Tamilnadu Board

Tamilnadu State Board Class 10 English Vocabulary Homophones & Confusables

♦ Homophones & Confusables (Text Book Page No.: 194,195)

Homophones

Homophones are words that sound the same but have different meanings and spellings. The text has many homophones such as: see-sea, hear-here, knew-new.

Confusables

Commonly confused words
English has a lot of commonly confused words. They either look alike or look and sound alike, but have completely different meanings and usage. Here are some examples from the text.

brought (v) –
past participle of bring.
E.g. Anitha had brought a book from the library,

bought (v) –
past participle of buy.
E.g. Lalitha had bought a new dress last week.

effect (v) – to have an effect on.
E.g. The pet’s death affected his master,

effect (n) –
anything brought about by a cause or agent; result.
E.g. Both El Nino and La Nina are opposite effects of the same phenomenon.

adopt	adapt
advice	advise
aloud	allowed
birth	berth
blew	blue
bore	boar
bow	bough
brake	break
by	buy
canvass	canvas
carat	carrot
career	carrier
ceased	seized
cent	sent
cereal	serial
check	cheque
chord	card
clause	claws
coarse	course
compliment	complement
corps	corpse
cot	caught
council	counsel
dairy	diary
dessert	desert
dew	due
die	dye
discord	discard
eyes	ice
fair	fare
feat	feet
flour	floor
form	farm
foul	fowl
groan	grown
hair	heir
hair	hare
heal	heel
heard	herd
here	hear
hole	whole
hour	our
inn	in
knight	night
knot	not
know	no
lakh	lack
lice	lies
line	lion
litter	litre
loan	lone
loss	laws
meat	meet
metal	mettle
navel	naval
new .	knew
none	nun
overseas	oversees
pale	pail
pass	pause
peace	piece
peal	peel
phase	face
pore	pour
prey	pray
principle	principal
prize	price
quiet	quite
rack	rake
rice	rise
riot	ryot
roll	role
root	route
sail	sale
sea	see
sealing	ceiling
sell	cell
site	cite
sole	soul
son	sun
sore	soar
sow	so
stare	stair
stationary	stationery
steel	steal
storey	story
strait	straight
sum	some
tale	tail
thrash	trash
tied	tide
too	two
tow	toe
troop	troupe
vacation	vocation
wait	weight
way	weigh
weak	week
wear	where
weary	wary
weather	whether
whole :	hole
wise	vice
write	right
year	ear

Exercises

Fill in the blank with correct Homophone.

Question 1.
Be careful around the figurines because they _______ easily, (brake/break)
Answer:
break

Question 2.
I can’t _______ this burden any longer, (bear/bare)
Answer:
bear

Question 3.
The poor man only had one _______ to his name, (scent/cent)
Answer:
cent

Question 4.
That poem is very _______ to me. (deer/dear)
Answer:
dear

Question 5.
I hope my plants don’t _______ in this heat, (die/dye)
Answer:
die

Question 6.
The recipe calls for two cups of _______ (flower/flour)
Answer:
flour

Question 7.
Have you read the story about the tortoise and the _______? (hair/hare)
Answer:
hare

Question 8.
Vegetarians do not eat _______ (meat/meet)
Answer:
meat

Question 9.
Let’s go buy a new _______ of shoes, (pear/pair)
Answer:
pair

Question 10.
My _______ is presently in the military, (son/sun)
Answer:
son

Question 11.
Why don’t the _______ of us go to town tomorrow? (too/two)
Answer:
two

Question 12.
My favourite _______ is cornflakes, (serial/cereal)
Answer:
cereal

Question 13.
I ordered the Mexican Hot Pizza, but couldn’t eat the _______ on top! (chilli/chilly)
Answer:
chilli

Question 14.
Chilled white _______ goes well with fish, (wine/whine)
Answer:
wine

Question 15.
He held the railing as he walked down the _______ (stairs/stares).
Answer:
stairs

Question 16.
The snowstorm will _______ thousands of travellers, (affect/effect)
Answer:
affect

Question 17.
The dog chased the _______ into the neighbour’s yard, (ball/bawl)
Answer:
ball

Question 18.
The contractors _______ the company for travel expenses, (billed/build)
Answer:
hilled

Question 19.
Seoul is the _______ of South Korea, (capital/capital)
Answer:
capital

Question 20.
The squirrels _______ the peanuts under the snow, (berry/bury)
Answer:
bury

Tamilnadu Board Class 10 English Solutions

Tamilnadu State Board Class 10 Maths Solutions Chapter 2 Numbers and Sequences Ex 2.2

Question 1.
For what values of natural number n, 4n can end with the digit 6?
Solution:
4ⁿ = (2 × 2)ⁿ = 2ⁿ × 2ⁿ
2 is a factor of 4ⁿ.
So, 4ⁿ is always even and end with 4 and 6.
When n is an even number say 2, 4, 6, 8 then 4ⁿ can end with the digit 6.
Example:

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

Question 4.
If 13824 = 2^a × 3^b then find a and b.
Solution:
If 13824 = 2^a × 3^b
Using the prime factorisation tree

13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
= 2⁹ × 3³ = 2^a × 3^b
∴ a = 9, b = 3.

Question 5.
If \(p_{1}^{x_{1}} \times p_{2}^{x_{2}} \times p_{3}^{x_{3}} \times p_{4}^{x_{4}}=113400\) 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.
Solution:
If \(p_{1}^{x_{1}} \times p_{2}^{x_{2}} \times p_{3}^{x_{3}} \times p_{4}^{x_{4}}=113400\)
p₁, p₂, p₃, P₄ are primes in ascending order, x₁, x₂, x₃, x₄ are integers.
using Prime factorisation tree.

