Samacheer Kalvi 10th Maths Solutions Chapter 1 Pdf
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TN Samacheer Kalvi 10th Maths Solutions Chapter 1 – Relations and Functions
Board |
TNSCERT Class 10th Maths |
Class |
10 Class |
Subject |
Maths |
Chapter |
1 (Exercise 1.1. 1.2, 1.3, 1.4,1.5,1.6,1.7,1.8,1.9,1.10) |
Chapter Name |
Relations and Functions |
TNSCERT Class 10th Maths Pdf
All Exercise Solution
Exercise- 1.1
(1) Find A×B, A×A and B×A
(i) A = {2, -2, 3} and B = {1, -4}
(ii) A = B = {p, q}
(iii) A = { m, n};
B = ⌀
Solution:
(i) Given that A = {2, -2, 3}
& B = {1, -4}
A×B = {2, -2, 3} × {1, -4}
= {(2, 1), (2, -4), (-2, 1), (-2, -4), (3,1), (3, -4)}
A×A = {2, -2, 3} × {2, -2, 3}
= {(2, 2), (2, -2) , (2,3), (-2, 2), (-2, -2),(-2,3), (3,2), (3,-2), (3,3)}
B×A = {1, -4} × {2, -2, 3}
= {(1,2), (1,-2), (1,3), (-4,2), (-4,-2), (-4,3)}
(ii) A = {p, q} and B = {p, q}
A×B = {p, q} × {p, q} = {(p, p), (p, q), (q, p), (q, q)}
A×A = {p, q} × {p, q} = {(p, p), (p, q),(q, p), (q, q)}
B×A = {p, q} × {p, q} = {(p, p), (p, q), (q, p), (q, q)}
(iii) A = {m, n} and B = ⌀
A × B = {m, n} × {⌀}
= ⌀
A × A = {m, n} × {m, n}
= {(m, m), (m, n), (n, m), (n, n)}
B × A = {⌀} × {m, n}
= ⌀
(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} and B = {x | x is a prime number less than 10}
B = {2, 3, 5, 7}
Thus, A × B = {1, 2, 3} × {2, 3, 5, 7}
= {(1, 2), (1, 3), (1,5), (1, 7), (2, 2), (2, 3),(2, 5), (2, 7), (3, 2), (3, 3), (3, 5) , (3, 7)}
B × A = {2, 3, 5, 7} × { 1, 2 ,3}
= {(2, 1),(2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (5, 1), (5, 2), (5, 3), (7, 1), (7, 2), (7, 3)}
(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)}
Every pair first element belong to set B and 2nd element belong to set A [∵ B×A]
∴ set of B = {-2, 0, 3}
Set of A = {3, 4}
(4) If A = {5, 6}, B = {4, 5, 6}, C = {5, 6, 7}, show that A × A = (B×B) ∩ (C×C).
Solution:
Given that A = {5, 6}
B = {4, 5, 6} and C = {5, 6, 7}
∴ A × A = {5, 6} × {5, 6} = {(5, 5), (5, 6), (6, 5), (6, 6)}….. (i)
∴ B × B = {4, 5, 6} × {4, 5, 6}
= {(4,4), (4,5) , (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5) , (6, 6)} …. (ii)
∴ C × C = {5, 6, 7} × {5, 6, 7}
= {(5, 5), (5, 6), (5, 7), (6, 5), (6, 6), (6, 7), (7, 5), (7, 6), (7, 7)}…. (iii)
From, (ii) and (iii),
(B × B) ∩ (C × C)
= {(4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)} ∩ {(5, 5), (5, 6), (5, 7), (6, 5),(6, 6), (6, 7),(7, 5), (7, 6), (7, 7)}
= {(5, 5), (5, 6), (6, 5), (6, 6)}
= A × A [By equation (i)]
Therefore, A×A = (B×B) ∩ (C×C) [Proved]
(5) Given A = {1, 2, 3}, B = {2, 3, 5}, C = {3, 4} and D = {1, 3, 5} check if (A∩C) × (B∩D) = (A×B) ∩ (C×D) is true?
Solution:
A = {1, 2, 3}, B = {2, 3, 5}, C = {3, 4} and D = {1, 3, 5}
∴ A∩C= {1, 2, 3} ∩ {3, 4} = {3} and B∩D {2, 3, 5} ∩ {1, 3, 5} = {3, 5}
∴ (A∩C) × (B∩D) = {3} × {3, 5}
= {(3, 3), (3, 5)} …. (i)
Now, A×B = {1, 2, 3} × {2, 3, 5}
= {(1,2), (1,3), (1,5), (2,2), (2,3), (2,5), (3,2), (3,3), (3,5)}…..(ii)
C×D = {3,4} × {1,3,5}
= {(3,1), (3,3), (3,5) , (4,1), (4,3), (4,5)}……(iii)
From (ii) and (iii),
(A×B) ∩ (C×D)
= {(1,2), (1,3), (1,5), (2,2),(2,3), (2,5), (3,2), (3,3), (3,5)}
∩ {(3,1), (3,3), (3,5), (4,3), (4,5)}
= {(3,3) , (3,5)} ……..(iv)
From (i) and (iv),
(A∩C) × (B∩D) = (A×B) ∩ (C×D)
It is true.
(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)
=> A = {x ∈ W | x < 2}
A = {0,1} [∵ W = whole numbers = {0,1,2,3,…..};]
B = {x∈ N | 1<x≤ 4}
B = { 2,3,4} [∵N = Natural numbers = {1,2,3,4,…..};]
And C = {3,5}.
(i) Now, B∪C = {2,3,4} ∪ {3,5} = {2,3,4,5}
A × (B∪C) = {0,1} × {2,3,4,5}
= {(0,2) , (0,3), (0,4) , (0,5), (1,2), (1,3), (1,4),(1,5)}
A × (B∪C) = {(0,2), (0,3), (0,4),(0,5), (1,2), (1,3),(1,4),(1,5)} ……..(a)
Now, A×B = {0,1} × {2,3,4}
= {(0,2), (0,3), (0,4) (1,2), (1,3), (1,4)}…..(b)
A × C = {0,1} × {3,5}
= {(0,3), (0,5), (1,3), (1,5)}……(c)
From (b) and (c) , equation,
(A×B) ∪ (A×C)
= {(0,2),(0,3), (0,4),(1,2),(1,3),(1,4)} ∪ {(0,3), (0,5), (1,3),(1,5)}
= {(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5)}
= A × (B∪C) by equation (a)
Thus, A × (B∪C) = (A×B) ∪ (A×C) .
