒ A = (2r, 0) and B = (0, 2c)
(i.e) x intercept = 2r and
y intercept = 2c .

Question 3.
Find the equation of the line passing through the point (1, 5) and also divides the co-ordinate axes in the ratio 3 : 10.
Solution:
Let x-intercept be 3a and y-intercept be 10a

Question 4.
If p is length of perpendicular from origin to the line whose intercepts on the axes are a and b, then show that \(\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}\)
Solution:

Question 5.
The normal boiling point of water is 100°C or 212°F and the freezing point of water is 0°C or 32°F.
(i) Find the linear relationship between C and F
(ii) Find the value of C for 98.6° F and
(iii) The value of F for 38°C.
Solution:
Given when C = 100, F = 212 and when C = 0, F = 32

Question 6.
An object was launched from a place P in constant speed to hit a target. At the 15th second, it was 1400m away from the target and at the 18th second 800m away. Find
(i) The distance between the place and the target
(ii) The distance covered by it in 15 seconds,
(iii) Time is taken to hit the target.
Solution:
Taking time = x and distance = y
We are given at x = 15, y = 1400 and at x = 18, y = 800

Question 7.
The population of a city in the years 2005 and 2010 are 1,35,000 and 1,45,000 respectively. Find the approximate population in the year 2015. (assuming that the growth of population is constant).
Solution:
Taking the year as x and population as y
We are given when x = 2005,
y = 1,35,000 and
when x = 2010,
y = 1,45,000

y – 135000 = 2000 (x – 2005)
y = 2000(x – 2005) + 135000
At x = 2015, y = 2000 (2015 – 2005) + 135000
(i.e) y = 2000 (10) + 135000 = 20000 + 135000 = 1,55,000
The approximate population in the year 2015 is 1,55,000

Question 8.
Find the equation of the line, if the perpendicular drawn from the origin makes an angle 30° with x – axis and its length is 12.
Solution:
The equation of the line is x cos α + y sin α = p

Question 9.
Find the equation of the straight lines passing through (8, 3) and having intercepts whose sum is 1.
Solution:
Given sum of the intercepts = 1 ⇒ when x-intercept = a then y-intercept = 1 – a

8 (1 – a) + 3a = a (1 – a)
8 – 8a + 3a = a – a²

Question 10.

Solution:

⇒ The points A, B, C lie on a line
⇒ The points A, B, C are collinear

Question 11.
A straight line is passing through the point A (1, 2) with slope \(\frac{5}{12}\). Find points on the line which are 13 units away from A.
Solution:

Question 12.
A 150m long train is moving with a constant velocity of 12.5 m/s. Find
(i) The equation of the motion of the train,
(ii) Time taken to cross a pole,
(iii) The time taken to cross the bridge of length 850 m is?
Solution:
(i) Now m = \(\frac{y}{x}\) = 12.5m / second,
The equation of the line is y = mx + c ….(1)
Put c = -150, m = 12.5 m,
The equation of motion of the train is y = 12.5x – 150

(ii) To find the time taken to cross a pole we take y = 0 in (1)
⇒ 0 = 12.5x – 150 ⇒ 12.5x = 150

(iii) When y = 850 in (1)
850 = 12.5 x – 150 ⇒ 12.5x = 850 + 150 = 1000

Question 13.
A spring was hung from a hook in the ceiling. A number of different weights were attached to the spring to make it stretch, and the total length of the spring was measured each time shown in the following table.

(i) Draw a graph showing the results.
(ii) Find the equation relating the length of the spring to the weight on it.
(iii) What is the actual length of the spring.
(iv) If the spring has to stretch to 9 cm long, how much weight should be added?
(v) How long will the spring be when 6 kilograms of weight on it?
Solution:
Taking weight (kg) as x values and length (cm) as y values we get (x₁, y₁) = (2, 3), (x₂, y₂) = (4, 4)

The equation of the line passing through the above two points is

(iii) When x = 0, 2y = 4 ⇒ y = 2 cm

(iv) When y = 9 cm, x – 18 = – 4
x = -4 + 18 = 14 kg

(v) When x = 6 (kg), 6 – 2y = – 4, -2y = -4 – 6 = -10
⇒ 2y = 10 ⇒ y = 10/2 = 5 cm.