113400 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 7
= 23 × 34 × 52 × 7

Question 6.
Find the L.C.M. and H.C.F. of 408 and 170 by applying the fundamental theorem of arithmetic.
Solution:
408 and 170.

408 = 2³ × 3¹ × 17¹
170 = 2¹ × 5¹ × 17¹

∴ H.C.F. = 2¹ × 17¹ = 34.
To find L.C.M, we list all prime factors of 408 and 170, and their greatest exponents as follows.

∴ L.C.M. = 2³ × 3¹ × 5¹ × 17¹
= 2040.

Question 7.
Find the greatest number consisting of i 6 digits which is exactly divisible by 24,15,36?
Solution:
To find L.C.M of 24, 15, 36

24 = 2³ × 3
15 = 3 × 5
36 = 2² × 3²

∴ L.C.M = 2³ × 3² × 5¹
= 8 × 9 × 5
= 360
If a number has to be exactly divisible by 24, 15, and 36, then it has to be divisible by 360. Greatest 6 digit number is 999999.
Common multiplies of 24, 15, 36 with 6 digits are 103680, 116640, 115520, …933120, 999720 with six digits.
∴ The greatest number consisting 6 digits which is exactly divisible by 24, 15, 36 is 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.
Solution:
The least number that is divisible by the first ten natural numbers is 2520.
Hint:
1,2, 3,4, 5, 6, 7, 8,9,10
The least multiple of 2 & 4 is 8
The least multiple of 3 is 9
The least multiple of 7 is 7
The least multiple of 5 is 5
∴ 5 × 7 × 9 × 8 = 2520.
L.C.M is 8 × 9 × 7 × 5
= 40 × 63
= 2520

Samacheer Kalvi 10th Maths Book Solutions

Tamilnadu State Board Class 9 Maths Solutions Chapter 1 Set Language Ex 1.4

Question 1.
If P= {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8}, then find
(i) (P ∪ Q) ∪ R
(ii) (P ∩ Q) ∩ S
(iii) (Q ∩ S) ∩ R
Solution:
(i) (P ∪ Q) ∪ R
(P ∪ Q) = {1, 2, 5, 7, 9} ∪ {2, 3, 5, 9, 11} = {1, 2, 3, 5, 7, 9, 11}
(P ∪ Q) ∪ R = {1, 2, 3, 5, 7, 9, 11} ∪ {3, 4, 5, 7, 9} = {1, 2, 3, 4, 5, 7, 9, 11}

(ii)(P ∩ Q) ∩ S
(P ∩ Q) = {1, 2, 5, 7, 9} ∩ {2, 3, 5, 9, 11} = {2, 5, 9}
(P ∩ Q) ∩ S = {2, 5, 9} ∩ {2, 3, 4, 5, 8} = {2, 5}

(iii) (Q ∩ S) ∩ R
(Q ∩ S) = {2, 3, 5, 9, 11} ∩ {2, 3, 4, 5, 8} = {2, 3, 5}
(Q ∩ S) ∩ R = {2, 3, 5} ∩ {3, 4, 5, 7, 9} = {3, 5}

Question 2.
Test for the commutative property of union and intersection of the sets
P = {x : x is a real number between 2 and 7} and
Q = {x : x is an irrational number between 2 and 7}
Solution:
Commutative Property of union of sets
(A ∪ B) = (B ∪ A)
Here P = {3, 4, 5, 6}, Q = {√3, √5, √6}
P ∪ Q = {3, 4, 5, 6} ∪ {√3, √5, √6} = {3, 4, 5, 6, √3, √5, √6} … (1)
Q ∪ P = {√3, √5, √6} ∪ {3, 4, 5, 6}= {√3, √5, √6, 3, 4, 5, 6} …(2)
(1) = (2)
∴ P ∪ Q = Q ∪ P
∴ It is verified that union of sets is commutative.
Commutative Property of intersection of sets (P ∩ Q) = (Q ∩ P)
P ∩ Q = {3, 4, 5, 6} ∩ {√3, √5, √6} = { } … (1)
Q ∩ P = {√3, √5, √6} ∩ {3, 4, 5, 6} = { } …(2)
From (1) and (2)
P ∩ Q = Q ∩ P
∴ It is verified that intersection of sets is commutative.

Question 3.
If A = {p, q, r, s}, B = {m, n, q, s, t} and C = {m, n,p, q, s}, then verify the associative property of union of sets.
Solution:
Associative Property of union of sets
A ∪ (B ∪ C) = (A ∪ B) ∪ C)
B ∪ C = {m, n, q, s, t} ∪ {m, n,p, q, s}= {m, n,p, q, s, t}
A ∪ (B ∪ C) = {p, q, r, s} ∪ {m, n,p, q, s, t} = {m, n,p, q, r, s, t} … (1)
(A ∪ B) = {p, q, r, s} ∪ {m, n, q, s, t} = {p, q, r, s, m, n, t}
(A ∪ B) ∪ C = {p, q, r, s, m, n, t} ∪ {m, n,p, q, s} = {p, q, r, s, m, n, t} … (2)
From (1) & (2)
It is verified that A ∪ (B ∪ C) = (A ∪ B) ∪ C

Question 4.
Verify the associative property of intersection of sets for A = {-11, √2 ,√5 ,7}, B = {√3, √5, 6, 13} and C = {√2, √3,√5, 9}.
Solution:
Associative Property of intersection of sets A ∩ (B ∩ C) = (A ∩ B) ∩ C)
B ∩ C = {√3, √5, 6, 13} ∩ {√2, √3, √5, 9} = {√3, √5}
A ∩ (B ∩ C) = {-11, √2, √5, 7} ∩ {√3, √5} = {√5} … (1)
A ∩ B = {-11, √2, √5, 7} ∩ {√3, √5, 6, 13} = {√5}
(A ∩ B) ∩ C = {√5} ∩ {√2, √3, √5 ,9} = {√5} …(2)
From (1) and (2), it is verified that A ∩ (B ∩ C) = (A ∩ B) ∩ C

