Question

If $ A=\left\{ 1,2,3,4 \right\} $ then which of the following are functions from A to itself?
(a) $ {{f}_{1}}=\left\{ \left( x,y \right):y=x+1 \right\} $
(b) $ {{f}_{2}}=\left\{ \left( x,y \right):x+y > 4 \right\} $
(c) $ {{f}_{3}}=\left\{ \left( x,y \right):y < x \right\} $
(d) $ {{f}_{4}}=\left\{ \left( x,y \right):x+y=5 \right\} $

Answer 1

Hint: In the above question, basically we have to find the function from any one of the options such that the functions given in the options should only take the values of x and y as $ \left\{ 1,2,3,4 \right\} $ . If we substitute any value of x in the function from 1, 2, 3, 4 and in any case if we get the value of y which is not from $ \left\{ 1,2,3,4 \right\} $ then that function is not possible. One more thing that should be kept in mind is for each value of x only one value of y is possible.

Complete step-by-step answer:
We have given a set A i.e.
 $ A=\left\{ 1,2,3,4 \right\} $
We have to check the options and see which option on plugging any value of x or y from A we will get the value of x or y from A only. It will be clearer when we check the options.
(a) $ {{f}_{1}}=\left\{ \left( x,y \right):y=x+1 \right\} $
When we put $ x=1 $ in the equation $ y=x+1 $ we will get the value of y as 2 and the number 2 lies in A.
When we put $ x=2 $ in the equation $ y=x+1 $ we will get the value of y as 3 and the number 3 lies in A.
When we put $ x=3 $ in the equation $ y=x+1 $ we will get the value of y as 4 and the number 4 lies in A.
 When we put $ x=4 $ in the equation $ y=x+1 $ we will get the value of y as 5 and the number 5 is not lying in A.
This option is incorrect because when we put $ x=4 $ in the equation $ y=x+1 $ we will get the value of y as 5 and the number 5 does not belong to set A.
(b) $ {{f}_{2}}=\left\{ \left( x,y \right):x+y > 4 \right\} $
The inequality given in the above function is:
  $ x+y > 4 $
The possible sets of x and y possible $ \left( x,y \right) $ for the above inequality is:
 $ \left( 2,3 \right),\left( 2,4 \right) $
In the above for $ x=2 $ , two values of y are possible that are 3 and 4 then the function is not following A to itself condition.
This option is incorrect because one value of x and two values of y are obtained.
(c) $ {{f}_{3}}=\left\{ \left( x,y \right):y < x \right\} $
The inequality given in the above function is:
 $ y < x $
The possible sets of x and y possible $ \left( x,y \right) $ for the above inequality is:
 $ \left( 2,1 \right),\left( 3,2 \right),\left( 3,1 \right),\left( 4,3 \right),\left( 4,2 \right),\left( 4,1 \right) $
In the above set as you can see that for $ x=3 $ , the possible values of y are 1, 2. Similarly, for $ x=4 $ three different values of y are obtaining 3, 2, 1 so the function is not following A to itself condition.
This option is incorrect because for one value of x more than one y are obtaining.
(d) $ {{f}_{4}}=\left\{ \left( x,y \right):x+y=5 \right\} $
When we put $ x=1 $ in the equation $ x+y=5 $ we will get the value of y as 4 and the number 4 lies in A.
When we put $ x=2 $ in the equation $ x+y=5 $ we will get the value of y as 3 and the number 3 lies in A.
When we put $ x=3 $ in the equation $ x+y=5 $ we will get the value of y as 2 and the number 2 lies in A.
When we put $ x=4 $ in the equation $ x+y=5 $ we will get the value of y as 1 and the number 1 lies in A.
The conditions that we mentioned in the hint are satisfying above, first one is all the values of x and y are lying in A and second is that for one value of x only one value of y is obtained.
Hence, the correct option is (d).

Note: The root of the above problem is to understand what the statement “functions from A to itself” means if you understand this statement properly which we have described above then the whole problem is solved so you need to understand its meaning correctly.
You might have wrongly understood this statement like you only got the understanding that every value of x and y of the given functions should lie in A but if you miss the point that for one value of x only one value of y is possible which we have shown in options (b), (c) and (d) in the above solution then you will get all the three options (b), (c) and (d) as correct.