in Set Theory & Algebra
1,961 views
1 vote
1 vote
The function f:[0,3]->[1,29] defined by f(x)=2*X^3 -15*X^2+36*X+1 is

a) injective and surjective

b) injective but not surjective

c) injective but not surjective

d) neither injective nor surjective
in Set Theory & Algebra
2.0k views

2 Comments

Bro B & C are same please correct
0
0
0
0

3 Answers

3 votes
3 votes
f(x) = 2x^3 – 15x^2 + 36x + 1
⇒ f′(x) = 6(x^2 − 5x + 6) = 6(x-2)(x-3)

f(x) has minima and maxima at x=3 & x=2 respectively.

Since f(x) is non monotonic in ∈ [0,3]
⇒ f(x) is not one-one.

And, f(x) is increasing in x ∈[0,2) and decreasing in ∈(2,3]
f(0) = 1, f(2) = 29 & f(3) = 28

since f(x) is continuous, so range of f(x) is minimum and maximum value b/w [0,3].

minimum value=1 at x=0 & maximum value=29 at x=2.

Therefore Range of f(x) is [1,29]
⇒ f(x) is onto.

Hence, f(x) is surjective(onto) but not injective(one-one). is the answer.
edited by
by

2 Comments

How did you find that function was surjective ?

please explain.
0
0

@Jhaiyam bro,

function is surjective for given interval [0,3] not for real number.

0
0
0 votes
0 votes
There is one to one relation :

f(0)=1

f(1)=2

f(2)=28

f(3)=29

ONE TO ONE MAPPING i.e. INJECTIVE

so answer is (B)
edited by

1 comment

You have only considered integer,what about fractions??
0
0
0 votes
0 votes

observe the graph. every element of the co-domain [1-29] is mapped with every element in domain [0-3] (function is continuous)

but in the range [2-3] all elements are mapped with some elements in the range [1-2]. so all elements in the co-domain have a pre-image(surjective) every every pre-image is not unique(not injective)

 

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true