(ii) Now, (B∪C) = {2,3,4} ∩ {3,5}
= {3}
∴ A × (B∩C) = {0,1} ×{3}
A × (B∩C) = {(0,3), (1,3)} ……(a)
Then, A×B = {0,1} × {2,3,4}
= {(0,2), (0,3), (0,4), (1,2), (1,3), (1,4)…..(b)
And A×C = {0,1} × {3,5}
= {(0,3), (0,5), (1,3), (1,5)}……(C)
From eqn (b) and (c)
(A×B) ∩ (A×C)
= {(0,2), (0,3), (0,4), (1,2), (1,3), (1,4)} ∩ {(0,3), (0,5), (1,3),(1,5)}
= {(0,3), (1,3)}
= A × (B∩C) by equation (a)
Thus, A × (B∩C) = (A×B) ∩ (A×C)
(iii) Now, A∪B = {0,1} ∪ {2,3,4} = {0,1,2,3,4}
(A∪B) × C = {0,1,2,3,4,} × {3,5}
= {(0,3), (0,5), (1,3), (1,5), (2,3), (2,5), (3,3), (3,5),(4,3),(4,5)}……..(a)
Then, A×C = {0,1} × {3,5}
= {(0,3), (0,5), (1,3), (1,5)}
B×C = {2,3,4} ×{3,5}
= {(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)} ={(0,3),(0,5),(1,3),(1,5),(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)}
= (A∪B) × C by equation (a)
∴ (A∪B) × C = (A×C) ∪ (B×C) .
(7) Let A = The set of all natural numbers less than 8,
B = The aet 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= The set of all natural numbers less than 8
= {1,2,3,4,5,6,7}
B = The set of all prime numbers less than 8,
= {2,3,5,7}
C = The set of even prime number.
= {2} [∵ 2 is the only even prime number]
(i) (A∩B) = {1,2,3,4,5,6,7} ∩ {2,3,5,7}
= {2,3,5,7}
∴ (A∩B) × C = {2,3,5,7} × {2}
= {(2,2) , (3,2),(5,2), (7,2)}…..(a)
Now, A×C = {1,2,3,4,5,6,7}×{2}
= {(1,2),(2,2),(3,2), (4,2), (5,2), (6,2),(7,2)}
B × C = {2,3,5,7} × {2}
= {(2,2),(3,2),(5,2),(7,2)}
(A×C) ∩ (B×C)
= {(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(7,2)} ∩ {(2,2),(3,2),(5,2),(7,2)}
= {(2,2),(3,2),(5,2),(7,2)}
= (A∩B) ×C by equation (a)
∴ (A∩B) × C = (A×C) ∩ (B×C).
(ii) B – C = {2,3,5,7} – {2}
= {3,5,7}
∴ A × (B-C) = {1,2,3,4,5,6,7} × {3,5,7}
= {(1,3), (1,5),(1,7),(2,3),(2,5),(2,7), (3,3), (3,5),(3,7),(4,3),(4,5),(4,7),(5,3),(5,5),(5,7),(6,3),(6,5),(6,7),(7,3),(7,5),(7,7)}…(a)
Now, A×B = {1,2,3,4,5,6,7} × {2,3,5,7}
={(1,2),(1,3),(1,5),(1,7),(2,2),(2,3),(2,5),(2,7),(3,2),(3,3),(3,5),(3,7),(4,2),(4,3),(4,5),(4,7),(5,2),(5,3),(5,5),(5,7),(6,2),(6,3),(6,5),(6,7),
(7,2),(7,3),(7,5),(7,7)}
A × C = {1,2,3,4,5,6,7} ×{2}
= {(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(7,2)}
∴ (A×B) – (A×C)
={(1,2),(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)} – {(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(7,2)}
= {(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)}
= A × (B-C) by equation (a)
∴ A × (B-C) = (A×B) – (A×C)
Exercise- 1.2
(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)}
=> A= {1,2,3,7} , B = {3,0,-1,7}
Now, A×B = {1,2,3,7} × {3,0,-1,7}
= {(1,3),(1,0),(1,-1),(1,7),(2,3),(2,0),(2,-1),(2,7),(3,3),(3,0),(3,-1),(3,7),(7,3),(7,0),(7,-1),(7,7)}
(i) R1 = {(2,1),(7,1)}
Then, (2,1) ,(7,1) ∈ R1 but (2,1),(7,1) ∉ A×B .
So, R1 is not a relation from A to B.
(ii) R2 = {(-1,1)}
Here, (-1,1) ∈ R1 but (-1,1) ∉ A× B
Therefore, R2 IS not a relation from A to B
(iii) R3 = {(2,-1) , (7,7), (1,3)}
We see that R3 ∈ A×B .
Then, R3 ≤ A × B
Therefore, R3 is a relation from A to B.
(iv) R4 = {(7,-1),(0,3),(3,3),(0,7)}
Now, (0,3) , (0,7), ∉ A×B
∴ Thus, R4 is not a relation from A to B .
(2) Let A = {1,2,3,4,……,45} and R be the relation defined as “is square of a number” on A . write R as a subset of A × A also, find the domain and range of R .
=> A = {1,2,3,4,…..,45}.
Now, R be the relation defined as “is square of a number” on A.
Then, 1 R 1 [∵ 12 = 1]
2 R 4 [∵ 22 =4]
Similarly, 3 R 9, 4 R 16, 5 R 25, 6 R 36 .
Thus, R = {(1,1), (2,4),(3,9),(4,16),(5,25),(6,36)}
Now, R ∈ A× A
Then, R ≤ A × A
R is a subset of A ×A .
∴ Domain of R = {1,2,3,4,5,6}
∴ Range of R = {1,4,9,16,25,36}
(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.
=> Let R = {(x, y)/y = x+3, x ∈ {0,1,2,3,4,5}}
Then, R = {(0,3), (1,4) ,(2,5), (3,6),(4,7), (5,8)}
Because, x=0, y = x+3 = 0+3 = 3
Similarly, x = 1, y= 4,
x =2, y=5
x = 3 , y = 6
x = 4, y = 7
x = 5, y = 8 .
Therefore, R = {(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}
(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}
=> (i) Let R = {(x, y)| x = 2y , x ∈ {2,3,4,5}, y ∈{1,2,3,4}}
When, y =1,then x = 2.1=2
y = 2, then x = 2.2 = 4
y = 3, then x = 2.3 = 6
y = 4, then x = 2.4 = 8
Thus, R = {(2,1) , (4,2)} .
(a) arrow diagram ,
(ii) Let R = {(x, y)|y = x +3, x, y are natural number < 10}
When, x =1, then y = 1+3= 4
x = 2 , then y = 2+3 = 5
x = 3, then y = 3+3= 6
x = 4 , then y = 4+3= 7
x =5, then y =5 +3= 8
x = 6, then y = 6+3= 9
x = 7 , then y = 7+3= 10
x = 8, then y = 8+3= 11
x = 9, then y = 9+3= 12
Thus, R = {(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)}
(a) arrow diagram
(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 A5 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 x R y , where x is the salary given to person Y , express the relation R through an ordered pair and an arrow diagram.
∴ R = {(1O000,A1), (10000,A2), (10000,A3), (10000,A4),(10000,A5),(25000,C1),(25000,C2),(25000,C3),(25000,C4),(50000,m1),(50000,m2), (50000,m3), (100000,E1),(100000,E2)}
R = {(x, y); x ∈ {10000,25000,50000,100000}; y ∈ {A1, A2, A3, A4, A5 , C1, C2, C3, C4, m1, m2, m3, E1, E2}}
Exercise – 1.3
(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?
=> f = {(x, y) |x, y ∈ ℕ and y = 2x}
Now, ℕ = {1,2,3,4,…..}
∴ x ∈ ℕ = {1,2,3,…….}
∴ y ∈ ℕ = {1,2,3,…..}
When, x = 1, then, y =2.1 = 2
Similarly, x= 2, y= 2.2 = 4
X = 3, y = 2.3 = 6
∴ Domain = {1,2,3,…..} = ℕ
∴ Co- domain = {1,2,3,……} = ℕ
∴ Range = {2,4,6,…}
= {2x,; x1 ∈ ℕ }
We see that each element of domain has unique image in co-domain.
Therefore, f is a function.
(2) Let X = {3,4,6,8} . Determine whether the relation
R = {(x, f (x))| x∈ X, f(x) = x2 +1} is function from X to N ?
=> X = {3,4,6,8,}
Now, R = {(x, f(x) | x ∈ X ; f(x) = x2 +1}
∴ f(x) = x2+1.
When x = 3, then, f(x) = 9+1= 10
x = 4, then f(x) = 16+1 =17
x = 6, then f(x) = 36 +1 = 37
x = 8, then f(x) = 64 +1 = 65
Now, (10,17,37,65,} ∈ ℕ .
∴ R = {(3,10),(4,17),(6,37),(8,65)}
∴ we see that every element in x has unique image in ℕ.
∴ f : x → N is a function.
(3) Given the function f:x →x2 -5x +6, evaluate
(i) f (-1)
(ii) f (2a)
(iii) f (2)
(iv) f (x-1)
=> f:x → x2-5x +6
∴ f(x) = x2– 5x +6
(i) when, x = -1, then, f(x) = (-1)2 – 5.(-1) +6
= 1+5+6
= 12
∴ f(-1) = 12
(ii) when, x= 2a , then f (x) = (2a)2 – 5.(2a) +6
= 4a2 -10a +6
∴ f(2a) = 4a2 -10a+ 6
(iii) when x = 2 , then f(x) = (2)2 -5.(2) +6
= 4-10+6
= 0
∴ f (2) = 0
(iv) when, x = x-1, then f(x) = (x-1)2 – 5.(x-1) +6
= x2 -2x+1 -5x +5+6
= x2 – 7x +12
∴ f (x-1) = x2 -7x +12
(4) A graph representing the function f(x) is given in this graph it is clear that f(9) = 2 .
(i) find the following value 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 ?
=> since f (9) = 2
(i) similarly,
(a) f (0) = 9 (photo)
(b) f(7) =6
(c) f(2) = 6
(d) f (10) = 0. [by given graph]
(ii) if f(x) = 1 then, x value is 9.5 [by given graph]
∴ f (x) = 1 ;
∴ x = 9.5
(iii) By given graph ,
Domain = {0,1,2,3,4,5,6,7,8,9,10}
And Range = {0,1,2,3,4,5,6,7,8,9,} .
(iv) Now, f(6) = 5.[by given graph]
So, The image of 6 under f is 5 .
(5) Let f(x) = 2x +5 . If x≠ 0 then find f(x+2) – f(2)/x
=> Here f(x) = 2x +5
Then, f(x+2) = 2(x+2) +5
= 2x + 4+5
= 2x +9
∴ when, x = 2, f(x) = 2.2+5 = 4+5 = 9
∴ f(2) = 9
Now, f(x+2) – f(2)/x
= 2x +9-9/x
= 2x/x
= 2
Thus, f(x+2) – f(2)/x = 2.
(6) A function f is defined by f(x) = 2x -3
(i) find 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) .
=> Here f (x) = 2x -3.
(i) when x = 0 , then
f (x) = 2.0-3 = -3
∴ f(0) = -3
Similarly f (1) = -1
∴ f(0) + f(1) / 2
= -3 +(-1)/2
= -4/2
= -2
Therefore f(0) + f(1)/2 = -2
(ii) Now, f(x) = 0
2x -3 = 0
x = 3/2
(iii) Now f (x) = x
2x -3 = x
2x – x = 3
X = 3
(iv) Now, f(x) = f(1-x)
∴ f(1-x) = 2.(1-x) -3
= 2 – 2x – 3
= – (2x +1)
∴ f(x) = f(1-x)
2x -3 = – (2x + 1)
2x +2x = -1 +3
4x = 2
X = ½
(7) An open box is to be made from a square piece of material, 24cm on a side, by cutting equal squares from the corners and turning up the sides as shown given image . Express the volume V of the box as a function of x.
=> We see that given image.
Length = 24 -2x
Breadth = 24 – 2x
Height = x
∴ volume of the box
= l × b × h
= (24 – 2x) ×(24-2x) × x
= (576 -96x + 4x2) × x
= 4x3 – 96x2 + 576x.
∴ Thus, volume of V is 4x3 – 96x2 + 576x.
