# 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) = x ^{2}**

**(ii) g(x) = sinx**

**(iii) h(x) = x ^{2} + 1**

**Find the range of each function.**

**Solution:**

(i)We have,f(x) = x

^{2}Range of f(x) = R+ (set of all real numbers greater than or equal to zero)

= {x ∈ R+ | x ≥ 0}

(ii)We haveg(x) = sinx

Range of g(x) = {x ∈ R : -1 ≤ x ≤ 1}

(iii)We haveh(x) = x

^{2}+ 1Range 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) f _{1} = {(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,f

_{1}= {(1, 1), (2, 11), (3, 1), (4, 15)}f

_{1}is a function from X to Y

(b)We have,f

_{2}= {(1, 1), (2, 7), (3, 5)}f

_{2}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,f

_{3}= {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}f

_{3}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) = x**^{2} + 1, then find f^{-1}{17} and f^{-1}{-3}.

^{2}+ 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⇒ x

^{2}+ 1 = 17⇒ x

^{2}= 17 – 1 = 16⇒ x = ±4

Let f

^{-1}{-3} = x. Then, f(x) = -3⇒ x

^{2}+ 1 = -3⇒ x

^{2}= -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) R _{1} = R_{1} = {(p, 1), (q, 2), (r, 1), (s, 2)}**

**(b) R _{2} = {(p, 1), (q, 1), (r, 1), (s, 2)}**

**(c) R _{3} = {(p, 1), (q, 2), (p, 2), (s, 3)}**

**(d) R _{4} = {(p, 2), (q, 3), (r, 2), (s, 2)}**

**Solution:**

We have

A = {p, q, r, s} and B = {1, 2, 3}

(a) Now,

R

_{1}= {(p, 1), (q, 2), (r, 1), (s, 2)}R

_{1}is a function(b) Now,

R

_{2}= {(p, 1), (q, 2), (r, 1), (s, 1)}R

_{2}is a function(c) Now,

R

_{3}= {(p, 2), (q, 3), (r, 2), (s, 2)}R

_{3}is not a function because an element p ∈ A is associated to two elements 1 and 2 in B.(d) Now,

R

_{4}= {(p, 2), (q, 3), (r, 2), (s, 2)}R

_{4}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 = 9and, g(2) = (2)

^{2}= 4 and g(2) = 3 × 2 = 6We 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) = x**^{2}, find

^{2}, find

**Solution:**

Given f(x) = x

^{2}f(1.1) = 1.21

f(1) = 1

= 2.1

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

^{3}+ 1 as set of ordered pairs, where x = {-1, 0, 3, 9, 7}.

**Solution:**

f : X ⇢ R given by f(x) = x

^{3}+ 1f(-1) = (-1)

^{3}+ 1 = -1 + 1 = 0f(0) = (0)

^{3}+ 1 = 0 + 1 = 1f(3) = (3)

^{3}+ 1 = 27 + 1 = 28f(9) = (9)

^{3}+ 1 = 81 + 1 = 82f(7) = (7)

^{3}+ 1 = 343 + 1 = 344Set of ordered pairs are {(-1, 0), (0, 1), (3, 28), (9, 82), (7, 344)}

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