evaluaciones-2015-0 - Maria Cristina Rodriguez Escalante

Vista previa del material en texto

Universidad del Pací!co
Manual de imagenLogotipo institucional
Primera Práctica Calificada
Matemáticas I Lunes 12 de Enero de 2015
Verifique que son 4 preguntas, 20 puntos. Duración: 100 minutos.
Está estrictamente prohibido el uso de cartucheras, calculadoras o notas y el préstamo
de materiales.
No hay consultas; si considera que alguna pregunta está errada o mal propuesta corrija
el enunciado y justifique su proceder.
Justifique su respuesta.
Son importantes el orden y la claridad en la presentación de su trabajo, caso contrario
se pueden restar puntos o invalidar completamente la respuesta.
APELLIDOS:
NOMBRES:
SECCIÓN:
1 2 3 4 NOTA
SOLICITUD DE RE-CALIFICACIÓN
Motivos
1. Puntaje sumado erróneamente
2. Pregunta no corregida
3. La respuesta incluye aspectos
o perspectivas alternativas no
consideradas. (Explicar)
Reglamento de solicitud
Solo se aceptan solicitudes el d́ıa de entrega.
Si dos solicitudes son declaradas “No procedentes” en un mismo ciclo, el alumno no podrá pre-
sentar solicitudes adicionales durante dicho ciclo.
Cualquier solicitud habilita al docente a realizar una revisión integral de la evaluación. Como
consecuencia, la nota puede mantenerse, subir o bajar.
Toda solicitud debe estar justificada adecuadamente en base a sus conocimientos del curso.
Se considera “No procedente” aquellas solicitudes donde el alumno sugiere el puntaje que
debe tener según su propio criterio.
Resultado
PROCEDENTE
NO PROCEDENTE
NOTA ANTERIOR NUEVA NOTA
1. (5 puntos) Justificar la falsedad de las siguientes proposiciones
a) Sea I ⊂ R, I es un intervalo si ∀a, b ∈ I, ∀c ∈ R, [a < b < c→ c ∈ I].
Solución. Sea I = {1} ∪ [2,+∞[, entonces se cumple que ∀a, b ∈ I, ∀c ∈ R,
[a < b < c→ c ∈ I], sin embargo, I no es un intervalo.
b) Si A ⊂ B, entonces SupA < SupB.
Solución. Sea A = B = {1}, luego A ⊂ B, sin embargo, SupA = SupB = 1.
c) Si A ⊂ B ∪ C, entonces A ⊂ B o A ⊂ C.
Solución. Sea A = {1, 2}, B = {1} y C = {2}, luego A ⊂ B ∪ C, sin embargo,
{1, 2} 6⊂ {1} ni {1, 2} ⊂ {2}.
d) El máximo entero de -1.5 es -1.
Solución. Como −2 ≤ −1,5 < −1, entonces b−1,5c = −2.
e) Sea a ∈ R y A = {a,−a}, entonces R = {(a, a), (a, |a|)} es una función en A×A.
Solución. Sea a = −1, luego R = {(−1,−1), (−1, 1)}, que no es función.
2. a) (2 puntos) Escribir por extensión los siguientes conjuntos
A = {x ∈ R : x = 2n, n ∈ N} y B =
{
x ∈ R :
⌊
bxc+ π
⌋
= bxc+ x
}
Solución. El conjunto A por extensión es A = {2, 4, 8 . . . }. Para B resolvemos la
ecuación, por propiedad by + nc = byc + n, cuando n es un entero e y cualquier
real, como bxc es un entero, entonces⌊
π + bxc
⌋
= bxc+ x
bπc+ bxc = bxc+ x
3 = x,
B = {3}.
b) (1 punto) Sea U el conjunto universo, dar la definición de A ⊂ B.
Solución. ∀x ∈ U, si x ∈ A→ x ∈ B.
c) (2 puntos) Demostrar que
Si A ⊂ B, entonces A ∩B = A.
Solución. La inclusión A ∩ B ⊂ A. es cierta ya que todo elemento que está en A
y en B está justamente en A. La otra inclusión A ⊂ A∩B se da, ya que, si x ∈ A
por hipótesis también esta en B, luego todo elemento de A esta en A y en B.
3. a) (3 puntos) Verificar la validez o falsedad del siguiente argumento.
Si el delf́ın es un mamı́fero entonces toma ox́ıgeno del aire. Si toma ox́ıgeno del
aire, entonces no necesita branquias. El delf́ın es un mamı́fero y vive en el océano.
Por lo tanto, el delf́ın no necesita branquias.
Solución. Sean las siguientes proposiciones
p : El delf́ın es un mamı́fero.
q : El delf́ın toma ox́ıgeno del aire.
r : El delf́ın no necesita branquias.
s : El delf́ın vive en el océano.
Luego en lógica formal tenemos
p→ q, q → r, p ∧ s ` r.
Supongamos que las premisas sean verdaderas, demostraremos que r ≡ V, Como
p ∧ s ≡ V, entonces p ≡ V. De p → q ≡ V, tenemos que q ≡ V. De q → r ≡ V
tenemos que r ≡ V, lo que queŕıamos demostrar.
b) (2 puntos) Sean A = {1, 2, 3} y B = {4, 5}, si la siguiente relación en A×B
R = {(x, 4), (2, 4), (x, x2 − 5x+ 10), (3, 5)}
es una función. Hallar el valor o los valores de x.
Solución. Como (x, 4) y (x, x2 − 5x + 10) están en R que es función, entonces
x2 − 5x+ 10 = 4,
x2 − 5x+ 10 = 4
x2 − 5x+ 6 = 0
(x− 2)(x− 3) = 0
De donde x = 2 o x = 3.
Si x = 2, R = {(2, 4), (3, 5)} que es función, sin embargo, si x = 3 R =
{(3, 4), (2, 4), (3, 5)}, que no es función, luego x = 2.