(8) A function f is defined by f(x) = 3 – 2x . Find x such that f(x2) = ( f (x))2 .
=> Here f (x) = 3-2x
Now, f(x2) = 3 -2x2
and (f(x))2 = (3-2x)2
= 9 – 12x +4x2
∴ f (x2) = (f(x))2
3 -2x2 = 9 -12x +4x2
4x2 + 2x2 -12x + 9 -3 = 0
6x2 – 12x +6 =0
x2 -2x +1 = 0
(x -1)2 = 0
x -1 = 0
x = 1 .
(9) A plane is flying at a speed of 500km per hour. Express the distance ‘d’ travelled by the plane as function of time t in hours .
=> We know that ,
Distance (d) = speed (s) × time (t)
∴ d = 500 × t
d = 500t
Let a function f is defined by
∴ d = f (t) = 500t
(10) The data in the adjacent table depicts the length of a person forehand and their corresponding height. Based on this data, a student find 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 person whose forehand length is 40cm.
(iv) Find the length of forehand of a person if the height is 53.3 inches.
Length ‘x’ of forehand (in cm) | Height ‘y’ (in inches) |
35 | 56 |
45 | 65 |
50 | 69.5 |
55 | 74 |
=> given that y = ax +b, where a, b are constant.
(i) Now, every length of forehand corresponding there exists a height .
So, This relation is a function.
Let y = f(x) = ax + b , where a, b are constant ……(i)
(ii) Now, x = 35 , y = 56 in equation (i)
Then, 35a + b = 56 …….(ii)
And x = 45 , y = 65 in equation….(i)
Then , 45a + b = 65 …….(iii)
Solve (ii) and (iii),
45a + b = 65
35a + b = 56
(-) (-) (-)
__________________________
10a = 9
∴ a = 0.9
From (ii) 35× 0.9 +b = 56
b = 56 – 31.5
b = 24. 5
∴ a = 0.9 and b = 24.5
(iii) Now, y = f(x) = ax +b
y = 0.9 x + 24. 5 [∵ a = 0.9, b = 24.5]
∴ when , x = 40cm , then ,
y = 0.9 × 40 + 24.5
= 36 + 24.5 = 60.5
∴ when forehand length is 40cm then height is 60.5 inches .
(iv) Here, y = 0.9 x + 24.5
When y = 53.3 inches then,
53.3 = 0.9x + 24.5
0.9x = 28.8
x = 288/0.9
x = 32
Therefore height is 53.3 inches then length of forehand is 32cm .
Exercise – 1.4
(1) Determine whether the graph given below represent function. Given reason for your answers concerning each graph .
(i)
(ii)
(iii)
(iv)
=> (i) if we draw the vertical line in graph then we will see that the vertical line meet the curve in two points.
So, This is not a represent the function.
(ii) If we draw the vertical line in the graph then we will see that the vertical line meet the curve in one point.
So, This graph is represent the function.
(iii) If we draw the vertical line in the graph we will see that the vertical line meet the curve in two points.
So, This graph is not a represent the function.
(iv) If we draw the vertical line in the graph then we will see that the vertical meet the curve in one point .
So, This graph is represent the function .
(2) Let f: A → B be a function defined by f(x) = x/2 -1, where A = {2,4,6,10,12} ,
B = {0,1,2,4,5,9} . Represent f by
(i) set of ordered pairs
(ii) a table
(iii) an arrow diagram
(iv) a graph
=> Here, f(x) = x/2 -1
When , x = 2, then f(x) = 2/2-1 = 1-1= 0
Similarly, f(4) = 1 , f(6) = 2 , f(10) = 4
f(12) = 5
(i) set of ordered pairs
f = {(2,0),(4,1),(6,2),(10,4),(12,5)}
(ii) a table
x | 2 | 4 | 6 | 10 | 12 |
f(x) | 0 | 1 | 2 | 4 | 5 |
(iii) an arrow diagram
(iv) a graph .
(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
=> given that
f = {(1,2),(2,2),(3,2),(4,3), (5,4)}
Let A = {1,2,3,4,5} , B ={2,3,4}
∴ f:A → B
(i) an arrow diagram
(ii) a table form
x | 1 | 2 | 3 | 4 | 5 |
f(x) | 2 | 2 | 2 | 3 | 4 |
(iii) a graph .
(4) Show that the function f: ℕ → ℕ defined by f(x) = 2x -1 is one – one but not onto .
=> f: ℕ → ℕ
∴ f(x) = 2x -1
When, x=1, then f(1) = 2.1 -1 = 2-1 = 1
∴ Similarly, f(2) = 2.2-1 = 4-1 = 3
f(3) = 3.2 -1= 5
f(4) = 4.2 -1 = 7
f(5) = 5.2 -1 = 9
∴ co-domain = {1,2,3,4,5,….} = ℕ
∴ Range = {1,3,5,7,……} ∈ ℕ .
Co-domain different elements in ℕ there are different image in ℕ .
So, f is one-one
But the range of f is not equal to the co-domain of f .
∴ so, f is not onto .
Hence, f is one- one but not onto. [proved]
(5) Show that the function f: ℕ → ℕ defined by f(m) = m2 + m +3 is one- one function .
=> ∴ f: ℕ → ℕ defined by
f(m) = m2 + m +3
f(1) = 12 +1 +3 = 5
f(2) = 22 +2+3 = 9
f(3) = 32 + 3 +3 = 15
∴ co-domain = {1,2,3,….} ∈ ℕ
∴ Range = {5,9,15,…..} ∈ ℕ
∴ Co-domain different elements in ℕ , there are different images in ℕ .
So, f: ℕ → ℕ is one-one . [proved]
(6) Let A = {1,2,3,4} and B= ℕ . Let f: ℕ → B be defined by f(x) = x3 then,
(i) find the range of f
(ii) identify the type of function .
=> f : A → B
A = {1,2,3,4} and B = ℕ .
f(x) = x3
f(1) = 13 = 1
f(2) = 23 = 8
f(3) = 33 = 27
f(4) = 43 = 64
∴ (i) Range of f = {1,8,27,64} .
(ii) Co-domain different elements in A ; there are different image in B and co-domain and Range are equal .
So, f is one-one and onto .
(7) In each of the following cases state whether the function is bijective or not. justify your answer .
(i) f: 1R → 1R defined by f (x) = 2x +1
(ii) f: 1R → 1R defined by f(x) = 3-4x2
=> (i) f:1R → 1R defined by
f (x) = 2x +1
f(0) = 2.0 +1 = 1
f(- ½) = 2 (- ½) +1 = 0
f(1/2) = 2. ½ +1 = 2
f (-1) = 2.(-1) +1 = -1
f(1) = 2.1 +1 = 3
∴ every elements in R , there are different image in B .
So, f is one-one .
∴ Now, This function co-domain and Range same, then this function is onto.
So, f is bijective.
(ii) Given, f(x) = 3-4x2
∴ f(0) = 3-4.0 = 3
f(-1) = 3 – 4(-1)2 = -1
f(1) = 3-4.1 = -1
so, -1 , 1 1R have same image -1 1R .
∴ f is not one-one .
So, f is not bijective.
(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 .
=> f(x) = ax + b
∴ f is onto function .
∴ f(-1) = 0 or 2 and f(1) = 2 or 0 [∵ f:A → B]
∴ f(-1) = a(-1) +b
f(-1) = 0
-a +b = 0 …..(i)
f(1) = a.1 +b
= a + b
f(1) = 2
a + b = 2 …..(ii)
solve (i) and (ii), we get
-a +b = 0
a+ b = 2
_________________
2b = 2
b = 1
from (i),
-a + b = 0
-a +1 = 0
a = 1
Now, f(-1) = 2
-a + b = 2 …..(iii)
And f(1) =0
a + b = 0 …..(iv)
solve (iii) and (iv) , we get
-a +b = 2
a + b = 0
_______________
2b = 2
b = 1
from (iv) , -a+b = 2
-a+1 = 2
-a = 2-1
-a = 1
a = -1
∴ Hence, a = 1 or -1 and b = 1 .
(9) If the function f is defined by f(x) = {x+2;x>1
2; -1 ≤ x≤ 1
x -1 ; -3< x < -1}
find the values of
(i) f (3) ,
(ii) f(0)
(iii) f(-1.5)
(iv) f(2) +f(-2)
=> Here f(3) ={x +2 ; x> 1
2 ; -1≤ x ≤ 1
x-1;-3<x< -1}
(i) when x = 3, then f(x) = x +2 [∵ x>1 ]
∴ f(3) = 3+2 = 5
(ii) when x = 0 , then f(x) = 2 [∵ -1≤ x≤1]
∴ f(0) = 2
(iii) when x = -1.5 , then f(x) = x -1 [∵-3 <x< -1]
f(-1.5) = -1.5 -1
= -2.5
(iv) when, x = 2 , then f(x) = x+2 [∵ x>1]
f(2) = 2+2 = 4
when, x = -2 then f(x) = x-1 [∵-3<x<-1]
f(-2) = -2-1 = -3
∴ f(2) + f(-2) = 4+(-3) = 1
(10) A function f; [-5,9] → R is defined as follow .
f(x) = {6x +1; -5≤ x<2
5x2 -1 ; 2 ≤ x < 6
3x -4 ; 6≤ x ≤ 9}
Find (i) f(-3) + f(2),
(ii) f(7) – f(1)
(iii) 2f (4) + f(8)
(iv) 2f (-2) – f(6)/f(4) + f(-2)
=> given, f(x) = { 6x + 1; -5 ≤ x<2
5x2 -1 ; 2≤ x < 6
3x -4 ; 6≤ x ≤9}
(i) when x = -3 , then f(x) = 6x +1 [∵ -5 ≤ x < 2]
f(-3) = 6(-3) +1= 17
when x =2 then f(x) = 5x2 -1 [∵ 2≤ x < 6]
f(2) = 5.(2)2 -1 = 19
Now, f(-3) + f(2) = -17 +19 = 2
(ii) when x = 7 , then f(x) = 3x -4 [∵ 6≤ x ≤ 9]
f(7) = 3.7 -4 = 17
when x =1 , then f(x) = 6x +1 [∵ -5≤ x<2]
f(1) = 6.1 +1 = 7
Now, ∴ f(7) – f(1) = 17 – 7 = 10
(iii) when x = 4 , then f(x) 5x2 -1 [∵ 2≤ x <6]
f(4) = 5.(4)2 -1 = 79
when x = 8 then f(x) = 3x -4 [∵ 6≤ x≤ 9]
f(8) = 3.8 -4 = 20
∴ 2f (4) + f(8) = 2 . 79 + 20
= 158 + 20 = 178
(iv) when x = 2 , then f(x) = 6x +1[∵ -5≤ x < 2]
f(-2) = 6 (-2) + 1 = -11
when x = 6 , then f(x) = 3x -4 [∵ 6≤ x ≤ 9]
f(6) = 18 -4 = 14
when x = 4, then f(x) = 5x2 -1 [∵ 2≤ x < 6]
f(x) = 79
Now, 2f (-2) – f(6) / f (4) + f(-2)
= 2.(-11) -14 / 79 + (-11)
= -22 -14/79 -11
= -36/68
= -36/68
= -9/17
∴ 2f(-2) – f (6)/f(4) + f(-2)
= – 9/17
(11) The distance S an object travele under the inflnence of gravity in time t seconds is given by.
S(t) = ½ gt2 +at + b , where, (g is the acceleration due to gravity), a, b are constants. Verify wheather the function S (t) is one-one or not.
=> Given that s(t) = ½ gt2 + at +b , where a, b constant .
Let t = t1, t1 , t3…..seconds
S(t1) = ½ g t12 + at1 + b
S(t2) = ½ g t22 + at2 + b
∴ for every different values of t , there will be different values as images .
Therefore s (t) is one-one function.
(12) The function ‘t’ which maps temperature in Celsius (c) into temperature in Fahrenheit (f) is defined by t(c) = f where f = 9/5c + 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 fahrenheit value .
