Tamilnadu Board Class 9 Maths Solutions Chapter 1 Set Language Ex 1.1

Tamilnadu State Board Class 9 Maths Solutions Chapter 1 Set Language Ex 1.1

Question 1.
Which of the following are sets?
(i) The Collection of prime numbers upto 100.
(ii) The Collection of rich people in India.
(iii) The Collection of all rivers in India.
(iv) The Collection of good Hockey players.
Solution:
(i) A = {2, 3, 5, 7,11, 13,17,19, 23, 29, 31, 37,41,43,47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97}
As the collection of prime numbers upto 100 is known and can be counted (well defined). Hence this is a set.
(ii) The collection of rich people in India. Rich people has no definition.
Hence, it is not a set.
(iii) A = {Cauvery, Sindhu, Ganga, }
Hence, it is a set.
(iv) The collection of good hockey players is not a well – defied collection because the criteria for determining a hockey player’s talent may vary from person to person.
Hence, this collection is not a set.

Question 2.
Listthe set of letters of the following words in Roster form.
(i) INDIA
(ii) PARALLELOGRAM
(iii) MISSISSIPPI
(iv) CZECHOSLOVAKIA
Solution:
(i) A = {I, N, D, A}
(ii) B = {P, A, R, L, E, O, G, M}
(iii) C = {M, I, S, P}
(iv) D = {C, Z, E, H, O, S, L, V, A, K, I}.

Question 3.
Consider the following sets A = {0, 3, 5, 8} B = {2, 4, 6, 10} C = {12, 14, 18, 20}
(a) State whether True or false.
(i) 18 ∈ C
(ii) 6 ∉A
(iii) 14 ∉ C
(iv) 10 ∈ B
(v) 5 ∈ B
(vi) 0 ∈ B

(b) Fill in the blanks?
(i) 3 ∈ ___
(ii) 14 e ___
(iii) 18 ___ B
(iv) 4 ___ B
Solution:
(a) (i) True (ii) True (iii) False (iv) True (v) False (vi) False,
(b) (i) A (ii) C (iii) ∉ (iv) ∈

Question 4.
Represent the following sets in Roster form.
(i) A = The set of all even natural numbers less than 20.
(ii) B = {y : y = \(\frac { 1 }{ 2n }\), n ∈N, n ≤ 5}
(iii) C = (x : x is perfect cube, 27 < x < 216}
(iv) D = {x : x ∈ Z, -5 < x ≤ 2}
Solution:
(i) A= {2,4, 6, 8,10, 12, 14,16, 18}
(ii) N = { 1, 2, 3, 4, 5}
Tamilnadu Board Class 9 Maths Solutions Chapter 1 Set Language Ex 1.1 1
(iii) C = {64, 125}
(iv) D = {-4,-3, -2, -1,0, 1, 2}

Question 5.
Represent the following sets in set builder form.
(i) B = The set of all Cricket players in India who scored double centuries in One Day Internationals.
(ii) C = { \(\frac { 1 }{ 2 }\), \(\frac { 2 }{ 3 }\), \(\frac { 3 }{ 4 }\) …..}.
(iii) D = The set of all tamil months in a year.
(iv) E = The set of odd Whole numbers less than 9.
Solution:
(i) B = {x : x is an Indian player who scored double centuries in one day internationals}
(ii) C = {x : x = \(\frac { n }{ n+1 }\), n ∈ N}
(iii) D = {x : x is a tamil month in a year}
(iv) E = {x : x is odd number, x ∈ W, x < 9, where W is the set of whole numbers}.

Question 6.
Represent the following sets in descriptive form.
(i) P = {January, June, July}
(ii) Q = {7, 11, 13, 17, 19, 23, 29}
(iii) R= {x : x ∈ N,x< 5}
(iv) S = {x : x is a consonant in English alphabets}
Solution:
(i) P is the set of English Months begining with J.
(ii) Q is the set of all prime numbers between 5 and 31.
(iii) R is the set of all natural numbers less than 5.
(iv) S is the set of all English consonants.

Samacheer Kalvi 9th Maths Book Answers

Tamilnadu Board Class 10 English Vocabulary Singular & Plural Forms

Tamilnadu State Board Class 10 English Vocabulary Singular & Plural Forms

Singular & Plural Forms (Text Book Page No.: 35)

I. Simple rules for plurals

a) add ‘-s’
words ending in consonants
boy – boys
desk – desks
book – books
cow – cows

b) add ‘-es’
words ending in
-s. -sh. -ch. -x
class – classes
brush – brushes
tax – taxes
dish – dishes

c) add ‘-es’
words ending in vowels
buffalo – buffaloes
negro – negroes
hero – heroes
echo – echoes

d) add ‘-s’
words ending in
vowels – exceptions
dynamo – dynamos
piano – pianos
kilo – kilos
ratio – ratios

e) Change ‘y’ into ‘ies’
baby – babies
army – armies
city – cities
story – stories

f) Change of inside vowel
man – men
woman – women
mouse – mice
goose – geese
tooth – teeth

