Because they are all oral calculations, there may be some calculation errors.

18.

From the expression of on, we can know that n is on the straight line AB. Naturally, when lamd is the same, the abscissa of n is the same as that of m, which is determined by the properties of straight lines. The problem is to find the absolute value of the maximum difference between the function value f(x) and the line segment connecting AB in the interval [1, 2]. This should also be the reason for the so-called linear approximate name. Let's go

For option a

Y = x^2, the straight line between point A (1, 1) and point B (2,4) can be written as y = 3x-2.

Then max | x2-3x+2 | = max | (x-3/2) 2-1/4 | is on [1, 2], and the maximum value is1/4.

For option b

Y = 2/x, A( 1, 2) B(2, 1) straight line can be written as y = -x+3.

Then 2/x+x-3 is derived from the basic inequality. When 2/x = x, the minimum value is taken and the maximum value is taken as the endpoint. Since the maximum value is zero and the absolute value is taken at the minimum point, when x = sqrt(2), the absolute value is taken as 3-2sqrt(2).

For option c

Y = sin (pi/3) x, a (1, sqrt (3)/2), b (2, sqrt (3)/2) so the straight line is y = sqrt(3)/2.

Max | sin(pi/3)x-sqrt(3)/2 | takes the maximum value at x = 3/2, that is, 1-sqrt(3)/2 (this conclusion can be obtained by looking at the image or taking derivative).

For option d

The line Y = x- 1/x, A( 1, 0) and b (2,3/2) can be written as y = 3/2 *x-3/2.

Max | x-1/x-3/2 * x+3/2 | = max |1/2 * x+1/x-3/2 | still uses the basic inequality, when1/2 * x =

So comparing the conclusions, we know that 3/2-sqrt(2) is the smallest, so the answer is D.

Question 13, I guess you don't know the direction of the vertical line. Actually, the straight line perpendicular to BB 1D 1D must be parallel to AC, because DD 1 is perpendicular to AC and BD is perpendicular to AC, so AC is perpendicular to BB1d1d.. Then after point p, take AC as the direction, and you can draw a conclusion. When p is located at the midpoint of BD 1, it happens to be a critical situation. I think the rest of the students can solve it by themselves.

14 I'm not sure. My personal opinion is that starting from the last an, an can only be one of n, n- 1, n-2. If an has been selected, consider an- 1, and the condition an- 1 can only be n, n-65448. So an- 1 can only choose one of three, and then consider an-2. You must choose from n, N- 1, N-2, N-3 and N-4, two of which can't be used any more, and you can only choose from three. By analogy, when considering a3, a4 needs to choose from 2, ***n- 1, n-4 have all been selected, so a4 has only three choices. As for a 1, A2 and A3, there is no limit, and the remaining three elements can be arranged at will, ***3! . So there are always 3 (n-2) * 2 kinds of * *.