Solucionario c.t. álgebra 5°
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Transcript of 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
ProYeCTo InGenIo SOLUCIONARIO - ÁLGEBRA 5°
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
ProYeCTo InGenIoSOLUCIONARIO - ÁLGEBRA 5°
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
ProYeCTo InGenIo SOLUCIONARIO - ÁLGEBRA 5°
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
ProYeCTo InGenIoSOLUCIONARIO - ÁLGEBRA 5°
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
ProYeCTo InGenIo SOLUCIONARIO - ÁLGEBRA 5°
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
ProYeCTo InGenIoSOLUCIONARIO - ÁLGEBRA 5°
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
ProYeCTo InGenIo SOLUCIONARIO - ÁLGEBRA 5°
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).
ProYeCTo InGenIoSOLUCIONARIO - ÁLGEBRA 5°
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
ProYeCTo InGenIo SOLUCIONARIO - ÁLGEBRA 5°
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