High School Math Problems Rush !!

Let u = k of log4 (i.e., the logarithm of k with 4 as the base).

f(x)=u^2(x-1)-6ux+x+1(u^2-6u+1)x-u^2+1

At [0,1] is always positive, so f(0)>0,f(1)>0 -u 2+1>0

u^2-6u+1-u^2+1>0

Solution: -11, so 1 4, do you have an answer? I guess I did nothing wrong, right?

Algebra first.

Geometry first. So geometry.

perpendicular to , then a perpendicular to any line within . Let b not be perpendicular to the intersection of the plane, then b is not perpendicular. So b is wrong.

c.If A is not perpendicular to , then A is not a straight line within , so A is not perpendicular to B, which contradicts the problem, so C is wrong.

d.There are an infinite number of planes that pass through A and B, of course, not necessarily hammers, and may intersect. So d is wrong.

The answer should be that at least one of A and B is perpendicular to the plane of the opposite side.

(1+sina)/(1-sina)-√1-sina)/(1+sina)

1+sina)^2/(1-sina)(1+sina)-√1-sina)^2/(1+sina)(1-sina)

1+sina-(1-sina)]/√(1-sin^2a)2sina/|cosa|

Because. a is the angle of the third quadrant, so cosa<0)-2tana

If you answer the algebra problem first, the probability of getting 30 points for the algebra problem is 15 points.

The 50 points for geometry questions can only be obtained if you answer both questions correctly, so the expected value is 50*.

So it's 25 points.

Instead, do geometry first and then algebra.

Same as above, 50* points.

So you should do geometry first.

For a quadratic equation of a single element, its basic form is ax2+bx+c=0, then for x2+(4m+1)x+2m-1=0.

A is 1 and B is (4m+1).

c is (2m-1).

It can also be said that a is the coefficient of the quadratic square, b is the coefficient of the first quadratic figure, and c is the constant term!

1) Find the maximum and minimum values of y (3sinx 1) (sinx 2).

Solution: From the known y (3sinx 1) (sinx 2), sinx (2y 1) (3 y) is obtained

sinx| ≤1

(2y－1)/(3－y)| 1

(2y－1)/(3－y)]²1

y 2) (3y 4) 0 and y≠3

Solution 2 y 4 3

The maximum value of the known function is 4 3 and the minimum value is 2

2) Find the range of the function y (tan x tanx 1) (tan x tanx 1).

Solution: From the known y (tan x tanx 1) (tan x tanx 1), obtained.

y－1)tan²x＋(y＋1)tanx＋y－1＝0

If y 1, then only tanx 0 is needed

If y≠1, then δ (y 1) 4(y 1) 0

Get (3y 1) (y 3) 0

Solution 1 3 y 3 (y≠1).

When x k 4 (k z), y 1 3

When x k 4 (k z), y 3

Based on the above, it can be seen that the range of the known function is [1 3,3].

3) Find the maximum and minimum values of the function y (2 sinx) (2 cosx).

Solution 1: Remove the denominator: sinx ycosx 2 2y

i.e. sin(x)2 2y) [ (1 y )].

sin(x－φ)1

2－2y)/[√(1＋y²)]1

i.e. 3y 8y 3 0

Solution: (4 7) 3 y (4 7) 3

Therefore, ymin (4 7) 3, ymax (4 7) 3

Solution 2: Let x cosx, y sinx, then x y 1, which denotes the unit circle, so that only the extreme value of the slope of y k(x 2) 2 is required for the system of straight lines that are common to (2,2), and it is obvious that the extreme value is obtained when the line and the circle are tangent.

By |－2k＋2|/[√(k²＋1)]＝1

Solution k (4 7) 3

Solution 3: Let t tan (x 2), and get y (2t 2t 2) (3t 1) from the universal formula

i.e. (2 3y) t 2t 2 y 0

by {δ 02 3y≠0

Get 3y 8y 3 0

Solution: (4 7) 3 y 4 7) 3,y≠2 3

Apparently when y 2 3, t 2 3 satisfies.

ymin＝(4－√7)/3，ymax＝(4＋√7)/3

Let y (3sinx+1) (sinx+2) then 3sinx+1=ysinx+2y

3-y)sinx=2y-1

sinx=-(2y-1)/(y-3)

Because -1 sinx 1

So -1 -(2y-1) (y-3) 1 gives -2 y 4 3

Maximum: 4 3, sinx=1

Minimum: -2, sinx=-1

Adopt it first, and then I'll answer it, I won't answer it here.

I'm a teacher, thank you for adopting.

1, original formula = 1+2tanx (tan x+tanx+1) = 1+2 (tanx+1 tanx+1).

tanx+1 tanx>=2 or less than =-2

Therefore, the value range [-1,1)u(1,3 5].

Right or wrong is not guaranteed.

2, can be converted to the slope problem of the point (2,2) to the point on the circle x 2+y 2=1.

The period with the same sign means that if the sign of x is the same, it is a periodic function, and if the sign of x is opposite, it is (you can make the formula out yourself, which means that you can really understand its mathematical meaning) Remember, me.

