第12类RD Sharma解决方案–第2章功能–练习2.2

问题1(i)。当f：R-> R和g：R-> R由f(x)= 2x + 3和g(x)= x ² + 5定义时，查找gof和雾

解决方案：

f: R -> R and g: R -> R

Therefore, f o g: R -> R and g o f: R -> R

Now, f(x) = 2x + 3 and g(x) = x² + 5

g o f(x) = g(2x + 3) =(2x + 3)² + 5

=> g o f(x) = 4x² + 12x + 14

f o g(x) = f(g(x)) = f(x² + 5) = 2(x² + 5) + 3

=> f o g(x) = 2x² + 13

问题1(ii)。当f：R-> R和g：R-> R由f(x)= 2x + x ²和g(x)= x ³定义时，查找gof和雾

解决方案：

f(x) = 2x + x² and g(x) = x³

g o f(x) = g(f(x)) = g(2x + x²)

g o f(x) =(2x + x²)³

f o g(x) = f(g(x)) = f(x³)

f o g(x) = 2x³ + x⁶

问题1(iii)。当f：R-> R和g：R-> R由f(x)= x ² + 8和g(x)= 3x ³ +1定义时，查找gof和雾

解决方案：

f(x) = x² + 8 and g(x) = 3x³ + 1

Thus, g o f(x) = g [f(x)]

=> g o f(x) = g [x² + 8]

=> g o f(x) = 3 [x² + 8]³ + 1

Similarly, f o g(x) = f [g(x)]

=> f o g(x) = f [3x³ + 1]

=> f o g(x) = [3x³ + 1]² + 8

=> f o g(x) = [9x⁶ + 1 + 6x³] + 8

=> f o g(x) = 9x⁶ + 6x³ + 9

问题1(iv)。当f：R-> R和g：R-> R由f(x)= x和g(x)= |定义时，查找gof和雾。 x |

解决方案：

f(x) = x and g(x) = | x |

Now, g o f = g(f(x)) = g(x)

g o f(x) = | x |

g o f(x) = | x |

and, f o g(x) = f(g(x)) = f(x)

f o g(x) = | x |

问题1(v)。当f：R-> R和g：R-> R由f(x)= x ² + 2x – 3和g(x)= 3x – 4定义时，查找gof和雾

解决方案：

f(x) = x² + 2x – 3 and g(x) = 3x – 4

Now, g o f(x) = g(f(x)) = g(x² + 2x – 3)

g o f(x) = 3(x² + 2x – 3) -4

g o f(x) = 3x² + 6x – 13

and, f o g(x) = f(g(x)) = f(3x – 4)

f o g(x) =(3x – 4)² + 2(3x – 4) – 3

= gx² + 16 – 24x + 6x – 8 – 3

f o g(x) = 9x² – 18x + 5

问题1(vi)。当f：R-> R和g：R-> R由f(x)= 8x ³和g(x)= x ^1/3定义时，查找gof和雾

解决方案：

f(x) = 8x³ and g(x) = x^1/3

Now, g o f(x) = g(f(x)) = g(8x³)

=(8x,3),1/3

g o f(x) = 2x

and, f o g(x) = f(g(x)) = f(x^1/3)

= 8(x^1/3)³

f o g(x) = 8x

问题2。令f = {(3，1)，(9，3)，(12，4)}，g = {(1，3)，(3，3)，(4，9)，(5， 9)}表明已经定义了gof和fog。另外，发现雾气并继续前进。

解决方案：

Let f = {(3, 1),(9, 3),(12, 4)} and

g = {(1, 3),(3, 3),(4, 9),(5, 9)

Now,

range of f = {1, 3, 4}

domain of f = {3, 9, 12}

range of g = {3, 9}

domain of g = {1, 3, 4, 5}

since, range of f ⊂ domain of g

Therefore g o f in well defined.

range of g ⊂ domain of g

g o f in well defined.