Question 14.
A family is using Liquefied petroleum gas (LPG) of weight 14.2 kg for consumption. (Full weight 29.5 kg includes the empty cylinders tare weight of 15.3 kg.). If it is used at a constant rate then it lasts for 24 days. Then the new cylinder is replaced
(i) Find the equation relating the quantity of gas in the cylinder to the days.
(ii) Draw the graph for the first 96 days.
Solution:

Question 15.
In a shopping mall, there is a hall of cuboid shape with dimension 800 × 800 × 720 units, which needs to be added the facility of an escalator in the path as shown by the dotted line in the figure. Find
(i) The minimum total length of the escalator,
(ii) The heights at which the escalator changes its direction,
(iii) The slopes of the escalator at the turning points.

Solution:
(i) the minimum total length of the escalator.
Shape of the hall in the shopping mall is cuboid. When you open out the cuboid, the not of the cuboid will be as shown in the following diagram.

The path of the escalator is from OA to AB to BC to CD

The minimum length = 3280 units

(ii) The height at which the escalator changes its direction.

(iii) Slope of the escalator at the turning points

Since ∆OAE = ∆ABB’ = ∆BCC’ = ∆CAD
Slope at the points B, C will be \(\frac{9}{40}\)

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.2 Additional Questions Solved

Question 1.
Find the equation of a straight line on which length of the perpendicular from the origin is four units and the line makes an angle of 120° with the positive direction of the x-axis.
Solution:

Question 2.
Find the equation of the line which passes through the point (- 4, 3), and the portion of the line intercepted between the axes is divided internally in the ratio 5 : 3 by this point.
Solution:
Let AB be a line passing through a point (-4, 3) and meets the x-axis at A (a, 0) and the y-axis at B (0, b).

Question 3.
If the intercept of a line between the coordinate axes is divided by the point (-5, 4) in the ratio 1 : 2, then find the equation of the line.
Solution:
Let a and b be the intercepts on the given line.


Hence, the required equation is 8x – 5y + 60 = 0

Question 4.
Find the equation of the straight line which passes through the point (1, -2) and cuts off equal intercepts from axes.
Solution:
Intercept form of a straight line is \(\frac{x}{a}+\frac{y}{b}\) = 1, where a and b are the intercepts on the axis

If equation (1) passes through the point (1, -2) we get

So, equation of the straight line is x v

Hence, the required equation x + y + 1 = 0

Question 5.
Find the distance of the line 4x – y = 0 from the point P(4, 1) measured along the line making an angle 135° with the positive x-axis
Solution:
The equation in distance form of the line passing through P(4, 1) and making an angle of 135° with the positive x-axis

Question 6.
The line 2x – y = 5 turns about the point on it, whose ordinate and abscissa are equal, through an angle of 45° in the anti-clockwise direction, find the equation of the line in the new position.
Solution:
If the line 2x – y = 5 makes an angle θ with the x-axis.
Then, tan θ = 2. Let P (α, α) be a point on the line 2x – y = 5.
Then, 2 α – α = 5 ⇒ α = 5

So, the coordinates of P are (5, 5). If the line 2x – y – 5 = 0 is rotated about point
P through 45° in an anti-clockwise direction, then the line in its new position makes angle θ + 45° with the x-axis.
Let m’ be the slope of the line in its new position. Then,

Thus, the line in its new pdsition passes through P (5, 5) and has slope m’ = -3
So, its equation y – 5 = m’ (x – 5) or, y – 5 = -3 (x – 5) or, 3x + y – 20 = 0

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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

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

Here m₁ = m₂ ⇒ the two lines are parallel.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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

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

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

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

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

Slope of the required line which is perpendicular to it

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

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

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

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

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

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

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

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

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

⇒ 28x + 7y – 74 = 0

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

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.4

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

Solution:

∴ P(k+ 1) is true.
Thus P(K) is true ⇒ (k + 1) is true.
Hence by principle of mathematical induction, P(n) is true for all n ∈ N.

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

Solution:

∴ P(1) is true
Let P(n) be true for n = k

∴ P(k + 1) is true
Thus P(k) is true ⇒ P(k + 1) is true. Hence by the principle of mathematical induction, P(k) is true for all n ∈ N.

