High Mathematics Freshman Mathematics O O Thank you

f(x) = (a-->x)f(t)dt+ (b-->x)f(t)dt is continuous in the interval [a,b].

The idea of proof is to prove that each term is continuous in [a,b], then their sum is continuous.

For any x0 [a,b], as long as lim(x-->x0) (a-->x)f(t)dt= (a-->x0)f(t)dt

That is, lim(x-->0) (a-->x)f(t)dt- (a-->x0)f(t)dt=0.

The above equation is equal to lim(x-->x0) (x0-->x)f(t)dt

Because f(x) is continuous, according to the integral median theorem there is a relationship between x and x0 such that (x0-->x)f(t)dt=f( )x-x0)--0

i.e. = (a-->x)f(t)dt is continuous at x0, and since the arbitrariness of x0 is known to be continuous at [a,b], similar to proving the other.

I'm too lazy to use my brain, I'll help you, I hope it can help you, this is the final exam question from my freshman year. Let f(x)=f(x)-g(x), according to the conditions, f(x) is continuous on [a,b] and f(x)dx=0, then there is x1 x2 [a,b], such that f(x1) 0, f(x2) 0. So there is x [x1,x2] such that f(x)=0

From the symmetry of the image of f(x)=x2+1 and g(x) with respect to x=1, it can be seen that when x=1 is g(x)=f(x)=2

Substituting x=1 into the option, lead selling tan a=5 4, b=2, c=5, d=2, so a and c can be excluded, and the matching case is selected in b and d.

But the answer I made was not b, but Huaitong d, yes ......Are you sure the correct answer is B?

f(x)=x 2+1 and g(x) are symmetrical with respect to the hidden search line x=1, so it must be Sun Chang's after translation.

According to f(x) vertex is (0,1), it can be seen that g (caleb x) fixed point is (2,1), so g(x) = (x 2) 2+1, simplified to d x 2-4x+5, your answer still has a problem

Landlord, I'm sorry, thank you for the one on the third floor, I know that there is a problem, this problem should be changed to siny, so that it is not a transcendent equation, and there will be no contradictions.

Because of the equal difference, 2lg(sinx-1 2)=lg3+lg(1-y), i.e., (sinx-1 2) 2=3(1-y).

Because sinx belongs to [-1,1].

When sinx=1, the minimum value of y is 11 12

When sinx = 1 2, y max is 1

According to the meaning of the question, Y is a fixed value, not a range, the question is wrong, right? Change a siny to sinx according to the 2nd floor solution:

2LG(sinx-1 2)=LG3+LG(1-y), i.e. (sinx-1, 2) 2=3(1-y).

Because sinx belongs to [-1,1].

When sinx=1, the minimum value of y is 11 12

When sinx = 1 2, y max is 1

In addition, it is not a transcendent equation.

ps: The 2nd floor solution has a common contradiction error in functions, siny=1, and y is 2k + 2 instead of 11 12 [2nd floor has now been corrected].

There is a maximum value of 1 and no minimum value.

There are minimum values of 11 12 and no maximum values.

There are minimum values of 11 and 12 and maximum values of 1

There is a minimum value of -1 and a maximum value of 1

Ziqing Muxue: Hello.

The distance from the center of the circle o of the semicircular tunnel to the wheel side a of the truck is:

The apex of the truck b, ab is the height of the truck.

The radius OA is the hypotenuse of the right-angled triangle ABO.

According to the Pythagorean law:

ab＝√（ab＝√

A: The height of the truck cannot exceed meters.

The truck walks in the middle and establishes a coordinate system with the center of the circle as the origin, r=x 2+y 2=

Let x= bring in to find y

y is approximately equal to b

Make a Cartesian coordinate system for a semicircular tunnel with a radius, and the coordinates of the tunnel in the Cartesian coordinate system are (,0), (0), (0, let y=ax2(x's square) + bx+c bring these three points into the equation to find the equation, the truck is wide, so bring (,y) in to find y, and then find the number less than or equal to y in the options!

The semicircular tunnel equation can be set to x 2 + y 2 =, and the width of the car is in meters, that is, the integer part of y obtained by substituting x = into the equation can be obtained by drawing the drawing, which is the maximum height of b

Solution: The tangent equation for the point (x,y) is y-y=y'(x-x), i.e. y=y'x+y-xy'

The intersection point of this tangent with the ordinate axis is (0,y-xy')==> tangent segment length from tangent (x,y) to the ordinate axis is [x +(xy.]'The tangent of a point (x,y) on a curve has a fixed length of 2 [x + (xy')²]=2

>x²+(xy')²=4

>x²+x²y'²=4

>x²(y'²+1)=4

So the differential equation that the curve should satisfy is x (y'²+1)=4

You can start by comparing f(x) = x 3, x (0,1) f(x) and y x images who is on top.

That is, f(x) x x 3 x x(x 2 1) is less than 0 at x (0,1).

So at x(0,1), f(x) is below the y x image, so its corresponding inverse must be above the y x image.

So f(x1).

f(x) is a concave function in the interval (0,1).

The inverse function of f(x) g(x) is symmetrical with respect to y=x, and g(x) is a convex function in the interval (0,1).

To put it simply, f(x) is in the (0,1) range, below y=x, and g(x) is above y=x.

so，f(x)

The easiest and most direct way for this type of question is to try it with a number, and the test is the most suitable ......

First, find the domain that defines it, where a>=0 and the domain is x>=0When a<0 defines the domain as x>=-aThis function is an increasing function, the smallest x is the minimum y, and the minimum value of y is 3 2, when a>= takes the minimum value when x=0 then a=3 2, and the same is < a

by s12=12a1+66d>0

s13=13a1+78d<0

Get a1+6d<0

a1+11/2d>0

Therefore, a7<0 a6>-d 2 Therefore, a6 absolute value and a7 absolute value are called powerful!

You should choose C, as the title, this number column is a decreasing series, S13=13A7<0, S12=6(A6+A7)>0, thus, A6>-A7>0, so choose C, I hope it will help you.

Take advantage of the equivalence term.

The first 13 terms<0 indicate that the sixth term is <0 as the middle term of the equal difference, and s12>0 means that s11>0 means that the fifth term is greater than 0, so the smallest contention is the fifth and sixth terms.

Let's look at s12>0 again, which shows that as the middle term of the difference (fifth term + sixth term), 2>0 strongly shows that the absolute value of the sixth term is less than the fifth term.

Because it is a periodic function, f(x)=f(x-t), because y=f(x), x (0,t), there is an inverse function y=g(x),x d, so the inverse function of y=f(x),x(t,2t) is y=g(x-t), x d