Solucionario c.t. álgebra 5°

15

SOLUCIONARIO CUADERNO DE TRABAJO

Cuaderno de TraBaJo CaP 01NÚMEROS REALES

01 I. (F) II. (V) III. (V)Clave D

02 a + b = c

•Asícumple

•Bnocumple;1+3=4B •Csícumple

Clave E

03 I.parpar×impar=par

II.parpar×impar=par

III.parpar×impar=par

IV.imparpar–1=impar

Clave A

04 I. (V) a q b = b q a esconmutativa

II. (F) a q e = e q a e=1

III. (V) a q (b q c) = (a q b) q c esasociativa

Clave C

05 I. (V) (Q)2 Q

II.(F)Sia = 2, a2 Qperoa Q

III.(F)Sia=–2 b=–1

|(–2)+(–1)|=|–2|+|–1| a, b 0

Clave D

06 Corrija: IV. z + y

x = z

x + y

x

III. x · y = x · y IV. z + y

x = z

x + y

x

Clave B

07 Corrija: Alternativa E) Notiene

I. (F) ( 3 + 2)( 3– 2)=1

II.(F)Noexisteelelementoneutro.

III.(V)Six=1secumple111=1

Clave E

08 a a b = 2a + 5b

•a a e = 2a + 5e = a e=– a5

•e a a = 2e + 5a = a e=–2a

\– a5=–2a a = 0 Notienea–1

Clave B

09 SoloKattydicelaverdad.Clave B

10 a * b = a + b +1

•a * e = e * a a + e+1=a e =–1

•a * a–1 = a–1 * a = e

4*4–1=1 4–1 + 4+1=–1 4–1 =–6

•EscorrectoClave E

Cuaderno de TraBaJo CaP 02MULTIPLICACIÓN ALGEBRAICA

01 a = 3+13–1

× 3+13+1

= 3 + 2

(a2–7)2=(4 3+7–7)2=48Clave A

02 M = (a + b)4 –(a –b)4

2(a2 + b2) = 8ab(a2 + b2)2(a2 + b2)

=4ab

Clave D

03 y(x) = x3 + a1x2 + a2x + a3 = (x + a)3

a1=3a, a2=3a2, a3 = a3

A=a12

a2 +

a13

a3 A=

9a2

3a2 + 27a2

a3=30

Clave C

04 a + b=5;a · b=1 b = a–1

a3 + a–3 = (a + a–1)3–3(a + a–1) = 53–3(5)

\ a3 + a–3=110Clave D

05 Corrija: F = ab

(a–b)2

Sia3 = b3 a3–b3 = (a–b)(a2 + ab + b2)

0

a2 + ab + b2 = 0 (a–b)2=–3ab

\ F = ab

(a–b)2=– 13 Clave B

06 Six + 1x=3E=x2 + 1

x2 + x3 + 1x3

E=x + 1

x

2–2+

x + 1

x

3–3

x + 1

x

E=32–2+33–3×3=25Clave C

07 P =

m–3 + n–3

m–3· n–3–1

= 1

m3 + n3

P = 1

(m + n)3–3mn(m + n) =

1–24

Clave C

08 x2 + y2 + z2 = 2(x–2y + 5z)–30

(x–1)2 + (y + 2)2 + (z–5)2 = 0

x=1 y=–2 z = 5

V = x2 + y2 + z2

x + z = 306

= 5Clave A

09 Sia b c

T = (a –b)3 + (b –c)3 + (c –a)3

(a–b)(b–c)(a–c)

T = 3(a –b)(b –c)(c –a)

(a –b)(b –c)(a –c)=–3

Clave A

10 Sia + b + c = 0

V = (a2 + b2)(b2 + c2)(a3 + b3 + c3)(2abc–a3)(2abc–c3)(a + c)

V = (a2 + b2)(b2 + c2)(a3 + b3 + c3)

ac(b2 + c2)(a2 + b2)(a + c)

V = 3abc(ac)(a + c)

= 3abc–abc

=–3Clave E

Cuaderno de TraBaJo CaP 03DIVISIÓN ALGEBRAICA

01 D(x)d(x)

= 2x3 +7x2+10x + p

x+1

Residuo=D(–1)=2

–2+7–10+p = 2 p=7

Clave E

02 P(x) = x3 + 2ax2–7ax2 + 2a3

P(a) = a3 + 2a3–7a3 + 2a3=1–2a3=1

\ a = 22

3

Clave C

03 P(x) = x3 + bx2–cx+3

P(a)=13 + b + c+3=0 b + c=–4

P(x) = (x–1)(x+1)(x + e)

Q(x)

Q(x) = (x+1)(x + e)

Q(1)=(2)(1+e)=– 32

e=– 14

Q(x) = x2 + 12

x–3Clave A

ÁLGEBRA 5°

ProYeCTo InGenIo SOLUCIONARIO - ÁLGEBRA 5°

2 5

04 P(x) = x3+3x2–Ax + B

P(–4)=(–4)3+3(–4)2–4A + B = 0

P(2) = (2)3+3(2)2–2A + B = 0

A=–6 B=–8 \ A–B = 2

Clave C

05 P(x) = R'(x)(2x–1)+6

P(x) = R''(x)(x+1)+3

P(x) = (2x–1)(x+1)R''(x) + R(x)

R(x) = ax + b

R

12

= a2

+ b=6R(–1)=–a + b=3

\ R(x) = 2x + 5Clave B

06 x4a +12–y4a –3

xa–8–ya–9

#tnos=4a+12a–8

= 4a–3a–9

a=15

Clave C

07 x240–y160

x3 + y2 tk = (x3)80–k(y2)k–1

240–3k + 2k–2=164 k=74

Clave E

08 Sea:4x

5y–9 = n a4x–b4x

a5x–9–b5y–9

t5 = (am)n–5(bp)4 = a176· b64

n(n–5)=176 (n+11)(n–16)=176

\ n=16Clave A

09 P(x) = (x2–3x + 2)(x2–8x+15)(x2–10x+24)

d(x) = x2–7x + 9

P(x) d(x)

R(x)=–9 Clave E

10 P(x) = M(x)(x3+1)+x2 + x–1

P(x) = M(x)(x+1)(x2–x+1)+x2 + x–1

Sedividex2–x+1 R(x) = 2x–2

Clave E

Cuaderno de TraBaJo CaP 04FACTORIZACIÓN

01 A(x) = x2+7x+6=(x+6)(x+1)

B(x) = x2–5x–6=(x–6)(x+1)

C(x) = x2+4xy–45y2 = (x + 9y)(x–5y)

D(x)=6x2+7xy + 2y2=(3x + 2y)(2x + y)

