2 Senior 1 Function Questions I want speed and quality

1+(a-b)/(x+b)

So this is a function similar to the inverse proportional function y=1 x divided into two monotonic intervals (-b) by x=-b, (b, )psss: you can't write the two intervals as unions, you can write them separately.

Proof of monotonicity: It must be proved separately in two monotonic intervals.

Any x1, x2 (-b], x1>x2

f(x1)-f(x2)

a-b) (x1+b)-(a-b) (x2+b)(a-b)(x2-x1)(x1+b)(x2+b) because x1<-b x2<-b

So x1+b<0 x2+b<0

So f(x1)-f(x2)<0

Any x1, x2 (-b], x1>x2

f(x1)-f(x2)

a-b) (x1+b)-(a-b) (x2+b)(a-b)(x2-x1)(x1+b)(x2+b) because x1<-b x2<-b

So x1+b<0 x2+b<0

So f(x1)-f(x2)<0

So the function decreases monotonically on (- b] and [-b,+).

2.Let x1, x2 (0, + and x1 x2.)

Then f(x1)-f(x2)=(x1+1 x1)-(x2+1 x2)(x1-x2)(x1x2-4) (x1x2).

When x1x2 4, i.e. x 2 4, a (0,2) is x1 x2, x1x2 0

Therefore, f(x1)-f(x2)=(x1-x2)(x1x2-4) (x1x2) 0

That is, when a (0,2) f(x) decreases monotonically;

When x1x2 4, i.e. x 2 4, a (2, + and x1 x2, x1x2 0

Therefore, f(x1)-f(x2)=(x1-x2)(x1x2-4) (x1x2) 0

i.e., when a (2,+, f(x) increases monotonically;

Finally, let's not forget that 2 is also in the domain of function definitions, in summary: when a (0,2) f(x) decreases monotonically;

When a[2,+, f(x) increases monotonically.

1. Because f(x)=(x+a) (x+b) (a>b>0), so the function image opening is upward, and the intersection point with the x-axis is (-a,0), (b,0) because a>b>0, so -a<-b<0The symmetry is called -(a+b) 2, so (-infinity, -(a+b) 2) single decrease, (-a+b) 2, +infinity) single increase.

2. Because x is not equal to 0So consider it in two parts, (-infinity, 0) and (0, + infinity). Make use of the definition of the monotonicity of the function. The process is complicated, write it yourself.

1. (1) a=-800°=280°-360° 3, a>270° in the fourth quadrant.

2) a is in the fourth quadrant, 280° is the same as the terminal edge of -80°, and y = -80° = 4 9 because y (- 2, 2).

2. If the point p(tana,cosa) is in the third quadrant, then tana<0 and cosa<0. The angle a (2, ) is obtained, so the terminal edge of the angle a is in the second quadrant.

3, 2<3< 4<3 2 3 2<5<2 So sin3>0, cos4<0, tan5<0, then sin3·cos4·tan5>0, the symbol is positive.

4. p is in the fourth quadrant, and the set of b is obtained from the straight line passing through the point p (root number three, -1).

The angle b in the range of -2 and 2 is: 11 6, - 6.

1(1) a=-800°=-80-720°=-80°-6 a=-80° in quadrant 4.

2) y (- 2, 2) 90° y 90° y is the same as the terminal edge of a y=-80°

2 p(tana, cosa) in the third quadrant tana 0. in the second or fourth quadrant. cosa 0. So a terminal side is in the second quadrant.

3 . sin3>0 cos4<0 tan5<0 sin3*cos4*tan5>0

4.The terminal point of b is ( 3,-1) b {2k -30°} if b (-2 ,2 ) then b = -30°

or -390°

Ask for adoption There may be a mistake in your own type, I haven't counted it for a long time.

1.It is known that a = -800°

1) Rewrite a to b+2k (k z,0 b<2 ) and indicate what quadrant a is? a=14 9 -6 quadrants.

2) Find the angle y so that y is the same as the terminal edge of the angle a, and y (- 2, 2)? Angle y=-4 9 2Knowing that the point p(tana, cosa) is in the third quadrant, then the terminal edge of the angle a is in what quadrant?

Tana O, Cosa O, A in the second quadrant.

3.Determine the symbol of sin3·cos4·tan5? π/2＜3＜π＜4＜3/2π＜5＜2πsin3＞0，cos4＜0，tan5＜0 3·cos4· tan5＞0

p(root number three, -1), thank you for the set of b and point out the angle b in the range of -2 and 2?

1. (1) a=280-3 2 4 quadrant (2) -80°

2 Because p(tana, cosa) is in the third quadrant, tana<0; cosa<0 ；Because tana = sina cosa<0, cosa<0, sina >0

So a is in the second quadrant.

1.(1) a=(14 9) +2*(-3) fourth quadrant (2) y=(-4 9).

2.Second quadrant.

