第12类NCERT解决方案-数学第I部分-第1章关系和功能-练习1.3

问题1.令f：{1，3，4}-> {1,2,5}和g：{1,2,5}-> {1，3}由f = {(1，2) ，(3、5)，(4、1)和g = {(1、3)，(2、5)，(5、1)}。写下gof。

解决方案：

f= {(1, 2), (3, 5), (4, 1)}

g= {(1, 3), (2, 3), (5, 1)}

f(1)= 2, g(2) = 3 => gof(1) = 3

f(3) = 5, g(5) = 1 => gof(3) = 1

f(4) =1, g(1) = 3 => gof(4) = 3 

=> gof = {(1,3), (3,1), (4,3)}

问题2。令f，g和h为从R到R的函数。证明(f + g)oh = foh + goh，(f * g)oh =(foh)*(goh)。

解决方案：

f: R-> R, g: R-> R, h: R-> R

(f+g) oh(x) = (f+g) oh(x)

= (f+g) [h(x)]

= f[h(x)] + g[h(x)]

= foh(x) + goh(x)

(f+g) oh = foh + goh  

(f * g) oh(x) = (f * g) oh(x)

= (f * g) [h(x)]

= f[h(x)] * g[h(x)]

= foh(x) * goh (x)

(f * g) oh = (foh) * (goh)

问题3：如果有，请检查gof和雾

(i)f(x)= | x |和g(x)= | 5x – 2 |

(ii)f(x)= 8x ³和g(x)= x ^1/3

解决方案：

(i) We have,        

f(x) = |x| and g (x) = | 5x – 2 |

gof(x) = g(f(x)) = g(|x|)

=> gof(x) = | 5 |x|-2 |

fog(x) = f(g(x)) = f(|5x-2|)

=> fog(x) = || 5x-2|| = | 5x -2 | 

(ii) We have,

f(x) = 8x³ and g(x) = x^1/3

gof(x) = g(f(x)) = g(8x³)

=> gof(x) = (8x³)^1/3 = 2x

fog(x) = f(g(x)) = f(x^1/3)

=> fog(x) = 8(x^1/3)³ = 8x

问题4.如果f(x)= $\frac{4x+3}{6x-4}x \neq \frac{2}{3}$ ，证明所有的fof(x)= x $x \neq \frac{2}{3}$ 。 f的逆是什么?

解决方案：

Given that,

$f (x) = \frac{4x+3}{6x-4}x \neq \frac{2}{3}$

Now, 

fof(x) = f(f(x)) = $f(\frac{4x+3}{6x-4})$

= $f(\frac{4(\frac{4x+3}{6x-4})+3}{6(\frac{4x+3}{6x-4})-4})$

On simplifying by taking LCM = (6x-4)

fof(x) =  $\frac{16x + 12 + 18x - 12}{24x + 18 -24x + 16}$

=> fof (x) = $\frac{34x}{34}$= x 

=> fof(x) = I_A (x) for all $x \neq \frac{2}{3}$

=> fof(x) = I_A such that A = R – ${\frac{2}{3}}$ which is the domain of f

=> f^-1 = f          

Hence, proved.

问题5.有理有据地说明以下函数是否具有反函数。找到逆，如果存在的话。

(i)f：{1、2、3、4}-> {10}

与f = {(1，10)，(2，10)，(3,10)，(4,10)}

(ii)g：{5，6，7，8}-> {1,2,3,4}

g = {(5，4)，(6，3)，(7，4)，(8，2)}

(iii)h：{2，3，4，5}-> {7，9，11，13}

与h：{(2，7)，(3，9)，(4，11)，(5，13)}

解决方案：

(i) We have f(1) = f(2) = f(3) = f(4) = 10 which means that f is many-one 

and not one-one, therefore inverse of f does not exist.

(ii) Here g(5) = g(7) =4 i.e. g is many-one, so inverse of g does not exist.

(iii) Since range of h = {7, 9, 11, 13} = co-domain, therefore h is onto,

Also, each element of domain has a unique image in h, therefore h is one-one.

Now, since h is both one-one and onto,thus inverse of h exists.

h^-1 = {(7, 2), (9, 3), (11, 4), (13, 5)}

问题6.证明f：[-1，1]-> R，由f(x)= $\frac{x}{x+2}$是一对一。找到函数f的反函数：{-1,1}->范围f。

解决方案：

Let x, y  $\in$ [-1, 1] 

f(x) =  $\frac{x}{x+2}$

f(y) = $\frac{y}{y+2}$

Now, 

Let f(x) = f(y) 

$\frac{x}{x+2} = \frac{y}{y+2}$  

=> x(x + 2) = y(x + 2)

=> x y + 2x = x y + 2y

=> 2x = 2y

=> x = y

=> f is one-one

Also,

X = [-1, 1] and,

Y = { $y \in R : y = \frac{x}{x+2} , x \in X$ } = range of f.

=> f is onto

Since f is one-one and onto, therefore inverse of f exists.

Let y = f(x) => x =f^-1(y)

=> y = $\frac{x}{x+2}$           

=> x y + 2y = x

=> 2y = x(1 – y)

=> x = $\frac{2y}{1 - y}, y \ne 1$

Therefore, f : Y-> X is defined by f(y) = $\frac{2y}{1-y}, y \ne 1$ .