Question 5.
If A = {x : x = 2ⁿ, n ∈ W and n < 4}, B = {x: x = 2 n, n ∈ N and n ≤ 4} and C = {0, 1, 2, 5, 6}, then verify the associative property of intersection of sets.
Solution:
A = {x : x = 2ⁿ, n ∈ W, n < 4}
⇒ x = 2°= 1
x = 2¹ = 2
x = 2² = 4
x = 2³ = 8
∴ A = {1, 2, 4, 8}

B = {x : x = 2n, n ∈ N and n ≤ 4}
⇒ x = 2 x 1 = 2
x = 2 x 2 = 4
x = 2 x 3 = 6
x = 2 x 4 = 8
∴ B = {2, 4, 6, 8}
C = {0, 1, 2, 5, 6}
Associative property of intersection of sets
A ∩ (B ∩ C) = (A ∩ B) ∩ C
B ∩ C = {2, 6}
A ∩ (B ∩ C) = {1, 2,4, 8} ∩ {2, 6} = {2} … (1)
A ∩ B = {1, 2, 4, 8} ∩ {2, 4, 6, 8} = {2, 4, 8}
(A ∩ B) ∩ C = {2, 4, 8} ∩ {0, 1, 2, 5, 6} = {2} …(2)
From (1) and (2), It is verified that A ∩ (B ∩ C) = (A ∩ B) ∩ C

Samacheer Kalvi 9th Maths Book Answers

Tamilnadu State Board Class 10 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:
The positive integers when divided by 3 leaves remainder 2.
By Euclid’s division lemma a = bq + r, 0 < r < b.
Here a = 3q + 2, where 0 < q < 3, a leaves remainder 2 when divided by 3.
∴ 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.
Solution:
Let n – 1 and n be two consecutive positive integers. Then their product is (n – 1)n.
(n – 1)(n) = n² – n.
We know that any positive integer is of the form 2q or 2q + 1 for some integer q. So, following cases arise.

Case I. When n = 2q.
In this case, we have
n² – n = (2q)² – 2q = 4q² – 2q = 2q(2q -1)
⇒ n² -n = 2r, where r = q(2q -1)
⇒ n² -n is divisible by 2.

Case II. When n = 2q + 1
In this case, we have
n² – n = (2q +1)² – (2q + 1)
= (2q +1)(2q + 1 – 1) = 2q(2q +1)
⇒ n² – n = 2r, where r = q (2q + 1).
⇒ n² – n is divisible by 2.
Hence, n² – n is divisible by 2 for every positive integer n.
Hence it is Proved

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 = 13 q + 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.
Solution:
Let x be any integer.
The square of x is x².
Let x be an even integer.
x = 2q + 0
then x² = 4q² + 0
When x be an odd integer
When x = 2k + 1 for some interger k.
x² = (2k + 1 )²
= 4k² + 4k + 1
= 4k (k + 1) + 1
= 4q + 1
where q = k(k + 1) is some integer.
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.
Solution:
The required number is the H.C.F. of the numbers.
1230 – 12 = 1218,
1926 – 12 = 1914
First we find the H.C.F. of 1218 & 1914 by Euclid’s division algorithm.
1914 = 1218 × 1 + 696
The remainder 696 ≠ 0.
Again using Euclid’s algorithm
1218 = 696 × 1 + 522
The remainder 522 ≠ 0.
Again using Euclid’s algorithm.
696 = 522 × 1 + 174
The remainder 174 ≠ 0.
Again by Euclid’s algorithm
522 = 174 × 3 + 0
The remainder is zero.
∴ The H.C.F. of 1218 and 1914 is 174.
∴ The required number 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
The remainder 4 ≠ 0.
Again applying division lemma
28 = 4 × 7 + 0
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?
Solution:
Let a (+ve) integer be x.
x = 88 × y + 61
61 = 11 × 5 + 6 (∵ 88 is multiple of 11)
∴ 6 is the remainder. (When the number is divided by 88 giving the remainder 61 and when divided by 11 giving the remainder 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.

Samacheer Kalvi 10th Maths Book Solutions

Tamilnadu State Board Class 9 Maths Solutions Chapter 1 Set Language Ex 1.3

Question 1.
Using the given venn diagram, write the elements of
(i) A
(ii) B
(iii) A ∪ B
(iv) A ∩ B
(v) A – B
(vi) B – A
(vii) A’
(viii) B’
(ix) U

Solution:
(i) A = {2, 4, 7, 8, 10}
(ii) B = {3, 4, 6, 7, 9, 11}
(iii) A ∪ B = {2, 3, 4, 6, 7, 8, 9, 10, 11}
(iv) A ∩ B = {4, 7}
(v) A – B = {2, 8, 10}
(vi) B – A = {3, 6, 9, 11}
(vii) A’ = {1, 3, 6, 9, 11, 12}
(viii) B’ = {1, 2, 8, 10, 12}
(ix) U = {1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12}.

Question 2.
Find A ∪ B, A ∩ B, A – B and B – A for the following sets.
(i) A = {2, 6, 10, 14} and B = {2, 5, 14, 16}
(ii) A = {a, b, c, e, u} and B = {a, e, i, o, u]
(iii) A = {x : x ∈ N, x ≤ 10} and B = {x : x ∈ W, x < 6}
(iv) A = Set of all letters in the word “mathematics” and B = Set of all letters in the word “geometry”
Solution:
(i) A= {2, 6, 10, 14} and B = {2, 5, 14, 16}
A ∪ B = {2, 6, 10, 14} ∪ {2, 5, 14, 16} = {2, 5, 6, 10, 14, 16}
A ∩ B = {2, 6, 10, 14} ∩ {2, 5, 14,16} = {2,14}
A – B = {2, 6, 10, 14} – {2, 5, 14, 16} = {6, 10}
B – A = {2, 5, 14,16} – {2, 6, 10, 14}  = {5,16}

(ii) A = {a, b, c, e, u} and B = {a, e, i, o, u}
A ∪ B = {a, b, c, e, u) ∪ {a, e, i, o, u) = {a, b, c, e, i, o, u}
A ∩ B = {a, b, c, e, u} ∩ {a, e, i, o, u} {a, e, u}
A – B = {a, b, c, e, u) – {a, e, i, o, u) = {b, c}
B – A = {a, e, i, o, u} – {a, b, c, e, u} =  {i, o}

(iii) x ∈ {1,2, 3, ……..} ; x ∈ {0,1,2, 3, 4, 5, ……..}
A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
B = {0, 1, 2, 3, 4, 5}
A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ∪ {0, 1, 2, 3, 4, 5} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A ∩ B = {1,2, 3,4, 5, 6, 7, 8, 9, 10} ∩ {0, 1, 2, 3,4, 5} = {1, 2, 3, 4, 5}
A – B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {0, 1, 2, 3, 4, 5} = {6, 7, 8, 9, 10}
B – A = {0, 1, 2, 3, 4, 5} – {1, 2, 3,4, 5, 6, 7, 8, 9, 10} = {0}

(iv) A= {m, a, t, h, e, i, c, s), B = {g, e, o, m, t, r, y)
A ∪ B = {m ,a, t, h, e, i, c, s} ∪ {g, e, o, m, t, r, y} = {m, a, t, h, e, i, c, s, g, o, r, y)
A ∩ B = {m, a, t, h, e, i, c, 5} ∩ {g, e, o, m, t,r,y} = {m, t, e}
A – B = {m ,a, t, h, e, i, c, s} ∪ {g, e, o, m, t, r, y} = {a, h, i, c, s)
B – A = {m, a, t, h, e, i, c, 5} ∩ {g, e, o, m, t,r,y} = {g, o, r, y}

Question 3.
If U = {a, b, c, d, e,f g ,h}, A = {b, d, f, h} and B = {a, d, e, h}, find the following sets.
(i) A’
(ii) B’
(iii) A’ ∪ B’
(iv) A’ ∩ B’
(v) (A ∪ B)’
(vi) (A ∩ B)’
(vii) (A’)’
(viii) (B’)’
Solution:
(i) A’ = U – A = {a, b, c, d,e, f, g, y} – {b, d, f, h} = {a, c, e, g}
(ii) B’ = U – B = {a, b, c, d, e, f, g, y) – {a, d, e, h] = {b, c, f, g}
(iii) A’ ∪ B’= {a, c, e, g} ∪ {b, c, f, g} = {a, b, c, e, f g}
(iv) A’ ∩ B’= {a, c, e, g} ∩ {b, c, f, g} = {c, g}
(v) (A ∪ B)’ = U – (A ∪ B) = {a, b ,c, d, e ,f , g, y) – {a, b, d, e, f, h} = {c, g}
(vi) (A ∩ B)’ = U – (A ∩B) = {a, b, c, d, e, f, g, y} – {d, h} = {a, b, c, e,f, g}
(vii) (A’)’ = U – A’ = {a, b, c, d, e, f, g, h} – {a, c, e, g} = {b, d, f, h)
(viii) (B’)’ = U – B’ = {a, b, c, d, e, f, g, h} – {b, c, f, g} = {a, d, e, h}

Question 4.
Let U = {0, 1, 2 , 3, 4, 5, 6, 7}, A = {1, 3, 5, 7} and B = {0, 2, 3, 5, 7}, find the following sets.
(i) A’
(ii) B’
(iii) A ‘ ∪ B’
(iv) A’ ∩ B’
(v) (A ∪ B)’
(vi) (A ∩ B)’
(vii) (A’)’
(viii) (B’)’
Solution:
(i) A’ = U – A = {0, 1 ,2, y, 4, 5, 6, 7} – {1, 3, 5, 7} = {0, 2, 4, 6}
(ii) B’ = U – B = {0, 1, 2, 3, 4, 5, 6 ,7} – {0, 2, 3, 5, 7} = {1, 4, 6}
(iii) A’ ∪ B’ = {0, 2, 4, 6} ∪ {1, 4, 6} = {0, 1, 2, 4, 6}
(iv) A’ ∩ B’ = {0, 2, 4, 6} ∩ {1, 4, 6} = {4, 6}
(v) (A ∪ B)’ = U – (A ∪ B) = {0, 1, 2, 3, 4, 5, 6, 7} – {0, 1, 2, 3, 5, 7} = {4, 6}
(vi) (A ∩ B)’ = U – (A ∩ B)= {0, 1, 2, 3, 4, 5, 6, 7} – {3,5,7} = {0, 1, 2, 4, 6}
(vii) (A’)’ = U – A’ = {0, 1, 2, 3, 4, 5, 6, 7} – {0, 2, 4, 6} = {1, 3, 5, 7}
(viii) (B’)’ = U – B’ = {0, 1, 2, 3, 4, 5, 6, 7} – {1, 4, 6} = {0, 2, 3, 5, 7}.

