Solucionarlo Examen de Admisión Uní - Traslado Externo
Solucionarlo Examen de Admisión Uní - Traslado Externo
Solucionarlo Examen de Admisión Uní - Traslado Externo
Calculamos R:
𝑥𝑥 − 2𝑦𝑦 + 3 = 0
𝑅𝑅 = 𝐿𝐿1 ˄𝐿𝐿2 � � 𝑅𝑅(13,8).
𝑥𝑥 − 𝑦𝑦 − 5 = 0
𝑥𝑥0 𝑥𝑥0 − 5
1
13 8
Calculamos el año: 𝐴𝐴 = 85 = 2 �� 𝑥𝑥 𝑥𝑥𝑥𝑥+3 ��
0
2
𝑥𝑥0 𝑥𝑥0 − 5
13𝑥𝑥0 + 39 𝑥𝑥02 + 3𝑥𝑥0
→ 50 = �8𝑥𝑥0 + + 𝑥𝑥02 − 5𝑥𝑥0 − 13𝑥𝑥0 + 65 − 8𝑥𝑥0 − �
2 2
Resolviendo:
𝐿𝐿: 3𝑥𝑥0 − 4𝑦𝑦° + 37 = 0 𝑂𝑂
𝐿𝐿: 3𝑥𝑥0 − 4𝑦𝑦° + 13 = 0
03. SOLUCION:
𝐶𝐶𝐶𝐶𝐶𝐶 𝑉𝑉 = Ѡ =
𝑉𝑉 . Ѡ 7−6 1 √5
√12 22
= =
Ѡ √5 5
√5
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅: 5
04. SOLUCION:
Cálculos pendientes:
𝐿𝐿3 : −𝑥𝑥 + 𝑦𝑦 − 1 = 0
→ 𝑚𝑚3 = 1
𝐿𝐿2 : −𝑥𝑥 + 2𝑦𝑦 − 4 = 0
1
→ 𝑚𝑚2 =
2
Sea 𝑚𝑚1 , pendiente de la 𝐿𝐿1 .
De la fórmula de tangente de un ángulo formado por 2 rectas, tenemos:
1
𝑚𝑚3 − 𝑚𝑚2 1−2 1
𝑇𝑇𝑇𝑇𝑇𝑇 = = =
1 + 𝑚𝑚2 𝑚𝑚3 1 − 1 3.
2
𝑚𝑚1 − 𝑚𝑚3 1
𝑇𝑇𝑇𝑇𝑇𝑇 = = 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑦𝑦 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑚𝑚1 = 2
1 + 𝑚𝑚2 𝑚𝑚3 3.
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅: 2
05. SOLUCION:
Nos pide el perímetro: 2𝑝𝑝 = 6𝑙𝑙.
3(−1)−4(3)−10 5√3
𝑑𝑑(1 − 1,3), 𝐿𝐿 = � √32 +42
� = 5 = 𝑙𝑙√3. → 𝑙𝑙 = 3.
06. SOLUCION:
𝐶𝐶1 : 𝑥𝑥 2 + 𝑦𝑦 2 + 8𝑦𝑦 = −12
𝐶𝐶1 : 𝑥𝑥 2 + (𝑦𝑦 + 4)2 = 4 → 𝐶𝐶1 (0, −4)
𝐶𝐶2 : 𝑥𝑥 2 8𝑥𝑥 + 𝑦𝑦 2 − 10𝑦𝑦 + 16 = 0
𝐶𝐶2 : (𝑥𝑥 − 4)2 + (𝑦𝑦 − 5)2 = 25 → 𝐶𝐶2 (4,5)
(−4)−(5) −9 9 9
𝑚𝑚 = (0)−(4)
= −4 = 4 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅: 4
07. SOLUCION:
𝑃𝑃: (𝑦𝑦 − 𝑘𝑘)2 = −4𝑃𝑃(𝑥𝑥 − ℎ) 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑞𝑞𝑞𝑞𝑞𝑞 𝑙𝑙𝑙𝑙 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑠𝑠𝑠𝑠 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎 𝑙𝑙𝑙𝑙 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖.
