STD - 10 Maths Exercise : 1.4 sum : 1 to 12 Full Solved Medium : English

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 Chapter - 1 : Relations & Functions EXERCISE : 1.4   அனைத்து கணக்குகளும் தனித்தனியாக தொகுக்கப்பட்டு விடைகளுடன் solved செய்து கீழே கொடுக்கப்பட்டுள்ளது.

 

       ஆசிரியர்களுக்கு பாடம் நடத்தும் வேலையில் ஏதேனும் சந்தேகங்கள் இருந்தால் கீழே கொடுக்கப்பட்டுள்ள கணக்குகளை refer செய்து கொள்ளலாம்.Chapter -1 பயிற்சி கணக்குகள் அனைத்தும் தனித்தனியாக தொகுக்கப்பட்டு கீழே கொடுக்கப்பட்டுள்ளது.

 மேலும் மாணவர்களுக்கும் சந்தேகங்கள் இருந்தால் கீழே கொடுக்கப்பட்டுள்ள கணக்குகளை பார்த்து சந்தேகங்களை சரி பார்த்துக் கொள்ளலாம்.

Question 1.

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

Answer:

The vertical line cuts the graph at A and B. The given graph does not represent a function.

The vertical line cuts the graph at most one point P. The given graph represent a function.

The vertical line cuts the graph at three points S,T and U. The given graph does not represent a function.

The vertical line cuts the graph at most one point D. The given graph represents a function.

Question 2.

Let f: A → B be a function defined by
f(x) = x2 – 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

Answer:

A = {2,4,6, 10, 12}
B = {0,1, 2, 4, 5, 9}
f(x) = x2 – 1
f(2) = 22 – 1 = 1 – 1 = 0
f(4) = 42 – 1 = 2 – 1 = 1
f(6) = 62 – 1 = 3 – 1 = 2
f(10) = 102 – 1 = 5 – 1 = 4
f(12) = 122 – 1 = 6 – 1 = 5

(i) Set of ordered pairs

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

(ii) Table

(iii) Arrow diagram

(iv) Graph

Question 3.

Represent the function f = {(1,2), (2,2), (3,2), (4,3),(5,4)} through (i) an arrow diagram (it) a table form (iii) a graph.

Answer:

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

(i) Arrow diagram

(ii) Table form

(iii) Graph

Question 4.

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

Answer:

f: N → N
N = {1,2,3,4,5,… }
f(x) = 2x – 1
f(1) = 2(1) – 1 = 2 – 1 = 1
f(2) = 2(2) – 1 = 4 – 1 = 3
f(3) = 2(3) – 1 = 6 – 1 = 5
f(4) = 2(4) – 1 = 8 – 1 = 7
f(5) = 2(5) – 1 = 10 – 1 = 9
f = {(1,1) (2, 3) (3, 5) (4, 7) (5,9) …..}

(i) Different elements has different images. This function is one to one function.
(ii) Here Range is not equal to co-domain. This function not an onto function.
∴ The given function is one-one but not an onto.

Question 5.

Show that the function f: N ⇒ N defined by f(m) = m2 + m + 3 is one-one function.

Answer:

N = {1,2,3, 4,5, ….. }
f(m) = m2 + m + 3
f(1) = 12 + 1 + 3 = 5
f(2) = 22 + 2 + 3 = 9
f(3) = 32 + 3 + 3 = 15
f(4) = 42 + 4 + 3 = 23
f = {(1,5) (2, 9) (3, 15) (4, 23)}

From the diagram we can understand different elements in (N) in the domain, there are different images in (N) co-domain.
∴ The function is a one-one function.

Question 6.

Let A = {1, 2, 3, 4) and B = N. Let f: A → B be
defined by f(x) = x3 then,
(i) find the range off
(ii) identify the type of function

Solution:

A = {1, 2, 3, 4}
B = N

f: A → B,f(x) = x3

(i) f(1) = 13 = 1
f(2) = 23 = 8
f(3) = 33 = 27
f(4) = 43 = 64
(ii) The range of f = {1, 8, 27, 64 )
(iii) It is one-one and into function.

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 – 4x2

Answer:

(i) f(x) = 2x + 1

f(0) = 2(0) + 1 = 0 + 1 = 1
f(1) = 2(1) + 1 = 2 + 1 = 3
f(2) = 2(2) + 1 = 4 + 1 = 5
f(3) = 2(3) + 1 = 6 + 1 = 7

Different elements has different images

∴ It is an one-one function.
It is also an onto function. The function is one-one and onto function.
∴ It is a bijective function.

(ii) f(x) = 3 – 4x2

f(1) = 3 – 4(1)2
= 3 – 4 = -1
f(2) = 3 – 4(2)2 = 3 – 16 = – 13
f(3) = 3 – 4(3)2 = 3 – 36 = – 33
f(4) = 3 – 4(42) = 3 – 64 = – 61

It is not a bijective function. The positive numbers “R” do not have negative pre – image in X in R.