Question 5.
Find the symmetric difference between the following sets.
(i) P = {2, 3, 5, 7, 11} and Q = {1, 3, 5, 11}
(ii) R = {l, m, n, o, p} and S = {j, l, n, q)
(iii) X = {5, 6, 7} and Y = {5, 7, 9, 10}
Solution:
(i) P = {2, 3, 5, 7, 11}
Q= {1, 3, 5, 11}
P – Q = {2, 3, 5, 7, 11} – {1, 3, 5, 11} = {2, 7}
Q – P = {1, 2, 3, 411} – {2, 3, 5, 7, 11} = {1}
P ∆ Q = (P – Q) ∪ (Q – P) = {2, 7} ∪ {1} = {1, 2, 7}

(ii) R = {l, m, n, o, p}
S = {j, l, n, q}
R – S = {l, m, n, o, p) – {j, l, n, q} = {m, o, p)
s – R = {j, l, y, q) – {l, m, n, o, p}= {j, q}
R ∆ S = (R – S) ∪ (S – R) = {m, o, p) ∪ {j, q} = {j, m, o, p, q)

(iii) X = {5, 6, 7}
Y = {5, 7, 9, 10}
X – Y = {5, 6, 7} – {5, 7, 9, 10} – {6}
Y – X = {5, 6, 9, 10} – {5, 6, 7} = {9, 10}
X ∆ Y = (X – Y) ∪ (Y – X)= {6} ∪ {9, 10} = {6, 9, 10}.

Question 6.
Using the set symbols, write down the expressions for the shaded region in the following



Solution:
(i) X – Y
(ii) (X ∪ Y)’
(iii) (X – Y) ∪ (X – Y)

Question 7.
Let A and B be two overlapping sets and the universal set U. Draw appropriate Venn diagram for each of the following,
(i) A ∪ B
(ii) A ∩ B
(iii) (A ∩ B)’
(iv) (B – A)’
(v) A’ ∪ B’
(vi) A’ ∩ B’
(vii)What do you observe from the diagram (iii) and (v)?
Solution:
(i) A ∪ B

(ii) A ∩ B

(iii) (A ∩ B)’

(iv) (B – A)’

(v) A’ ∪ B’

(vi) A’ ∩ B’

(vii) From the diagram (iii) and (v) we observe that (A ∩ B)’ = A’ ∪ B’.

Samacheer Kalvi 9th Maths Book Answers

Tamilnadu State Board Class 10 Maths Solutions Chapter 1 Relations and Functions Additional Questions

Question 1.
Let A = {1, 2, 3, 4} and B = {-1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} Let R = {(1, 3), (2, 6), (3,10), (4, 9)} ⊂ A × B be a relation. Show that R is a function and find its domain, co-domain and the range of R.
Answer:
Domain of R = {1, 2, 3, 4}
Co-domain of R = B = {-1,2, 3, 4, 5, 6, 7, 9, 10, 11,12}
Range of R= {3,6,10,9}

Question 2.
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 and (iv) a graph.
Solution:
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) A set of ordered pairs.
f = {(0,1), (1,3), (2, 5), (3, 7)}
(ii) A table

(iii) An arrow diagram

(iv) A Graph f = {(x,f(x)/x ∈ A}
= {(0,1), (1,3), (2, 5), (3, 7)}

Question 3.
State whether the graph represent a function. Use vertical line test.

Solution:
It is not a function as the vertical line PQ cuts the graph at two points.

Question 4.
Let f = {(2, 7), (3, 4), (7, 9), (-1, 6), (0, 2), (5,3)} be a function from A = {-1,0,2,3,5,7} to B = {2, 3, 4, 6, 7, 9}. Is this (i) an one-one function (ii) an onto function, (iii) both one- one and onto function?
Solution:
It is both one-one and onto function.

All the elements in A have their separate images in B. All the elements in B have their preimage in A. Therefore it is one-one and onto function.

Question 5.
A function f: (-7,6) → R is defined as follows.

Solution:

(i) 2f(-4)+ 3f(2)
f(-4) = x + 5 = -4 + 5 = 1
2f(-4) = 2 × 1 = 2
f(2) = x + 5 = 2 + 5 = 7
3f(2) = 3(7) = 21
∴ 2f(-4) + 3f(2) = 2 + 21 = 23
f(-7) = x² + 2x + 1
= (-7)² + 2(-7) + 1
= 49 – 14 + 1 = 36
f(3) = x + 5 = -3 + 5 = 2
f(-7) – f(-3) = 36 – 2 = 34

Question 6.
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 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 pre¬images in A. Therefore it is not on-to function.

Question 8.
A function f: [1, 6] → R is defined as follows:

Find the value of (i) f(5), (ii) f(3), (iii) f(2) – f(4).
Solution:

Question 9.
The following table represents a function from A = {5, 6, 8, 10} to B = {19, 15, 9, 11}, where f(x) = 2x – 1. Find the values of a and b.
Solution:

A = {5,6,8, 10}, B = {19, 15,9, 11}
f(x) = 2x – 1
f(5) = 2(5) – 1 = 9
f(8) = 2(8) – 1 = 15
∴ a = 9, b = 15

Question 10.
If R = {(a, -2), (-5,6), (8, c), (d, -1)} represents the identity function, find the values of a,b,c and d.
Solution:
R = {(a,-2), (-5,b), (8,c), (d,-1)} represents the identity function.
a = -2, b = -5, c = 8, d = -1.

Samacheer Kalvi 10th Maths Book Solutions

Tamilnadu State Board Class 10 English Vocabulary Phrase & Clauses

♦ Phrase & Clauses (Text Book Page No.: 108)

A phrase is a related group of words. The words work together as a “unit,” but they do not have a subject and a verb.

Examples of Phrases:

1. the boy on the bus (noun phrase)
2. will be running (verb phrase)
3. in the kitchen (prepositional phrase)
4. very quickly (adverb phrase)
5. Rama and Sita (noun phrase)

A clause is a group of words that does have both a subject and a verb. Some clauses are independent, meaning that they express a complete thought. An independent clause is the same as a complete sentence. Some clauses are dependent, meaning that they cannot stand alone. They do have a subject and a verb, but they do not express a complete thought. Another word for dependent is subordinate.

