Show that the Modulus Function f : R → R, given by f(x) = | x |, is neither oneone nor onto, where | x | is x, if x is positive or 0 and | x

Show that the Modulus Function f : R → R, given by f(x) = | x |, is neither oneone nor onto, where | x | is x, if x is positive or 0 and | x | is – x, if x is negative

Class 12, NCERT Chapter 1, Exercise 1.2, Q4

f: R → R is given by,

It is seen that
f(-1)= |-1|=1, f(1)=|1|=1

∴f(-1) = f(1), but - 1 ≠ 1.

∴ f is not one-one.

Now, consider -1 ∈ R.

It is known that f(x) = |x| is always non-negative. Thus, there does not exist any element x in domain R such that f(x) = |x| = -1.

∴ f is not onto.

Hence, the modulus function is neither one-one nor onto.

Comments

Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y) : x and y have same number of pages} is an equivalence relation.

Class 12, NCERT Chapter 1, Exercise 1.1, Q7 Set A is the set of all books in the library of a college. Given, R = { x , y ): x and y have the same number of pages} Now, R is reflexive since ( x , x ) ∈ R as x and x has the same number of pages. Let ( x , y ) ∈ R ⇒ x and y have the same number of pages. ⇒ y and x have the same number of pages. ⇒ ( y , x ) ∈ R ∴R is symmetric. Now, let ( x , y ) ∈R and ( y , z ) ∈ R. ⇒ x and y and have the same number of pages and y and z have the same number of pages. ⇒ x and z have the same number of pages. ⇒ ( x , z ) ∈ R ∴R is transitive. Hence, R is an equivalence relation.

Let A be the set of all 50 students of Class X in a school. Let f : A → N be function defined by f(x) = roll number of the student x. Show that f is one-one but not onto.

Class 12, NCERT Chapter 1, Example7 No two different students in the class can have the same roll number. Therefore, f must be one-one. We can assume without any loss of generality that roll numbers of students are from 1 to 50. This implies that 51,52,53... in N is not roll number of any student of the class, so that 51,52,53... can not be an image of any element of X under f. Hence, f is not onto.

Let f : N → N be defined by f (n) ={ (n+1)/2, if n is odd and (n-1)/2, if n is even, for all n ∈ N.State whether the function f is bijective. Justify your answer

Class 12, NCERT Chapter 1, Exercise 1.2, Q9 Thus it is bijective.

KnowledgeLess

Search This Blog