Show that the function f : N → N defined by f ( x ) = x 3 is one-to-one function.

Graph of f ( x ) = x 3 Since the function f(x) is monotonically increasing from the domain N → N ∴ f ( x ) is one-one Range of f ( x ) = ( - ∞ , ∞ ) ≠ N (co-domain) ∴ f ( x ) is into ∴ f ( x ) is one - one into.

Show that the function f : R → R defined by f ( x ) = 2 x 2 + 1 is neither one one nor onto function.

Given function f : R → R defined as f ( x ) = 2 x 2 + 1 . - 1 , 1 ∈ R , - 1 ≠ 1 but f ( - 1 ) = 1 = f ( 1 ) ⇒ f is many-one. ∴ f is not injective i.e., not one-one. Also - 2 ∈ R , the co-domain but there exists no element x in the domain R such that f ( x ) = - 2 . i . e . , 2 x 2 + 1 = - 2 since the square of any real number is always non-negative. ( ∵ 2 x 2 + 1 = - 2 ⇒ x = ± - 3 2 = ± 3 2 i ∉ R ) Therefore, f is not surjective i.e., not onto. Hence, f is neither one-one nor onto.

Show that the function f : Z → Z defined by f ( x ) = 3 x 3 is a bijective function.

Given: f : Z → Z : f ( x ) = 3 x 3 Solution: We have, f ( x ) = 3 x 3 For, f x 1 = f x 2 ⇒ 3 x 1 3 = 3 x 2 3 ⇒ x 1 = x 2 When, f x 1 = f x 2 ) then x 1 = x 2 ∴ f ( x ) is one-one Let f ( x ) = y such that y ∈ z ⇒ y = 3 x 3 ⇒ x = y 3 3 If y = 2 , as y ∈ z Then we will get an irrational value of x, but x ∈ Z Hence f ( x ) is into Hence we can say that the function f : Z → Z defined by f ( x ) = 3 x 3 is a bijective function.

Show that the function f : Q → Q defined by f ( x ) = 4 x + 3 is one-one and onto function.

The given function is f : Q → Q and f = 2 x + 3 Take x 1 , x 2 ∈ Q such that f x 1 = f x 2 2 1 + 3 = 2 x 2 + 3 ⇒ x 1 = x 2 Hence, f x is one-one. The given function is f : Q → Q and f x = 2 x + 3 Onto : take y = 2 x + 3 ⇒ x = y - 3 2 For every rational number y , y - 3 2 is a rational number. Therefore, f x is onto.

If A = R - ( 3 ) and B = R - ( 1 ) consider the function f : A → B defined by f ( x ) = x + 4 x - 3 , show that its bijective function.

Given, f : R - { 3 } defined by f ( x ) = x + 4 x - 3 . First, we shall check if f is a one-one function (i.e. injective function) or not. For any x , y ∈ R , f ( x ) = f ( y ) ⇒ x + 4 x - 3 = y + 4 y - 3 ⇒ x + 4 y - 3 = x - 3 y + 4 ⇒ x y - 3 x + 4 y - 12 = x y + 4 x - 3 y - 12 ⇒ - 3 x - 4 x = - 3 y - 4 y ⇒ - 7 x = - 7 y ⇒ x = y . i.e., f ( x ) = f ( y ) ⇒ x = y for all x , y ∈ R . Hence, f is a one-one or injective function. Next, we shall check if f is an onto function or not. Let y be an arbitrary element of R . Then, y = x + 4 x - 3 and f ( x ) = x + 4 x - 3 ⇒ f ( x ) = y = x + 4 x - 3 ⇒ y x - 3 = x + 4 ⇒ x y - 3 y = x + 4 ⇒ x y - x = 3 y + 4 ⇒ x y - 1 = 3 y + 4 ⇒ x = 3 y + 4 y - 1 ∈ R - 3 since x = 3 ⇒ y is undefined. Hence, f is an onto or surjective function. A function that is both injective and surjective is bijective. So, f is a bijective function. Therefore, the given function is a bijective function. Hence, proved.

E M B I B E

I.S.C MATHEMATICS FOR CLASS XI>Chapter 2 - Relations and Functions>EXERCISE-2(III)

SURANJAN SAHA and SABITA SAHA Solutions for Chapter: Relations and Functions, Exercise 3: EXERCISE-2(III)

Author:SURANJAN SAHA & SABITA SAHA

SURANJAN SAHA Mathematics Solutions for Exercise - SURANJAN SAHA and SABITA SAHA Solutions for Chapter: Relations and Functions, Exercise 3: EXERCISE-2(III)

Attempt the free practice questions on Chapter 2: Relations and Functions, Exercise 3: EXERCISE-2(III) with hints and solutions to strengthen your understanding. I.S.C MATHEMATICS FOR CLASS XI solutions are prepared by Experienced Embibe Experts.

Questions from SURANJAN SAHA and SABITA SAHA Solutions for Chapter: Relations and Functions, Exercise 3: EXERCISE-2(III) with Hints & Solutions

EASY

11th ICSE

IMPORTANT

I.S.C MATHEMATICS FOR CLASS XI>Chapter 2 - Relations and Functions>EXERCISE-2(III)>Q 5

Draw arrow diagram for each of the following functions: $\{(1, 2), (2, 7), (3, 5), (4, 3), (5, 2), (6, 3)\}$

EASY

11th ICSE

IMPORTANT

I.S.C MATHEMATICS FOR CLASS XI>Chapter 2 - Relations and Functions>EXERCISE-2(III)>Q 15

Show that the function $f : N \to N$ defined by $f (x) = x^{3}$ is one-to-one function.

EASY

11th ICSE

IMPORTANT

I.S.C MATHEMATICS FOR CLASS XI>Chapter 2 - Relations and Functions>EXERCISE-2(III)>Q 16

Show that the function $f : R \to R$ defined by $f (x) = 2 x^{2} + 1$ is neither one one nor onto function.

EASY

11th ICSE

IMPORTANT

I.S.C MATHEMATICS FOR CLASS XI>Chapter 2 - Relations and Functions>EXERCISE-2(III)>Q 17

Show that the function $f : Z \to Z$ defined by $f (x) = 3 x^{3}$ is a bijective function.

HARD

11th ICSE

IMPORTANT

I.S.C MATHEMATICS FOR CLASS XI>Chapter 2 - Relations and Functions>EXERCISE-2(III)>Q 18

Show that the function f: NR defined by $f (x) = 3 x - 2$ is injective and surjective function.

HARD

11th ICSE

IMPORTANT

I.S.C MATHEMATICS FOR CLASS XI>Chapter 2 - Relations and Functions>EXERCISE-2(III)>Q 19

Show that the function $f : Q \to Q$ defined by $f (x) = 4 x + 3$ is one-one and onto function.

HARD

11th ICSE

IMPORTANT

I.S.C MATHEMATICS FOR CLASS XI>Chapter 2 - Relations and Functions>EXERCISE-2(III)>Q 20

If $A = R - (3)$ and $B = R - (1)$ consider the function $f : A \to B$ defined by $f (x) = \frac{x + 4}{x - 3}$ , show that its bijective function.

HARD

11th ICSE

IMPORTANT

I.S.C MATHEMATICS FOR CLASS XI>Chapter 2 - Relations and Functions>EXERCISE-2(III)>Q 21

Consider a function $f : R \to R$ defined by $f (x) = p x ² + q x + r$ , where $p, q, r$ are integers. Given that $f (- 2) = 15, f (0) = 5 a n d f (1) = 3$ . Find the value of $p, q, r .$

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