Show that the function f : R∗ → R∗ defined by f(x) = 1/x is one-one and onto, where R∗ is the set of all non-zero real numbers. Is the result true, if the domain R∗ is replaced by N with co-domain being same as R∗ ?

Class 12, NCERT Chapter 1, Exercise 1.2, Q1

It is given that f: R∗ →R∗ is defined by f(x) =1/x
For one – one:
Let x, y ∈ R∗
such that f(x) = f(y)
⇒1/x =1/y
⇒ x = y
Therefore, f is one–one.
For onto:
For y ∈ R∗,

there exists x =1/y ∈ R∗ [as y ≠ 0]

such that
⇒f(x) = 1/ [1/y]= y
Therefore, f is onto.
Thus, the given function f is one-one and onto.
Now, consider function g:N→R∗ defined by g(x) =1/x
We have,
⇒g(x) = g(y)
⇒1/x=1/y
⇒x = y

Therefore, g is one–one.
Further, it is clear that g is not onto as for 1.2 ∈ R∗,

there does not exit any x in N
such that g(x) =1/1.2

Comments

Let L be the set of all lines in a plane and R be the relation in L defined as R = {(L1 , L2 ) : L1 is perpendicular to L2 }. Show that R is symmetric but neither reflexive nor transitive.

RELATIONS AND FUNCTIONS Class 12, NCERT Chapter 1, Example 3 Solution Figure R is not reflexive, as a line L1 can not be perpendicular to itself, i.e., (L1 , L1 ) ∉ R. R is symmetric as (L1 , L2 ) ∈ R ⇒ L1 is perpendicular to L2 ⇒ L2 is perpendicular to L1 ⇒ (L2 , L1 ) ∈ R. R is not transitive. Indeed, if L1 is perpendicular to L2 and L2 is perpendicular to L3 , then L1 can never be perpendicular to L3 . In fact, L1 is parallel to L3 , i.e., (L1 , L2 ) ∈ R, (L2 , L3 ) ∈ R but (L1 , L3 ) ∉ R. Example1 Example2 Example3 ←you are here

Show that the function f : N → N, given by f(x) = 2x, is one-one but not onto

Class 12, NCERT Chapter 1, Example8 The function f is one-one, for f ( x 1 ) = f ( x 2 ) ⇒ 2 x 1 = 2 x 2 ⇒ x 1 = x 2 Further, f is not onto, as for 1 ∈ N , there does not exist any x in N , such that f(x) = 2x = 1.

Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by f(x) = (x-2)/(x-3) Is f one-one and onto? Justify your answer.

Class 12, NCERT Chapter 1, Exercise 1.2, Q10 A = R - {3}, B = R - {1} f : A → B is defined as Let x,y∈A such that f(x)=f(y). ⇒ (x-2)/(x-3)=(y-2)/(y-3) ⇒ (x-2)(y-3)=(y-2)(x-3) ⇒ xy-3x-2y+6=xy-3y-2x+6 ⇒ -3x-2y=-3y-2x ⇒ 3x-2x=3y-2y ⇒ x=y ∴ f is one-one. Let y ∈B = R - {1}. Then, y ≠ 1. The function f is onto if there exists x ∈A such that f ( x ) = y. Now, f(x)=y ⇒(x-2)(x-3)=y ⇒x-2=xy-3y ⇒x(1-y)=-3y+2 ⇒ x=2-3y/1-y ∈A Thus, for any y ∈ B, there exists x=2-3y/1-y ∈A such that Hence, function f is one-one and onto.

KnowledgeLess

Search This Blog