Clave A

02 K=25a4–109a2+36

K=(5a–3)(5a+3)(a–2)(a + 2)

Clave B

03 3x2 14x2 d

8x4 + bx2–(2+d) = (2x2+1)(4x2–1)

= (2x2+1)(2x+1)(2x–1)

d=–2–d d=–1Clave A

04 2x2–3x–2–y2+3y + xy

(x–1)2–(y–1)2 + y(x+1)+(x–2)(x+1)

(x + y–2)(x–y) + (x+1)(x + y–2)

(x + y–2)(2x–y+1)

\Sumadefactores:3x–1Clave D

05 P(x) = x3 + 2x2–2x–1

P(x) = (x–1)(x2+3x+1)

1 D > 0 2factores

\3factoresClave C

06 2x3–x2–x–3 x3–1+x3–x2–x–2

2(x3–1)–(x2 + x+1)

2(x–1)(x2 + x+1)–(x2 + x+1)

(x2 + x+1)(2x–3)

Sumadecoeficientesdelfactorlineales:–1

Clave B

07 3x –5y –54x 2y 2

12x2–14x–14xy–10y2–20y–10

(3x–5y–5)(4x + 2y + 2) 7x–3y–3

Clave A

08 11x –3 x 4

11x2+41x–12=(11x–3)(x+4)

\(11+3)–(1+4)=9Clave C

09 (x–5)(x+4)(x–7)(x+6)–504

(x2–x–20)(x2–x–42)–504

(a–20)(a–42)–504

a2–62a+336=(a–6)(a–56)

(x2–x–6)(x2–x–56)=(x–3)(x + 2)(x–8)(x+7)

\ x +7Clave B

10 x2 –5x +6x2 –5x 4

x4–10x3+35x2–50x+24

(x2–5x+6)(x2–5x+4)

(x–3)(x–2)(x–4)(x–1)Clave A

Cuaderno de TraBaJo CaP 05MÉTODOS DE FACTORIZACIÓN

01 A(x) = x2+7x+6=(x+6)(x+1)

B(x;y) = x2y–5xy–6y = (xy–6y)(x+1)

= y(x–6)(x+1)

\Factoresprimos:2y3.Clave C

02 P(x;y) = 2x2 + xy–x–y2+4y–3

= (2x–y+3)(x + y–1)

\Sumadefactoresprimos:3x + 2.

Clave B

03 P(x;y) = 2x4–3x3y–4x2y2+3xy3 + 2y4

= 2(x4–2x2y2 + y4)–3xy(x2–y2)

= 2(x2–y2)2–3xy(x2–y2)

= (x + y)(x–y)(2x + y)(x–2y)

Clave E

04 P(x) = x3 + 2x2–2x–1=x3–1+2x2–2x

P(x) = (x–1)(x2+3x+1)

D > 0

\Factoresprimos:3.Clave C

05 x3–6x2–x+6=x(x2–1)–6(x2–1)

= (x–1)(x+1)(x–6) \–6

Clave B

06 P(x;y) = (x + y+1)2–2x–2y–10

= (x + y+1)2–2(x + y+1)–8

= (x + y–3)(x + y+3)

Sumadefactoresprimosmónicos:2(x + y)

Clave B

07 P(x) = x5 + x4–x2–x = x(x4–1)+x2(x2–1)

P(x) = (x2–1)[x(x2+1)+x2]

P(x) = (x–1)(x+1)(x)[x2 + x+1]

\3xClave D

08 2a4–a2b2–b4 = (a2–b2)(2a2 + b2)

a2 –b2

2a2 b2

Clave D

ProYeCTo InGenIoSOLUCIONARIO - ÁLGEBRA 5°

35

09 x3+6x2+11x+6=x3 + x2 + 5x2+11x+6

(x+1)(x+3)(x + 2)

\Factoreslineales:3. Clave C

10 B(x) = x3 + 2x2–5x–6

= (x+1)(x+3)(x–2)

\Sumadefactoreslineales:3x.

Clave B

Cuaderno de TraBaJo CaP 06NÚMERO COMBINATORIO

01

9!+8!8!

10!+9!9!

= 8!(9+1)

8!×9!(10+1)

9!

=110 Clave E

02 (x+4)!(x+6)!

(x+4)!+(x + 5)! =

(x + 5)!(x+6)(x+6)

= 20

(x + 5)! = 20! x=15Clave A

03 Si3!1!

+ 4!2!

+ 5!3!

+ 6!4!=3(n!)–4

3(n!)=6+12+20+30+4 n=4

\ (n+1)!=5!=120Clave E

04 Cx14 = 14C2x

–

1 x + 2x–1=14 x = 5

Clave C

05 C1n + C2

n + n + 1C3 =35 n + 1C2 + n + 1C3 =35

n + 2C2 =35 n = 5 \ 2C25 = 20

Clave B

06 Corrija: Alternativa E) 13

C08 + C1

8 + C29 + C3

10 = m + 2Cn –

1

C19 + C2

9 + C310 = m + 2Cn

–

1 C2

10 + C310 = m + 2Cn

–

1

C311 = m + 2Cn

–

1 m = 9 n=4

\ m + n=13Clave E

07 V3x=60 x!

(x–3)!=60

(x–2)(x–1)(x)=3×4×5 x = 5

Clave B

08 5H 4M C24× C4

5=6×5=30

Clave D

09 V = {a;e;i;o;u} B={1;2;3;4;5;6}

Placa:V1V2 b1b2 b3b4 (5)(4)(6)(5)(4)(3)

\7200Clave C

10 2(1!)+4(2!)+6(3!)+...+50(25!)=2(n!)–2!

1!+1!+2(2!)+3(3!)+...+25(25!)=n!

Sea:1!+2!+3!+4!+...+25!=S

1!+2!+3!+4!+...+25!

S+26!=n! + S

\26!=n! n=26Clave E

Cuaderno de TraBaJo CaP 07BINOMIO DE NEWTON

01 (2x + y3)14

t10 = C914(2x)14–9· (y3)9 t10 = C9

14 25 · x5 · y27

\Grado=27+5=32Clave B

02

x3 + 1

x

10 Existen11términos.

tC=t6 = C510(x3)5

1x

5 t6 = C5

10x10

Clave A

03

1x

+ x2n t4 = C3

n

1x

n – 3(x2)3

6–(n–3)=–8 n=17

Entoncestiene17términos:

t16+1 = 17C16

1x

1(x2)16

\Coeficiente: 17C16 = 17C1 =17 Clave E

04 (x+3)n

Coeficientelugar9=Coeficientelugar10

C8n · 38 = C9

n · 39 C8n=3C9

n

C8n=3 (n–8)

9 · C8

n n=11

Clave E

05 Sia Z a 0;5

ti+1 = Ci7(x9)7–i

1x3

i a(7–i)=3i

Sia 0;5 a=4 i=4

\Términoindependiente:C47 = 7!

