Category: Class 10

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.4 Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.4

Question 1.
Find the next three terms of the following sequence.
(i) 8, 24, 72,…
(ii) 5, 1, -3, …
(iii) \(\frac { 1 }{ 4 } \), \(\frac { 2 }{ 9 } \), \(\frac { 3 }{ 16 } \)
(i) 216, 648, 1944 (This sequence is multiple of 3)
Next three terms are 216, 648, 1944
(ii) Next three terms are -7, -11, -15
(adding -4 with each term)
(iii) Next three terms are \(\frac { 4 }{ 25 } \),\(\frac { 5 }{ 36 } \) and \(\frac { 6 }{ 49 } \)
[using \(\frac{n}{(n+1)^{2}}\)]

Question 2.
Find the first four terms of the sequences whose nth terms are given by
(i) a_n = n³ – 2
(ii) a_n = (-1)ⁿ⁺¹ n(n+1)
(iii) a_n = 2n² – 6
Solution:
t_n = a_n = n³ -2
(i) a₁ = 1³ – 2 = 1 – 2 – 1
a₂ = 2³ – 2 = 8 – 2 = 6
a₃ = 3³ – 2 = 27 – 2 = 25
a₄ = 4³ – 2 = 64 – 2 = 62
∴ The first four terms are -1, 6, 25, 62, ……….

(ii) a_n = (-1)ⁿ⁺¹ n(n + 1)
a₁ = (-1)¹⁺¹ (1) (1 +1)
= (-1)² (1) (2) = 2
a₂ = (-1)²⁺¹ (2) (2 + 1)
= (-1)³ (2) (3)= -6
a₃ = (-1)³⁺¹ (3) (3 + 1)
= (-1)⁴ (3) (4) = 12
a₄ = (-1)⁴⁺¹ (4) (4 + 1)
= (-1)⁵ (4) (5) = -20
∴ The first four terms are 2, -6, 12, -20,…

(iii) a_n = 2n² – 6
a₁ = 2(1)² – 6 = 2 – 6 = -4
a₂ = 2(2)² – 6 = 8 – 6 = 2
a₃ = 2(3)² – 6 = 18 – 6 = 12
a₄ = 2(4)² – 6 = 32 – 6 = 26
∴ The first four terms are -4, 2, 12, 26, …

Question 3.
Find the n^th term of the following sequences
(i) 2, 5, 10, 17, ……
Answer:
(1² + 1);(2² + 1),(3² + 1),(4² + 1)….
n^th term is n² + 1
a_n = n² + 1

(ii) 0,\(\frac { 1 }{ 2 } \),\(\frac { 2 }{ 3 } \) ……
Answer:
(\(\frac { 1-1 }{ 1 } \)), (\(\frac { 2-1 }{ 2 } \)), (\(\frac { 3-1 }{ 3 } \)) …..
n^th term is \(\frac { n-1 }{ n } \)
a_n = \(\frac { n-1 }{ n } \)

(iii) 3,8,13,18,…….
Answer:
[5(1) -2], [5(2) – 2], [5(3) – 2], [5(4) – 2] ….
The n^th term is 5n – 2
a_n = 5n – 2

Question 4.
Find the indicated terms of the sequences whose n^th terms are given by

(i) a_n = \(\frac { 5n }{ n+2 } \) ; a₆ and a₁₃
Answer:
a_n = \(\frac { 5n }{ n+2 } \)
a₆ = \(\frac { 5(6) }{ 6+2 } \) = \(\frac { 30 }{ 8 } \) = \(\frac { 15 }{ 4 } \)
a₁₃ = \(\frac { 5(13) }{ 13+2 } \) = \(\frac{5 \times 13}{15}\) = \(\frac { 13 }{ 3 } \)
a₆ = \(\frac { 15 }{ 4 } \), a₁₃ = \(\frac { 13 }{ 3 } \)

(ii) a_n = – (n² – 4); a₄ and a₁₁
Answer:
a_n = -(n² – 4)
a₄ = -(4² – 4)
= – (16 – 4)
= -12
a₁₁ = -(11² – 4)
= – (121 – 4)
= – 117
a₄ = -12 and a₁₁ = -117

Question 5.
Find a₈ and a₁₅ whose n^th term is a_n

Answer:

Question 6.
If a₁ = 1, a₂ = 1 and a_n = 2a_n-1 + a_n-2, n > 3, n ∈ N, then find the first six terms of the sequence.
Solution:
a₁ = 1, a₂ = 1, a_n = 2a_n-1 + a_n-2
a₃ = 2a_(3-1) + a_(3-2)
= 2a₂ + a₁
= 2 × 1 + 1 = 3
a₄ = 2a_(4-1) + a_(4-2)
= 2a₃ + a₂
= 2 × 3 + 1 = 7
a₅ = 2a_(5-1) + a_(5-2)
= 2a₄ + a₃
= 2 × 7 + 3 = 17
a₆ = 2a_(6-1) + a_(6-2)
= 2a₅ + a₄
= 2 × 17 + 7
= 34 + 7
= 41
∴ The first six terms of the sequence are 1, 1, 3, 7, 17, 41 ………..

