Answer

Verified

432.6k+ views

Hint: Use the conditions for surjectivity for taking the right examples.

Complete step-by-step answer:

Consider examples of \[f\left( x \right)=x+1\]and \[g\left( x \right)=\left\{ \begin{matrix}

x-1,x>1 \\

1,x=1 \\

\end{matrix} \right.\].

First prove that f is not onto by proving the digit formed is not a natural number. As mentioned in \[f:N\to N\]. Then prove gof is onto. Find the composite of g and f\[\Rightarrow \left( gof \right)\left( x \right)=g\left( f\left( x \right) \right)\]. Thus prove gof is onto.

Given to us two functions of \[f:N\to N\]and \[g:N\to N\].

We need to prove that gof is onto and f is not onto.

Onto function could be explained by considering two sets, set A and set B which consist of elements. If for every element of B there is at least one or more than one element matching with A, then the function is said to be onto.

Let us consider \[f:N\to N\]be \[f\left( x \right)=x+1\].

And \[g:N\to N\]be taken as \[g\left( x \right)=\left\{ \begin{matrix}

x-1,x>1 \\

1,x=1 \\

\end{matrix} \right.\].

We will first show that f is not onto.

Let \[f:N\to N\]be \[f\left( x \right)=x+1\]

Let us consider\[y=f\left( x \right)\], where \[y\in N\].

\[\begin{align}

& f\left( x \right)=x+1\Rightarrow y=x+1 \\

& \therefore x=y-1 \\

\end{align}\]

Let us put \[y=1\],

\[x=1-1=0\]

We get x = 0, which is not a natural number.

It is said that \[f:N\to N\]i.e. natural numbers.

But zero is not a natural number. So f is not onto and it doesn’t belong in the elements of natural numbers.

If given functions f and g with domains and target sets consisting sets consisting of real numbers, the composite of g and f is the new function gof defined as,

\[\left( gof \right)\left( x \right)=g\left( f\left( x \right) \right)\]

Here, we need to find gof.

Let us consider \[f\left( x \right)=x+1\]and \[g\left( x \right)=\left\{ \begin{matrix}

x-1,x>1 \\

1,x=1 \\

\end{matrix} \right.\].

\[f\left( x \right)=x+1,g\left( x \right)=1\]

Since, \[g\left( x \right)=1\], for x = 1.

\[\begin{align}

& g\left( f\left( x \right) \right)=1 \\

& \therefore gof=1 \\

\end{align}\]

Since, \[g\left( x \right)=x-1,x>1\]

\[g\left( f\left( x \right) \right)=f\left( x \right)-1\], put, \[x=f\left( x \right)\].

\[\begin{align}

& gof=\left( x+1 \right)-1 \\

& \Rightarrow gof=x \\

& \therefore gof=\left\{ \begin{matrix}

x,x>1 \\

1,x=1 \\

\end{matrix} \right. \\

\end{align}\]

Let us consider, \[gof=y\], where \[y\in N\].

\[\therefore y=\left\{ \begin{matrix}

x,x>1 \\

1,x=1 \\

\end{matrix} \right.\]

Here, y is a natural number, as y = x.

So, x is also a natural number.

\[\therefore \]gof is onto, as y is a natural number and y = gof.

Hence, we proved that gof is onto and f is not onto.

Note: Remember what is onto function. We can only prove a function is not onto, if the elements found don’t belong to it. In the case of \[f:N\to N\], the element was zero, which is not a natural number and doesn’t belong to f. If it was any number greater than zero, then it would belong to \[f:N\to N\]and f would be an onto function.

Complete step-by-step answer:

Consider examples of \[f\left( x \right)=x+1\]and \[g\left( x \right)=\left\{ \begin{matrix}

x-1,x>1 \\

1,x=1 \\

\end{matrix} \right.\].

First prove that f is not onto by proving the digit formed is not a natural number. As mentioned in \[f:N\to N\]. Then prove gof is onto. Find the composite of g and f\[\Rightarrow \left( gof \right)\left( x \right)=g\left( f\left( x \right) \right)\]. Thus prove gof is onto.

Given to us two functions of \[f:N\to N\]and \[g:N\to N\].

We need to prove that gof is onto and f is not onto.

Onto function could be explained by considering two sets, set A and set B which consist of elements. If for every element of B there is at least one or more than one element matching with A, then the function is said to be onto.

Let us consider \[f:N\to N\]be \[f\left( x \right)=x+1\].

And \[g:N\to N\]be taken as \[g\left( x \right)=\left\{ \begin{matrix}

x-1,x>1 \\

1,x=1 \\

\end{matrix} \right.\].

We will first show that f is not onto.

Let \[f:N\to N\]be \[f\left( x \right)=x+1\]

Let us consider\[y=f\left( x \right)\], where \[y\in N\].

\[\begin{align}

& f\left( x \right)=x+1\Rightarrow y=x+1 \\

& \therefore x=y-1 \\

\end{align}\]

Let us put \[y=1\],

\[x=1-1=0\]

We get x = 0, which is not a natural number.

It is said that \[f:N\to N\]i.e. natural numbers.

But zero is not a natural number. So f is not onto and it doesn’t belong in the elements of natural numbers.

If given functions f and g with domains and target sets consisting sets consisting of real numbers, the composite of g and f is the new function gof defined as,

\[\left( gof \right)\left( x \right)=g\left( f\left( x \right) \right)\]

Here, we need to find gof.

Let us consider \[f\left( x \right)=x+1\]and \[g\left( x \right)=\left\{ \begin{matrix}

x-1,x>1 \\

1,x=1 \\

\end{matrix} \right.\].

\[f\left( x \right)=x+1,g\left( x \right)=1\]

Since, \[g\left( x \right)=1\], for x = 1.

\[\begin{align}

& g\left( f\left( x \right) \right)=1 \\

& \therefore gof=1 \\

\end{align}\]

Since, \[g\left( x \right)=x-1,x>1\]

\[g\left( f\left( x \right) \right)=f\left( x \right)-1\], put, \[x=f\left( x \right)\].

\[\begin{align}

& gof=\left( x+1 \right)-1 \\

& \Rightarrow gof=x \\

& \therefore gof=\left\{ \begin{matrix}

x,x>1 \\

1,x=1 \\

\end{matrix} \right. \\

\end{align}\]

Let us consider, \[gof=y\], where \[y\in N\].

\[\therefore y=\left\{ \begin{matrix}

x,x>1 \\

1,x=1 \\

\end{matrix} \right.\]

Here, y is a natural number, as y = x.

So, x is also a natural number.

\[\therefore \]gof is onto, as y is a natural number and y = gof.

Hence, we proved that gof is onto and f is not onto.

Note: Remember what is onto function. We can only prove a function is not onto, if the elements found don’t belong to it. In the case of \[f:N\to N\], the element was zero, which is not a natural number and doesn’t belong to f. If it was any number greater than zero, then it would belong to \[f:N\to N\]and f would be an onto function.

Recently Updated Pages

Assertion The resistivity of a semiconductor increases class 13 physics CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE

Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE

What are the possible quantum number for the last outermost class 11 chemistry CBSE

Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Write two gases which are soluble in water class 11 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

What organs are located on the left side of your body class 11 biology CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths