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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.4

Exercise 1.4 Class 10 Maths Samacheer Question 1.

Determine whether the graph given below represent functions. Give reason for your answers concerning each graph.

Solution:

(i) It is not a function. The graph meets the vertical line at more than one points.

(ii) It is a function as the curve meets the vertical line at only one point.

(iii) It is not a function as it meets the vertical line at more than one points.

(iv) It is a function as it meets the vertical line at only one point.

10th Maths Exercise 1.4 Samacheer Kalvi Question 2.

Let f :A → B be a function defined by f(x) = \(\frac{x}{2}\) – 1, Where A = {2, 4, 6, 10, 12},

B = {0, 1, 2, 4, 5, 9}. Represent f by

(i) set of ordered pairs;

(ii) a table;

(iii) an arrow diagram;

(iv) a graph

Solution:

f: A → B

A = {2, 4, 6, 10, 12}, B = {0, 1, 2, 4, 5, 9}

(i) Set of ordered pairs

= {(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)}

(ii) a table

(iii) an arrow diagram;

Ex 1.4 Class 10 Samacheer Question 3.

Represent the function f = {(1, 2),(2, 2),(3, 2), (4,3), (5,4)} through

(i) an arrow diagram

(ii) a table form

(iii) a graph

Solution:

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

(i) An arrow diagram.

10th Maths Exercise 1.4 Question 4.

Show that the function f : N → N defined by f{x) = 2x – 1 is one – one but not onto.

Solution:

f: N → N

f(x) = 2x – 1

N = {1, 2, 3, 4, 5,…}

f(1) = 2(1) – 1 = 1

f(2) = 2(2) – 1 = 3

f(3) = 2(3) – 1 = 5

f(4) = 2(4) – 1 = 7

f(5) = 2(5) – 1 = 9

In the figure, for different elements in x, there are different images in f(x).

Hence f : N → N is a one-one function.

A function f: N → N is said to be onto function if the range of f is equal to the co-domain of f

Range = {1, 3, 5, 7, 9,…}

Co-domain = {1, 2, 3,..}

But here the range is not equal to co-domain. Therefore it is one-one but not onto function.

10th Maths 1.4 Exercise Question 5.

Show that the function f: N → N defined by f (m) = m^{2} + m + 3 is one – one function.

Solution:

f: N → N

f(m) = m^{2} + m + 3

N = {1, 2, 3, 4, 5…..}m ∈ N

f{m) = m^{2} + m + 3

f(1) = 1^{2} + 1 + 3 = 5

f(2) = 2^{2} + 2 + 3 = 9

f(3) = 3^{2} + 3 + 3 = 15

f(4) = 4^{2} + 4 + 3 = 23

In the figure, for different elements in the (X) domain, there are different images in f(x). Hence f: N → N is a one to one but not onto function as the range of f is not equal to co-domain.

Hence it is proved.

10th Maths Ex 1.4 Question 6.

Let A = {1,2, 3, 4} and B = N .

Let f: A → B be defined by f(x) = x^{3} then,

(i) find the range of f

(ii) identify the type of function

Answer:

A = {1,2, 3,4}

B = {1,2, 3, 4, 5,….}

f(x) = x^{3}

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

f(2) = 2^{3} = 8

f(3) = 3^{3} = 27

f(4) = 4^{3} = 64

(i) Range = {1,8, 27, 64}

(ii) one -one and into function.

10th Standard Maths Exercise 1.4 Question 7.

In each of the following cases state whether the function is bijective or not. Justify your answer.

(i) f: R → R defined by f(x) = 2x + 1

(ii) f: R → R defined by f(x) = 3 – 4x^{2}

Solution:

(i) f : R → R

f(x) = 2x + 1

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

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

f(-1) = 2(-1) + 1 = -1

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

It is a bijective function. Distinct elements of A have distinct images in B and every element in B has a pre-image in A.

