第12类NCERT解决方案-数学第I部分-第1章关系和函数-第1章的其他练习|套装2

问题11：设S = {a，b，c}，T = {1,2,3}。从S到T查找以下函数F中的F ^–1 (如果存在)。

(i)F = {(a，3)，(b，2)，(c，1)}

解决方案：

As, F = {(a, 3), (b, 2), (c, 1)} and S = {a,b,c} and T={1,2,3}

F: S→T is defined as

F(a) = 3, F(b) = 2 and F(c) = 1

F is one-one and onto.

Taking F^-1, so F^-1: T→S

a = F^-1(3), b = F^-1(2) and c = F^-1(1)

F^-1 = {(3,a),(2,b),(1,c)}

(ii)F = {(a，2)，(b，1)，(c，1)}

解决方案：

As, F = {(a, 2), (b, 1), (c, 1)}

F: S→T is defined as

F(a) = 2, F(b) = 1 and F(c) = 1

Here, F(b) = F(c) but b ≠ c

Hence, F is not one-one.

So, F is not invertible and F^-1 doesn’t exists.

问题12.考虑二元运算∗：R×R→R和o：R×R→R定义为a ∗ b = | a – b | aob = a，∀a，b∈R。证明∗是可交换的而不是关联的，o是可关联的但不是可交换的。此外，证明∀a，b，c∈R，a ∗(boc)=(a ∗ b)o(a ∗ c)。 [如果是这样，我们说操作∗分布在操作o上]。 o是否分布在∗上?证明你的答案。

解决方案：

Binary operations ∗ : R × R → R defined as a ∗b = |a – b|

a*b = |a-b|

b*a = |b-a| = |-(a-b)| = |a-b|

a*b = b*a

Hence, ∗ is commutative.

Now, let’s take a=1, b=2 and c=3 for better understanding

a*(b*c) = a*|b-c| = |a-|b-c|| = |1-|2-3|| = 0

(a*b)*c = |a-b|*c = ||a-b|-c| = ||1-2|-3| = 2

a*(b*c) ≠ (a*b)*c

Hence, ∗ is not associative.

Binary operations o : R × R → R defined as a o b = a, ∀ a, b ∈ R

a o b = a

b o a = b

a o b ≠ b o a

Hence, o is not commutative.

a o (b o c) = a o b = a

(a o b) o c = a o c = a

a o (b o c) ≠ (a o b) o c

Hence, o is associative.

Let’s check for a ∗ (b o c) = (a ∗ b) o (a ∗ c) a, b, c ∈ R

a ∗ (b o c) = a * b = |a-b|

(a ∗ b) o (a ∗ c) = |a-b| o |a-c| = |a-b|

Hence, a ∗ (b o c) = (a ∗ b) o (a ∗ c)

Now, let’s check for a o (b * c) = (a o b) * (a o c)

a o (b * c) = a

(a o b) * (a o c) = a * a = |a-a| = 0

Hence, a o (b * c) ≠ (a o b) * (a o c)

o does not distribute over ∗

问题13。给定一个非空集合X，令∗：P(X)×P(X)→P(X)定义为A * B =(A – B)∪(B – A)，∀A， B∈P(X)。证明空集φ是运算*的标识，并且P(X)的所有元素A在A – ¹ = A时都是可逆的。

(提示：(A –φ)∪(φ– A)= A和(A – A)∪(A – A)= A * A =φ)。

解决方案：

Set X, such that P(X) × P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), ∀ A, B ∈ P(X)

φ*A = (φ-A) U (A-φ) = φ U A = A

A*φ = (A-φ) U (φ-A) = A U φ = A

Hence, φ is the identity element for the operation * on P(X)

A*A = (A-A) U (A-A) = φ U φ = φ

⇒ A = A^-1

Hence, all the elements A of P(X) are invertible with A^–1 = A. 

