Category: Class 10

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.9 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 2 Numbers and Sequences Ex 2.9

Question 1.
Find the sum of the following series
(i) 1 + 2 + 3 + …….. + 60
(ii) 3 + 6 + 9 + …….. +96
(iii) 51 + 52 + 53 + …….. + 92
(iv) 1 + 4 + 9 + 16 + …….. + 225
(v) 6² + 7² + 8² + …….. + 21²
(vi) 10³ + 11³ + 12³ + …….. + 20³
(vii) 1 + 3 + 5 + …… + 71
Solution:
(i) 1 + 2 + 3 + …….. + 60 = \(\frac{60 \times 61}{2}\)
[Using \(\frac{n(n+1)}{2}\) formula]
= 1830

(ii) 3 + 6 + 9 + …….. + 96 = 3(1 + 2 + 3 + ……… + 32)
= \(\frac{3 \times 32 \times 33}{2}\)
= 1584

(iii) 51 + 52 + 53 + …….. + 92 = (1 + 2 + 3 + ……. + 92) – (1 + 2 + 3 + …… + 50)

= 4278 – 1275
= 3003

(iv) 1 + 4 + 9 + 16 + …….. + 225 = 1² + 2² + 3² + 4² + ………… + 15²
\(\frac{15 \times 16 \times 31}{6}\)
[using \(\frac{n(n+1)(2 n+1)}{6}\)] formula
= 1240

(v) 6² + 7² + 8² + …….. + 21² = 1 + 2² + 3² + 4² + ………… + 21² – (1 + 2² + ………… + 5²)
= \(\frac{21 \times 22 \times 43}{6}\) – \(\frac{5 \times 6 \times 11}{6}\)
= 3311 – 55
= 3256

(vi) 10³ = 11³ + 12³ + …….. + 20³ = 1³ + 2³+ 3³ + ………… + 20³ – (1³ + 2³ + 3³ + …………. + 9³)

[Using (\(\frac{n(n+1)}{2}\))² formula]
= 210² – 45² = 44100 – 2025
= 42075

(vii) 1 + 3 + 5+ … + 71
Here a = 1; d = 3 – 1 = 2; l = 71

Question 2.
If 1 + 2 + 3 + …. + k = 325 , then find 1³ + 2³ + 3³ + …………. + k³
Answer:
1 + 2 + 3 + …. + k = 325
\(\frac{k(k+1)}{2}\) = 325 ……(1)

= 325² (From 1)
= 105625

Question 3.
If 1³ + 2³ + 3³ + ………… + K³ = 44100 then find 1 + 2 + 3 + ……. + k
Answer:
1³ + 2³ + 3³ + ………….. + k³ = 44100

\(\frac{k(k+1)}{2}\) = \(\sqrt { 44100 }\) = 210
1 + 2 + 3 + …… + k = \(\frac{k(k+1)}{2}\)
= 210

Question 4.
How many terms of the series 1³ + 2³ + 3³ + …………… should be taken to get the sum 14400?
Answer:
1³ + 2³ + 3³ + ……. + n³ = 14400

\(\frac{n(n+1)}{2}\) = \(\sqrt { 14400 }\)
\(\frac{n(n+1)}{2}\) = 120 ⇒ n² + n = 240

n² + n – 240 = 0
(n + 16) (n – 15) = 0
(n + 16) = 0 or (n – 15) = 0
n = -16 or n = 15 (Negative will be omitted)
∴ The number of terms taken is 15

Question 5.
The sum of the squares of the first n natural numbers is 285, while the sum of their cubes is 2025. Find the value of n.
Answer:
1² + 2² + 3² + …. + n² = 285

Question 6.
Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm, …, 24 cm. How much area can be decorated with these colour papers?
Answer:
Area of 15 square colour papers
= 10² + 11² + 12² + …. + 24²
= (1² + 2² + 3² + …. + 24²) – (1² + 2² + 9²)
= \(\frac{24 \times 25 \times 49}{6}-\frac{9 \times 10 \times 19}{6}\)
= 4 × 25 × 49 – 3 × 5 × 19
= 4900 – 285
= 4615
Area can be decorated is 4615 cm²

Question 7.
Find the sum of the series (2³ – 1)+(4³ – 3³) + (6³ – 15³) + …….. to
(i) n terms
(ii) 8 terms
Answer:
Sum of the series = (2³ – 1) + (4³ – 3³) + (6³ – 15³) + …. n terms
= 2³ + 4³ + 6³ + …. n terms – (1³ + 3³ + 5³ + …. n terms) …….(1)
2³ + 4³ + 6³ + …. n = ∑(2³ + 4³ + 6³ + ….(2n)³]
∑ 2³ (1³ + 2³ + 3³ + …. n³)
= 8 (\(\frac{n(n+1)}{2}\))²
= 2[n (n + 1)]²
1³ + 3³ + 5³ + ……….(2n – 1)³ [sum of first 2n cubes – sum of first n even cubes]

Substituting (2) and (3) in (1)
Sum of the series = 2n² (n + 1)² – n² (2n + 1)² + 2n²(n + 1)²
= 4n² (n + 1)² – n² (2n + 1)²
= n² [(4(n + 1)² – (2n + 1)²]
= n² [4n² + 4 + 8n – 4n² – 1 – 4n]
= n² [4n + 3]
= 4n³ + 3n²

(ii) when n = 8 = 4(8)³ + 3(8)²
= 4(512) + 3(64)
= 2240

Students can download Maths Chapter 1 Relations and Functions Ex 1.5 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 1 Relations and Functions Ex 1.5

Question 1.
Using the functions f and g given below, find fog and gof Check whether fog = gof.

(i) f(x) = x – 6, g(x) = x²
Answer:
f(x) = x – 6, g(x) = x²
fog = fog (x)
= f(g(x))
fog = f(x)²
= x² – 6
gof = go f(x)
= g(x – 6)
= (x – 6)²
= x² – 12x + 36
fog ≠ gof

(ii) f(x) = \(\frac { 2 }{ x } \), g(x) = 2x² – 1
Answer:
f(x) – \(\frac { 2 }{ x } \); g(x) = 2x² – 1
fag = f[g (x)]
= f(2x² – 1)
= \(\frac{2}{2 x^{2}-1}\)
gof = g [f(x)]
= g (\(\frac { 2 }{ x } \))
= 2 (\(\frac { 2 }{ x } \))² – 1
\(=2 \times \frac{4}{x^{2}}-1\)
\(=\frac{8}{x^{2}}-1\)
fog ≠ gof

(iii) f(x) = \(\frac { x+6 }{ 3 } \), g(x) = 3 – x
Answer:
f(x) = \(\frac { x+6 }{ x } \), g(x) = 3 – x
fog = f[g(x)]
= f(3 – x)

(iv) f(x) = 3 + x, g(x) = x – 4
Answer:
f(x) = 3 + x ;g(x) = x – 4
fog = f[g(x)]
= f(x – 4)
= 3 + x – 4
= x – 1
gof = g[f(x)]
= g(3 + x)
= 3 + x – 4
= x – 1
fog = gof

(v) f(x) = 4x² – 1,g(x) = 1 + x
Answer:
f(x) = 4x² – 1 ; g(x) = 1 + x
fog = f[g(x)]
= 4(1 + x)
= 4(1 + x)² – 1
= 4[1 + x² + 2x] – 1
= 4 + 4x² + 8x – 1
= 4x² + 8x + 3
gof = g [f(x)]
= g (4x² – 1)
= 1 + 4x² – 1
= 4x²
fog ≠ gof

Question 2.
Find the value of k, such that fog = gof
(i) f(x) = 3x + 2, g(x) = 6x – k
(ii) f(x) = 2x – k, g(x) = 4x + 5
Solution:
(i) f(x) = 3x + 2, g(x) = 6x – k
fog(x) = f(g(x)) = f(6x – k) = 3(6x – k) + 2
= 18x – 3k + 2 …………… (1)
gof(x) = g(f(x)) = g(3x + 2) = 6(3x + 2) – k
= 18x + 12 – k ……………. (2)
(1) = (2)
⇒ 18x – 3k + 2 = 18x + 12 – k
2k = -10
k = -5

(ii) f(x) = 2x – k, g(x) = 4x + 5
fog(x) = f(g(x)) = f(4x + 5) = 2(4x + 5) – k
= 8x + 10 – k ……………… (1)
gof(x) = g(f(x)) = g(2x – k) = 4(2x – k) + 5
= 8x – 4k + 5 ……………. (2)
(1) = (2)
⇒ 8x + 10 – k = 8x – 4k + 5
3k = -5
k = \(\frac{-5}{3}\)

Question 3.
If f(x) = 2x – 1, g(x) = \(\frac { x+1 }{ 2 } \), show that f o g = g o f = x
Answer:
f(x) = 2x – 1 ; g(x) = \(\frac { x+1 }{ 2 } \)
fog = f[g(x)]

∴ fog = gof = x
Hence it is proved.

Question 4.
(i) If f (x) = x² – 1, g(x) = x – 2 find a, if gof(a) = 1.
(ii) Find k, if f(k) = 2k – 1 and fof (k) = 5.
Solution:
(i) f(x) = x² – 1, g(x) = x – 2
Given gof(a) = 1
gof(x) = g(f(x)
= g(x² – 1) = x² – 1 – 2
= x² – 3
gof(a) ⇒ a² – 3 = 1 =+ a² = 4
a = ± 2
(ii) f(k) = 2k – 1
fo f(k) = 5
f(f(k)m = f(2k – 1) = 5
⇒ 2(2k – 1) – 1 = 5
4 k – 2 – 1 = 5 ⇒ 4k = 8
k = 2

Question 5.
Let A,B,C ⊂ N and a function f: A → B be defined by f(x) = 2x + 1 and g: B → C be defined by g(x) = x² . Find the range of fog and gof.
Answer:
f(x) = 2x + 1 ; g(x) = x²
fog = f[g(x)]
= f(x²)
= 2x² + 1
2x² + 1 ∈ N
g o f = g [f(x)]
= g (2x + 1)
g o f = (2x + 1)²
(2x + 1)² ∈ N
Range = {y/y = 2x² + 1, x ∈ N};
{y/y = (2x + 1)², x ∈ N)

Question 6.
If f(x) = x² – 1. Find (i)f(x) = x² – 1, (ii)fofof
Solution:
(i) f(x) = x² – 1
fof(x) = f(fx)) = f(x² – 1)
= (x² – 1 )² – 1;
= x⁴ – 2x² + 1 – 1
= x⁴ – 2x²
(ii) fofof = f o f(f(x))
= f o f (x⁴ – 2x²)
= f(f(x⁴ – 2x²))
= (x⁴ – 2x²)² – 1
= x⁸ – 4x⁶ + 4x⁴ – 1

Question 7.
If f : R → R and g : R → R are defined by f(x) = x⁵ and g(x) = x⁴ then check if f, g are one – one and fog is one – one?
Answer:
f(x) = x⁵ – It is one – one function
g(x) = x⁴ – It is one – one function
fog = f[g(x)]
= f(x⁴)
= (x⁴)⁵
fag = x²⁰
It is also one-one function.

Question 8.
Consider the functions f(x), g(x), h(x) as given below. Show that (fog)oh = fo(goh) in each case.
(i) f(x) = x – 1, g(x) = 3x + 1 and h(x) = x²
(ii) f(x) = x², g(x) = 2x and h(x) = x + 4
(iii) f(x) = x – 4, g(x) = x² and h(x) = 3x – 5
Solution:
(i) f(x) = x – 1, g(x) = 3x + 1 and h(x) = x²
f(x) = x – 1
g(x) = 3x + 1
f(x) = x²
(fog)oh = fo(goh)
LHS = (fog)oh
fog = f(g(x)) = f(3x + 1) = 3x + 1 – 1 = 3x
(fog)oh = (fog)(h(x)) = (fog)(x² ) = 3²  ……………. (1)
RHS = fo(goh)
goh = g(h(x)) = g(x²) = 3x² + 1
fo(goh) = f(3x² + 1) = 3x² + 1 – 1= 3x² ………… (2)
LHS = RHS Hence it is verified.