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.8 Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.8

Question 1.
Find the sum of first n terms of the G.P.
(i) 5, -3, \(\frac { 9 }{ 5 } \),-\(\frac { 27 }{ 25 } \), ……
(ii) 256,64,16,…….
Answer:
(i) 5,-3,\(\frac { 9 }{ 5 } \),\(\frac { 27 }{ 55 } \), ….. n terms

(ii) 256,64,16,…….
Answer:
Here a = 256, r = \(\frac { 64 }{ 256 } \) = \(\frac { 1 }{ 4 } \)

Question 2.
Find the sum of first six terms of the G.P. 5,15,45,…
Answer:
Here a = 5, r = \(\frac { 15 }{ 3 } \) = 3, n = 6

Sum of first 6 terms = 1820

Question 3.
Find the first term of the G.P. whose common ratio 5 and whose sum to first 6 terms is 46872.
Answer:
Common ratio (r) = 5
S₆ = 46872

The first term of the G.P. is 12.

Question 4.
Find the sum to infinity of (i) 9 + 3 + 1 + ….(ii) 21 + 14 + \(\frac { 28 }{ 3 } \) ……
Answer:
(i) 9 + 3 + 1 + ….
a = 9, r = \(\frac { 3 }{ 9 } \) = \(\frac { 1 }{ 3 } \)
Sum of infinity term = \(\frac { a }{ 1 – r } \) = \(\frac{9}{1-\frac{1}{3}}\)

Question 5.
If the first term of an infinite G.P. is 8 and its sum to infinity is \(\frac { 32 }{ 3 } \) then find the common ratio.
Answer:
Here a = 8, S∞ = \(\frac { 32 }{ 3 } \)
\(\frac { a }{ 1-r } \) = \(\frac { 32 }{ 3 } \)
\(\frac { 8 }{ 1-r } \) = \(\frac { 32 }{ 3 } \)
32 – 32 r = 24 ⇒ 32 r = 8
r = \(\frac { 8 }{ 32 } \) = \(\frac { 1 }{ 4 } \)
Common ration = \(\frac { 1 }{ 4 } \)

Question 6.
Find the sum to n terms of the series
(i) 0.4 + 0.44 + 0.444 + …… to n terms
Answer:


(ii) 3 + 33 + 333 + ………… to n terms
Answer:
S_n = 3 + 33 + 333 + …. to n terms
= 3[1 + 11 + 111 + …. to n terms]
= \(\frac { 3 }{ 9 } \) [9 + 99 + 999 + …. n terms]
= \(\frac { 1 }{ 3 } \) [(10 – 1) + (100 – 1) + (1000 – 1) + …… n terms]
= \(\frac { 1 }{ 3 } \) [10 + 100 + 1000 + ….. n terms – (1 + 1 + 1 ….. n terms)]

Question 7.
Find the sum of the Geometric series 3 + 6 + 12 + …….. + 1536
Answer:
3 + 6 + 12 …. +1536
a = 3, r = \(\frac { 6 }{ 3 } \) = 2
t_n = 1536


∴ Sum of the series is 3069

Question 8.
Kumar writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with the instruction that they continue the process similarly. Assuming that the process is unaltered and it costs ₹2 to mail ong letter, find the amount spent on postage when 8th set of letters is mailed.
Answer:
When kumar writes a letter to his friend.Friend writes a letter to another person.
It form a G.P
The G.P is 4, 16, 64,………
Here a = 4, r = 4
The last term is 4 (4)^8-1 = 4(4)⁷

Question 9.
Find the rational form of the number 0.123 .
Answer:
Let x = \(\overline { 0.123 } \)
= 0.123123123….
= 0.123 + 0.000123 + 000000123 + ….
This is an infinite G.P

Question 10.
If S_n = (x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + ………… n terms then prove that
Answer:
S_n = (x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + …….. n
Multiply by x
x S_n = x(x + y) + x(x² + xy + y²) + x(x³ + x²y + xy² + y³) + ……….. n
= x² + xy + x³ + x²y + xy² + x⁴ + x³y + x²y² + xy³ + …… n terms ……(1)
Multiply by y
yS_n = y(x + y) + y(x² + xy + y²) + y(x³ + x²y + xy² + y³) + ….. n
= xy + y² + x²y + xy² + y³ + x³y + x²y² + xy³ + y⁴ + ….. n terms
Subtract (1) and (2)
x S_n – y S_n = x² + xy + x³ + x²y + xy² + x⁴ + x³y + x²y² + xy³ + …….
– xy + y² + x²y + xy² + y³ + x³y + x²y² + xy³ + y⁴ + ……
(x – y) S_n = (x² + x³ + x⁴ + ……) – (y² + y³ + y⁴ + ……)
[ a = x²; r = x and a = y²; r = y, S_n = \(\frac{a\left(r^{n}-1\right)}{r-1}\)]

Hence it is proved.