问题7.考虑f：R-> R由f(x)= 4x + 3给出。证明f是可逆的。求f的逆。

解决方案：

It is given that,

f(x) = 4x + 3     where f : R -> R

Let, 

f(x) = f(y) 

=> 4x + 3 = 4y + 3

=> 4x = 4y

=> x = y

=> f is one-one function 

Also,

Let y  = 4x + 3     where y $\in$  R

=> x = $\frac{y-3}{4}$ $\in R$

Since for any $y \in R$. there exists $x = \frac{y-3}{4} \in R$ such that

f(x) = $f(\frac{y-3}{4})$= 4 $(\frac{y-3}{4})$ +3 = y

=> f is onto

Since f is both one-one and onto, therefore f^-1 exists

=> f^-1(y) = $\frac{y-3}{4}$

问题8.考虑f：R ₊ -> [4， $\infty$ )由f(x)= x ² + 4给出。表明f是可逆的，而f的反f ^-1由f ^-1 (y)= $\sqrt{ y -4}$ ，其中R ₊是所有非负实数的集合。

解决方案：

Let f(x) = f(y)

=> x + 4 = y + 4

=> x² = y²

=> x = y          [ x,y $\in$ R₊ ]

=> f is one-one

Let y = x² + 4             where y $\in [4, \infty )$

=>  x² = y – 4 $\geq$ 4        [ y $\geq 4 ]$

=> x = $\sqrt{y -4}$

Therefore, for any y $\in R$ , there exists x = $\sqrt{y-4}$ $\in R$

=> f is onto

Since, f is both one-one and onto, f^-1 exists for every $y \geq 4$,

=> f^-1(y) = $\sqrt{y-4}$

问题9：考虑R ₊ -> [-5， $\infty$ )由f(x)= 9x ² + 6x -5给出。证明在f ^-1 (y)=时f是可逆的$(\frac{(\sqrt{(y+6)}-1}{3})$

解决方案：

Let f(x) = f(y)

=> 9x² + 6x -5 = 9y² + 6y – 5

=> 9x² + 6x = 9y² + 6y

=> 9(x² – y²) + 6 (x – y) = 0

=> (x – y) [9 (x + y) + 6] = 0

=> x – y =0

=> x = y

=> f is one-one

Now, let y = 9x² + 6x – 5

=> 9x² + 6x – 5 (x + y) = 0

=> x = $\frac{-6 \pm \sqrt{6*6 + 4*9(5+y)} }{18} = \frac{\sqrt{(y+6)}-1}{3}$  

=> f(x) = $f(\frac{\sqrt{(y+6)}-1}{3}) = 9 (\frac{\sqrt{(y+6)}-1}{3})^{2}+(\frac{\sqrt{(y+6)}-1}{3}) -5$

On simpliying, we have f (x) = y 

=> f is onto

Since f is both one-one and onto. f^-1 exists

f^-1(y) = $\frac{\sqrt{(y+6)}-1}{3}$

问题10.让f：X-> Y是一个可逆函数。证明f具有唯一的逆。

解决方案：

We have,

f : X -> Y is an invertible function

Let g and h be two distinct inverses of f.

Then, for all y$\in$ Y,

fog (y) = I (y) = foh (y)

=> f g (y)) = f(h (y))

=> g(y) = h(y)        [f is one-one]

=> g = h                  [g is one-one]

which contradicts our supposition. 

Hence, f has a unique inverse.

问题11。考虑f：{1,2,3}-> {a，b，c}由f(1)= a，f(2)= b，f(3)= c给出。找到f并证明(f ^-1 )f ^-1 = f。

解决方案：

Given that,

f(1) = a, f(2) = b, f(3) = c

We have,

f = {(1, a), (2, b), (c, 3)}

which shows that f is both one-one and onto and thus f is invertible.

Therefore,

f^-1 = {(a, 1), (b, 2), (c, 3)} 

Also,

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

=> (f^-1)^-1 = f 

Hence proved.

问题12.让f：X-> Y是一个可逆函数。证明f ^-1的倒数是f，即(f ^-1 ) ^-1 = f。

解决方案：

Since, f is an invertible function,

=> f is both one-one and onto

Also,

Let g : Y -> X , where g is a one-one and onto function such that

gof (x) = I_x and fog (y) = I_y => g = f^-1

=> f^-1 o (f^-1)^-1 = I

=> f o [f^-1 o (f^-1)^-1] = f o I

=> (f o f^-1) o (f^-1)^-1 = f 

=> I o (f^-1)^-1 = f

Hence, (f^-1)^-1 = f

问题13.如果f：R-> R由f(x)=(3 – x ³ ) ^1/3给出，则fof(x)为：

(A)x ^1/3 (B)x ³ (C)x。 (D)(3 – x ³ )

解决方案：

Answer: (C)

We have,

f(x) = (3 – x³)^1/3           where f : R -> R

Now,

fof(x) = f(f(x))

=> fof(x) = f((3 – x³)^1/3) 

=> fof(x) = [3 – ((3 – x³)^1/3 )³]^1/3  

=> fof(x) = [3 – (3 – x³)]^1/3              

=> fof(x) = (x³)^1/3

=> fof(x) = x

Hence, option C is correct.

问题14.让f：R-{ $-\frac{4}{3}$ }-> R是定义为f(x)=的函数$\frac{4x}{3x+4}$ 。 f的倒数是映射g：范围f-> R – { $-\frac{4}{3}$ }由

(A)g(y)= $\frac{3y}{3-4y}$ (B)g(y)= $\frac{4y}{4-3y}$

(C)g(y)= $\frac{4y}{3-4y}$ (D)g(y)= $\frac{3y}{4-3y}$

解决方案：

Answer: (B)

Let y = f(x)

=> y = $\frac{4x}{3x+4}$

=> 3xy + 4y = 4x

=> x( 4 – 3y) = 4y

=> x = $\frac{4y}{4-3y}$ 

f^-1(y) = g (y) = $\frac{4y}{4-3y}$