II. Other Plurals:

1. No Change:
apparatus / apparatuses chassis, poultry, cattle headquarters, people, shorts aircraft, trousers,eans

2. sis – ses, xis – xes
analysis – analyses
basis – bases;
crisis – crises
diagnosis – diagnoses;
thesis – theses
synopsis – synopses
hypothesis – hypotheses
axis – axes

3. us – i
fungus – fungi
stimules – stimuli
radius – radii
alumnus – alumni
cactus – cacti / cactuses
terminus – termini / terminuses
syllabus – syllabi

4. on – a
criterion – criteria
phenomenon – phenomena
memorandum – memoranda

5. lum – la, turn -ta
curriculum – curricula;
erratum – errata

6. us – ora, era
corpus – corpora;
genus – genera

7. eau – eaux
beau – beaux;
bureau – bureaux / bureaus

8. la – lae, bra – brae
formula – formulae
vertebra – vertebrae

9. ex – ices / exes
apex – apices / apexes
index – indices / indexes

10. ix- ices / ixes
appendix – appendices / appendixes

11. cus – ci
focus – foci / focuses

12. Nouns take ‘s’
commander-in-chief:
commanders-in-chief
son-in-law: sons-in-law
on-looker: on-lookers
step-son: step-sons
maid-servant: maid-servants
passer-by: passers-by
man servant: men servants

13. chief – chiefs
gulf – gulfs
proof – proofs
handkerchief – handkerchiefs

14. dwarf – dwarfs, dwarves
hoof – hoofs, hooves
scarf – scarfs, scarves

15. ox – oxen
child – children

Singular Plural
aircraft aircraft
analysis analyses
baby babies
boat boats
box boxes
bus buses
cactus cacti
cat cats
child children
city cities
crisis crises
criterion criteria
daisy daisies
datum data
deer deer
diagnosis diagnoses
elf elves
fish fish
focus foci
foot feet
fungus fungi
goose geese
half halves
house houses
knife knives
leaf leaves
life lives
loaf loaves
man men
mouse mice
nucleus nuclei
oasis oases
penny pennies
person people
phenomenon phenomena
pitch pitches
potato potatoes
river rivers
sheep sheep
species species
spy spies
syllabus syllabi/ syllabuses
thesis theses _
tomato tomatoes
tooth teeth
wife wives
wish wishes
woman women

Singular Form

Singular Form Exercises

What is the singular form of the word ……………………….
1. What is the singular form of the word days?
(a) day (b) dae (c) doy
Answer:
(a) day

2. What is the singular form of the word data
(a) date (b) datum (c) datae
Answer:
(b) datum

3. What is the singular form of the word taxis?
(a) taxie (b) taxies (c) taxi
Answer:
(c) taxi

4. What is the singular form of the word ladies?
(a) lady (b) ladices (c) ladys
Answer:
(a) lady

5. What is the singular form of the word oxen?
(a) axe (b) oxe (c) ox
Answer:
(c) ox

6. What is the singular form of the word choices?
(a) choice (b) choices (c) choiceies
Answer:
(a) choice

7. What is the singular form of the word boxes?
(a) boxs (b) boxies (c) box
Answer:
(c) box

8. What is the singular form of the word thieves?
(a) thiefies (b) thieve (c) thief
Answer:
(c) thief

9. What is the singular form of the word armies?
(a) armie (b) army (c) armices
Answer:
(b)army

10. What is the singular form of the word owls?
(a) owl (b) owlies (c) owelds
Answer:
(a) owl

Plural Form

Plural Form Exercises

What is the plural form of the word …………………………..
1. What ¡s the plural form of the word qiarium?
(a) aquariums (b) aquari (c) aquaria
Answer:
(c) aquaria

2. What is the plural form of the word memorandum?
() memorundam (b) memoranda (c) memorandums
Answer:
(b) memoranda

3. What is the plural form of the word stratum?
(a) stratas (b) stratii (c) strata
Answer:
(c) strata

4. What is the plural form of the word deer?
(a) deer (b) deers (c) deeres
Answer:
(a) deer

5. What ¡s the plural form of the word erratum?
(a) errates (b) errata (c) errati
Answer:
(b) errata

6. What is the plural form of thief?
(a) thiefs (b) thiefes (c) thieves
Answer:
(c) thieves

7. What is the plural form of the word alumnus?
(a) alumna (b) alumni (c) alum
Answer:
(b)alumnil

8. What is the plural form of the words man servant?
(a) man servants (b) men servant (c) men servants
Answer:
(c) men servants