Examples of Clauses

1. When I get home (dependent or subordinate clause)
2. The lights are not on. (independent clause)
3. Put it on the shelf, (independent clause-understood subject “you)

A Phrase, on the other hand, is a small group of words put together as a conceptual unit. It does not take a figurative meaning. The meaning of the word is literal. It can be long or short but it does not include the subject-verb pairing, necessary to make a clause.

For example, ‘looking stunning’; to live and breathe’; ‘comfortable bed’.

Example:

1. Next-week, Prasanth has planned to visit the countryside. In the above sentence, to visit the countryside is a phrase with the conceptual meaning of going on a visit to the countryside.
2. The child hid under the stairs when the mother called her for a bath. In the above sentence, under the stairs is a phrase.

Exercises
In the following sentences, state whether the underlined part is a phrase or a clause.

Question 1.
In spite of the heavy rains, we reached on time.
Answer:
In spite of the heavy rains – phrase

Question 2.
She passed her test because of her teacher.
Answer:
because of her teacher – phrase i

Question 3.
She got low marks in her exams because she was sick.
Answer:
Because she was sick – clause

Question 4.
Though she performed well in the interview, she didn’t get the job.
Answer:
Though she performed well in the interview – clause

Question 5.
In spite of his good grades, he couldn’t secure admission in a good college.
Answer:
In spite of his good grades – phrase

Question 6.
Driven by the storm, we took shelter under a bridge.
Answer:
Driven by the storm – phrase

Question 7.
Having delivered the message, he departed.
Answer:
Having delivered the message – phrase

Question 8.
Not knowing what to do, she stood there motionless.
Answer:
Not knowing what to do – phrase

Question 9.
Now that he got a medical seat, he has become more responsible.
Answer:
Now that he… – clause

Question 10.
Despite her protests, they took her jewels away.
Answer:
Despite her protests – phrase

Phrasal Verb

A Phrasal, verb is an idiomatic phrase consisting of a verb and another element, typically an adverb or a preposition or both, the meaning of which is different from the meaning of its separate parts.

For example, see to, or a combination of both, such as look down on.

Example:

1. The crew ran out of water and food before they could complete their expedition. In the above sentence, ran out is a Phrasal Verb. which means to use completely.
2. The Police personnel instructed the mob to go from the place during the strike. In the above sentence, go away is a Phrasal Verb which means to leave from the place.

Learning Points:

• A phrasal verb is made up of a verb and an adverbial or prep Prepositional particle.
• It is used idiomatically to convey a special meaning completely different from the meaning expressed by the verb or the particle.
• The same verb followed by different particles conveys different meanings.
• A list of phrasal verbs and their meanings selected from the prose lessons is given below. Learn them very carefully before doing the. exercises that follow.

Example:
get – acquire
get together – meeting

For Reading:

Phrasal verb	Meaning	Usage
look up	refer	I look up the word in the dictionary.
look after	take care of	I look after the child.
get on	continue	I want you to get on with the work.
get up	rise from bed	I always get up early.
break down	cease to function	My car is break down at present.
put off	postpone	The manager put off the meeting.
turn down	refuse	They turned down my offer.
set Out	displays	His work is very well set out.
bring in	introduce	New traffic rules are brought in.
look up to	admire	Cricketers always look up to Kohli.
hit on	think of	He hit on a brilliant idea.
reel off	recite	He reeled off lines after line.
look into	investigate	The police look into the matter.
get over	recover	She got over her grief.
take after	resemble	He takes after his mother.
take on	accept	He is willing to take on the job.
put up with	tolerate	I can’t put up with such a noisy class.
put aside	save	I put aside a little every month.
give up	abandon / stop	I give up smoking.
call out	summon	He called out an engineer.
put up ‘	build	We are putting up a new compound.
pass away	die	His mother passed away last month.
break into	enter (by force)	The thief broke into the house.
give away	free	He gives away a sample pack to all.
go after	to chase	The police went after the criminals.
call up	invite	The teacher calls up the SPL.
call on	visit	She called on me yesterday.
get out	leave	I will get out soon.
carry out	fulfil	We carry out a successful attack.
take away	remove	The robbers took away the jewels.

More Phrasal Verbs:

Phrasal Verb	Meaning
back up	support
bear with	endure, tolerate
break up	disperse, shatter
break down	fall, stop working, collapse
bring forth	produce
bring about	make something happen
bring out	publish
bring up	educate, rear
call for	demand
call off	cancel
call on	to pay a short visit to a person
carry on	continue
carryout	execute, perform
come on	hurry
dealt with	managed
drop-in	visit
drop out	discontinue
get away	escape
get on	have a friendly relationship (with), cope with
get over	overcome, recover
get through	pass
get up	rise
give in	yield
give up	abandon
give way	yield
give away	present, distribute
go after	chase, follow
go through	endure; examine; study something
hold on	wait, stop
hold up	delay
keep on	continue
keep up	continue, maintain
lay by –	keep for future use
look after	take care of
look for	to try to find/search
look over/into	examine, investigate
look up (book)	search for, refer
look up to	admire
make out	understand
make over	redo, renovate
pass away	die
pass-through	undergo
put off	postpone, delay
put on	wear
put up with	endure, tolerate
run after	chase, seek
set back	to slow down the Droeress
set out	begin
set up	establish
set aside	discard
stand up	rise
stand by	support, wait
stand out	continue to resist
take down	write
take after	resemble
take up	to deal with
take off	leave the ground (aeroplane), remove
throw out	discard, reject
throw away	to get rid of as useless
turn away	refuse admission
turn down ‘	reject, refuse

Exercises

Replace the bold word in the sentence with one of the phrasal verbs given below to convey the same meaning :

Question 1.
The champion yielded to the strength of his opponent.
(a) gave on
(b) gave back
(c) gave in
(d) gave up
Answer:
(c) gave in