(ii) f(x) = x², g(x) = 2x, h(x) = x + 4
(fog)oh = fo(goh)
LHS = (fog)oh
fog = f(g(x)) = f(2x) = (2x)² = 4x²
(fog)oh = (fog) h(x) = (fog) (x + 4)
= 4(x + 4)² = 4(x² + 8x+16)
= 4x² + 32x + 64 ………….. (1)
RHS = fo(goh) goh = g(h(x)) = g(x + 4)
= 2(x + 4) = (2x + 8)
fo(goh) = f(goh) = f(2x + 8) = (2x + 8)²
= 4x² + 32x + 64 ……………… (2)
(1) = (2)
LHS = RHS
∴ (fog)oh = fo(goh) It is proved.

(iii) f(x) = x – 4, g(x) = x², h(x) = 3x – 5
(fog)oh = fo(goh)
LHS = (fog)oh
fog = f(g(x)) = f(x²) = x² – 4
(fog)oh = (fog)(3x – 5) = (3x – 5)² – 4
= 9x² – 30x + 25 -4
= 9x² – 30x + 21 ………….. (1)
∴ RHS = fo(goh)
(goh) = g(h(x)) = g(3x – 5) = (3x – 5)²
= 9x² – 30x + 25
fo(goh) = f(9x² – 30 x + 25)
= 9x² – 30x + 25 – 4
= 9x² – 30x + 21 …………… (2)
(1) = (2)
LHS = RHS
∴ (fog)oh = fo(goh)
It is proved.

Question 9.
Let f = {(-1, 3), (0, -1), (2, -9)} be a linear function from Z into Z. Find f(x).
Answer:
The linear equation is f(x) = ax + b
f(-1) = 3
a(-1) + b = 3
-a + b = 3 ….(1)
f(0) = -1
a(0) + b = -1
0 + b = -1
b = -1
Substitute the value of b = -1 in (1)
-a – 1 = 3
-a = 3 + 1
-a = 4
a = -4
∴ The linear equation is -4(x) -1 = -4x – 1 (or) – (4x + 1)

Question 10.
In electrical circuit theory, a circuit C(t) is called a linear circuit if it satisfies the superposition principle given by C(at₁ + bt₂) = aC(t₁) + bC(t₂), where a,b are constants. Show that the circuit C(t) = 31 is linear.
Solution:
Given C(t) = 3t. To prove that the function is linear
C(at₁) = 3a(t₁)
C(bt₂) = 3 b(t₂)
C(at₁ + bt₂) = 3 [at₁ + bt₂] = 3at₁ + 3bt₂
= a(3t₁) + b(3t₂) = a[C(t₁) + b(Ct₂)]
∴ Superposition principle is satisfied.
Hence C(t) = 3t is linear function.

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.2 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 2.2

Question 1.
For what values of natural number n, 4th can end with the digit 6?
Answer:
4n = (2²)ⁿ = 2²ⁿ
= 2ⁿ × 2ⁿ
2 is a factor of 4ⁿ
∴ 4ⁿ is always even.

Question 2.
If m, n are natural numbers, for what values of m, does 2ⁿ × 5ⁿ ends in 5?
Solution:
2ⁿ × 5^m
2ⁿ is always even for all values of n.
5^m is always odd and ends with 5 for all values of m.
But 2ⁿ × 5^m is always even and ends in 0.
∴ 2ⁿ × 5^m cannot end with the digit 5 for any values of m. No value of m will satisfy 2ⁿ × 5^m ends in 5.

Question 3.
Find the H.C.F. of 252525 and 363636.
Answer:
To find the HCF of 252525 and 363636 by using Euclid’s Division algorithm.
363636 = 252525 × 1 + 111111
The remainder 111111 ≠ 0
By division of Euclid’s algorithm
252525 = 111111 × 2 + 30303
The remainder 30303 ≠ 0
Again by division of Euclid’s algorithm
111111 = 30303 × 3 + 20202
The remainder 20202 ≠ 0
Again by division of Euclid’s algorithm.
30303 = 20202 + 10101
The remainder 10101 ≠ 0
Again by division of Euclid’s algorithm.
20202 = 10101 × 2 + 0
The remainder is 0
∴ The H.C.F. is 10101

Question 4.
If 13824 = 2^a × 3^b then find a and b?
Answer:
Using factor tree method factorise 13824


13824 = 2⁹ × 3³
Given 13824 = 2a × 3b
Compare we get a = 9 and b = 3

Aliter:
13824 = 2⁹ × 3³
Compare with
13824 = 2^a × 3^b
The value of a = 9 b = 3

Question 5.
If p₁^x1 × p₂^x2 × p₃^x3 × p₄^x4 = 113400 where p₁ p₂, p₃, p₄ are primes in ascending order and x₁, x₂, x₃, x₄, are integers, find the value of p₁,p₂,p₃,p₄ and x₁,x₂,x₃,x₄.
Answer:
Given 113400 = p₁^x1 × p₂^x2 × p₃^x3 × p₄^x4
Using tree method factorize 113400

113400 = 23 × 34 × 52 × 7
compare with
113400 = p₁^x1 × p₂^x2 × p₃^x3 × p₄^x4
P₁ = 2, x₁ = 3
P₂ = 3, x₂ = 4
P₃ = 5, x₃ = 2
P₄ = 7, x₄ = 1

Question 6.
Find the L.C.M. and H.C.F. of 408 and 170 by applying the fundamental theorem of Arithmetic.
Answer:
Factorise 408 and 170 by factor tree method


408 = 2³ × 3 × 17
170 = 2 × 5 × 17
To find L.C.M. list all prime factors of 408 and 170 of their greatest exponents.
L.C.M. = 2³ × 3 × 5 × 17
= 2040
To find the H.C.F. list all common factors of 408 and 170.
H.C.F. = 2 × 17 = 34
L.C.M. = 2040 ; HCF = 34

Question 7.
Find the greatest number consisting of 6 digits which is exactly divisible by 24,15,36?
Answer:
The greatest number of 6 digits is 999999.
The greatest number must be divisible by L.C.M. of 24, 15 and 36

24 = 2³ × 3
15 = 3 × 5
36 = 2² × 3²
L.C.M = 2³ × 3² × 5
= 360
To find the greatest number 999999 must be subtracted by the remainder when 999999 is divided by 360
The greatest number in 6 digits = 999999 – 279
= 999720

Question 8.
What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves remainder 7 in each case?
Solution:
35 = 5 × 7
56 = 2 × 2 × 2 × 7
91 = 7 × 13
LCM of 35, 56, 91 = 5 × 7 × 2 × 2 × 2 × 13 = 3640
∴ Required number = 3647 which leaves remainder 7 in each case.

Question 9.
Find the least number that is divisible by the first ten natural numbers?
Answer:
Find the L.C.M of first 10 natural numbers

The least number is 2520

Modular Arithmetic
Two integers “a” and “b” are congruence modulo n if they differ by an integer multiple of n. That b – a = kn for some integer k. This can be written as a = b (mod n).

Euclid’s Division Lemma and Modular Arithmetic

Let m and n be integers, where m is positive. By Euclid’s division Lemma we can write n = mq + r where 0 < r < m and q is an integer.
This n = r (mod m)

Students can download Maths Chapter 1 Relations and Functions Unit Exercise 1 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 1 Relations and Functions Unit Exercise 1

Question 1.
If the ordered pairs (x² – 3x, y² + 4y) and (-2, 5) are equal, then find x and y.
Answer:
(x² – 3x, y² + 4y) = (-2, 5)
x² – 3x = -2
x² – 3x + 2 = 0
(x – 2) (x – 1) = 0
x – 2 = 0 or x – 1 = 0
x = 2 or 1

y² + 4y = 5
y² + 4y – 5 = 0
(y + 5) (y – 1) = 0
y + 5 = 0 or y – 1 = 0
y = -5 or y = 1
The value of x = 2, 1
and 7 = -5, 1

Question 2.
The Cartesian product A × A has 9 elements among which (-1, 0) and (0, 1) are found. Find the set A and the remaining elements of A × A.
Solution:
A = {-1, 0, 1}, B = {1, 0, -1}
A × B = {(-1, 1), (-1, 0), (-1, -1), (0, 1), (0, 0), (0, -1), (1, 1), (1, 0), (1, -1)}

Question 3.
Given that f(x) = \(\left\{\begin{array}{rl}
{\sqrt{x-1}} & {x \geq 1} \\
{4} & {x<1}
\end{array}\right.\).
Find
(i) f(0) (ii)f (3) (iii) f(a + 1) in terms of a.(Given that a > 0)
Answer:
f(x) = \(\sqrt { x-1 }\) ; f(x) = 4
(i) f(0) = 4
(ii) f(3) = \(\sqrt { 3-1 }\) = \(\sqrt { 2 }\)
(iii) f(a + 1) = \(\sqrt { a+1-1 }\) = \(\sqrt { a }\)

Question 4.
Let A = {9, 10, 11, 12, 13, 14, 15, 16, 17} and let f: A → N be defined by f(n) = the highest prime factor of n ∈ A. Write f as a set of ordered pairs and find the range of f.
Solution:
A = {9, 10, 11, 12, 13, 14, 15, 16, 17}
f: A → N
f(n) = the highest prime factor of n ∈ A
f = {(9, 3), (10, 5), (11, 11), (12, 3), (13, 13), (14, 7), (15, 5), (16, 2), (17, 17)}
Range = {3, 5, 11, 13, 7, 2, 17}
= {2, 3, 5, 7, 11, 13, 17}

Question 5.
Find the domain of the function

Answer:

Domain of f(x) = {-1, 0, 1}

Question 6.
If f(x)= x², g(x) = 3x and h(x) = x – 2 Prove that (fog)oh = fo(goh).
Solution:
f(x) = x²
g(x) = 3x
h(x) = x – 2
(fog)oh = x – 2
LHS = fo(goh)
fog = f(g(x)) = f(3x) = (3x)² = 9x²
(fog)oh = (fog) h(x) = (fog) (x – 2)
= 9(x – 2)² = 9(x² – 4x + 4)
= 9x² – 36x + 36 ……………. (1)
RHS = fo(goh)
(goh) = g(h(x)) = g(x – 2)
= 3(x – 2) = 3x – 6
fo(goh) = f(3x – 6) = (3x – 6)²
= 9x² – 36x + 36 ………….. (2)
(1) = (2)
LHS = RHS
(fog)oh = fo(goh) is proved.

Question 7.
Let A= {1,2} and B = {1,2,3,4}, C = {5,6} and D = {5,6,7,8}. Verify whether A × C is a subset of B × D?
Answer:
Given A = {1, 2}
B = {1, 2, 3, 4}
C = {5,6}
D = {5,6, 7,8}
A × C = {1,2} × {5,6}
= {(1,5) (1,6) (2, 5) (2, 6)}
B × D = {1,2, 3, 4} × {5, 6, 7, 8}
= {(1,5) (1,6) (1,7) (1,8)
(2, 5) (2, 6) (2,7) (2, 8)
(3, 5) (3, 6) (3, 7) (3, 8)
(4, 5) (4, 6) (4, 7) (4, 8)}
∴ A × C ⊂ B × D
Hence it is verified

Question 8.
If f(x) = \(\frac{x-1}{x+1}, x \neq 1\) Show that
f(f(x)) = – \(\frac { 1 }{ x } \), Provided x ≠ 0.
Answer:

Question 9.
The functions f and g are defined by f{x) = 6x + 8; g(x) = \(\frac { x-2 }{ 3 } \)
(i) Calculate the value of gg [latex]\frac { 1 }{ 2 } [/latex]
(a) Write an expression for gf (x) in its simplest form.
Answer:
f(x) = 6x + 8 ; g(x) = \(\frac { x-2 }{ 3 } \)

Question 10.
Write the domain of the following real functions

Answer:
(i) f (x) = \(\frac { 2x+1 }{ x-9 } \)
If the denominator vanishes when x = 9
So f(x) is not defined at x = 9
∴ Domain is x ∈ [R – {9}]

(ii) if p(x) = \(=\frac{-5}{4 x^{2}+1}\)
p(x) is defined for all values of x. So domain is x ∈ R.

