1.

Solution:

For. The arrow diagram defines a function from A to B because each element of A has a unique image in B.

B. The arrow diagram defines a function from A to B, as each element of A has a unique image in B.

C. The arrow diagram does not define a function from A to B because element c of A has no image in B.

D. The arrow diagram defines a function from A to B because each element of A has a unique image in B.

My. The arrow diagram does not define a function from A to B because element b of A has no image in B.

F. The arrow diagram defines a function from A to B because each element of A has a unique image in B.

2.

Sol:

For. The reason R_{1}is not a function from X to Y because element a of x has two images p and r in y, but p ≠ r.

B. The reason R_{2}is not a function from X to Y because element c of x has no image in y.

C. The R relationship_{3}is a function from X to Y since each element of x has a unique image in y.

D. The reason R_{4}is a function from X to Y since each element of x has a unique image in y.

My. The reason R_{5}is a function from X to Y since each element of x has a unique image in y.

3.

Sol:

Here f(x) = 1 – x and x = 1, 3/2, –1, 2 ε A.

So image, f(1) = 1 – 1 = 0

f$\left( {\frac{3}{2}} \right)$ = 1 – $\frac{3}{2}$ = $ - \frac{1}{2}$

f(–1) = 1 – (–1) = 2

Yf(2) = 1 – 2 = –1

4.

Sol:

Here,

(EU)

f(x) = x + 2, onde –1 ≤ x ≤ 2.

Now,

- f(–1) = –1 + 2 = 1
- f(0) = 0 + 2 = 2
- f(1) = 1 + 2 = 3
- f(2) = 2 + 2 = 4
- f(-2) and f(3) are undefined because -2 and 3 are not at -1 ≤ x ≤ 2.

(ii)

f(x) = 2|x| + 3x, onde –1 ≤ x ≤ 2.

Now,

A. f(–1) = 2|–1| + 3(–1) = 2 – 3 = –1

B. f(0) = 2|0| + 3,0 = 0 + 0 = 0

B. f(1) = 2|1| + 3,1 = 2 + 3 = 5

D. f(2) = 2|2| + 3,2 = 4 + 6 = 10

My. f(-2) and f(3) are undefined because -2 and 3 are not at -1 ≤ x ≤ 2.

5.

(EU)

Sol:

A. f$\left( { - \frac{1}{2}} \right)$ = 3 + 2 $\left( { - \frac{1}{2}} \right)$ = 3 – 1 = two

B. f(0) = 3 – 2,0 = 3

B. f$\left( {\frac{1}{2}} \right)$ = –3 – 2 $\left( { - \frac{1}{2}} \right)$ = – 3 – 1 =-4.

(d) $\frac{{{\rm{f}}\left( {\rm{h}} \right) - {\rm{f}}\left( 0 \right)}}{{\rm{ h}}}$ = $\frac{{\esquerda( {3 - 2{\rm{h}}} \direita) - \esquerda( {3 - 2,0} \direita)}}{{\rm{h} }}$ [0≤h<$\frac{1}{2}$]

= – 2

(ii)

Sol:

A. f(2) = 4,2 – 2 = 6

B. f(1) = 4,1 – 2 = 2

Z. f(0) = 2,0 = 0

D. f(–1) = 2(–1) = –2

mi. $\frac{{{\rm{f}}\left( {\rm{h}} \right) - {\rm{f}}\left( 1 \right)}}{{\rm{h}} }$ = $\frac{{\left( {4{\rm{h}} - 2} \right) - \left( {4.1 - 2} \right)}}{{\rm{h}}}$ [puesto que, 1<h]

= 4(h-1) / hour

6.

Sol:

Here A = {-1, 0, 2, 4, 6}

(EU)

y = f(x) = $\frac{{\rm{x}}}{{{\rm{x}} + 2}}$

So, Interval of f = {f(–1), f(0), f(2), f(4), f(6)}

= $\izquierda\{ { - \frac{1}{{ - 1 + 2}},\frac{0}{{0 + 2}},{\rm{\: }}\frac{2}{{ 2 + 2}},{\rm{\: }}\frac{4}{{4 + 2}},{\rm{\: }}\frac{6}{{6 + 2}}} \right p.s

= $\izquierda\{ { - 1,{\rm{\: }}0,{\rm{\: }}\frac{1}{2},{\rm{\: }}\frac{2} {3},{\rm{\: }}\frac{3}{4}} \right\}$

