Relations and Functions Test - 4

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.

Here, f(1) = 8, f(2) = 7, f(3) = 6.

Since, different points of domain have the different f-image in the range, therefore f is a one-one function.

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.

A bijective function is also known as an invertible function. It is a function that has both one-to-one and onto properties.

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.

It 's because f(x) has a constant value, and range will always be 5.

The function f(x) = 2x - 1 is surjective.

Explanation:

A function is surjective (also called onto) if for every element y in the codomain, there exists an element x in the domain such that f(x) = y.

In this case, the function f(x) = 2x - 1 is a linear function. For any given y value, we can find an x value such that f(x) = y:

y = 2x - 1
x = (y + 1) / 2

So, for any y value, there exists an x value that satisfies the equation. Therefore, the function 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