(iii) g(x) = \(\sqrt { x-2 }\)
When x < 2 g(x) becomes complex. But given “g” is real valued function.
So x > 2
Domain x ∈ (2, α)

(iv) h (x) = x + 6
For all values of x, h(x) is defined. Hence domain is x ∈ R.

Students can download Maths Chapter 1 Relations and Functions Ex 1.6 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 1 Relations and Functions Ex 1.6

Multiple Choice Questions

Question 1.
If n(A × B) = 6 and A= {1, 3} then n (B) is ………….
(1) 1
(2) 2
(3) 3
(4) 6
Answer:
(3) 3
Hint: n(A × B) = 6
n(A) = 2
n(A × B) = n(A) × n(B)
6 = 2 × n(B)
n(B) = \(\frac { 6 }{ 2 } \) = 3

Question 2.
A = {a, b, p}, B = {2, 3}, C = {p, q, r, s} then n[(A ∪ C) × B] is
(1) 8
(2) 20
(3) 12
(4) 16
Answer:
(3) 12
Hint:
A = {a, b, p}, B = {2, 3}, C = {p, q, r, s}
n (A ∪ C) × B
A ∪ C = {a, b, p, q, r, s}
(A ∪ C) × B = {{a, 2), (a, 3), (b, 2), (b, 3), (p, 2), (p, 3), (q, 2), (q, 3), (r, 2), (r, 3), (s, 2), (s, 3)
n [(A ∪ C) × B] = 12

Question 3.
If A = {1,2}, B = {1,2, 3, 4}, C = {5,6} and D = {5, 6, 7, 8} then state which of the following statement is true ……………….
(1) (A × C) ⊂ (B × D)
(2) (B × D) ⊂ (A × C)
(3) (A × B) ⊂ (A × D)
(4) (D × A) ⊂ (B × A)
Answer:
(1) (A × C) ⊂ (B × D)
Hint: n(A × B) = 2 × 4 = 8
(A × C) = 2 × 2 = 4
n(B × C) = 4 × 2 = 8
n(C × D) = 2 × 4 = 8
n(A × C) = 2 × 2 = 4
n(A × D) = 2 × 4 = 8
n(B × D) = 4 × 4 = 16
∴ (A × C) ⊂ (B × D)

Question 4.
If there are 1024 relations from a set A = {1, 2, 3, 4, 5} to a set B, then the number of elements in B is
(1) 3
(2) 2
(3) 4
(4) 6
Answer:
(2) 2
Hint:
n(A) = 5
n(B) = x
n(A × B) = 1024 = 2¹⁰
2^5x = 2¹⁰
⇒ 5x = 10
⇒ x =2

Question 5.
The range of the relation R = {(x, x²) a prime number less than 13} is ……………………
(1) {2, 3, 5, 7}
(2) {2, 3, 5, 7, 11}
(3) {4, 9, 25, 49, 121}
(4) {1, 4, 9, 25, 49, 121}
Answer:
(3) {4, 9, 25, 49, 121}
Hint:
Prime number less than 13 = {2, 3, 5, 7, 11}
Range (R) = {(x, x²)}
Range = {4, 9, 25, 49, 121} (square of x)

Question 6.
If the ordered pairs (a + 2, 4) and (5, 2a + b)are equal then (a, b) is
(1) (2, -2)
(2) (5, 1)
(3) (2, 3)
(4) (3, -2)
Answer:
(4) (3, -2)
Hint:
(a + 2, 4), (5, 2a + b)
a + 2 = 5
a = 3
2a + b = 4
6 + b = 4
b = -2

Question 7.
Let n(A) = m and n(B) = n then the total number of non-empty relations that can be defined from A to B is ……………..
(1) mⁿ
(2) n^m
(3) 2^mn – 1
(4) 2^mn
Answer:
(4) 2^mn

Question 8.
If {(a, 8),(6, b)}represents an identity function, then the value of a and b are respectively
(1) (8, 6)
(2) (8, 8)
(3) (6, 8)
(4) (6, 6)
Answer:
(1) (8, 6)
Hint:
{{a, 8), (6, b)}
a = 8
b = 6

Question 9.
Let A = {1, 2, 3, 4} and B = {4, 8, 9, 10}.
A function f: A → B given by f = {(1, 4), (2, 8),(3,9),(4,10)} is a ……………
(1) Many-one function
(2) Identity function
(3) One-to-one function
(4) Into function
Answer:
(3) One-to-one function
Hint:

Different elements of A has different images in B.
∴ It is one-to-one function.

Question 10.
If f (x) = 2x² and g(x) = \(\frac { 1 }{ 3x } \), then fog is …………..
(1) \(\frac{3}{2 x^{2}}\)
(2) \(\frac{2}{3 x^{2}}\)
(3) \(\frac{2}{9 x^{2}}\)
(4) \(\frac{1}{6 x^{2}}\)
Answer:
(3) \(\frac{2}{9 x^{2}}\)
Hint:

Question 11.
If f: A → B is a bijective function and if n(B) = 7, then n(A) is equal to
(1) 7
(2) 49
(3) 1
(4) 14
Answer:
(1) 7
Hint:
In a bijective function, n(A) = n(B)
⇒ n(A) = 7

Question 12.
Let f and g be two functions given by
f = {(0,1),(2, 0),(3-4),(4,2),(5,7)}
g = {(0,2),(1,0),(2, 4),(-4,2),(7,0)}
then the range of f o g is …………………
(1) {0,2,3,4,5}
(2) {-4,1,0,2,7}
(3) {1,2,3,4,5}
(4) {0,1,2}
Answer:
(4) {0,1,2}
Hint: f = {(0, 1)(2, 0)(3, -4) (4, 2) (5, 7)}
g = {(0,2)(l,0)(2,4)(-4,2)(7,0)}

fog = f[g(x)]
f [g(0)] = f(2) = 0
f [g(1)] = f(0) = 1
f [g(2)] = f(4) = 2
f[g(-4)] = f(2) = 0
f[g(7)] = f(0) = 1
Range of fog = {0,1,2}

Question 13.
Let f(x) = \(\sqrt{1+x^{2}}\) then
(1) f(xy) = f(x),f(y)
(2) f(xy) ≥ f(x),f(y)
(3) f(xy) ≤ f(x).f(y)
(4) None of these
Answer:
(3) f(xy) ≤ f(x).f(y)
Hint:
\(\sqrt{1+x^{2} y^{2}} \leq \sqrt{\left(1+x^{2}\right)} \sqrt{\left(1+y^{2}\right)}\)
⇒ f(xy) ≤ f(x) . f(y)

Question 14.
If g= {(1,1),(2,3),(3,5),(4,7)} is a function given by g(x) = αx + β then the values of α and β are
(1) (-1,2)
(2) (2,-1)
(3) (-1,-2)
(4) (1,2)
Answer:
(2) (2, -1)
Hint: g (x) = αx + β
g(1) = α(1) + β
1 = α + β ….(1)
g (2) = α (2) + β
3 = 2α + β ….(2)
Solve the two equations we get
α = 2, β = -1

Question 15.
f(x) = (x + 1)³ – (x – 1)³ represents a function which is
(1) linear
(2) cubic
(3) reciprocal
(4) quadratic
Answer:
(4) quadratic
Hint:
f(x) = (x + 1)³ – (x – 1)³
= x³ + 3x² + 3x + 1 -[x³ – 3x² + 3x – 1]
= x³ + 3x² + 3x + 1 – x³ + 3x² – 3x + 1 = 6x² + 2
It is a quadratic function.

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.3 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 2.3

Question 1.
Find the least positive value of x such that

(i) 71 = x (mod 8)
Answer:
71 = 7 (mod 8)
∴ The value of x = 7

(ii) 78 + x = 3 (mod 5)
78 + x – 3 = 5n (n is any integer)
75 + x = 5n
(Let us take x = 5)
75 + 5 = 80 (80 is a multiple of 5)
∴ The least value of x is 5

(iii) 89 = (x + 3) (mod 4)
89 – (x + 3) = 4n
(n may be any integer)
89 – x – 3 = 4n
89 – x = 4n
86 – x is a multiple of 4
(84 is a multiple of 4)
86 – 2 = 4n
84 = 4n
The value of x is 2

(iv) 96 = \(\frac { x }{ 7 } \) (mod 5)
96 – \(\frac { x }{ 7 } \) = 5n (n may be any integer)
672 – x = 35n (multiple of 35 is 665)
672 – 7 = 665
∴ The value of x = 7

(v) 5x = 4 (mod 6)
5x – 4 = 6n (n may be any integer)
5x = 6n + 4
x = \(\frac { 6n+4 }{ 5 } \)
Substitute the value of n as 1, 6, 11, 16 …. as n values in x = \(\frac { 6n+4 }{ 5 } \) which is divisible by 5.
2, 8, 14, 20,…………
The least positive value is 2.

Question 2.
If x is congruent to 13 modulo 17 then 7x -3 is congruent to which number modulo 17?
Solution:
x ≡ 13 (mod 17)
Let p be the required number …………. (1)
7x – 3 ≡ p (mod 17) ………….. (2)
From (1),
x – 13 = 17n for some integer M.
x – 13 is a multiple of 17.
x must be 30.
∴ 30 – 13 = 17
which is a multiple of 17.
From (2),
7 × 30 – 3 ≡ p (mod 17)
210 – 3 ≡ p (mod 17)
207 ≡ p (mod 17)
207 ≡ 3 (mod 17)
∴ P ≡ 3

Question 3.
Solve 5x ≡ 4 (mod 6)
5x – 4 = 6n (n may be any integer)
5x = 6n + 4
x = \(\frac{6 n+4}{5}\)
The value of n 1, 6, 11, 16 ……..
∴ The value of x is 2, 8, 14, 20 …………..

Question 4.
Solve 3x – 2 = 0 (mod 11)
Answer:
Given 3x – 2 = 0(mod 11)
3x – 2 = 11n (n may be any integer)
3x = 2 + 11n
x = \(\frac { 11n+2 }{ 3 } \)
Substitute the value of n = 2, 5, 8, 11 ….
When n ≡ 2 ⇒ x = \(\frac { 22+2 }{ 3 } \) = \(\frac { 24 }{ 3 } \) = 8
When n = 5 ⇒ x = \(\frac { 55+2 }{ 3 } \) = \(\frac { 57 }{ 3 } \) = 19
When n = 8 ⇒ x = \(\frac { 88+2 }{ 3 } \) = \(\frac { 90 }{ 3 } \) = 30
When n = 11 ⇒ x = \(\frac { 121+2 }{ 3 } \) = \(\frac { 123 }{ 3 } \) = 41
∴ The value of x is 8, 19, 30,41

Question 5.
What is the time 100 hours after 7 a.m.?
Answer:
100 ≡ x (mod 12) Note: In a clock every 12 hours
100 ≡ 4 (mod 12) the numbers repeats.

The time repeat after 7 am is 7 + 4 = 11 o’ clock (or) 11 am.

Question 6.
What is time 15 hours before 11 p.m.?
Solution:
15 ≡ x (mod 12)
15 – x = 12n
15 – x is a multiple of 12 x must be 3.

∴ The time 15 hrs before 11 O’clock is 11 – 3 = 8 O’ clock i.e. 8 p.m

Question 7.
Today is Tuesday. My uncle will come after 45 days. In which day my uncle will be coming?
Answer:
Number of days in a week = 7
45 ≡ x (mod 7)
45 ≡ 3 (mod 7)

The value of x must be 3.
Three days after tuesday is friday uncle will come on friday.

Question 8.
Prove that 2ⁿ + 6 × 9ⁿ is always divisible by 7 for any positive integer n.
Solution:
2¹ + 6 × 9¹ = 2 + 54 = 56 is divisible by 7
When n = k,
2^k + 6 × 9^k = 7 m [where m is a scalar]
⇒ 6 × 9^k = 7 m – 2^k …………. (1)
Let us prove for n = k + 1
Consider 2^k+1 + 6 × 9^k+1 = 2^k+1 + 6 × 9^k × 9
= 2^k+1 + (7m – 2^k)9 (using (1))
= 2^k+1 + 63m – 9.2^k = 63m + 2^k.2¹ – 9.2^k
= 63m – 2^k (9 – 2) = 63m – 7.2^k
= 7 (9m – 2^k) which is divisible by 7
∴ 2ⁿ + 6 × 9ⁿ is divisible by 7 for any positive integer n

Question 9.
Find the remainder when 2⁸¹ is divided by 17?
Answer:
2⁸¹ ≡ x(mod 17)
2⁴⁰ × 2⁴⁰ × 2¹ ≡ x(mod 17)
(2⁴)¹⁰ × (2⁴)¹⁰ × 2¹ ≡ x(mod 17)
(16)¹⁰ × (16)¹⁰ × 2¹ ≡ x(mod 17)
(16²)⁵ × (16²)⁵ × 2¹ ≡ x(mod 17)
= 1 × 1 × 2 (mod 17)
[(16)² = 256 = 1 (mod 17)]
= 2 (mod 17)
2⁸¹ = 2(mod 17)
∴ x = 2
The remainder is 2

Question 10.
The duration of flight travel from Chennai to London through British Airlines is approximately 11 hours. The airplane begins its journey on Sunday at 23:30 hours. If the time at Chennai is four and half hours ahead to that of London’s time, then find the time at London, when will the flight lands at London Airport?
Answer:
Duration of the flight time = 11 hours
(Chennai to London)
Starting time on Sunday = 23 : 30 hour

Time difference is 4 \(\frac { 1 }{ 2 } \) horns ahead to london
The time to reach London airport = (10.30 – 4.30)
= 6 am
The first reach the london airport next day (monday) at 6 am

Students can download Maths Chapter 1 Relations and Functions Ex 1.1 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 1 Relations and Functions Ex 1.1

1. Find A × B, A × A and B × A
(i) A = {2, -2, 3} and B = {1, -4}
(ii) A = B = {p, q}
(iii) A – {m, n} ; B = Φ
Answer:
(i) A = {2, -2, 3} and B = {1, -4}
A × B = {2,-2, 3} × {1,-4}
= {(2, 1), (2, -4)(-2, 1) (-2, -4) (3, 1) (3,-4)}
A × A = {2,-2, 3} × {2,-2, 3}
= {(2, 2)(2, -2)(2, 3)(-2, 2)
(-2, -2)(-2, 3X3,2) (3,-2) (3,3)}
B × A = {1,-4} × {2,-2, 3}
= {(1, 2)(1, -2)( 1, 3)(-4, 2) (-4,-2)(-4, 3)}

(ii) A = B = {p, q}
A × B = {p, q) × {p, q}
= {(p,p),(p,q)(q,p)(q,q)}
A × A = {p,q) × (p,q)
= {(p,p)(p,q)(q,p)(q,q)
B × A = {p,q} × {p,q}
= {(p,p)(p,q)(q,p)(q,q)

(iii) A = {m, n} × B = Φ
Note: B = Φ or {}
A × B = {m, n) × { }
= { )
A × A = {m, n) × (m, n)}
= {(m, m)(w, w)(n, m)(n, n)}
B × A = { } × {w, n}
= { }

Question 2.
Let A = {1, 2, 3} and B = {x | x is a prime number less than 10}. Find A × B and B × A.
Solution:
A = {1, 2, 3}, B = {2, 3, 5, 7}
A × B = {(1, 2), (1, 3), (1, 5), (1, 7), (2, 2), (2, 3) , (2, 5), (2, 7), (3, 2), (3, 3), (3, 5), (3, 7)}
B × A = {(2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3) , (5, 1), (5, 2), (5, 3), (7, 1), (7, 2) , (7, 3)}

Question 3.
If B × A = {(-2, 3),(-2, 4),(0, 3),(0, 4), (3,3) ,(3, 4)} find A and B.
Answer:
B × A = {(-2, 3)(-2, 4) (0, 3) (0, 4) (3, 3) (3,4)}
A = {3,4}
B = {-2,0,3}

Question 4.
If A ={5, 6}, B = {4, 5, 6} , C = {5, 6, 7}, Show that A × A = (B × B) ∩ (C × C).
Solution:
A = {5,6}, B = {4, 5, 6},C = {5, 6, 7}
A × A = {(5, 5), (5, 6), (6, 5), (6, 6)} ……….. (1)
B × B = {(4, 4), (4, 5), (4, 6), (5, 4),
(5, 5), (5, 6), (6, 4), (6, 5), (6, 6)} …(2)
C × C = {(5, 5), (5, 6), (5, 7), (6, 5), (6, 6),
(6, 7), (7, 5), (7, 6), (7, 7)} …(3)
(B × B) ∩ (C × C) = {(5, 5), (5,6), (6, 5), (6,6)} …(4)
(1) = (4)
A × A = (B × B) ∩ (C × C)
It is proved.

Question 5.
Given A = {1,2,3}, B = {2,3,5}, C = {3,4} and D = {1,3,5}, check if
(A ∩ C) × (B ∩ D) = (A × B) ∩ (C × D) is true?
Answer:
A = {1,2, 3}, B = {2, 3, 5}, C = {3,4} D = {1,3,5}
A ∩ c = {1,2,3} ∩ {3,4}
= (3}
B ∩ D = {2,3, 5} ∩ {1,3,5}
= {3,5}
(A ∩ C) × (B ∩ D) = {3} × {3,5}
= {(3, 3)(3, 5)} ….(1)
A × B = {1,2,3} × {2,3,5}
= {(1,2) (1,3) (1,5) (2, 2) (2, 3) (2, 5) (3, 2) (3, 3) (3, 5)}
C × D = {3,4} × {1,3,5}
= {(3,1) (3, 3) (3, 5) (4,1) (4, 3) (4, 5)}
(A × B) ∩ (C × D) = {(3, 3) (3, 5)} ….(2)
From (1) and (2) we get
(A ∩ C) × (B ∩ D) = (A × B) ∩ (C × D)
This is true.

Question 6.
Let A = {x ∈ W | x < 2}, B = {x ∈ N |1 < x < 4} and C = {3, 5}. Verify that
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
(iii) (A ∪ B) × C = (A × C) ∪ (B × C)
(iv) A × (B ∪ C) = (A × B) ∪ (A × C)
Solution:
A = {x ∈ W|x < 2} = {0,1}
B = {x ∈ N |1 < x < 4} = {2,3,4}
C = {3,5}
LHS =A × (B ∪ C)
B ∪ C = {2, 3, 4} ∪ {3, 5}
= {2, 3, 4, 5}
A × (B ∪ C) = {(0, 2), (0, 3), (0,4), (0, 5), (1, 2) , (1, 3), (1, 4),(1, 5)} ………. (1)
RHS = (A × B) ∪ (A × C)
(A × B) = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
(A × C) = {(0, 3), (0, 5), (1, 3), (1, 5)}
(A × B) ∪ (A × C)= {(0, 2), (0, 3), (0,4), (1, 2), (1, 3), (1, 4), (0, 5), (1, 5)} ….(2)
(1) = (2), LHS = RHS
Hence it is proved.

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
LHS = A × (B ∩ C)
(B ∩ C) = {3}
A × (B ∩ C) = {(0, 3), (1, 3)} …(1)
RHS = (A × B) ∩ (A × C)
(A × B) = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
(A × C) = {(0, 3), (0, 5), (1, 3), (1, 5)}
(A × B) ∩ (A × C) = {(0, 3), (1, 3)} ……….. (2)
(1) = (2) ⇒ LHS = RHS.
Hence it is verified.

(iii) (A ∪ B) × C = (A × C) ∪ (B × C)
LHS = (A ∪ B) × C
A ∪ B = {0, 1, 2, 3, 4}
(A ∪ B) × C = {(0, 3), (0, 5), (1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5), (4, 3), (4, 5)} …………. (1)
RHS = (A × C) ∪ (B × C)
(A × C) = {(0, 3), (0, 5), (1, 3), (1, 5)}
(B × C) = {(2, 3), (2, 5), (3, 3), (3, 5), (4, 3), (4, 5)}
(A × C) ∪ (B × C) = {(0, 3), (0, 5), (1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5), (4, 3), (4, 5)} ………… (2)
(1) = (2)
∴ LHS = RHS. Hence it is verified.

Question 7.
Let A = The set of all natural numbers less than 8, B = The set of all prime numbers less than 8, C = The set of even prime number. Verify that
(i) (A ∩ B) × C = (A × c) ∩ (B × C)
(ii) A × (B – C) = (A × B) – (A × C)
Answer:
A = {1,2, 3, 4, 5,6, 7}
B = {2, 3, 5,7}
C = {2}

(i) (A ∩ B) × C = (A × C) ∩ (B × C)
A ∩ B = {1, 2, 3, 4, 5, 6, 7} ∩ {2, 3, 5, 7}
= {2, 3, 5, 7}
(A ∩ B) × C = {2, 3, 5, 7} × {2}
= {(2, 2) (3, 2) (5, 2) (7, 2)} ….(1)
A × C = {1,2, 3, 4, 5, 6, 7} × {2}
= {(1,2) (2, 2) (3, 2) (4, 2)
(5.2) (6, 2) (7, 2)}
B × C = {2, 3, 5, 7} × {2}
= {(2, 2) (3, 2) (5, 2) (7, 2)}
(A × C) ∩ (B × C) = {(2, 2) (3, 2) (5, 2) (7, 2)} ….(2)
From (1) and (2) we get
(A ∩ B) × C = (A × C) ∩ (B × C)

(ii) A × (B – C) = (A × B) – (A × C)
B – C = {2, 3, 5, 7} – {2}
= {3,5,7}
A × (B – C) = {1,2, 3, 4, 5, 6,7} × {3,5,7}
= {(1, 3) (1, 5) (1, 7) (2, 3) (2, 5)
(2, 7) (3, 3) (3, 5) (3, 7) (4, 3)
(4, 5) (4, 7) (5, 3) (5, 5) (5, 7)
(6, 3) (6, 5) (6, 7) (7, 3) (7, 5) (7, 7)} ………….(1)
A × B = {1,2, 3, 4, 5, 6, 7} × {2, 3, 5,7}
= {(1, 2) (1, 3) (1, 5) (1, 7) (2, 2) (2, 3)
(2, 5) (2, 7) (3, 2) (3, 3) (3, 5) (3, 7)
(4, 2) (4, 3) (4, 5) (4, 7) (5, 2) (5, 3) (5, 5)
(5, 7) (6, 2) (6, 3) (6, 5) (6, 7)
(7, 2) (7, 3) (7, 5) (7, 7)}
A × C = {1,2, 3,4, 5, 6, 7} × {2}
= {(1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6.2) (7,2)}
(A × B) – (A × C) = {(1, 3) (1, 5) (1, 7)
(2, 3) (2, 5) (2, 7) (3, 3) (3, 5)
(3, 7) (4, 3) (4, 5) (4, 7) (5, 3) (5, 5)
(5, 7) (6, 3) (6, 5) (6, 7) (7, 3) (7, 5) (7, 7)} ….(2)
From (1) and (2) we get
A × (B – C) = (A × B) – (A × C)

Relations
Let A and B be any two non-empty sets. A “relation” R from A to B is a subset of A × B satisfying some specified conditions.

Note:

1. The domain of the relations R = {x ∈ A/xRy, for some y ∈ B}
2. The co-domain of the relation R is B
3. The range of the ralation

R = (y ∈ B/xRy for some x ∈ A}

Students can download Maths Chapter 2 Numbers and Sequences Ex 2.1 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 2.1

Question 1.
Find all positive integers which when divided by 3 leaves remainder 2.
Answer:
All the positive integers when divided by 3 leaves remainder 2
By Euclid’s division lemma
a = bq + r, 0 < r < b
a = 3q + r where 0 < q < 3
a leaves remainder 2 when divided by 3
∴ The positive integers are 2, 5, 8, 11,…

Question 2.
A man has 532 flower pots. He wants to arrange them in rows such that each row contains 21 flower pots. Find the number of completed rows and how many flower pots are left over.
Solution:
Using Euclid’s division algorithm,
a = 21q + r, we get 532 = 21 × 25 + 7.
The remainder is 7.
No. of completed rows = 25, left over flower pots = 7 pots.

Question 3.
Prove that the product of two consecutive positive integers is divisible by 2.
Answer:
Let n – 1 and n be two consecutive positive integers, then the product is n (n – 1)
n(n – 1) = n² – n
We know that any positive integer is of the form 2q or 2q + 1 for same integer q

Case 1:
when n = 2 q
n² – n = (2q)² – 2q
= 4q² – 2q
= 2q (2q – 1)
= 2 [q (2q – 1)]
n² – n = 2 r
r = q(2q – 1)
Hence n² – n. divisible by 2 for every positive integer.

Case 2:
when n = 2q + 1
n² – n = (2q + 1 )² – (2q + 1 )
= (2q + 1) [2q + 1 – 1]
= 2q (2q + 1)
n² – n = 2r
r = q (2q + 1)
n² – n divisible by 2 for every positive integer.

Question 4.
When the positive integers be a, b and c are divided by 13, the respective remainders are 9, 7 and 10. Show that a + b + c is divisible by 13.
Solution:
Let the positive integers be a, b, and c.
a = 13 q + 9
b = 13q + 1
c = 13 q + 10
a + b + c = 13q + 9 + 13q + 7 + 13q + 10
= 39q + 26
= 13 (3q + 2)
which is divisible by 13.

Question 5.
Prove that square of any integer leaves the remainder either 0 or 1 when divided by 4.
Answer:
Let the integer be ” x ”
The square of its integer is “x² ”
Let x be an even integer
x = 2q + 0
x2 = 4q²
When x is an odd integer
x = 2k + 1
x² = (2k + 1)²
= 4k² + 4k + 1
= 4k (k + 1) + 1
= 4q + 1 [q = k(k + 1)]
It is divisible by 4
Hence it is proved

Question 6.
Use Euclid’s Division Algorithm to find the Highest Common Factor (H.C.F) of
(i) 340 and 412
(ii) 867 and 255
(iii) 10224 and 9648
(iv) 84, 90 and 120
Solution:
To find the H.C.F. of 340 and 412. Using Euclid’s division algorithm.
We get 412 = 340 × 1 + 72
The remainder 72 ≠ 0
Again applying Euclid’s division algorithm
340 = 72 × 4 + 52
The remainder 52 ≠ 0.
Again applying Euclid’s division algorithm
72 = 52 × 1 + 20
The remainder 20 ≠ 0.
Again applying Euclid’s division algorithm,
52 = 20 × 2 + 12
The remainder 12 ≠ 0.
Again applying Euclid’s division algorithm.
20 = 12 × 1 + 8
The remainder 8 ≠ 0.
Again applying Euclid’s division algorithm
12 = 8 × 1 + 4
The remainder 4 ≠ 0.
Again applying Euclid’s division algorithm
8 = 4 × 2 + 0
The remainder is zero.
Therefore H.C.F. of 340 and 412 is 4.

(ii) To find the H.C.F. of 867 and 255, using Euclid’s division algorithm.
867 = 255 × 3 + 102
The remainder 102 ≠ 0.
Again using Euclid’s division algorithm
255 = 102 × 2 + 51
The remainder 51 ≠ 0.
Again using Euclid’s division algorithm
102 = 51 × 2 + 0
The remainder is zero.
Therefore the H.C.F. of 867 and 255 is 51.

(iii) To find H.C.F. 10224 and 9648. Using Euclid’s division algorithm.
10224 = 9648 × 1 + 576
The remainder 576 ≠ 0.
Again using Euclid’s division algorithm
9648 = 576 × 16 + 432
Remainder 432 ≠ 0.
Again applying Euclid’s division algorithm
576 = 432 × 1 + 144
Remainder 144 ≠ 0.
Again using Euclid’s division algorithm
432 = 144 × 3 + 0
The remainder is zero.
There H.C.F. of 10224 and 9648 is 144.

(iv) To find H.C.F. of 84, 90 and 120.
Using Euclid’s division algorithm
90 = 84 × 1 + 6
The remainder 6 ≠ 0.
Again using Euclid’s division algorithm
84 = 6 × 14 + 0
The remainder is zero.
∴ The H.C.F. of 84 and 90 is 6. To find the H.C.F. of 6 and 120 using Euclid’s division algorithm.
120 = 6 × 20 + 0
The remainder is zero.
Therefore H.C.F. of 120 and 6 is 6
∴ H.C.F. of 84, 90 and 120 is 6.

Question 7.
Find the largest number which divides 1230 and 1926 leaving remainder 12 in each case.
Answer:
Find the HCF of ( 1230 – 12) and (1926- 12)
i.e HCF of 1218 and 1914
By Euclid’s division algorithm
1914 = 1218 × 1 + 696
The remainder 696 ≠ 0
By Euclid’s division algorithm
1218 = 696 × 1 + 522
The remainder 522 ≠ 0
Again by Euclid’s division algorithm
696 = 522 × 1 + 174
The remainder 174 ≠ 0 Again by Euclid’s division algorithm
522 = 174 × 3 + 0
The remainder is zero
∴ HCF of 1218 and 1914 is 174
The largest value is 174

Question 8.
If d is the Highest Common Factor of 32 and 60, find x and y satisfying d = 32x + 60y.
Solution:
Applying Euclid’s divison lemma to 32 and 60, we get
60 = 32 × 1 + 28 ……………. (i)
The remainder is 28 ≠ 0.
Again applying division lemma
32 = 28 × 1 + 4 ……………. (ii)
The remainder 4 ≠ 0.
Again applying division lemma
28 = 4 × 7 + 0 ………….. (iii)
The remainder zero.
∴ H.C.F. of 32 and 60 is 4.
From (ii), we get
32 = 28 × 1 + 4
⇒ 4 = 32 – 28 × 1
⇒ 4 = 32 – (60 – 32 × 1) × 1
⇒ 4 = 32 – 60 + 32
⇒ 4 = 32 × 2+(-1) × 60
∴ x = 2 and y = -1

Question 9.
A positive integer, when divided by 88, gives the remainder 61. What will be the remainder when the same number is divided by 11?
Answer:
Let the positive integer be “x”
x = 88 × y + 61 (a = pq + r)
since 88 is a multiple of 11
61 = 11 × 5 + 6
∴ The remainder is 6

Question 10.
Prove that two consecutive positive integers are always coprime.
Solution:
Let the numbers be I, I + 1:
They are co-prime if only +ve integer that divides both is 1.
I is given to be +ve integer.
So I = 1, 2, 3, ….
∴ One is odd and the other one is even. Hence H.C.F. of the two consecutive numbers is 1. Hence the result.

Students can download Maths Chapter 1 Relations and Functions Additional 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 1 Relations and Functions Additional Questions

I. Multiple Choice Questions.

Question 1.
If n(A × B) = 15 and B = {1, 3, 7} then n(A) is ……………
(1) 3
(2) 5
(3) 1
(4) 15
Answer:
(2) 5
Hint: B(A × B) = 15
n(A) × n(B) = 15
n(A) × 3 = 15
n(A) = \(\frac { 15 }{ 3 } \) = 5

Question 2.
If A = {a, b,c) B = {b, d, e}
C = {a, e, i, o, u} then n [A ∩ C] × B] is
(1) 18
(2) 36
(3) 9
(4) 3
Answer:
(4) 3
Hint:
A ∩ C = {a,b,c} ∩ {a, e, i, o, u}
= {a}
n(A ∩ C) = 1
n[(A ∩ C) × B] = n(A ∩ C) × n(B)
= 1 × 3
= 3

Question 3.
If there are 28 relation from a set A = {2,4, 6, 8} to a set B, then the number of elements in B is ………………
(1) 7
(2) 14
(3) 5
(4) 4
Answer:
(1) 7
Hint: n(A) = 4
n(A × B) = 28
n(A) × n(B) = 28
4 × n(B) = 28
n(B) = \(\frac { 28 }{ 4 } \) = 7

Question 4.
The ordered pairs (a + 1, 4) (3, 4a + b) are equal then (a, b) is ………………..
(1) (4, 20)
(2) (20, 4)
(3) (-4, 20)
(4) (20, -4)
Answer:
(3) (-4, 20)
Hint: (a + 7, 4) = (3, 4a + b)
a + 7 = 3
a = 3 – 7
= – 4
4a + b = 4
4(-4) + b = 4
-16 + b = 4
b = 4 + 16 = 20
The pair (a, 6) is (-4, 20)

Question 5.
The range of the relation R = {(x, x³) / x} is a prime number less than 13} is …………………
(1) (2,3,5,7,11)
(2) (4,9,25,49,121)
(3) (8,27, 125,343, 1331)
(4) (1,8,27, 125,343, 1331)
Answer:
(3) (8, 27, 125,343, 1331)
Hint: x = {2, 3, 5, 7, 11}
Range (x³) = {8, 27, 125, 343, 1331}

Question 6.
If {( x, 2), (4, y) } represents an identity function, then (x, y) is
(1) (2, 4)
(2) (4, 2)
(3) (2, 2)
(4) (4, 4)
Answer:
(1) (2, 4)
Hint: In an identity function each element is associated with itself.

Question 7.
If {(7, 11), (5, a)} represents a constant
function, then the value of ‘a’ is
(1) 7
(2) 11
(3) 5
(4) 9
Answer:
(2) 11
Hint: All the images are same in a constant function.

Question 8.
Given f(x) = (- 1)^x is a function from N to Z. Then the range of f is
(1) {1}
(2) N
(3) { 1,- 1 }
(4) Z
Answer:
(3) {1, – 1}
Hint: f(x) = (- 1)^x = ± 1

Question 9.
If f = { (6, 3), (8, 9), (5, 3), (-1, 6) }, then the pre-images of 3 are
(1) 5 and-1
(2) 6 and 8
(3) 8 and-1
(4) 6 and 5
Answer:
(4) 6 and 5.
Hint: The Pre images of 3 are 6 and 5

Question 10.
Let A= { 1, 3, 4, 7, 11 }, B = {-1, 1, 2, 5, 7, 9 } and f : A → B be given by
f = {(1, -1), (3,2), (4, 1), (7, 5), (11, 9)}.
Then f is ………………….
(1) one-one
(2) onto
(3) bijective
(4) not a function
Answer:
(1) one – one
Hint:

Question 11.
The given diagram represents
(1) an onto function
(2) a constant function
(3) an one-one function
(4) not a function

Answer:
(4) not a function
Hint: 2 has two images 4 and 2.
It is not a function.

Question 12.
If A = { 5, 6, 7 }, B = { 1, 2, 3, 4, 5 }and f: A → B is defined by f(x) = x – 2, then the range of f is …………….
(1) {1,4, 5}
(2) {1,2, 3, 4, 5}
(3) { 2, 3, 4 }
(4) { 3, 4, 5 }
Answer:
(4) {3, 4, 5}
Hint: f(x) = x – 2
f(5) = 5 – 2 = 3
f(6) = 6 – 2 = 4
f(7) = 7 – 2 = 5
Range of f = {3, 4, 5}

Question 13.
If f(x) = x² + 5, then f(-4) = ………
(1) 26
(2) 21
(3) 20
(4) – 20
Answer:
(2) 21
Hint: f(x) = x² + 5
f(- 4) = (-4)² + 5 = 16 + 5 = 21

Question 14.
If the range of a function is a singleton set, then it is ……………..
(1) a constant function
(2) an identity function
(3) a bijective function
(4) an one-one function
Answer:
(1) a constant function
Hint: Every element of the first set has same image in the second set.

Question 15.
If f : A → B is a bijective function and if n(A) = 5 , then n(B) is equal to ………………
(1) 10
(2) 4
(3) 5
(4) 25
Answer:
(3) 5
Hint: If A and B are Bijective (one-one and onto) function then n (A) = n (B)

Question 16.
If f: R → R defined by f(x) = 3x – 6 and g : R → R defined by g(x) = 3x + k if fog – gof then the value of k is …………………..
(1) – 5
(2) 5
(3) 6
(4) -6
Answer:
(4) – 6
Hint: f(x) = 3x – 6 ;g(x) = 3x + k
fog = f[g(x)]
= f(3x + k)
= 3 (3x + k) – 6
= 9x + 3k – 6
g o f = g[f(x)]
= g(3x – 6)
= 3(3x – 6 ) + k
= 9x – 18 + k
But fog = gof
9x + 3k – 6 = 9x – 18 + k
3k – k = -18 + 6
2k = -12
k = \(\frac { -12 }{ 2 } \) = -6

Question 17.
If f(x) = x² – x then f (x – 1) – f(x + 1) is ……………….
(1) 4x
(2) 4x + 2
(3) 2 – 4x
(4) 4x – 2
Answer:
(3) 2 – 4x
Hint: f(x – 1) = (x – 1)² – (x – 1)
= x² – 2x + 1 – x + 1
= x² – 3x + 2
f(x + 1) = (x + 1)² – (x + 1)
= x² + 2x + 1 – x – 1
= x² + x
f(x – 1) – f(x + 1) = x² – 3x + 2 – (x2 + x)
= x² – 3x + 2 – x² – x
= -4x + 2 = 2 – 4x

Question 18.
If K(x) = 3x – 9 then L (x) = 7x – 10 then LOK is ……………..
(1) 21x + 73
(2) – 21x + 73
(3) 21x – 73
(4) 22x – 73
Answer:
(3) 21x – 73
Hint: K (x) = 3x – 9 ; L(x) = 7x – 10
LOK = L[K(x)]
= L (3x – 9)
= 7(3x – 9) – 10
= 21x – 63 – 10
= 21x – 73

Question 19.
Composition of function is ……………..
(1) commutative
(2) associative
(3) commutative and associative
(4) not associative
Answer:
(2) associative

Question 20.
A comet is heading for Jupiter with acceleration a = 50 kms^-2. The velocity of the comet at time ”t” is given by f(t) = at² – at + 1. Then the velocity at time t = 5 seconds is …………..
(1) 900kms^-1
(2) 1001 kms^-1
(3) 2001 kms^-1
(4) 50 kms^-1
Answer:
(2) 1001 kms^-1
Hint: f(t) = at² – at + 1
m = 50(5)² – 50(5) + 1
= 1250 – 250 + 1
= 1001 kms^-1

II. Answer the following questions.

Question 1.
f(x) = (1 + x)
f(x) = (2x – 1)
Show that fo(g(x)) = gof(x)
Solution:
f(x) = 1 + x
g(x) = (2x – 1)
fog(x) = f(g(x)) = f(2x – 1)
= 1 + 2x – 1 = 2x ………….. (1)
gof(x) = g(f(x)) = g(1 + x) = 2(1 + x) = 1
= 2 + 2x – 1
= 2x + 1 ……………. (2)
(1) ≠ (2)
∴ fog(x) + gof(x) It is verified.

Question 2.
If A × B = {(a, x) (a, y) (b, x) (b, y) (c, x) (c, y)} then find A and B
Answer:
A × B = {(a, x) (a, y) (b, x) (b, y) (c, x) (c, y)}
A = {a, b, c}
B = {x,y}

Question 3.
Let A = {x ∈ w/3 < x < 7},
B = {x ∈ N/0 < x < 3}, C = {x ∈ w/x < 2}
verify A × (B ∩ C) = (A × B) ∩ (A × C)
Answer:
A = {4,5,6} ; B = {1,2} C = {0, 1}
B ∩ C = {1,2} ∩ {0, 1}
= {1}
A × (B ∩ C) = {4,5,6} × {1}
= {(4, 1) (5, 1) (6, 1)} …. (1)
A × B = {4,5,6} × {1,2}
= {(4, 1) (4, 2) (5, 1) (5, 2) (6, 1) (6, 2)}
A × C = {4,5,6} x {0, 1}
= {(4,0) (4,1) (5,0)
(5, 1) (6, 0) (6, 1)}
(A × B) ∩ (A × C) = {(4, 1) (5, 1) (6, 1)}…. (2)
From (1) and (2) we get
A × (B ∩ C) = (A × B) ∩ (A × C)

Question 4.
Let A = {10, 11, 12, 13, 14}; B = {0, 1, 2, 3, 5} and f_i: A → B, i = 1, 2, 3. State the type of function for the following (give reason):
(i) f₁ = {(10,1), (11,2), (12,3), (13,5), (14,3)}
Answer:
The element 12 and 14 in A have same image 3 in B.

∴ It is not one-one function.
The element ‘0’ in B has no preimage in A
∴ It is not onto function
So the given function is neither one – one nor onto function.

(ii) f₂ = {(10,1), (11,1), (12,1), (13,1), (14,1)}
Answer:
f₂ is a constant function

(iii) f₃ = {(10,0), (11,1), (12,2), (13,3), (14,5)}
Answer:
f₃ is one-one and onto function (or) bijective function.

Question 5.
If X = {1, 2, 3, 4, 5}, Y = {1, 3, 5, 7, 9} determine which of the following relations from X to Y are functions? Give reason for your answer. If it is a function, state its type.
(i) R1 = {(x,y)| y = x + 2,x ∈ X,y ∈ Y}
Answer:
Given y = x + 2
When x = 1 ; y = 1 + 2 = 3
When x = 2 ; y = 2 + 2 = 4
When x = 3 ; y = 3 + 2 = 5
When x = 4 ; y = 4 + 2 = 6
When x = 5 ; y = 5 + 2 = 7

R₁ = {1,3), (2,4), (3, 5), (4, 6), (5,7)}
R₁ is not a function ; 2 and 4 has no image in Y.

(ii) R₂ = {(1,1), (2,1), (3,3), (4,3), (5,5)}
Answer:
R₂ is a function.
Every element of X has unique image in Y.
1 and 2 have same image 1
3 and 4 have same image 3
It is not one – one function …. (1)
7 and 9 has no pre image in X
It is not an onto function …. (2)

From (1) and (2) we know that, it is
neither one – one nor onto function.

(iii) R₃ = {(1,1), (1,3), (3,5), (3,7), (5,7)}
Answer:
R₃ is not a function.
1 has two images 1 and 3
3 has two images 5 and 7

(iv) R₄ = {(1,3), (2,5), (4,7), (5,9), (3,1)}
Answer:
Every element of X has unique image in
Y. Range = Co-domain
R₄ is a function.

It is an one-one and onto function (or) bijective function

Question 6.
A= {-2,-1, 1, 2} and f = {(x,\(\frac { 1 }{ x } \)) ; x ∈ A}
Write down the range of f. Is f a function from A to A?
Answer:
Given, f = (x,\(\frac { 1 }{ x } \)) ; So f(x) = \(\frac { 1 }{ x } \)
f (-2) = \(\frac { 1 }{ -2 } \) = – \(\frac { 1 }{ 2 } \) ; f(-1) = \(\frac { 1 }{ -1 } \) = -1
f(1) = \(\frac { 1 }{ 1 } \) = 1 ; f(2) = \(\frac { 1 }{ 2 } \) = \(\frac { 1 }{ 2 } \)
Range of f = {\(\frac { -1 }{ 2 } \), -1, 1, \(\frac { 1 }{ 2 } \)}
It is not a function from A to A since – \(\frac { 1 }{ 2 } \) ,\(\frac { 1 }{ 2 } \) ∈ A

Question 7.
Let A = {1, 2, 3, 4, 5}, B = N and f: A → B be defined by f(x) = x². Find the range of f. Identify the type of function.
Solution:
A = {1, 2, 3, 4, 5}
B = {1, 2, 3, 4 ….}
f: A → B, f(x) = x²
∴ f(1) = 1² = 1
f(2) = 2² = 4
f(3) = 3² = 9
f(4) = 4² = 16
f(5) = 5² = 25
∴ Range of f = {1, 4, 9, 16, 25)
Elements in A have been different elements in B. Therefore it is one-one function. But not all the elements in B have preimages in A. Therefore it is not on-to function.

Question 8.
Let A = { 1, 2, 3, 4, 5 }, B = N and f: A → B be defined by f(x) = x².
Find the range of f. Identify the type of function.
Answer:
Now, A = { 1, 2, 3, 4, 5 };
B = { 1, 2, 3, 4, … }
Given f: A → B and f(x) = x²
f(1) = 1² = 1;
f(2) = 4;
f(3) = 9;
f(4) = 16;
f(5) = 25.
Range of f = {1, 4, 9, 16, 25}
Since distinct elements are mapped into distinct images, it is a one-one function.
However, the function is not onto, since 3 ∈ B but there is no x ∈ A
such that
f(x) = x² = 3.

Question 9.
Identify the type of function.


Answer:
(i) Many – one into
(ii) One – one onto
(iii) Constant function
(iv) One – one into

Question 10.
Find the domain and range of the following
(i) f = {(1, 2), (2, 3), (3, 4), (4, 5) (5, 6)}
(ii) R = {(-2, 4), (-1,1), (2,4), (1,1) (-3, 9)}
Answer:
(i) f = {( 1,2), (2, 3), (3, 4), (4, 5) (5, 6)}
Domain = {1,2, 3,4, 5}
Range = {2, 3, 4, 5, 6}

(ii) R = {(-2,4), (-1, 1),(2,4), (1,1) (-3,9)}
Domain = {-2, -1,2, 1,-3} (or)
= {-3,-2,-1, 1,2}
Range = {4, 1, 9} (or) {1, 4, 9}

Question 11.
Given P ={-2,-1, 0,1}
Q = {1,-2, 6,-3}
R = {x,y/y = x² – 3 x ∈ P,y ∈ Q}
(i) List the elements of R
(ii) Is the relation a function? If so identity the function
Answer:
P = {-2, -1, 0, 1}; Q = {1, -2,6, -3}
y = x2- 3 x ∈ P, y ∈ Q
When x = -2 ⇒ y = (-2)² – 3 = 4 – 3 = 1
When x = -1 ⇒ y = (-1 )² – 3 = 1 – 3 = -2
When x = 0 ⇒ y = (0)² – 3 = 0 -3 = -3
When x = 1 ⇒ y = 1² – 3 = 1 – 3 = -2
(i) R = {(-2,1), (-1,-2), (0,-3), (1,-2)}
(ii) Yes the relation is a function many – one into function.

Question 12.
Given f(x) = 3x – 2; g(x) = 2x² find
(i) fog and
(ii) gof what do you find
Answer:
f(x) = 3x – 2 ; g(x) = 2x²
(i) f o g = f[g(x)]
= f(2x²)
= 3(2x²) – 2
= 6x² – 2

(ii) g o f = g [f(x)]
= g (3x – 2)
= 2(3x – 2)²
= 2(9x² + 4 – 12x)
= 18x² – 24x + 8
we find that fog ≠ gof
Composition of function is not commutative.