4! 3!=35

Clave C

06 ti+1 = Ci10(x2)10–i

1x2

i

20–2i = 2i i = 5 \ C510 = 252

Clave B

07 (x3 + y5)12

t8+1 = C812 · (x3)4· (y5)8 t9=495x12· y40

\Coeficiente=495grado=52

Clave A

08 Sibespar;términocentral

tb/2+1 = bCb/2

xa

yb–5

b2

yb

x

b2 = ax3y15

b2

(b–b+5)=15 b=6

b2

(a–1)=3 a = 2 \ ab=12

Clave B

09 (x2–x–3)55 ti+1 = Ci55 · (x2)55–i · (x–3)i

2(55–i)=3i 55 · 2 = 5i i = 22

\Lugar:i+1=23Clave C

10

x

13 + x

30

tk+1 = Ck30 ·

x

13

30–k· ( x)k

tk+1 = Ck30 · (x)

k2

30 – k3

–

Términosracionales:

k2

30 – k3

–

5términos

Z

Clave B

Cuaderno de TraBaJo CaP 08CANTIDADES IMAGINARIAS

01 M = –9 + 2 –16– –1

M=3i+8i–i M=10iClave E

02 E=i + i2 + i3 + i4 + ... + i4k + 2

E=i + i2 + i3 + i4 +... ... ...+ i4k + i4k+1 + i4k + 2

0 00 ... E=i–1

Clave E

03 M = i573 + 5i1259 +3i1415

i781 =

i + 5i3 +3i3

i=–7

Clave B

04 Y=(1+i)48–(1+i)49

Y = 224 · i24–224 · i24(1+i)

Y = 224(1–1–i)=–224iClave A


4 5

05 A=(1+i)2

i21–(1–i)

2

i9 A=

2ii––2ii=4

Clave D

06 R = 1+

1+1+i1–i

1+i1+i =

1+1+i1–i

1+i = 1+i1–i

= i

Clave D

07 A=

1+i1–i

4 +

1–i1+i

3 +

1+i1–i

2

A=i4+(–i)3 + i2 = iClave D

08 M3 = 2i 172 2i i− = 2i 22 (1 ) i i− +

M3 = 2i 2i = 2i(1+i)=(1+i)3

\M=1+iClave A

09 S = i129 + i73 + i251

2i75 –i81 =

i + i +(–i)2i –i

=1

Clave A

10 M=16

1+2i4–2i

4 + 4+2i1–2i

=16

i2

4 + 2i

\ M = 2i+1Clave A

Cuaderno de TraBaJo CaP 09NÚMEROS COMPLEJOS

01 x–2y + xi–yi = 2 + 5i

(x–2y) + (x–y)i = 2 + 5i x=8 y=3

\ x + y=11Clave C

02 2 + ai1–i

= m 2 + ai = m–mi a=–2

Clave B

03 z=4–3i z=4+3i

\ |z|=|z|= 42+32 = 5Clave C

04 Corrija: Alternativa A) 390

z=(3+4i)(5–12i)(2 2 + i)(1+ 3i)

|z|=|3+4i||5–12i||2 2 + i||1+ 3i|

|z|=(5)(13)(3)(2)=390Clave A

05 Z = 1+ai1–ai

= mi 1+ai = mi + am

n = a am=1 a2=1 a = 1

Clave C

06 a + 2ib+3i

= m b+(8+a)ia + bi

= ni

a=–4 b=–6 \ m · ni=– 23

i

Clave C

07 1+i = (x + yi)2

1+i = x2 + 2xyi–y2 = (x2–y2) + (2xy)i

\E=x2 –y2

xy = 1

12

= 2

Clave E

08 Z = a + bi 1–Z1+Z

=1

|1–a–bi|=|1+a + bi|

(1–a)2 + b2=(1+a)2 + b2

a = 0 Z = bi

I. (F) Z=1+i II. (F) Z = 2

III. (V) Z = biClave A

09 Z=cos6°+isen6°=e6i Z15 = e90i

|Z90i|=|cos90°+isen90°|=1

Clave A

10 Z = (1+itanq)7

cos7q–isen7q = cosq7

cos7q ·

(1+itanq)7

cos7q–isen7q

(isenq+cosq)7

(isen7q+cos7q) · sec7q =

sec7q (eqi)7

e7qi

\sec7qClave E

Cuaderno de TraBaJo CaP 10ECUACIONES

01

x+12

+1

2 =

x+13

+1

3 x+3

4 = x+4

9

\ x=–2,2Clave D

02 xa–a

b–1=b

a–x

b+1 x

1a

+ 1b

= ba

+ ab

+ 2

x (a + b)ab

= (a + b)2

ab x = a + b

Clave B

03 6x#denaranjas

3x x 2x 2x = 20 Juan Pedro Resto x=10

\6x=60Clave C

04 Corrija: Alternativas D) 36y E) 21

Padre:P

Hijo:H

Hoy +12

P=6H P +12=2(H+12)

P=18 H=3 \18+3=21

Clave E

05 x2 + x+1=0 x1;2 = 2

–1 –3

A)(V) B)(V) C)(V)

D)(F) E)(V)Clave D

06 Para:m + n + p = a m + n + q = b

x–ab

+ x–ba

= 2xa + b

xb–a

b + x

a + b

a = 2x

a + b

x1b

+ 1a– 2

a + b

= a

b + b

a

x = a + b = p + q + 2(m + n)

\E=p + q + 2(m + n)–2(m + n) = p + q

Clave C

07 A={2}B={1;2}AB={2}

Clave D

08 f(x) = x2 + (n+1)x + n = 0

Six=–2:(–2)2 + (n+1)(–2)+n = 0 n = 2

f(x) = x2+3x + 2 x1=–2 x2=–1

\ x2+2=1 Clave B

09 x + 2– xx + 2 + x

= a x + 2x

= a+11–a

xx + 2

= (a–1)2

(a+1)2 x = (a–1)

2

2aClave D

10 1–13

+ 13–1

5 + 1

5–17

+...+ 12n – 1

– 12n + 1

= 2nn + 9

1– 12n + 1

= 2nn + 9

2n2n + 1

= 2nn + 9

n=8

P(x)=16x2+8x–8=(2x–1)(x+1)

\ x1 = 12

x2=–1 Clave A

Cuaderno de TraBaJo CaP 11ECUACIONES DE 2° GRADO CON UNA INCÓGNITA

01 Corrija: Alternativas D) Las raíces son imaginarias

P(x) = x2–2x+1=(x–1)2 = 0 x=1

\Productoderaíces=1Clave B


55

02 x2 + x–1=0 –13

x2+8x–8=0 –23

x2+27x–27=0–33

x2+103x–103 = 0 –103

\–103=–1000Clave C

03 3k2x2–6kx–(k + 2) = 0 k 0

x1 + x2 = 2x1x2 6k3k2 = –2(k + 2)

3k2 k=– 12

Clave C

04 (2k + 2)x2+(4–4k)x + (k–2)=0

x1 = 1x2

x1x2=1 k–22k–2

=1 k=–4

x12 + x2

2 = (x1 + x2)2–2x1x2 = 102

32 –2(1)

\ x12 + x2

2 = 829

Clave D

05 x2–6x + c=0 •a + b=6 • ab = c

\ a2 + b2 + 2c

9 = (a + b)2

9 = (6)

2

9=4

Clave D

06 •(5a–2)x2 + (a+1)x + 2 = 0

•(2b–1)x2 + 5x+3=0

\ a+15

= 23

a = 73 Clave D

07 2x2–(k+1)x+8=0

Raícesiguales:D = 0

D = (k+1)2–4(2)(8)=0

\ k=7 k=–9Clave B

08 P(x) = x2 + mx+3

x1 = 2

–m + m2–12 =1+ –2 m=–2

\2–m=4Clave D

09 3x2 + (m + 2)x + (m + 2) = 0 D = 0

D = (m + 2)2–4(3)(m + 2) = 0

m=10 m=–2

Máximovalordem:+–+

–2 10 \ mmáx = 9

Clave A

10 5x2 + bx + 20 raíces:r1yr2 = r1+3

(r1)(r1+3)=205=4 r1=1 r2=4

•r1 + r2=5=– b5

b=–25

\ r1 + r2–b=5–(–25)=30 Clave D

Cuaderno de TraBaJo CaP 12ECUACIONES DE GRADO SUPERIOR

01 x3–x = 0 x(x2–1)=0 x = 0 x2=1

x1=0;x2=1;x3=–1

\ x12 + x2

2 + x32 = 2

Clave B

02

–2–210–281–3428

2–5140

x3–3x2+4x+28=(x + 2)(x2–5x+14)

\ x2–5x+14Clave C

03 2x3–x2–2x + a = 0

121–12–1–2+a

21–10a=1

(x–1)(2x2 + x–1)=(x–1)(x+1)(2x–1)

Clave B

04 x3 + 0x2 + 0x–10=0

\Sumaderaíces=0Clave C

05 2x3–9x–3x2 = x(9x–3x2–4x)

5x3 + x2–18x = 0

\Sumaderaíces=– 15=–0,2

Clave A

06 x3–6x2 + x+30=0

x1 = 5, x2 = ?, x3 = ?

x1 + x2 + x3=6 x2 + x3=1Clave C

07 x3–ax2 + ax–1

Six1 = 2 + 3 x2=2– 3

•x1· x2 · x3=1

(2 + 3)(2– 3)(x3)=1 x3=1

\ x1 + x2 + x3 = a 5 = aClave E

08 x2–6x2+6+3–4 x2–6x+6

a2+3–4a = 0 a2–4a+3=0

(a–3)(a–1)=0

• x2–6x+6=3 • x2–6x+6=1

x2–6x+3=0 x2–6x + 5 = 0

x1 + x2=6 x1' + x2'=6

\6+6=12Clave C

09 x3–30x2 + 0x + (m+1)2 = 0

x1 = n–3rx2 = n–rx3 = n + rx4 = n+3r

x1 + x2 + x3 + x4 = 0

4n = 0

n = 0

x1x2 + x1x3 + x1x4 + x2x3 + x2x4 + x3x4=–30

3r2–3r2–9r2–r2–3r2+3r2=–30 r2=3

x1x2x3x4 = 9r4 = (m+1)2 (m+1)2 = (9)2

m=8 m=–10

\8–10=–2Clave A

10 x3–px2 + qx–r = 0

Raícesdea, byc a + b + c = pab + bc + ac = q abc=–r

\ b2c2 + a2c2 + a2b2

a2b2c2 = q2 + 2pr

r2

Clave A

Cuaderno de TraBaJo CaP 13INECUACIONES

01 A= 1–2x2

< x + 23

< 3–2x4

/ x Z

12–x < x

3 + 23

< 34–x

2

– 18

< x x < 110

\SumadeelementodeA=0

Clave A

02 n5+14 3n+24

4 20

13 n

n={2;3;...;75}

n+14

–29–10 n 75

\Sumadevaloresden=2849

Clave A

03 7–3x2

;–3 x<8–24<–3x 9

–17<7–3x 16

\– 172

< 7–3x2

8

– 172;8

Clave C

04 Six [–5;3 –5 x<3

–4 x+1<4– 14

1x+1

(1)

1x+1

> 14

(2)

De(1):– 24

2x+1

32

2 + 2x+1


6 5

De(2): 2x+1

> 24

2 + 2x+1

> 52

\

–;32

52;+

Clave E

05 x2 + x+1 x + 50 < x2–3x + 50

x2 + x+1 x + 50 x + 50 < x2–3x + 50

x2 49 0 < x(x–4)

–7 x x 7 x < 0 x >4

[–7;0 4;7] \4 · 7=28

Clave C

06 x2–2bx–c < 0 C.S.=–3;5

(x+3)(x–5)<0 x2–2x–15<0

\ b + c=1+15=16Clave A

07 3x(x+1)2–2x 3x2 + 5x–2 0

(3x–1)(x + 2) 0

a=–2 b=3+–+

–2 1/3 \ ab=–6

Clave D

08 3x2–5x + A 13x2–5x + (A–1) 0

D < 0 52–4(3)(A–1)<0 A 3,08 Amín=3

Clave B

09 A=–{1} B= C={2}

D = – – 15

[(AB)–D]C

[(–{1})–D]C

\ – 15;2

Clave B

10 x1+x + 2 1– x 0

2 x + x + x x–2 x + x 4 x 0 (1)

1– x 0 1 x (2)

De(1)y(2):x [0;1]Clave D

Cuaderno de TraBaJo CaP 14INECUACIONES DE GRADO SUPERIOR

01 4x–1x+3

–2 9x+3

x –3

4x–1–2x–6x+3

– 9x+3

0

2x–16x+3

0

P.C.={–3;8}

+–+–3 8

\C.S.=–;–3 [8;+ Clave D

02 (x2–x–2)(1–x) 0

(x–2)(x+1)(x–1) 0

\C.S.=[–1;1][2;+Clave E

03 (x)(x + 5)19(x–3) 0 – ++––5 0 3

\C.S.=[–5;0][3;+ Clave A

04 (x+1)2(x2+1)(x+3)7

x(x–2)5 0

\C.S.=–;3] 0;2 {–1}

Clave A

05 x4–4x3+4x2–4x+3>0

(x2+1)(x–1)(x–3)>0 +–+1 3

\C.S.=–;1 3;+Clave D

06 (x–1)(x–2)

