Test: Introduction To Functions

If y is expressed in terms of a variable x as y=f(x), then y is called explicit function.

Here, we are given a function f ,which is defined from R to R. Also, f is said to be the Identity Function.
The Identity Function returns the same output as given in the input. 
That is, if you input some x into f(x), you get the result as x itself. So, the Identity Function is:

As we can see, the Domain of the function is R , and the co-domain as well as range is also R.

Clearly, mappings given in options (a),(b) and (c ) satisfy the given conditions and are one-one onto.

F(n) = n+2 if n is odd
so if n=1, then f(1)= 3
if n=3 , f(3)=5
On the other hand
F(n)= n+3 if n is even
if n= 2,f(2)=5
if n=4 ,f(4)= 7
Clearly f(3) = f(2) = 5
but 3 does not equals to 3, so it is not one one i.e. injective also 1€N
 i.e. codomain
but here 1€ range does not equals to codomain so it is not surjective.

 f: R → R is defined as f(x) = x⁴
Let x, y ∈ R such that f(x) = f(y).
=> x⁴ = y⁴
=> x =+-y
Therefore, f(x¹) = f(x²) does not imply that x¹=x².
For instance,f(1) = f(-1) = 1
∴ f is not one-one.
Consider an element 2 in co-domain R. It is clear that there does not exist any x in domain R such that f(x) = 2.
∴ f is not onto.
Hence, function f is neither one-one nor onto.

Let X, Y be sets, and let f : X → Y be a function. We say that f is injective (sometimes called one-to-one) if ∀x₁, x₂ ∈ X, f(x₁) = f(x₂) 
⇒ x₁ = x₂.

Suppose x1 and x2 are real numbers such that f(x1) = f(x2). (We need to show x1 = x2 .)
5x1  = 5x2 
Dividing by 5 on both sides gives
x1 = x2  (function is one - one)
Let y ∈R. (We need to show that  x in R such that f(x) = y.)
If such a real number x exists, then 5x -2 = y and x = (y + 2)/5. x is a real number since sums and quotients (except for division by 0) of real numbers are real numbers. It follows that
f(x) = 5((y + 2)/5)          
by the substitution and the definition of f
       = y + 2 
       = y                
by basic algebra
Hence, f is onto.

A function f (from set X to Y) is surjective if and only if for every y in Y, there is at least one x in X such that f(x) = y, in other words f is surjective if and only if f(X) = Y.

As x is square of every real number, so the f(x) = x² cannot be the negative.

Let the domain of x be R
 For surjectivity-
Let f(x) = Y 
=> 2x +1 = Y
=> x = (Y - 1)/2
It is clear that Y takes all the values from (-∞, +∞) 
So the range of f(x) is the same as the domain of x.
 So it is surjective.

(i) f(1)=1,f(2)=2,f(3)=3
(ii) f(1)=1,f(2)=3,f(3)=2
(iii) f(1)=2,f(2)=3,f(3)=1
(iv) f(1)=2,f(2)=1,f(3)=3
(v) f(1)=3,f(2)=1,f(3)=2
(vi) f(1)=3,f(2)=2,f(3)=1
Since, f is onto, all elements of {1,2,3} have unique pre-image.
The above cases are possible.
Since, every element 1,2,3 has either of image 1,2,3 and that image is unique.
∴  f is one-one.
∴ Function f:A→A is one-one.

f(x) = x³
Checking one-one f(x1) = (x1)³
f(x2) = (x2)³
Comparing f(x1) = f(x2)
(x1)³ = (x2)³
x1 = x2   => x1 = -x2
Since, x1 & x2 are natural number, they are always positive.
Hence, x1 = x2  ( it is one - one)
Checking onto : f(x) = (x)³
Let f(x) = y such that y implies N
x³ = y
x = (y)^⅓
x = 1.71, hence it is not a natural number
Thus function is not onto