
Give examples of two one-one functions \[{{f}_{1}}\] and \[{{f}_{2}}\] from R to R such that \[{{f}_{1}}+{{f}_{2}}:R\to R\] defined by \[\left( {{f}_{1}}+{{f}_{2}} \right)\left( x \right)={{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)\] is not one-one.
Answer
513k+ views
Hint: For this question make \[{{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)=\text{constant}\] as y = k is the easiest function which is not one-one. Take \[{{f}_{1}}\left( x \right)\] and \[{{f}_{2}}\left( x \right)\] as linear function in x such that \[{{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)\] is constant.
Here we have to find two one-one functions \[{{f}_{1}}\] and \[{{f}_{2}}\] R to R such that \[\left( {{f}_{1}}+{{f}_{2}} \right)x={{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)\] is not one-one.
We know that one-one function is a function that maps distinct elements of its domain to distinct elements of its co-domain that is for a particular value of x, there is a particular value of y and that value of y should not repeat for any other value of x.
Now, we have to make \[{{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)\] such that it is not one-one.
We know that \[f\left( x \right)=\text{constant}\] is the easiest function which is not one-one because its value of y keeps getting repeated for all values of x.
Therefore, we will choose \[{{f}_{1}}\left( x \right)\] and \[{{f}_{2}}\left( x \right)\] such that
\[{{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)=k\]
Now, we are given that \[{{f}_{1}}\left( x \right)\] and \[{{f}_{2}}\left( x \right)\] must be one-one.
We know that the easiest one-one function is \[y=ax+b:R\to R\] where a and b are constants as it gives different values of y for different values of x.
Therefore, we take \[{{f}_{1}}\left( x \right)=9x+5\].
Now to make \[{{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)\] constant, 9x must disappear.
Therefore, we take \[{{f}_{2}}\left( x \right)=-9x+8\].
Therefore, we get \[{{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)=\left( 9x+5 \right)+\left( -9x+8 \right)\].
\[\Rightarrow {{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)=13\]
Therefore, finally we get
\[{{f}_{1}}\left( x \right)=9x+5\]
\[{{f}_{2}}\left( x \right)=-9x+8\]
which are one-one functions.
Also, we get \[{{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)=13\] which is not a one-one function.
Note: Students could also check if a function is one-one or not by making the line on the graph of the function which is parallel to the x axis. If this line cuts the graph just 1 time then, it is one-one function, otherwise it is not one-one.
Here we have to find two one-one functions \[{{f}_{1}}\] and \[{{f}_{2}}\] R to R such that \[\left( {{f}_{1}}+{{f}_{2}} \right)x={{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)\] is not one-one.
We know that one-one function is a function that maps distinct elements of its domain to distinct elements of its co-domain that is for a particular value of x, there is a particular value of y and that value of y should not repeat for any other value of x.
Now, we have to make \[{{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)\] such that it is not one-one.
We know that \[f\left( x \right)=\text{constant}\] is the easiest function which is not one-one because its value of y keeps getting repeated for all values of x.

Therefore, we will choose \[{{f}_{1}}\left( x \right)\] and \[{{f}_{2}}\left( x \right)\] such that
\[{{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)=k\]
Now, we are given that \[{{f}_{1}}\left( x \right)\] and \[{{f}_{2}}\left( x \right)\] must be one-one.
We know that the easiest one-one function is \[y=ax+b:R\to R\] where a and b are constants as it gives different values of y for different values of x.
Therefore, we take \[{{f}_{1}}\left( x \right)=9x+5\].
Now to make \[{{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)\] constant, 9x must disappear.
Therefore, we take \[{{f}_{2}}\left( x \right)=-9x+8\].

Therefore, we get \[{{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)=\left( 9x+5 \right)+\left( -9x+8 \right)\].
\[\Rightarrow {{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)=13\]
Therefore, finally we get
\[{{f}_{1}}\left( x \right)=9x+5\]
\[{{f}_{2}}\left( x \right)=-9x+8\]
which are one-one functions.
Also, we get \[{{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)=13\] which is not a one-one function.
Note: Students could also check if a function is one-one or not by making the line on the graph of the function which is parallel to the x axis. If this line cuts the graph just 1 time then, it is one-one function, otherwise it is not one-one.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

The correct order of melting point of 14th group elements class 11 chemistry CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

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