Page 6 - mathsvol1ch1to3ans

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Exercise - 1.3

1. Suppose that 120 students are studying in 4 sections of eleventh standard in a school. Let A denote
the set of students and B denote the set of the sections. Define a relation from A to B as “x related
to y if the student x belongs to the section y”. Is this relation a function? What can you say about
the inverse relation? Explain your answer.
Solution:
Since a student cannot belongs to two different sections this relation is a function. Since the section
consist of more than one student, inverse function does not exist.


−x + 4 if − ∞ < x ≤ −3
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

x + 4 if − 3 < x < −2


2
2. Write the values of f at −4, 1, −2, 7, 0 if f(x) = x − x if − 2 ≤ x < 1
 2
x − x if 1 ≤ x < 7



0 otherwise

Solution:
f(−4) = 8 because −4 ≤ −3
f(1) = 0 because 1 ≤ 7
f(−2) = 6 because −2 ≤ 1
f(7) = 0 because 7 ≥ 7
f(0) = 0 because 0 ≤ 1
 2
x + x − 5 if x ∈ (−∞, 0)


 2
x + 3x − 2 if x ∈ (3, ∞)
3. Write the values of f at −3, 5, 2, −1, 0 if f(x) =
x 2 if x ∈ (0, 2)


x − 3 otherwise
 2
Solution:
f(−3) = 1 because −3 ≤ 0
f(5) = 38 because 5 ≥ 3
2
f(2) = 1 because 2 − 3 = 1
f(−1) = −5 because −1 ≥ 0
2
f(0) = −3 because 0 − 3 = −3
4. State whether the following relations are functions or not. If it is a function check for one–to–
oneness and ontoness. If it is not a function, state why?
(i) If A = {a, b, c} and f = {(a, c), (b, c), (c, b)}; (f : A → A).
Solution:
It is a function and it is not one–to–one and not onto.
(ii) If X = {x, y, z} and f = {(x, y), (x, z), (z, x)}; (f : X → X).
Solution:
It is not a function because x has two images y and z.
5. Let A = {1, 2, 3, 4} and B = {a, b, c, d}. Give a function from A → B for each of the following:
(i) neither one–to–one nor onto. (ii) not one–to–one but onto.

(iii) one-to–one but not onto. (iv) one-to–one and onto.

Solution:
(a) neither one–to–one nor onto. f(x) = a where x ∈ A
f = {(1, a), (2, a), (3, a), (4, a)}
(b) not one–to–one but onto. This is not possible since both A and B has same number of
elements. If the function is not one–to–one then it has more than one element has same image.