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In this article, we will be going to solve the entire exercise 3.1 of our NCERT textbook. Functions are fundamental concepts in mathematics that describe relationships between sets of elements. They provide a way to map or associate each element from one set (the domain) to exactly one element in another set (the codomain).
A function f from a set X to a set Y is a rule or correspondence that assigns each element x in X exactly one element f(x)in Y. This can be denoted as f: X→Y.
Let's learn about the solution of set 2 for NCERT Exercise 3.1 class 11 in this article below:
(i) f(x) = x2
(ii) g(x) = sinx
(iii) h(x) = x2 + 1
Find the range of each function.
Solution:
(i) We have,
f(x) = x2
Range of f(x) = R+ (set of all real numbers greater than or equal to zero)
= {x ∈ R+ | x ≥ 0}
(ii) We have
g(x) = sinx
Range of g(x) = {x ∈ R : -1 ≤ x ≤ 1}
(iii) We have
h(x) = x2 + 1
Range of h(x) = {x ∈ R : x ≥ 1}
Determine which of the following sets are functions from X to Y
(a) f1 = {(1, 1), (2, 11), (3, 1), (4, 15)}
(b) f = {(1, 1), (2, 7), (3, 5)}
(c) f = {(1,5), (2, 9), (3, 1), (4, 5), (2, 11)}
Solution:
(a) We have,
f1 = {(1, 1), (2, 11), (3, 1), (4, 15)}
f1 is a function from X to Y
(b) We have,
f2 = {(1, 1), (2, 7), (3, 5)}
f2 is not a function from X to Y because there is an element 4 ∈ x which is not associated to any element of Y.
(c) We have,
f3 = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}
f3 is not a function from X to Y because an element 2 ∈ x is associated to two elements 9 and 11 in Y.
Solution:
We have,
f(x) = highest prime factor of x.
Therefore,
12 = 3 × 4,
13 = 13 × 1,
14 = 7 × 2,
15 = 5 × 3,
16 = 2 × 8,
17 = 17 × 1
Therefore,
f = {(12, 3), (13, 3), (14, 7), (15, 5), (16, 2), (17, 17)}
Range (f) = {3, 13, 7, 5, 2, 17}
Solution:
We know that,
if f : A ⇢ 13
such that y ∈ 3. Then,
f-1 (y) = {x ∈ A : f(x) = y}. In other words, f-1 (y) is the set of pre-images of y.
Let f-1 (17) = x. Then, f(x) = 17
⇒ x2 + 1 = 17
⇒ x2 = 17 - 1 = 16
⇒ x = ±4
Let f-1 {-3} = x. Then, f(x) = -3
⇒ x2 + 1 = -3
⇒ x2 = -3 - 1 = -4
⇒ x =
Therefore, f-1 {-3} = 0
(a) R1 = R1 = {(p, 1), (q, 2), (r, 1), (s, 2)}
(b) R2 = {(p, 1), (q, 1), (r, 1), (s, 2)}
(c) R3 = {(p, 1), (q, 2), (p, 2), (s, 3)}
(d) R4 = {(p, 2), (q, 3), (r, 2), (s, 2)}
Solution:
We have
A = {p, q, r, s} and B = {1, 2, 3}
(a) Now,
R1 = {(p, 1), (q, 2), (r, 1), (s, 2)}
R1 is a function
(b) Now,
R2 = {(p, 1), (q, 2), (r, 1), (s, 1)}
R2 is a function
(c) Now,
R3 = {(p, 2), (q, 3), (r, 2), (s, 2)}
R3 is not a function because an element p ∈ A is associated to two elements 1 and 2 in B.
(d) Now,
R4 = {(p, 2), (q, 3), (r, 2), (s, 2)}
R4 is a function
Solution:
We have,
f(n) = the highest prime factor of n.
Now,
9 = 3 × 3,
10 = 5 × 2,
11 = 11 × 1,
12 = 3 × 4,
13 = 13 × 1
Therefore,
f = {(9, 3), (10, 5), (11, 11), (12, 3), (13, 13)}
Clearly, range(f) = {3, 5, 11, 13}
The relation f is defined by
Show that f is a function and g is not a function
Solution:
We have,
and,
Now, f(3) = (3)2 = 9 and f(3) = 3 × 3 = 9
and, g(2) = (2)2 = 4 and g(2) = 3 × 2 = 6
We observe that f(x) takes unique value at each point in its domain [0,10]. However, g(x) does not take unique value at each point in its domain [0, 10].
Hence, g(x) is not a function.
Solution:
Given f(x) = x2
f(1.1) = 1.21
f(1) = 1
= 2.1
Solution:
f : X ⇢ R given by f(x) = x3 + 1
f(-1) = (-1)3 + 1 = -1 + 1 = 0
f(0) = (0)3 + 1 = 0 + 1 = 1
f(3) = (3)3 + 1 = 27 + 1 = 28
f(9) = (9)3 + 1 = 81 + 1 = 82
f(7) = (7)3 + 1 = 343 + 1 = 344
Set of ordered pairs are {(-1, 0), (0, 1), (3, 28), (9, 82), (7, 344)}
Here are 10 practice questions, a summary, and FAQs for Class 11 RD Sharma Solutions - Chapter 3 Functions - Exercise 3.1 | Set 2:
