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

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 = 

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 

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.

Question 17. If f(x) = x², find 

Solution:

Given f(x) = x²

f(1.1) = 1.21

f(1) = 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)}