Question 13.
If f: R → R is defined by f(x) = ax + 3 and g: R → R is defined by g (x) = 4x – 3 find a so that fog = gof
Answer:
f(x) = ax + 3 ; g(x) = 4x -3
fog = f[g(x)]
= f(4x – 3)
= a (4x – 3) + 3
= 4ax – 3a + 3
gof = g [f(x)]
= g (ax + 3)
= 4 (ax + 3) – 3
= 4 ax + 12 – 3
= 4ax + 9
But fog = gof
4ax – 3a + 3 = 4ax + 9
-3a + 3 = 9
– 3a = 6
a = – 2
The value of a = – 2

Question 14.
Given f(x) = 3 + x ; g(x) = x² ;
h(x) = \(\frac { 1 }{ x } \) find fo (goh)
Answer:
f(x) = 3 + x ; g (x) = x², h(x) = \(\frac { 1 }{ x } \)
goh = g[h(x)]
= g (\(\frac { 1 }{ x } \))
= (\(\frac { 1 }{ x } \))²
goh = \(\frac{1}{x^{2}}\)
fo(goh) = f (\(\frac{1}{x^{2}}\))
= 3 + \(\frac{1}{x^{2}}\)

Question 15.
If f(x) = x + 3 where A = {4, 6, 8,10} B = {7, 9,11,13} and f: A → B
(i) Draw the arrow diagram
(ii) Why type of function is f.
Answer:
A= {4, 6, 8, 10}
f(x) = x + 3
f(4) = 4 + 3 = 7
f(6) = 6 + 3 = 9
f(8) = 8 + 3 = 11
f(10) = 10 + 3 = 13

(ii) one – one onto function

III. Answer the following Questions

Question 1.
Given A = {2,3, 5}, B = {1,2,3}
C = {2, 5}, D = {2,3, 5} check if
(A ∩ C) × (B ∩ D) = (A × B) ∩ (C × D)
Answer:
A ∩ C = {2, 3, 5} ∩ (2, 5}
= (2,5}
B ∩ D = {1,2,3} ∩ {2,3,5}
= {2,3}
(A ∩ C) × (B ∩ D) = {2, 5} × {2, 3}
= {(2, 2) (2, 3) (5, 2) (5, 3)} …. (1)
A × B = {2,3,5} × {1,2,3}
= {(2,1) (2, 2) (2, 3)
(3, 1) (3, 2) (3, 3)
(5, 1) (5, 2) (5, 3)}
C × D = {2, 5} × {2, 3, 5}
= {(2, 2) (2, 3) (2, 5) (5, 2) (5, 3) (5, 5)}
(A × B) ∩ (C × D) = {(2,2) (2, 3) (5, 2) (5, 3)} …. (2)
From (1) and (2) we get
(A ∩ C) × (B ∩ D) = (A × B) ∩ (C × D)

Question 2.
Study the relation given below an set- builder form. Represent each of them by
(a) an arrow diagram
(b) a graph
(c) a set in roster.
If {{x,y}/y = 2x + 1; x < 10 and y < 12 x ∈ N, y ∈ N}
Answer:
y = 2x + 1
when x = 1 ⇒ y = 2(1) + 1 = 2 + 1 = 3
when x = 2 ⇒ y = 2(2) + 1 = 4 + 1 = 5
when x = 3 ⇒ y = 2(3) + 1 = 6 + 1 = 7
when x = 4 ⇒ y = 2(4) + 1 = 8 + 1 = 9
when x = 5 ⇒ y = 2(5) + 1 = 10 + 1 = 11
f = {(1,3) (2, 5) (3, 7) (4, 9) (5, 11)}

(a) Arrow diagram

(b) A graph

(c) Roster form: R = {(1,3) (2,5) (3,7) (4,9) (5,11)}

Question 3.
State whether the following graphs represents a function. Give reason for your answer.





Answer:
(i) The given graph represents a function. The vertical line cuts the graph at most one point R

(ii) The vertical line cuts the graph at most one point Q. The given graph represents a function.

(iii) The vertical line cuts the graph at A and B. The given graph does not represents a function.

(iv) The vertical line cuts the graph at A and B. The given graph does not represents a function.

(v) The vertical line cuts the graph at most one point R. The given graph represents a function.

Question 4.
Let A = {6, 9,15,18, 21}; B = {1, 2, 4, 5, 6} and f: A → B be defined by f(x) = \(\frac { x-3 }{ 3 } \) Represent f by, (i) an arrow diagram, (ii) a set of ordered pairs, (iii) a table, (iv) a graph.
Given, A = {6, 9, 15, 18, 21}, B = {1, 2, 4, 5, 6}
f(x) = \(\frac { x-3 }{ 3 } \)
f(6) = \(\frac { 6-3 }{ 3 } \) = \(\frac { 3 }{ 3 } \) = 1
f(9) = \(\frac { 9-3 }{ 3 } \) = \(\frac { 6 }{ 3 } \) = 2
f(15) = \(\frac { 15-3 }{ 3 } \) = \(\frac { 12 }{ 3 } \) = 4
f(18) = \(\frac { 18-3 }{ 3 } \) = \(\frac { 12 }{ 3 } \) = 4
f(18) = \(\frac { 18-3 }{ 3 } \) = \(\frac { 15 }{ 3 } \) = 5
f(21) = \(\frac { 21-3 }{ 3 } \) = \(\frac { 18 }{ 3 } \) = 6

(i) an arrow diagram

(ii) a set of ordered pairs
f = {(6,1), (9, 2), (15, 4), (18, 5), (21,6)}

(iii) a table

(iv) a graph

Question 5.
Let A = {4,6,8,10} and B = {3,4,5,6,7}. If f: A → B is defined by f(x) = \(\frac { 1 }{ 2 } \) x + 1 then represent f by (i) an arrow diagram, (ii) a set of ordered pairs and, (iii) a table.
Answer:
Given, A = {4, 6, 8, 10}
B = {3, 4, 5, 6, 7}
f(x) = \(\frac { x }{ 2 } \) + 1
f(4) = \(\frac { 4 }{ 2 } \) + 1 = 2 + 1 = 3
f(6) = \(\frac { 6 }{ 2 } \) + 1 = 3 + 1 = 4
f(8) = \(\frac { 8 }{ 2 } \) + 1 = 4 + 1 = 5
f(10) = \(\frac { 10 }{ 2 } \) + 1 = 5 + 1 = 6

(i) an arrow diagram

(ii) a set of ordered pairs
f = {(4, 3), (6, 4), (8, 5), (10, 6)}

(iii) a table

Question 6.
A function f[- 3, 7 ) → R is defined as follows f(x) =

Find (i) f(5) + f(6)
(ii) f(1) – f(-3)
(iii) f(-2) – f(4)
(iv) \(\frac{f(3)+f(-1)}{2 f(6)-f(1)}\)
Answer:
Given, f(x) = 4x² – 1; x = {-3, -2, -1, 0, 1}
f(x) = 3x – 2; x = {2,3,4}
f(x) = 2x – 3; x = {5,6}
(i) f(5) + f(6)
f(x) = 2x – 3
f(5) = 2(5) – 3 = 10 – 3 = 7
f(6) = 2(6) – 3 = 12 – 3 = 9
∴ f(5) + f(6) = 7 + 9 = 16

(ii) f(1) – f(-3)
f(x) = 4x² – 1
f(1) = 4(1)² – 1 = 4 – 1 = 3
f(-3) = [4(-3)² – 1]
= 4 (9) – 1
= 36 – 1 = 35
∴ f(1) – f(-3) = 3 – (35) = -32

(iii) f(-2) – f(4)
f(x) = 4x² – 1
f(-2) = 4(-2)² – 1 = 4(4) – 1 = 16 – 1 = 15
f(x) = 3x – 2
f(4) = [3(4) – 2] = 12 – 2 = 10
∴ f(-2) – f(4) = 15 – 10 = 5

Question 7.
A function f : [- 7, 6) → R is defined as follows


Answer:
Given, f(x) = x² + 2x + 1 ; x = {-7, -6}
f(x) = x + 5 ; x = {-5, -4, -3, -2, -1, 0, 1, 2}
f(x) = x – 1; x{3, 4, 5}

(i) 2f(- 4) + 3f(2)
f(x) = x + 5
f(-4) = -4 + 5 = 1
f(2) = 2 + 5 = 7
∴ 2f(-4) + 3 f(2) = 2(1) + 3(7) = 2 + 21 = 23

(ii) f(-7) – f(-3)
f(x) = x² + 2x + 1
f(-7) = (-7)² + 2(-7) + 1 = 49 – 14 + 1 = 36
f(x) = x + 5
f(-3) = -3 + 5 = 2
∴ f(-7) – f(-3) = 36 – 2 = 34

(iii) \(\frac{4 f(-3)+2 f(4)}{f(-6)-3 f(1)}\)
f(x) = x + 5
f(-3) = -3 + 5 = 2
f(x) = x – 1
f(4) = 4 – 1 = 3
f(x) = x² + 2x + 1
f(-6) = (-6)² + 2(-6) + 1 = 36 – 12 + 1 = 25
f(x) = x + 5
f(1) = 1 + 5 = 6

Question 8.
Let A= { 0,1, 2, 3 } and B = {1, 3, 5, 7, 9 } be two sets. Let f: A → B be a function given by f (x) = 2x + 1. Represent this function as
(i) a set of ordered pairs
(ii) a table
(iii) an arrow diagram and
(iv) a graph.
Answer:
A = {0, 1, 2, 3}, B = { 1, 3, 5, 7, 9 },f(x) = 2x + 1
f(0) = 2(0) + 1 = 1, f(1) = 2(1) + 1 = 3 ,f(2) = 2(2) + 1 = 5, f(3) = 2(3) + 1 = 7

(i) Set of ordered pairs
The given function/can be represented as a set of ordered pairs as
f = {(0, 1), (1, 3), (2, 5), (3,7)}

(ii) Table form
Let us represent f using a table as shown below.

(iii) Arrow Diagram
Let us represent f by an arrow diagram.
We draw two closed curves to represent the sets A and B. Here each element of A and its unique image element in B are related with an arrow.

(iv) Graph
We are given that
f = {(x,f(x)) | x ∈ A} = {(0,1), (1, 3), (2, 5), (3, 7)}. Now, the points (0, 1), (1, 3), (2, 5) and (3, 7) are plotted on the plane as shown below.
The totality of all points represent the graph of the function.

Question 9.
A. function f: [1, 6) → R is defined as follows

(Here, [1, 6) = {x ∈ R : 1 ≤ x < 6})
Find the value of
(i) f(5)
(ii) f(3)
(iii) f(1)
(iv) f(2) – f(4)
(v) 2f(5) – 3f(1).
Answer:
(i) Let us find f(5). Since 5 lies between 4 and 6, we have to use f(x) = 3x² – 10.
Thus, f(5) = 3(5²) – 10 = 65.

(ii) To find f(3), note that 3 lies between 2 and 4.
So, we use f(x) = 2x – 1 to calculate f(3).
Thus, f(3) = 2(3) – 1 = 5.

(iii) Let us find f(1).
Now, 1 is in the interval 1 < x < 2
Thus, we have to use f(x) = 1 + x to obtain f(1) = 1 + 1 = 2.

(iv) f (2) – f(4)
Now, 2 is in the interval 2 < x < 4 and so, we use f(x) = 2x – 1.
Thus, f(2) = 2(2) -1 = 3.
Also, 4 is in the interval 4 < x < 6. Thus, we use f(x) = 3x² – 10
Therefore, f(4) = 3(4²) – 10 = 3(16) – 10 = 48 – 10 = 38.
Hence, f(2) – f(4) = 3 – 38 = -35.

(v) To calculate 2 f (5) – 3f (1), we shall make use of the values that we have already calculated in (i) and (iii). Thus, 2f(5) – 3f(1) = 2(65) – 3(2) = 130 – 6 – 124.

Question 10.
Given f(x) = 5x + 2; g(x) = 2x – 3;
h(x) = 3x + 1. Verify fo (goh) = (fog) oh
Answer:
f(x) = 5x + 2 ; g(x) = 2x – 3; h(x) = 3x + 1
L.H.S. = fo (goh)
goh = g[h(x)]
= g(3x + 1)
= 2(3x + 1) – 3
= 6x – 1
fo (goh) = f[goh (x)]
= f(6x – 1)
= 5 (6x – 1) + 2
= 30 x – 5 + 2
fo (goh) = 30x – 3 ….(1)
R.H.S. = (fog) oh
fog = f[g(x)]
= f(2x – 3)
= 5(2x – 3) + 2
= 30x – 5 + 2
fo (goh) = 30x – 3 …..(1)
R.H.S. = (fog) oh
fog = f[g(x)]
= f(2x – 3)
= 5 (2x – 3)
= 5 (2x – 3) + 2
= 10x – 15 + 2
= 10x – 13
(fog) oh = fog [h(x)]
= fog (3x + 1)
= 10 (3x + 1) – 13
= 30x + 10 – 13
= 30x – 3 ….(2)
From (1) and (2) we get L.H.S. = R.H.S.
fo(goh) = (fog) oh

Question 11.
Given f(x) = x² + 4; g(x) = 3x – 2;
h(x) = x – 5. Show that the composition of functions is associative.
Answer:
f(x) = x² + 4 ; g(x) – 3x – 2; h(x) = x – 5
To prove fo (goh) = (fog) oh
L.H.S. fo (goh)
goh = g[h(x)]
= g(x – 5)
= 3(x – 5) – 2
= 3x – 15 – 2
goh = 3x – 17
fo (goh) = f [goh (x)]
= f(3x – 17)
= (3x – 17)² + 4
= 9x² + 289 – 102 x + 4
= 9x² – 102x + 293 ….(1)
R.H.S. = (fog) oh
fog – f[g(x)]
= f(3x-2)
= (3x – 2)² + 4
= 9×2 + 4 – 12x + 4
= 9×2 – 12x + 8
(fog) oh = fog [h(x)]
= fog (x – 5)
= 9(x – 5)² – 12 (x – 5) + 8
= 9(x² + 25 – 10x) – 12x + 60 + 8
= 9x² + 225 – 90x – 12x + 60 + 8
= 9x² – 102x + 293 ….(2)
From (1) and (2) we get fo (goh) = (fog) oh.
Composition of function is associative

Question 12.
Given f(x) = x – 2; g(x) = 3x + 5; h(x) = 2x – 3. Verify that (goh) of = go (hof)
Answer:
f(x) = x – 2 ; g(x) = 3x + 5; h(x) = 2x – 3
L.H.S. (goh) of
goh = g[h(x)]
= g(2x – 3)
= 3(2x – 3) + 5
= 6x – 9 + 5
= 6x – 4
(goh) of = goh [f(x)]
= goh (x – 2)
= 6(x – 2) – 4
= 6x – 12 – 4
= 6x – 16 ….(1)

R.H.S. go(hof)
hof = h[f(x)]
= h(x- 2)
= 2(x – 2) – 3
= 2x – 4 – 3
= 2x – 7
go(hof) = g [hof (x)]
= g (2x – 7)
= 3(2x – 7) + 5
= 6x – 21 + 5
= 6x – 16 ….(2)
From (1) and (2) we get (goh) of = go(hof)

Tamilnadu State Board New Syllabus Samacheer Kalvi 10th English Guide Pdf Supplementary Chapter 2 Zigzag Questions and Answers, Zigzag Supplementary Summary, Notes.

