Let L be the set of all lines in X Y plane and R be the relation in L defined as R = { ( L 1 , L 2 ) : L 1 is parallel to L 2 } . Show that R is an equivalence relation. Find the set of all lines related to the line y = 2 x + 4 .

R = ( L 1 , L 2 ) : L 1 is parallel to L 2 R is reflexive as any line L 1 is parallel to itself i.e., ( L 1 , L 1 ) ∈ R . Now , Let ( L 1 , L 2 ) ∈ R . ⇒ L 1 is parallel to L 2 . ⇒ L 2 is parallel to L 1 . ( L 2 , L 1 ) ∈ R ∴ R is symmetric. Now, Let ( L 1 , L 2 ) , ( L 2 , L 3 ) ∈ R . ⇒ L 1 is parallel to L 2 . Also, L 2 is parallel to L 3 . ⇒ L 1 is parallel to L 3 ∴ R is transitive. Hence, R is an equivalence relation. The set of all lines related to the line y = 2 x + 4 is the set of all lines that are parallel to the line y = 2 x + 4 . Slope of line y = 2 x + 4 is m = 2 It is known that parallel lines have the same slopes. The line parallel to the given line is of the form y = 2 x + c , where c ∈ R . Hence, the set of all lines related to the given line is given by y = 2 x + c , where c ∈ R .

Show that f : [ - 1 , 1 ] → R , given by f ( x ) = x ( x + 2 ) is one-one. Find the inverse of the function f : [ - 1 , 1 ] → Range f .

f : [ 1 , 1 ] → R is given as f ( x ) = x x + 2 . let f ( x ) = f ( y ) . ⇒ x x + 2 = y y + 2 ⇒ x y + 2 x = x y + 2 y ⇒ 2 x = 2 y ⇒ x = y ∴ f is a one-one function. It is clear that f : [ - 1 , 1 ] → range f is onto. ∴ f : [ - 1 , 1 ] → Range f is one-one and onto and therefore, the inverse of the function: f : [ - 1 , 1 ] → Range f exists. Let g : Range f → [ - 1 , 1 ] → be the inverse of f . Let y be an arbitrary element of range. f . Since f : [ - 1 , 1 ] → Range f is onto, we have: y = f ( x ) for same x ∈ [ - 1 , 1 ] ⇒ y = x x + 2 ⇒ x y + 2 y = x ⇒ x ( 1 - y ) = 2 y ⇒ x = 2 y 1 - y , y ≠ 1 Now, let us define g : Range f → [ - 1 , 1 ] as g ( y ) = 2 y 1 - y , y ≠ 1 . Now, ( g o f ) ( x ) = g ( f ( x ) ) = g ( x x + 2 ) = 2 ( x x + 2 ) 1 - x x + 2 = 2 x x + 2 - x = 2 x 2 = x ( f o g ) ( y ) = f ( g ( y ) ) = f ( 2 y 1 - y ) = 2 y 1 - y 2 y 1 - y + 2 = 2 y 2 y + 2 - 2 y = 2 y 2 = y ∴ g o f = I - 1 , 1 and f o g = I Range f ∴ f - 1 = g ⇒ f - 1 ( y ) = 2 y 1 - y , y ≠ 1

Let f : R → R is defined by f ( x ) = 3 x - 2 and g : R → R is defined by g ( x ) = x + 2 3 . Show that f · g = g · f .

Given that, f ( x ) = 3 x - 2 & g ( x ) = x + 2 3 f · g = f ( x ) · g ( x ) = 3 x - 2 x + 2 3 = x 2 + 4 x 3 - 4 3 g · f = g ( x ) · f ( x ) = x + 2 3 3 x - 2 = x 2 + 4 x 3 - 4 3 Hence, proved.

Determine the relation R on the set of whole numbers ≤ 10 defined by R = { ( x , y ) ∣ 2 x + 3 y = 12 } .

Given, R = ( x , y ) ; 2 x + 3 y = 12 , { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 910 } x = 0 , 2 × 0 + 3 y = 12 ⇒ y = 4 ⇒ ( 0 , 4 ) x = 1 , 2 × 1 + 3 y = 12 ⇒ y = 10 3 x = 2 , 2 × 2 + 37 = 12 ⇒ y = 8 3 x = 3 , 2 × 3 + 3 y = 12 ⇒ y = 2 ⇒ ( 3 , 2 ) x = 4 , 2 × 4 + 3 y = 12 ⇒ y = 4 3 x = 5 , 2 × 5 + 3 y = 12 ⇒ y = 2 3 x = 6 , 2 × 6 + 3 y = 12 ⇒ y = 0 ⇒ ( 6 , 0 ) ∴ R = { ( 0 , 4 ) , ( 3 , 2 ) , ( 6 , 0 ) }

Let f : ℝ - { 3 } → ℝ - { 1 } is defined by f ( x ) = x - 2 x - 3 . Show that $f$ is a bijective function.

