Classical problems in high school mathematics

consolidate

1. The image of the function f (x) = 1x-x is about ().

A.y axis symmetry B. straight line y =-x symmetry

C. coordinate origin symmetry D. straight line y = x symmetry

Analysis: select the domain {x∈R|x≠0} where c .∫f(x) is symmetrical about the origin.

f(-x)= 1-x-(-x)=-( 1x-x)=-f(x)，

∴f(x) is odd function, and its image is symmetrical about the origin, so c 。

2. The image of function y = ln (1-x) is roughly ().

Analysis: C. In this question, because we are familiar with the image of Y = lnx, its image is located at the right intersection of the Y axis (1, 0) and has an upward trend. Then fold the image with Y = lnx to the left of Y axis to get the image with Y = ln (-x). Then move the folded image to the right by one unit to get Y = ln. if

Analysis: We can get the image containing f(x) from the characteristics of odd function image, and we can get the result from the image.

Answer: {x |-2 < x < 0 or 2 < x ≤ 5}

6.( 1) image with function y = | x-x2 |;

(2) Make an image with the function y = x2-| x |.

Solution: (1) y = x-x2, 0≤x≤ 1, -(x-x2), x > 1 or x < 0,

That is, y =-(x- 12) 2+ 14, 0≤x≤ 1, (x- 12) 2- 14, x > 1 or x.

(2)y=x2-x，x≥0，x2+x，x & lt0,

That is, y = (x- 12) 2- 14, x≥0, (x+ 12) 2- 14, and x < 0, as shown in figure ②.

practise

1. There is an empty container with a water pipe hanging from it, and water is evenly injected into it until the container is full. In the process of water filling, the height change curve of the water surface is as shown in the figure, where PQ is a line segment and the shape of the container corresponding to this figure is ().

Analysis: C. It can be judged from the function image that the container must be irregular, then PQ is a straight line, and the upper end of the container must be a straight line, so ABD, C can be excluded.

2. let A < B and the image of function Y = (x-a) 2 (x-b) be ().

Analysis: choose C. when x > b, y > 0, when x < b, y≤0. So choose C.

3. The image of function y = f (x) is as shown in the figure, so the image of function y = log0.5f (x) is roughly ().

Analysis: Choose C. According to the monotonicity principle of the same increase but different decrease, y = log0.5f (x) is increasing function when x∈(0, 1), y < 0 when x∈( 1, 2), and y = log0.5f (x).

4. In order to get the image of function Y = LGX+3 10, just put all the points () on the image of function Y = LGX.

A. translate 3 unit lengths to the left, and then translate 1 unit length upward.

B. Translate 3 unit lengths to the right, and then translate 1 unit length upward.

C. Translate 3 unit lengths to the left, and then translate 1 unit length downward.

D. Translate 3 unit lengths to the right, and then translate 1 unit length downward.

Analysis: select C. ∫ y = lgx+310 = LG (x+3)-1,∴ move the point on the image of y = lgx to the left by 3 unit lengths to get the image of Y = LG (x+3), and then move y = LG.

5. The image of the following function, after translation or folding, cannot coincide with the image of the function y = log2x. The function is ().

A.y=2x B.y=log 12x

C.y= 12？ 4x D.y=log2 1x+ 1

Analysis: choose C.y=log2x, and y = 2x is symmetrical about y = x; Y = log2x and y = log 12x are symmetric about x; The image of y = log2 1x+ 1 can be obtained by folding and translating the image of y = log2x.

6. The image of function f(x) is a part of two straight lines (as shown in the figure), and the image whose domain is as shown in the figure, then f (x)+f (x)+f (-x) = _ _ _ _ _ _

Analysis: From the image, we can know that f(x) is the odd function on the field.

∴f(x)+f(-x)=f(x)-f(x)=0.

Answer: 0

9. It is known that the function f (x) = 2-x2 and g (x) = X. If f(x)*g(x) = m is in {f (x), g(x)}, the maximum value of f(x)*g(x) is _ _ _ _ _.

Analysis: Draw a schematic diagram.

When f(x)*g(x)= 2-x2, x≤-2, x, -2 134 or a≤ 1, the original equation has no solution.

12. It is known that the image of function f (x) = m (x+ 1x) and the image of function h (x) = 14 (x+ 1x)+2 are about point A (0, 1).

(1) Find the value of m;

(2) If g (x) = f (x)+A4x is the subtraction function in [0,2], find the range of number A. 。

Solution: (1) Let P(x, y) be a point on the h(x) image, and the symmetry point of point p relative to A(0, 1) is Q(x0, y0), then x0 =-x, y0 = 2-y. 。

∴2-y =m(-x- 1x)，

Y = m (x+1x)+2, so m = 14.

(2)g(x)= 14(x+ 1x)+a4x = 14(x+a+ 1x)。

Set 0

Then g (x1)-g (x2) =14 (x1+a+1)-14 (x2+a+/kloc-0

= 14(x 1-x2)+ 14(a+ 1)？ x2-x 1x 1x2

= 14( x 1-x2)？ x 1x 2-(a+ 1)x 1x 2 & gt； 0,

And it holds on x 1, x2 ∈ (0 0,2).

∴x 1x2-(a+ 1)<； 0,∴ 1+a>； x 1x2, 1+a≥4,∴a≥3.

Integration (2)

1.(20 10) Let the function f(x) be the odd function defined on R, and f (-3) =-2, then f (3)+f (0) = ().

a3 B- 3

C.2 D.7

Analysis: choose C. F (3)+F (0) =-F (-3)+F (0) = 2+0 = 2. So choose C.

