第11类RD Sharma解决方案–第1集–练习1.2

问题1.以花名表形式描述以下几组：

(i){x：x是英语字母中e之前的字母}

(ii){x∈N：x2 <25}

(iii){x∈N：x是质数，10

(iv){x∈N：x = 2n，n∈N}

(v){x∈R：x> x}

(vi){x：x是一个质数，它是60的除数}

(vii){x：x是一个两位数，因此其位数之和为8}

(viii)“三角”一词中所有字母的集合

(ix)“更好”一词中所有字母的集合。

解决方案：

(i) {x : x is a letter before e in the English alphabet}

So, when we read whole sentence it becomes x is such that, x is a letter before ‘e’ in the English alphabet. Now letters before ‘e’ are a, b, c, d.

∴ Roster form will be {a, b, c, d}.

(ii) {x ∈ N: x2 < 25}

x ∈ N that implies x is a natural number.

x² < 25

x < ±5

As x belongs to the natural number that means x < 5.

All numbers less than 5 are 1, 2, 3, 4.

∴ Roster form will be {1, 2, 3, 4}.

(iii) {x ∈ N: x is a prime number, 10 < x < 20}

X is a natural number and is between 10 and 20.

X is such that X is a prime number between 10 and 20.

Prime numbers between 10 and 20 are 11, 13, 17, 19.

∴ Roster form will be {11, 13, 17, 19}.

(iv) {x ∈ N: x = 2n, n ∈ N}

X is a natural number also x = 2n

∴ Roster form will be {2, 4, 6, 8……}.

This an infinite set.

(v) {x ∈ R: x > x}

Any real number is equal to its value it is neither less nor greater.

So, Roster form of such real numbers which has value less than itself has no such numbers.

∴ Roster form will be ϕ. This is called a null set.

(vi) {x : x is a prime number which is a divisor of 60}

All numbers which are divisor of 60 are = 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.

Now, prime numbers are = 2, 3, 5.

∴ Roster form will be {2, 3, 5}.

(vii) {x : x is a two-digit number such that the sum of its digits is 8}

Numbers which have sum of its digits as 8 are = 17, 26, 35, 44, 53, 62, 71, 80

∴ Roster form will be {17, 26, 35, 44, 53, 62, 71, 80}.

(viii) The set of all letters in the word ‘Trigonometry’

As repetition is not allowed in a set, then the distinct letters are

Trigonometry = t, r, i, g, o, n, m, e, y

∴ Roster form will be {t, r, i, g, o, n, m, e, y}

(ix) The set of all letters in the word ‘Better.’

As repetition is not allowed in a set, then the distinct letters are

Better = b, e, t, r

∴ Roster form will be {b, e, t, r}

问题2.以集构建器形式描述以下集：

(i)A = {1,2,3,4,5,6}

(ii)B = {1，1/2，1/3，1/4，1/5，…..}

(iii)C = {0、3、6、9、12 ….}

(iv)D = {10，11，12，13，13，14，15}

(v)E = {0}

(vi){1、4、9、16，…，100}

(vii){2，4，6，8，….}

(viii){5，25，125，625}

解决方案：

(i) A = {1, 2, 3, 4, 5, 6}

{x : x ∈ N, x<7}

This is read as x is such that x belongs to natural number and x is less than 7. It satisfies all condition of roster form.

(ii) B = {1, 1/2, 1/3, 1/4, 1/5, …}

{x : x = 1/n, n ∈ N}

This is read as x is such that x =1/n, where n ∈ N.

(iii) C = {0, 3, 6, 9, 12, ….}

{x : x = 3n, n ∈ Z+, the set of positive integers}

This is read as x is such that C is the set of multiples of 3 including 0.

(iv) D = {10, 11, 12, 13, 14, 15}

{x : x ∈ N, 9

This is read as x is such that D is the set of natural numbers which are more than 9 but less than 16.

(v) E = {0}

{x : x = 0}

This is read as x is such that E is an integer equal to 0.

(vi) {1, 4, 9, 16,….,100}

Where,

1² = 1

2² = 4

3² = 9

4² = 16

.

.

.

10² = 100

So, above set can be expressed in set-builder form as {x²: x ∈ N, 1≤ x ≤10}

(vii) {2, 4, 6, 8….}

{x: x = 2n, n ∈ N}

This is read as x is such that the given set are multiples of 2.