9. Which is the plural form of the word index?
(a) indices (b) indexs (c) indice
Answer:
(a) indices

10. What is the plural form of the word criterion?
(a) criterions (b) criteriae (c) criteria
Answer:
(c) criteria

11. What is the plural form of the word crisis?
(a) crisae (b) crises (c) crise
Answer:
(b) crisesi

12. What is the plural form of the word focus?
(a) focuses (b) focae (c) foci
Answer:
(c) foci

13. What is the plural form of the word medium?
(a) mediums (b) media (c) mediae
Answer:
(b) media

14. What is the plural form of the word fungus?
(a) fungi (b) fungae (c) funguss
Answer:
(a) fungi

15. What is the plural form of the word locus?
(a) locuses (b) bd (c) locae
Answer:
(b) bd

16. What is the plural for son-in-law?
(a) son-in-laws (b) sons-in-law (c) sons-in-laws
Answer:
(b) sons-in-law

17. What is the plural for species?
(a) specie (b) specy (c) species
Answer:
(c) species

18. What is the plural for piece of furniture?
(a) pieces of furniture (b) pieces of furnitures (c) piece of furnitures
Answer:
(a) pieces of furniture

19. What is the plural for buffalo?
(a) buffalos (b) buffaloes (c) baffles
Answer:
(b) buffaloes

20. What is the plural for goose?
(a) geese (b) gooses (c) geeses
Answer:
(a) geese

Tamilnadu Board Class 10 English Solutions

Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.6

Tamilnadu State Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.6

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:
If n(A × B) = 6
A = {1, 1}, n(A) = 2
n(B) = 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:
A = {1, 2}, B = {1, 2, 3, 4},
C = {5, 6}, D ={5,6, 7, 8}
A × C ={(1,5), (1,6), (2, 5), (2, 6)}
B × D = {(1,5),(1,6),(1,7),(1,8),(2,5),(2,6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8)}
∴ (A × C) ⊂ B × D it is true

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
25x = 210
⇒ 5x = 10
⇒ x =2

Question 5.
The range of the relation R = {(x, x2)|x is 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:
R = {(x, x2)/X is a prime number < 13}
The squares of 2, 3, 5, 7, 11 are
{4, 9, 25,49,121}

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) nm
(3) 2mn-1
(4) 2mn
Answer:
(4) 2mn
Hint:
n(A) = m,n(B) = n
n(A × B) = 2mn

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:
A = {1, 2, 3, 4), B = {4, 8, 9,10}
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.6 1

Question 10.
If f(x) = 2x2 and g (x) = \(\frac { 1 }{ 3x } \), Then fog is
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.6 2
Answer:
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.6 3
Hint:
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.6 4

Queston 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 fog 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:
gof = g(f(x))
fog = f(g(x))
= {(0,2),(1,0),(2,4),(-4,2),(7,0)}
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:
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.6 5

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

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]
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.6 6
It is a quadratic function.

Samacheer Kalvi 10th Maths Book Answers

Tamilnadu Board Class 10 English Vocabulary Parts of Speech

Tamilnadu State Board Class 10 English Vocabulary Parts of Speech

♦ Parts of Speech (Text Book Page No: 6)

In this question, the student has to rewrite the sentence changing the given noun into a verb or a verb into a noun, or a verb into an adjective, etc.
Tamilnadu Board Class 10 English Vocabulary Parts of Speech 1
1. The young seagull uttered a joyful scream. (adjective)
2. The young seagull screamed with joy. (noun)
3. The young seagull screamed joyfully. (adverb)

Set 2
1. The young bird pretended to be falling asleep, (verb)
2. The young bird made a pretension of falling asleep, (noun)
3. The young bird made a pretentious posture of falling asleep, (adjective)

Note that in set 1, the adjective ‘joyful’ is changed to its noun form ‘joy’ and to its adverb form ‘joyfully’.

In the Set 2, the verb ‘pretend’ has been transformed to its noun form ‘pretension and to its adjective form ‘pretentious’.

We can transform a sentence by interchanging parts of speech without changing its meaning.

♦ Exercises

Reframe the sentence using the underlined word as instructed.

1. Reframe the sentence using the underlined word as a verb :
His behaviour was that of a gentleman.
Answer:
He behaves like a gentleman.

2. Reframe the sentence using the underlined word as noun:
She is tolerant towards mischievous children.
Answer:
She tolerates the mischiefs of children.

3. Reframe the sentence using the underlined word as noun :
I decided to join the Arts college.
Answer:
My decision is to join Arts college.

4. Reframe the sentence using the underlined word as noun :
Eve-teasers should be severely punished.
Answer:
Punishment should be given to Eve-teasers severely.

5. Reframe the sentence using the underlined word as noun:
He responded correctly to my question.
Answer:
His response was correct to my question.

6. Reframe the sentence using the underlined word as adverb:
Please handle the glasses with care.
Answer:
Please handle the glass carefully.

