CPhill

Set the euations   equal

x^3 - x^2 + x + 1 =  x^3 + x^2 + x + 1    rearrage as

2x^2  =  0                    divide through by  2

x^2   = 0

x = 0

And using either equation  y  = (0)^3 + (0)^2  + 0 + 1     =   1

So......the intersection point  is (0,1)

This means that we need to find  the x coordinate of  (x , 4)

When y = 4 on the graph, x = 5

So......

f(5) = 4

And  

f^-1 (4)  = 5

If it contains the point (1,0)  then 

a + b + c  = 0

If the point   (3,4)  is the vertex  than

-b/(2a) = 3

-b= 6a

b = -6a

Then

a(3)^2 + -6a(3) + c  =  4

9a  - 18a + c = 4

-9a + c = 4

c = 4 + 9a

So

a + b + c  = 0

a + ( - 6a)  +  (4 + 9a) =  0

4a =  -4

a = -1

b = -6(-1) =  6

c =  4 + 9(-1)  =    -5

{a , b , c}   = { -1, 6 , - 5 }

It appears that   the point  (2, -1) is on the graph

So   f(2)  = -1

And c = 2

If a^2=  49  then either a = 7 or a  = -7

If  a = 7, the parabola turns  upward   ....either the parbola lies entirely above the x axis or it intersects the x axis so it is impossible that a = 7

Then a  = -7.....the parabola turns downward  and lies completely below the x axis

So  we have     -7x^2  + bx - 18

Let's  suppose that this parabola has  just one root   (its x coordinate of its vertex is the root)

Then the discriminant must =  0....so.....

b^2 - 4(-7)(-18)  = 0

b^2  - 504 = 0

b^2 = 504

b ≈ 22.45

If b were larger than this  we would have two roots   and if b is smaller than this we have no real roots (the parabola lies completely below the x axis....what we want)

So.....the largest  interger value for b that puts the parabola entirely below the x axis is when  b= 22

3/x + x/9   = b      multiply through by x

3 + x^2/9  = bx       rearrange  as

(1/9)x^2  - bx  + 3  = 0

If this has one solution , the  discriminant   =  0 

So

b^2  - 4(1/9)(3)  = 0

b^2  = 12/9

b^2  =  4/3     take the positive root

b =  2 / sqrt (3)    

We can write the above series in a slightly different manner

(1/7) ( 1 + 1/7^3 +  1/7^6 + .......+ 1/7^(3n-3))   + (2/7^2) (1 + 1/7^3 + 1/7^6 +......) +  (3/7^3) ( 1 + 1/7^3 + 1/7^6 + ....)  =

(1 + 1/7^3  +1/7^6 +.....+ 1/7^(3n - 3) )  ( 1/7  + 2/7^2 + 3/7^3 +......+ n / 7^n )

The first series is a geometic sum with r =  1/7^3   and a first term of 1

The sum of this series =  1 / ( 1 - 1/7^3)  =  343/342

For the second series.. let the sum = S and multiply multiply each term  by r  = 1/7

So

Sr  =   1/7^2  + 2/7^3  +   ... +  n /7^n 

So

     S  =  1/7 + 2/7^2 + 3/7^3   +  n / 7^n

   Sr =            1/7^2  +  2/7^3     + (n-1) / 7^n +  n / 7^n

S - Sr  =  1/7  + 1/7^2 + 1/7^3  +  1/7^n -  n/7^n

The term in red  approaches 0  as n approaches infinity so we can  ignore it

So

S - Sr  =  1/7 + 1/7^2 + 1/7^3 + 1/7^n

S(1 - r)  =  the sum of  a geometric  series  with r = 1/7  and a first term of 1/7

S ( 1  - 1/7)  =  (1/7) / ( 1 - 1/7)

S (6/7) =  (1/7)/ (6/7) =     1/6

S =   (1/6)(7/6) = 7 /36

So.....the sum of the  given  series =  (343/342) ( 7/36) =   2401 / 12312

1/x  + 1/y =  3

1/x  -  1/y  = 7          add these

2 / x =  10

x = 2/10  = 1/5

And  

1 / (1/5) + 1/y = 3

5 + 1/y=  3

1/y =  3 -5

1/y =  -2

y  = -1/2

x + y =   1/5  - 1/ 2 =   -3 /  10

a@b  =  ab - 2b                   a#b = a + b  - ab

So

(6 @ 2) /   (6 # 2)  =

(6*2 - 2*2) / ( 6 + 2  - 6*2)  =

(8 ) / ( -4)  =

-2

Getting a common  denominator we have

30/60      20/60       15/60     4/60

We need the result to be  < 3/60  (but positive)

We have 8 possibilities

 + + +           reject  as too large

+ + -            reject  as too large

+ - -             reject as too large

 - -  -              reject  (negative result)

- - +               reject  (negative result)

- + +              reject as too large

+ - +            reject as too large

- + -              reject as too large

The probability is  0