𝑃𝑃: (𝑦𝑦 − (−11)2 = −4(𝑥𝑥 − 1)
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅: (𝑦𝑦 + 1)2 = −4(𝑥𝑥 − 1)
08. SOLUCION:
𝑘𝑘: 25𝑥𝑥 2 + 16𝑦𝑦 2 − 200𝑥𝑥 − 160𝑦𝑦 + 400 = 0
25[𝑥𝑥 2 − 8𝑥𝑥] + 16[𝑦𝑦 2 − 10𝑦𝑦] − 400 = 0
25[(𝑥𝑥 − 4)2 − 16] + 16[(𝑦𝑦 − 5)2 − 25] + 400 = 0
(𝑥𝑥−4)2 (𝑦𝑦−5)2
+ =1
16 25
09. SOLUCION:
𝐸𝐸𝐸𝐸𝐸𝐸 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 = 2𝑎𝑎 = 12 → 𝑎𝑎 = 6,
𝐸𝐸𝐸𝐸𝐸𝐸 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 = 2𝑏𝑏 = 16 → 𝑏𝑏 = 8,
𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁: 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑎𝑎 𝑞𝑞𝑞𝑞𝑞𝑞 𝑒𝑒𝑒𝑒 𝑒𝑒𝑒𝑒𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑒𝑒𝑒𝑒 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑞𝑞𝑞𝑞𝑞𝑞 𝑒𝑒𝑒𝑒 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡, 𝑙𝑙𝑙𝑙 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒ó𝑛𝑛 𝑠𝑠𝑠𝑠𝑠𝑠á 𝑑𝑑𝑑𝑑
𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
(𝑦𝑦 − 𝑘𝑘)2 (𝑥𝑥 − 𝑏𝑏)2
− =1
𝑏𝑏 2 𝑎𝑎2
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑙𝑙𝑙𝑙𝑙𝑙 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑:
(𝑦𝑦 + 3)2 (𝑥𝑥 + 2)2
− =1 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅.
64 36
10. SOLUCION:
𝐸𝐸𝑛𝑛 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑚𝑚𝑚𝑚 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑒𝑒𝑒𝑒 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑑𝑑𝑑𝑑 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖ó𝑛𝑛 𝑑𝑑𝑑𝑑 2 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡. 𝑢𝑢𝑢𝑢𝑢𝑢 𝑑𝑑𝑑𝑑 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
𝑒𝑒𝑒𝑒 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑙𝑙𝑙𝑙 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑦𝑦 𝑙𝑙𝑙𝑙 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑎𝑎 𝑙𝑙𝑙𝑙 𝑝𝑝𝑝𝑝𝑝𝑝á𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏. 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑎𝑎 ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑑𝑑𝑑𝑑𝑑𝑑ℎ𝑎𝑎𝑎𝑎
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡.
𝐶𝐶: 𝑥𝑥 2 + 𝑦𝑦 2 − 2𝑥𝑥 + 𝑦𝑦 − 5 = 0
2 − 2𝑥𝑥 7 1
𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖: 𝑦𝑦21 = → 𝑦𝑦11 = 𝑚𝑚1 = ∞ 𝑦𝑦 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑒𝑒𝑒𝑒 � , − �
⊥ +2𝑦𝑦 2 2
7 1
𝑙𝑙𝑙𝑙 𝑞𝑞𝑞𝑞𝑞𝑞 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑞𝑞𝑞𝑞𝑞𝑞 𝑒𝑒𝑒𝑒 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 � , − � 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑎𝑎 𝑙𝑙𝑙𝑙 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑥𝑥
2 2
7
= , 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑠𝑠𝑠𝑠 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
2
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑒𝑒𝑒𝑒 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑑𝑑𝑑𝑑 𝑙𝑙𝑙𝑙 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒ó𝑛𝑛 𝑑𝑑𝑑𝑑 𝑙𝑙𝑙𝑙 𝑐𝑐ó𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑘𝑘𝑗𝑗 :
𝑘𝑘: 𝑦𝑦 2 − 4𝑥𝑥 + 8 = 0
2
𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖: 𝑦𝑦2 = → 𝑦𝑦21 = 𝑚𝑚2 = −1 𝑦𝑦 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑒𝑒𝑒𝑒 (3, −2)
𝑦𝑦
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑙𝑙𝑙𝑙 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒ó𝑛𝑛 𝑑𝑑𝑑𝑑 𝑙𝑙𝑙𝑙 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑒𝑒𝑒𝑒 (3, −2)
𝑦𝑦 + 2
−⊥= → 𝐿𝐿⎾ : 𝑥𝑥 + 𝑦𝑦 − 1 = 0
𝑥𝑥 − 3
7
𝑥𝑥 = 7 −5
𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓: } 2 { � , � 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅.