4. a) (2 puntos) Dar el conjunto solución de
|3x− 1| < 2x+ 5.
Solución. Para que la inecuación posea solución 2x + 5 > 0, es decir x > −5
2
,
además −2x− 5 < 3x− 1 y 3x− 1 < 2x+ 5. Luego −4
5
< x y x < 6. Por lo tanto
C.S.=
]
−4
5
, 6
[
.
b) (1 punto) Enunciar la desigualdad triangular.
Solución. Para todo a, b ∈ R, |a+ b| ≤ |a|+ |b|.
c) (2 puntos) Demostrar que ∀a, b ∈ R∣∣∣|a| − |b|∣∣∣ ≤ |a− b|.
Solución.
|a||b| ≥ ab
−2|a||b| ≤ −2ab
a2 − 2|a||b|+ b2 ≤ a2 − 2ab+ b2
|a|2 − 2|a||b|+ |b|2 ≤ a2 − 2ab+ b2
(|a| − |b|)2 ≤ (a− b)2√
(|a| − |b|)2 ≤
√
(a− b)2∣∣∣|a| − |b|∣∣∣ ≤ |a− b|.
[ESTA PÁGINA PUEDE USARSE COMO BORRADOR O PARA COMPLETAR UNA
PREGUNTA INDICÁNDOLO DEBIDAMENTE]
[ESTA PÁGINA PUEDE USARSE COMO BORRADOR O PARA COMPLETAR UNA
PREGUNTA INDICÁNDOLO DEBIDAMENTE]
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Universidad del Pací!co
Manual de imagenLogotipo institucional
Práctica Calificada 2
Matemáticas I Lunes 19 de enero de 2015
Verifique que son 4 preguntas, 20 puntos. Duración: 100 minutos.
Está estrictamente prohibido el uso de cartucheras, calculadoras o notas y el préstamo
de materiales.
No hay consultas; si considera que alguna pregunta está errada o mal propuesta corrija
el enunciado y justifique su proceder.
Justifique su respuesta.
Son importantes el orden y la claridad en la presentación de su trabajo, caso contrario
se puede invalidar completamente la respuesta.
APELLIDOS:
NOMBRES:
SECCIÓN:
1 2 3 4 NOTA
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SOLICITUD DE RECALIFICACIÓN
Motivos
1. Puntaje sumado erróneamente
2. Pregunta no corregida
3. La respuesta incluye aspectos
o perspectivas alternativas no
consideradas. (Explicar)
Reglamento de solicitud
Solo se aceptan solicitudes el d́ıa de entrega.
Si dos solicitudes son declaradas “No procedentes” en un mismo ciclo, el alumno no podrá pre-
sentar solicitudes adicionales durante dicho ciclo.
Si un alumno solicita reconsideración por “Puntaje sumado erroneamente”, sólo se revisará la
suma de los puntajes obtenidos por cada pregunta.
Cualquier solicitud cuyo argumento sea diferente de “Puntaje sumado erróneamente” habilita
al docente a realizar una revisión integral de la evaluación. Como consecuencia, la nota
puede mantenerse, subir o bajar.
Toda solicitud debe estar justificada adecuadamente en base a sus conocimientos del curso.
Se considera “No procedente” aquellas solicitudes donde el alumno sugiere el puntaje que
debe tener según su propio criterio.
Resultado
PROCEDENTE
NO PROCEDENTE NOTA ANTERIOR NUEVA NOTA
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1. (6 pts) Conteste:
a) Justifique por qué las siguientes proposiciones son falsas:
El ĺımite de una sucesión cuyos términos son todos negativos siempre es un
número negativo.
Sea (xn) = (−
1
n
), entonces ĺım
n→∞
(− 1
n
) = 0
Dada la sucesión (an)n∈N =
(
8n
1− 2n
)
, la distancia entre el término a100 y el
ĺımite de la sucesión (an)n∈N es menor que una centésima.
Si ĺım
n→∞
(
8n
1− 2n
) = −4. Por definición: |a100 − (−4)| = |a100 + 4|
Se sabe que: a100 =
8(100)
1− 2(100)
→ | − 4, 02 + 4| < 1
100
Pero | − 4, 02 + 4| = | − 0, 02| = |0, 02| = | 2
100
| = 2
100
<
1
100
(��).
b) Dada la sucesión (an)n∈N =
( √
n
n+ 1
)
; demostrar que la sucesión es acotada.
∀n ∈ N :
√
n− 1 ≥ 0→ (
√
n− 1)2 ≥ 0→ n+ 1 ≥ 2
√
n→ 0 <
√
n
n+ 1
<
1
2
∴ (an)n∈N es acotada.
c) Sea (an)n∈N una sucesión tal que a1 = 2 y an+1 =
1
4
· an; probar que la sucesión
es monótona.
Dado que
1
4
< 1→ 1
4
· an < an → ∀n ∈ N : an+1 < an
∴ (an)n∈N es estrictamente decreciente.
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2. (4 pts) Usando el álgebra de ĺımites, calcular:
a) ĺım
n→∞
(n−
√
n+ 1
√
n+ 2) · (n+
√
n+ 1
√
n+ 2)
(n+
√
n+ 1
√
n+ 2)
= ĺım
n→∞
−(3n+ 2)
(n+
√
n+ 1
√
n+ 2)
= − ĺım
n→∞

3n+ 2
n
n+
√
n+ 1
√
n+ 2
n
 = −( 3 + 01 + (1)(1)) = −32
b) ĺım
n→∞
(
7 · 2n + π · 3n
3n
)
= ĺım
n→∞
(
7 · 2n
3n
+
π · 3n
3n
)
= ĺım
n→∞
(
7 · (2
3
)n + π(1)
)
= 7 · (0) + π = π
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3. (4 pts) Sea ϕ la ráız positiva de la ecuación x2 − x− 1 = 0.
a) Probar que ϕ3 = 2ϕ+ 1.
ϕ3 = ϕ2 · ϕ = (ϕ+ 1) · ϕ = ϕ2 + ϕ = (ϕ+ 1) + ϕ = 2ϕ+ 1
b) Se define la sucesión (an)n∈N con término general an = ϕ
n. Demostrar, por in-
ducción, que cada término de la sucesión - a partir del tercero - se puede obtener
sumando los dos anteriores.
Paso 01: Para n = 3, se verifica que an = an−1 + an−2;n > 2.
Paso 02: Asumiendo que el enunciado es cierto para P (k):
ak = ak−1 + ak−2
Debemos demostrar que P (k + 1) también lo es.
ak+1 = ϕ
k+1 = ϕk · ϕ = (ϕk−1 + ϕk−2) · ϕ = ϕk + ϕk−1 = ak + ak−1
∴ P (k + 1) : V
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4. (6 pts) Conteste:
a) La competencia entre las empresas de Telefońıa Móvil se ha intensificado a partir
del mes de enero de este año. Según la sección económica del diario E-Digital, se
proyecta que n meses después el precio de un smartphone, con sistema Lollipop,
será de D(n) =
126 + 100n2
n2 + 1
dólares.
¿Cuántos meses - como mı́nimo - debo esperar para comprar un smartphone
si pienso pagar a lo más 101 dólares?
126 + 100n2
n2 + 1
≤ 101→ 126 + 100n2 ≤ 101n2 + 101→ (n ≥ 5 ∨ n ≤ −5)
Rpta: Al menos cinco meses.
¿Cuál será el precio del smartphone a muy largo plazo?
ĺım
n→∞
126 + 100n2
n2 + 1
= ĺım
n→∞
 126n2 + 100
1 +
1
n2
 = (0 + 100
1 + 0
) = 100
Rpta: 100 dólares.
b) Evaluar:
99∑
n=1
log
(
1 +
1
n
) 1
10
99∑
n=1
1
10
log
(
1 +
1
n
)
=
1
10
99∑
n=1
log(
n+ 1
n
) = − 1
10
99∑
n=1
[log(n)− log(n+ 1)]
= − 1
10
[log(1)− log(2) + log(2)− log(3) + · · ·+ log(99)− log(100)] = 0, 2
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[ESTA PÁGINA PUEDE USARSE COMO BORRADOR O PARA COMPLETAR UNA
PREGUNTA INDICÁNDOLO DEBIDAMENTE]
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Tercera Práctica Calificada
Matemáticas I Lunes 26 de enero de 2015
Está estrictamente prohibido el uso de cartucheras, calculadoras o notas y el préstamo
de materiales.
No hay consultas; si considera que alguna pregunta está errada o mal propuesta corrija
el enunciado y justifique su proceder.
Justifique su respuesta.
Son importantes el orden y la claridad en la presentación de su trabajo, caso contrario
se puede invalidar completamente la respuesta.
APELLIDOS:
NOMBRES:
SECCIÓN:
1 2 3 4 NOTA
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1. (4 puntos) Complete con V (verdadero) o F(falso) según corresponda. Justifique su
respuesta.
a) Si m ∈ R̂ y P ∈ R2 entonces l(m,P ) = {Q ∈ R2 : pend(P,Q) = m} F
l(m,P ) = {Q ∈ R2 : pend(P,Q) = m} ∪ {P}
b) ¿Si una recta tiene x-intercepto igual a su y-intercepto, su pendiente debe ser
siempre igual a −1? F
Contraejemplo: Cualquier recta que pase por el origen y de pendiente diferente
de −1
c) Las pendientes de dos rectas perpendiculares siempre tienen signos opuestos.
F
Contraejemplo: Una recta de pendiente horizontal y otra recta de pendiente ver-
tical.
d) Si la pendiente de la oferta aumenta y su p-intercepto se mantiene constante, la
cantidad de equilibrio aumenta. F
pe
qe
p
q
O
O
qe ←
D
Del gráfico vemos que a medida que la pendiente de la oferta aumenta la cantidad
de equilibrio disminuye.
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2. a) (2 puntos) Si Rd ◦Rh ◦Rd(x, y) = (x′, y′), calcule x′ e y′.
∀(x, y) ∈ R2, Rd ◦Rh ◦Rd(x, y) = Rd(Rh(Rd(x, y)))
= Rd(Rh(y, x)))
= Rd(−y, x))
= (x,−y)
Luego: x′ = x e y′ = y.
b) (1 punto) Determine cuál de todas las transformaciones hechas en clase corres-
ponde a la transformación Rd ◦Rh ◦Rv.
Reflexión vertical (Rv).
c) (2 puntos) Sean h1, h2, k1, k2 constates reales positivas. Pruebe que
E(h1,k1) ◦ E(h2,k2) = E(h1·h2 , k1·k2)
∀(x, y) ∈ R2, E(h1,k1) ◦ E(h2,k2)(x, y) = E(h1,k1)(E(h2,k2)(x, y))
= E(h1,k1)
(
x
h2
,
y
k2
)
=
(
x
h1 · h2
,
y
k1 · k2
)
= E(h1·h2 , k1·k2)(x, y)
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3. Sean A = (0, 4), B = (5, 0) y O el origen de coordenadas. Sobre los catetos del triángulo
AOB se construyen los cuadrados AOCD y OBEF, donde C es de abscisa negativa y
pertenece al eje X y F es de ordenada negativa y pertenece al eje Y. Sea la recta L1
que pasa por B y D y sea L2 la recta que pase por A y E.
a) (1 punto) Represente graficamente el triángulo, los cuadrados y las rectas L1 y
L2 en el plano cartesiano.
