Tamilnadu Board

Tamilnadu State Board Class 10 English Vocabulary Idioms

♦ Idioms (Text Book Page No.: 66-68,167)

Idioms are groups of words put together as a unit with a particular meaning. The meaning of the word is not literal.

For example, if one says that the cat is out of the bag then it does not literally mean the cat is out of the bag but it has a figurative meaning which means the secret is out.

That’s why the meaning of idioms cannot be assumed based on the individual meaning of the words but by studying the words as a unit.

Examples:

1. Lalitha takes a late-night walk in the beach once in a blue moon. In the above sentence ‘once in a blue moon’ is an idiom which means an event that happens rarely.
2. The women cricketers were on the ball in their last over of the match. In the above sentence bn, the ball’ is an idiom which means when someone understands the situations well.

Some more idioms and meanings:

1	a bolt from the blue	unexpected event; complete surprise (usually unwelcome)
2	a drop in the ocean	a very small amount compared with what is needed or expected
3	a penny for your thoughts	a way of asking what someone is thinking.
4	a stone’s throw	a very short distance
5	a wild goose chase	a worthless hunt or chase
6	a yellow streak	cowardice in one’s character
7	above board	honest, not a secret
8	add insalt to injury	to worsen an unfavourable situation
9	armchair expert	one who gives advice in an area in which he was not actively involved
10	at close quarters	very near
11	at hand	very near
12	at loggerheads	to disagree strongly
13	at snails pace	very slowly
14	at the drop of the hat	without any hesitation
15	at the eleventh hour	at the last moment
16	at the end of one’s tether	to have no power, patience or endurance left
17	barking up the wrong tree	accusing the wrong person
18	be armed with	be equipped with
19	beat around the bush	avoiding the main topic
20	bed of roses	comfortable position
21	best of both worlds	all the advantages
22	bite the bullet	to get something over with because it is inevitable
23	blessing in disguise	something good that isn’t recognized at first.
24	break even	make no profit or loss
25	break the ice	make people feel more comfortable
26	call it a day	stop working on something
1 27	can’t judge a book by its cover	cannot judge something primarily on appearance.
28	clean slate	a past record without discredit
29	comparing apples to oranges	comparing two things that cannot be compared
30	costs an arm and a leg	very expensive
31	curiosity killed the cat	being inquisitive can lead you into an unpleasant situation.
32	devil’s advocate	to present a counter argument
33	draw a blank	unable to get information
34	every cloud has a silver lining	good-things come after bad things
35	eyewash	something to deceive
36	fair and square	in a fair way
37	fall a prey to	become a victim
38	far cry from	very different from
39	feel the pinch	feeling unpleasant change in one’s standard of living
40	fit as a fiddle	in good health
41	fortune favours the bold	take risks
42	give (someone) a piece of one’s mind	to tell someone frankly what one thinks especially when one disapproves of the other’s behaviour
43	go down in flames	fail spectacularly
44	go on a wild goose chase	to do something pointless
45	going from bad to worse	deteriorate further
46	hard to come by	difficult to find
47	have a hand	to get involved
48	have no hand in                     ,	does not take part in an activity
49	head back	return
50	herculean task	difficult task
51	hit the nail on the head	do or say something exactly right
52	hit the sack	go to sleep
53	holds good	valid at the time of discussion
54	honour bound (to do something)	required to do something as a moral duty but not by law
55	in a big way	on a large scale
56	in a nutshell	briefly
57	in all walks of life	all social groups / all the aspects
58	in deep waters	in trouble
59	in hot pursuit	following closely
60	in short supply	Not enough / scarce
61	in the service of	available for
62	it is a piece of cake	it is easy
63	it’s raining cats and dogs	it’s raining heavily
64	keep pace with	to move with same speed
65	keep something at bay	keep something away
66	kicked the bucket	passed away
67	leave no stone unturned	look everywhere
68	let the cat out of the bag	give away a secret
69	lion’s share	major share
1 70	loud and clear	very clearly
71	make both ends meet	live within means
72	make fun of	ridicule
73	make up one’s mind	decide, determine
74	matter of concern	something to worry about
75	miss the boat	it’s too late
76	muffle up	to cover
77	not playing with a full deck	someone who lacks intelligence
78	note of hand	promissory note
79	null and void	invalid
80	<m cloud nine	to be extremely happy
81	once and for all	completely and finally
82	once in a blue moon	very rarely
83	one thing leads to another	series of events in which each event was caused by the previous one.
84	pink of health	extremely healthy, in perfect condition
85	play an important role	to have a significant position
86	pull yourself together	calm down
87	put on airs	behave in an unnatural way to impress others
88	shadow of one’s	not having the strength, former self influence, etc., that one once had
89	side by side	along with
90	speak volumes	to express something very clearly and completely
91	spill the beans	give away a secret
92	take to one’s heels	to run away
93	take a very hard line	not giving in
94	the ball is in your court	it’s your decision
95	the burning question	a crucial issue
96	the whys and wherefores	the reasons for something
97	thick and fast	in large numbers
98	tit for tat	revenge
99	told him flat	expressed opinion directly
100	tread on	walk with difficulty
101	trial and error	to try many times to succeed
102	tricks of the trade	the expertise of doing business
103	whole nine yards	everything, all of it
104	with a bang	in a very exciting way