Question 2.
I will not yield to pressure, I shall face the challenge bravely.
(a) give away
(b) give in
(c) give out
(d) give off
Answer:
(b) give in

Question 3.
Our workers perform their jobs well.
(a) Carry off
(b) Carryover
(c) Carryout
(d) Carry for
Answer:
(c) Carryout

Question 4.
He renounced his wealth and became a social worker.
(a) give in
(b) give on
(c) give up
(d) give out
Answer:
(c) give up

Question 5.
I shall search the information using the Net.
(a) look at
(b) look in
(c) look for
(d) look after
Answer:
(c) look for

Question 6.
The plane left as scheduled.
(a) took away
(b) took off
(c) took apart
(d) took in
Answer:
(b) took off

Question 7.
The meeting was postponed due to bad weather.
(a) put off
(b) put in
(c) put up
(d) put on
Answer:
(a) put off

Question 8.
Sometimes we must continue to resist fof our own view.
(a) stand on
(b) stand out
(c) stand back
(d) standoff
Answer:
(b) stand out

Question 9.
You must keep some money for future use.
(a) lay off
(b) lay about
(c) lay by
(d) layover
Answer:
(c) lay by

Question 10.
Students should have a friendly relationship with their peers.
(a) get off
(b) get on
(c) get up
(d) get in
Answer:
(b) get on

Question 11.
School life is an enjoyable one. So you must have friendly relationship with your classmates.
(a) get up
(b) give up
(c) get on
(d) get back
Answer:
(c) get on

Question 12.
The lawyer managed the case cleverly.
(a) dealt in
(b) dealt with
(c) dealt out
(d) dealt on
Answer:
(b) dealt with

Question 13.
Smoking is injurious to health. So everyone must stop the habit.
(a) give away
(b) give in
(c) give up
(d) give out
Answer:
(c) give up

Question 14.
The match was postponed due to heavy rain.
(a) put off
(b) put up with
(c) put on
(d) put in
Answer:
(a) put off

Question 15.
I can’t tolerate your laziness.
(a) put off
(b) put forward
(c) put up with
(d) put out
Answer:
(C) put up with

Question 16.
Students should know how to have a friendly relationship with their class.
(a) get up
(b) get back
(c) get on
(d) get into
Answer:
(c) get on

Question 17.
I will discard these letters.
(a) throw out
(b) throw off
(c) take off
(d) take up
Answer:
(a) throw out

Question 18.
twill search the dictionary for the meaning of this word.
(a) look upon
(b) look up to
(c) look up
(d) look at
Answer:
(c) look up

Question 19.
The manager will cancel the meeting.
(a) call on
(b) call out
(c) call in
(d) call off
Answer:
(d) call off

Question 20.
He will surely pass the exam with flying colours.
(a) get out
(b) get in
(c) get through
(d) get on
Answer:
(c) get through

Tamilnadu Board Class 10 English Solutions

Tamilnadu State Board Class 9 Maths Solutions Chapter 1 Set Language Ex 1.2

Question 1.
Find the cardinal number of the following sets.
(i) M = {p, q, r, s, t, u}
(ii) P = {x : x = 3n + 2, n ∈ W and x < 15}
(iii) Q = {v : v = \(\frac { 4 }{ 3n }\) ,n ∈ N and 2 < n ≤ 5}
(iv) R = {x : x is an integers, x ∈ Z and -5 ≤ x < 5}
(v) S = The set of all leap years between 1882 and 1906.
Solution:
(i) n(M) = 6
(ii) W = {0, 1, 2, 3, ……. }
if n = 0, x = 3(0) + 2 = 2
if n = 1, x = 3(1) + 2 = 5
if n = 2, x = 3(2) + 2 = 8
if n = 3, x = 3(3)+ 2 =11
if n = 4, x = 3(4) + 2=14
∴ P= {2, 5, 8, 11, 14}
n(P) = 5

(iii) N = {1,2, 3, 4, …..}
n ∈ {3, 4, 5}

n(Q) = 3

(iv) x ∈ z
R = {-5, – 4, -3, -2, -1, 0, 1, 2, 3, 4}
n(R)= 10.

(v) S = {1884, 1888, 1892, 1896, 1904}
n (S) = 5.

Question 2.
Identify the following sets as finite or infinite.
(i) X = The set of all districts in Tamilnadu.
(ii) Y = The set of all straight lines passing through a point.
(iii) A = {x : x ∈ Z and x < 5}
(iv) B = {x : x² – 5x + 6 = 0, x ∈ N}
Solution:
(i) Finite set
(ii) Infinite set
(iii) A = { ……. , -2, -1, 0, 1, 2, 3, 4}
∴ Infinite set

(iv) x² – 5x + 6 = 0
(x – 3) (x – 2) = 0
B = {3, 2}
∴ Finite set.

Question 3.
Which of the following sets are equivalent or unequal or equal sets?
(i) A = The set of vowels in the English alphabets.
B = The set of all letters in the word “VOWEL”
(ii) C = {2, 3, 4, 5}
D = {x : x ∈ W, 1 < x < 5}
(iii) X = A = { x : x is a letter in the word “LIFE”}
Y = {F, I, L, E}
(iv) G = {x : x is a prime number and 3 < x < 23}
H = {x : x is a divisor of 18}
Solution:
(i) A = {a, e, i, o, u}
B = {V, O,W, E, L}
The sets A and B contain the same number of elements.
∴ Equivalent sets

(ii) C ={2, 3, 4, 5}
D = {2, 3, 4}
∴ Unequal sets

(iii) X = {L, I, F, E}
Y = {F, I, L, E}
The sets X and Y contain the exactly the same elements.
∴ Equal sets.

(iv) G = {5, 7, 11, 13, 17, 19}
H = {1, 2, 3, 6, 9, 18}
∴ Equivalent sets.

