t & lt 1:S = 3t；
t≥ 1:S=4t-t？ = t(4-t)；
T ∈- 1, 3, t= 1 In the interval, it is necessary to study by stages:
T ∈- 1, 1),-1 ≤ t < 1, multiplied by 3, -3 ≤ 3t < 3, S=3t substituted,-3 ≤ s < 3;
T ∈ 1，3。 At this time, S is a parabola with downward opening, the zero point is t=0 and t=4, the vertex is t=2 (the middle of the two zeros), and the interval 1 3 is between the two zeros, which is symmetrical about the vertex t=2, and the minimum value is t = 65438+. The maximum value is at the vertex, t=2, S=4, 3 ≤ s ≤ 4;
Combining the above two paragraphs, -3≤S ≤4, the answer is a.
7. This flow chart is a cycle. The key is to understand which is a loop variable, which is a loop function and what is the initial value. This cycle problem can be summed up as a mathematical sequence problem.
Variables: k, s, t, the value of each periodic variable is different, so we add subscripts to distinguish them, which is also the standard practice of sequences.
Initial value: K 1= 1(K= 1), s0=0(s=0), T0= 1(T= 1). When the initial values of s and t are given, K= 1.
Action: kn+1= kn+1(k = k+1), Tn=Tn- 1/n (T=T/k), sn = s (n-1)+
K is a cyclic variable, and every time you increase 1, it is the common n in the sequence. T n is the general term of the sequence, and sn is the sum of the first n terms of the sequence Tn.
With these mathematical formulas, we can make a list:
K initial value 1 2 3 4
t 1 1/ 1 = 1 1/2 = 1/( 1×2) 1/2/3 = 1/( 1×2×3) 1/2/3/4
s 0 0+T = 0+ 1 = 1 1+ 1/2 = 3/2 1+ 1/2+ 1/2/3 1+ 1/2+ 1/2+ 1/2+ 1/2/3+ 1/2/3+ 1/2/3+0/2/3
Generally speaking, you can understand this rule by listing 3 or 4 items:
Tn= 1/n！
sn= 1/ 1！ + 1/2! +...+ 1/n！
The loop starts at K= 1 and ends at K= 10. When K= 1 1, there is no action. So, the answer: B.
4. It is also a cyclic problem, but note that the initial value of k =2, not 1, so =1. There is something wrong with the title, so S and S should be regarded as the same variable. List:
K initial value 2 3 4
s 1 1×log2(2+ 1)= log2(3)log2(3)log3(4)log2(3)log3(4)log4(5)
= lg3/lg2 = lg3/lg2×lg4/lg3 = lg4/lg2×lg5/lg4
=lg4/lg2=2 =lg5/lg2
Law: s = LG (k+1)/lg2 = log2 (k+1). When s = 3, K+ 1=2? =8，K=7