(ii)

y = f(x) = $\frac{{{\rm{x}}\left( {{\rm{x}} + 1} \right)}}{{{\rm{x}} + 2} p.s

So, Interval of f = {f(–1), f(0), f(2), f(4), f(6)}

= $\left\{ {\frac{{ - 1\left( { - 1 + 1} \right)}}{{ - 1 + 2}},{\rm{\: }}\frac{{0\ left( {0 + 1} \right)}}{{0 + 2}},{\rm{\: }}\frac{{2\left( {2 + 1} \right)}}{{2 + 2}},{\rm{\: }}\frac{{4\left( {4 + 1} \right)}}{{4 + 2}},{\rm{\: }}\frac{{ 6\left( {6 + 1} \right)}}{{6 + 2}}} \right\}{\rm{\: \: }}$

= $\left\{ {0,{\rm{\: }}0,{\rm{\: }}\frac{3}{2},{\rm{\: }}\frac{{10} }{3},{\rm{\: }}\frac{{21}}{4}} \right\}$ = $\left\{ {0,{\rm{\: }}\frac{3 {2},{\rm{\: }}\frac{{10}}{3},{\rm{\: }}\frac{{21}}{4}} \right\}$

7.

Sol:

Hier gilt f(x) = x^{2}where A = {-1, -2, -3}

Also f(1) = (–1)^{2}= 1

f(–2) = (–2)^{2}= 4

f(–3) = (–3)^{2}= 9

So area f = {-1, -2, -3} and area f = {1, 4, 9}

E: From the Venn diagram we have:

g(–1) = 1

g(–2) = 4

g(–3) = 9

So area g = {-1, -2, -3} and area g = {1, 4, 9}

Then f and g have the same domain {–1, –2, –3} and f(x) = g(x) for all x ϵ A, then functions f and g are equal.

8.

A.

Solution:

Here A = {-2, -1, 0, 1, 2}, B = {4, 0, 1} and f:A→B is f(x) = x^{2}, then f is not a – one, because f(–2) = 4 = f(2) but –2 ≠ 2.

B.

here A = set of positive integers.

Also, A = {1, 2, 3, ….}

E B = set of squares of positive integers.

Then B = {1, 4, 9, …} and the function f:A→B is defined by f(x) = x^{2}.

To show: f is a - one.

Let x ≠ y with x, y ϵ A, then f(x) ≠ f(y).

If x ≠ y.

= x^{2}≠ j^{2}[x, yϵz^{+}]

= f(x) ≠ f(y) [f(x) = x^{2}]

So f is one-one.

C.

Here f:[–2, –2] → R is equal to f(x) = x^{2}.

To show: f is a - one.

Let x, y ϵ [–2, –2] as x ≠ y.

So, to show: f(x) ≠ f(y)

If x ≠ y.

plus x^{2}= j^{2}[–2, 2 ϵ [–2, –2], –2≠2 Aber (–2)^{2}= 2^{2}= 4]

Also f(x) = f(y)

Then x ≠ y→ f(x) = f(y). So it's not a--a.

D.

Here f:[0, 3] → R is equal to f(x) = x^{2}is a – one because x, y ϵ [0, 3], x ≠ y implies that f(x) ≠ f(y).

9.

Trial:

(EU)

Here A = {1, -3, 3}, B = {1, 9}

And f:A→B we have f(x) = x^{2}is not a – one, because –3 ≠ 3, but f(–3) = 9 = f(3). Furthermore, the function f is an because every element of B has a preimage in A. So f is not one, but an.

(ii)

Here f: N → N equals f(x) = 2x.

To show: f is one - one and more.

A.

f is a - one.

Let x, y ϵ N such that x ≠ y. So, to show: f(x) ≠ f(y).

If x ≠ y

= 2x ≠ 2 years

= f(x) ≠ f(y) [ logo,f(x) = 2x]

So f is one to one.

B.

f is about.

Set y ε N such that y = f(x) = 2x

Also y = 2x.

= x = $\frac{{\rm{y}}}{2}$ ϵ N if y is odd.

So f is not over.

Therefore, the one-one function is not over..

(iii)

Here f:(–2, 2) → R is equal to f(x) = x^{2}

A.

The function is not a why.

–1 ≠ 1 pero f(–1) = 1 = f(1)

B.

The function is not one because

f(–2, 2) = (0, 4) ≠ R.

So f is neither a nor over.

(4)

Aqui f:Q→Q f(x) = 6x + 5.

To show: f is a–one and more

A.

f is one - one and more.

Let x,y ε Q (that is, area Q)

So x ≠ y. So, to show that f(x) ≠ f(y)

If x ≠ y.

= 6x ≠ 6 years

= 6x + 5 ≠ 6a + 5

= f(x) ≠ f(y)

So f is a – one.

Ö

B.

f is one–one: Let x, y ϵ Q with f(x) = f(y). Then show: x = y.

