第 12 类 RD Sharma 解决方案 – 第 2 章函数 – 练习 2.1 |设置 2

问题 12：证明由 f(x) = e ^x给出的指数函数f: R → R 是一但不是上。如果共域被 Ro+ 取代会发生什么。

解决方案：

We have f: R → R, given by f(x) = e^x.

Let x,y ϵ R, such that

=> f(x) = f(y)

=> e^x = e^y

=> e^(x-y) = 1 = e⁰

=> x – y = 0

=> x = y

Hence,

f is one-one.

Clearly range of f = (0, INFINITY) not equals to R.

Hence, f is not onto.

When co-domain is replaced by Ro+ i.e. (0, INFINITY) then f becomes an onto function.

问题 13：证明由 f(x) = log(a)x, a>0 给出的对数函数f: Ro+→ R 是双射。

解决方案：

We have f: Ro+→ R given by f(x) = log(a)x, a>0.

let x,y ϵ Ro+ such that,

f(x) = f(y)

=> log(a)x = log(a)y

=> log(a)x (x/y) = 0 [log(a)x = 0]

=> x/y = 1

=> x = y

Hence, f is not one-one.

Now, let y ϵ R be arbitrary, then f(x) = y

=> log (a)x = y

=>x = a^y ϵ Ro+

Thus for all yϵR there exist x = a^y such that f(x) = y.

Hence,

f is onto.

f is one-one.

=> f is bijective.

问题 14：如果 A = {1,2,3}，证明一个一对一的函数f: A → A 必须在。

解决方案：

Since f is one-one, three elements of {1,2,3} must be taken to the 3 different elements of the co-domain {1,2,3} under f. Hence f has to be onto.

问题 15：如果 A={1, 2, 3}，证明一个上函数f: A→A 必须是一一的。

解决方案：

A={1,2,3}

Possible onto functions from A to A can be the following:

(0){(1,1),(2,2).(3,3)

(ii) (1,1),(2,3),(3,2)

(ii){(1,2),(2,2),(3,3)}

(iv){(1,2),(2,1),(3,3)}

(v){(1,3), (2,2),(3,1)]

(vi){(1,3),(2,1),(3,2)

Here, in each function, different elements of the domain have different images.

Therefore,

All the functions are one-one

问题 16：从集合 A = { 1, 2, 3, ....n} 到它自己，找到所有到函数的数量。

解决方案：

We know that every onto function from A to itself is one-one.

Therefore,

The number of one-one functions=number of bijections =n!

问题 17：给出两个一对一函数 f1 和 f2 的例子。从 R 到 R 使得 f1 + f2：由 (f1 + f2)(x) = f1(x) + f2(x) 定义的 R→ R 不是一对一的。

解决方案：

We know that f1: R→ R, given by f1 (x)= x, and f2 (x)=-x are one-one.

Proving f1, is one-one:

Let f1(x)= f1(y)

Implies that x = y

Therefore,

f1 is one-one.

Proving f2 is one-one:

Let f2(x) = f2(y)

Implies that – x=-y

Implies that x = y

Therefore,

f2 is one-one.

Proving (f1 + f2) is not one-one:

Given:(f1+f2)(x)= f1(x)+ f2 (x)= x+(-x) = 0

Therefore,

For every real number x,(f1 + f2)(x)=0

Therefore,

The image of ever number in the domain is same as 0.

Thus, (f1 + f2) is not one-one.

问题 18：给定从 Z 到 Z 的两个满射函数 f1 和 f2 的示例，使得 f1 + f2 不是满射的。

解决方案：

We know that f1: R → R, given by f1(x) = x, and f2(x)=-x are surjective functions.

Proving f1 is surjective:

Let y be an element in the co-domain (R), such that f1(x)= y.

f1(x)= y

Implies that x = y, which is in R.

Therefore,

for every element in the co-domain, there exists some pre-image in the domain.

Therefore,

f1 is surjective

Proving f2 is surjective:

Let f2 (x)= y

x = y, which is in R.

Therefore,

for every element in the co-domain, there exists some pre-image in the domain.

Therefore,

f2 is surjective.

Proving (f1+f2) is not surjective:

Given:(f1 + f2)(x) = f1(x)+ f2(x)=x+(-x)=0

Therefore, for every real number x, (f1 + f2)(x) = 0

Therefore, the image of every number in the domain is same as 0.

Implies that Range = {0}

Co-domain = R

Therefore, both are not same.

Therefore, f1+f2 is not surjective.

