11类RD Sharma解决方案–第3章功能–练习3.1 |套装1

问题1.将一个函数定义为一组有序对。

解决方案：

Let A and B be two non-empty sets. A relation from A to B, i.e., a subset of A×B, is called a function (or a mapping) from A to B, if

(i) for each a ∈ A there exists b ∈ B such that (a, b) ∈ f

(ii) (a, b) ∈ f and (a, c) ∈ f ⇒ b = c

问题2.将函数定义为两组之间的对应关系。

解决方案：

Let A and B be two non-empty sets. Then a function ‘f’ from set A to B is a rule or method or correspondence which associates elements of set A to elements of set B such that:

(i) all elements of set A are associated to elements in set B.

(ii) an element of set A is associated to a unique element in set B.

问题3.关系和函数之间的根本区别是什么?每个关系都是函数吗?

解决方案：

Let ‘f’ be a function and R be a relation defined from set X to set Y.

The domain of the relation R might be a subset of the set X, but the domain of the function f must be equal to X. This is because each element of the domain of a function must have an element associated with it, whereas this is not necessary for a relation.

In relation, one element of X might be associated with one or more elements of Y, while it must be associated with only one element of Y in a function.

Thus, not every relation is a function. However, every function is necessarily a relation.

问题4.令A = {–2，–1，0，1，2}和f：A→Z是由f(x)= x ² – 2x – 3定义的函数。查找：

(i)fie f(A)的范围

(ii)6，-3和5的原像

解决方案：

Given:

A = {–2, –1, 0, 1, 2}

f : A → Z such that f(x) = x² – 2x – 3

(i) Range of f i.e. f (A)

A is the domain of the function f. Hence, range is the set of elements f(x) for all x ∈ A.

Substituting x = –2 in f(x), we get

f(–2) = (–2)² – 2(–2) – 3

= 4 + 4 – 3

= 5

Substituting x = –1 in f(x), we get

f(–1) = (–1)² – 2(–1) – 3

= 1 + 2 – 3

= 0

Substituting x = 0 in f(x), we get

f(0) = (0)² – 2(0) – 3

= 0 – 0 – 3

= – 3

Substituting x = 1 in f(x), we get

f(1) = 1² – 2(1) – 3

= 1 – 2 – 3

= – 4

Substituting x = 2 in f(x), we get

f(2) = 2² – 2(2) – 3

= 4 – 4 – 3

= –3

Thus, the range of f is {-4, -3, 0, 5}.

(ii) pre-images of 6, –3 and 5

Let x be the pre-image of 6 

⇒ f(x) = 6

x² – 2x – 3 = 6

x² – 2x – 9 = 0

x = [-(-2) ± √ ((-2)² – 4(1) (-9))] / 2(1)

= [2 ± √ (4+36)] / 2

= [2 ± √40] / 2

= 1 ± √10

However, 1 ± √10 ∉ A

Thus, there exists no pre-image of 6.

Now, let x be the pre-image of –3 

⇒ f(x) = –3

x² – 2x – 3 = –3

x² – 2x = 0

x(x – 2) = 0

x = 0 or 2

Clearly, both 0 and 2 are elements of A.

Thus, 0 and 2 are the pre-images of –3.

Now, let x be the pre-image of 5 

⇒ f(x) = 5

x² – 2x – 3 = 5

x² – 2x – 8= 0

x² – 4x + 2x – 8= 0

x(x – 4) + 2(x – 4) = 0

(x + 2)(x – 4) = 0

x = –2 or 4

However, 4 ∉ A but –2 ∈ A

Thus, –2 is the pre-images of 5.

∴ Ø, {0, 2}, -2 are the pre-images of 6, -3, 5

问题5.如果将函数f：R→R定义为

$f(x)=\begin{cases}3x-2,x<0\\\ \ \ \ \ \ \ \ \ 1,x=0\\4x+1,x>0\end{cases}$

查找：f(1)，f(–-1)，f(0)，f(2)。

解决方案：

Given:

Let us find f (1), f (–1), f (0) and f (2).

When x > 0, f (x) = 4x + 1

Substituting x = 1 in the above equation, we get

f (1) = 4(1) + 1

= 4 + 1

= 5

When x < 0, f(x) = 3x – 2

Substituting x = –1 in the above equation, we get

f (–1) = 3(–1) – 2

= –3 – 2

= –5

When x = 0, f(x) = 1

Substituting x = 0 in the above equation, we get

f (0) = 1

When x > 0, f(x) = 4x + 1

Substituting x = 2 in the above equation, we get

f (2) = 4(2) + 1

= 8 + 1

= 9

∴ f (1) = 5, f (–1) = –5, f (0) = 1 and f (2) = 9.

问题6.函数f：R→R由f(x)= x ^2定义。决定

(i)f的范围

(ii){x：f(x)= 4}

(iii){y：f(y)= –1}

解决方案：

Given:

f : R → R and f(x) = x².

(i) range of f

Domain of f = R (set of real numbers)

We know that the square of a real number is always positive or equal to zero.

∴ range of f = R⁺∪ {0}

(ii) {x: f(x) = 4}

Given:

f(x) = 4

we know, x² = 4

x² – 4 = 0

(x – 2)(x + 2) = 0

∴ x = ± 2

∴ {x: f(x) = 4} = {–2, 2}

(iii) {y: f(y) = –1}

Given:

f(y) = –1

y² = –1

However, the domain of f is R, and for every real number y, the value of y² is non-negative.

Hence, there exists no real y for which y² = –1.

∴{y: f(y) = –1} = ∅

问题7。令f：R ⁺ →R，其中R ⁺是所有正实数的集合，使得f(x)= log _e x。决定

(i)f域的图像集

(ii){x：f(x)= –2}

(iii)f(xy)= f(x)+ f(y)是否成立。

解决方案：

Given f: R+→ R and f(x) = log_e x.

(i) the image set of the domain of f

Domain of f = R⁺ (set of positive real numbers)

We know the value of logarithm to the base e (natural logarithm) can take all possible real values.

∴ The image set of f = R

(ii) {x: f(x) = –2}

Given f(x) = –2

log_e x = –2

∴ x = e^-2 [since, log_b a = c ⇒ a = bc]

∴ {x: f(x) = –2} = {e^-2}

(iii) Whether f (xy) = f (x) + f (y) holds.

We have f (x) = log_e x ⇒ f (y) = log_e y

Now, let us consider f (xy)

F (xy) = log_e (xy)

f (xy) = log_e (x × y) [since, log_b (a×c) = log_b a + log_b c]

f (xy) = log_e x + log_e y

f (xy) = f (x) + f (y)

∴ the equation f (xy) = f (x) + f (y) holds.

问题8.将以下关系写为有序对的集合，并找出其中的哪些是函数：

(i){(x，y)：y = 3x，x∈{1，2，3}，y∈{3，6，9，12}}

(ii){(x，y)：y> x + 1，x = 1，2和y = 2，4，6}

(iii){(x，y)：x + y = 3，x，y∈{0，1，2，3}}

解决方案：

(i) {(x, y): y = 3x, x ∈ {1, 2, 3}, y ∈ {3, 6, 9, 12}}

When x = 1, y = 3(1) = 3

When x = 2, y = 3(2) = 6

When x = 3, y = 3(3) = 9

∴ R = {(1, 3), (2, 6), (3, 9)}

Hence, the given relation R is a function.

(ii) {(x, y): y > x + 1, x = 1, 2 and y = 2, 4, 6}

When x = 1, y > 1 + 1 or y > 2 ⇒ y = {4, 6}

When x = 2, y > 2 + 1 or y > 3 ⇒ y = {4, 6}

∴ R = {(1, 4), (1, 6), (2, 4), (2, 6)}

Hence, the given relation R is not a function.

(iii) {(x, y): x + y = 3, x, y ∈ {0, 1, 2, 3}}

When x = 0, 0 + y = 3 ⇒ y = 3

When x = 1, 1 + y = 3 ⇒ y = 2

When x = 2, 2 + y = 3 ⇒ y = 1

When x = 3, 3 + y = 3 ⇒ y = 0

∴ R = {(0, 3), (1, 2), (2, 1), (3, 0)}

Hence, the given relation R is a function.

问题9.令f：R→R和g：C→C是定义为f(x)= x ²和g(x)= x ^2的两个函数。他们是平等的职能吗?

解决方案：

Given:

f: R → R ∈ f(x) = x² and g : R → R ∈ g(x) = x²

f is defined from R to R, the domain of f = R.

g is defined from C to C, the domain of g = C.

Two functions are equal only when the domain and codomain of both the functions are equal.

In this case, the domain of f ≠ domain of g.

∴ f and g are not equal functions.