0cos4>0

tan5<0

is ((-1 6) +2k ,k z), (1 6) ,11 6).

The spatial Cartesian coordinate system is established with CP as the Z-axis, Ca as the Y-axis, and CB as the X-axis. From the known data, pc=2, ad=4, ac=2*root number 3. The normal vector of the clear wide surface abp is (2, root number 3.).

3) And the normal vector of the ACP of the Taisho Zen face, i.e., the rolling dust is BC(3,0,0), so the cosine value of the dihedral angle is, so the dihedral angle is 30 degrees.

1 solution: cosx≠ 3+1 2 and cosx 3 2 know: x (2k - 6, 2k + 6) k n

2 solution: y= sinx+lg(cosx) tanx knows: cosx>0 and tanx≠0; Then: x [2k - 0) (0,2k +.)

3 solution: 2sina + cosa = 7 13;sin²a+cos²a=1；

obtained: sina = (14 + 2 199) 65, cosa = (7-4 199) 65;or sina = (14-2 199) 65, cosa = (7 + 4 199) 65

and a (0, ) so sina>0, then: sina = (14 + 2 199) 65, cosa = (7-4 199) 65

Then: tana=sina cosa=(14+2 199) (7-4 199)=-2(845+35 199) 3135

4 proof: from tan x=2tan y+1, obtain: sin x cos x=2sin y cos y+1 ;

i.e. (sin x-cos x) cos x=2sin y cos y, there is (sin x-cos x)=2sin x-1=2sin ycos x cos y;

Therefore, only when cos x cos y=1, sin 2(y)=2sin 2(x)-1

However, the known condition is that cos x cos y=1 cannot be obtained, so the problem is problematic and the condition is missing.

i.e. f'(x)=2ax+1 x=0 has a solution in x>0, obviously a is not equal to 0, so that 2ax 2+1=0, (x>0, a is not equal to 0), so that g(x)=2ax 2+1, then the quadratic function g(x)=0 has a solution in x>0, and g(0)=1>0, g(x) is the axis of symmetry y, so 2a<0, i.e., a<0

Or after the second row, it becomes a=-1 (2x 2), (x>0), which gives a<0

The vertical y-axis, i.e., the slope, is 0

The slope of the tangent is the value of the derivative.

So it's f'(x)=2ax+1 x=0 has a solution, so (2ax +1) x=0

2ax²+1=0

a=0, then 1=0, no solution.

a≠0 then x =-1 2a>0

So a<0

To sum up, a<0

1. y=x (a 2-4a-9) is an even function, then a 2-4a-9 is an even number y is a subtraction function on (0, +, then a 2-4a-9<0 can be solved by the inequality u=a 2-4a-9<0 to obtain 2- 130, y(0)->

If the maximum value over the interval [-1 2,2] is + (or no maximum) 3, f(x)=x (n 2-2n-3) and the image is symmetric with respect to the y axis, then f(x) is an even function.

If there is no common point between the image and the coordinate axis, then f(0)-> and f(x)≠0 can be seen as u=n 2-2n-3<0 and is an even number.

Solve -1 (with the bonus process, the money is given casually, the more the merrier).

A -4a-9 is even, and a -4a-9 = (a-2) 2-13<0, a = 3, a -4a-9 = -12, or a = 5, a -4a-9 = -4

The maximum value is infinity, and the minimum value should be required, which is 1 4

n -2n-3 = (n-1) 2-4<0, with respect to y-axis symmetry, then n -2n-3 is even, i.e. n-1 is even and n is odd, so n=1, n -2n-3=-4

f(x·y)=f(x)+f(y), you can think of it as an abstract function, you can think of it as a logarithmic function, e.g. f(x)=lnx.

As long as you figure out that abstract function, the rest of the questions won't be difficult. f(x+ y)=f(x)*f(y), which can be seen as an exponential function.

1., x-5)(x-2)(x-3)<0x<2.

20 Not established.

35 (x-5)(x-2)(x-3)>0 is not true, so it is solved as x<2 or 30, b>0, c<0).

a-c/(b-x)<0 a+c/(x-b)<0a(x-b+c/a)/(x-b)<0

A>0 then [x-(b-c a)] (x-b)<0a>0 b>0 c<0 then b-c a>b, so the solution is b1 x>-b

1) a>1/x 1/x-a<0 (1-ax)/x<0(ax-1)/x>0

a>0 (x-1/a)/x>0

So x<0 or x>1 a

2) 1/x>-b 1/x+b>0(bx+1)/x>0

b>0 (x+1 b) x>).

x<-1 b or x>0

In summary, x<-1 b or x>1 a

Hope it can help you, good luck with your learning progress o(o

1.Use the root piercing method. On the coordinate axis, when x=2,3,5 is 0, so odd and even, but from the top right through five, and then up through three, down through two, below the coordinate axis, even is found, x <2 and 3b-ac