问题 19：证明如果 f1 和 f2 是一对一映射，从 R 到 R，则由 (f1 X f2)(x) = f1(x)f2(x) 定义的积 f1 X f2: R→R 不需要一对一。

解决方案：

We know that f: R→ R, given by f1(x) = x, and f2(x) = x are one-one.

Proving f1 is one-one:

Let x and y be two elements in the domain R, such that f1(x) = f1(y)

f1(x) = f1(y)

x = y

Therefore,

f1 is one-one.

Proving f2 is one-one:

Let x and y be two elements in the domain R, such that f2 (x) = f2(y)

f2(x)= f2(y)

Implies that x = y

Therefore,

f2 is one-one.

Proving f1 X f2 is not one-one:

Given:

(f1 X f2)(x) = f1(x) X f2(x) = x * x = x²

Let x and y be two elements in the domain R, such that

(f1 X f2)(x) = (f1 X f2)(y)

Implies that x² = y²

Implies that x = (+-)y

Therefore,

(f1 X f2) is not one-one.

问题 20：假设 f1 和 f2 是从 R 到 R 的非零的一对一函数。 (f1/f2) 一定是一对一的吗？证明合法。

解决方案：

We know that f1: R→R given by f1(x) = x³ and f2(x)= x are one-one.

Injectivity of f1:

Consider x and y be two elements in the domain R, such that

f1(x)= f1(y)

Implies that x³ = y

x=3√y belongs to R

Therefore,

f1 is one-one.

Injectivity of f2:

Consider x and y be two elements in the domain R, such that

f2(x)= f2(y)

Implies that x = y

x belongs to R

Therefore,

f2 is one-one.

Providing (f1 / f2) is not one-one:

Given that (f1/f2)(x)= = f1(x)/f2(x) = (x³ / x) = x²

Consider x and y be two elements in the domain R, such that

(f1/f2)(x) = (f1/f2)(y)

f2 f2

x² = y²

x= (+-)y

Therefore,

(f1/f2) is not one-one.

问题 21：给定 A = {2, 3, 4}，B = {2,5,6,7}。构建以下每个示例。

(i) 从 A 到 B 的单射映射。

(ii) 非单射的从 A 到 B 的映射。

(iii)从 A 到 B 的映射。

解决方案：

Given A={1,2,3,4}, B = {2,5,6,7}

Let f: A → Bf: A → B be a mapping from A to B f = {(2,5)(3,6)(4,7)}

f is an injective mapping.

Since for every element a € A there is an unique element b € B

Let us define a mapping: A→B given by g = {(2,2)(2,5)(3,6)(4,7)}

g is not an injective mapping.

since the element 2 € A is not uniquely mapped

Since (2,2) and (2,5) both belong to the mapping g, g is not injective

Let us define a mapping h: A→B

h: A → B given by h = {(2,2),(5,3),(7,4)}

h is a mapping from A to B

B to A since the every ordered puts {2,5,7} € B to elements in {2,3,4} € A

问题 22：证明 f: R → R, 由 f(x) = x – [x] 既不是一一也不是上。

解决方案：

f:R → R, given by f (x)= x-[x]

Injectivity:

f(x)=0 for all x belongs to Z,

Therefore,

f is not one-one.

Surjectivity:

Range of f = (0,1) not equals to R.

Co-domain of f = R

Both are not same.

Therefore,

f is not onto.

问题 23：让 f:N→N 定义为

f(n) = n + 1，如果 n 是奇数。

f(n) = n – 1，如果 n 是偶数。证明如果 f 是双射。

解决方案：

Injectivity:

Let x and y be any two elements in the domain (N).

Case-1: Let both x and y be even and

Let x, y belongs to N such that f (x) = f(y)

As, f (x) = f(y)

Implies that x – 1= x-1

Implies that x = y

Case-2: Let both x and y be odd and

Let x, y belongs to N such that f (x)= f (y)

As, f (x)= f (y)

Implies that x +1 = y +1

Implies that x = y

Case-3:Let x be even and y be odd then, x y.

Then,

x+1 is odd and y-1 is even.

Implies that x+1≠ y-1

Implies that f(x) ≠ f(y)

Therefore,

x ≠ y

= f(x) ≠ f(y)

In all the 3 cases,

Therefore,

f is one-one.

Surjectivity:

Co-domain off = {1, 2,3,4,…}

Range of f = {1+1,2 – 1,3+1,4 – 1,…} = {2,1,4, 3,…}={1, 2, 3, 4,…}

Both are same.

Implies that f is onto.

Therefore,

f is a bijection.