=> t(c) = f
f(c) = 9/5 c + 32
(i) when , c = 0 , then
t(0) = 9/5 .0 +32 = 32
(ii) when , c = 28 , then , f(28) = 9/5 × 28 +32
= 252/5 + 32 = 50.4 +32
82.4
(iii) when c = -10 , then t (-10) = 9/5 × (-10) + 32
= 18 +32
= 24
(iv) t(c) = 212
9/5c +32 = 212
9 c + 160 = 1060
9c = 1060 – 160
9c = 900
C = 100° c
∴ c = 100° c
(v) f = 9/5 c + 32
C = 9/5 c + 32 [∵ The temperature is Celsius value and fahrenheit value same ]
5c = 9c + 160
9c – 5c = – 160
4c = 160
C = -40
Exercise – 1.5
(1) Using the function f and f given below find f o g and g o f. check whether f o g = g o f
(i) f(x) = x -6 , g(x) = x2
(ii) f(x) = 2/x , g(x) = 2x2 -1
(iii) f(x) = x+6/ 3 , g(x) = 3 –x
(iv) f(x) = 3+x , g(x) = x -4
(v) f(x) = 4x2 -1 , g(x) = 1 + 4
=> (i) f(x) = x -6 , g (x) = x2
∴ fog(x) = f (g (x)) = f(x2)
= x2 -6
∴ gof (x) = g(f(x)) = g (x-6) = (x-6)2
= x2 -12x + 36
∴ fog ≠ gof .
(ii) f(x) = 2/x , g(x) = 2x2 -1
∴ fog (x) = f(g(x)) = f (2x2 -1) = 2/2x2 -1
∴ gof (x) = g (f(x)) = g(2/x) = 2 (2/x)2 -1
= 8/x2 -1
∴ fog (x) ≠ gof (X)
∴ fog ≠ gof.
(iii) f(x) = x+6/3 , g(x) = 3-x
∴ fog (x) = f(g(x)) = f(3-x) = 3-x +6/3 = 1- x/3 +2
= 3-x/3
∴ gof (x) = g(f(x)) = g (x+6/3) = 3- x+6/3 = 3- x/3 -2
= 1- x/3
∴ fog ≠ gof
(iv) f(x) = 3+x , g(x) = x -4
fog (x) = f(g(x)) = f(x -4) = 3 + (x-4) = x -1
gof (x) = g(f(x)) = g(3+x) = 3+x-4 = x-1
∴ fog ≠ gof
(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 .
=> (i) f(x) = 3x +2 , g(x) = 6x –k
∴ fog(x) = f(g(x)) = f(6x-k) = 3(6x – k ) +2
= 18x -3k +2
∴ gof (x) = g(f(x)) = g(3x +2) = 6(3x +2) –k
= 18x + 12 –k
Now, fog (x) = gof (x)
18x – 3k+ 2 = 18x +12 -k
-3k +k = 12-2
-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
∴ gof (x) = g(f(x)) = g(2x – k) = 4(2x – k) +5
= 8x – 4k +5
Now,
∴ fog (x) = gof (x)
8x +10 – k = 8 x – 4k + 5
4k –k = 5-10
3k = -5
K = -5/3
(3) If f(x) = 2x -1 , g(x) = x+1/2 , show that fog = gof = x
=> f(x) = 2x – 1 , g(x) = x+1/2
fog (x) = f(g(x)) = f(x+1/2) = 2 (x+1 /2) -1
= x+1-1 = x
gof (x) = g(f(x)) = g(2x -1) = 2x – 1 +1 / 2
= x
∴ fog = gof = x [proved]
(4) If f(x) = x2 -1 , g(x) = x -2 . find a , if gof (a)=1.
=> f (x) = x2 -1 g(x) = x -2
f(a) = a2-1 g(a)= a-2
gof (a) = g (f(a)) = g(a2 -1 ) = a2-1 -2
=a2 -3
∴ gof (a) = 1
a2 -3 = 1
a2 = 4
a = ± 2
(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 .
=> given , f(x) = 2x +1 , g(x) = x2
∴ fog(x) = f(g(x)) = f(x2) = 2x2 +1
∴ gof (x) = g(f(x)) = g(2x+1) = (2x +1)2
∴ rang of fog = { 2x2 +1 ; x ∈ ℕ}
∴ rang of gof = {(2x +1)2; x ∈ ℕ }
(6) Let f(x) = x2 -1 . find (i) fofof
=> f(x) = x2 -1
(i) fof (x) = f(f(x)) = f(x2-1)
= (x2 -1)2 -1
= x4 – 2x2 +1 -1
= x4 -2x2
(ii) fofof(x) = fof(f(x))
= f(f(x2 -1))
= f (x4 -2x2)
= (x4 – 2x2)2 -1
= x4(x2-2)2 -1
= x4(x4-2x2+4) -1
= x8 – 2x6 + 4x4-1
(7) If f:1R → 1R and g: 1R → 1R are defined by f(x) = x5 and g(x) = x4 then check if f , g are one – one and fog is one- one ?
=> Given that ,
f(x) = x5 and g(x) = x4
Now, Let x1 ,x2 ∈ 1R
Then , f(x1) = f(x2)
x15 = x25
x1 = x2
∴ f is one-one .
And y1,y2 ∈ 1R
g(y1) = g(y2)
y14 = y24
y1 = y2
∴ g is one- one .
Now, fog (x) = f(g(x)) = f(x4) = (x4)5
= x20
Now, fog (p) = fog(q)
P20 = q20
p=q
so, fog is one-one function .
(8) Consider the function 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
=> (i) Here f(x) = x -1 , g(x) = 3x +1 and h(x) = x2
Now, fog(x) = f(g(x)) = f(3x +1)
= 3x +1 -1 = 3x
goh (x) = g(h(x)) = g(x2) = 3x2 +1
∴ (fog) oh (x) = fog (h(x))
= fog (x2)
= 3x2
∴ fo (goh) (x) = f (goh(x)) = f (3x2 +1) = 3x2 +1 -1
= 3x2
∴ (fog) oh = fo (goh) [proved].
(ii) f(x) = x2 , g(x) = 2x and h(x) = x+4
Fog (x) = f(g(x)) = f(2x) = (2x)2 = 4x2
goh (x) = g(h(x)) = g(x +4) = 2 (x +4) = 2x +8
Now, (fog) oh (x) = fog (h(x) = fog (x +4)
= 4(x+4)2
∴ fo(goh)(x) = f(goh(x) = f(2x+8)
= (2x +8 )2
= 4(x+4)2
∴ (fog) oh = fo (goh) [proved] .
(iii) f(x) = x -4 , g (x) = x2 and h(x) = 3x -5
fog (x) = f(g(x)) = f(x2) = x2 -4
goh (x) = g(h(x)) = g(3x -5) = (3x -5)2
∴ (fog) oh (x) = fog (h(x)) = fog (3x -5)
= (3x -5)2 – 4
∴ fo(goh) (x) = f(goh(x)) = f((3x -5)2)
= (3x -5)2 – 4
∴ (fog) oh = fo (goh) [proved] .