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

Step 1:
Let us verify the statement for n = 1
P (1 ) = 2 = 1² + 1 = 1 + 1 = 2.
∴ The given result is true for n = 1.

Step 2:
Let us assume that the given result is true for n = k
P ( k) = 2 + 4 + 6 + ………… + 2k = k² + k

Step 3:
Let us prove the result for n = k + 1
P (k+ 1 ) = 2 + 4 + 6+ + 2k + (2k + 2 )
P(k+ 1 ) = P(k) + (2k + 2)
= k² + k + 2k + 2
= k² + 3k + 2
= k² + 2k + k + 2
= k(k +2) + 1(k + 2)
P(k+ 1 ) = (k+ 1) (k + 2) ……….. (1)
P (k) = k² + k
= k (k + 1 )
P(k+ 1) = (k + 1) (k + 1 + 1)
= (k + 1) (k + 2)
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
2 + 4 + 6 + ………….. + 2n = n² + n
is true for all natural numbers n.

Question 4.
By the principle of Mathematical induction, prove that, for n ≥ 1.

Solution:


∴ P(k + 1) is true
Thus P(k) is true ⇒ P(k + 1) is true
Hence by the principle of mathematical induction, P(n) is true for all n ∈ N

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

Solution:


⇒ P(k + 1) is true when P(k) is true so by the principle of mathematical induction P(n) is true.

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

Solution:


⇒ P(k + 1) is true when P(k) is true so by the principle of mathematical induction P(n) is true for n ≥ 2.

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

Solution:


∴ P(k + 1) is true
Thus p(k) is true ⇒ P(k + 1) is true
Hence by the principle of mathematical induction,
p(n) is true for all n ∈ z

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

Solution:


∴ P(k + 1) is true
Thus P(k) is true ⇒ P(k + 1) is true. Hence by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 9.
Prove by Mathematical Induction that
1! + (2 × 2!) + (3 × 3!) + … + (n × n!) = (n + 1)! – 1
Solution:
Let p(n) = 1! + (2 × 2 !) + (3 × 3!) +…. .+ (n × n!) = (n + 1)! – 1

Step 1:
First let us verify the result for n = 1
P(1) = 1! = (1 + 1)! – 1
P(1) = 1! = 2! – 1
P(1) = 1 = 2 – 1 = 1
∴ We have verified the result for n = 1.

Step 2:
Let us assume that the result is true for n = k
P(k) = (1 × 1 !) + (2 × 2!) + (3 × 3!) + …………. + (k × k!) = (k + 1)! – 1

Step 3:
Let us prove the result for n = k + 1
P(k + 1)=(1 × 1!) + (2 × 2!) + (3 × 3!) + ………….. + (k × k!) + ((k + 1) × (k + 1)!)
P(k + 1) = P(k) + ((k + 1) × (k + 1)!)
P(k + 1) = (k + 1)! – 1 + (k + 1) × (k + 1)!
= (k + 1)! + (k + 1) (k + 1)! – 1
= (k + 1)! (1 + k + 1) – 1
= (k + 1)! (k + 2) – 1
= (k + 2)! – 1
P(k + 1) = ((k + 1) + 1)! – 1
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
(1 × 1!) + (2 × 2!) + (3 × 3!) + …………… + (n × n!) = (n + 1)! – 1
is true for all natural numbers n.

Question 10.
Using the Mathematical induction, show that for any natural number n, x²ⁿ – y²ⁿ is divisible by x +y.
Solution:
Let P(n) = x²ⁿ – y²ⁿ is divisible by x + y
Step 1:
First, let us verify the result for n = 1.
P ( 1 ) = x²⁽¹⁾ – y²⁽¹⁾ = x² – y²
P(1) = (x + y) (x – y) which is divisible by x + y
∴ The result is true for n = 1

Step 2:
Let us assume that the result is true for n = k
P(k) = x^2k – y^2k which is divisible by x + y
∴ P (k) = x^2k – y^2k = λ (x + y) where λ ∈ N ——— (1)

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

∴ P ( k + 1) is divisible by x + y
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
x²ⁿ – y²ⁿ is divisible by x + y
for all natural numbers n.