X
Y
-1
1
O
A
B
EF
C
D
L1
L2
b) (2 puntos) Determine las ecuaciones de las rectas L1 y L2.
L1 : 4x + 9y = 20
L2 : 9x + 5y = 20
c) (1 punto) Determine el punto de intersección P de las rectas L1 y L2.
Resolviendo el sistema de ecuaciones se obtiene: P =
(
80
61
,
100
61
)
d) (1.5 puntos) Si L3 es la recta que pasa por O y por P, pruebe que dicha recta es
perpendicular al segmento AB.
Dado que pend(O,P ) =
5
4
, pend(A,B) = −4
5
y como pend(O,P )× pend(B,D) =
−1, L3 es perpendicular al segmento AB.
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4. Sea (qe, pe) el punto de equilibrio entre la oferta y la demanda de cierto producto, donde
pe, qe ∈ R son constantes positivas. Si consideramos que la oferta y la demanda son
relaciones lineales, que el excedente del consumidor es EC y el excedente del productor
es EP.
a) (2 puntos) Determine la ecuación de la oferta, en términos de EP, pe y qe.
p
q
b1
O
b2
D
pe
qe
EC
EP
área: EP =
qe
2
[pe − b1]
b1 = pe −
2EP
qe
pendiente: mO =
pe − b1
qe
=
2EP
q2e
Oferta: p =
2EP
q2e
q + pe −
2EP
qe
b) (1.5 puntos) Determine la ecuación de la demanda, en términos de EC, pe y qe.
área: EC =
qe
2
[b2 − pe]→ b2 =
2EC
qe
+ pe
pendiente: mD =
b2 − pe
−qe
= −2EC
q2e
Demanda: p = −2EC
q2e
q +
2EC
qe
+ pe.
c) (2 puntos) Cuando el gobierno cobra, junto al productor, un impuesto de I
unidades monetarias por unidad vendida, la oferta se traslada verticalmente I
unidades y la demanda permanece estable. Determine las coordenadas del nuevo
punto de equilibrio en términos de pe, qe, EC, EP e I.
Demanda: p = −2EC
q2e
q +
2EC
qe
+ pe.
Nueva Oferta: p =
2EP
q2e
q + pe −
2EP
qe
+ I
Resolviendo el sistema:
q∗e = qe −
q2e
2(EP + EC)
· I
p∗e = pe +
EC
EP + EC
· I
Universidad del Pací!co
Manual de imagenLogotipo institucional
Cuarta Práctica Calificada
Matemáticas I Lunes 09 de Febrero de 2015
Verifique que son 4 preguntas, 20 puntos. Duración: 100 minutos.
Está estrictamente prohibido el uso de cartucheras, calculadoras o notas y el préstamo
de materiales.
No hay consultas; si considera que alguna pregunta está errada o mal propuesta corrija
el enunciado y justifique su proceder.
Justifique su respuesta.
Son importantes el orden y la claridad en la presentación de su trabajo, caso contrario
se pueden restar puntos o invalidar completamente la respuesta.
APELLIDOS:
NOMBRES:
SECCIÓN:
1 2 3 4 NOTA
SOLICITUD DE RE-CALIFICACIÓN
Motivos
1. Puntaje sumado erróneamente
2. Pregunta no corregida
3. La respuesta incluye aspectos
o perspectivas alternativas no
consideradas. (Explicar)
Reglamento de solicitud
Solo se aceptan solicitudes el d́ıa de entrega.
Si dos solicitudes son declaradas “No procedentes” en un mismo ciclo, el alumno no podrá pre-
sentar solicitudes adicionales durante dicho ciclo.
Cualquier solicitud habilita al docente a realizar una revisión integral de la evaluación. Como
consecuencia, la nota puede mantenerse, subir o bajar.
Toda solicitud debe estar justificada adecuadamente en base a sus conocimientos del curso.
Se considera “No procedente” aquellas solicitudes donde el alumno sugiere el puntaje que
debe tener según su propio criterio.
Resultado
PROCEDENTE
NO PROCEDENTE
NOTA ANTERIOR NUEVA NOTA
1. (5 puntos) Justificar la falsedad de las siguientes proposiciones
a) Sean A y B dos matrices cuadradas, si AB = 0, entonces A = 0 o B = 0.
Solución. Por ejemplo A =
[
1 0
0 0
]
y B =
[
0 0
0 1
]
, son tales que AB = 0, pero
ninguna de ellas es la matriz nula.
b) Si A y B son dos matrices cuadradas, entonces (AB)T = ATBT .
Solución. Por ejemplo A =
[
1 2
0 0
]
y B =
[
1 0
0 2
]
, son tales que (AB)T=
[
1 0
4 0
]
,
sin embargo ATBT =
[
1 0
2 0
]
c) Sea A = (ai,j)n×n una matriz tal que ai,j 6= 0 para todo i, j ∈ {1, 2 . . . , n}, entonces
detA 6= 0.
Solución. Por ejemplo A =
[
1 1
1 1
]
es tal que aij 6= 0 pero detA = 0.
d) Sea A una matriz de orden 2 × 2, si rango(A) = 2, entonces A es una matriz
triangular superior o inferior.
Solución. Por ejemplo la matriz A =
[
1 2
1 1
]
no es triangular superior ni inferior
y como su matriz reducida es la identidad, entonces tiene rango 2.
e) Dado el sistema de inecuaciones
y ≤ 2x,
x2 + y2 ≤ 1.
su conjunto solución es un punto del plano.
Solución. Por ejemplo, los puntos (0, 0) y (1/2, 1/2) satisfacen ambas inecuacio-
nes.
2. a) (1 punto) Sea A =
1 2 10 1 0
0 0 1
 , calcular A−1.
