Class 11 RD Sharma Solutions - Chapter 3 Functions - Exercise 3.1 | Set 2

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).

What is a Function?

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.

Chapter 3 Functions - Exercise 3.1 | Set 2

Let's learn about the solution of set 2 for NCERT Exercise 3.1 class 11 in this article below:

Question 10. If f, g, h are three functions defined from R to R as follows:

(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}

Question 11. Let X = {1, 2, 3, 4} and Y = {1, 5, 9, 11, 15, 16}

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.

Question 12. Let A = {12, 13, 14, 15, 16, 17} and f : A ⇢  Z be a function given by f(x) = highest prime factor of x. Find range of f.

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}

Question 13. If f : R ⇢ R be defined by f(x) = x² + 1, then find f^-1{17} and f^-1{-3}.

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 = \sqrt{-4}

Therefore, f-1 {-3} = 0

Question 14. Let A = {p, q, r, s} and B = {1, 2, 3}. Which of the following relations form A to B is not a function?

(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

Question 15. Let A = {9, 10, 11, 12, 13} and let f : A ⇢ N be defined by f(n) = the highest prime factor of n. Find the range of f.

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}

Question 16. The function f is defined by 

f(x)=\begin{cases}x^2,\ 0≤ x≤ 3\\ 3x,\ 3≤ x≤ 10\end{cases}

The relation f is defined by 

g(x)=\begin{cases}x^2,\ 0≤ x≤ 2\\ 3x,\ 2≤ x≤ 10\end{cases}

Show that f is a function and g is not a function

Solution:

We have,

f(x)=\begin{cases}x^2,\ 0≤ x≤ 3\\ 3x,\ 3≤ x≤ 10\end{cases}

and, 

g(x)=\begin{cases}x^2,\ 0≤ x≤ 2\\ 3x,\ 2≤ x≤ 10\end{cases}

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.

Question 17. If f(x) = x², find 

\frac{f(1.1)-f(1)}{(1.1)-1}

Solution:

Given f(x) = x²

f(1.1) = 1.21

f(1) = 1

\frac{f(1.1)-f(1)}{(1.1)-1}=\frac{1.21-1}{1.1-1}\\ =\frac{0.21}{0.1}

= 2.1

Question 18. Express the function f : X ⇢ R given by f(x) = x³ + 1 as set of ordered pairs, where x = {-1, 0, 3, 9, 7}.

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: