gre math solutions

5/20/2018 GRE Math Solutions

1/11

1

15

()

20 2013 - :

: (5)

1. f(x)=x

f ( ) x =1, x

7

2. f . f 0x ;

4

3. () .

4

4. , ,

, , , , .

) f ( ) , 0= 1

x xx

f ( ) =2

1x

x ( 2)

) f , g

( )f ( ) g( ) f ( ) g( ) f ( ) g ( ) = +x x x x x x

( 2)

) . ( 2)

) , . ( 2)

) , )()( > ( 2)

10


2/11

2

25

{ } = 1 2 3 4, , ,

{ } = 1 4, { } = 1 3,

{ }1 { }3 :

+ + =

+

2

1 3 2x 1

1 x x 1 1P( ) lim

2 x x

H 3P( ) f ( )x x , =x 1,

f = >x

(x) ln x, x 03

1. =11

P( )4

=31

P( )3

10

2. 1 3

P(A )3 4

, A A .

7

3. =3

P(A )

4

, 2P( ) , 4P( ) , [ ] P (A B) (B A)

P( - ) , .

8

X , 4 .

:

50

4 85=x

= 75

=x 74


3/11

3

35

1. =c 10

4

2.

K

ix

if

[, )

[, )

[, )

[, )

8

3. = = = =1 2 3 4f 0,1 , f 0,3 , f 0,2 f 0,4

,

80, 200

3

7

4. f(x) x ln x , x 0, > 1

() f ( )( )1,f 1 , , E , E< 2

1. = 2

5


4/11

4

45

2. 1 2 50x , x , ..., x 50 ()

=y 31

) =x 30 ( 2)

) :

1 2 20x , x , ..., x 3,

15

> 0 .

, 31(4)

6

3. < < < + t : f(t) f (t) 1 ,

= + f(t) t ln t 2

:

) (3)

) ( 4)

7


5/11

5

55

( )

1. . - . - .

.2.

. . .

3. . , , , ..

4. .5. : (3) .6. : 10.30 ..

K


6/11

Tech and ath

-

www.techandmath.gr: 6974473322

1. . , 28.

2. . , 14.

f

0 f(x) f(x0) 0.

3. . , 87.

() , ,

() .

4. () -

() -

() -

() -

() -

1. :

( )( )( )

( ) ( )

( )

( ) ( )

2 22

1 3 2x 1 x 1 3 2 2

2 2

x 1 x 13 2 2 3 2 2

x 1 x 12 2 2

x x 1 1 x x 1 11 x x 1 1 1P( ) lim lim

2 x x 2 (x x ) x x 1 1

1 x x 1 1 1 x xlim lim2 2(x x ) x x 1 1 (x x ) x x 1 1

x x 11 1 1 1 1lim lim

2 2 2 2x (x 1) x x 1 1 x x x 1 1

+ + + + ++ + = = =

+ + + + +

+ + + = =+ + + + + + + +

+ = = + + + + + + +

1

4=


7/11

Tech and ath

-


( ) ( )x x x 1 x 1 1

f (x) ln x ln x ln x ln x 1 ln x3 3 3 3 3 x 3

= = + = + = +

3 1P( ) f (1)3

= = .

2. : { }2 3A , = , : { }3 31

A P( ) P(A ) P(A )3

(1)

: { }1 4A ,= : { }1 11

A P( ) P(A) P(A)4

:

1 3P(A ) 1 P(A) 1

4 4 =

(2) .

, (1) (2) :1 3

P(A )3 4

.

3. 1

1P( )

4

1 4 4

3 1 1P(A ) P(A) P( ) P( ) P( ) 0

4 4 4

=

= = + = = .

: { }( )2 1 3 4 1 3 41 1 5

P( ) 1 P , , 1 P( ) P( ) P( ) 1 0

4 3 12

= = = = .

( ) ( ) { } { }4 3 4 31

P A B B A P P( ) P( )3

= = + =

( ) 31

P A B P( )3

= =

1. 24 .

: 85=50+2+2+2+ =5 c=2=10

.

2. =75 : 3 31 2 4f f

f f 0,5 (1) , f 0,5 (2)2 2

+ + = + =

: 4 3f 2f (3)=

(2) (3) :3

f 0,2= 4

f 0,4= , (1) :

1 2f f 0,4 (4)+ =


8/11

Tech and ath

-


:(4 )

1 1 2 2 3 3 4 4 1 1 1 2 3 3 4 4

1 1 1

1

x f x f x f x f x x f x (0,4 f )x f x f x

74 55 f 65(0,4 f ) 0,2 75 0,4 85 74 10f 26 15 34

f 0,1

= + + + = + + +

= + + + = + + +

=

:

xi

fi[50,60) 55 0,1

[60,70) 65 0,3

[70,80) 75 0,2

[80,90) 85 0,4

1

3. y 80. :

( )

( )

31 21 1 2 2 3 3 1 2 31 1 2 2 3 3

31 21 2 31 2 3

1 1 2 2 3 3

1 2 3

v1 v vx v x v x v x x xx v x v x v v v v vy1 vv vv v v

v v vv v v v

x f x f x f 55 0,1 65 0,3 75 0,2 40 200

f f f 0,6 0,6 3

+ + + ++ += = =+ + + + + +

+ + + + = = = =

+ +

4. 2,5% 74 x

s : x 2s 74 + = (1).

16% 68

x s : x s 68 = (2).

(1) (2) : x 70= s 2= .

:s 2

CV 10%x 70

= = <

.


9/11

Tech and ath

-


1. f x=1

f (1) . : ( )f (x) x ln x 1 ln x = + = + : f (1) 1 = .

: y x ,= + ,

, : f (1) 1 1 1= + = + =

: y x 1, 1= + > .:

x 0 y 1= = , A(0, 1) y 0 x 1= = , A(1 ,0)

:2

22

(OA)(OB) ( 1)E

2 2

( 1)2 ( 1) 4 2 1 2 1 3

2

= =

< < < < < : =2.

2. ) 50 :i i

y x 1= +

:

y x 1 31 x 1 x 30= + = + = .

) :50 x 20 3 15 2

31 50 30 20 3 15 155050 3

+ = + = = .

3. : f (x) 1 ln x = + f(x)

: x 01

e +

f (x) - 0 +

f(x)


10/11

Tech and ath

-


f :1

,e

+ :

1 ee < < < .

{ }iA t :11 i 30= it ,

:N(A) 20 2

P(A)N( ) 30 3

= = =

.

) : f (t) f (t) 1 t ln t 2 ln t 1 1 (t 1) ln t 0> + + > + + > .

g(t) (t 1) ln t , t 0= > :


11/11

Tech and ath

-


{ }iB t :1 i 29=

{ } ( )

( )i

N A B 19

A B t :11 i 29 P A B N( ) 30

= = =

t 0 1 +

t-1

lnt

g(t)

- +

- +

+ +

0

0

0

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