Students can download Maths Chapter 3 Algebra Ex 3.9 Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 3 Algebra Ex 3.9

Question 1.
Determine the quadratic equations, whose sum and product of roots are
(i) -9, 20
Answer:
Sum of the roots = -9 and Product of the roots = 20
The Quadratic equation is
x² – (sum of the roots) x + product of the roots = 0
x² – (-9) x + 20 = 0 ⇒ x² + 9x + 20 = 0

(ii) \(\frac { 5 }{ 3 } \), 4
Answer:
Sum of the roots = \(\frac { 5 }{ 3 } \); Product of the roots = 4
The Quadratic equation is
x² – (sum of the roots) x + product of the roots = 0
x² – (\(\frac { 5 }{ 3 } \)) x + 4 = 0 ⇒ x² – \(\frac { 5 }{ 3 } \) x + 4 = 0
3x² – 5x + 12 = 0

(iii) \(\frac { -3 }{ 2 } \), -1
Answer:
Sum of the roots = \(\frac { -3 }{ 2 } \); Product of the roots = -1
The Quadratic equation is
x² – (sum of the roots) x + product of the roots = 0
x² – (-\(\frac { 3 }{ 2 } \)) x + (-1) = 0 ⇒ x² + \(\frac { 3 }{ 2 } \) x – 1 = 0
2x² + 3x – 2 = 0

(iv) – (2 – a)², (a + 5)²
Answer:
Sum of the roots = – (2 – a)²; Product of the roots = (a + 5)²
x² – (sum of the roots) x + product of the roots = 0
x² – [-(2 – a)²] x + (a + 5)² = 0
x² + (2 – a)² x + (a + 5)² = 0

Question 2.
Find the sum and product of the roots for each of the following quadratic equations
(i) x² + 3x – 28 = 0
(ii) x² + 3x = 0
(iii) 3 + \(\frac{1}{a}=\frac{10}{a^{2}}\)
(iv) 3y² – y – 4 = 0
Solution:
(i) x² – (-3)x + (-28) = 0.
Comparing this with x² – (α + β)x + αβ = 0.
(α + β) = Sum of the roots = -3
αβ = product of the roots = -28

(ii) x² + 3x = 0 = x² – (-3)x + 0 = 0
x² – (α + β)x + αβ = 0
Sum of the roots α + β = -3
Products of the roots αβ =0

(iii) 3 + \(\frac { 1 }{ a } \) = \(\frac{10}{a^{2}}\)
Answer:
Multiply by a²
3a² + a = 10
3a² + a – 10 = 0
Sum of the roots (α + β) = \(\frac { -1 }{ 3 } \)
Product of the roots (α β) = \(\frac { -10 }{ 3 } \)

(iv) 3y² – y – 4 = 0
Answer:
Sum of the roots (α + β) = \(\frac { -(-1) }{ 3 } \) = \(\frac { 1 }{ 3 } \)
Product of the roots (α β) = \(\frac { -4 }{ 3 } \)

Students can download Maths Chapter 3 Algebra Ex 3.8 Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 3 Algebra Ex 3.8

Question 1.
Find the square root of the following polynomials by division method
(i) x⁴ – 12x³ + 42x² – 36x + 9
Answer:

(ii) 31 x² – 28x³ + 4x⁴ + 42x + 9
Answer:
Rearrange the order we get
4x⁴ – 28x³ + 37x² + 42x + 9

(iii) 16x⁴ + 8x² + 1
Answer:

(iv) 121 x⁴ – 198x³ – 183x² + 216x + 144
Answer:

Question 2.
Find the square root of the expression
\(\frac{x^{2}}{y^{2}}\) – \(\frac { 10x }{ y } \) + 27 – \(\frac { 10y }{ x } \) + \(\frac{y^{2}}{x^{2}}\)
Answer:

Question 3.
Find the values of a and b if the following polynomials are perfect squares.
(i) 4x⁴ – 12x³ + 37x² + bx + a
Answer:

Since it is a perfect square
b + 42 = 0
b = – 42
a – 49 = 0
a = 49
∴ The value of a = 49 and b = – 42

(ii) ax⁴ + bx³ + 361x² + 220x + 100
Answer:
Re-arrange the order we get
100 + 220x + 361x² + bx³ + ax⁴

Since it is a perfect square
b – 264 = 0
b = 264
a – 144 = 0
a = 144
∴ The value of a = 144 and b = 264

Question 4.
Find the values of m and n if the following expressions are perfect sqaures.
(i) \(\frac{1}{x^{4}}\) – \(\frac{6}{x^{3}}\) + \(\frac{13}{x^{2}}\) + \(\frac { m }{ x } \) + n
Answer:

Since it is a perfect square
\(\frac { 1 }{ x } \) (m + 12) = 0
m + 12 = 0
m = -12
n – 4 = 0
n = 4
∴ The value of m = -12 and n = 4

(ii) x⁴ – 8x³ + mx² + nx + 16
Answer:

Since it is a perfect square
m – 16 – 8 = 0
m – 24 = 0
m = 24
n + 32 = 0
n = -32
∴ The value of m = 24 and n = -32

Students can download Maths Chapter 3 Algebra Ex 3.16 Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 3 Algebra Ex 3.16

1. In the

write (i) The number of elements
(ii) The order of the matrix
(iii) Write the elements a₂₂, a₂₃, a₂₄, a₃₄, a₄₃, a₄₄.
Answer:
(i) The number of elements is 16
(ii) The order of the matrix is 4 × 4
(iii) Elements corresponds to