(ii) f: R → R; f(x) = 3 – 4x^{2}

f(1) = 3 – 4(1^{2}) = 3 – 4 = -1

f(2) = 3 – 4(2^{2}) = 3 – 16 = -13

f(-1) = 3 – 4(-1)^{2} = 3 – 4 = -1

It is not bijective function since it is not one-one

10th Maths 1.4 Question 8.

Let A = {-1, 1} and B = {0, 2}. If the function f: A → B defined by f(x) = ax + b is an onto function? Find a and b.

Solution:

A= {-1, 1},B = {0, 2}

f: A → B, f(x) = ax + b

f(-1) = a(-1) + b = -a + b

f(1) = a(1) + b = a + b

Since f(x) is onto, f(-1) = 0

⇒ -a + b = 0 …(1)

& f(1) = 2

⇒ a + b = 2 …(2)

-a + b = 0

10th Maths Exercise 1.4 10th Sum Question 9.

If the function f is defined by

(i) f(3)

(ii) f(0)

(iii) f(-1.5)

(iv) f(2) + f(-2)

Solution:

(i) f(3) ⇒ f(x) = x + 2 ⇒ 3 + 2 = 5

(ii) f(0) ⇒ 2

(iii) f (- 1.5) = x – 1

= -1.5 – 1 = -2.5

(iv) f(2) + f(-2)

f(2) = 2 + 2 = 4 [∵ f(x) = x + 2]

f(-2) = -2 – 1 = -3 [∵ f(x) = x – 1]

f(2) + f(-2) = 4 – 3 = 1

Samacheer Kalvi 10th Maths Exercise 1.4 Question 10.

A function f: [-5,9] → R is defined as follows:

Solution:

f : [-5, 9] → R

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

f(-3) = 6x + 1 = 6(-3) + 1 = -17

f(2) = 5 × 2 – 1 = 5(2^{2}) – 1 = 19

∴ f(-3) + f(2) = -17 + 19 = 2

(ii) f(7) – f(1)

f(7) = 3x – 4 = 3(7) – 4 = 17

f(1) = 6x + 1 = 6(1) + 1 = 7

f(7) – f(1) = 17 – 7 = 10

(iii) 2f(4) + f(8)

f(4) = 5x^{2} – 1 = 5 × 4^{2} – 1 = 79

f(8) = 3x – 4 = 3 × 8 – 4 = 20

∴ 2f(4) + f(8) = 2 × 79 + 20 = 178

10th Maths Relation And Function Question 11

The distance S an object travels under the influence of gravity in time t seconds is 1 2 given by S(t) = \(\frac { 1 }{ 3 } \)gt^{2} + at + b, where, (g is the acceleration due to gravity), a, b are constants. Check if the function S(t) is one-one.

Answer:

S(t) = \(\frac { 1 }{ 2 } \)gt^{2} + at + b

Let the time be 1, 2, 3 …. n seconds

S(1) = \(\frac { 1 }{ 2 } \)g(1)^{2} + a(1) + b

= \(\frac { g }{ 2 } \) + a + b

S(2) = \(\frac { 1 }{ 2 } \) g(2)^{2} + a(2) + b

= \(\frac { 4g }{ 2 } \) + 2a + b

= 2g + 2a + b

S(3) = \(\frac { 1 }{ 2 } \) g(3)^{2} + a(3) + 6

= \(\frac { 9 }{ 2 } \) g + 3a + b

For every different value of t, there will be different distance.

∴ It is a one-one function.

Samacheer Kalvi Guru 10th Maths Question 12.

The function ‘t’ which maps temperature in Celsius (C) into temperature in Fahrenheit (F) is defined by t(C)= F where F = \(\frac{9}{5}\) C + 32 . Find,

(i) t(0)

(ii) t(28)

(iii) t(-10)

(iv) the value of C when t(C) = 212

(v) the temperature when the Celsius value is equal to the Farenheit value.

Solution: