Prob adicionales subidas 2020_II_GRÁFICAS

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𝑇 =
7𝜋
2
− −
𝜋
2
= 4𝜋 →
2𝜋
𝑏
= 4𝜋 → 𝑏 =
1
2
𝑓 𝑥 = 𝑎𝑐𝑜𝑠 𝑏𝑥 + 𝑐 + 𝑑
No hay desplazamiento vertical: d=0
𝑓 𝑥 = 3𝑐𝑜𝑠
𝑥
2
+ 𝑐
𝐿𝑢𝑒𝑔𝑜: 3 = 3𝑐𝑜𝑠 −
𝜋
4
+ 𝑐 → 𝑐 =
𝜋
4
∴ 𝑓 𝑥 = 3𝑐𝑜𝑠
𝑥
2
+
𝜋
4
𝐴𝑀𝑃𝐿𝐼𝑇𝑈𝐷: 𝑎 =3
𝑓 𝑥 = 𝑎𝑠𝑒𝑛 𝑏𝑥 + 𝑐 + 𝑑
𝑇 = 2
5𝜋
4
−
𝜋
2
=
3𝜋
2
→
2𝜋
𝑏
=
3𝜋
2
→ 𝑏 =
4
3
𝐷𝑒𝑠𝑝𝑙𝑎𝑧𝑎𝑚𝑖𝑒𝑛𝑡𝑜 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙: 𝑑 =
3 − 5
2
= −1
𝐴𝑀𝑃𝐿𝐼𝑇𝑈𝐷: 𝑎 =
3 − −5
2
= 4
𝑓 𝑥 = 4𝑠𝑒𝑛
4𝑥
3
+ 𝑐 − 1
𝐿𝑢𝑒𝑔𝑜: 3 = 4𝑠𝑒𝑛
2𝜋
3
+ 𝑐 − 1 → 𝑐 = −
𝜋
6
→ 𝑎 + 𝑏𝜋 + 𝑐 + 𝑑 =
18 + 7𝜋
6
𝑓 𝑥 = 𝐴𝑐𝑠𝑐 𝑀𝑥 → 𝑇 =
𝜋
2
→
2𝜋
𝑀
=
𝜋
2
→ 𝑀 = 4
g 𝑥 = 𝐴𝑐𝑠𝑐 𝑁𝑥 → 𝑇 =
𝜋
3
→
2𝜋
𝑁
=
𝜋
3
→ 𝑁 = 6
Por lo tanto: 𝑀 +𝑁 = 10
𝑓 𝑥 = 𝑎𝑠𝑒𝑛 𝑏𝑥 + 𝑐 + 𝑑
𝑇 = 2
7𝜋
32
−
𝜋
32
=
3𝜋
8
→
2𝜋
𝑏
=
3𝜋
8
→ 𝑏 =
16
3
𝐷𝑒𝑠𝑝𝑙𝑎𝑧𝑎𝑚𝑖𝑒𝑛𝑡𝑜 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙: 𝑑 =
8 − 2
2
= 3
𝐴𝑀𝑃𝐿𝐼𝑇𝑈𝐷: 𝑎 ==
8 − −2
2
= 5
𝑓 𝑥 = 5𝑠𝑒𝑛
16𝑥
3
+ 𝑐 + 3
𝐿𝑢𝑒𝑔𝑜: 8 = 5𝑠𝑒𝑛
16𝜋
3(32)
+ 𝑐 + 3 → 𝑐 =
𝜋
3
𝑓 𝑥 = 5𝑠𝑒𝑛
16𝑥
3
+
𝜋
3
+ 3
𝑆𝑖 𝑥 = 0 → 𝐶 = 0; 1
𝑆𝑖 𝑓 𝑥 = 0 → 𝑡𝑎𝑛
𝑥
2
= −1 →
𝑥
2
= −
𝜋
4
→ 𝐴 = −
𝜋
2
; 0
𝑇 =
𝜋
1
2
= 2𝜋
Ahora 𝑓 𝐵𝑥 = 𝑡𝑎𝑛
𝐵𝑥
2
+ 1 = 3 + 3 → 𝑡𝑎𝑛
𝐵𝑥
2
= 2 + 3
Luego: 
𝐵𝑥
2
=
5𝜋
12
→ 𝐵 =
5𝜋
6
; 0
El área de la región sombreada es:
S=
1
2
5𝜋
6
+
𝜋
2
=
2𝜋
3
𝑇 = 5𝜋 →
2𝜋
𝑏
= 5𝜋 → 𝑏 =
2
5
𝑓 𝑥 = 𝑎𝑠𝑒𝑛
2𝑥
5
− 6 = 𝑎𝑠𝑒𝑛
2
5
10𝜋
3
− 6 = 𝑎𝑠𝑒𝑛
4𝜋
3
= −
3
2
𝑎
a = 2 2
∴ a. 𝑏 =
4 2
5
𝑇 = 4 →
2𝜋
𝐵
= 4 → 𝐵 =
𝜋
2
𝑓 𝑥 = 3𝑠𝑒𝑐
𝜋𝑥
2
− 1
𝐴 =
2 − −4
2
= 3 ; 𝐶 =
2 − 4
2
= −1
𝐵. 𝐶
𝐴
= −
𝜋
6
𝑇1 = 2𝜋 → 𝑓 𝑥 = 𝑎𝑠𝑒𝑛 𝑥
𝑇2 =
2𝜋
3
→ 𝑔 𝑥 = 𝑎𝑠𝑒𝑛 3𝑥
Si 𝑠𝑒𝑛 3𝑥 = 𝑠𝑒𝑛 𝑥 →
𝑠𝑒𝑛 3𝑥
𝑠𝑒𝑛 𝑥
= 1 → 2𝑐𝑜𝑠 2𝑥 + 1 = 1 → 𝑐𝑜𝑠 2𝑥 = 0
Luego: 𝑥 =
2𝑘 + 1 𝜋
4
→ 𝑀𝑥 =
3𝜋
4
𝑦 𝑁𝑥 = −
3𝜋
4
y = 𝑠𝑒𝑛 2𝑥 → 𝑇 =
2𝜋
2
= 𝜋
y = 𝑠𝑒𝑛 𝐵𝑥 → =
2𝜋
𝐵
=
𝜋
2
→ 𝐵 = 4 → 𝑦 = 𝑠𝑒𝑛 4𝑥
Ahora: 𝑠𝑒𝑛 2𝑥 = 𝑠𝑒𝑛 4𝑥
cos 2𝑥 =
1
2
→ 𝑥 =
𝜋
6
h
h= 𝑠𝑒𝑛 2
𝜋
6
=
3
2
𝐸𝑙 á𝑟𝑒𝑎 𝑑𝑒 𝑙𝑎 𝑟𝑒𝑔𝑖ó𝑛 𝑠𝑜𝑚𝑏𝑟𝑒𝑎𝑑𝑎 𝑒𝑠:
3
2
𝜋
2
−
𝜋
6
=
3𝜋
6
𝑢2