Question 2.
If a matrix has 18 elements, what are the possible orders it can have? What if it has 6 elements?
Answer:
The possible orders of the matrix having 18 elements are

The possible orders of the matrix having 6 elements are

Question 3.
Construct a 3 × 3 matrix whose elements are given by
(i) a_ij = |i – 2j|
Answer:
a_ij = |i – 2j|
The general 3 × 3 matrices is

a₁₁ = |1 – 2(1)| = |1 – 2| = | – 1| = 1
a₁₂ = |1 – 2(2)| = |1 – 4| = | – 3| = 3
a₁₃ = |1 – 2(3)| = |1 – 6| = | – 5| = 5
a₂₁ = |2 – 2(1)| = |2 – 2| = 0 = 0
a₂₂ = |2 – 2(2)| = |2 – 4| = | – 2| = 2
a₂₃ = |2 – 2(3)| = |2 – 6| = | – 4| = 4
a₃₁ = |3 – 2(1)| = |3 – 2| = | 1 | = 1
a₃₂ = |3 – 2(2)| = |3 – 4| = | – 1 | = 1
a₃₃ = |3 – 2(3)| = |3 – 6| = | – 3 | = 3
The required matrix

(ii) a_ij = \(\frac{(i+j)^{3}}{3}\)
Answer:
a₁₁ = \(\frac{(1+1)^{3}}{3}\) = \(\frac{2^{3}}{3}\) = \(\frac { 8 }{ 3 } \)
a₁₂ = \(\frac{(1+2)^{3}}{3}\) = \(\frac { 27 }{ 3 } \) = 9
a₁₃ = \(\frac{(1+3)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₂₁ = \(\frac{(2+1)^{3}}{3}\) = \(\frac { 27 }{ 3 } \) = 9
a₂₂ = \(\frac{(2+2)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₂₃ = \(\frac{(2+3)^{3}}{3}\) = \(\frac { 125 }{ 3 } \) = \(\frac { 125 }{ 3 } \)
a₃₁ = \(\frac{(3+1)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)
a₃₂ = \(\frac{(3+2)^{3}}{3}\) = \(\frac { 125 }{ 3 } \) = \(\frac { 125 }{ 3 } \)
a₃₃ = \(\frac{(3+3)^{3}}{3}\) = \(\frac { 216 }{ 3 } \) = 72
The required matrix

Question 4.
If  then find the tranpose of A.
Answer:

transpose of A = (A^T)

Question 5.
If then find the tranpose of – A
Answer:

Transpose of – A = (-A^T) =

Question 6.
If A = then verify (AT)T = A
Answer:

Hence it is verified

Question 7.
Find the values of x, y and z from the following equations
(i)

Answer:
Since the given matrices are equal then all the corresponding elements are equal.
y = 12, z = 3, x = 3
The value of x = 3, y = 12 and z = 3

(ii)

Answer:
x + y = 6 ……(1)
5 + z = 5
z = 5 – 5 = 0
xy = 8
y = \(\frac { 8 }{ x } \)
Substitute the value of y = \(\frac { 8 }{ x } \) in (1)
x + \(\frac { 8 }{ x } \) = 6
x² + 8 = 6x
x² – 6x + 8 = 0
(x – 4) (x – 2) = 0
∴ x – 4 = 0 or x – 2 = 0
x = 4 or x = 2
y = \(\frac { 8 }{ 4 } \) = 2 or y = \(\frac { 8 }{ 2 } \) = 4

∴ The value of x, y and z are 4, 2, 0 (or) 2, 4, 0

(iii)

Solution:
x + y + z = 9 ……….(1)
x + z = 5 ……….(2)
y + z = 7 ……….(3)

Substitute the value of y = 4 in (3)
y + z = 7
4 + z = 7
z = 7 – 4
= 3
Substitute the value of z = 3 in (2)
x + 3 = 5
x = 5 – 3
= 2
∴ The value of x = 2 , y = 4 and z = 3

Students can download Maths Chapter 3 Algebra Ex 3.7 Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 3 Algebra Ex 3.7

Question 1.
Find the square root of the following.
(i) \(\frac{400 x^{4} y^{12} z^{16}}{100 x^{8} y^{4} z^{4}}\)
Answer:

(ii) \(\frac{7 x^{2}+2 \sqrt{14} x+2}{x^{2}-\frac{1}{2} x+\frac{1}{16}}\)
Answer:

(iii) \(\frac{121(a+b)^{8}(x+y)^{8}(b-c)^{8}}{81(b-c)^{4}(a-b)^{12}(b-c)^{4}}\)
Answer:

Question 2.
Find the square root of the following
(i) 4x² + 20x + 25
Answer:

(ii) 9x² – 24xy + 30xz – 40yz + 25z² + 16y²
Answer:

(iii) \(1+\frac{1}{x^{6}}+\frac{2}{x^{3}}\)
Answer:

(iv) (4x² – 9x + 2)(7x² – 13x – 2)(28x² – 3x – 1)
Answer:
4x² – 9x +2 = 4x² – 8x – x + 2
= 4x(x – 2)-1 (x – 2)
= (x – 2)(4x – 1)

7x² – 13x – 2 = 7x² – 14x + x – 2
= 7x (x – 2) + 1 (x – 2)
= (x – 2) (7x + 1)

28x² – 3x – 1 = 28x² – 7x + 4x – 1
= 7x (4x – 1) + 1 (4x – 1)
= (4x – 1) (7x + 1)

(v) (2x² + \(\frac { 17 }{ 6 } \)x + 1) (\(\frac { 3 }{ 2 } \) x² + 4x + 2) (\(\frac { 4 }{ 3 } \) x² + \(\frac { 11 }{ 3 } \) x + 2)
Answer:

Students can download Maths Chapter 3 Algebra Ex 3.18 Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 3 Algebra Ex 3.18

Question 1.
Find the order of the product matrix AB if

Answer:
Given A = [a_ij]_p×q and B = [a_ij]_q×r
Order of product of AB = p × r
Order of product of BA is not defined. Number columns in r is not equal to the number of rows in P.
∴ Product BA is not defined.

Question 2.
A has ‘a’ rows and ‘a + 3 ’ columns. B has ‘6’ rows and ‘17 – b’ columns, and if both products AB and BA exist, find a, b?
Solution:
A has a rows, a + 3 columns.
B has b rows, 17 – b columns
If AB exists a × a + 3
b × 17 – b
a + 3 = 6 ⇒ a – 6 = -3 ………… (1)
If BA exists 6 × 17-6
a × a + 3
17 – 6 = a ⇒ a + 6 = 17 …………. (2)
(1) + (2) ⇒ 2a = 14 ⇒ a = 7
Substitute a = 7 in (1) ⇒ 7 – b = -3 ⇒ b = 10
a = 7, b = 10

Question 3.
A has ‘a’ rows and ‘a + 3 ’ columns. B has rows and ‘b’ columns, and if both products AB and BA exist, find a,b?
Answer:

1. Order of matrix AB = 3 × 3
2. Order of matrix AB = 4 × 2
3. Order of matrix AB = 4 × 2
4. Order of matrix AB = 4 × 1
5. Order of matrix AB = 1 × 3

Question 4.

find AB, BA and check if AB = BA?
Answer:

Question 5.
Given that


verify that A(B + C) = AB + AC
Answer:

From (1) and (2) we get
A (B + C) = AB + AC

Question 6.
Show that the matrices

satisfy commutative property AB = BA
Answer:

From (1) and (2) we get
AB = BA. It satisfy the commutative property.

Question 7.

Show that (i) A(BC) = (AB)C
(ii) (A-B)C = AC – BC
(iii) (A-B)^T = A^T – B^T
Answer:



From (1) and (2) we get
A(BC) = (AB)C

From (1) and (2) we get
(A – B) C = AC – BC


From (1) and (2) we get
(A-B)^T = A^T – B^T

Question 8.

then snow that A² + B² = I.
Answer:

Question 9.

prove that AA^T = I.
Answer:

AA^T = I
∴ L.H.S. = R.H.S.

Question 10.
Verify that A² = I when

Answer:

∴ L.H.S. = R.H.S.

Question 11.

show that A² – (a + d)A = (bc – ad)I₂.
Answer:

L.H.S. = R.H.S.
A² – (a + d) A = (bc – ad)I₂

Question 12.

verify that (AB)^T = B^T A^T
Answer:


From (1) and (2) we get, (AB)^T = B^T A^T

Question 13.

show that A² – 5A + 7I₂ = 0
Answer:


L.H.S. = R.H.S.
∴ A² – 5A + 7I₂ = 0

Students can download Maths Chapter 3 Algebra Ex 3.15 Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 3 Algebra Ex 3.15

Question 1.
Graph the following quadratic equations and state their nature of solutions.
(i) x² – 9x + 20 = 0
(ii) x² – 4x + 4 = 0
(iii) x² + x + 7 = 0
(iv) x² – 9 = 0
(v) x² – 6x + 9 = 0
(vi) (2x – 3) (x + 2) = 0

(i) x² – 9x + 20 = 0
Answer:
Let y = x² – 9x + 20
(i) Prepare the table of values for y = x² – 9x + 20

(ii) Plot the points (-1, 30) (0,20) (1, 12) (2, 6) (3,2), (4, 0), (5, 0), (6,2) (omit the high value)
(iii) Join the points by a free hand smooth curve.
(iv) The roots of the equation are the X-coordinates of the intersecting points of the curve with X-axis (4, 0) and (5, 0)
There are two points of intersection with the X-axis at 4 and 5. The solution set is 4 and 5. The quadratic equation has real and unequal roots.
(v) Since there is two point of intersection with X-axis (different solution)
∴ The equation x² – 9x + 20 = 0 has real and unequal roots.

(ii) x² – 4x + 4 = 0
Answer:
Let y = x² – 4x + 4
(i) Prepare the table of values for y = x² – 4x + 4

(ii) Plot the points (-3,25) (-2,16) (-1, 9) (0,4) (1,-1) (2, 0), (3,1) and (4, 4)
(iii) Join the points by a free hand smooth curve.
(iv) The roots of the equation are the X-coordinates of the intersecting points of the curve with X-axis (2, 0) which is 2.
(v) Since there is only one point of intersection with X-axis (2, 0).
∴ The solution set is 2.
The Quadratic equation x² – Ax + 4 = 0 has real and equal roots.

