answer:1. Show that ( n = 13 ) does not work: - Suppose the first player removes the card with number ( 4 ). - The second player has the option to remove two consecutive cards. Consider the following scenarios: - If the second player removes the pair ( (8, 9) ) or ( (9, 10) ): - If the second player removes ( (8, 9) ), the first player can then remove ( (10, 11, 12) ). - If the second player removes ( (9, 10) ), the first player can then remove ( (5, 6, 7) ). - In both cases, the second player cannot find four consecutive numbers to remove in their next move. - If the second player does not remove ( (8, 9) ) or ( (9, 10) ): - The first player can then remove ( (8, 9, 10) ). - The remaining cards are ( (1, 2, 3), (5, 6, 7), (11, 12, 13) ), and the second player cannot find four consecutive numbers to remove in their next move.2. Show that ( n = 14 ) works: - Let the first player remove the card with number ( k ). - We have two groups of consecutive numbers: ( k-1 ) numbers on one side and ( 14-k ) numbers on the other side. - Without loss of generality, assume ( k-1 leq 14-k ), which implies ( k leq 7 ). - Consider the following cases: - If ( k-1 geq 4 ): - The second player can remove the pair ( (k+1, k+2) ). - This leaves two groups: ( (1, 2, 3, ldots, k) ) and ( (k+3, k+4, ldots, 14) ), each with more than 4 elements. - The second player can always find four consecutive numbers to remove in their next move. - If ( k = 4 ) or ( k = 3 ): - The second player can remove the pair ( (1, 2) ). - This leaves at least 10 consecutive numbers, and no matter how the first player removes 3 consecutive numbers, there will still be 4 consecutive numbers left for the second player to remove. - If ( k = 1 ) or ( k = 2 ): - The second player can remove the pair ( (3, 4) ). - This leaves at least 10 consecutive numbers, and no matter how the first player removes 3 consecutive numbers, there will still be 4 consecutive numbers left for the second player to remove.Therefore, the smallest value of ( n ) for which the second player can ensure that he completes both his moves is ( n = 14 ).The final answer is ( boxed{14} ).

answer:Since f^{-1}(x) = x^2 for x > 0, it means that for f(x), we have x = f^{-1}(f(x)) = (f(x))^2. Given that we need to find f(4), we set f(4) = y. Therefore, according to the definition of the inverse function, we have 4 = y^2. Solving this equation for y when y > 0 (since the domain of the inverse function is x > 0), we get y = 2. Thus, f(4) = boxed{2}.

answer:To approach this problem using proof by contradiction, we start by understanding the statement we need to prove: "If a+bgeqslant 0, then at least one of a or b is not less than 0." Proof by contradiction involves assuming the opposite of what we want to prove and then showing that this assumption leads to a contradiction.Step 1: Understand the statement to be proved. We need to show that if the sum of a and b is non-negative, then at least one of them must be non-negative as well.Step 2: Formulate the assumption for proof by contradiction. In this context, the opposite of "at least one of a or b is not less than 0" is "both a and b are less than 0". This is because, for proof by contradiction, we assume the negation of the conclusion to show it leads to a contradiction with the given premise.Step 3: Identify the correct assumption to make. Based on Step 2, the assumption we make for proof by contradiction is that both a and b are less than 0. This assumption will be used to show that it contradicts the given premise that a+bgeqslant 0.Therefore, the correct choice for the first step in using proof by contradiction to prove the given statement is to assume that a and b are both less than 0.Hence, the correct answer is boxed{text{A}}.

answer:Using only digits 1,2,3,4, and 5 , a sequence is created as follows: one 1 , two 2 's, three 3 's, four 4's, five 5 's, six 1's, seven 2's, and so on.The sequence appears as: 1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,1,1,1,1,1,1,2,2, ldots.The 100th digit in the sequence is(A) 1(B) 2(C) 3(D) 4(E) 5## SolutionThe total number of digits in n groups of the sequence is given by 1+2+3+ldots+n. In order to determine the group containing the 100th digit in the sequence, we must find the positive integer n such that 1+2+3+ldots+(n-1)100. By examining a few of these sums we find that 1+2+3+ldots+13=91 and 1+2+3+ldots+13+14=105. Thus the 100th digit in the sequence is in the 14th group. The 100th digit is a 4.ANSWER: (D)

answer:From the given, we have a=4, b=3. The equation of the asymptotes is y=± dfrac {3}{4}x.Therefore, the correct option is boxed{C}.From the given, a=4, b=3, we can determine the equation of the asymptotes of the hyperbola.This question tests the equation and properties of hyperbolas, assessing the student's computational skills, which is quite basic.

answer:Solution:(1) Since when a=1, f(x) geqslant 2x+1,Therefore, |2x-1| geqslant -6x+1,which means begin{cases} & 2x-1geqslant -6x+1, & 2x-1geqslant 0 end{cases}or begin{cases} & 1-2xgeqslant -6x+1, & 2x-1 -2, the inequality f(x)-7x-a^{2}+3 geqslant 0 always holds,which means f(x)-7x geqslant a^{2}-3,Let F(x)=f(x)-7x,Therefore, F(x)=f(x)-7x=|2x-a|=begin{cases}3x-a, & xgeqslant frac{a}{2} a-x, & x 0 and xin(-2,+infty),Therefore, when x= frac{a}{2}, F(x) has its minimum value Fleft( frac{a}{2}right)= frac{a}{2},For the original inequality to always hold, we only need frac{a}{2}geqslant {a}^{2}-3,Therefore, we get boxed{ain(0,2]}.