MATHEMATICS - RELATIONS AND FUNCTIONS - Answers to Exercise 1.2
MATHEMATICS
CLASS 12
RELATIONS AND FUNCTIONS
SOLUTIONS TO EXERCISE 1.2
This exercise is about the types of functions. To identify the type of function, either elements like x, y can be used or random values can also be used.
A function f: X -> Y is said to be one-one (or injective), if every element of the domain has one and only one image in the codomain.
A function f: X -> Y is said to be onto (surjective) if for every y ∈ Y (codomain) there exists x ∈ X (domain).
A function is said to be bijective if it is both one-one and onto.
1. Show that the function f: R∗ -> R∗, defined by f(x) =1/x is one - one and onto where R∗ is the set of all non zero real numbers. Is the result true, if the domain R∗ is replaced by N with codomain being the same R∗ Let x, y ∈ R∗ such that f(x) = f(y)
f (x)= f(y)
1/x = 1/y
⇒ x = y
Therefore f is one - one.
Let t be an element of the codomain such that f(x) = t.
f(x) = t
1/x = t
⇒ x = 1/t
Thus for each t belonging to codomain, there exists x belonging to the domain.
Therefore f : R∗ -> R∗ is onto.
If f : N -> R∗ , then
for x, y ∊ N, we get
f(x) = f(y)
1/x = 1/y
⇒ x = y
Therefore f is one - one.
But for elements like 1/3, 1/6 belonging to the codomain (R∗) there are no pre-images in the domain N.
Therefore f : N -> R∗ is not onto.
2. Check the injectivity and surjectivity of the following functions:
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3. Prove that the Greatest Integer Function f: R -> R given by f(x)=[x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Let us take the interval [1, 2)
f(x) = [x]
So, f(x) = 1 for all x є [1,2)
which implies f : R -> R is not one - one.
The greatest integer function consists of only integers as the range. So elements other than integers in the codomain do not have pre-images in the domain.
Therefore f : R -> R is not onto.
Hence f: R-> R is neither one - one nor onto.
4. Show that the Modulus Function f: R-> R , given by f(x)=|x|, is neither one - one nor onto, where |x| is x, if x is positive or 0 and |x| is -x, if x is negative.
Given f: R -> R given by f(x) = | x |
Let us take f( 5 ) = 5
f(-5) = 5
which implies f(5) = f(-5)
So f is not one - one.
The range of the Modulus function consists of non-negative values only. Therefore all the negative real numbers belonging to the codomain R do not have pre-images in the domain R. So f is not onto.
Hence f : R -> R is neither one - one nor onto.
5. Show that the Signum Function is neither one - one nor onto.
Given signum function f: R -> R
f(x) = 1 , if x>0
0 , if x=0
-1 , if x<0
All x > 0 in the domain, have the same image 1 in the codomain R.
So f is not one - one.
The range of signum function is { -1, 0,1 }. All other elements in the codomain R do not have pre-images in the domain R. So f is not onto.
Hence the signum function f: R -> R is neither one - one nor onto.
6. Let A = { 1,2,3 } and B = { 4,5,6,7 } and let f = {(1,4), (2,5), (3,6)} be a function from A to B. Show that f is one - one.
Given A = {1,2,3} B = {4,5,6,7}
f = { (1,4), (2,5), (3,6) } is a function from A to B.
All the elements belonging to A have one and only image in B. Therefore f is one - one.
7. In each of the following cases, state whether the function is one - one , onto or bijective. Justify your answer.
( i ) f: R -> R f(x) = 3 - 4x
Let x, y ∊ R such that f(x) = f(y)
3 - 4x = 3 - 4y
-4x = -4y
⇒ x = y
Therefore f is one - one
Let y = 3 - 4x
4x = 3 - y
x = ( 3 - y )/4
which implies for every y belonging to the codomain R, there exists a pre-image in the domain R. Therefore f is onto.
Hence f: R -> R is one - one and onto.
(ii) f: R -> R f(x) = 1 + x^2
Let x, y є R such that f(x) = f(y)
1 + x^2 = 1 + y^2
x^2 = y^2
⇒ x = +y or -y
which says f is not one - one.
Let y = 1 + x^2
x^2 = y - 1
x = √(y - 1)
which says for negative values of y belonging to the codomain R, there exist no pre-images in the domain R. So f is not onto.
Hence f : R -> R is neither one - one nor onto.
8. Let A and B be sets. Show that f: A x B -> B x A such that f(a, b) = (b, a) is a bijective function.
Given f: A x B -> B x A such that f (a, b) = ( b, a )
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Let (b, a) be an element of B x A
b є B, a є A ⇒ (a, b) є A x B
So, for every (b, a) belonging to B x A, there exists (a, b) belonging to A x B such that f(a, b) = (b, a). So f is onto.
Hence f : A x B -> B x A is a bijective function.
9. Let f: N -> N
defined by f(n) = (n+1)/2 if n is odd
n/2 if n is even
for all n ∈ N. State whether the function f is bijective. Justify your answer.
f(1) = (1 + 1)/ 2 = 2/2 = 1 (since n is odd)
f(2) = 2/2 = 1 (since n is even)
So f(1) = f(2) but 1 is not equal to 2. So f is not one - one.
Let m be an element of N
If m is odd, then 2m - 1 is also odd
So f(2m-1) = ( 2m - 1 + 1 )/2 = 2m/2 = m
If m is even, then 2m is also even
So f(2m) = 2m/2 = m
This implies for every m (even or odd), there exists a pre-image in N. So f is onto.
Hence f is not bijective.
10. Given f: A -> B defined by f(x) = (x-2)/(x-3)
A = R-{3} B = R-{1}. Is f one - one and onto. Justify your answer.
Let x, y ∊ R - {3} such that f(x) = f(y)
(x-2)/(x-3) = (y-2)/(y-3)
(x-2) (y-3) = (y-2) (x-3)
x y - 3x - 2y + 6 = x y - 3y - 2x + 6
-3x + 2x = -3y + 2y
-x = -y
x = y
So f is one - one.
Let y = (x-2)/(x-3)
y(x-3) = (x-2)
y x - 3y = x - 2
x y - x = -2 + 3y
x(y-1) = 3y - 2
x = (3y - 2) / ( y-1 )
So for every y ∈ R - {1) there exists x ∈ R - {3} and so f is onto.
Hence f : A -> B is both one - one and onto.
11. Let f: R-> R be defined as f(x)= x^4 (MCQ)
Answer: Option D : f is neither one-one nor onto.
12. Let f: R-> R be defined as f(x) = 3x. (MCQ)
Answer: Option A : f is one - one onto.
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