A high school one two math problem 20

Even function, the image is symmetrical with respect to the y-axis, option 2

If you lease X sets of A units, then lease equipment B (50-5x) 6 sets will have 10x+(50-5x) 6*20=14060x+1000-100x=840

40x=160

x=4（50-5x）/6=5

Therefore, the minimum cost is 4 * 200 + 5 * 300 = 800 + 1500 = 2300 yuan.

1 There is no other way, because this problem is to convert f(1 x) away.

The specific method is af(x)+f(1 x)=ax af(1 x)+f(x)=a x

Multiply 1 or so by a- at the same time to get a*af(x)-f(x)=a*ax-a x=a(ax -1) [preceding a -1)x] (a≠ 1).

2 It is known that x belongs to n*, f(x)={x-5, x 6 f(x+2), x 6

Seek f(3)=? When x=3, f(x)=f(x+2)=f(5)=f(x+2)=f(7)= x-5=2

The answer should be 2, and the answer you give is 6, which is wrong.

In the first question, find the first and second derivatives, you can know the approximate shape of the function in the coordinate axis, and you can estimate the zero point.

In the second question, because (0,+ decreasing and even function, there must be a function image opening downward, so m2-m-2 < 0 and m2-m-2 should be negative even m=0 or 1

1. 4/x+9/y=1

(4y+9x) xy=1

xy=4y+9x=(4y+9x)(4/x+9/y)=72+16y/x+81x/y

From the mean inequality, we get xy 72+2 (16, 81)=144, so there is a minimum value of 144

Choose C2Let the two right-angled sides be x,y

Then the title is: x+y+ (x+y)=4

x+y≥2√xy

x²+y²≥2xy

So x+y+ (x +y)=4 2 xy+ 2xy, simplified to xy 24-16 2

That is, the maximum value of the area is 12-8 2

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10.4 x+9 y>=2 root number 36 xy=1 Simplification: xy<=12 choose b

11.When the circumference of the right triangle is given, the area of the isosceles right triangle is the largest, so let the right angle side length be x (2 + root number 2) x = 4 x = 4-2 root number 2 area is 1 2x square = 12-8 root number 2 choose d

The first one is C

Let 4 x=cos2a and 9 y=sin2a, then xy=36 cos2asin2a=144 sin22a, since the maximum value of sin22a is 1, the minimum value of xy is 144

When the circumference of the second right triangle is given, when the two right-angled sides are equal, the area is maximum, and the right-angled side length is a, then (2 + root number 2) a=4, a=4-2 root number 2 area is 1 2a2=12-8 root number 2, so choose d

10.Using the important inequality: a+b>= ab (both a,b are greater than zero), we get: 1=4 x+9 y>=2 (36 xy)simplify: xy<=144 choose c

11. Let the two right-angled sides of the right-angled triangle be x, y

From the title: x+y+ (x 2+y 2)=4 due to x+y>=2 xy

x^2+y^2>=2xy

Substituting the above formula to simplify, we get: 4>=(2+ 2) (xy) So xy<=24-16 2

The area of the triangle is s=1 2xy<=12-8 2 Choose d

4y+9x=xy

xy=4y+9x>=2√36xy

xy>=12

xy>=144

So the minimum value = 144

Therefore, C2Solution: The two right-angled sides of a right-angled triangle can be set to be a and b, and the hypotenuse is c

a²+b²=c²

a+b+c=4

From a+b+c=4: 4-c=a+b, both sides are squared at the same time, and the :

16-8c+c²=a²+b²+2ab

16-8c+c²=c²+2ab

c²+8c≥16

c²+8c+16≥32

c+4)²≥32

Because c>0, the solution obtains: c 4 2-4, from the know: ab=8-4c, so

Area s=1 2 ab=1 2 (8-4c)=4-2c, it can be seen that to make area s largest, then c must be the smallest, from the above, the minimum value of hypotenuse c is c=4 2-4, then the maximum value of area is:

s max = 4-2 (4 2-4) = 12-8 2 so choose d