(iii) x² + x + 7 = 0
Answer:
Let y = x² + x + 7
(i) Prepare the table of values for y = x² + x + 7

(ii) Plot the points (-4,19) (-3,13) (-2, 9) (-1, 7) (0, 7) (1, 9), (2,13) (3,19) and (4,27)
(iii) Join the points by a free hand smooth curve.
(iv) The solution of the given quadratic equation are the X-coordinates of the intersecting points of the parabola with the X-axis.
(v) The curve does not intersecting the X-axis. There is no solution set.
The equation x² + x + 7 = 0 has no real roots.

(iv) x² – 9 = 0
Answer:
Let y = x² – 9
(i) Prepare the table of values for y = x² – 9

(ii) Plot the points (-4, 7) (-3, 0) (-2, -5) (-1, -8) (0, -9) (1, -8), (2, -5) (3, 0) (4, 7)
(iii) Join the points by a free hand smooth curve.
(iv) The curve intersect the X-axis at -3 and 3.
The solution is (-3, 3).
(v) Since there are two points of intersection -3 and 3 with the X-axis the quadratic equation has real and unequal roots.

(v) x² – 6x + 9 = 0
Answer:
Let y = x² – 6x + 9
(i) Prepare a table of values for y = x² – 6x + 9

(ii) Plot the points (-2,25) (-1,16) (0,9) (1,4) (2,1) (3,0), (4,1) and (5,4) on the graph using suitable scale (omit the points (-4, 49) and (-3, 36)
(iii) Join the points by a free hand smooth curve.
(iv) The X – coordinates of the point of intersection of the curve with X-axis are the roots of the , given equation, provided they intersect.
The solution is 3.
(v) Since there is only one point of intersection with X-axis the quadratic equation x² – 6x + 9 = 0 has real and equal roots.

(vi) (2x – 3) (x + 2) = 0
Answer:
y = (2x – 3) (x + 2)
= 2x² + 4x – 3x – 6
= 2x² + x – 6

(i) Prepare a table of values for y from x – 4 to 4

(ii) Plot the points (-4, 22) (-3, 9) (-2, 0) (-1, -5) (0, -6) (1, -3), (2, 4), (3, 15) and (4, 30).
(iii) Join the points by a free hand smooth curve.
(iv) The curve intersect the X – axis at (-2, 0) and (1\(\frac { 1 }{ 2 } \), 0)
∴ The solution set is (-2,1\(\frac { 1 }{ 2 } \))
(v) Since there are two points of intersection with X – axis, the quadratic equation has real and un – equal roots.

Question 2.
Draw the graph of y = x² – 4 and hence solve x² – x – 12 = 0
Answer:
(i) Draw the graph of y = x² – 4 by preparing the table of values as below.

(ii) Plot the points for the ordered pairs (-4, 12) (-3, 5) (-2, 0) (-1, -3) (0, -4) (1, -3), (2, 0), (3, 5) and (4, 12). Draw the curve with the suitable scale.
(iii) To solve x² – x – 12 = 0 subtract x² – x – 12 from y = x² – 4

The equation y = x + 8 represents a straight line. Prepare a table for y = x + 8

(iv) Mark the point of intersection of the curve and the straight line is (-3, 5) and 4, 12)
∴ The solution set is (-3, 4) for x² – x – 12 = 0.

Question 3.
Draw the graph of y = x² + x and hence solve x² + 1 = 0
Answer:
Let y = x² + x
(i) Draw the graph of y = x² + x by preparing the table.

(ii) Plot the points (-4, 12), (-3, 6), (-2, 2), (-1, 0), (0, 0), (1, 2), (2, 6), (3, 12) and (4, 20).
(iii) Join the points by a free hand to get smooth curve.
(iv) To solve x² + 1 = 0, subtract x² + 1 = 0 from x² + x we get.

The equation represent a straight line. Draw a line y = x – 1

Observe the graph of y = x² + 1 does not interset the parabola y = x² + x.
This x² + 1 has no real roots.

Question 4.
Draw the graph of y = x² + 3x + 2 and use it to solve x² + 2x + 1 = 0.
(i) Draw the graph of y = x² + 3x + 2 preparing the table of values as below.

(ii) Plot the points (-4, 6), (-3, 2), (-2, 0), (-1, 0), (0, 2), (1, 6), (2, 12), (3, 20) (4, 30).
(iii) To solve x² + 2x + 1 = 0 subtract x² + 2x + 1 = 0 from y = x² + 3x + 2

(iv) Draw the graph of y = x + 1 from the table

The equation y = x + 1 represent a straight line.
This line intersect the curve at only one point (-1, 0). The solution set is (-1).