Let A = R - { 3 } and B = R - { 1 } f : A → B Given, f ( x ) = x - 2 x - 3 f x 1 = f x 2 x 1 - 2 x 1 - 3 = x 2 - 2 x 2 - 3 ⇒ x 2 - 3 x 1 - 2 = x 2 - 2 x 1 - 3 ⇒ x 1 x 2 - 3 x 1 - 2 x 2 + 6 = x 1 x 2 - 3 x 2 - 2 x 1 + 6 ⇒ - 3 x 1 - 2 x 2 = - 3 x 2 - 2 x 1 ⇒ - x 1 = - x 2 ⇒ x 1 = x 2 So, f ( x ) is one-one f ( x ) = x - 2 x - 3 ⇒ y = x - 2 x - 3 ⇒ y ( x - 3 ) = x - 2 ⇒ y x - 3 y = x - 2 ⇒ y x - x = 3 y - 2 ⇒ x ( y - 1 ) = 3 y - 2 ⇒ x = 3 y - 2 ( y - 1 ) f ( x ) = x - 2 x - 3 = 3 y - 2 y - 1 - 2 3 y - 2 y - 1 - 3 = 3 y - 2 - 2 ( y - 1 ) y - 1 3 y - 2 - 3 ( y - 1 ) y - 1 = 3 y - 2 - 2 y + 2 3 y - 2 - 3 y + 3 = 3 y - 2 y - 2 + 3 = y f ( x ) = y f ( x ) is onto Hence, f ( x ) is bijective

If f : ℝ → ℝ is defined by f ( x ) = x 2 - 3 x + 2 , find f ( f ( x ) ) .

Given f : R → R f ( x ) = x 2 - 3 x + 2 f f ( x ) = f x 2 - 3 x + 2 = x 2 - 3 x + 2 2 - 3 x 2 - 3 x + 2 + 2 = x 2 - 3 x 2 + 2 x 2 x 2 - 34 + 4 - 3 x 2 + 9 x - 6 + 2 = x 4 - 6 x 3 + 9 x 2 + 4 x 2 - 12 x - 3 x 2 + 9 x = x 4 - 6 x 3 + 10 x 2 - 3 x f f ( x ) = x 4 - 6 x 3 + 10 x 2 - 3 x

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Relations and Functions>Assam Board-2020

Embibe Experts Solutions for Chapter: Relations and Functions, Exercise 1: Assam Board-2018

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Relations and Functions, Exercise 1: Assam Board-2018

Attempt the free practice questions on Chapter 1: Relations and Functions, Exercise 1: Assam Board-2018 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Relations and Functions, Exercise 1: Assam Board-2018 with Hints & Solutions

HARD

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Relations and Functions>Assam Board-2018>Q 1

Let $L$ be the set of all lines in $X Y$ plane and $R$ be the relation in $L$ defined as $L_{2}$ . Show that $R$ is an equivalence relation. Find the set of all lines related to the line $y = 2 x + 4 .$

HARD

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Relations and Functions>Assam Board-2018>Q 2

Show that $f : [- 1, 1] \to R,$ given by $f (x) = \frac{x}{(x + 2)}$ is one-one. Find the inverse of the function $f : [- 1, 1] \to$ Range $f$ .

MEDIUM

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Relations and Functions>Assam Board-2019>Q 3

Let $A = {x : 1 < x < 10, x is an odd natural number}$ and $B = {y : 90 < y < 100, y is a prime number}$ . Write the number of relations from $A$ to $B$ .

EASY

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Relations and Functions>Assam Board-2019>Q 4

Let $f : R \to R$ is defined by $f (x) = 3 x - 2$ and $g : R \to R$ is defined by $g (x) = \frac{x + 2}{3}$ . Show that $f \cdot g = g \cdot f$ .

EASY

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Relations and Functions>Assam Board-2020>Q 5

Determine the relation $R$ on the set of whole numbers $\leq 10$ defined by $R = {(x, y) ∣ 2 x + 3 y = 12}$ .

MEDIUM

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Relations and Functions>Assam Board-2020>Q 6

Let $f : ℝ - {3} \to ℝ - {1}$ is defined by $f (x) = \frac{x - 2}{x - 3} .$ Show that $f$ is a bijective function.

EASY

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Relations and Functions>Assam Board-2018>Q 7

If $f : ℝ \to ℝ$ is defined by $f (x) = x^{2} - 3 x + 2$ , find $f (f (x))$ .

EASY

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Relations and Functions>Assam Board-2020>Q 8

Show that the relation $R$ in the set $A = {x \in Z : 0 \leq x \leq 12},$ given by $R = {(a, b) : | a - b |$ is a multiple of $4}$ is an equivalence relation. Find the set of all elements related to $1$ .

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