问题14.在集合{0，1，2，3，4，5}上定义一个二进制运算*

$a*b= \begin{cases} a+b, \hspace{0.2cm}a+b<6\\ a+b-6,\hspace{0.2cm}a+b\geq6 \end{cases}$

证明零是该操作的标识，并且集合中每个元素a≠0都可以用6求逆– a是a的倒数。

解决方案：

Let the set x = {0, 1, 2, 3, 4, 5}

Let’s take i as identity element, where a*i = a = i*a ∀ a ∈ x

a*0 = a

0*a = a, when (a+0<6)

Hence, zero is the identity for this operation

An element a ∈ x is invertible if there exists b ∈ x such that a*b = b*a = 0

$i.e.,\begin{cases} a+b=0=b+a, \hspace{0.2cm}a+b<6\\ a+b-6=0=b+a-6,\hspace{0.2cm}a+b\geq6 \end{cases}$

From above equations, we have

a = -b or b = 6-a

But, as x = {0, 1, 2, 3, 4, 5} and a,b∈ x. Then a≠-b

Hence, b = 6-a is the inverse of an element a∈ x

a≠0

a^-1 = 6-a

问题15.设A = {– 1，0，1，2}，B = {– 4，– 2，0，2}并且f，g：A→B是由f(x)= x ² –定义的函数x，x∈A和$g(x) = 2|x-\frac{1}{2}|-1$ x∈A。f和g相等吗?证明你的答案。

(提示：可能注意到两个函数f：A→B和g：A→B使得f(a)= g(a)∀a∈A，被称为相等函数)。

解决方案：

Given, f, g : A → B be functions defined by f(x) = x² – x, x ∈ A and g(x) = $2|x-\frac{1}{2}|-1$  x ∈ A

At x = -1

f(0) = (-1)² – (-1) = 2

g(0) = $2|-1-\frac{1}{2}|-1$ = 2

Here, f(-1) = g(-1) and 2=2

At x = 0

f(0) = 0² – 0 = 0

g(0) = $2|0-\frac{1}{2}|-1$ = 0

Here, f(0) = g(0) and 0=0

At x = 1

f(1) = 1² – 1 = 0

g(1) = $2|1-\frac{1}{2}|-1$ = 0

Here, f(1) = g(1) and 1=1

At x = 2

f(1) = 2² – 2 = 2

g(1) = $2|2-\frac{1}{2}|-1$ = 2

Here, f(2) = g(2) and 2=2

For, every c∈ A, f(c) = g(c)

Hence, f and g are equal functions.

问题16。令A = {1,2,3}。那么包含(1，2)和(1，3)的关系是自反且对称但不传递的关系的数量为

(A)1

(B)2

(C)3

(D)4

解决方案：

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

Reflexive : (1,1), (2,2), (3,3) ∈ R

Symmetric: (1,2), (2,1)∈ R and (1,3), (3,1) ∈ R

R is not Transitive because, (1,2), (1,3) ∈ R but (3,2) ∉R

So, if we will add (3,2) and (2,3) or both, then R will become transitive.

New, R = {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)}

Hence, A is the correct option.

问题17。令A = {1,2,3}。则包含(1，2)的等价关系的数量为

(A)1

(B)2

(C)3

(D)4

解决方案：

Smallest equivalence relations containing (1, 2):

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

or R = {(1,1),(2,2),(1,2),(2,1),(3,2),(2,3)(3,3)}

Hence, B is the correct option.

问题18.令f：R→R为定义为的Signum函数

$f(x)= \begin{cases} 1, \hspace{0.2cm}x<0\\ 0,\hspace{0.2cm}x=0\\ -1\hspace{0.2cm}x>0 \end{cases}$

g：R→R是g(x)= [x]给出的最大整数函数，其中[x]是小于或等于x的最大整数。然后，雾和果蝇在(0，1]中重合吗?

解决方案：

Given, f : R → R and g : R → R

when x ∈ (0,1]

[x] = 1, when x=1

[x] = 0, when 0

$g(x)= \begin{cases} 1, \hspace{0.2cm}x=1\\ 0,\hspace{0.2cm}0<x<1 \end{cases}$

Now, fog(x)=f(g(x)) = f([x])

$f([x])= \begin{cases} f(1), \hspace{0.2cm}x=1\\ f(0),\hspace{0.2cm}0<x<1 \end{cases} =$ $\begin{cases} 1, \hspace{0.2cm}x=1\\ 1,\hspace{0.2cm}0<x<1 \end{cases}$

And, Now gof(x) = g(f(x))

g(1) = [1] = 1

g(0) = [0] = 0

g(-1) = [-1] = -1

When x ∈ (0,1), fog = 0 and gof = 1. fog(1) ≠ gof(1)

Hence, fog and gof do not coincide in (0, 1].

问题19：在集合{a，b}上的二进制运算数量为

(A)10

(B)16

(C)20

(D)8

解决方案：

Let A = {a,b}

A x A = {a,b} x {a,b}

R = {(a,a),(a,b),(b,a),(b,b)}

Number of elements are 4.

Hence, the number of binary operations on the set will be 2⁴ = 16

Hence, B is the correct option.