Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

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

  • Last Updated : 21 Feb, 2021

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

(i) f(x) = x2

(ii) g(x) = sinx

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

(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\le x\le 3\\ 3x,\ 3\le x\le 10\end{cases}

The relation f is defined by 

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

and, 

g(x)=\begin{cases}x^2,\ 0\le x\le 2\\ 3x,\ 2\le x\le 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)}




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!