Set f(x) = f(y)

= 6x + 5 = 6y + 5 →x = y. [f is one-one]

B.

f is over

let y ε Q (codominium Q)

So y = f(x) = 6x + 5

So to show: x ε Q.

So y = 6x + 5

= 6x = y – 5

Then x = $\frac{{{\rm{y}} - 5}}{6}$ ϵ Q [y ϵ Q]

E f$\left( {\frac{{{\rm{y}} - 5}}{6}} \right)$ = $6\left( {\frac{{{\rm{y}} - 5} {6}} \right)$ + 5 = y – 5 + 5 = y

So y is the image of $\frac{{\left( {{\rm{y}} - 5} \right)}}{6}.$

So f is approx.

So f is one-one plus.

(v)

Aqui f:R → R f(x) = x^{3}.

To show: f is one-one and more.

A.

f is one-one.

Let x, y ϵ R(that is, region R) with x ≠ y. So, to show: f(x) ≠ f(y).

If x ≠ y.

= x^{3}≠ j^{3}[Cube of two different real numbers is different]

= f(x) ≠ f(y)

So f is one-one.

B.

f is over

Sei y ε R (d. h. Kodomäne R)

So y = f(x) = x^{3}

To show: x ε R

If y = x^{3}

So x = $\sqrt[3]{{\rm{y}}}$ ϵ R [The cube root of different real numbers is different]

So x ε R.

And f($\sqrt[3]{{\rm{y}}}$) = ${\left( {{{\rm{y}}^3}} \right)^{\frac{1}{ 3}}}$ = e. So f is approx.

So f is one-one plus.

10

A.

Sol:

Aqui A = {–2, –1, 0, 1, 2}, B = {1, $\frac{1}{6}$, $\frac{2}{3}$} e f(x) = $ \frac{{{{\rm{x}}^2}}}{6}$.

Então, rango de f = {f(–2), f(–1), f(0), f(1), f(2)} = $\left\{ {\frac{4}{6}. \frac{1}{6}.0.\frac{1}{6},{\rm{\: }}\frac{4}{6}} \right\}$

So area of f = $\left\{ {0,{\rm{\: }}\frac{1}{6},{\rm{\: }}\frac{2}{3}} \ right \ }{\rm{\: }}$

The function is not one-one because f(–2) = $\frac{2}{3}$ = f(2) but –2 ≠ 2.

However, the function is complete, since f(A) = B.

B.

Sol:

Aqui, A = {–1, 0, 1, 2, 3, 4}, B = $\left\{ { - 1,{\rm{\: }}0,{\rm{\: }}\frac {4}{5},{\rm{\: }}\frac{5}{7},{\rm{\: }}1,{\rm{\: }}2,{\rm{\: }}3} \right\}$ y f:A→B es f(x) = $\frac{{\left( {{\rm{x}} + 1} \right)}}{{2{\ rm {x}} - 1}}$.

Now,

(EU)

Faixa de f = {f(–1), f(0), f(1), f(2), f(3), f(4)}

= $\{ \frac{{ - 1 + 1}}{{2\esquerda( { - 1} \direita) - 1}},{\rm{\: }}\frac{{0 + 1}}{ {2,0 - 1}},{\rm{\: }}\frac{{1 + 1}}{{2,1 - 1}},{\rm{\: }}\frac{{2 + 1}}{ {2,2 - 1}},{\rm{\: }}\frac{{3 + 1}}{{2,3 - 1}},{\rm{\: }}\frac {{4 + 1}}{ {2,4 - 1}}$

= $\esquerda\{ {0,{\rm{\: }} - 1,{\rm{\: }}2,{\rm{\: }}1,{\rm{\: }}\frac {4}{5},{\rm{\: }}\frac{5}{7}} \right\}$

So, Rank of f = $\left\{ { - 1,{\rm{\: }}0,{\rm{\: }}\frac{4}{5},{\rm{\: }} \frac{5}{7},{\rm{\: }}\frac{1}{2}} \right\}$

(ii)

The function f is one-one because if A has any other image in B, each element is not activated, but the function is not activated because element 3 ϵ B has no previous image in A.

(iii)

The function can be done one-one and over both is 3 ϵ B. That is, if

A = {–1, 0, 1, 2, 3, 4}, B = $\esquerda\{ { - 1,{\rm{\: }}0,{\rm{\: }}\frac{4 {5},{\rm{\: }}\frac{5}{7},{\rm{\: }}1,{\rm{\: }}2} \right\}$ y

f:A→ B es f(x) = $\frac{{{\rm{x}} + 1}}{{2{\rm{x}} - 1}}$

So the function f is one and over.