Solución. 1 2 1 1 0 00 1 0 0 1 0
0 0 1 0 0 1
 −→
f1 − f3
 1 2 0 1 0 −10 1 0 0 1 0
0 0 1 0 0 1
 −→
f1 − 2f2
 1 0 0 1 −2 −10 1 0 0 1 0
0 0 1 0 0 1
 ,
luego A−1 =
1 −2 −10 1 0
0 0 1
 ,
b) (2 punto) Sean A y B dos matrices del orden apropiado, tales que AB = BA,
demostrar que A(A + B) = (A + B)A.
Solución.
A(A + B) = AA + AB = AA + BA,
donde la última igualdad es debido a que AB = BA, finalmente factorizando A
por la derecha tenemos que
A(A + B) = AA + BA = (A + B)A.
c) (2 puntos) Sea A =
[
1 2
0 1
]
, deducir una expresión para An donde n ∈ N, y
demostrar por inducción que esta expresión es válida.
Solución. Calculando las potencias de A tenemos que A2 =
[
1 4
0 1
]
, A3 =
[
1 6
0 1
]
y A4 =
[
1 8
0 1
]
, deduciendo que An =
[
1 2n
0 1
]
. Probaremos por inducción que
esto es válido, para n = 1, A1 =
[
1 2
0 1
]
= A. Suponiendo que es válido para un
k ∈ N, tenemos que
Ak+1 = AkA =
[
1 2k
0 1
]
·
[
1 2
0 1
]
=
[
1 2(k + 1)
0 1
]
.
3. Decimos que una matriz cuadrada A es nilpotente cuando existe un número natural n
tal que An = 0.
a) (2 puntos) Sea A = (ai,j)3×3, donde ai,j =
{
1; si i < j
0; en otro caso
, probar que A es
nilpotente.
Solución. A =
0 1 10 0 1
0 0 0
 , calculando las potencias de A tenemos que A2 =0 0 10 0 0
0 0 0
y A3 =
0 0 00 0 0
0 0 0
 , luego A es nilpotente.
b) (1 punto) Calcular |A|.
Solución. Como la primera columna de A es nula, entonces |A| = 0.
c) (2 punto) Siendo A como en el item anterior, calcular B = I+A+A2+A3+A4+A5.
Solución. B =
1 0 00 1 0
0 0 1
+
0 1 10 0 1
0 0 0
+
0 0 10 0 0
0 0 0
 =
1 1 20 1 1
0 0 1
 .
4. a) Una cadena de supermercados vende carne molida del tipo popular y selecta. Un
lote de carne molida popular contiene 3 kg de grasa y 17 kg de carne roja, un lote
de carne molida selecta contiene 2 kg de grasa y 18 kg de carne roja.
i) (1 punto) Si en un momento dado se cuenta con 10 kg de grasa y 90 kg de
carne roja, escriba el sistema de ecuaciones lineales en la forma A2×2 · x = b,
donde x =
[
x1
x2
]
, siendo x1 y x2 las cantidades de lotes de carne molida
popular y selecta que se deben producir respectivamente.
Solución. Las ecuaciones son 3x1 + 2x2 = 10 y 17x1 + 18x2 = 90, luego[
3 2
17 18
]
·
[
x1
x2
]
=
[
10
90
]
ii) (2 puntos) Calcular x1 y x2.
Solución. El sistema es equivalente a 27x1 + 18x2 = 90 y 17x1 + 18x2 = 90,
restando ambas ecuaciones, tenemos que x1 = 0 y x2 = 5.
b) (2 punto) Graficar el conjunto solución del siguiente sistema de inecuaciones
4x2 + 25y2 ≤ 100
y ≥ 2x2
x2 + (y − 2)2 ≥ 1
Solución. La región es
[ESTA PÁGINA PUEDE USARSE COMO BORRADOR O PARA COMPLETAR UNA
PREGUNTA INDICÁNDOLO DEBIDAMENTE]
[ESTA PÁGINA PUEDE USARSE COMO BORRADOR O PARA COMPLETAR UNA
PREGUNTA INDICÁNDOLO DEBIDAMENTE]
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Universidad del Pací!co
Manual de imagenLogotipo institucional
Práctica Calificada 5
Matemáticas I Lunes 16 de febrero de 2015
Verifique que son 4 preguntas, 20 puntos. Duración: 100 minutos.
Está estrictamente prohibido el uso de cartucheras, calculadoras o notas y el préstamo
de materiales.
No hay consultas; si considera que alguna pregunta está errada o mal propuesta corrija
el enunciado y justifique su proceder.
Justifique su respuesta.
Son importantes el orden y la claridad en la presentación de su trabajo, caso contrario
se puede invalidar completamente la respuesta.
APELLIDOS:
NOMBRES:
SECCIÓN:
1 2 3 4 NOTA
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SOLICITUD DE RECALIFICACIÓN
Motivos
1. Puntaje sumado erróneamente
2. Pregunta no corregida
3. La respuesta incluye aspectos
o perspectivas alternativas no
consideradas. (Explicar)
Reglamento de solicitud
Solo se aceptan solicitudes el d́ıa de entrega.
Si dos solicitudes son declaradas “No procedentes” en un mismo ciclo, el alumno no podrá pre-
sentar solicitudes adicionales durante dicho ciclo.
Si un alumno solicita reconsideración por “Puntaje sumado erroneamente”, sólo se revisará la
suma de los puntajes obtenidos por cada pregunta.
Cualquier solicitud cuyo argumento sea diferente de “Puntaje sumado erróneamente” habilita
al docente a realizar una revisión integral de la evaluación. Como consecuencia, la nota
puede mantenerse, subir o bajar.
Toda solicitud debe estar justificada adecuadamente en base a sus conocimientos del curso.
Se considera “No procedente” aquellas solicitudes donde el alumno sugiere el puntaje que
debe tener según su propio criterio.