7. Reframe the sentence using the underlined word as verb :
Birds migration occurs every year from one place to another place.
Answer:
Birds migrate every year from one place to another place.

8. Reframe the sentence using the underlined word as a verb:
Most Labour laws face the challenge of implementation.
Answer:
The government implemented many labour laws.

9. Reframe the sentence using the underlined word as noun :
You performed gracefully.
Answer:
Your performance was graceful.

10. Reframe the sentence using the underlined word as noun :
He was appreciated for his best performance.
Answer:
He had an appreciation for his best performance.

11. Reframe the sentence using the underlined word as a verb :
The professor gave a definition of the formula.
Answer:
The professor defined the formula.

12. Reframe the sentence using the underlined word as a noun :
He was punished for his disobedience.
Answer:
He received punishment for his disobedience.

13. Reframe the sentence using the underlined word as a verb :
The construction of this building has been done well.
Answer:
This building has been constructed well.

14. Reframe the sentence using the underlined word as a noun:
Be courageous to stand up for the truth.
Answer:
Take courage even at crucial moments.

15. Reframe the sentence using the underlined word as a verb :
The declaration of the local holiday was given by our Collector.
Answer:
Our Collector declared it to be a local holiday.

16. Reframe the sentence using the underlined word as a noun:
He was penalised for his disobedience.
Answer:
His disobedience leads to penalisation.

Tamilnadu Board Class 10 English Solutions

Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.5

Tamilnadu State Board Class 10 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
(ii) f(x) = \(\frac { 2 }{ x } \), g(x) = 2x2– 1
(iii) f(x) = \(\frac { x+6 }{ 3 } \)g(x) = 3 – x
(iv) f(x) = 3 + x, g(x) = x – 4
(v) f(x) = 4x2– 1,g(x) = 1 + x
Solution:
(i) f(x) = x – 6, g(x) = x2
fog(x) = f(g(x)) = f(x2) = x2 – 6 …(1)
gof(x) = g(f(x)) = g(x – 6) = (x – 6)2
= x2 + 36 – 12x = x2 – 12x + 36…(2)
(1) ≠ (2)
∴ fog(x) ≠ gof(x)

(ii) f(x) = \(\frac { 2 }{ x } \),g(x) = 2x2 – 1
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.5 1

(iii) f(x) = \(\frac { x+6 }{ 3 } \),g(x) = 3 – x
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.5 2

(iv) f(x) = 3 + x, g(x) = x – 4
fog(x) = f(g(x)) = f(x -4) = 3 + x – 4
= x – 1 …(1)
gof(x) = g(f(x)) = g(3 + x) = 3 + x – 4
= x – 1 …(2)
Here fog(x) = gof(x)

(v) f(x) = 4x2 – 1, g(x) = 1 + x
fog(x) = f(g(x)) = f(1 + x) = 4(1 + x)2 – 1
= 4(1 + x2 + 2x) – 1 = 4 + 4x2 + 8x – 1
= 4x2 + 8x + 3 …(1)
gof(x) = g(f(x)) = g(4x2 – 1)
= 1 + 4x2 – 1 = 4x2 …(2)
(1) ≠ (2)
∴ fog(x) ≠ gof(x)

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)
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.5 3
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)
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.5 4
3k = -5
k = \(\frac { -5 }{ 3 } \)

Question 3.
if f(x) = 2x – 1, g(x) = \(\frac { x+1 }{ 2 } \), show that fog = gof = x
Solution:
f(x) = 2x – 1, g(x) = \(\frac { x+1 }{ 2 } \), fog = gof = x
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.5 5

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
Solution:
f(x) = 2x + 1
g(x) = x2
fog(x) = fg(x)) = f(x2) = 2x2 + 1
gof(x) = g(f(x)) = g(2x + 1) = (2x + 1)2
= 4x2 + 4x + 1
Range of fog is
{y/y = 2x2 + 1, x ∈ N}
Range of gof is
{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?
Solution:
f(x) = x5
g(x) = x4
fog = fog(x) = f(g(x)) = f(x4)
= (x4)5 = x20
f is one-one, g is not one-one.
∵ g(1) = 14 = 1
g(-1) = ( -1)4 = 1
Different elements have same images
fog is not one-one. [∵ fog (1) = fog (-1) = 1]

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) = 3×2 …(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).
Solution:
f ={(-1,3), (o,-1), 2,-9)
f(x) = (ax) + b ….(1)
is the equation of all linear functions.
∴ f(-1) = 3
f(0) = -1
f(2) = -9
f(x) = ax + b
f(-1) = -a + b = 3 …(2)
f(0) = b = -1
-a – 1 = 3 [∵ substituting b = – 1 in (2)]
-a = 4
a = -A
The linear function is -4x – 1. [From (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] = 3 at1 + 3 bt2
= a(3t1) + b(3t2) = a[C(t1) + b(Ct62)]
∴ Superposition principle is satisfied.
Hence C(t) = 3t is linear function.