(viii) {5, 25, 125, 625}

Where,

5¹ = 5

5² = 25

5³ = 125

5⁴ = 625

So, above set can be expressed in set-builder form as {5n: n ∈ N, 1≤ n ≤ 4}

问题3.列出以下集合的所有元素：

(ⅰ)A = {X：X ^{2≤10，X∈Z}}

(ii)B = {x：x = 1 /(2n-1)，1≤n≤5}

(iii)C = {x：x是一个整数，-1/2

(iv)D = {x：x是“等式”一词中的元音}

(v)E = {x：x是一年中没有31天的月份}

(vi)F = {x：x是“ MISSISSIPPI”一词的字母}

解决方案：

(i) A = {x : x²≤ 10, x ∈ Z}

First, x is an integer hence it can be positive and negative also.

x² ≤ 10

(-3)² = 9 < 10

(-2)² = 4 < 10

(-1)² = 1 < 10

0² = 0 < 10

1² = 1 < 10

2² = 4 < 10

3² = 9 < 10

Square root of next integers is greater than 10.

x ≤ √10

x = 0, ±1, ±2, ±3

A = {0, ±1, ±2, ±3}

(ii) B = {x : x = 1/(2n-1), 1 ≤ n ≤ 5}

Let us substitute the value of n to find the values of x.

At n = 1, x = 1/(2(1)-1) = 1/1

At n = 2, x = 1/(2(2)-1) = 1/3

At n = 3, x = 1/(2(3)-1) = 1/5

At n = 4, x = 1/(2(4)-1) = 1/7

At n = 5, x = 1/(2(5)-1) = 1/9

x = 1, 1/3, 1/5, 1/7, 1/9

∴ B = {1, 1/3, 1/5, 1/7, 1/9}

(iii) C = {x : x is an integer, -1/2 < x < 9/2}

Given, x is an integer between -1/2 and 9/2

So, all integers between -0.5

∴ C = {0, 1, 2, 3, 4}

(iv) D = {x : x is a vowel in the word “EQUATION”}

All vowels in the word ‘EQUATION’ are E, U, A, I, O

∴ D = {A, E, I, O, U}

(v) E = {x : x is a month of a year not having 31 days}

A month has 28, 29, 30, 31 days.

Out of 12 months in a year which are not having 31 days are:

February, April, June, September, November.

∴ E: {February, April, June, September, November}

(vi) F = {x : x is a letter of the word “MISSISSIPPI”}

Letters in word ‘MISSISSIPPI’ are M, I, S, P.

∴ F = {M, I, S, P}.

问题4.将花名表左侧的每个集合与set-builder表格中描述的右侧的相同集合进行匹配：

(i){A，P，L，E}(i){x：x + 5 = 5，x∈z}

(ii){5，-5}(ii){x：x是一个质数自然数和10的除数}

(iii){0}(iii){x：x是“ RAJASTHAN”一词的字母}

(iv){1,2,5,10}(iv){x：x是10的自然数和除数}

(v){A，H，J，R，S，T，N}(v){x：x ² – 25 = 0}

(vi){2，5}(vi){x：x是单词“ APPLE”的字母}

解决方案：

(i) {A, P, L, E} ⇔ {x: x is a letter of word “APPLE”}

(ii) {5, -5} ⇔ {x: x² – 25 =0}

The solution set of x² – 25 = 0 is x = ±5

(iii) {0} ⇔ {x: x + 5 = 5, x ∈ z}

The solution set of x + 5 = 5 is x = 0.

(iv) {1, 2, 5, 10} ⇔ {x: x is a natural and divisor of 10}

The natural numbers which are divisor of 10 are 1, 2, 5, 10.

(v) {A, H, J, R, S, T, N} ⇔ {x: x is a letter of the word “RAJASTHAN”}

The distinct letters of word “RAJASTHAN” are A, H, J, R, S, T, N.

(vi) {2, 5} ⇔ {x: x is a prime natural number and a divisor of 10}

The prime natural numbers which are divisor of 10 are 2, 5.

问题5.用英语字母在q之前写下所有元音的集合。

解决方案：

Set of all vowels which precede q are

A, E, I, O these are the vowels they come before q.

∴ B = {A, E, I, O}.

问题6.写出所有正整数，其立方为奇数。

解决方案：

Every odd number has an odd cube

Odd numbers can be represented as 2n+1.

{2n+1: n ∈ Z, n>0} or

{1, 3, 5, 7……}

问题7.以集构建器形式写集{1/2，2/5，3/10，4/17，5/26，6/37，7/50}。

解决方案：

Now,

2 = 12 + 1

5 = 22 + 1

10 = 32 + 1

.

.

50 = 72 + 1

Here we can see denominator is square of numerator +1.

So, we can write the set builder form as

{n/(n²+1): n ∈ N, 1≤ n≤ 7}