Question 4.
Identify the following sets as null set or singleton set.
(i) A = (x : x ∈ N, 1 < x < 2}
(ii) B = The set of all even natural numbers which are not divisible by 2.
(iii) C = {0}. (iv) D = The set of all triangles having four sides.
Solution:
(i) A = { } ∵ There is no element in between 1 and 2 in Natural numbers.
∴ Null set
(ii) B = { }∵ All even natural numbers are divisible by 2.
∴ B is Null set
(iii) C = {0} ∴ Singleton set (iv) D = { }
∵No triangle has four sides.
∴ D is a Null set.

Question 5.
State which pairs of sets are disjoint or overlapping?
(i) A = {f, i, a, s} and B = {a, n, f, h, s)
(ii) C = {x : x is a prime number, x > 2} and D = {x : x is an even prime number}
(iii) E = {x: x is a factor of 24} and F = {x : x is a multiple of 3, x < 30}
Solution:
(i) A = {f, i, a, s}
B = {a, n, f, h, s}
A ∩ B ={f, i, a, s} ∩ {a, n,f h, s} = {f, a, s}
Since A ∩ B ≠ φ , A and B are overlapping sets.

(ii) C = {3, 5, 7, 11, ……}
D = {2}
C ∩ D = {3, 5, 7, 11, …… } ∩ {2} = { }
Since C ∩ D = φ, C and D are disjoint sets.

(iii) E = {1, 2, 3, 4, 6, 8, 12, 24}
F = {3, 6, 9, 12, 15, 18, 21, 24, 27}
E ∩ F = {1, 2, 3, 4, 6, 8, 12, 24} ∩ {3, 6, 9, 12, 15, 18, 21, 24, 27}
= {3, 6, 12, 24}
Since E ∩ F ≠ φ, E and F are overlapping sets.

Question 6.
If S = {square,rectangle,circle,rhombus,triangle}, list the elements of the following subset of S.
(i) The set of shapes which have 4 equal sides.
(ii) The set of shapes which have radius.
(iii) The set of shapes in which the sum of all interior angles is 180°
(iv) The set of shapes which have 5 sides.
Solution:
(i) {Square, Rhombus}
(ii) {Circle}
(iii) {Triangle}
(iv) Null set.

Question 7.
If A = {a, {a, b}}, write all the subsets of A.
Solution:
A= {a, {a, b}} subsets of A are { } {a}, {a, b}, {a, {a, b}}.

Question 8.
Write down the power set of the following sets.
(i) A = {a, b}
(ii) B = {1, 2, 3}
(iii) D = {p, q, r, s}
(iv) E = φ
Solution:
(i) The subsets of A are φ, {a}, {b}, {a, b}
The power set of A
P(A ) = {φ, {a}, {b}, {a,b}}

(ii) The subsets of B are φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}
The power set of B
P(B) = {φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}

(iii) The subset of D are φ, {p}, {q}, {r}, {s}, {p, q}, {p, r}, {p, s}, {q, r}, {q, s}, {r, s},{p, q, r}, {q, r, s}, {p, r, s}, {p, q, s}, {p, q, r, s}}
The power set of D
P(D) = {φ, {p}, {q}, {r}, {s}, {p, q}, {p, r}, {p, s}, {q, r}, {q, s}, {r, s}, {p, q, r}, {q, r, s}, {p, r, s}, {p, q, s}, {p, q, r, s}

(iv) The power set of E
P(E) = { }.

Question 9.
Find the number of subsets and the number of proper subsets of the following sets.
(i) W = {red,blue, yellow}
(ii) X = { x² : x ∈ N, x² ≤ 100}.
Solution:
(i) Given W = {red, blue, yellow}
Then n(W) = 3
The number of subsets = n[P(W)] = 2³ = 8
The number of proper subsets = n[P(W)] – 1 = 2³ – 1 = 8 – 1 = 7

(ii) Given X ={1,2,3, }
X² = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
n(X) = 10
The Number of subsets = n[P(X)] = 2¹⁰ = 1024
The Number of proper subsets = n[P(X)] – 1 = 2¹⁰ – 1 = 1024 – 1 = 1023.

Question 10.
(i) If n(A) = 4, find n[P(A)].
(ii) If n(A) = 0, find n[P(A)].
(iii) If n[P(A)] = 256, find n(A).
Solution:
(i) n( A) = 4
n[ P(A)] = 2ⁿ = 2⁴ = 16
(ii) n(A) = 0
n[P(A)] = 2⁰ = 1
(iii) n[P(A)] = 256

n[P(A)] = 2⁸
∴ n(A) = 8.

Samacheer Kalvi 9th Maths Book Answers

Tamilnadu State Board Class 10 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.
Solution:
(x² – 3x, y² + 4y) = (-2,5)
x² – 3x = -2
x² – 3x + 2 = 0
(x – 2)(x – 1) = 0

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

(i) f(0)
(ii) f(3)
(iii) f(a + 1) in terms of a.(Given that a > 0)
Solution:
(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 = {9,1O,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

Solution:

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.
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?
Solution:
A = {1,2),B = (1,2,3,4)
C = {5,6},D = {5,6,7,8)

(A × C) ⊂ (B × D) It is proved.

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

Hence it is proved.

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

Question 10.
Write the domain of the following real functions

Solution:
(i) f(x) = \(\frac { 2x+1 }{ x-9 } \)
The denominator should not be zero as the function is a real function.
∴ The domain = R – {9}
(ii) p(x) = \(\frac{-5}{4 x^{2}+1}\)
The domain is R.
(iii) g(x) = \(\sqrt { x-2 }\)
The domain = (2, ∝)
(iv) h(x) = x + 6
The domain is R.

Samacheer Kalvi 10th Maths Book Solutions

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