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.

Answer:

A = {-1, 1}; B = {0,2}
f(x) = ax + b
f(-1) = a(-1) + b
0 = -a + b
a – b = 0 ….(1)
f(1) = a(1) + b
2 = a + b
a + b = 2 ….(2)
Solving (1) and (2) we get

Substitute a = 1 in (1)
The value of a = 1 and b = 1

Question 9.

If the function f is defined by

find the value of
(i) f(3)
(ii) f(0)
(iii) f(1. 5)
(iv) f(2) + f(-2)

Answer:

f(x) = x + 2 when x = {2,3,4,……}
f(x) = 2
f(x) = x – 1 when x = {-2}

(i) f(x) = x + 2

f(3) = 3 + 2 = 5

(ii) f(x) = 2

f(0) = 2

(iii) f(x) = x – 1

f(-1.5) = -1.5 – 1 = -2.5

(iv) f(x) = x + 2

f(2) = 2 + 2 = 4
f(x) = x – 1
f(-2) = – 2 – 1 = – 3
f(2) + f(-2) = 4 – 3
= 1

Question 10.

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

Answer:

f(x) = 6x + 1 ; x = {-5,-4,-3,-2,-1,0,1}
f(x) = 5x2 – 1 ; x = {2, 3, 4, 5}
f(x) = 3x – 4 ; x = {6, 7, 8, 9}

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

f(x) = 6x + 1
f(-3) = 6(-3) + 1 = -18 + 1 = -17
f(x) = 5x2 – 1
f(2) = 5(2)2 – 1 = 20 – 1 = + 19
f(-3) + f(2) = – 17 + 19
= 2

(ii) f(7) – f(1)

f(x) = 3x – 4
f(7) = 3(7) – 4 = 21 – 4 = 17
f(x) = 6x + 1
f(1) = 6(1) + 1 = 6 + 1 = 7
f(7) – f(1) = 17 – 7
= 10

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

f(x) = 5x2 – 1
f(4) = 5(4)2 – 1 = 5(16) – 1
= 80 – 1 = 79
f(x) = 3x – 4
f(8) = 3(8) – 4 = 24 – 4 = 20
2f(4) + f(8) = 2(79) + 20
= 158 + 20
= 178

f(x) = 6x + 1
f(-2) = 6(-2) + 1 = -12 + 1 = -11
f(x) = 3x – 4
f(6) = 3(6) – 4 = 18 – 4 = 14
f(x) = 5x2 – 1
f(4) = 5(4)2 – 1 = 5(16) – 1
= 80 – 1 = 79
f(x) = 6x + 1
f(-2) = 6(-2) + 1 = -12 + 1 = -11

Question 11.

The distance S an object travels under the influence of gravity in time t seconds is given by S(t) = 12 gt2 + at + b where, (g is the acceleration due to gravity), a, b are constants. Check if the function S (t)is one-one.

Solution:

S(t) = 12 gt2 + at + b
Let t be 1, 2, 3, ………, seconds
S(1) = 12 g(12) + a(1) + b = 12 g + a + b
S(2) = 12 g(22) + a(2) + b = 2g + 2a + b
Yes, for every different values of t, there will be different values as images. And there will be different preimages for the different values of the range. Therefore it is one-one function.

Question 12.

The function ‘t’ which maps temperature in Celsius (C) into temperature in Fahrenheit (F) is defined by
t(C) = F where F = 95 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 Fahrenheit value.

Answer:

Given t(C) = 9C5 + 32

(i) t(0) = 9(0)5 + 32

= 32° F

(ii) t(28) = 9(28)5 + 32

= 2525 + 32
= 50.4 + 32
= 82.4° F

(iii) t(-10) = 9(−10)5 + 32

= -18 + 32
= 14° F

(iv) t(C) = 212

9C5 + 32 = 212
9C5 = 212 – 32
= 180
9C = 180 × 5
C = 180×59
= 100° C

(v) consider the value of C be “x

t(C) = 9C5 + 32
x = 9x5 + 32
5x = 9x + 160
-160 = 9x – 5x
-160 = 4x
x = −1604 = -40
The temperature when the Celsius value is equal to the fahren heit value is -40°

Composition of two Functions

Let f: A → B and g: B → C be two functions. Then the composition of f and g denoted by go f is defined as the function go f (x) = g[f(x)] for all x ∈ A.

Composition of three Functions

Let A, B, C, D be four sets and let f: A → B; g : B → C and h : C → D be three functions, using composite functions fog and goh, we get two new functions like (fog) oh and        fo (goh).
Note: Composition of three function is always associative.

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