Question

The function f : A → B defined by f(x) = 4x + 7, x ∈ R is

Answer 1

Given:

A function function f : A → B defined by $f(x) = 4x + 7$, x ∈ R.

To Find:

f(x) is one-one or onto or both?

Solution:

Here, function f : A → B defined by $f(x) = 4x + 7$, x ∈ R

Claim: f is one-one ?

Consider ,

       $f(x)=f(y)\\4x+7=4y+7\\ 4x=4y\\x=y$

Therefore, function f is one-one.

Now,

Claim: f is onto?

         let, y ∈ B , such that

           $y=f(x)\\ y=4x+7\\4x=y-7\\x=(y-7)/4$

Since, y ∈ B

and $x=(y-7)/4$ also belongs to B.

i.e., x ∈ B

Therefore, f is onto.

Hence, f is both one-one and onto (bijective).

Answer 2

$Given \: that \: the \: function \:f: A \rightarrow B \\defined \: f(x) = 4x + 7$

/* We know that */

$\blue { \: f : A \rightarrow B \:is \: one - one }\\\blue {Let \: x_{1} ,x_{2} \in A\: If \: f(x_{1}) = f(x_{2}) }\\\blue{ then \: x_{1} = x_{2}, \: where \: x_{1} ,x_{2} \in A }$

$Now, f(x_{1} ) = f_(x_{2})$

$\implies 4x_{1} + 7 = 4x_{2} + 7$

$\implies 4x_{1} = 4x_{2}$

$\implies x_{1} = x_{2}$

$\therefore \green {f : A \rightarrow B\: is \: One - One \: function\: --(1)}$

/* We know that */

$\blue{ f: A \rightarrow B \: is \: onto \: function }\\\blue { Let \: b \in B \: \exists a \in A }\\\blue{ such \:that \: f(a) = b }$

$Now, Let \: y = f(x)$

$\implies y = 4x + 7$

$\implies 4x = y - 7$

$\implies x = \frac{y-7}{4}$

$\therefore f(x) = f\Big( \frac{y-7}{4}\Big) \\= 4 \Big( \frac{y-7}{4}\Big) + 7 \\= y - 7 + 7 \\= y$

$\green { f: A \rightarrow \: is \: Onto \: function \: --(2) }$

/* From (1) and (2) */

$\pink{ f : A \rightarrow B\: is \: bijective \: function}$

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