Tamilnadu Samacheer Kalvi 10th English Solutions Supplementary Chapter 2 Zigzag

10th English Guide Supplementary Zigzag Textbook Questions and Answers

A. Identify the speaker / character.

Question 1.
Even though I clearly said “no”!
Answer:
Dr. Krishnan

Question 2.
The one that spits deadly poison straight into its opponent’s eyes.
Answer:
Maya

Question 3.
Remember the tiny penknife he gave me last year.
Answer:
Maya

Question 4.
It’s Somu’s thoughtless ways that reduce me to tears.
Answer:
Mrs. Krishnan

Question 5.
‘Come in, zigzag, come in dear!’
Answer:
Visu, the old cook

B. Read the story again and write how these characters reacted in these situations.

Question 1.
“You’re both quite mistaken”.
Answer:
Dr. Krishnan hastened to explain
Mrs. Krishnan was horrified on hearing about zigzag.

Question 2.
“It’s Somu’s thoughtless ways that reduce me to tears”.
Answer:
Mrs. Krishnan spoke irritably.
Dr. Krishnan was hurrying to his clinic.

Question 3.
“Just wait till zigzag settles down in this new home”.
Answer:
Visu comforted everyone.
Aravind and Maya were disappointed as the bird didn’t talk.

Question 4.
“Zigzag hardly ever sleeps”.
Answer:
Somu sent an e-mail to Dr.Krishnan about Zigzag.
Dr.Krishnan predicted it as ridiculously simple.

Question 5.
“You are an absolute treasure…”
Answer:
Dr. Krishnan sighed and spoke to zigzag Zigzag didn’t bother to reply.

C. Complete the given tabular columns.

Answer:

D. Answer the following questions in one or two sentences.

Question 1.
Why did Dr. Ashok’s cousin call him?
Answer:
Dr. Ashok Krishnans cousin Somu called him to shelter Zigzag, his pet bird, when he left for Alaska.

Question 2.
Mention atleast …. zigzag at home?
Answer:

1. I’m going crazy with the sound of zigzag snoring, plus all these telephone calls.
2. “And my beautiful painting…”

Question 3.
What are pets did Somu have?
Answer:
Somu had a giant green and gold fighting beetle and an African snake.

Question 4.
What was Mrs. Krishnan busy with?
Answer:
Mrs. Krishnan was busy with her paintings to be displayed for sale the following week.

Question 5.
What commotion did the boomerang cause in the neighbourhood?
Answer:
Boomerang sliced through all the TV aerials in the neighborhood. It caused permanent damage to several cars in the parking lot. It also knocked out their watchman cold with the force thrown by Arvind.

Question 6.
What happened when Somu left zigzag with the Krishnan’s?
Answer:
Zigzag deposited the walnuts and fruits to the Chandelier and transferred them all the blades of the ceiling fan. Then it perched on the curtain rod and went off to sleep.

Question 7.
How did zigzag communicate with the Krishnan’s?
Answer:
Zigzag did not communicate with the Krishnan’s although everyone tried several times and in several languages to speak to him, he only slept and snored.

Question 8.
What was the e-mail message sent to Somu by Dr. Krishnan?
Answer:
Dr. Krishnan sent an email to Somu, asking for instructions on how to stop zigzag from snoring.

Question 9.
What did Aravind confess?
Answer:
Arvind confessed that for the first time in his life, he was actually looking forward to go to school. The school was as calm as a monastery compared to their house.

Question 10.
Why did Mrs. Jhunjhunwala buy the painting?
Answer:
Mrs. jhunjhunwala bought the painting because she liked the new technique of- painting.

E. Answer the following questions in about 100-150 words:

Question 1.
Write a passage in your own words the various commotions caused by zigzag at Dr. Krishnan’s
Answer:
When Visu, the old cook of Dr. Somu brought Zigzag to Dr. Krishnan’s house, all were surprised to see such a strange and weird bird. It was a foot and a half tall. It was a multilingual talking- singing bird. It could talk and sing in twenty-one languages.

When it refused to say a word, Arvind brought some juicy fruits and nuts for the bird. It did not eat them but transferred one ‘ by one to the chandelier and on to the blades of the ceiling fan. Then it perched on a curtain rod, it sank his beak into plumpy guava and then went off to sleep. Then he began to snore. The snore became louder and very louder.

His snoring pounded their eardrums till their heads ached. When the fan was switched on by their maid, the painting of Mrs. Krishnan got spoilt by streaks of orange pulp spreading on it. His snoring troubled all the neighbors and they were complaining about it.

(OR)

Title: Zigzag
Author: Asha Nehemiah
Characters: Zigzag, Dr. Krishnan, Mrs. Krishnan, Dr. Somu, Aravind, the old-cook Visu and Lakshmi.
Theme: Things are not always as we think.

Dr. Somu left for Alaska. He had a pet bird. He sent it to Dr. Krishnan’s house through his old cook Visu. His name was Zigzag. All were surprised to see the strange bird. He was one and a half foot tall. He could talk and sing in 21 languages. But he did not speak a word. Aravind brought juicy fruits and nuts for zigzag.

He transferred them to the chandelier and on to the fan plates. He perched on the curtain rod and started sleeping and snoring. The snore became louder. It pounded their eardrums causing headaches. When the fan was switched on, the streaks of orange pulp spread on Mrs. Krishnan’s painting. It was spoilt. His snoring troubled all the neighbours and they were complaining about it.

(OR)

1. The cook of Dr. Somu brought Zigzag to Dr. Krishnan’s house.
2. All were surprised to see him
3. He was a strange and wierd bird
4. He could talk and sing in 21 languages
5. But he did not say a word
6. Aravind gave juicy fruits and nuts to him
7. He did not eat but transferred them to the chandelier and fan plate
8. He perched on the curtain rod and slept with snoring.
9. When the maid Lakshmi switched on the fan the painting of Mrs. Krishnan got spoilt by the streaks of orange pulp
10. The neighbours complained about the loud snoring of the bird as they were troubled.

Question 2.
What happened when zigzag was taken to the clinic?
Answer:
Introduction:
Zigzag created lot of misunderstandings in Dr. Krishnan’S house. So he wanted to take zigzag to his clinic. Let us see below what happens in the clinic.

Zigzag’s arrival to clinic:
Dr. Krishnan took zigzag to the clinic. He warned him not to sleep. Zigzag went to the clinic and perched on the reception table. Dr. Krishnan heard some strange sound in the clinic.

Zigzag’s behaviour:
The strange sound was zigzag’s clear and commanding voice. There was complete silence in the clinic. Dr, Krishnan was amazed. Zigzag’s bored and grumpy expression was gone. He went about the job.

Conclusion:
Thus zigzag made the clinic completely silent. Zigzag efficiently comforted the frightened y patients. Dr. Krishnan thought that zigzag was an absolute treasure.

(OR)

Title: Zigzag
Author: Asha Nehemiah
Characters: Dr. Krishnan and Zigzag.
Theme: Change is the law of life.

Dr. Krishnan took Zigzag to the clinic. He asked him to wait in the car. He warned him not to sleep. But he was not used to being kept waiting. So he went into the clinic and perched on the reception table. Dr. Krishnan heard a voice when he hardly walked into his clinic. It was Zigzag’s clear and commanding voice.

There was silence in the room. Everyone waited, open mouth for his next sentence. Dr. Krishnan was amazed. Zigzag’s bored and grumpy expression was gone. Instead he looked happy and alert. He went about the job. He had been trained for the job with the doctors. The clinic became a calm and orderly place without any confusion. Zigzag efficiently comforted the frightened patients.

(OR)

1. Dr. Krishnan took Zigzag to the clinic.
2. He asked zigzag to wait in the car.
3. But he went into the clinic and perched on the reception table.
4. Dr. Krishnan heard a voice even before he went into his room
5. There was a silence and everyone waited for Zigzag’s next sentence.
6. Dr. Krishnan was amazed
7. Zigzag’s bored and grumpy expression was gone
8. Instead Zigzag looked happy and alert to do his job
9. He comforted the frightened patients.
10. Dr. Krishnan’s clinic became a calm place without confusion.

Question 3.
Narrate the story zigzag in your own words.
Answer:
This story is about a multilingual weird bird, named Zigzag. Dr. Krishnan’s clinic usually sounded noisy because of the shouting and crying of children. His friend Dr. Somu requested him to shelter Zigzag, his pet as he was about to leave for Alaska.

When Zigzag was brought to his house, it did not speak to anyone. It could sing and talk in 21 different languages. But it did not speak any word to them. Instead, he transferred the fruits and nuts given to him to a Chandelier and to the blades of the ceiling fan. It perched on a curtain rod and went off to sleep. When the fan was switched on, all the fruits and nuts fell down.

The papaya slice splattered on the painting of Mrs. Krishnan and spoiled it. Mrs. Krishnan was irritated and persuaded Dr. Krishnan to send the bird away to Visu s house. So Dr. Krishnan took the bird to his clinic. There it commanded everyone in an orderly way. He brought peace, calmness and a disciplined atmosphere in the clinic. The painting which had been spoilt by Zigzag was sold out for ₹ 5,000. It was bought because the new technique in painting was liked by the buyer. Thus Mr. Krishnan’s family invited Zigzag to stay with them for another week.

(OR)

This is a story about Zigzag, a multilingual bird. Dr. Somu who left for Alaska asked Dr. Krishnan to shelter him. Visu brought him to Dr. Krishnan’s house. He did not speak to anyone though he could talk and sing in 21 languages. Instead, he transferred the fruits and nuts to a chandelier and the fan plates.

Then he perched on a curtain rod and went to sleep. When the fan was switched on the fruit slices fell off spoiling the painting of Mrs. Krishnan. She got irritated and told her husband to send Zigzag to Visu’s house. Dr. Krishnan took him to his clinic.

Zigzag commanded the patients and the clinic became a calm and orderly place. The spoilt painting was sold for Rs. 5000. The buyer considered it as a new technique of painting. Mrs. Krishnan wanted to keep Zigzag for another week.

(OR)

1. Zigzag was a bird.
2. Dr. Krishnan was a child specialist.
3. Vishu brought zigzag to Dr. Krishnan’s house.
4. Zigzag could speak 21 languages.
5. But the bird didn’t speak a word.
6. Zigzag fell asleep and began to snore.
7. Its loud snore disturbed everyone.
8. Mrs. Krishnan’s painting was spoiled
9. Mrs. Krishnan was very angry.
10. Dr. Krishnan took zigzag to the clinic.
11. At the clinic it helped to maintain calmness.
12. It made the children laugh.
13. It recited the French poetry.
14. So the family decided to keep him with them.
15. Everyone is happy now.