𝑥𝑥 + 𝑦𝑦 − 1 = 0 2 2
11. SOLUCION:
Calcular el determinante de la matriz.
𝑎𝑎 𝑏𝑏 𝑐𝑐 𝑓𝑓2 +(−1)𝑓𝑓1 𝑎𝑎 𝑏𝑏 𝑐𝑐
�𝑎𝑎 𝑎𝑎 + 𝑏𝑏 𝑎𝑎 + 𝑏𝑏 � �⎯⎯⎯⎯⎯� �0 𝑎𝑎 𝑎𝑎 + 𝑏𝑏 �
𝑎𝑎 𝑎𝑎 + 2𝑏𝑏 3𝑎𝑎 + 2𝑏𝑏 + 𝑐𝑐 𝑓𝑓3 + (−1)𝑓𝑓1 0 𝑎𝑎 + 𝑏𝑏 3𝑎𝑎 + 2𝑏𝑏
𝑎𝑎
𝑓𝑓3 +(−1)𝑓𝑓2 𝑏𝑏 𝑐𝑐 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
�⎯⎯⎯⎯⎯� �0 𝑎𝑎 𝑎𝑎 + 𝑏𝑏� 𝑦𝑦 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑒𝑒𝑒𝑒
0 𝑏𝑏 2𝑎𝑎 + 𝑏𝑏 𝑙𝑙𝑙𝑙 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓.
𝑎𝑎 𝑎𝑎 + 𝑏𝑏 0 𝑎𝑎 + 𝑏𝑏 0 𝑎𝑎
= 𝑎𝑎(−1)1+1 � � + 𝑏𝑏(−1)1+2 � � + 𝐶𝐶(−1)1+3 � �
𝑏𝑏 2𝑎𝑎 + 𝑏𝑏 0 2𝑎𝑎 + 𝑏𝑏 0 𝑏𝑏
= 𝑎𝑎|2𝑎𝑎2 + 𝑎𝑎𝑎𝑎 − 𝑎𝑎𝑎𝑎 − 𝑏𝑏 2 | = 𝑎𝑎|2𝑎𝑎2 − 𝑏𝑏 2 | 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅.
𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁: 𝑛𝑛𝑛𝑛 ℎ𝑎𝑎𝑎𝑎 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐.
12. SOLUCION:
𝑘𝑘 0 0 1 𝑓𝑓2 −𝑓𝑓1 𝑘𝑘 0 0 1 𝑓𝑓3 −𝑓𝑓2 𝑘𝑘 0 0 1
� 4 𝑘𝑘 4 2 � �⎯⎯� � 0 𝑘𝑘 4 1 � �⎯⎯� �0 𝑘𝑘 4 1�
2𝑘𝑘 𝑘𝑘 𝑘𝑘 + 4 3 𝑓𝑓3 − 2𝑓𝑓1 0 𝑘𝑘 𝑘𝑘 + 4 1 ���������
0 0 𝑘𝑘 0
𝑒𝑒𝑒𝑒 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑒𝑒𝑒𝑒 𝑒𝑒𝑒𝑒 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 𝑙𝑙𝑙𝑙 ran(A)=3.
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 . : 𝑘𝑘 ≠ 0
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅: 𝑘𝑘 𝑒𝑒 𝑅𝑅−}0{
13. SOLUCION:
1 0 0
𝐴𝐴 = �0 −1 0�
0 0 𝑎𝑎
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑙𝑙𝑙𝑙𝑙𝑙 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑑𝑑𝑑𝑑 𝐴𝐴:
𝑃𝑃(𝑥𝑥) = |𝐴𝐴 − 𝜆𝜆𝜆𝜆| = 0
1 − 𝜆𝜆 0 0
→� 0 −1 − 𝜆𝜆 0 � = (1 − 𝜆𝜆)(−1 − 𝜆𝜆)(𝑎𝑎 − 𝜆𝜆) = 0
0 0 𝑎𝑎 − 𝜆𝜆
𝑃𝑃(𝜆𝜆) = 𝜆𝜆3 − 𝑎𝑎𝜆𝜆2 − 𝜆𝜆 + 𝑎𝑎 = 0
−𝑎𝑎
Ƹ𝑥𝑥 = − � � = 𝑎𝑎 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅.