Question 5.
Draw the graph of y = x² + 3x – 4 and hence use it to solve x² + 3x – 4 = 0
Answer:
Let y = x² + 3x – 4
(i) Draw the graph of y = x² + 3x – 4

(ii) Plot the points (-5, 6), (-4, 0), (-3, -4), (-2, -6), (-1, -6), (0, -4), (1, 0), (2, 6), (3, 14) on the graph using suitable scale.
(iii) Join the points by a free hand smooth curve.
The smooth curve is the graph of y = x² – 4x + 4
(iv) To solve x + 3x – 4 = 0, subtract x² + 3x – 4 = 0 from y = x² + 3x – 4.
y = 0
∴ The point of intersection with the x – axis is the solution set.
The solution set is -4 and 1.

Question 6.
Draw the graph of y = x² – 5x – 6 and hence solve x², – 5x – 14 = 0
Answer:
Let y = x² – 5x – 6
(i) Draw the graph of y = x² – 5x – 6 by preparing the table of values as below.

(ii) Plot the points (-3, 18), (-2, 8), (-1, 0), (0, -6), (1, -10), (2, -12), (3, -12), (4, -10), (5,-6), (6, 0) and (7, 8).
(iii) Join the points by a free hand to get smooth curve.
(iv) To solve x² – 5x – 14 = 0, subtract x² – 5x – 14 = 0 from y = x² – 5x – 6.

The equation y = 8 represent a straight line draw a straight line through y = 8 intersect the curve at two places. From the two points draw perpendicular line to the X – axis it will intersect at -2 and 7.
The solution is -2 and 7

Question 7.
Draw the graph of y = 2x² – 3x – 5 and hence solve 2x² – 4x – 6 = 0
Answer:
(i) Draw the graph of y = 2x² – 3x – 5 by preparing the table of values given below.

(ii) Plot the points (-3, 22), (-2, 9), (-1, 0), (0, -5), (1,-6), (2, -3), (3, 4), (4, 15) on the graph sheet using suitable scale.
(iii) To solve 2x² – 4x – 6 = 0 subtract 2x² – 4x – 6 = 0 from y = 2x² – 3x – 5

(iv) y = x + 1 represent a straight line.

The straight line intersect the curve at (-1, 0) and (3, 4). From the two point draw perpendicular lines to the X – axis it will intersect at -1 and 3.
The solution set is (-1, 3)

Question 8.
Draw the graph of y = (x – 1) (x + 3) and hence solve x² – x – 6 = 0
Answer:
y = (x – 1) (x + 3)
y = x² + 2x – 3
(i) Draw the graph of y = x² + 2x – 3 by preparing the table of values given below

(ii) Plot the points (-4, 5), (-3, 0), (-2, -3), (-1, -4), (0, -3), (1, 0), (2, 5), (3, 12) and (4, 21) on the graph sheet using suitable scale.
(iii) To solve x² – x – 6 = 0 subtract x² – x – 6 = 0 from y = x2 + 2x – 3
(iv) Draw the graph of y = 3x + 3 by preparing the table.

(v) The straight line cuts the curve at (-2, -3) and (3, 12). Draw perpendicular lines from the point to X – axis.
The line cut the X – axis at -2 and 3.
The solution set is (-2, 3)

Students can download Maths Chapter 3 Algebra Ex 3.6 Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 3 Algebra Ex 3.6

Question 1.
Simplify
(i) \(\frac{x(x+1)}{x-2}+\frac{x(1-x)}{x-2}\)
Answer:
\(\frac{x(x+1)}{x-2}+\frac{x(1-x)}{x-2}\) = \(\frac{x(x+1)+x(1-x)}{x-2}\)
= \(\frac{x^{2}+x+x-x^{2}}{x-2}\)
= \(\frac { 2x }{ x-2 } \)

(ii) \(\frac { x+2 }{ x+3 } \) + \(\frac { x-1 }{ x-2 } \)
Answer:
\(\frac { x+2 }{ x+3 } \) + \(\frac { x-1 }{ x-2 } \) = \(\frac{(x+2)(x-2)+(x-1)(x+3)}{(x+3)(x-2)}\)

(iii) \(\frac{x^{3}}{x-y}+\frac{y^{3}}{y-x}\) = \(\frac{x^{3}}{x-y}+\frac{y^{3}}{-1(x-y)}\)
Answer:
= \(\frac{x^{3}}{x-y}-\frac{y^{3}}{x-y}\)
= \(\frac{x^{3}-y^{3}}{x-y}\) (using a³ – b³ = (a – b) (a² + ab + b²))
= \(\frac{(x-y)\left(x^{2}+x y+y^{2}\right)}{x-y}\)
= x² + xy + y²

Question 2.
Simplify
(i) \(\frac{(2 x+1)(x-2)}{x-4}\) – \(\frac{\left(2 x^{2}-5 x+2\right)}{x-4}\)
Answer:

(ii) \(\frac{4 x}{x^{2}-1}-\frac{x+1}{x-1}\)
Answer:

Question 3.
Subtract \(\frac{1}{x^{2}+2}\) from \(\frac{2 x^{3}+x^{2}+3}{\left(x^{2}+2\right)^{2}}\)
Answer:

Question 4.
Which rational expression should be subtracted from \(\frac{x^{2}+6 x+8}{x^{3}+8}\)
Answer:

Question 5.
If A = \(\frac{2x+1}{2 x-1}\), B = \(\frac{2x-1}{2x+1}\) find \(\frac{1}{A-B}\) – \(\frac{2 \mathbf{B}}{\mathbf{A}^{2}-\mathbf{B}^{2}}\)
Answer:

Question 6.
If A = \(\frac { x }{ x+1 } \) B = \(\frac { 1 }{ x+1 } \) prove that \(\frac{(\mathbf{A}+\mathbf{B})^{2}+(\mathbf{A}-\mathbf{B})^{2}}{\mathbf{A}+\mathbf{B}}=\frac{2\left(x^{2}+1\right)}{x(x+1)^{2}}\)
Answer:

Question 7.
Pari needs 4 hours to complete a work. His friend Yuvan needs 6 hours to complete the same work. How long will it take to complete if they work together?
Solution:
Pari: time required to complete the work = 4 hrs.
∴ In 1 hr. he will complete = \(\frac{1}{4}\) of the work.
= \(\frac{1}{4}\) w.
Yuvan: Time required to complete the work = 6 hrs.
∴ In 1 hr. he will complete the \(\frac{1}{6}\) of the work
= \(\frac{1}{6}\) w
Working together, in 1 hr. they will complete \(\frac{w}{4}+\frac{w}{6}\) of the work.
= \(\frac{6 w+4 w}{24}=\frac{5}{12} w\)
∴ To complete the total work time taken
= \(\frac{w}{\frac{5}{12} w}=\frac{12}{5}\) = 2.4 hrs. [∵ (4) hrs = 4 × 60 = 24 min]
= 2 hrs 24 minutes.

Question 8.
Iniya bought 50 kg of fruits consisting of apples and bananas. She paid twice as much per kg for the apple as she did for the banana. If Iniya bought ? 1800 worth of apples and ₹ 600 worth bananas, then how many kgs of each fruit did she buy?
Let the weight of applies be a kg.
Let the weight of bananas be b kg.
a + b = 50
ax = ₹ 1800 ………….. (1)
by = ₹ 600 ………… (2)
x = 2y …………… (3)
Use (3) in (1) ⇒ a(2y) = 1800
y = \(\frac{900}{a}\) ………….. (4)
Use (4) in (2) ⇒ \(b\frac{900}{a}\) = 600
∵ 3b = 2a …………….. (5)
∵ a + b = 50
a + \(\frac{2 a}{3}\) = 50 ⇒ \(\frac{5 a}{3}\) = 50
⇒ a = 50 × \(\frac{3}{5}\)
= 30
∴ b = 20
∴ Iniya bought 30 kg of applies and 20 kg of bananas
= 30
.’. b = 20.
.’. Iniya bought 30 kg of applies and 20 kg of bananas.

Students can download Maths Chapter 3 Algebra Ex 3.17 Questions and Answers, Notes, Samacheer Kalvi 10th 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 10th Maths Solutions Chapter 3 Algebra Ex 3.17

Question 1.
If then

verify that (i) A + B = B + A
(ii) A + (-A) = (-A) + A = O.
Answer:

From (1) and (2) we get A + B = B + A

From (1) and (2) we get
A + (-A) = (-A) + A = 0

Question 2.
If
then verify that
A + (B + C) = (A + B) + C.
Answer:


From (1) and (2) we get
A + (B + C) = (A + B) + C

Question 3.
Find X and Y if and
Answer:

Question 4.
If
find the value of (i) B – 5A (ii) 3A – 9B
Answer:

Question 5.
Find the values of x, y, z if

Answer:
(i) x – 3 = 1 ⇒ x = 1 + 3 ⇒ x = 4
3x – z = 0 (substitute the value of x)
3(4) – z = 0
12 – z = 0
∴ z = 12
x + y + z = 6
4 + y + 12 = 0
y + 16 = 6
y = 6- 16
∴ y = -10
The value of x = 4, y = -10 and z = 12

(ii) [x y – z z + 3] + [y 4 3] = [4 8 16]
x + y = 4 ….(1)
y – z + 4 = 8
Substitute the value
of z in (2)
(2) ⇒ y – 10 = 4
Substitute the value of y in (1)
z + 3 + 3 = 16
z + 6 = 16
z = 16 – 16 = 10
y = 14
(1) ⇒ x + 14 = 4
x – 4 – 14 = -10
The value of x = -10, y = 14 and z = 10

Question 6.
Find x and y if x
Answer:

4x – 2y = 4
(1) ⇒ 2x – y = 2
(2) ⇒ 3x – y = 2
– x + y = 2
Add (1) and (2)

Substitute the value of x = 4 in (2)
– 4 + y = 2
y = 2 + 4 = 6
The value of x = 4 and y = 6

Question 7.
Find the non-zero values of x satisfying the matrix equation

Answer:

The value of x = 4

Question 8.
Solve for x,y :

Answer:

x² – 4x = -5
x² – 4x + 5 = 0
(x – 5) (x + 1) = 0
x – 5 = 0 or x + 1 = 0
x = 5 or x = – 1

The value of x = -1 and 5
y² – 2y = 8
y² – 2y – 8 = 0
(y – 4) (y + 2) = 0
y – 4 = 0 or y + 2 = 0

y = 4 or y = -2
The value of y = -2 and 4