Question 11.
By the principle of mathematical induction, prove that, for n ≥ 1,

Solution:

Question 12.
Use induction to prove that n³ – 7n + 3, is divisible by 3, for all natural numbers n.
Solution:
Let P ( n) = n³ – 7n + 3 is divisible by 3

Step 1:
First let us verify the results for n = 1
P(I) = 1³ – 7 × 1 + 3
= 1 – 7 + 3
P (1) = – 3
which is divisible by 3
∴ The result is true for n = 1

Step 2:
Let us assume that the result is true for n = k
P(k) = k³ – 7k + 3 is divisible by 3
P(k) = k³ – 7k + 3 = 3λ where λ ∈ N

Step 3:
Let us prove the result for n = k + 1
P(k + 1) = (k + 1)³ – 7(k + 1 ) + 3
= k³ + 3k² + 3k + 1 – 7k – 7 + 3
= k³ + 3k² – 4k – 3
= k³ – 4k – 3k + 3k – 3 + 6 – 6 + 3k²
= k³ – 7k + 3 + 3k – 6 + 3k²
= (k³ – 7k + 3) + 3(k² + k – 2)
= 3λ + 3 (k² + k – 2)
P(k + 1) = 3 (λ + k² + k – 2 )
which is a multiple of 3, hence divisible by 3
This implies P (k + 1) is true.
∴ Thus, we have proved the result for n = k + 1.
Hence by the principle of mathematical induction, the result is true for all natural numbers n.
n³ – 7n + 3 is divisible by 3 for all natural numbers n.

Question 13.
Use induction to prove that 5^{n + 1} + 4 × 6ⁿ when divided by 20 leaves a remainder 9, for all natural numbers n.
Solution:
P(n) is the statement 5^{n + 1} + 4 × 6ⁿ – 9 is ÷ by 20
P(1) = 5^{1 + 1} + 4 × 6¹ – 9 = 5² + 24 – 9
= 25 + 24 – 9 = 40 ÷ by 20
So P(1) is true
Assume that the given statement is true for n = k
(i.e) 5^{k + 1} + 4 × 6ⁿ – 9 is ÷ by 20
P(1) = 5^{1 + 1} + 4 × 6¹ – 9
= 25 + 24 – 9
So P(1) is true
To prove P(k + 1) is true
P(k + 1) = 5^{k + 1 + 1} + 4 × 6^{k + 1 + 1} – 9
= 5 × 5 ^{k + 1} + 4 × 6 × 6^k – 9
= 5[20C + 9 – 4 × 6^k] + 24 × 6^k – 9 [from(1)]
= 100C + 45 – 206^k + 246^k – 9
= 100C + 46^k + 36
= 100C + 4(9 + 6^k)
Now for k = 1 ⇒ 4(9 + 6^k) = 4(9 + 6)
= 4 × 15 = 60 ÷ by 20 .
for k = 2 = 4(9 + 6²) = 4 × 45 = 180 ÷ 20
So by the principle of mathematical induction 4(9 + 6^k) is ÷ by 20
Now 100C is ÷ by 20.
So 100C + 4(9 + 6^k) is ÷ by 20
⇒ P(k + 1) is true whenever P(k) is true. So by the principle of mathematical induction P(n) is true.

Question 14.
Use induction to prove that 10ⁿ + 3 × 4^{n + 2} + 5, is divisible by 9, for all natural numbers n.
Solution:
P(n) is the statement 10ⁿ + 3 × 4^{n + 2} + 5 is ÷ by 9
P(1) = 10¹ + 3 × 4² + 5 = 10 + 3 × 16 + 5
= 10 + 48 + 5 = 63 ÷ by 9
So P(1) is true. Assume that P(k) is true
(i.e.) 10^k + 3 × 4^{k + 2} + 5 is ÷ by 9
(i.e.) 10^k + 3 × 4^{k + 2} + 5 = 9C (where C is an integer)
⇒ 10^k = 9C – 5 – 3 × 4^{k + 2} ……(1)
To prove P(k + 1) is true.
Now P(k + 1) = 10^{k + 1} + 3 × 4^{k + 3} + 5
= 10 × 10^k + 3 × 4^{k + 2} × 4 + 5
= 10[9C – 5 – 3 × 4^{k + 2}] + 3 × 4^{k + 2} × 4 + 5
= 10[9C – 5 – 3 × 4^{k + 2}] + 12 × 4^{k + 2} + 5
= 90C – 50 – 30 × 4^{k + 2} + 12 × 4^{k + 2} + 5
= 90C – 45 – 18 × 4^{k + 2}
= 9[10C – 5 – 2 × 4^{k + 2}] which is ÷ by 9
So P(k + 1) is true whenever P(K) is true. So by the principle of mathematical induction P(n) is true.

