函数类型| 12年级数学

在本文中，我们将讨论函数的类型。因此，在跳到该主题之前，您首先必须了解这些功能。那么，让我们看看什么是函数。

Note: In the whole article X refers to Domain and Y refers to Codomain.

有什么函数?

令X和Y为两个非空集。从X到Y的函数或映射f记作f：X-> Y是一个规则，通过该规则，每个元素x∈X与唯一元素y∈Y相关联。

域，共域和函数范围

X的元素称为f的域，Y的元素称为f的域。 X元素的图像称为范围，它是Y的子集。下图演示了函数的域，共域和范围。

该图像演示了函数的域，共域和范围。请记住，仅被映射的元素将被计入图像中所示的范围内。如果我们编写上述函数的域，共域和范围，则

Domain = {a, b, c}

Codomain = {1, 2, 3, 4, 5}

Range = {1, 2, 3}

什么时候函数不存在?函数不存在的条件，当所有域元素都未与共域元素映射时，如果其中任何一个元素将与共域映射，则该函数将不存在。

从上面的例子中，我们可以看到，该域的所有元素都没有被映射B元素是留给被映射所以，当这情况出现，我们的函数将不存在。让我们跳到这个话题。

功能类型

基本上有6种功能。

1. 一对一(内射)函数
2. 多对一函数
3. 上(形容词)函数
4. 入函数
5. 一对一的函数(双射函数)
6. 多对一函数

一对一(内射)函数

A function f: X -> Y is said to be a one to one function if the images of distinct elements of X under f are distinct. Thus, f is one to one iff f(x₁) = f(x₂).

属性：函数f ：如果对于任何f (x ₁ )= f (x ₂ )表示x ₁ = x ₂ ，即，即f映射(函数)下A的不同元素的图像为a，则A-> B是一对一的清楚的。

成为一对一函数：映射后，域的每个元素都有一个具有共域的单个图像。

一对一(内射)函数的样本示例

示例1：取f(x)= 2x + 3，将1，2，1/2代替x。

• 因此，域= {1，2，1/2}
• 共域=(5，7，4}

从上面的图像中，我们可以得出结论，函数f(x)是一对一的，因为域中的每个元素都具有一个图像。

示例2：检查函数是否一对一：f(x)= 3x – 2

Solution: 

To check the function is one to one or not, we have to check that elements of the domain are having only a single pre-image in codomain or not. For checking, we can write the function as,

f(x₁) = f(x₂) 

3x₁ – 2 = 3x₂ – 2 

3x₁ = 3x₂ 

x₁ = x₂  

Since both x₁ =  x₂ means that elements of the domain having a single pre-image in its codomain. Hence the function f(x) = 3x – 2 is one to one function.

示例3：检查函数是否为一对一：f(x)= x ² + 3。

Solution: 

To check whether the function is One to One or not, we will follow the same procedure. Now let’s check, we can write the function as,

f(x₁) = f(x₂) 

(x₁)² + 3 = (x₂)² + 3 

(x₁)² = (x₂)²  

Since (x₁)² ≠ (x₂)²

Hence the function f(x) = x² + 3 is not one to one function.

示例4：如果N-> N，f(x)= 2x +1，则检查该函数是否具有内射性?

Solution:

In question N -> N, where N belongs to Natural Number, which means that the domain and codomain of the function is a natural number. For checking whether the function is injective or not, we can write the functions as,

Let, f(x₁) = f(x₂)

2x₁ + 1= 2x₂ + 1

2x₁ = 2x₂ 

x₁ = x₂

Since x₁ = x₂, means all elements of the domain are mapped with a single element of the codomain. Hence function f(x) = 2x + 1 is Injective (One to One).

示例5：如果f：R-> R乘f(x)= 1 / x。然后检查给定的函数是否为一对一?

Solution: 

In question R -> R, where R belongs to Non-Zero Real Number, which means that the domain and codomain of the function are non zero real number. We will check this too as we had done the above question. We can write the function as,

f(x₁) = f(x₂)

1/x₁ = 1/x₂

x₁ = x₂

Since x₁ = x₂, means all elements of the domain are mapped with a single element of the codomain. Hence the function f(x) = 1/x is One to One.

多对一函数

If the function is not one to one function then it should be many to one function means every element of the domain has more than one image at codomain after mapping.

属性：一个或多个在共域中具有相同图像的元素

多对一函数：域中的一个或多个元素在共域中具有单个图像。

多对一函数的示例示例

范例1：f(x)= x ² 。检查函数是否为多对一?

Solution: 

Domain = {1, -1, 2, -2}, let’s put the elements of the domain in the function

f(1) = 1² = 1

f(-1) = (-1)² = 1

f(2) = (2)² = 4

f(-2) = (-2)² = 4

因此，我们的共域= {1，4}。映射后：

From the above mapping, we can see that one than one element of the domain having a single image after mapping. So this is our Many to One function.

范例2：A = {1,2,3,4}，B = {a，b，c，d，e}函数定义为f = {((1，a)，(2，a)，(3， b)，(4，c)}。检查函数是否为多对一?

Solution: 

Let’s represent the function through mapping.

As we know the condition of the function to be Many to One is, one or more than one element of domain having same image in the codomain. As we see in the mapping that the element of domain { 1, 2 } is having same image in codomain { a }  after mapping. So the function is Many to One function.

上(形容词)函数

A function f: X -> Y is said to be an onto function, if every element of Y is an image of some element of set X under f, i.e for every y ∈ Y there exists an element x in X such that f(x) = y.

特性：

• 功能范围应等于共域。
• B的每个元素都是A的某个元素的图像。

条件是走上函数：函数范围应该等于陪域。

从上两幅图中可以看出，范围等于共域，这意味着共域的每个元素都与域的元素映射，因为我们知道在共域中映射的元素称为范围。因此，这些是Onto函数的示例。

Onto(形容词)函数的样本示例

示例1：如果R-> R由f(x)= 2x + 1定义。则检查以下函数是否为Onto。

Solution: 

For checking the function is Onto or not, Let’s first put the function f(x) equal to y

f(x) = y

y = 2x + 1

y – 1 = 2x

x = (y – 1) / 2

Now put the value of x in the function f(x), we get,

f((y – 1) / 2) = 2 . [(y – 1) / 2] +1

Taking LCM 2, we get

= [2(y – 1) + 2] / 2

= (2y – 2 + 2) / 2

= y

Since we get back y after putting the value of x in the function. Hence the given function f(x) = 2x + 1 is Onto function.

示例2：如果f：R – {1}-> R – {1}由f(x)=(x + 1)/(x – 1)定义。检查函数是否为Onto。

Solution: 

As we did in the first step put function f(x) equal to y and solve

(x + 1) / (x – 1) = y

y (x – 1) = (x + 1)

yx – y = x + 1

Let us transfer x to the left-hand side and y to the right-hand side, we get

yx – x = y + 1

x(y – 1) = y + 1

x = (y + 1) / (y – 1)

Now let’s put the value of x in the function, we get

f((y + 1) / (y – 1)) = [{(y + 1) / (y – 1)} + 1] / [{(y + 1) / (y – 1)} – 1]

= [{(y +1 + y – 1) / (y – 1)} / {(y + 1 – y + 1) / (y – 1)}]

Cancel (y – 1) with (y – 1), we get

= (2y + 1 – 1) /  (2 + y – y)

= 2y / 2

= y

Since we get back y after putting the value of x in the function. Hence the given function f(x) = (x + 1) / (x – 1) is Onto function.

示例3：如果N-> N由f(x)= 3x + 1定义，那么证明函数f(x)是Surjective吗?

Solution: 

To prove that the function is Surjective or not, firstly we put the function equal to y. Then find out the value of x and then put that value in the function. So let’s start solving it.

Let f(x) = y

3x + 1 = y

3x = y – 1

x = (y – 1) / 3

Now put the value of x in the function f(x), we get

f((y – 1) / 3) = {3 (y – 1) / 3} + 1

= y – 1 + 1

= y

Since we get back y after putting the value of x in the function. Hence the given function f(x) = (3x + 1) is Onto function.

示例4：如果N-> N由f(x)= 1 / x定义。然后检查函数f(x)是否为Onto?

Solution: 

For every y there exist 1 / y such that,

f(1 / y) = 1 / (1 / y)

then we get only y

means that y belongs to codomain and 1 / y belongs to the domain. So for every 1 / y, we are getting y. Since we get y, this means the function is onto.

示例5：如果N-> N由f(x)= 5x + 3定义。然后检查函数f(x)是否为Onto?

Solution: 

To check we have to put the function equal to y

f(x) = y

5x + 3 = y

x = (y – 3) / 5

Put the value of x in the function

f(x) = {5 . (y – 3) / 5} + 3

Cancel 5 with 5, we get

= y – 3 + 3

= y 

Since we get back y after putting the value of x in the function. Hence the given function f(x) = 5x + 3 is Onto function.

入函数

A function f: X -> Y is said to be an into a function if there exists at least one element or more than one element in Y, which do not have any pre-images in X, which simply means that every element of the codomain are not mapped with elements of the domain.

从上图可以清楚地看到，共域的每个元素都没有与域的元素映射，这意味着共域的第10个元素未映射。因此，这种类型的函数称为Into函数。

特性：

• 函数范围是B的适当子集
• 函数的范围不应等于B，其中B是共域。

函数样例

示例1：如果集合A = {1,2,3}，B = {x，y，z}，则函数定义为f = {(1，x)，(1，y)，(2，z )}。然后检查函数是否为Into?

Solution: 

We have f = {(1, x), (1, y), (2, z)} 

If we map this function, then we get

In the above mapping we clearly see that the range of function is the proper subset of codomain and also it is not equal to the codomain. Hence we can say that the function is Into function.

示例2：如果集合A = {1,2,3,4}，B = {a，b，c，d}，则函数定义为f = {((1，d)，(2，a)， (3，c)，(4，b)}。然后检查函数是否为Into?

Solution: 

Let’s first represent the function through mapping.

As we know the condition for Into function is, that the Range of the function is the proper subset of codomain and also should not equal to the codomain. In the mapping off the function we clearly see that all the elements of the codomain is mapped with elements of domain.

• So, Range of the function = {a, b, c, d}
• Codomain of the function = {a, b, c, d}

Since the Range of the function is equal to codomain of the function. It is proved that the function is not Into function. 

一对一的函数(双射函数)

A function f: X-> Y is said to be a bijective function if it is both One to One and Onto.

因此，让我们看一下Bijective函数的示例，以便更好地理解。

仔细看一下上面的示例，我们可以看到该函数既是一对一的又是Onto的，因为域的元素在共域具有单个图像，这是我们上面讨论的一对一的条件以及其他条件，其中范围等于共域。因此，一对一和Onto的条件都得到满足。所以我们的函数是一对一。

特性：

• 函数f：如果对于任何f(x ₁ )= f(x ₂ )均意味着x ₁ = x ₂ ，即在f映射(函数)下A的不同元素的图像是不同的，则f：A-> B是一对一的。
• 函数f：如果f = B的范围，即f(A)= B，则A-> B是一对一的
• 功能范围应等于共域。
• B的每个元素都是A的某个元素的图像。

一对一On函数(双射函数)的示例示例

示例1：如果A = R – {3}且B = R – {1}。考虑函数f：对于所有x∈A，由f(x)=(x – 2)/(x – 3)定义的A->B。然后证明函数f是双射的。

Solution: To show the function is bijective we have to prove the given function both One to One and Onto.

Let’s first check for One to One:

Let x₁, x₂ ∈  A such that f(x₁) = f(x₂) 

Then, (x₁ – 2) / (x₁ – 3) = (x₂ – 2) / (x₂ – 3)

(x₁ – 2) ( x₂ – 3) = (x₂ – 2) (x₁ – 3)

x₁ . x₂ – 3x₁ – 2x₁ + 6 = x₁ . x₂ – 3x₂ -2x₁ + 6

-3x₁ – 2x₂ = -3x₂ – 2x₁

-3( x₁ – x₂) + 2( x₁ – x₂) = 0

-( x₁ – x₂) = 0

x1 – x₂  = 0 

⇒ x₁ = x₂

Thus, f(x₁) = f(x₂) ⇒ x₁ = x₂, ∀ x1, x2  ∈ A

So, the function is a One to One

Now let us check for Onto:

Let y ∈ B = R – {1} be any arbitrary element.

Then, f(x) = y

⇒ (x – 2) / (x – 3) = y

⇒ x – 2 = xy – 3y

⇒ x – xy = 2 – 3y

⇒ x(1 – y) = 2 – 3y

⇒ x =  (2 – 3y) / (1 – y) or x = (3y – 2) / (y – 1)

Now put the value of x in the function f(x) 

f((3y – 2) / (y – 1)) = { (3y – 2) / (y – 1) } – 2 / { (3y – 2) / (y – 1) – 3 }

= (3y – 2 – 2y + 2) / (3y – 2 – 3y + 3)

= y

Hence f(x) is Onto function. Since we proved both One to One and Onto this implies that the function is bijective.

示例2：A = {1,2,3,4}，B = {a，b，c，d}，则函数定义为f = {(1，a)，(2，b)，(3，c)，(4，d)}。检查函数是否为一对一。

Solution: To check whether the function is One to One Onto or not. We have to check for both one by one. Let’s first represent the function through mapping.

Let’s check for One to One:

As we know the condition for One to One that all the elements of the domain are having a single image in the codomain. As we see in the mapping that all the elements of set A are mapped with set B and each having a single image after mapping. So the function is One to One.

Now let’s check for Onto:

As we know the condition for the function to be Onto is that, Range = Codomain means all the elements of codomain are mapped with domain elements, in this case, codomain will equal to the domain. As we see in the mapping that the condition of the function to be Onto is satisfied. So the function is Onto. Since we had proved that the function is both One to One and Onto. Hence function is One to One Onto (Bijective).

多对一函数

A function f: X-> Y is said to be a Many to One Into function if it is both Many to One and Into.

因此，让我们看一下“多对一”函数示例以更好地理解。

因此，我们可以在上面的示例中看到两个条件都得到满足。如我们所见，在共域{1，6，10，12}中，这些是要映射的元素。首先，如上所述，域中的每个元素在映射后的共域具有多于一个图像的条件是多对一的，以及其他条件是，在Y(共域)中存在至少一个元素或多于一个元素，这在X(Domain)中没有任何原像，这仅意味着共域的每个元素都没有映射到域元素，这是Into函数的条件。因此，这两个条件都得到满足，这就是为什么我们的函数是“一对多”函数。

特性：

• 一对多的共同域中具有相同图像的一个或多个元素。
• 函数的范围是B的正确子集。
• 函数的范围应不等于B为in。

关于多对一函数的示例示例

示例1： A = {1,2,3,4}，B = {a，b，c}。函数定义为f = {((1，a)，(2，b)，(3，c)，(4，c) }。

Solution: 

Let’s first represent the function through mapping.

To check the function is Many to One Into or not. We have to check for both one by one.

Let’s first check for Many to One function:

As we know the condition for Many to One function is that more than one element of domain should have more same image in codomain. From the above mapping we can see that the elements of A {3, 4 } are having same image in B { c }, so the function is Many to One.

Now let’s check for Into function:

As we know the condition for Into function is that the Range of function should be the subset of codomain and also not equal to codomain. Let’s check both the conditions are satisfied or not.

• Range of function = { a, b, c }
• Codomain of function = { a, b, c }

Range of function = Codomain of function

As we see that the range of function is equal to codomain of the function. Hence we can say that the function is not Into function. As we see that the function is Many to One but not Into, Hence the function is not Many to One Into.

示例2：A = {1,2,3}，B = {a，b，c，d}。函数定义为f = {((1，a)，(2，b)，(3，c)}

Solution: 

Let’s first represent the function through mapping.

As we had done in the above question same we will going to do in this question too.

Let’s first check for Many to One function:

As we know the condition for Many to One function is that more than one element of domain should have more same image in codomain. From the above mapping we can see that every element of A having single image in B , so the function is not Many to One.

Now let’s check for Into function:

As we know the condition for Into function is that the Range of function should be the subset of codomain and also not equal to codomain. Let’s check both the conditions are satisfied or not.

• Range of function = {a, b, c}
• Codomain of function = {a, b, c, d}

Since, Range of function ≠ Codomain of function

As we see that the range of function is not equal to codomain of the function and also Range of the function is the proper subset of codomain. Hence we can say that the function is Into function. As we prove that the function is not Many to One but function is Into,  Since both the conditions are not satisfied. Hence the function is not Many to One Into.

示例3：A = {1,2,3,4}，B = {a，b，c，d}。函数定义为f = {(1，a)，(2，b)，(3，c)，(4，c)}。

Solution:

Let’s represent the function through mapping.

To check the function is Many to One Into or not. We have to check for both one by one.

Let’s first check for Many to One function:

As we know the condition for Many to One function is that more than one element of domain should have more same image in codomain. From the above mapping we can see that the elements of A {3, 4 } are having same image in B { c }, so the function is Many to One.

Now let’s check for Into function:

As we know the condition for Into function is that the Range of function should be the subset of codomain and also not equal to codomain. Let’s check both the conditions are satisfied or not.

• Range of function = {a, b, c}
• Codomain of function = {a, b, c, d}

Range of function ≠ Codomain of function

As we check that the range of function is not equal to codomain of the function. Hence we can say that the function is Into function. As we prove that the function is Many to One and Into. Hence the function is Many to One Into.