Now g o f = {(3, 3),(9, 3),(12, 9)}

f o g = {(1, 1),(3, 1),(4, 3),(5, 3)}

问题3。令f = {((-1，-1)，(4，-2)，(9，-3)，(16，4)}}，而g = {(-1，-2)，(-2， -4)，(-3，-6)，(4、8)}表示已定义gof，而未定义雾。另外，找到f。

解决方案：

We have,

f = {(1, -1),(4, -2),(9, -3),(16, 4)} and

g = {(-1, -2),(-2, -4),(-3, -6),(4, 8)}

Now,

Domain of f = {1, 4, 9, 16}

Range of f = {-1, -2, -3, 4}

Domain of g = {-1, -2, -3, 4}

Range of g = {-2, -4, -6, 8}

Clearly range of f = domain of g

Therefore, g o f is defined.

but, range of g != domain of f

Therefore, f o g is not defined.

Now,

g o f(1) = g(-1) = -2

g o f(4) = g(-2) = -4

g o f(g) = g(-3) = -6

g o f(16) = g(4) = 8

Therefore, g o f = {(1, -2),(4, -4),(9, -6),(16, 8)}

问题4。令A = {a，b，c}，B = {u，v，w}，令f和g分别是从A到B和从B到A的两个函数，定义为：f = {( a，v)，(b，u)，(c，w)}，g = {(u，b)，(v，a)，(w，x)}。证明f和g都是双射，并发现雾，然后走f。

解决方案：

A = {a, b, c}, B = {u, v, w} and

f = A -> B and g: B -> A defined by

f = {(a, v),(b, u),(c, w)} and

g = {(u, b) .(v, a),(w, c)}

For both f and g, different elements of domain have different images

Therefore, f and g are one-one

Again for each element in co-domain of f and g, there in a preimage in domain

Therefore, f and g are onto

So f and f are bijectives,

Now,

g o f = {(a, a),(b, b),(c, c)} and

f o g = {(u, u),(v, v),(w, w)}

问题5.在以下情况下找到fog(2)和gof(1)：f：R-> R; f(x)= x ² + 8和g：R-> R; g(x)= 3 x ³ + 1。

解决方案：

We have,

f: R -> R given by f(x) = x² + 8 and

g: R -> R given by g(x) = 3x³ + 1

Therefore,

f o g(x) = f(g(x)) = f(3x³ + 1)

=(3x³ + 1)² + 8

Therefore, f o g(2) =(3 * 8 + 1)² + 8 = 625 + 8 = 633

Again

g o f(x) = g(f(x)) = g(x² + 8)

= 3(x² + 8)³ + 1

g o f(1) = 3(1 + 8)³ + 1 = 2188

问题6.让R ⁺为所有非负实数的集合。如果：R ⁺ -> R ⁺和g：R ⁺ -> R ⁺定义为f(x)= x ²和g(x)= +√x，则求出fog和gof。他们是平等的职能吗?

解决方案：

We have, f: R⁺ -> R⁺ given by

f(x) = x²

g: R⁺ -> R⁺ given by

g(x) = √x

f o g(x) = f(g(x)) = f(√x) =(√x)² = x

Also,

g o f(x) = g(f(x)) = g(x² ) = √x² = x

Thus,

f o g(x) = g o f(x)

问题7.设f：R-> R和g：R-> R由f(x)= x ²和g(x)= x + 1定义。

解决方案：

We have f: R -> R and g: R -> R are two functions defined by f(x) = x² and g(x) = x + 1

Now,

f o g(x) = f(g(x)) = f(x + 1) =(x + 1)²

f o g(x) = x² + 2x + 1 —> eq(i)

g o f(x) = g(f(x)) = g(f(x)) = g(x,2) = x² + 1 —->(ii)

from(i),(ii)

f o g != g o f

问题8.令f：R-> R和g：R-> R由f(x)= x + 1和g(x)= x – 1定义。证明雾= gof = IR _。

解决方案：

Let f: R -> R and g: R -> R are defined as .

f(x) = x +1 and g(x) = x – 1

Now,

f o g(x) = f(g(x)) = f(x – 1) = x – 1 + 1

= x = I_R —>(i)