1
14. SOLUCION:
𝐴𝐴𝐴𝐴 + 𝐵𝐵𝐵𝐵 = 𝐶𝐶 𝐶𝐶 + 𝐷𝐷 −1
→ 𝑋𝑋 = � � 𝐴𝐴 … . . (𝑋𝑋)
𝐴𝐴𝐴𝐴 + 𝐵𝐵𝐵𝐵 = 𝐷𝐷 2
𝐶𝐶 + 𝐷𝐷 −1
→ 𝑌𝑌 = � � 𝐵𝐵 … . . (𝐵𝐵)
2
Cálculo de la inversa de A:
1 1 1 1
1 1 𝑓𝑓 3𝑓𝑓
𝑓𝑓1 � � 1 1
21 10 1 0 2− 1 1 2 2 0 𝑓𝑓2 (2) 1
2 0 𝑓𝑓1 −2𝑓𝑓2 1 0 −1 −1
� � � �⎯⎯� � 2� 2 � �⎯⎯⎯� � � � �⎯� � 2� 2 � �⎯⎯⎯� � � �
32 01 3 1 0 1 −3 1 0 2 0 1 �����
−3 2
2 0 1 −3
2 2
2 −1
→ 𝐴𝐴−1 = � � 𝐴𝐴−1
−3 2
Calculo de la inversa de B.
𝐵𝐵 −1
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅, 𝑦𝑦 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜, 𝑒𝑒𝑒𝑒 (𝑥𝑥)𝑦𝑦 𝑙𝑙𝑙𝑙):
9 7 7
− 7 −3 15
𝑥𝑥 = � 2 2� , 𝑦𝑦 = � 1 2 � → 𝑥𝑥 + 𝑦𝑦 = �− 2� 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅.
11 1 2 7
− 4 2 − −5
2 2 2
15. SOLUCION:
𝑎𝑎� = (3 − ʈ, 2 − ʈ, 0)
𝑏𝑏� = (0, ʈ − 1, ʈ − 1) } 𝑏𝑏�𝑥𝑥𝑐𝑐̅ = (−ʈ2 + 3ʈ − 2, ʈ − 1,
𝑐𝑐̅ = (1,0,2−⊥) 𝐼𝐼 − ʈ)
𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑞𝑞𝑞𝑞𝑞𝑞 𝑠𝑠𝑠𝑠𝑠𝑠 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑:
ʈ3 − 7ʈ2 + 14ʈ − 8 = 0
Ƹ𝑣𝑣𝑣𝑣𝑣𝑣. ʈ = 7 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅.
16. SOLUCION:
𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑞𝑞𝑞𝑞𝑞𝑞 𝑉𝑉�1 , 𝑉𝑉�2 ∈ 𝑉𝑉 𝑦𝑦 𝑉𝑉�1 ⊥ 𝑉𝑉�2
𝑉𝑉�1 . 𝑉𝑉�2 = 0
→ 𝑎𝑎𝑎𝑎 + 1 − 𝑎𝑎𝑎𝑎 + 𝑎𝑎 = 0 → 𝑎𝑎 = −1
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑉𝑉�2 ∈ 𝑉𝑉: 2(𝑏𝑏) − 1 + 3(𝑎𝑎) = 0
2𝑏𝑏 − 1 − 3 = 0
→ 𝑏𝑏 = 2
𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓: 𝑉𝑉�1 . 𝑉𝑉�2 = (1,2, −2) 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅.
17. SOLUCION:
𝐿𝐿𝐿𝐿 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑄𝑄(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) 𝑎𝑎 𝑢𝑢𝑢𝑢 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑃𝑃: 𝐴𝐴𝐴𝐴 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐𝑐𝑐 + 𝐷𝐷 = 0
𝑒𝑒𝑒𝑒𝑒𝑒á 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑝𝑝𝑝𝑝𝑝𝑝.
𝐴𝐴𝐴𝐴 + 𝐵𝐵𝐵𝐵 + 𝐶𝐶𝐶𝐶 + 𝐷𝐷
𝑑𝑑(𝑄𝑄, 𝑃𝑃) = � �
√𝐴𝐴2 + 𝐵𝐵 2 𝐶𝐶 2
𝑒𝑒𝑒𝑒 𝑒𝑒𝑒𝑒 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝. 𝑄𝑄(1, −2,4)
𝑃𝑃. 2𝑥𝑥 − 𝑦𝑦 + 3𝑧𝑧 = 10
6
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟: 𝑑𝑑(𝑄𝑄, 𝑃𝑃) = 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅.
