# Tamilnadu State board 10th Maths Answers Chapter 1 Relations and Functions Exercise 1.3 ( EX 1.3 )

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# Tamilnadu State board 10th Maths Solutions Chapter 1 Relations and Functions Exercise 1.3 ( EX 1.3 )

- Question 1.

Let f = {(x, y)|x, y ∈ N and y = 2x} be a relation on N. Find the domain, co-domain and range. Is this relation a function?

- Solution:

F = {(x, y)|x, y ∈ N and y = 2x}

x = {1, 2, 3,…}

y = {1 × 2, 2 × 2, 3 × 2, 4 × 2, 5 × 2 …}

R = {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10),…}

Domain of R = {1, 2, 3, 4,…},

Co-domain = {1, 2, 3…..}

Range of R = {2, 4, 6, 8, 10,…}

Yes, this relation is a function.

- Question 2.

Let X = {3, 4, 6, 8}. Determine whether the relation R = {(x, f(x))|x ∈ X, f(x) = x² + 1} is a function from X to N ?

- Solution:

x = {3,4, 6, 8}

R = ((x, f(x))|x ∈ X, f(x) = X² + 1}

f(x) = x² + 1

f(3) = 3² + 1 = 10

f(4) = 4² + 1 = 17

f(6) = 6² + 1 = 37

f(8) = 8² + 1 = 65

R = {(3, 10), (4, 17), (6, 37), (8, 65)}

R = {(3, 10), (4, 17), (6, 37), (8, 65)}

Yes, R is a function from X to N.

- Question 3.

Given the function

f : x → x2 – 5x + 6, evaluate

(i) f(-1)

(ii) f(2 a)

(iii) f(2)

(iv) f(x – 1)

- Answer:

f(x) = x2 – 5x + 6

(i) f (-1) = (-1)2 – 5 (-1) + 6 = 1 + 5 + 6 = 12

(ii) f (2a) = (2a)2 – 5 (2a) + 6 = 4a2 – 10a + 6

(iii) f(2) = 22 – 5(2) + 6 = 4 – 10 + 6 = 0

(iv) f(x – 1) = (x – 1)2 – 5 (x – 1) + 6

= x2 – 2x + 1 – 5x + 5 + 6

= x2 – 7x + 12

- Question 4.

A graph representing the function f(x) is given in figure it is clear that f(9) = 2.

(i) Find the following values of the function

(a) f(0)

(b) f(7)

(c) f(2)

(d) f(10)

(ii) For what value of x is f (x) = 1?

(iii) Describe the following

(i) Domain

(ii) Range.

(iv) What is the image of 6 under f?

- Solution:

From the graph

(a) f(0) = 9

(b) f(7) = 6

(c) f(2) = 6

(d) f(10) = 0

(ii) At x = 9.5, f(x) = 1

(iii) Domain = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

= {x |0 < x < 10, x ∈ R}

Range = {x|0 < x < 9, x ∈ R}

= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

(iv) The image of 6 under f is 5.

- Question 5.

Let f(x) = 2x + 5. If x ≠ 0 then

find `frac{f(x+2)-f(2)}{x}`

- $\frac{\mathrm{Question\; 6.}}{}$

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