Answers:
Question #1
100
∑(5𝑖 − 5𝑖−1 ) = 51 − 1 + 52 − 51 + ⋯ + 599 − 598 + 5100 − 5999 = 5100 − 1
𝑖=1
Answer: D.
Question #2
As the function is decreasing:
𝑥𝑎 > 𝑥𝑏
Thus, we will have that:
𝐿𝑛 > 𝐴 > 𝑅𝑛 => 𝑅𝑛 < 𝐴 < 𝐿𝑛
Answer: E.
Question #3
𝑓(𝑥) = ∫(3𝑥 2 + 2𝑥)𝑑𝑥 = 𝑥 3 + 𝑥 2 + 𝑐.
If 𝑓(1) = 13 + 12 + 𝑐 = 2 + 𝑐 = 4. Thus, 𝑐 = 4 − 2 = 2:
𝑓(𝑥) = 𝑥 3 + 𝑥 2 + 2
𝑓(2) = 23 + 22 + 4 = 8 + 4 + 4 = 16
Answer: D.
Question #4
Answer: C.
Question #5
𝑛
∑(2(𝑖 + 2)(3𝑖 − 2) − 6𝑖
𝑖=1
𝑛
2)
= ∑(8𝑖) = 4𝑛2 − 4𝑛
𝑖=1
Answer: D.
Question #6
∫ 𝑓 ′ (𝑥)𝑑𝑥 = ∫(𝑔′ (𝑥) − 4)𝑑𝑥 = 𝑔(𝑥) − 4𝑥 + 𝑐
Where 𝑐 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.
Thus, the answer E could be true.
Answer: E.