Samacheer Kalvi 10th Maths Book Answers

Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4

Tamilnadu State Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4

Question 1.
Determine whether the graph given below represent functions. Give reason for your answers concerning each graph.
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 1
Solution:
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 2
(i) It is not a function. The graph meets the vertical line at more than one points.
(ii) It is a function as the curve meets the vertical line at only one point.
(iii) It is not a function as it meets the vertical line at more than one points.
(iv) It is a function as it meets the vertical line at only one point.

Question 2.
Let f :A → B be a function defined by f(x) = \(\frac { x }{ 2 } \) – 1, Where A = {2,4,6,10,12},
B = {0,1, 2, 4, 5, 9}. Represent/by
(i) set of ordered pairs;
(ii) a table;
(iii) an arrow diagram;
(iv) a graph
Solution:
f: A → B
A = {2,4, 6, 10, 12}, B = {0,1, 2, 4,5,9}
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 3Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 3
(i) Set of ordered pairs
= {(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)}
(ii) a table
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 4
(iii) an arrow diagram;
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 5
(iv) a graph
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 6

Question 3.
Represent the function f = {(1, 2),(2, 2),(3, 2), (4,3), (5,4)} through
(i) an arrow diagram
(ii) a table form
(iii) a graph
Solution:
f = {(1,2), (2, 2), (3, 2), (4, 3), (5, 4)}
(i) An arrow diagram.
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 7
(ii) a table form
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 8
(iii) A graph representation.
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 9

Question 4.
Show that the function f : N → N defined by f{x) = 2x – 1 is one – one but not onto.
Solution:
f: N → N
f(x) = 2x – 1
N = {1,2, 3, 4, 5,…}
f(1) = 2(1) – 1 = 1
f(2) = 2(2) – 1 = 3
f(3) = 2(3) – 1 = 5
f(4) = 2(4) – 1 = 7
f(5) = 2(5) – 1 = 9
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 10
Hence f : N → N is a one-one function.
A function f: N → N is said to be onto function if the range of f is equal to the co-domain of f
Range = {1,3, 5, 7, 9,…}
Co-domain = {1, 2,3,..}
But here the range is not equal to co-domain. Therefore it is one-one but not onto function.

Question 5.
Show that the function f: N → N defined by f (m) = m2 + m + 3 is one – one function.
Solution:
f: N → N
f(m) = m2 + m + 3
N = {1,2, 3,4, 5…..}m ∈ N
f{m) = m2 + m + 3
f(1) = 12 + 1 + 3 = 5
f(2) = 22 + 2 + 3 = 9
f(3) = 32 + 3 + 3 = 15
f(4) = 42 + 4 + 3 = 23
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 11
In the figure, for different elements in the (X) domain, there are different images in f(x). Hence f: N → N is a one to one but not onto function as the range of f is not equal to co-domain.
Hence it is proved.

Question 6.
Let A = {1, 2,3,4) and B = N. Letf: A → B be
defined by f(x) = x3 then,
(i) find the range off
(ii) identify the tpe of function
Solution:
A = {1,2,3,4}
B = N
f: A → B,f(x) = x3
(i) f(1) = 13 = 1
f(2) = 23 = 8
f(3) = 33 = 27
f(4) = 43 = 64
(ii) Therange of f = {1,8,27,64 )
(iii) It is one-one and into function.

Question 7.
In each of the following cases state whether the function is bijective or not. Justify your answer.
(i) f: R → R defined by f(x) = 2x + 1
(ii) f: R → R defined by f(x) = 3 – 4x2
Solution:
(i) f : R → R
f(x) = 2x + 1
f(1) = 2(1) + 1 = 3
f(2) = 2(2) + 1 = 5
f(-1) = 2(-1) + 1 = -1
f(0) = 2(0) + 1 = 1
It is a bijective function. Distinct elements of A have distinct images in B and every element in B has a pre-image in A.
(ii) f: R → R; f(x) = 3 – 4x2
f(1) = 3 – 4(12) = 3 – 4 = -1
f(2) = 3 – 4(22) = 3 – 16 = -13
f(-1) = 3 – 4(-1)2 = 3 – 4 = -1
It is not bijective function since it is not one-one

Question 8.
Let A = {-1, 1} and B = {0, 2}. If the function f: A → B defined by f(x) = ax + b is an onto function? Find a and b.
Solution:
A= {-1, 1},B = {0,2}
f: A → B, f(x) = ax + b
f(-1) = a(-1) + b = -a + b
f(1) = a(1) + b = a + b
Since f(x) is onto, f(-1) = 0
⇒ -a + b = 0 …(1)
& f(1) = 2
⇒ a + b = 2 …(2)
-a + b = 0
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 13

Question 9.
If the function f is defined by
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 14
(i) f(3)
(ii) f(0)
(iii) f(-1.5)
(iv) f(2) + f(-2)
Solution:
(i) f(3) ⇒ f(x) = x + 2 ⇒ 3 + 2 = 5
(ii) f(0) ⇒ 2
(iii) f (- 1.5) = x – 1
= -1.5 – 1 = -2.5
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 154

Question 10.
A function f: [-5,9] → R is defined as follows:
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 15
Solution:
f : [-5, 9] → R
(i) f(-3) + f(2)
f(-3) = 6x + 1 = 6(-3) + 1 = -17
f(2) = 5×2 – 1 = 5(22) – 1 = 19
∴ f(-3) + f(2) = -17 + 19 = 2
(ii) f(7) – f(1)
f(7) = 3x – 4 = 3(7) – 4 = 17
f(1) = 6x + 1 = 6(1) + 1 = 7
f(7) – f(1) = 17 – 7 = 10
(iii) 2f(4) + f(8)
f(4) = 5x2 – 1 = 5 × 42 – 1 = 79
f(8) = 3x – 4 = 3 × 8 – 4 = 20
∴ 2f(4) + f(8) = 2 × 79 + 20 = 178
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 16

Question 11
The distance S an object travels under the influence of gravity in time t seconds is given by S(t) = \(\frac { 1 }{ 2 } \) gt2 + at + b where, (g is the 2 acceleration due to gravity), a, b are constants. Check if the function S (t)is one-one.
Solution:
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 17
Yes, for every different values of t, there will be different values as images. And there will be different preimages for the different values of the range. Therefore it is one-one function.

Question 12.
The function ‘t’ which maps temperature in Celsius (C) into temperature in Fahrenheit (F) is defined by ty(C)= F where F = \(\frac { 9 }{ 5 } \) C +32 . Find,
(i) t(0)
(ii) t(28)
(iii) t(-10)
(iv) the value of C when t(C) = 212
(v) the temperature when the Celsius value is equal to the Farenheit value.
Solution:
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 18
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.4 19

Samacheer Kalvi 10th Maths Book Answers

Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.3

Tamilnadu State Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.3

Question 1.
Let f = {(x,y)|x,y ∈ N and y = 2x} be a relation on N. Find the domain, co-domain and range. Is this relation a function?
Solution:
F = {(x, y)|x, y ∈ N and y = 2x}
x = {1,2,3,…}
y = {1 × 2, 2 × 2, 3 × 2, 4 × 2, 5 × 2 …}
R = {(1,2), (2,4), (3, 6), (4, 8), (5, 10),…}
Domain of R = {1, 2, 3, 4,…},
Co-domain = {1,2,3…..}
Range of R = {2, 4, 6, 8, 10,…}
Yes, this relation is a function.

Question 2.
Let X = {3, 4, 6, 8}. Determine whether the relation R = {(x,fx))|x ∈ X, f(x) = x2 + 1} is a function from X to N ?
Solution:
x = {3,4, 6, 8}
R = ((x,f(x))|x ∈ X,f(x) = X2 + 1}
f(x) = x2 + 1
f(3) = 32 + 1 = 10
f(4) = 42 + 1 = 17
f(6) = 62 + 1 = 37
f(8) = 82 + 1 = 65
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.3 1
R = {(3, 10), (4,17), (6,37), (8, 65)}
R = {(3, 10), (4,17), (6,37), (8, 65)}
Yes, R is a function from X to N.

Question 3.
Given the function f: x → x2 – 5x + 6, evaluate
(i) f(-1)
(ii) A(2a)
(iii) f(2)
(iv) f(x – 1)
Solution:
Give the function f: x → x2 – 5x + 6.
(i) f(-1) = (-1)2 – 5(1) + 6 = 1 + 5 + 6 = 12
(ii) f(2a) = (2a)2 – 5(2a) + 6 = 4a2 – 10a + 6
(iii) f(2) = 22 – 5(2) + 6 = 4 – 10 + 6 = 0
(iv) f(x- 1) = (x – 1)2 – 5(x – 1) + 6
= x2 – 2x + 1 – 5x + 5 + 6
= x2 – 7x + 12

Question 4.
A graph representing the function f(x) is given in figure it is clear that f (9) = 2.
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.3 2
(i) Find the following values of the function
(a) f(0)
(b) f(7)
(c) f(2)
(d) f(10)
(ii) For what value of x is f (x) = 1?
(iii) Describe the following
(i) Domain
(ii) Range.
(iv) What is the image of 6 under f?
Solution:
From the graph
(a) f(0) = 9
(b) f(7) = 6
(c) f(2) = 6
(d) f(10) = 0
(ii) Atx = 9.5, f(x) = 1
(iii) Domain = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
= {x |0 < x < 10, x ∈ R}
Range = {x|0 < x < 9, x ∈ R}
= {0, 1,2, 3,4, 5,6, 7, 8,9}
(iv) The image of 6 under f is 5.

Question 5.
Let f(x) = 2x + 5. If x ≠ 0 then find
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.3 3
Solution:
Given f(x) = 2x + 5, x ≠ 0.
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.3 4

Question 6.
A function fis defined by f(x) = 2x – 3
(i) find \(\frac{f(0)+f(1)}{2}\)
(ii) find x such that f(x) = 0.
(iii) find x such that f(x) = x.
(iv) find x such that f(x) = f(1 – x).
Solution:
Given f(x) = 2x – 3
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.3 5

Question 7.
An open box is to be made from a square piece of material, 24 cm on a side, by cutting equal squares from the corners and turning up the sides as shown in figure. Express the volume V of the box as a function of x.
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.3 6
Solution:
Volume of the box = Volume of the cuboid
= l × b × h cu. units
Here l = 24 – 2x
b = 24 – 2x
h = x
∴ V = (24 – 2x) (24 – 2x) × x
= (576 – 48x – 48x + 4×2)x
V = 4x3 – 96x2 + 576x

Question 8.
A function f is defined bv f(x) = 3 – 2x . Find x such that f(x2) = (f(x))2.
Solution:
f(x) = 3 – 2x
f(x2) = 3 – 2x2
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.3 7

Question 9.
A plane is flying at a speed of 500 km per hour. Express the distance d travelled by the plane as function of time t in hours.
Solution:
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.3 8

Question 10.
The data in the adjacent table depicts the length of a woman’s forehand and her corresponding height. Based on this data, a student finds a relationship between the height (y) and the forehand length(x) as y = ax + b, where a, b are constants.
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.3 9
(i) Check if this relation is a function.
(ii) Find a and b.
(iii) Find the height of a woman whose forehand length is 40 cm.
(iv) Find the length of forehand of a woman if her height is 53.3 inches.
Solution:
(i) Given y = ax + b …(1)
The ordered pairs are R = {(35, 56) (45, 65) (50, 69.5) (55, 74)}
∴ Hence this relation is a function.
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.3 10
Substituting a = 0.9 in (2) we get
⇒ 65 = 45(.9) + b
⇒ 65 = 40.5 + b
⇒ b = 65 – 40.5
⇒ b = 24.5
∴ a = 0.9, b = 24.5
∴ y = 0.9x + 24.5
(iii) Given x = 40 , y = ?
∴ (4) → y = 0.9 (40) + 24.5
⇒ y = 36 + 24.5
⇒ y = 60.5 inches
(iv) Given y = 53.3 inches, x = ?
(4) → 53.3 = 0.9x + 24.5
⇒ 53.3 – 24.5 = 0.9x
⇒ 28.8 = 0.9x
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.3 11
∴ When y = 53.3 inches, x = 32 cm

Samacheer Kalvi 10th Maths Book Answers

Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.2

Tamilnadu State Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.2

Question 1.
Let A = {1,2,3,7} and B = {3,0,-1,7}, which of the following are relation from A to B ?
(i) R1 = {(2,1), (7,1)}
(ii) R2 = {(-1,1)}
(iii) R3 = {(2,-1), (7,7), (1,3)}
(iv) R4 = {(7,-1), (0,3), (3,3), (0,7)}
(i) A = {1,2, 3,7}, B = {3, 0,-1, 7}
Solution:
R1 = {(2,1), (7,1)}
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.2 1
It is not a relation there is no element as 1 in B.
(ii) R2 = {(-1, 1)}
It is not [∵ -1 ∉ A, 1 ∉ B]
(iii) R3 = {(2,-1), (7, 7), (1,3)}
It is a relation.
R4 = {(7,-1), (0, 3), (3, 3), (0, 7)}
It is also not a relation. [∵ 0 ∉ A]

Question 2.
Let A = {1, 2, 3, 4,…,45} and R be the relation defined as “is square of ” on A. Write R as a subset of A × A. Also, find the domain and range of R.
Solution:
A = {1, 2, 3, 4, . . . 45}, A × A = {(1, 1), (2, 2) ….. (45,45)}
R – is square of’
R = {(1,1), (2,4), (3, 9), (4, 16), (5,25), (6,36)}
R ⊂ (A × A)
Domain of R = {1, 2, 3, 4, 5, 6}
Range of R = {1,4, 9, 16, 25, 36}