2. In the following function f(x), it is true that "for any x 1, x2∈(0, +∞), when X 1 < X2, there is F (X 1) > F (X2)".

a . f(x)= 1x b . f(x)=(x- 1)2

C.f(x)=ex D.f(x)=ln(x+ 1)

Analysis: Choose A. From the meaning in the question, we know that the function f(x) is a decreasing function on (0, +∞).

In a, f ′ (x) =-1x 2 < 0 is a decreasing function on (-∞, 0) and (0,+∞);

In b, when f ′ (x) = 2 (x-1) < 0, x is less than 1, so f(x) is a decreasing function at (-∞,1).

In c, f ′ (x) = ex > 0 indicates that f(x) is a increasing function on R. 。

In d, from f' (x) = 1x+ 1 and x+ 1 > 0, we know that f' (x) > 0, so f(x) is a decreasing function at (-1, +∞).

3. Given that the function f(x) is a decreasing function on R, the range where the real number x satisfies f (| 1x |) < f (1) is ().

A.(- 1， 1) B.(0， 1)

C.(- 1，0)∩(0， 1) D.(-∞，- 1)∩( 1，+∞)

Analysis: choose c .∫f(x) as the subtraction function on r and f (| 1x |) < f (1).

∴| 1x|> 1，

That is | x | < 1 and x≠0, get- 1 < x < 0 or 0 < x < 1.

4. (Original problem) If f (x) = x2+x, then f (a+1a) _ _ _ _ _ _ f (1). (Fill in "≤" ≥ ").

Analysis: ∫A+ 1A≥2 or A+ 1A ≤-2,

The symmetry axis of f(x) is x =- 12.

∴f(x) is the increasing function on (-12, +∞),

In (-∞,-12), it is a decreasing function.

f(2)= 22+2 = 6 >； 2=f( 1)，

f(-2)=(-2)2+(-2)=2=f( 1)，

∴f(a+ 1a)≥f( 1).

Answer: ≥

5. If the function f (x) = (x+a) (bx+2a) (constants A and b∈R) is an even function, and its range is (-∞, 4), then the analytic formula of the function f (x) = _ _ _ _ _ _ _ _ _ _ _

Analysis: Since the definition domain of f(x) is r and the value domain is (-∞, 4),

We know that b≠0 and f (x) are quadratic functions,

f(x)=(x+a)(bx+2a)

=bx2+(2a+ab)x+2a2。

F (x) is an even function,

∴ Its symmetry axis is x = 0, ∴-2a+ab2b = 0,

∴ 2a+ab = 0, ∴ a = 0 or b =-2.

If a = 0, then f (x) = bx2 and the range (-∞, 4) is contradictory, ∴a≠0.

If b =-2 and its maximum value is 4,

∴4b×2a24b=4,∴2a2=4，

∴f(x)=-2x2+4.

Answer: -2x2+4

6. The function f (x>0) is known. = 1A- 1x (a > 0，x > 0)。

(1) proves that f(x) is a increasing function on (0, +∞);

(2) If f(x) is a decreasing function, then f(x) ()

A. In the interval, it is an increasing function.

B. it is a decreasing function in the interval.

C. it is a decreasing function in the interval.

Analysis: Choose B. From f(x) = f (2-x), we can know that the image of function f(x) is symmetrical about the straight line x = 1, and make the characteristic property diagram of the function as follows.

A.- 1

c . 6d 12

Analysis: C. As can be seen from the meaning of the question,

When -2 ≤ x ≤ 1 and f (x) = x-2,

When 1 < x ≤ 2 and f (x) = x3-2,

And ∵ f (x) = x-2, and f (x) = x3-2 are all increasing function in the domain.

The maximum value of f (x) is f (2) = 23-2 = 6.

5. Part of the image of even function f(x) defined on R is as shown on the right, but on (-2,0), the following function is different from the monotonicity of f(x) in ().

A.y=x2+ 1 B.y=|x|+ 1

C.y=2x+ 1, x≥0x3+ 1, x 0, f (x) = x+ 1, then when x < 0, f (x) = _ _ _ _.

Analysis: ∫f(x) is odd function, when x > 0, f (x) = x+ 1,

When x < 0 and -x > 0,

f(x)=-f(-x)=-(-x+ 1)

That is, when x < 0, f (x) =-(-x+1) =-x-1.

Answer: -X- 1

8. The increasing interval of the function y =-(x-3) | x | is _ _ _ _ _.

Analysis: y =-(x-3) | x |

=-x2+3x，x>0，x2-3x，x≤0。

Do the image of this function, and observe the image to know that the increasing interval is f (MX-2)+f (x) < 0, and the value range of x is _ _ _ _ _.

Analysis: It is easy to know that the original function is monotonically increasing on R, which is odd function, so f (MX-2)+f (x) < 0? Female (MX-2)

Answer: (-2,23)

10. It is proved that f (x) = 1+xx is a decreasing function on (0, 1).

Proof: let x 1, x2∈(0, 1), X 1

Then f (x1)-f (x2) =1+x1x1-kloc-0/+x2x2.

= x2+x 1x 2-x 1-x2 x 1x 1？ x2

= x2-x 1+x 1x 2(x 1-x2)x 1？ x2

=(x2-x 1)( 1-x 1x 2)x 1x 2。

∫x 1, x2∈(0, 1] and x 1

∴x2-x 1>； 0， 1-x 1x 2 & gt； 0,

∴f(x 1)-f(x2)>； 0, namely f(x 1)>f(x2).

So f (x) = 1+xx is the subtraction function on (0, 1).