Resultado
PROCEDENTE
NO PROCEDENTE NOTA ANTERIOR NUEVA NOTA
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1. a) (3 pts) Justifique por qué las siguientes proposiciones son falsas:
Sean las funciones f : A→ B y g : B → A, entonces f ◦ g = g ◦ f .
Si f(x) = x2 y g(x) = x+ 1 entonces f ◦ g 6= g ◦ f
La función f : R→ R definida por f(x) = |x− π| es sobreyectiva.
∃y = −2 ∈ R tal que −2 /∈ Rang(f) ya que |x− π| 6= −2
Si f : A→ B es una función real constante, entonces su gráfica es una recta
horizontal.
Contraejemplo: f(x) =
√
5 ; −3 ≤ x ≤ 3.
b) (2 pts) Sean Mn×m el conjunto de las matrices de orden n × m. Probar que la
función f : Mn×m →Mm×n definida por f(A) = AT es inyectiva.
Sean A,B ∈Mn×m:
f(A) = f(B)
AT = BT
(AT )T = (BT )T
A = B
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2. (4 pts) Dada la gráfica de la función f :
y
x−6 −5 −4 −3 −2 −1 1 2 3 4 5 6
−1
−2
−3
−4
−5
1
2
3
4
graf(f)
a) Determine el dominio y rango de f .
Dom(f) = [−5, 4]; Ran(f) = [−4, 2]
b) Determine el/los intervalo(s) dónde f es estrictamente creciente.
I1 = [−5,−2]; I2 = [2, 4]
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3. (5 pts) Sea f(x) una regla de correspondencia definida por f(x) =
√
16− x2.
a) Determine el máximo dominio de definición de f(x).
16− x2 ≥ 0→ (x+ 4)(x− 4) ≤ 0→ −4 ≤ x ≤ 4.
b) Esboce la gráfica de f en el plano cartesiano.
−5. −4. −3. −2. −1. 1. 2. 3. 4. 5.
−1.
1.
2.
3.
4.
5.
0
f
c) ¿Es f inyectiva? Justifique.
No, porque −4 6= 4 y f(−4) = 0 = f(4).
d) Si se restringe f al intervalo [0, 4] es invertible; halle la regla de correspondencia
de f−1 : [0, 4]→ [0, 4].
Si f(x) = y =
√
16− x2 → y2 = 16 − x2 → x2 = 16 − y2 → x =
√
16− y2 →
f−1(y) =
√
16− y2.
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4. (6 puntos) Conteste:
Se adjunta la gráfica de la función definida por f(x) = x3 − x2 − 2x
−3 −2 −1 1 2 3 4
−4
−3
−2
−1
1
2
3
0
f
a) Si f(x) = x · (x− α) · (x− β). Determine los valores de α y β, si α < β.
f(x) = x3 − x2 − 2x = x(x2 − x− 2) = x(x+ 1)(x− 2) = x(x− (−1))(x− 2)
Dado que α < β → α = −1; β = 2.
b) ¿Para qué valores reales x ∈ dom(f), f(x) > 0?
Si f(x) > 0→ (−1 < x < 0) ∨ (x > 2)
c) ¿Para qué valores reales x ∈ dom(f), f(x) < 0?
Si f(x) < 0→ (x < −1) ∨ (0 < x < 2)
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[ESTA PÁGINA PUEDE USARSE COMO BORRADOR O PARA COMPLETAR UNA
PREGUNTA INDICÁNDOLO DEBIDAMENTE]
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Universidad del Pací!co
Manual de imagenLogotipo institucional
Matemática I
Ciclo Verano 2015
Examen Parcial
Sábado 31 de Enero de 2015
DURACIÓN: 120 Minutos
Está estrictamente prohibido el uso de cartucheras, calculadoras o notas y el préstamo de
materiales.
No hay consultas; si considera que alguna pregunta está errada o mal propuesta corrija el
enunciado y justifique su proceder.
Justifique su respuesta.
Son importantes el orden y la claridad en la presentación de su trabajo, caso contrario se
pueden restar puntos o invalidar completamente la respuesta.
APELLIDOS:
NOMBRES:
SECCIÓN:
1 2 3 4 5 NOTA
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1. a) i) Dé la definición de máximo entero. (1 pto)
Solución.
x ∈ R, bxc = n ∈ Z, donde n ≤ x < n+ 1.
ii) Pruebe que bx+ nc = bxc+ n ∀ n ∈ Z, x ∈ R. (1 pto)
Solución.
De la definición de máximo entero tenemos bxc ≤ x < bxc+ 1 , luego
bxc+ n ≤ x+ n < bxc+ n+ 1, bxc+ n ∈ Z
Por lo tanto
bx+ nc = bxc+ n
b) i) De la definición de una sucesión acotada superiormente. (1 pto)
Solución.
(an)n∈N es acotada superiormente si ∃M ∈ R,∀ n ∈ R[an ≤M ]
ii) Demuestre que el producto de dos sucesiones acotadas también es una sucesión
acotada. (1 pto)
Solución.
(an)n∈N es acotada si ∃M > 0, |an| ≤M , (bn)n∈N es acotada si ∃M1 > 0, |bn| ≤M1,
luego
|an · bn| ≤M ·M1
Por lo tanto (an · bn)n∈N está acotado.
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2. Dada la siguiente sucesión (an)n∈N tal que an =
n+ 1
n+ 6
.
a) Pruebe que es creciente. (1 pto)
Solución.
La sucesión es an =
n+ 1
n+ 6
= 1 − 5
n+ 6
. De n < n + 1 tenemos n + 6 < n + 7, luego
− 5
n+ 6
< − 5
n+ 7
de esto 1 − 5
n+ 6
< 1 − 5
(n+ 1) + 6
. Por lo tanto an < an+1 para
todo n ∈ N, la sucesión es estŕıctamente y en consecuencia es creciente.
b) Pruebe que es acotada superiormente. (0.5 pto)
Solución.