Question 15.
Prove that using the Mathematical induction

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.4 Additional Questions

Question 1.
Prove by induction the inequality (1 + x)ⁿ ≥ 1 + nx, whenever x is positive and n is a positive integer.
Solution:
P(n) : (1 +x)ⁿ ≥ 1 +nx
P(1): (1 + x)¹ ≥ 1 + x
⇒ 1 + x ≥ 1 + x, which is true.
Hence, P(1) is true.
Let P(k) be true
(i.e.) (1 + x)^k ≥ 1 + kx
We have to prove that P(k + 1) is true.
(i.e.) (1 + x)^{k + 1} ≥ 1 + (k + 1)x
Now, (1 + x)^{k + 1} ≥ 1 + kx [∵ p(k) is true]
Multiplying both sides by (1 + x), we get
(1 + x)^k(1 + x) ≥ (1 + kx)(1 + x)
⇒ (1 + x)^{k + 1} ≥ 1 + kx + x + kx²
⇒ (1 + x)^{k + 1} ≥ 1 + (k + 1)x + kx² ….. (1)
Now, 1 + (k + 1) x + kx² ≥ 1 + (k + 1)x …… (2)
[∵ kx² > 0]
From (1) and (2), we get
(1 + x)^{k + 1} ≥ 1 + (k + 1)x
∴ P(k + 1) is true if P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all values, of n.

Question 2.
3²ⁿ – 1 is divisible by 8.
Solution:
P(n) = 3²ⁿ – 1 is divisible by 8
For n = 1, we get
P(1) = 3^2.1 – 1 = 9 – 1 = 8
P(1) = 8, which is divisible by 8.
Let P(n) be true for n = k
P(k) = 3^2k – 1 is divisible by 8 ….. (1)
Now, P(k + 1) = 3^{(2k + 2)} – 1 = 3^2k.3² – 1
= 3²(3^2k – 1) + 8
Now, 3^2k – 1 is divisible by 9. [Using (1)]
∴ 3² (3^2k – 1) + 8 is also divisible by 8.
Hence, 3²ⁿ – 1 is divisible by 8 ∀ n E N

Question 3.
Prove by the principle of mathematical induction if x and y are any two distinct integers, then xⁿ – yⁿ is divisible by x – y. [OR]
xⁿ – yⁿ is divisible by x – y, where x – y ≠ 0.
Solution:
Let the given statement be P(n).
(i.e.) P(n): xⁿ – yⁿ = M(x – y), x – y ≠ 0

Step I.
When n = 1,
xⁿ – yⁿ = x – y = M(x – y) ….(1)
⇒ P(1) is true.

Step II.
Assume that P(k) is true.
(i.e.) x^k – y^k = M(x – y), x – y ≠ 0
We shall now show that P(k + 1) is true
Now, x^{k + 1} – y^{k + 1} = x^{k + 1} – x^ky + x^{k + 1}y – y^{k + 1}
= x^k(x – y) + y(x^k – y^k)
= x^k(x – y) + yM(x – y) [Usng ….. (1)]
= (x – y)(x^k – yM)
∴ By the principle of mathematical induction, P(n) is true for all n ∈ N

Question 4.
Prove by the principle of mathematical induction that for every natural number n, 3^{2n + 2} – 8n – 9 is divisible by 8.
Solution:
Let P(n): 3^{2n + 2} – 8n – 9 is divisible by 8.
Then, P(1): 3^{2.1 + 2} – 8.1 – 9 is divisible by 8.
(i.e.) 3⁴ – 8 – 9 is divisible by 8 or 81 – 8 – 9 is divisible by 8
(or) 64 is divisible by 8, which is true.
Suppose P(k) is true, then
P(k) : 3^{2k + 2} – 8k – 9 is divisible by 8
(i.e.) 3^{2k + 2} – 8k – 9 = 8m, where m ∈ N (or)
3^{2k + 2} = 8m + 8k + 9
P(k + 1) is the statement given by, …(1)
P(k + 1) : 3^{2(k + 1) + 2} – 8(k + 1) – 9

∴ P(k + 1) is true
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N

Question 5.
Use the principle of mathematical induction to prove that for every natural number n.

