2021年高考數(shù)學壓軸專題解法分析與強化訓練遞推函數(shù)型(含答案與解析)

f(

)

3)

2

07

［］

1(2019

n)f(x)Rf(x1)2f(x)x(0,1］f(x)

8

D.

C—oo

5

8

oo—

3

B

【解

x(0,1］f(x)x(x

1)

f(x)

(1,2］x1(0,1］f(x)f((x1)1)2f(x1)2(x1)(x2)

f

x

(x)

f(

3

2)

x(x1).x(

o，m］f(x)A?，m

()

x(2,3］

B

f(x)2(x2)(x

2

3)

f(x)

5

f(-)1

x(k,k1］(kZ)f(x)2k(xk)［x(k1)］

1

f(x)f(k+-)

x(

,2

f(X)

min

2

18

3

22

x(2,3］f(x)22(x2)(x3)

,m］f(x)

f(x)2(x2)(x

Umm

8

B.

11

【解

1<xW0

0<x+

1

W1,

f(x)

=

2

f(x+1)=2(x+

1)x;

1<X

W2

0<x

—

1

<1

f(x)

=

2f(x—1)

=

2(x

—1)(x—2)

;

2<XW3

0<x—2

<1,

f(x)

=

2f(x—1)

=

2f(x

—

2)22(x2)(x3)

-V

—

*

12

—

-

“

"7

—

x

--

-

-

?n22,n

n*1

n1

n1

x+

1x,—

1<xW0

f(x)

=

xx—

1

,

0<x<1

f(x)

?

2x—1x—2

,1<x<2

2

2

2

x

—

x—3,2<x<3

2

2(x2)xe

3)9

(3x7)(3x8)0x3x3

933

?

x(

g,

m］f(x)A9mm3

B.

B

12J5534

.aua,&JN

y=?

-2

2(201714

f(x)

［1,+x

f(X)

1|2x3|1$<211

22,x>2

Uy2xf(x)3(1,2015)

11

3

x

2x211$<2f(x)3

x[2

n1,2n)(nN)

2

—

n1

42x2<x<2.

11

[1,2)f(x)

2

-

n

1f2

—

n1

x

2

x

—€

*1

3

Vx

€

[2

n

—

3

2

11112]f(x)2f

2

—

*1x

2

—

*1

2

2

——

"1x2

11

——

2xf(x)32x^22x230x2

2

2

2x3

4

0x3

2

—

2

x2

n2.x[2

—

n1'

3

-

n2

]x3

n—

-

2

2

3

-

n1

2

2

x(3

1111

—

2)f(x)gf2Xg422X

12x

2xf(x)32x2

n-1

—

4

2

-

*1

—

30x2

4

—

2

x3

4

0?

—

x3

2

2

x2

n2

n

.

x(3

—

2

x3

2

2

2,2n)

x3

2

—

2

(1,2015)n11y2xf(x)3(1,2

2x

2

^

2

—

3

—

—

—

-

“1

?2

2

2

2x

12

22

—

2

3

n

015)11.

X[2

_

n1,2n)f(X)

11

Xf(X)maxf2'21

3

—

?

1

33_3g(x)?x-n^^1g(x)g

2

-

n-1

1_3-x[2n12n)x?21

y2xf(x)3

x[2

n1,2n)y2xf(x)3

x[2

—

n1,

?22]yf(x)yg(x)f(3-

n22

)g(3-

n

f2)x

11x

[2

n1,

3

n2

]x3

2

x(3

-

n2,2n)

—

yf(x)2

-—

n1*2

3

k

1

f'x)

2

n-3

g(x)

2x

33

kg'x)

3

1

3

2

23+1

k<k.

f(3-

n

f

2)g(3-

n)x[2

n1,2n)y2xf(x)3x3

-2

2

(1,2015)nW11y2xf(x)3(1,2015)11.

33

yf(x)y*=^X3X6

x3

-n

—

2

X3

2—

2

(1,2015)nW11y2xf(x)3(1,2015)

11.

［］

1.f(x)Rf(x)

Ua

(1)

2x—1,xW0

fx1x0

f(x)xa

xWOf(x)2

—

x

1

4

0X

W1

1X1

WQ

——

f(x)f(x1)2(X1}x0f(x)

1.

f(x)xaUf(x)yxaav1

a

(

1)

x—［x］,x>0

2.

f(X)

［X］

x

f(x+1),xv0,

yk(x1)(k0)yf(x)Uk

.

11

41

［x］

0$v

1f(x)x1$v2f(x)x1

2$v3f(x)x2k

(

Vk1f(x)xkkZ

1<xv0f(x)x1f(x)yk(x1)(10)

(31)(21)

11

3.20194

f(x)

X2

—2x,x0f(x

—1)—1,x

[0,100]

1

k

43

.

x

f

(x)x-5

10000

1|x1,