Exercises

Choose a suitable meaning for the idiom found in the following sentence.
1. Orders for the new product are coming in thick and fast.
(a) large numbers (b) small volumes (c) limited quantity (d) appropriate level
Answer:
(a) large numbers

2. How can anyone make ends meet with just Rs. 2000/- a month?
(a) join the two ends (b) arrange for a meeting (c) manage with the money (d) account for the meeting
Answer:
(c) manage with the money

3. My uncle is an armchair expert and often tells us how to play cricket.
(a) best cricket player (b) best expert in making armchairs (c) best adviser but without any practical knowledge (d) best adviser with excellent practical knowledge
Answer:
(c) best adviser but without any practical knowledge

4. I told him flat that I had no intention of lending him any money.
(a) telling in humorous way (b) expressed opinion directly (c) expressed insignificantly (d) expression to confuse
Answer:
(b) expressed opinion directly

5. He invited all and sundry for a function, but could not manage it all.
(a) important people only (b) friends only (c) relatives only (d) everyone
Answer:
(d) everyone

6. His progress at school was at a snail’s pace.
(a) very slow (b) in fast pace (c) moderately (d) satisfactorily
Answer:
(a) very slow

7. In the middle of the beautiful scenery, the building stood as an eyesore
(a) excellent icon (b) outstanding one (c) ugly sight (d) important monument
Answer:
(c) ugly sight

8. He performed the Herculean task of converting barren land into fertile land.
(a) easy task (b) difficult task (c) uncomplicated task (d) rituals
Answer:
(b) difficult task

9. He kicked the bucket last Monday at the Adyar Cancer hospital.
(a) fell down (b) played a game (c) was completely cured (d) passed away
Answer:
(d) passed away

10. I call upon our chief guest to address the gathering.
(a) appeal (b) summon (c) beg (d) invite
Answer:
(d) invite

Tamilnadu Board Class 10 English Solutions

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}

(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 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
2^5x = 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) n^m
(3) 2^mn-1
(4) 2^mn
Answer:
(4) 2^mn
Hint:
n(A) = m,n(B) = n
n(A × B) = 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:
A = {1, 2, 3, 4), B = {4, 8, 9,10}

Question 10.
If f(x) = 2x² and g (x) = \(\frac { 1 }{ 3x } \), Then fog is

Answer:

Hint:

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:

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]

It is a quadratic function.

Samacheer Kalvi 10th Maths Book Answers

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.

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 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

(iii) f(x) = \(\frac { x+6 }{ 3 } \),g(x) = 3 – x

(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)

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)

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

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 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.

Solution:

(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}

(i) Set of ordered pairs
= {(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)}
(ii) a table

(iii) an arrow diagram;

(iv) a graph

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.

(ii) a table form

(iii) A graph representation.

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

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

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

Question 9.
If the function f is defined by

(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

Question 10.
A function f: [-5,9] → R is defined as follows:

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

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:

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:

Samacheer Kalvi 10th Maths Book Answers

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

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.

(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

Solution:
Given f(x) = 2x + 5, x ≠ 0.

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

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.

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

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:

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.

(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.

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

∴ When y = 53.3 inches, x = 32 cm

Samacheer Kalvi 10th Maths Book Answers

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)}

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

⇒ 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

(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)}


(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)}

Samacheer Kalvi 10th Maths Book Answers

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