Again,

f o g(x) = f(g(x)) = g(x + 1) = x + 1 – 1

= x = I_R —>(ii)

from i and ii

f o g = g o f = I_R

问题9.验证以下三个映射的关联性：f：N-> Z ₀ (非零整数的集合)，g：Z _0- > Q和h：Q-> R由f(x)= 2x给出，g(x)= 1 / x和h(x)= e ^x

解决方案：

We have f: N -> Z₀, g: Z₀ ->Q and

h: Q -> R

Also, f(x) = 2x, g(x) = 1 / x and h(x) = e^x

Now, f: N -> Z₀ and h o g: Z₀ -> R

(h og) of: N -> R

also, g o f: N -> Q and h: Q -> R

h o(g o f): N -> R

Thus,(h o g) o f and h o(g o f) exist and are function from N to set R.

Finally,(h o g) o f(x) =(h o g)(f(x)) =(h o g)(2x)

= h(1 / 2x)

= e^1/2x

Now, h o(g o f)(x) = h o(g(2x)) = h(1 / 2x)

= e^1/2x

Associativity verified.

问题10.考虑f：N-> N，g：N-> N和h：N-> R当f(x)= 2x，g(y)= 3y + 4且h(z)= sin z x，y，z∈N。证明ho(gof)=(hog)of。

解决方案：

We have,

h o(g o f)(x) = h(g o f(x)) = h(g(f(x)))

= h(g(2x)) h(3(2x) + 4)

= h(6x + 4) = sin(6x + 4) x ∈ N

((h o g) o f)(x) =(h o g)(f(x)) =(h o g)(2x)

= h(g(2x)) == h(3(2x) + 4)

= h(6x + 4) = sin(6x + 4) x ∈ N

This shows, h o(g o f) =(h o g) o f

问题11。举两个函数f：N-> N和g：N-> N的示例为例，gof在上面，而f在上面。

解决方案：

Define f: N -> N by, f(x) = x + 1

And, g: N -> N by,

g(x) = x – 1 if x > 1

1 if x = 1

first show that f is not onto.

for this, consider element 1 in co – domain N . It is clear that this element is not an image of any of the elements in domain N .

Therefore, f is not onto.

Now, g o f: N -> N is defined.

问题12.举两个函数f的示例：N-> Z和g：Z-> Z，这样gof是单射的，而g不是单射的。

解决方案：

Define f: N -> Z as f(x) = x and g: z -> z as g(x) = | x | .

We first show that g is not injective.

It can be observed that:

g(-1) = | -1 | = 1

g(1) = | 1 | = 1

Therefore, g(-1) = g(1), but -1 != 1 .

Therefore, g is not injective.

Now, g o f: N -> Z is defined as g o f(x) = g(f(x)) = g(x) = | x |.

let x, y E N such that g o f(x) = g o f(y) .

=> | x | = | y |

Since x and y ∈ N, both are positive.

| x | = | y | => x = y

Hence, g o f is injective

问题13.如果f：A-> B和g：B-> C是一一函数，则表明gof是一一函数。

解决方案：

We have f: A -> B and g: B -> C are one – one functions

Now we have to prove: g o f: A -> C in one – one

let x, y ∈ A such that

g o f(x) = g o f(y)

=> g(f(x)) = g(f(y))

=> f(x) = f(y)

x = y

Therefore, g o f is one – one function.

问题14.如果f：A-> B和g：B-> C在函数上，则表明gof是on函数。

解决方案：

We have, f: A -> B and g: B -> C are onto functions

Now, we need to prove: g o f: A -> C in onto

let y ∈ C, then,

g o f(x) = y

g(f(x)) = y –>(i)

Since g is onto, for each element in c, then exists a preimage in B.

g(x) = y —->(ii)

from(i) and(ii)

f(x) = ∞

Since f is onto, for each element in B there exists a preimage in A

f(x) = ∞ —>(iii)

From(ii) and(iii) we can conclude that for each y ∈ c there exists a pre image in A

Such that g o f(x) = y

Therefore, g o f is onto