ĐỀ THI GIỮA KÌ GT3 HỌC KÌ 20193 – NHÓM NGÀNH 2 Câu 1
Trang 1ĐỀ THI GIỮA KÌ GT3 HỌC KÌ 20193 – NHÓM NGÀNH 2
Câu 1:
𝑎) ∑ 1
𝑛 √ln 𝑛3
∞
𝑛=2
𝐶ó 𝑢𝑛 > 0∀𝑛 ≥ 2 → 𝑐ℎ𝑢ỗ𝑖 𝑑ươ𝑛𝑔
𝑋é𝑡 𝑓(𝑥) = 1
𝑥 √ln 𝑥3 +) 𝑓′(𝑥) = −
1
3 ln
−23𝑥 (𝑥 √ln 𝑥3 )2
< 0 → 𝑓(𝑥) đơ𝑛 đ𝑖ệ𝑢 𝑔𝑖ả𝑚
+) lim
𝑥→∞
1
𝑥 √ln 𝑥3 = 0
𝑥 √ln 𝑥 3 𝑑𝑥 = ∫ 𝑡
−13
𝑑𝑡 =3
2𝑡
2
ln 2= ∞ → 𝑇𝑝 𝑝ℎâ𝑛 𝑘ì
∞
ln 2
∞
2
→ 𝐶ℎ𝑢ỗ𝑖 đã 𝑐ℎ𝑜 𝑝ℎâ𝑛 𝑘ì 𝑡ℎ𝑒𝑜 𝑇𝐶 𝑡í𝑐ℎ 𝑝ℎâ𝑛
𝑏) ∑ √𝑛3 (𝑒𝑛12 − 1)
∞
𝑛=1
𝑢𝑛 > 0∀𝑛 ≥ 1
𝑢𝑛 = √𝑛3 (𝑒𝑛12 − 1) ~√𝑛3 1
𝑛2 = 1
𝑛53 ( 𝑑𝑜 𝑒𝑡− 1 ~ 𝑡 𝑘ℎ𝑖 𝑡 → 0 )
𝑀à ∑ 1
𝑛53
ℎộ𝑖 𝑡ụ (5
3> 1)
∞
𝑛=1
→ 𝐶ℎ𝑢ỗ𝑖 đã 𝑐ℎ𝑜 ℎộ𝑖 𝑡ụ 𝑡ℎ𝑒𝑜 𝑡𝑖ê𝑢 𝑐ℎ𝑢ẩ𝑛 𝑠𝑜 𝑠á𝑛ℎ
Câu 2:
√𝑛(
2𝑥3
𝑥6+ 4)
𝑛
∞
𝑛=1
𝑢𝑛(𝑥) = 1
√𝑛(
2𝑥3
𝑥6+ 4)
𝑛
= 1
√𝑛(
𝑥3 2
𝑥6
4 + 1
)
𝑛
≤ 1
√𝑛 (
1
2)
𝑛
2𝑛 √𝑛
𝑋é𝑡 ∑ 1
2𝑛 √𝑛 𝑙à 𝑐ℎ𝑢ỗ𝑖 𝑑ươ𝑛𝑔 𝑑𝑜
1
2𝑛 √𝑛 > 0∀𝑛 ≥ 1
∞
𝑛=1
lim
𝑛→∞
2𝑛 √𝑛
2.2𝑛 √𝑛 + 1=
1
2 < 1 → 𝐶ℎ𝑢ỗ𝑖 ℎộ𝑖 𝑡ụ 𝑡ℎ𝑒𝑜 𝑇𝐶 𝐷
′𝐴𝑙𝑒𝑚𝑏𝑒𝑟𝑡
→ 𝐶ℎ𝑢ỗ𝑖 đã 𝑐ℎ𝑜 ℎộ𝑖 𝑡ụ đề𝑢 𝑡𝑟ê𝑛 𝑅 𝑡ℎ𝑒𝑜 𝑇𝐶 𝑊𝑒𝑖𝑒𝑟𝑠𝑡𝑟𝑎𝑠𝑠
Câu 3:
Trang 2∑ ( 𝑛 − 1
2𝑛 + 1)
𝑛 (2𝑥 + 1)𝑛
∞
𝑛=0
𝑋é𝑡:
lim
𝑛→∞ | 1
√𝑢𝑛(𝑥)
𝑛→∞ |2𝑛 + 1
𝑛 − 1 | |
1 2𝑥 + 1| =
2
|2𝑥 + 1| < 1
→ −2 < 2𝑥 + 1 < 2 → −3
2 < 𝑥 < −
1 2
𝑉ậ𝑦 𝑚𝑖ề𝑛 ℎộ𝑖 𝑡ụ 𝑐ầ𝑛 𝑡ì𝑚 𝑙à (−3
2; −
1
2) Câu 4:
𝑓(𝑥) = 𝑥𝑙𝑛(2 − 𝑥) = 𝑥 ln 2 + 𝑥 ln (1 −𝑥
2)
→ 𝑓(𝑥) = 𝑥 ln 2 + 𝑥 ∑ −(
𝑥 2) 𝑛+1
𝑛 + 1
∞
𝑛=0
∀ |𝑥
2| < 1
= 𝑥 ln 2 + ∑ −1
2.