Solution:
Let P(n) be the given statement, i.e.

⇒ P(1) is true.
We note that P(n) is true for n = 1.
Assume that P(k) is true

Now, we shall prove that P(k + 1) is true whenever P(k) is true. We have,

∴ P(k + 1) is also true whenever P(k) is true
Hence, by the principle of mathematical induction, P(n) is also true for all n ∈ N.

Question 6.
n³ – n is divisible by 6, for each natural number n ≥ 2.
Solution:
Let P(n) : n³ – n

Step 1 :
P(2): 2³ – 2 = 6 which is divisible by 6. So it is true for P(2).

Step 2 :
P(A): k³ – k = 6λ. Let it is be true for k ≥ 2
⇒ k³ = 6λ + k …(i)

Step 3 :
P(k + 1) = (k + 1)³ – (k + 1)
= k³ + 1 + 3k² + 3k – k – 1 = k³ – k + 3(k² + k)
= k³ – k + 3(k² + k) = 6λ + k – k + 3(k² + k)
= 6λ + 3(k² + k) [from (i)]
We know that 3(k² + k) is divisible by 6 for every value of k ∈ N.
Hence P(k + 1) is true whenever P(k) is true.

Question 7.
For any natural number n, 7ⁿ – 2ⁿ is divisible by 5.
Solution:
Let P(n) : 7ⁿ – 2ⁿ

Step 1:
P(1) : 7¹ – 2¹ = 5λ which is divisible by 5. So it is true for P(1).

Step 2:
P(k): 7^k – 2^k = 5λ. Let it be true for P(k)

Step 3:
P(k + 1) = 7^{k + 1} – 2^{k + 1}

So, it is true for P(k + 1)
Hence, P(k + 1) is true whenever P(k) is true.

Question 8.
n² < 2ⁿ, for all natural numbers n ≥ 5.
Solution:
Let P(n) : n² < 2ⁿ for all natural numbers, n ≥ 5

Step 1 :
P(5) : 1⁵ < 2⁵ ⇒ 1 < 32 which is true for P(5)

Step 2 :
P(k): k² < 2k. Let it be true for k ∈ N

Step 3 :
P(k + 1): (k + 1)² < 2^{k + 1}
From Step 2, we get k² < 2^k
⇒ k² < 2^{k + 1} < 2k + 2k + 1


From eqn. (i) and (ii), we get (k + 1)² < 2^{k + 1}
Hence, P(k + 1) is true whenever P(k) is true for k ∈ N, n ≥ 5.

Question 9.
In 2n < (n + 2)! for all natural number n.
Solution:
Let P(n) : 2n < (n + 2)! for all k ∈ N.

Hence, P(k + 1) is true whenever P(k) is true.

Question 10.
1 + 5 + 9 + … + (4n – 3) = n(2n – 1), ∀ n ∈ N.
Solution:
Let P(n) : 1 + 5 + 9 + … + (4n – 3) = n(2n – 1), ∀ n ∈ N

Step 1:
P(1) : 1 = 1(2.1 – 1) = 1 which is true for P(1)

Step 2:
P(k) : 1 + 5 + 9 + … + (4k – 3) = k(2k – 1). Let it be true.

Step 3:
P(k + 1) : 1 + 5 + 9 + … + (4k – 3) = k(4k + 1)
= k(2k – 1) + (4k + 1) = 2k² – k + 4k + 1
= 2k² + 3k + 1 = 2k² + 2k + k + 1
= 2k(k + 1) + 1 (k + 1) = (2k + 1)(k + 1)
= (k+ 1) (2k + 2 – 1) = (k + 1) [2(k + 1) – 1]
Which is true for P(k + 1).
Hence, P(k + 1) is true whenever P(k) is true.