√14
18. SOLUCION:
𝑋𝑋 − 1 2𝑦𝑦 − 4 1 − 𝑍𝑍
∗ 𝐿𝐿: = = =ʈ
3 2 2
𝑋𝑋 = 3ʈ + 1
� 𝑦𝑦 = ʈ + 2
𝑍𝑍 = 1 − 2ʈ
∗ 𝑃𝑃: 2𝑥𝑥 − 𝑦𝑦 + 3𝑧𝑧𝑧𝑧 = 1
𝐿𝐿˄𝑃𝑃: 2(3ʈ + 1) − (ʈ + 2) + 3(1 − 2ʈ) = 1
→ʈ=2
: . 𝑋𝑋 + 𝑦𝑦 + 2𝑧𝑧 = 2ʈ + 4 = 8 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅.
19. SOLUCION:
20. SOLUCION:
21. SOLUCION:
22. SOLUCION:
𝑙𝑙𝑙𝑙𝑙𝑙 𝑓𝑓(𝑥𝑥+ℎ)−𝑓𝑓(𝑥𝑥)
I. 𝑃𝑃𝑃𝑃𝑃𝑃 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑ó𝑛𝑛: 𝑓𝑓 1 (𝑥𝑥) =
ℎ→0 ℎ
1
II. 𝑆𝑆𝑆𝑆𝑆𝑆 𝑓𝑓(𝑥𝑥) = 4 𝑥𝑥 4 → 𝑓𝑓 1 (𝑥𝑥) = 𝑥𝑥 3
III. 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑢𝑢𝑢𝑢𝑢𝑢 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓ó𝑛𝑛 𝑓𝑓 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑒𝑒𝑒𝑒 𝑒𝑒𝑒𝑒 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 < 𝑎𝑎, 𝑏𝑏 >
𝑦𝑦 𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐 ∈ < 𝑎𝑎, 𝑏𝑏 >, 𝑠𝑠𝑠𝑠 𝑓𝑓(𝑐𝑐)𝑒𝑒𝑒𝑒 𝑢𝑢𝑢𝑢 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑑𝑑𝑑𝑑 𝑓𝑓, 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑓𝑓 1 (𝑐𝑐) =
0 𝑜𝑜 𝑓𝑓 1 (𝑐𝑐)𝑛𝑛𝑛𝑛 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒. 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅. 𝑌𝑌𝑌𝑌𝑌𝑌
23. SOLUCION:
24. SOLUCION:
Grafica de t.
Rpta: D
26. SOLUCION:
𝑓𝑓(𝑥𝑥) = 𝑥𝑥 2 − 𝑥𝑥 + 1. ˄ 𝑓𝑓(3) = 7
𝑓𝑓 1 (𝑥𝑥) = 2𝑥𝑥 − 1
𝑓𝑓 1 (3) = 5.
𝑦𝑦 − 7
𝐿𝐿⎾ : = 5 → 𝐿𝐿⎾ : 𝑦𝑦 = 5𝑥𝑥 − 8 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅.
𝑥𝑥 − 3
27. SOLUCION:
1 1
⩝ 𝑥𝑥 ≠ 0, −1 ≤ 𝑠𝑠𝑠𝑠𝑠𝑠 , −𝑥𝑥 ≤ 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 ≤ 𝑥𝑥, 𝑥𝑥 > 0
𝑥𝑥 𝑥𝑥
𝑙𝑙𝑙𝑙𝑙𝑙 (−𝑥𝑥) 𝑙𝑙𝑙𝑙𝑙𝑙 1 𝑙𝑙𝑙𝑙𝑙𝑙
≤ 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 ≤ (𝑥𝑥)
𝑥𝑥 → 0 𝑥𝑥 → 0 𝑥𝑥 𝑥𝑥 → 0
𝑙𝑙𝑙𝑙𝑙𝑙 1
𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 = 0 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅. 0; 0
𝑥𝑥 → 0 𝑥𝑥
28. SOLUCION:
𝑙𝑙𝑙𝑙𝑙𝑙 𝑙𝑙𝑙𝑙𝑙𝑙
𝐿𝐿1 = 1 𝑥𝑥 − [𝑥𝑥] = 1 1
𝑥𝑥 → 𝑥𝑥 → (𝑥𝑥 − 0) =
2 2 2
𝑙𝑙𝑙𝑙𝑙𝑙 𝑥𝑥 𝑙𝑙𝑙𝑙𝑙𝑙 𝑥𝑥
𝐿𝐿2 = 𝑥𝑥 → 0+ = =1
[𝑥𝑥] 𝑥𝑥 → 0+ 𝑥𝑥
𝑥𝑥 > 0
𝑙𝑙𝑙𝑙𝑙𝑙 𝑥𝑥|𝑥𝑥| 𝑙𝑙𝑙𝑙𝑙𝑙 |𝑥𝑥|
𝐿𝐿3 = = =0
𝑥𝑥 → 0 𝑥𝑥 𝑥𝑥 → 0+
3
: . 𝐿𝐿1 + 𝐿𝐿2 + 𝐿𝐿3 = 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅.