Question 3.
A Relation R is given by the set {(x, y) /y = x + 3, x ∈ {0, 1, 2, 3, 4, 5}}. Determine its domain and range.
Solution:
x = {0,1,2,3,4,5}
y = x + 3
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.2 2
⇒ y = {3, 4, 5, 6, 7, 8}
R = {(x,y)}
= {(0, 3),(1, 4),(2, 5),(3, 6), (4, 7), (5, 8)}
Domain of R = {0, 1, 2, 3, 4, 5}
Range of R = {3, 4, 5, 6, 7, 8}

Question 4.
Represent each of the given relation by (a) an arrow diagram, (b) a graph and (c) a set in roster form, wherever possible.
(i) {(x,y)|x = 2y,x ∈ {2,3,4,5},y ∈ {1, 2,3,4)
(ii) {(x, y)y = x + 3, x, y are natural numbers <10}
Solution:
(i){(x,y)|x = 2y,x ∈ {2,3,4,5},y ∈ {1,2,3,4}} R = (x = 2y)
2 = 2 × 1 = 2
4 = 2 × 2 = 4
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.2 3
(c) {(2,1), (4, 2)}
(ii) {(x, v)[y = x + 3, x,+ are natural numbers <10}
x = {1,2, 3, 4, 5, 6, 7, 8,9} R = (y = x + 3)
y = {1,2, 3, 4,5,6, 7, 8,9}
R = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.2 4
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.2 5
(c) R = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}

Question 5.
A company has four categories of employees given by Assistants (A), Clerks (C), Managers (M) and an Executive Officer (E). The company provide ₹10,000, ₹25,000, ₹50,000 and ₹1,00,000 as salaries to the people who work in the categories A, C, M and E respectively. If A1, A2, A3, A4 and As were Assistants; C1, C2, C3, C4 were Clerks; M1, M2, M3 were managers and E1,E2 were Executive officers and if the relation R is defined by xRy, where x is the salary given to person y, express the relation R through an ordered pair and an arrow diagram.
Solution:
A – Assistants → A1, A2, A3, A4, A5
C – Clerks → C1, C2, C3, C4
D – Managers → M1, M2, M3
E – Executive officer → E1, E2
(a) R = {(10,000, A1), (10,000, A2), (10,000, A3),
(10,000, A4), (10,000, A5), (25,000, C1),
(25,000, C2), (25,000, C3), (25,000, C4),
(50,000, M1), (50,000, M2), (50,000, M3),
(1,00,000, E1), (1,00,000, E2)}
Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.2 6

Samacheer Kalvi 10th Maths Book Answers

Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.1

Tamilnadu State Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.1

Question 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)
Solution:
(i) A = {2,-2,3}, B = {1,-4}
A × B = {(2, 1), (2, -4), (-2, 1), (-2, -4), (3,1) , (3,-4)}
A × A = {(2, 2), (2,-2), (2, 3), (-2, 2), (-2, -2), (-2, 3), (3, 2), (3, -2), (3,3) }
B × A = {(1, 2), (1, -2), (1, 3), (-4, 2), (-4, -2), (-4,3)}

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

(iii) A = {m,n} × Φ
A × B = { }
A × A = {(m,m), (m,n), (n, m), (n, n)}
B × A = { }

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.
Solution:
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) x (B ∩ D) = (A × B) ∩ (C × D) is true?
Solution:
LHS = {(A∩C) × (B∩D)
A ∩C = {3}
B ∩D = {3,5}
(A ∩ C) × (B ∩ D) = {(3, 3) (3, 5)} …(1)
RHS = (A × B) ∩ (C × D)
A × B = {(1,2), (1,3), (1,5), (2,2), (2,3), (2, 5), (3, 2), (3,3), (3,5)}
C × D = {(3,1), (3,3), (3,5), (4,1), (4, 3), (4,5)}
(A × B) ∩ (C × D) = {(3, 3), (3, 5)} …(2)
∴ (1) = (2) ∴ It 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)
(i) 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)
A = {1,2, 3, 4, 5, 6, 7}
B = {2, 3, 5, 7}
C = {2}
Solution:
(i)(A ∩ B) × C = (A × c) ∩ (B × C)
LHS = (A ∩ B) × C
A ∩ B = {2, 3, 5, 7}
(A ∩ B) × C = {(2, 2), (3, 2), (5, 2), (7, 2)} …(1)
RHS = (A × C) ∩ (B × C)
(A × C) = {(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (7,2)}
(B × C) = {2,2), (3,2), (5,2), (7,2)}
(A × C) ∩ (B × C) = {(2,2), (3,2), (5,2), (7,2)} …(2)
(1) = (2)
∴ LHS = RHS. Hence it is verified.

(ii) A × (B – C) = (A × B) – (A × C)
LHS = A × (B – C)
(B – C) = {3,5,7}
A × (B – 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)} …(1)
RHS = (A × B) – (A × C)
(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),
(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), (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)
(1) = (2) ⇒ LHS = RHS.
Hence it is verified.

Samacheer Kalvi 10th Maths Book Answers