(9) Let f = [(-1,3) , (0,-1) , (2,-9)} be a linear function from z into z . Find f(x) .
=> We know that, linear equation is f(x) = ax + b
Now, f(-1) = 3
-a + b = 3 ….(i)
And f (0) = -1
a . 0 + b = -1
b = -1 ……(ii)
from (i), we get ,
-a +b = 3
-a -1 = 3
a = -4
∴ The linear equation is
f(x) = -4x – 1
= – (4x +1) .
(10) In electrical circuit theory, a circuit c (t) is called a linear circuit if it satisfies the super position principle given by C (at1+b + bt2) = ac (t1) + b c +(t2) , where a, b are constants . Show that the circuit c (t) = 3t is linear .
=> Given that c(t) = 3t
Then, c(at1 + bt2) = 3(at1 + bt2)
Now, c(t1) = 3t1 and c(t2) = 3t2
∴ a c (t1) + bc (t2)
= a3 t1 + b3 t2
= 3 (at1 + bt2)
= c (at1 + bt2)
∴ c (at1 + bt2) = ac (t1) + b c (t2)
Therefore , c(t) = 3t is linear.
Exercise – 1.6
Multiple choice questions
(1) If n(A×B) = 6 and A = {1,3} then n(B) is
(A) 1
(B) 2
(C) 3
(D) 6
=> n(B) = 3 (C)
[∵ n(A× B) = 6
n(A) × n (B) = 6
2 × n(B) = 6
n (B) = 3
(2) A = {a, b, p} , B= {2, 3} , c = {p, q, r, s} then n[(A∪C) × B] is
(A) 8
(B) 20
(C) 12
(D) 16
=> B = {2, 3} and A∪C = { a, b, p, q, r, s}
n [(A∪C) × B] = n(A∪C) × n(B)
= 6×2
= 12 (C)
(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 .
(A) (A×C) ⊂ (B×D)
(B) (B×D) ⊂ (A× C)
(C) (A×B) ⊂ (A× D)
(D) (D×A) ⊂ (B×A)
=> 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),(4,5),(4,6),(4,7), (4,8)}
(A×C) ⊂ (B×D) (A)
(4) if There are 1024 relations from a set A = {1,2,3,4,5} to a set B , then the number of element in B is
(A) 3
(B) 2
(C) 4
(D) 8
=> 2AB = 1024
25B = 1024 = (2)10
5B = 10
B = 2 (B)
(5) The range of the relation R = {(x , x2) |x is a prime number less than 13} is –
(A) {2,3,5,7}
(B) {2, 3, 5,7, 11}
(C) {4, 9,25,49,121}
(D) 1,4,9,25,49,121}
=> Range of 1R={4,9,25,49,121} (C)
(6) if the ordered pairs (a+2, 4 ) and (5, 2a+b) are equal then (a, b) is –
(A) (2,-2)
(B) (5,1)
(C) (2,3)
(D) (3,-2)
=> a+2 = 5
a = 3
2a +b = 4
6+b = 4
b = -2
(3,-2) (D)
(7) Let n(A) = m and n(B) = n then the total number of non-empty relation that can be defined from A to B is –
(A) mn
(B) nm
(C) 2mn -1
(D) 2mn
=> 2mn (D)
(8) If {(a, b) ,(6,7)} represent an identify function, then the value of a and b are respectively –
(A) (8, 6)
(B) (8, 8)
(C) (6, 8)
(D) (6, 6)
=> (a, 8), (6, b)
a=8 b=6
(8, 6) (A)
(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
(A) Many –one function
(B) Identify function
(C) one-to-one function
(D) Into function
=> (C) one-to-one function .
(10) If f(x) = 2x2 and g(x) = 1/3x, then fog is
(A) 3/2x2
(B) 2/3x2
(C) 2/4x2
(D) 2/6x2
=> fog = f(g(x)) = f(1/3x) = 2.(1/3x)2 = 2/9x2 (C)
(11) If f:A → B is a bijective function and if n(B) = 7 , then n(A) is equal to –
(A) 7
(B) 49
(C) 1
(D) 14
=> bijective function same co-domain and image
So, n(A) = n(B) = 7(A)
(12) Let f and g be two function 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
(A) {0,2,3,4,5}
(B) {-4,1,0,2,7}
(C) {1, 2, 3, 4, 5,}
(D) {0,1,2} .
=> (D) {0, 1, 2}
fog (0) = f(g(0)) = f(2) = 0
similarly, fog(1)= 1, fog(2) = 2
fog (2) = 0, fog(7) = 1
∴ range of fog = {0, 1, 2}
(13) Let f(x) = √1+x2 then
(A) f(x y) = f(x) . f(y)
(B) f(x, y) ≥ f(x) . f(y)
(C) f(x y) ≤ f(x) . f(y)
(D) None of these .
=> f(x) = √1+x2 and f(x y) = √1+ x2 y2
∴ f(y) = √1+y2
Now, f(x) . f(y) = √1+x2 . √1+y2 ≥ √1+x2 y2
∴ f(x , y ) ≤ f(x) . f(y) (c)
(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
(A) (-1 , 2)
(B) (2, -1)
(C) (-1, -2)
(D) (1, 2)
=> g(x) = ∝x + β , g(2) = 3
g(1) = 1 2∝ + β = 3 2∝ + β = 3
∝ + β = 1 ∝ + β = 1
∝ + β = 1 (-) (-) (-)
β = 1- ∝ = 1-2 = -1 ____________________
(2, -1) (B) ∝ = 2
(15) f(x) = (x+ 1)3 – (x-1)3 represents a function which is
(A) linear
(B) cubic
(C) reciprocal
(D) quadratic
=> (D) quadratic .
Unit Exercise – 1
(1) If the ordered pairs (x2– 3x , y2 + 4y) and (-2, 5) are equal , then find x and y .
=> Given ordered pairs (x2-3x, y2+4y) and (-2, 5) are equal .
So, (x2– 3x , y2 + 4yt) = (-2, 5)
x2 – 3x = -2
x2 – 3x + 2 = 0
x = 1, 2
y2 + 4y = 5
y2 + 4y -5 = 5
y = 1, -5
∴ x = 1 or 2
y = 1 or -5
(2) The cartesian product A× A has 9 element among which (-1 , 0) and (0, 1) are found . Find the set A and the remaining elements of A × A .