2
29. SOLUCION:
𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑒𝑒𝑒𝑒 𝑥𝑥 = +1. (𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷)
30. SOLUCION:
𝑠𝑠𝑠𝑠𝑠𝑠 𝑦𝑦 = 𝑓𝑓(𝑥𝑥) 𝑔𝑔(𝑥𝑥)
𝑓𝑓 1 (𝑥𝑥)
→ 𝑦𝑦´ = 𝑓𝑓(𝑥𝑥) 𝑔𝑔(𝑥𝑥) �𝑔𝑔´(𝑥𝑥)𝐿𝐿𝐿𝐿𝐿𝐿(𝑥𝑥) + 𝑔𝑔(𝑥𝑥). �
𝑓𝑓(𝑥𝑥)
31. SOLUCION:
𝑒𝑒
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐼𝐼 = � 𝑥𝑥𝑥𝑥𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 .
0
𝑥𝑥 2 1 𝑥𝑥 2 𝑒𝑒
𝐼𝐼 = �(𝐿𝐿𝐿𝐿𝐿𝐿) − �
2 2 2 0
𝑥𝑥 2 𝑥𝑥 2 𝑒𝑒
𝐼𝐼 = � 𝐿𝐿𝐿𝐿𝐿𝐿 − �
2 4 0
𝑒𝑒 2 𝑒𝑒 2 𝑒𝑒 2
𝐼𝐼 = − −0 = 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅.
2 4 4
32. SOLUCION:
𝐿𝐿𝐿𝐿 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝ó𝑛𝑛 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑑𝑑𝑑𝑑 𝑢𝑢𝑢𝑢𝑢𝑢 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑒𝑒𝑒𝑒𝑒𝑒á 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑝𝑝𝑝𝑝𝑝𝑝 𝑙𝑙𝑙𝑙 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓ó𝑛𝑛:
𝑓𝑓(Ƚ) = 100 + 100Ƚ − 100Ƚ2
𝑁𝑁𝑁𝑁𝑁𝑁 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑒𝑒𝑒𝑒 𝑞𝑞𝑞𝑞𝑞𝑞 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠á 𝑚𝑚á𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 𝑙𝑙𝑙𝑙 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝ó𝑛𝑛 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑎𝑎 ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠
3
𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑎𝑎 𝑙𝑙𝑙𝑙𝑙𝑙 14 ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑦𝑦 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡é𝑛𝑛 𝑛𝑛𝑛𝑛𝑛𝑛 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑓𝑓( )
4
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆ó𝑛𝑛:
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑢𝑢𝑢𝑢 𝑚𝑚á𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 𝑑𝑑𝑑𝑑 𝑓𝑓.
1
𝑓𝑓 1 (Ƚ) = 100 − 200Ƚ = 0 → Ƚ = ℎ𝑜𝑜𝑜𝑜𝑜𝑜
2
1
𝑃𝑃𝑃𝑃𝑃𝑃 𝑙𝑙𝑙𝑙 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡, 𝑙𝑙𝑙𝑙 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝ó𝑛𝑛 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑚𝑚á𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜á 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑑𝑑𝑑𝑑 ℎ𝑜𝑜𝑜𝑜𝑜𝑜. 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐ó
2
𝑎𝑎 𝑙𝑙𝑙𝑙𝑙𝑙 14 ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜, 𝑠𝑠𝑠𝑠𝑠𝑠á 𝑚𝑚á𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 𝑎𝑎 𝑙𝑙𝑙𝑙𝑙𝑙 14: 30 ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
3
𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴á𝑠𝑠 𝑓𝑓(45𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚) = 𝑓𝑓( ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜)
4
3 3 3
𝑓𝑓 � � = 100 + 100 � � − 100( )2
4 4 4
3
𝑓𝑓 � � = 118,75. 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅: 14: 30 ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜; 118,75𝓊𝓊
4
33. SOLUCION:
34. SOLUCION:
35. SOLUCION:
𝑦𝑦 = −𝑥𝑥 + 2
2 2
𝑉𝑉 = 𝜋𝜋 � [2𝑥𝑥 − 𝑥𝑥 ] − 𝜋𝜋 � (−𝑥𝑥 + 2)2 𝑑𝑑𝑑𝑑
2 2
1 1
𝜋𝜋
𝑉𝑉 = 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅.
