Question 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 :
A be the set of students and B be the set of sections.
Every student in set A will be belonging at least any one of the classes of set B. Hence the above relation is a function.
Now let us consider about inverse relation. During the inverse relation, each section will have more than 1 students, which are given one to many relation.
That is, the section C1 will have more than 1 students. So we have to draw arrow mark from C1 to more than one student in set A. Hence it is not a function.
Question 2 :
Write the values of f at −4, 1,−2, 7, 0 if
Solution :
(i) x = -4,
-4 lies in the interval (-∞, -3]
f(x) = -x + 4
x = -4
f(-4) = - (-4) + 4 = 8
(ii) x = 1,
1 lies in the interval [1, 7)
f(x) = x - x^{2 }
f(1) = 1 - 1^{2 } = 0
(iii) x = -2,
-2 lies in the interval [-2, 1)
f(x) = x^{2 }- x
f(-2) = (-2)^{2 }- (-2)
= 4 + 2
= 6
(iv) x = 7,
Do not lie in the given intervals. Hence f(7) is 0.
(v) x = 0,
0 lies in the interval [-2, 1)
f(x) = x^{2 }- x
f(0) = 0^{2 }- 0 = 0
Question 3 :
Write the values of f at −3, 5, 2,−1, 0 if
Solution :
(i) x = -3,
-3 lies in the interval (-∞, 0)
f(x) = x^{2} + x - 5
f(-3) = (-3)^{2} + (-3) - 5
= 9 - 3 - 5
f(-3) = 1
(ii) x = 5
5 lies in the interval (3, ∞)
f(x) = x^{2} + 3x - 2
f(5) = 5^{2} + 3(5) - 2
= 25 + 15 - 2
= 38
(iii) x = 2
Do not lie in the given intervals. Hence f(2) is 0.
(iv) x = -1
-1 lies in the interval (-∞, 0)
f(x) = x^{2} + x - 5
f(-1) = (-1)^{2} + (-1) - 5
= 1 - 1 - 5
= 1 - 6
= -5
(v) x = 0
Do not lie in the given intervals. Hence f(0) is 0.
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