If the landlord's parentheses refer to the number of terms, then it is easy to solve the problem and bring n 1 into the formula: a1-a2=a3

If the sequence of fruit numbers an is a series of equal differences, then you can get it.

d<0 so it's not a series of equal differences.

It's because it's a positive series.

So it's a proportional series.

by a1-a2=a3

Yield: a1-a1xq = a1xq squared.

Because a1 = 1

So we get: 1-q=q squared.

is a binary equation.

So it's a solution.

q squared + q-1 = 0

Let's do the math for the landlords below.

Finally, just remove the negative root value.

If you can't even figure out the equation, I can't help it.

Let b>0, from known, f(a+b)-f(a)=f(b)-1>1-1=0, so f(a+b)>f(a)

That is, f(x+b) > f(x) are true.

So, f(x) is an additive function over r.

Take either x1 > x2

From f(a+b)= f(a)+ f(b)-1 let a=x2, b=x1-x2

There is f(x1) = f(x2) + f(x1-x2)-1b=x1-x2>0 f(x1-x2)>1

f(x1) = f(x2) + f(x1-x2)-1> f(x2) so increase the function.

a=0,f(b)=f(0)+f(b)-1

f(0)=1

And because x>0, f(x)>1

So f(x) is an increasing function.

There can be two ways to do it.

1 Cosine theorem.

cosa=(b 2+c 2-a) 2bc find the length of each side of ab1c, and substitute to find cos =1 5

2 space vectors.

D1 is the origin, D1D is the x-axis, D1C1 is the Y-axis, and D1A1 is the z-axis, and the number product of the matching vector is used.

a·b=|a||b|cos still Texas refers to cos = 1 5 attention angles.

Connect AB1 and CB1, as thin as sock AC

ab1=cb1=√（ab²+bb1²）=5ac=√（ab²+bc²）=2√2

cos ab1c=(ab1 +cb1 rubber-ac ) 2ab1cb1=1 5

The figure should be a one-to-one correspondence between the points of a1b1c1d1 above abcd below.

Is there a specific diagram, the distribution of specific points is still not very clear, and it is much easier if there is a diagram.

1) [Note: Easy to know, take 4 RT ABC can be spelled into a diamond, the diamond has exactly one high for 2] Solution: easy to know, the problem can be turned into:

In a rhomboid, where one of its heights is 2, find the minimum of the sum of the two diagonals and the sum of the side lengths. Half the length of the two diagonals can be set to be a and bthen its side length is (a2+b2).

and p=a+b+ (a2+b2)It is easy to know that the rhomboid area s=2ab=2 (a2+b2)===>ab=√(a2+b2).

=>(√2)ab=√[2(a2+b2)]≥a+b≥2√(ab).===>√(ab)≥√2,ab≥2.The equal sign is obtained only when a=b=2.

Therefore p=a+b+ (a2+b2)=a+b+ab 2 (ab)+ab 2+2 2The equal sign is obtained only when a=b=2, so pmin=2+2 2(2) The original equation can be reduced to [sinx-(1 2)]2=(4m-3) 4

=>|sinx-(1/2)=[√(4m-3)]/2.From the question setting, 1 2 [ 4m-3)] 2 3 2and 4m-3 0

=>1≤m＜3.Hence m [1,3).(3) Solution:

It is easy to know that the original equation can be reduced to sin(x+t)=a 5.(where 0 x 2, t is an acute angle, sint=4 5, cost=3 5.)Therefore, the number form is combined with knowledge, -1 a 5 4 5

=>-5＜a＜4.In this case, one axis of symmetry of the curve in (t,2 +t) is x=(3 2)-t.Therefore, a+b=3 -2t=3 -2arcsin(4 5)

4) Solution: easy to know, k≠-1(Otherwise there is -4cosx-3-5=0.)

=>cosx=-2.The original equation can be reduced to (k+1)(1-cos2x)-4cosx+3k-5=0===>(k+1)cos2x+4cosx+4(1-k)=0.

=>[(k+1)cosx-2(k-1)](cosx+2)=0.===>cosx=2(k-1)/(k+1).cosx=-2 (rounded).

by -1 [2(k-1)] (k+1) 1(k∈z)===>k=1.At this point, cosx=0

=>x=mπ+(/2).(m∈z).

1 2 * 2 (1 2) + 2 symbol is not easy to play, it is (2 * root number 2) + 22 2 3>m> 1 2

3 -50<=a<5,a+b=3π

4 k = 1 or 2

1: Convert to a function image problem, i.e., y=x +mx+2 has an intersection with the source y=x+1. Combined with the system of cubic equations.

dux +(m-1)x+3=0 has a solution, b -4ac 0 can get m -2m-11 0 solution zhi (1-2 root 3) m (1 + 2 root 3).

2: From the question, we know that the value range of daob -a 0 a -b, -x is [-a,-b] f(x)-f(x) is the domain of f(x) and f(-x), that is, [-a,a].

3: Because f(x) is an even function, f(-x)=f(x) gives k=1, so f(x)=-x +3 axis of symmetry-2a is equal to b. And because a=-1 0 so the function opening is downward, so the decreasing interval is (,+

Whew, it's finally done, and if you think it's good, add points.