(x–3)(x+1) 0

\C.S.: – ++ +––1 1 2 3

–1;1][2;3Clave C

07 1x+1

–1x>1C.S.={a;b}

x–(x+1)

(x)(x+1) –1>0C.S.=–1;0

\2012(a)+2014(b)=–2012Clave B

08 2x–1x+1

1 x–2x+1

0

\C.S.=–1;2]Clave D

09 • 6x–1

+ 51–x

<–2 • 1x–1

+ 2 < 0

•2x–1x–1

< 0 +–+1/2 1

\Ningúnvalorentero.Clave A

10 •1–xx+1

–1x>0 •

–x2–1(x)(x+1)

> 0

•(x2+1)

(x)(x+1) < 0 +–+

–1 01

\Longituddelconjuntosolución=1

Clave A

Cuaderno de TraBaJo CaP 15ECUACIONES E INECUACIONES CON VALOR ABSOLUTO

01 |18–3x|=6

18–3x=6 18–3x=–6 x1=4 x2=8

\ x1· x2=32 Clave E

02 ||x–3|–5|=2

|x–3|–5=2 |x–3|–5=–2

|x–3|=7 |x–3|=3

(x–3)=7 (x–3)=–7 (x–3)=3 (x–3)=–3

x1=10 x2=–4 x3=6 x4 = 0

\ a + b + c = x1 + x3 + x4=16Clave B

03 |x–a + b|=|x + a–b|

x–(a–b)=–x–(a–b) 2x = 0 x = 0

x–(a–b) = x + a–b a = b

Clave D

04 • x / 25

x 5 x

25;5 =A

•{x /|x+1|=3x} x 12;– 14

=C

\AC= 12

Clave C

05 2 x + 12

2–7 x + 1

2 +6=0

2 x + 1

2 –3

x + 1

2 + 2

0

= 0

x + 12

= 32

x + 12=– 3

2 x=1 x=–2

Clave A

06 •Six 3 x2–6x + 2x–x+3–6<0 x2–5x–3<0

2

5 + 37

x–

25– 37

x– < 0

x [–0,54;5,54]=2elementos

•Six<3

x2–6x + 2x + x–3–6<0 x2–3x–9<0

2

3+3 5

x–

23–3 5

x– < 0

x [–1,85;4,85]=6elementosenteros

\#deelementos:n=6 2n = 26=64

Clave E


75

07 Corrija: Alternativa E) – 32

[1;+

|3x+2|–|x–1|=2x+3

I II

– 32

1+– +

+–

C.S.= – 32

[1;+Clave E

08 1

|x|–3 +

1|x|–4

< |x|–12

x2–7|x|+12

Sea|x|= a

(a–4)+(a–3)

(a–3)(a–4)–

(a–12)(a–4)(a–3)

< 0

(a + 5)

(a–3)(a–4) < 0

(|x|+ 5)

(|x|–3)(|x|–4) < 0 3<|x|<4

C.S.:–4;–3 3;4Clave C

09 |x–6+|x–5|+|4–x|| 3–x

Como3–x 0 x 3

|x–6+5–x+4–x| 3–x

|–x+3| 3–x |x–3| 3–x

3–x 3–x x2–6x + 9 3–x x2–5x+6 0 (x–3)(x–2) 0

x [2;3]

\ mn–1=23–1=7Clave E

10 (x2–4)2 > x2–4 |x2–4|>x2–4

0 < 0

(x2–4)<0 (x–2)(x + 2) < 0

\C.S.=–2;2Clave E

Cuaderno de TraBaJo CaP 16MATRICES Y DETERMINANTES

01 A=

61–20

yB=

2–154

2A+3B=

122–40

+

6–31512

=

18–11112

\18+11+12–1=40Clave E

02 M =

x 14 – 3x x

Si|M|=0

x2+3x–4=0 (x+4)(x–1)=0

\ xmín=–4 Clave B

03 a m ba m xx m b

= 0

0=–m(ab–x2) + m(ab–bx)–m(ax–ab)

0 = m(x2–(a + b)x + ab)

\ x1 + x2 = a + bClave C

04 SeaAmatrizsimétrica

•x + y=–3 •3y=–3 •x2 + z = 5

y=–1;x=–2yz=1 \ xyz = 2

Clave C

05 c 2c c 5a a 3bb + 5c b + d b + 3c

=–4

Operandoconfilasycolumnas:

= 20 c cb a 0c d b

=–4–2c 0 ca b 0d c b

=–4

\ c 0 ca b 0d c b

= 2

Clave C

06 A=1 4 k1 k 41 k k

ComoAesinvertible,entoncesA 0

A=0 k=4

\Aesinvertiblesik –{4}.

Clave C

07 A=

1–112–10100

,comoA3 =

100010001

= I

\(A100)=(A3)33· A=I · A=A

Clave E

08 Corrija: Alternativa D) 196

A=

–35–22

yB=

2–345

3x–6A=5B+2x–4A x=5B+2A

x =

4–51629

|x|=196

Clave D

09 Corrija: En III cambie A1 porAT.

I. (F) Paraquedosmatricessepuedan sumarorestartienenqueserdel mismoorden.

II.(V)A4 = 0 k=4

III.(F)NocumpleA=–AT

Clave C

10 A=

a b cd e fg h i

PAP=

a c bg i hd f e

M

P=(PA)–1· M P=A–1P–1· M

\ P =

–1101–10001 Clave C

Cuaderno de TraBaJo CaP 17SISTEMA DE ECUACIONES I

01 •x + y=500 •3x=7y

x=350 y=150

(k)(y)=1200 k=8Clave D

02 •ax + by=–11 •cx–dy=1

Seaxlasoluciónúnica:

(x1)(a + b)(x1)(c–d)

= –111

a + bd–c

=11

Clave E

03 Para 1x + y

= a 1x–y

= b

2b+3a = 2 a = 13

x + y=3

8b–9a=1

b = 12

x–y = 2

x = 52

y = 12

\ (2x–6y)5=(5–3)5=32Clave D

04 •y=–5 •x=–3

\ m = 2x–y m=–1Clave C

05 xyz = 2(x + y) = 65

(y + z) = 32

(x + z)

•(x + z) = 2xyz3

•(x + y) = xyz2

•(y + z) = 5xyz6

(x + y + z) = xyz

\ P = 8xyzxyz

= 2 2Clave E

06

a + b=94

a + c = 205

b + c=187

a + b + c=243

a=56 b=38c=149(máx)