5
36. SOLUCION:
→ 𝐴𝐴𝑇𝑇 2𝐴𝐴1
𝐿𝐿𝐿𝐿2 𝐿𝐿𝐿𝐿2
1
𝐴𝐴1 = � 𝑒𝑒 −𝑥𝑥 𝑑𝑑𝑑𝑑 − � 𝑑𝑑𝑑𝑑
0 0 2
1 1
𝐴𝐴1 = − 𝐿𝐿𝐿𝐿2
2 2
: . 𝐴𝐴𝑇𝑇 = 𝐼𝐼 − 𝐿𝐿𝐿𝐿2 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅.
37. SOLUCION:
1 𝑒𝑒
𝑉𝑉 = 𝜋𝜋(1)2 − (� (−𝐿𝐿𝐿𝐿𝐿𝐿)𝑑𝑑𝑑𝑑 + � 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿
𝑒𝑒−1 1
𝜋𝜋 2
𝑉𝑉 = (𝑒𝑒 + 𝑒𝑒 −2 − 2 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅.
2
38. SOLUCION:
𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑦𝑦 = 𝐿𝐿𝐿𝐿(𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠). → 𝑦𝑦´ = = 𝑡𝑡𝑡𝑡𝑡𝑡
𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
→ 𝑦𝑦´2 = +𝑠𝑠 2 𝑥𝑥 → 𝑦𝑦´2 + 1 = 𝑠𝑠𝑠𝑠𝑠𝑠 2 𝑥𝑥
𝜋𝜋 𝜋𝜋
4 4
𝜋𝜋
𝐿𝐿 = � �1 + 𝑦𝑦´2 = � 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = [𝐿𝐿𝐿𝐿|𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑡𝑡𝑡𝑡𝑡𝑡|]4
0 0 0
𝐿𝐿 = 𝐿𝐿𝐿𝐿�√2 + 1� − 𝐿𝐿𝐿𝐿|1 + 0| = 𝐿𝐿𝐿𝐿�√2 + 1� 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅.
39. SOLUCION:
1
𝓊𝓊 = 𝑥𝑥 𝑑𝑑𝑑𝑑 = 𝑒𝑒 𝑥𝑥 𝑑𝑑𝑑𝑑
� 𝑥𝑥 𝑒𝑒 𝑑𝑑𝑑𝑑
0
𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑 𝑣𝑣 = 𝑒𝑒 𝑥𝑥
1 1
1 1
� 𝑥𝑥 𝑒𝑒 𝑑𝑑𝑑𝑑 = �𝑥𝑥𝑒𝑒 𝑥𝑥 − � 𝑒𝑒 𝑥𝑥 𝑑𝑑𝑑𝑑 � = [ 𝑥𝑥𝑥𝑥 𝑥𝑥 − 𝑒𝑒 𝑥𝑥 ]
0 0 0 0
40. SOLUCION:
−𝑥𝑥 2
⎧
𝑓𝑓(𝑥𝑥) = 2 , −≤ 𝑥𝑥 < 0
⎨ 𝑥𝑥 , 0 ≤ 𝑥𝑥 < 1
2
⎩ 2
𝐿𝐿𝑇𝑇 = 𝐿𝐿1 + 𝐿𝐿2 , 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝐿𝐿1 = 𝐿𝐿2
1 1
→ 𝐿𝐿𝑇𝑇 = 2𝐿𝐿2 = 2 � �1 + 𝑦𝑦´2 𝑑𝑑𝑑𝑑 = 2 � �1 + 𝑥𝑥 2 𝑑𝑑𝑑𝑑
0 0