=> given that A×A has 9 elements.
So, n(A) = 3 [∵ 3×3 = 9]
So, A has 3 elements .
Let A = {x, y, z}
Now, A × A = {x, y, z} × {x, y, z}
= {(x, x ),(x, y) , (x, z) , (y, x), (y, y) , (y, z) , (z, x), (z, y), (z , z)}
Now, (-1, 0) ∈ A× A , then, taking (x, y) = (-1, 0)
∴ x = -1 , y = 0 .
(0, 1) ∈ A×A, then taking (y, z) = (0, 1)
∴ y = 0 , z = 1
∴ A = {-1, 0, 1}
A × A = {-1, 0, 1} × { -1, 0, 1}
= {(-1, -1) , (-1,0) , (-1,1) ,(0,-1), (0, 0) , (0, 1) , (1, -1), (1, 0) , (1, 1)}
Thus, remaining elements of A× A
= {(-1, -1) , (-1, 1) , (0, -1) , (0, 0), (1, -1), (1, 0), (1, 1)}
(3) Given that f(x) = {√x -1 4 x ≥ 1 x<1} . Find
(i) f(0),
(ii) f(3)
(iii) f(a+1) in terms of a(Given that a ≥ 0)
=> Here, f(x) = {√x -1 4 x ≥ 1 x<1}
When x = 0 , then f(x) = 4 [∵ x < 1]
f(0) = 4
(ii) when x = 3, then f(x) = √x -1 [∵ x ≥ 1]
f(3) = √3-1 = √2
(iii) when x =a +1, then f(x) = √x -1 [∵ x ≥ 1 and a≥o
So, a+1 ≥ a]
f(a+1) = √a+1-1 = √ a
(4) Let A = {9, 10, 11, 12, 13, 14, 15, 16, 17} and f:A → ℕ be defined by f(x)= the highest prime factor of n ∈ A . Write f as asset of ordered pairs and find the rang of f.
=> f(x) = the highest prime factor of n ∈ A
f(9) = 3 [∵ 9 = 3×3, highest prime factor 3]
similarly, f(10) = 5, f(11) = 11, f(12) = 3 , f(13)= 13
f(14) = 7 , f(15) = 15, f(16)= 2, f(17) = 17
∴ f = {(10,5),(11,11),(12, 3),(13,13),(14, 7), (15,5),(16, 2), (17, 17),(9,3)
∴ Range of f = {2, 3, 5, 7, 11, 13, 17}
(5) Find the domain of the function f(x) = √ 1+√1-√1-x2
=> Let y = f(x)
y = √1+√1-√1-x2
y2 = 1 + √1- √1-x2[Group by both side]
∴ y2≥ o then 1 + √1-√1-x2≥ 0
Let z =√1 – √1-x2
z2 = 1-√1-√1-x2
z2 ≥ 0 then 1- √1-x2 ≥ 0
Let p = √1- x2
P2 = 1 –x2[Group by both side]
photo
(6) If (x) = x2, g(x) = 3x and h(x) = x -2 , prove that (fog) oh = fo (goh) .
=> Here f(x) = x2, g(x) = 3x h(x)= x-2
fog(x) = f(g(x))= f(3x) = (3x)2 = 9x2
goh (x) = g(h(x))= g(x-2) = 3(x -2) = 3x – 6
∴ (fog) oh (x)
= fog (h(x)) = fog(x-2)
= 9 (x-2)2= (3x-6)2
∴ fo (goh)(x) = f(goh(x)) = f(3x – 6)
= (3x – 6)2
∴ (fog)oh = fo(goh) [proved]
(7) Let A = {1,2} and B = {1,2, 3,4} , C = {5, 6}
And D = {5, 6,7,8}. Verify whether A × C is a subset of B×D?
=> A = {1, 2} , B={1,2,3,4}, C = {5,6}, D = {5,6,7,8}
A×C = {1, 2} × {5, 6}
= {(1, 5) , (1, 6) , (2, 5) , (2 , 6)}
∴ B×D = {1, 2, 3, 4} × {5, 6, 7, 8}
= {(1,5), (1, 6), (1, 7), (1, 8) ,(2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8) , (4, 5) ,(4, 6), (4, 7), (4, 8)}
Now, {(1, 5),(1, 6), (2, 5), (2, 6)} ∈ B× D
So, A×C ⊂ B×D
Thus, A × C is a subset of B×D [ Proved]
(8) If f(x)= x- 1/ x +1 , x≠ -1 show that f(f(x)) = – 1/x, provided x ≠ 0
=> f(x) = x -1/x+1
∴ f(f(x)) = f (x-1/x+1)
= x -1/ x+1 -1/x-1/x+1 +1
= x-1 –x -1/x+1 /x-1+x+1/x+1
= -2/2x = – 1/x
∴ f(f(x)) = – 1/x, x ≠ 0 . [proved]
(9) The function f and g are defined by f(x) = 6x +8 ;
g(x) = x -2/3
(i) calculate the value of gg(1/2)
(ii) write an expression for gf(x) in its simplest form .
=> (i) Here g(x) = x-2/3
g g(x) = g(x -2/3) = x-2/3 -2 /3 = x-2-6/3×3 = x-8/9
∴ gg(1/2) = ½ – 8/9 = 1-16/9×2 = – 15/18
= -5/6
(ii) f(x) = 6x +8 and g(x) = x -2/3
g f (x)
= g(6x +8)
= 6x +8 – 2/3
= 6x +6 /3 = 2x + 2
∴ g f (x) = 2x +2
(10) Write the domain of the following real function .
(i) f(x) = 2x +1/x – 9
(ii) p(x) = -5/4x2 +1
(iii) g(x) = √x – 2
(iv) h(x) = x +6
=> (i) f(x) = 2x +1/x -9
x –q ≠ 0
x ≠ q ∴ domain of f = 1R – {q} .
(ii) p(x) = -5/4x2 +1
∴ domain of p = 1R
(iii) g(x) = √ x – 2
Let L = √x -2
L2 = X -2 ≥ 0 [∵ both side group]
∴ x – 2 ≥ 0
x ≥ 2 ∴ domain of g = [2, ∝] .
(iv) h(x) = x +6
∴ domain of h = 1R .
Useful
excellent /Useful