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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Additional Questions

Question 1.

Let A = {1, 2, 3, 4} and B = {-1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} Let R = {(1, 3), (2, 6), (3,10), (4, 9)} ⊂ A × B be a relation. Show that R is a function and find its domain, co-domain and the range of R.

Answer:

Domain of R = {1, 2, 3, 4}

Co-domain of R = B = {-1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12}

Range of R= {3, 6, 10, 9}

Question 2.

Let A = {0, 1, 2, 3} and B = {1, 3, 5, 7, 9} be two sets. Let f: A → B be a function given by f(x) = 2x + 1. Represent this function as (i) a set of ordered pairs (ii) a table (iii) an arrow and (iv) a graph.

Solution:

A = {0, 1, 2, 3}, B = {1, 3, 5, 7, 9}

f(x) = 2x + 1

f(0) = 2(0) + 1 = 1

f(1) = 2(1) + 1 = 3

f(2) = 2(2) + 1 = 5

f(3) = 2(3) + 1 = 7

(i) A set of ordered pairs.

f = {(0, 1), (1, 3), (2, 5), (3, 7)}

(ii) A table

(iii) An arrow diagram

Question 3.

State whether the graph represent a function. Use vertical line test.

Solution:

It is not a function as the vertical line PQ cuts the graph at two points.

Question 4.

Let f = {(2, 7), (3, 4), (7, 9), (-1, 6), (0, 2), (5, 3)} be a function from A = {-1, 0, 2, 3, 5, 7} to B = {2, 3, 4, 6, 7, 9}. Is this (i) an one-one function (ii) an onto function, (iii) both one- one and onto function?

Solution:

It is both one-one and onto function.

All the elements in A have their separate images in B. All the elements in B have their preimage in A. Therefore it is one-one and onto function.

Question 5.

A function f: (-7,6) → R is defined as follows.

Find (i) 2f(-4) + 3 f(2)

(ii) f(-7) – f(-3)

Solution:

(i) 2f(-4) + 3f(2)

f(-4) = x + 5 = -4 + 5 = 1

2f(-4) = 2 × 1 = 2

f(2) = x + 5 = 2 + 5 = 7

3f(2) = 3(7) = 21

∴ 2f(-4) + 3f(2) = 2 + 21 = 23

(ii) f(-7) = x^{2} + 2x + 1

= (-7)^{2} + 2(-7) + 1

= 49 – 14 + 1 = 36

f(3) = x + 5 = -3 + 5 = 2

f(-7) – f(-3) = 36 – 2 = 34

Question 6.

If A = {2,3, 5} and B = {1, 4} then find

(i) A × B

(ii) B × A

Answer:

A = {2, 3, 5}

B = {1, 4}

(i) A × B = {2,3,5} × {1,4}

= {(2, 1) (2, 4) (3, 1) (3, 4) (5,1) (5, 4)}.

(ii) B × A = {1,4} × {2,3,5}

= {(1,2) (1,3) (1,5) (4, 2) (4, 3) (4, 5)}

Question 7.

Let A = {5, 6, 7, 8};

B = {- 11, 4, 7, -10, -7, – 9, -13} and

f = {(x,y): y = 3 – 2x, x ∈ A, y ∈ B}.

(i) Write down the elements of f.

(ii) What is the co-domain?

(iii) What is the range?

(iv) Identify the type of function.

Answer:

Given, A = {5, 6, 7, 8},

B = {- 11,4, 7,-10,-7,-9,-13}

y = 3 – 2x

ie; f(x) = 3 – 2x

f(5) = 3 – 2 (5) = 3 – 10 = – 7

f(6) = 3 – 2 (6) = 3 – 12 = – 9

f(7) = 3 – 2(7) = 3 – 14 = – 11

f(8) = 3 – 2 (8) = 3 – 16 = – 13

(i) f = {(5, – 7), (6, – 9), (7, – 11), (8, – 13)}

(ii) Co-domain (B)

= {-11,4, 7,-10,-7,-9,-13} i

(iii) Range = {-7, – 9, -11,-13}

(iv) It is one-one function.

Question 8.

A function f: [1, 6] → R is defined as follows:

Find the value of (i) f(5)

(ii) f(3)

(iii) f(2) – f(4).

Solution:

(i) f(5) = 3x^{2} – 10

= 3 (5^{2}) – 10 = 75 – 10 = 65

(ii) f(3) = 2x – 1

= 2(3) – 1 = 6 – 1 = 5

(ii) f(2) – f(4)

f(2) = 2x – 1

= 2(2) – 1 = 3

f(4) = 3x^{2} – 10

= 3(4^{2}) – 10 = 38

∴ f(2) – f(4) = 3 – 38 = 35

Question 9.

The following table represents a function from A = {5, 6, 8, 10} to B = {19, 15, 9, 11}, where f(x) = 2x – 1. Find the values of a and b.

Solution:

A = {5, 6, 8, 10}, B = {19, 15, 9, 11}

f(x) = 2x – 1

f(5) = 2(5) – 1 = 9

f(8) = 2(8) – 1 = 15

∴ a = 9, b = 15

Question 10.

If R = {(a, -2), (-5, 6), (8, c), (d, -1)} represents the identity function, find the values of a,b,c and d.

Solution:

R = {(a, -2), (-5, b), (8, c), (d, -1)} represents the identity function.

a = -2, b = -5, c = 8, d = -1.