Question

Define a function f : R → R by the formula f (x) = 3x − 5.
(a) Prove that f is one-to-one.
(b) Prove that f is onto.

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:f(x) = 3x - 5$

One - one

A function f(x) defined from A to B, is said to be one - one if corresponds to one element of A, there is one image in B or no two elements of A have same image in B.

Let us consider two elements,

$\rm :\longmapsto\:x, \: y \: \in \: R \: such \: that \: f(x) = f(y)$

$\rm :\longmapsto\:3x - 5 = 3y - 5$

$\rm :\longmapsto\:3x = 3y$

$\bf\implies \:x = y$

Hence,

$\bf\implies \:f(x) \: is \: one \: - \: one.$

Onto :-

A function f(x) defined from A to B is called onto iff every element of B has a pre - image in A.

Let if possible there exist an element y belongs to B, such that

$\rm :\longmapsto\:y = f(x)$

$\rm :\longmapsto\:y = 3x - 5$

$\rm :\longmapsto\:y + 5 = 3x$

$\rm :\longmapsto\:x = \dfrac{y + 5}{3}$

$\rm :\longmapsto\:As \: y \: \in \: R \:$

So,

$\rm :\longmapsto\:\dfrac{y + 5}{3} \: \in \: R$

$\bf\implies \:x \: \in \: R$

Hence,

$\bf\implies \:f(x) \: is \: onto.$

Additional Information :-

1. Let us consider two sets A and B such that n(A) = n and n(B) = m and n(B), then number of one - one functions from A to B is given by

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:Number \: of \: one - one \: {f}^{n} -\begin{cases} &\sf{0 \: \: if \: n > m} \\ &\sf{P(m,n) \: if \: n \leqslant m} \end{cases}\end{gathered}\end{gathered}$

2. One - one function is also called injective function.

3. Onto function is also called subjective function.

4. If function is both one - one and onto, then function is called bijective function.

Answer 2

Solution−

Given that,

$\rm :\longmapsto\:f(x) = 3x - 5:⟼f(x)=3x−5$

One - one

A function f(x) defined from A to B, is said to be one - one if corresponds to one element of A, there is one image in B or no two elements of A have same image in B.

Let us consider two elements,

$\rm :\longmapsto\:x, \: y \: \in \: R \: such \: that \: f(x) = f(y)$

$\rm :\longmapsto\:3x - 5 = 3y - 5$

$\rm :\longmapsto\:3x = 3y$

$\bf\implies \:x = y$

Hence,

$\bf\implies \:f(x) \: is \: one \: - \: one$

Onto :-

A function f(x) defined from A to B is called onto iff every element of B has a pre - image in A.

Let if possible there exist an element y belongs to B, such that

$\rm :\longmapsto\:y = f(x)$

$\rm :\longmapsto\:y = 3x - 5$

$\rm :\longmapsto\:y + 5 = 3x$

$\rm :\longmapsto\:x = \dfrac{y + 5}{3}$

$\rm :\longmapsto\:As \: y \: \in \: R$

So,

$\rm :\longmapsto\:\dfrac{y + 5}{3} \: \in \: R$

$\bf\implies \:x \: \in \: R$

Hence,

$\bf\implies \:f(x) \: is \: onto.$