De n + 1 < n + 6 tenemos
n+ 1
n+ 6
< 1 , luego
n+ 1
n+ 6
< 1 ≤ 2 para todo n ∈ N. Por lo
tanto (an)n∈N es acotada superiormente.
c) Si ĺım
n→+∞
an = L, calcule L. (0.5 ptos)
Solución.
Tenemos ĺım
n→+∞
n+ 1
n+ 6
= ĺım
n→+∞
1 + 1n
1 + 6n
= 1.
d) Pruebe ĺım
n→+∞
an = L, usando la definición de ĺımite. (2 ptos)
Solución.
∀ � > 0 ∃N ∈ N, ∀n ∈ N [n > N → |an − L| < �].
Dado � > 0 consideremos N = b5
�
c+ 1; si n > N = b5
�
c+ 1 → n > 5� → n+ 6 >
5
�
, de
donde | 5
n+ 6
| < � → |1− 5n+6 − 1| < � → |an − 1| < �.
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3. En cada caso calcule el conjunto solución respecto a la variable x.
a) bb2x+ 1c+ 1c = x7 + 5 (1.5 ptos)
Solución.
Como
x
7
+ 5 ∈ Z se tiene x7 ∈ Z, de esto x ∈ Z. Luego la ecuación queda
2x+ 2 = x7 + 5
2x− x
7
= 5− 2
13x
7
= 3
x = 2113
CS = ∅
b) |4|x− 2| − 10| ≤ 2x− 4 (1.5 ptos)
Solución.
De la desigualdad vemos que 2x − 4 ≥ 0, de donde x ≥ 2. De esto la inecuación queda
|4x− 18| ≤ 2x− 4 → |2x− 9| ≤ x− 2 → 2− x ≤ 2x− 9 ≤ x− 2
Resolviendo 2− x ≤ 2x− 9:
2− x ≤ 2x− 9
−3x ≤ −11
x ≥ 113
Luego x ∈ [113 ,+∞]
Resolviendo 2x− 9 ≤ x− 2:
2x− 9 ≤ x− 2
x ≤ 7
Luego x ∈ [−∞, 7]
El conjunto solución es [113 , 7]
c) (x+ 3a)(2b− x) ≥ 0. (a < b < 0) (1 pto)
Solución.
CS = [2b,−3a]
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4. Pedro debe construir el gráfico del costo, ingreso y utilidad de la empresa donde trabaja, la
cual se encarga de la venta de pasadores deportivos. Olvidó trazar una de las rectas, por lo
cual obtuvo el gráfico mostrado. Si la empresa vende 100 pasadores más, sus utilidades se
incrementan en 8 soles.
a) Determine las ecuaciones de costo, ingreso y utilidad. (2 ptos)
b) Calcule m y n (1 pto)
c) Determine la cantidad en la cual la utilidad es el 100 % del costo. (1 pto)
Solución.
a) El Ingreso es I = 10q. Si el costo es C = cf + cu · q entonces la utilidad es U =
(10− cu)︸ ︷︷ ︸
8
·q − cf → cu = 2. De U(60) = 0 tenemos que cf = 480. Por lo tanto
C = 2q + 480 , U = 8q − 480
b) U = 320, q = 100 entonces m = 100 y n = 1000.
c) Se tiene U = C y I = 2C, entonces 10q = 4q+960; de donde q = 160 cientos de unidades.
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5. a) Hallar el área del triángulo formado por la aśıntota de pendiente positiva de la hipérbola
9x2 − 4y2 = 36, la recta 9x+ 2y = 24 y el eje x. (1.5 ptos)
Solución.
El área del triángulo es A = 83 ·
3
2 = 4.
b) Dada la ecuación de la elipse 25x2 + 16y2 − 100x+ 96y − 156 = 0 determine:
i) La ecuación de la circunferencia cuyo centro coincide con el de la elipse, pasa por
uno de los vértices y está inscrita. (1.5 ptos)
Solución.
Completando cuadrados la ecuación queda (x−2)
2
16 +
(y+3)2
25 = 1, luego la ecuación de
la circunferencia es
(x− 2)2 + (y + 3)2 = 42
ii) Determine una transformación de coordenadas que convierta a la circunferencia del
párrafo anterior en unitaria y centrada en el origen. (1pto)
Solución.
La trasformación de coordenadas es
E(4,4) ◦ T (2,−3)
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Universidad del Pací!co
Manual de imagenLogotipo institucional
EXAMEN FINAL
Matemáticas I Lunes 23 de Febrero de 2015
DURACIÓN: 120 Minutos
Está estrictamente prohibido el uso de cartucheras, calculadoras o notas y el préstamo
de materiales.
No hay consultas; si considera que alguna pregunta está errada o mal propuesta corrija
el enunciado y justifique su proceder.
Justifique cada respuesta.
Son importantes el orden y la claridad en la presentación de su trabajo, caso contrario
se pueden restar puntos o invalidar completamente la respuesta.
Puede utilizar el reverso de cada hoja tanto para responder como hoja de borrador,
indicándolo claramente.
1 2 3 4 5 NOTA
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1. (4 puntos) A continuación se muestra el gráficodel conjunto solución de un sistema de
inecuaciones. Se sabe que P representa a una función cuadrática y que A se obtiene
de la gráfica del arco tangente después de aplicar otra transformación de coordenadas.