1
2𝑛(𝑛 + 1) 𝑥
𝑛+2 ∀|𝑥| < 2
∞
𝑛=0 Câu 5:
𝑎) (1 − 𝑥) + 𝑥𝑦′𝑦 = 0
→ (1 − 𝑥) = −𝑥𝑑𝑦
𝑑𝑥𝑦 →
𝑥 − 1
𝑥 𝑑𝑥 = 𝑦𝑑𝑦 → 𝑥 − ln|𝑥| + 𝐶 =
𝑦2 2
→ 𝑦 = ±√2𝑥 − 2 ln|𝑥| + 𝐶
𝑏)(𝑥 − 2𝑦)𝑑𝑥 + 𝑥𝑑𝑦 = 0
+) 𝑥 = 0 𝑙à 𝑛𝑔ℎ𝑖ệ𝑚 𝑘ì 𝑑ị
+) 𝑥 ≠ 0 𝐶ó
𝑥 − 2𝑦 + 𝑥 𝑦′ = 0 → 𝑦′ −2
𝑥𝑦 = 1 (𝑃𝑇𝑉𝑃 𝑡𝑢𝑦ế𝑛 𝑡í𝑛ℎ )
→ 𝑦 = 𝑒∫ 𝑑𝑥(∫ 1 𝑒∫ −2𝑥𝑑𝑥𝑑𝑥 + 𝐶 )
= 𝑥2(−1
𝑥 + 𝐶) = −𝑥 + 𝐶𝑥
2 𝑐)(3𝑥2𝑦 + 2 cos 𝑦)𝑑𝑥 + (𝑥3 − 2𝑥𝑠𝑖𝑛 𝑦)𝑑𝑦 = 0
𝑃(𝑥; 𝑦) = 3𝑥2𝑦 + 2 cos 𝑦 → 𝑃𝑦′ = 3𝑥2− 2 sin 𝑦
𝑄(𝑥; 𝑦) = 𝑥3− 2𝑥𝑠𝑖𝑛 𝑦 → 𝑄𝑥′ = 3𝑥2− 2 sin 𝑦
→ 𝑃𝑇𝑉𝑃 𝑡𝑜à𝑛 𝑝ℎầ𝑛
Trang 3𝐶 = ∫ 𝑃(𝑥; 0)𝑑𝑥 + ∫ 𝑄(𝑥; 𝑦)𝑑𝑦 = ∫ 2𝑑𝑥 + ∫ 𝑥3− 2𝑥𝑠𝑖𝑛 𝑦 𝑑𝑦
𝑦 0
𝑥 0
𝑦 0
𝑥
0
= 2𝑥 + 𝑥3𝑦 + 2𝑥𝑐𝑜𝑠 𝑦
Câu 6:
𝑓(𝑥) = {0 𝑛ế𝑢 − 2 < 𝑥 < 0
1 𝑛ế𝑢 0 < 𝑥 < 2 𝑡𝑢ầ𝑛 ℎ𝑜à𝑛 𝑐ℎ𝑢 𝑘ỳ 4
𝑎0 = 1
2∫ 𝑓(𝑥)𝑑𝑥 =
1
2(∫ 0𝑑𝑥 + ∫ 1𝑑𝑥
2 0
0
−2
) = 1
2 2 = 1
2
−2
𝑎𝑛 = 1
2∫ 𝑓(𝑥) cos
𝑛𝜋𝑥
2 𝑑𝑥 =
1
2(∫ 0 cos
𝑛𝜋𝑥
2 𝑑𝑥 + ∫ cos
𝑛𝜋𝑥
2 𝑑𝑥
2 0
0
−2
) 2
−2
= 1
2 sin
𝑛𝜋𝑥
2 .
2
𝑛𝜋|
𝑥 = 2
𝑥 = 0 =
1
𝑛𝜋sin 𝑛𝜋 = 0 𝑣ớ𝑖 𝑛 = 1,2,3, …
𝑏𝑛 = 1
2∫ 𝑓(𝑥) sin
𝑛𝜋𝑥
2 𝑑𝑥 =
1
2(∫ 0 sin
𝑛𝜋𝑥
2 𝑑𝑥 + ∫ sin
𝑛𝜋𝑥
2 𝑑𝑥
2 0
0
−2
) 2
−2
= −1
2 cos
𝑛𝜋𝑥
2 .
2
𝑛𝜋|
𝑥 = 2
𝑥 = 0
= − 1
𝑛𝜋cos 𝑛𝜋 +
1
𝑛𝜋 =
(−1)𝑛+1 + 1
𝑛𝜋 𝑣ớ𝑖 𝑛 = 1,2,3, …
→ 𝐾ℎ𝑎𝑖 𝑡𝑟𝑖ể𝑛 𝐹𝑜𝑢𝑟𝑖𝑒𝑟 𝑐ủ𝑎 ℎà𝑚 𝑙à:
𝑓(𝑥) =
{
1
2+ ∑
(−1)𝑛+1 + 1
𝑛𝜋 sin
𝑛𝜋𝑥 2
∞
𝑛=1
𝑣ớ𝑖 𝑥 ≠ 0 𝑓(0 + 0) + 𝑓(0 − 0)
1
2 𝑣ớ𝑖 𝑥 = 0 ( Đị𝑛ℎ 𝑙í 𝐷𝑖𝑟𝑖𝑐ℎ𝑙𝑒𝑡 )