Laxmi

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 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