Determine las inecuaciones que generan dicho conjunto solución.
x
y
1
−1
2
−2
2
P
A
Solución:
y ≥ −(x− 2)2 + 2
y < 2
y ≥ − 4
π
arctanx
x < 2
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2. La función de Gompertz, f : [0,+∞[ → B ⊂ R, f(x) = ae−ce
−kx
donde a, c, k
son constantes positivas, se usa a veces para describir las ventas de un nuevo producto
cuyas ventas son inicialmente grandes pero luego de un tiempo x se nivelan hacia un
nivel máximo de saturación. A continuación se muestra la representación gráfica de f.
x
y
y = f(x)
a
a
y = f−1(x)
a) (0.5 puntos) Calcule la venta inicial.
Solución:
f(0) = ae−ce
0
= ae−c
b) (1 punto) Calcule el momento en que la venta es
a
e
.
Solución:
ae−ce
−kx
=
a
e
→ e−ce−kx = e−1 → ce−kx = 1→ e−kx = 1
c
→ −kx = ln
(
1
c
)
→ x = ln(c)
k
c) (1 punto) Determine el conjunto B de forma que f sea sobreyectiva.
Solución:
B =
[
ln(c)
k
, a
[
d) (1.5 puntos) Determine f−1 indicando el dominio, conjunto de llegada y regla de
correspondencia.
Solución: f−1 :
[
ln(c)
k
, a
[
→ [0,∞[
y = f(x)↔ y = ae−ce−kx ↔ y
a
= e−ce
−kx ↔ ln
(y
a
)
= −ce−kx ↔ −1
c
ln
(y
a
)
= e−kx
↔ −kx = ln
(
−1
c
ln
(y
a
))
↔ x = −1
k
ln
(
−1
c
ln
(y
a
))
↔ f−1(y) = −1
k
ln
(
−1
c
ln
(y
a
))
e) (1 punto) En el mismo plano cartesiano superior, grafique f−1 y su aśıntota.
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3. a) (1 punto) Dé la definición de función estrictamente creciente.
Solución: Una función real f : A→ R es estrictamente creciente cuando
∀x1, x2 ∈ A, [x1 < x2 → f(x1) < f(x2)].
b) (2 puntos) Pruebe anaĺıticamente que la función de Gompertz es estrictamente
creciente.
Solución:
∀x1, x2 ∈ [0,∞[ , x1 < x2 → −kx1 > −kx2
→ e−kx1 > e−kx2
→ −ce−kx1 < −ce−kx2
→ e−ce−kx1 < e−ce−kx2
→ ae−ce−kx1 < ae−ce−kx2
→ f(x1) < f(x2)
c) (1 punto) Grafique una función f : [1, 3]→ [−2,−1] ∪ [2, 3] que sea sobreyectiva.
x
y
1
−1
−3
2
3
4
−2
1 2 3 4−1
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4. Sobre una compañ́ıa que produce gas, aceite y gasolina, se sabe que para producir una
unidad de gas requiere 1/5 del mismo, 2/5 de aceite y 1/5 de gasolina. Para producir
una unidad de aceite, requiere de 2/5 de gas y 1/5 de aceite. Para producir una unidad
de gasolina usa 1 unidad de gas y una de aceite. Finalmente se tiene una demanda del
mercado de 100 unidades de cada producto.
a) (2 puntos) Halle la matriz aumentada del sistema lineal que determine la produc-
ción bruta de cada industria para cumplir con su mercado.
Solución: Sean x, y, z el número total de unidades de gas, aceite y gasolina,
respectivamente. Tenemos que:
x
5
+
2y
5
+z +100 = x
2x
5
+
y
5
+z +100 = y
x
5
+ 100 = z
⇒
4x
5
−2y
5
−z = 100
−2x
5
+
4y
5
−z = 100
−x
5
+ z = 100
⇒

4
5
−2
5
−1 100
−2
5
4
5
−1 100
−1
5
0 1 100

b) (1.5 puntos) Calcule la matriz reducida de la matriz aumentada.
Solución:
 4/5 −2/5 −1 100−2/5 4/5 −1 100
−1/5 0 1 100
 5f15f2
−5f3
−→
 4 −2 −5 500−2 4 −5 500
1 0 −5 −500
 f1 × f3
−→
 1 0 −5 −500−2 4 −5 500
4 −2 −5 500

f2 + 2f1
f3 − 4f1
−→
 1 0 −5 −5000 4 −15 −500
0 −2 15 2500
 f2 + 2f3
−→
 1 0 −5 −5000 0 15 4500
0 −2 15 2500
 115f2
−→
 1 0 −5 −5000 0 1 300
0 −2 15 2500

f2 × f3
−→
 1 0 −5 −5000 −2 15 2500
0 0 1 300
 f1 + 5f3f2 − 15f3
−→
 1 0 0 10000 −2 0 −2000
0 0 1 300
 −12f2
−→
 1 0 0 10000 1 0 1000
0 0 1 300

c) (0.5 puntos) Determine el conjunto solución del sistema de ecuaciones.
Solución:
CS = {(1000, 1000, 300)}
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5. (3 puntos) Dada la siguiente regla de correspondencia
f(x) =
√
π − A arc cos
(√
2x
)
+
√
B arc cos
(√
2x
)
− π.
Si A y B son constantes positivas y el máximo dominio de definición de f es
[
−1
2
,
1
2
]
,
determine el valor de A y B.
Solución: Tenemos las siguientes restricciones:
π − A arc cos
(√
2x
)
≥ 0 ∧B arc cos
(√
2x
)
− π ≥ 0
lo que equivale a:
π
B
≤ arc cos
(√
2x
)
≤ π
A
(1)
Por otro lado:
−1
2
≤ x ≤ 1
2
⇔ −
√
2
2
≤
√
2x ≤
√
2
2
⇔ arc cos
(√
2
2
)
≤ arc cos
(√
2x
)
≤ arc cos
(
−
√
2
2
)
⇔ π
4
≤ arc cos
(√
2x
)
≤ 3π
4
(2)
Igualando (1) con (2) se obtiene: A =
4
3
, B = 4.