\149Clave E


8 5

07

2x + y+3z = 5

3x + y + z = 0

x+3y + 2z=6

x=–1

y=1

z = 2

\ x02 + y0

2 + z02=6

Clave E

08 (1)–(2):x–5y– z = 2 I. (V)

(1)+(2):3x+3y– 3z=0(3) II. (F)

(1)+(3):5x + 2y + 4z=1 III. (V)

Clave C

09

•xz=6

•(x + y)x=103

•(x + y)z=102

x=3

y=7

z = 2

\ y=7Clave C

10 •x2–4x + y2=64

•x3–6x2+12x + y=8

•(x–2)2 + y2=68

•(x–2)3 + y=4

x1=0;y1=8

x2=4;y2=–8

\C.S.={(0;8);(4;–8)}Clave B

Cuaderno de TraBaJo CaP 18SISTEMA DE ECUACIONES II

01

2x + y = 5

x + y=1x=4 y=–3

\ xy=–12Clave E

02 I.B II.C III.AClave A

03 Corrija: En la gráfica, una recta pasa por (–3; 2) y (0; 3) y la otra por (–3; 2) y (0; 0).

Porlagráfica:

y = x3+3 y=– 2

3 x

Clave C

04 –(m + 2)

4 = 1–1

m = 2Clave B

05 Corrija: ¿Para qué peso el costo es igual en ambas compañías?

Compañía: "Compraenlínea" "Vida"

x+2=Costo1 x+5=Costo2

Siendox:pesoenkilogramos

\ Nuncaseránigualesyaquesonrectasparalelasdelcosto.

Clave E

06 (a + b)x + (a–b)y = 72

b

3(a + b)x+(3a–7b)y = a + 2b

•(a + b)3(a–b)

= (a–b)3a–7b

= 7b

2(a + 2b)

•a

3a–5b = 1

2 a = 5b

\ H = 11b+3(5b)

13b = 2

Clave A

07 (m2–17)x–(6–n)y=16 nx + 2y=1

m2–17

n =

n–62

= 161

n=38 m = 25

\ m + n=63Clave B

08 (a–1)x + 2y=7b 2x + (a–1)y=14

a–1

2 =

2a–1

= 7b14

a=3 b = 2

\ a + b = 5Clave E

09 32

= 6a

a1

a=4Clave E

10 y + mx=2(1) x + y=10(2)

y + my=3(3)

(1)+(3):(m+1)(x + y

10

) = 5 \ m=– 12

Clave B

Cuaderno de TraBaJo CaP 19SISTEMA DE INECUACIONES I

01 y+3 2x

A)(3;1) B)(1;2)

C)(2;1) D)(2;–2)

E)(2;–3)Clave B

02 y 5–x

5

5y

x Clave A

03 2x + y>5–x

y

x y>5–3x

Clave D

04 5x 4y

y

xClave C

05 x 4 y>–1Clave D

06 I.c II.a III.bClave E

07 2x–3y<6 x 0

y

x 2x–6<3y x 0

Clave E

08 2x + 5y 10 y<4

y

x y 2–2x5

y<4

Clave C

09 y<3+|x|

y

x

Clave E

10 y –|x–1|+2

3

2

–1

y

xClave A

Cuaderno de TraBaJo CaP 20SISTEMA DE INECUACIONES LINEALES CON 2 INCÓGNITAS

01 Corrija: Cambie y–x<3por y + x<3

3

3

y

x Clave D

02 y + 2x 4 (I)

y–x > 0 (II)

2

4

III

III

y

x x 0 (III)

Clave B

03 I. 3x + y 6 II. y–2x 1

–2

4

III

IV

II

I

y

x III. x –2 IV. y 4

Clave C


95

04 •y < 2x–3

•y<12–3x

3

12 Soluciones positivas

43–3

y

x32

\(3;1),(3;2)

Clave A

05 y<–x+4

43

4

–4

y

x y x+4

0 x<3

Clave D

06

3y–2x>5(1) x + y > 5 (2)x + 2y<11(3)

(1)+2(2):y>3 (4)

De(4)y(2):x 2

en(3):3<y<4,5

\ y=4 x = 2 yx xx + y = 23=8

Clave B

07

\7soluciones

–2

–2

(1; 1)

(2; 0)

y

x

Clave C

08 x–y 3 x + y 3 x–y –3

–3

–3

3

3

y

x

x –3

\27Clave B

09 x + y>76 x–y<10 x + 2y<112

Soluciónentera(43;34)

76

76 11210

–10

56

\ 2m–5=–3Clave A

10

2

y

xiiiii

i

Clave E

Cuaderno de TraBaJo CaP 21PROGRAMACIÓN LINEAL

01 •f(5;3)=2(5)+3=13máx

•f(0;0)=0mínimovalorClave A

02 Z(x;y)=3x + 2y

1

1

y

x máxZ(1;0)=3

Clave C

03 4x + y 240 2x+3y 420 x+3y 300

300

240 (1)

(2)(3)

(4)

y

x x 0;y 0

Mínf(x;y)=6x+8y

1.f(0;240)=1920 2.f(30,120)=1140

3.f(120;60)=1200 4.f(300;0)=1800

\Siendo,mínimovalor=1140

Clave D

04 x=vacas y=carneros

x + y 30 10x +15y 400

Ganancia=15x + 20y 30

30 24

26,6 (10; 20)

\Siendolaganancia máxima=550.

Clave E

05 Vértices:(1;2),(3;6),(4;3),(3;0),(4;1)

•Smáx = S(3;6)=–15 •Smín = S(3;0) = 9

\–15 9Clave C

06

Traje VestidoTela x 2y 80Lana 3x 2y 120

x + 2y 803x + 2y 120

\20y30Clave E

07 Corrija: Alternativa E) 6336

50x+30y 1200 x=18

x + y 28

y=10

Máximo:252x+180y \6336

Clave E

08 x=económicosy=supereconómicos

30000x + 20 000y 1800000 (I)

x + y 80 (II)

Beneficio:4 000x+3 000y

De(I)y(II):x = 20 y=60Clave D

09 Six:hectáreasdemaíz, y:hectáreasdetrigo

•2x + y 800 x=320

•x + y 480

y=160

Umáx:40x+30y=40(320)+30(160)=17600

Clave C

10 x:texto1 y:texto2

•4x + 5y 200 x = 25

•6x+3y 210

y = 20

Umáx:2x+3y=2(25)+3(20)=110

Clave B

Cuaderno de TraBaJo CaP 22FUNCIONES

01 f(x) = ax + b

f(4)=4a + b=7 a=6

f(3)=3a + b=1

b=–17

\ f(x)=–29Clave C

02 20;0 t 60

Ct20+0,4(t–20);60<t

88=20+20+0,4(t–20)48=0,4(t2–20)

t2=100 t1=60

\ t1 + t2=160 Clave C

03 f(x) = 12–x–x2

|2x–5|

Dominio

12–x–x2 0

x2 + x–12 0

(x–3)(x+4) 0

2x–5=0 x = 52

\[–4;3]– 52

Clave D

04 f(x) = m(x–2)2–p

f(0)=4m–p = 0 m=3

f(1)=m–p=–9

p=+12

\ m + p=15Clave D

05 f(x) = x2+4x+1 f(x) = (x + 2)2–3

Mínimovalor=–3

Máximovalor=33

\Rangof=[–3;33]Clave D

06 G(x)=–4x2+8x G(x)=–4x2+8x –4+4

G(x)=4–(2x–2)

0

2

\MáximovalorG(x)=4Clave D


10 5

07 f(x) = (x2–1)

0

1/2

x2–1 0 (x–1)(x +1) 0

\Dominiof(x) = –;–1][1;+

Clave E

08 f(x) = –x2

x2+1=1–

1x2+1

1

Mínf(x) = 0

Máxf(x)=1yaquex + 1

x2+1 @ 0

\Rangof=[0;1Clave C

09 f(x) = x2–8–1+1

x2–9 = 1

x2–9

Mínimo valor(x = 0)

1+

f(0) = 89

\ –;89

Clave A

10 f(x–2 x) = 2(x–4 x);x 4

Domf = [4;+ Ranf = [–8;+

\Ranf Domf = [4;+Clave D

Cuaderno de TraBaJo CaP 23TRAZADO DE GRÁFICOS

01 Corrija: Alternativa D) I,IIyIII

Los tresgráficoscorrespondena funcio-nes.

Clave D

02 Porlagráfica,eldominioes:

\[–5;–1 1;4Clave A

03 Tienependientepositiva f(x) = x+4

Clave C

04 Haydesplazamientoenelejeyde1,seveelespejo,esoindicavalorabsoluto.

\ f(x) =|x|+1Clave E

05 f(x)=–x2+6x

f(x)=9–(x–3)2, 0 < x <6

\Rangof = 0;9]Clave A

06 Seaf(x) = x2–(x0 + y0)x + x0y0

f(0) = 2 x0y0 = 2

x0 + y0

2=3 x0 + y0=6

\ f(x) = x2–6x + 2Clave A

07 y

x

f(x) =||x+1|–2|

Clave D

08 Seaf(x) = a(x–3)2–4

f(0) = 5 a(–3)2–4=5 a=1

\ f(x) = (x–3)2–4=x2–6x + 5

Clave E

09 Deacuerdoalagráfica:

\Dominio:–;2 2,5;+

Clave B

10 Seaf(x) = a(x–1)2–1

f(0) = 0 a(–1)2–1=0 a=1

f(x) = (x–1)2–1

•f(3)=3 •f(2)=0 •g(2)=3·2–0=6

\ f(3)+g(2) =3+6=9Clave C

Cuaderno de TraBaJo CaP 24FUNCIÓN PAR E IMPAR

01 Funciónpar:f(x) = f(–x)

I. f(x) = x2 = f(–x) (V)

II. g(x) g(–x) (F)

III. h(x) = h(–x) (V)

\SoloIInoespar.Clave B

02 Funciónimpar:f(–x)=–f(x)

I. f(–x)=–x=–f(x) (Sí)

II. g(–x)=(–x)3=–x3=–g(–x) (Sí)

III. h(–x)=–x+1–h(x) (Sí)

III. I(–x)=(–x)2+3=x2+3–I(x)(No)

\2sonfuncionesimpares.Clave C

03 a-2;b-3;c-1Clave E

04 Periodo:|7–3|=4 \4

Clave B

05 G(x) = x

x3 + x3 + x

G(–x) = –x

–x3–x3 –x = G(x)

I. G(x) = G(–x) (V)

II. G(–x) –G(x) (F)

III.Esparperonoimpar (F)

\SoloI.Clave A

06 T=3periodo k=3T k=12

Clave B

07 •f(x) = x

x3+1 +

xx3–1

•g(x) = x3 + x3

I. f(x) = f(–x) fespar (F)

II. g(x) –g(–x) gesimpar (F)

III. fespar (V)

\ FFV.Clave D

08 Funciónpar1

Funciónimpar 2

a-1;b-2;c-2 \122Clave B

09 f(x) = 2x2–kespar f(–x) = f(x)

2(–x)2–k = 2x2–k k

I. (V) II. (F)

III. (V)Sik = x2 f(x) = x2par

\SoloIyIII.Clave E

10 f(x)=4x2–2x + kespar

f(x) = f(–x) 4x2–2x + k=4x2–2(–x)–k

2k=4x k = 2x \ 2x

Clave D

Cuaderno de TraBaJo CaP 25FUNCIONES ESPECIALES

01 Funcióninyectiva crecienteodecreciente

f(x1) = f(x2) x1 = x2

\SoloIClave A

02 •Función inyectiva:Paraunvalordex hayunsolof(x).


115

•Funciónsuryectiva:Todoelementodelconjuntodellegadaesimagendealme-nosunelementodeldominio.

•Funciónbiyectiva:Tienequeserinyec-tivaysuryectivaalavez.

\SoloIClave A

03 Elgráficodealternativa B)correspondeauna

y

x funciónbiyectiva.

Clave B

04 •f(x)=3x+1 y=3y–1+1 y–1 = x–13

•G(x) = (x + 2)2–3 y = (y–1 + 2)2–3

y–1 = x+3–2Clave A

05 Segenerareflejando respectoalarecta y = x.

y

x

Clave E

06 f •funciónbiyectiva •funcióndecreciente

f(1)=–3

f(–2)=2a = 11

5 b = 3

5

\ 25

115

35=33

Clave E

07 I. F(x) Noesinyectiva

y

x

II. H(x) = 3(x–1)2; x –;–1 H(x) =|x – 1| 3para esedominio la funciónesinyectivadecreciente.

III. G(x) = x3

Funcióninyectiva

y

x

\IIyIIIClave E

08 I. (V)

II.(V)Comoestadefinidaen[–2;4 escontinuaen[–2;2 essobreyectiva.

III. (F) (x)(x–2)(x + 2) Funciónimpar ynoesunivalente.

Clave B

09 P(x) = ax2 + bx + c

P(1)=–2

P(2)=3 a = 43

, b =1,c=– 133

P(5)=34

P(x) = 43

x2 + x–133

P(x) = 0

\ x1 = 8

–3+ 217Clave B

10 f(x)=3 + 11

x–2

I. fesfuncióndecreciente, alavezinyectiva.(V)

y

xf3

2

II.Paratododominiohayunrango.(V)

III.Siunafunciónesinyectivaysuryecti- vaalaveztieneinversa. (V)

Clave A

Cuaderno de TraBaJo CaP 26ÁLGEBRA DE FUNCIONES

01 Losquecumple:

f = {(2;18),(3;6),(5;4)}

g = {(2;3),(3;2),(5;0)}

\Ran(f + g)=21+8+4=33

Clave E

02 f={(6;1),(3;5),(2;4)}

g = {(4;3), (2;2),(5;1)}

\Ran(fog)=5+4=9Clave C

03 •F(x) = (x + 5)(x–5)=x2–25;x

•G(x) = x2–55 x2–25;x

I. (V) II. (F) III. (V)Clave E

04 •F(x) = x–2 •G(x) = x–4

I. (F) II. (V) III. (F)Clave D

05 F={(2;5),(3;6),(4;7),(5;8)}

G={(2;1),(3;2),(4;3),(5;4)}

\ F + G={(2;6),(3;8),(4;10),(5;12)}

Clave A

06 •F(x) = x–5DomF:[5;+

•G(x) = 9–x DomG:–;9]

\Dom(F · G)=DomF DomG=[5;9]

Clave D

07 F={(3;5),(5;7),(7;9),(9;11)}

G={(5;2),(7;4),(9;6),(11;8)}

\ G(F(x))={(3;2),(5;4),(7;6),(9;8)}

Clave C

08 •F(x) = x2+1;x –2;5] •G(x) = 2x;x 0;6]

F–G = x2–2x+1=(x–1)2

Dom(F–G)=DomF DomG = 0;5]

\ (F–G)(x) = (x–1)2;x 0;5]

Clave B

09 Corrija: Alternativa B) VVV

•F(x) = 1

x–2 •G(x) =

x–2x–2

I. (V) II. (V) III. (V)

Clave B

10 •F(x) = 5 x–2DomF:[2;+

•G(x) = 2

5–2x4DomG:

–;52

\Dom(3F –2G)=DomF DomG

=

2;52

Clave B

Cuaderno de TraBaJo CaP 27LOGARITMOS

01 3log2x–4log2x=1log2x=–1 x = 12

Clave B

02 •45 = a2b3

16•b=42 b=16•a = 2

\ (a+1)2=32 = 9Clave E

03 E=1

logr(pq) + 1 +

1logq(pr) + 1

+ 1

logp(rq) + 1+1

E=1

logr(pq) + logrr +

1logq(pr) + logqq

+ 1

logp(rq) + logpp+1

E=1

logrpqr +

1logqpqr

+ 1

logppqr+1

E=logpqrr+logpqrq+logpqrp+1

E=logpqrpqr+1E=1+1=2

Clave E


12 5

04 Dato:logbxb = 1x

logbb = 1x

M=1+2+3+4+...+20

M = 20(21)

2=210

Clave A

05 •Log2(2x)=log2(y) x=log2y

•x2–2(2–x) = 0 (x–2)2 = 0

x = 2 y=16

\ (x + y)=16+2=18Clave C

06 SiLn(6)=a 2Ln

32

= 2b

Ln(3)–Ln(2)=b Ln(3)=a + b

2

Ln(3)+Ln(2)=a

Ln(2)=a–b

2

\Ln(3) · Ln(2)=a2–b2

4 Clave D

07 1–logx(27)1/2 · (log3x)=1 1–32

k = k

k2 + 52

k–1=0Logx3=–2(*)

Logx3=12

(**)

Solo(**):x = 3Clave E

08 •log(x2 + y2)=log130

•log(x + y)=log[8(x–y)]9y=7x

x2 + y2=130 x = 9 y=7

\ x · y=63Clave A

09 •2x = x + y (*)

•log(x + y)+log3x=log27+log37

En(*):log2x+log3x=log(2 · 3)7

log(2 · 3)x=log(2 · 3)7 x=7 y=121

\log(y–3x) = 2Clave C

10 6+log2x2=(log2x)2+3

(log2x)2–2(log2x)–3=0

x=8 x = 12

\ x=8

Clave A

Cuaderno de TraBaJo CaP 28FUNCIÓN EXPONENCIAL Y LOGARÍTMICA

01 f(t) = 5 · 2t+4 f(5) = 5 · 29=2560

Clave B

02 Corrija: f(x) = 2x–1

Sif(x) = 2x–1

\ M = 2x + 2–2x–1

2x =4–12

= 72=3,5

Clave E

03 (|x|+1)2x2–5x + 2 > (|x|+1)14

2x2–5x–12>0 (2x+3)(x–4)>0

\C.S.= –;– 32

4;+Clave C

04 21/x–2x 0 21/x 2x x 0

log221/x log22x 1x

x (x2–1)

x 0

C.S.=[–1;0 [1;+ a=1

\ 2a+1=3Clave E

05 f(x) =

2x–3x + 5

Ln

2x–3x + 5

1 x–8x + 5

0 +–+–5 8

\C.S.:–[5;8Clave E

06 Corrija: Alternativa E) –;– 32

3;+

log3|4x –3|>log39 log3

4x–39

> 0

4x–39

>1 3–4x9

>1 x>3 x<– 32

\C.S= –;– 32

3;+Clave E

07 f(x)=|logx–5|+|1+logx|

•Silogx<–1 f(x)=logx+5–1–logx

f(x)=4–2logx –2logx > 2

4–2logx >6 f(x)>6 (1)

•Si–1logx < 5

f(x)=–logx+5+1+logx=6 (2)

•Silogx 5 f(x)=logx–5+1+logx

f(x)=2logx–42logx 10 2logx–4 6 f(x) 6 (3)

\De(1),(2)y(3):Ranf=[6;+

Clave A

08 f(x)=ln(–x2+3x+10)+ x–1x2(x–4)

6

f(x)=ln((x–5)(–x–2))+ x–1x2(x–4)

6

Dominio:

•(x–5)(x + 2) < 0 C.S.:–2;5

•x–1

x2(x–4) 0 C.S.:+;1][4;+

Intersectandosoluciones:–2;1][4;5

\Productodevaloresenteros: (–1)(0)(1)(4)=0

Clave C

09 x +log(1+2x) < xlog5+log6

log[(10x)(1+2x)]<log[5x · 6]

5x · 6>10x(1+2x) 5x[6–2x–22x] > 0

\C.S.:–;1Clave D

10 3x–1·3 1

22x

x−

+ =13x–1·2( 1)

22x

x−+ =1

\ x=1Clave A

Science

Solucionario c.t. álgebra 5°