Given that Pn(x)=(x-1)(x-2)...(x-n)

P5(x) = (x - 1)(x - 2)(x - 3)(x - 4)(x - 5)

P6(x) = (x - 1)(x - 2)(x - 3)(x - 4)(x - 5)(x - 6)

So P5(x)/ P6(x) = [(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)]/[(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)(x - 6)]

cancelling the common terms

=> 1/(x - 6) >= 2

Now if we take that x - 6 is positive or x > 6

=> 1 >=2(x - 6)

=> 1 >= 2x - 12

=> 13 >= 2x

=> 13/2 >=x

=> 6.5 >=x

So x can lie between 6.5 and 6. But as x only takes on integer values as the function indicates, there is no solution.

If Pn(x)=(x-1)(x-2)...(x-n), then P5(x) = (x-1)(x-2)...(x-5) and P6(x) = (x-1)(x-2)...(x-6)

P5(x)/P6(x) = (x-1)(x-2)...(x-5)/(x-1)(x-2)...(x-5)(x-6)

We'll simplify and we'll get:

P5(x)/P6(x) = 1/(x-6)

But, from enunciation, we know that P5(x)/P6(x) > =2

1/(x-6) >= 2

We'll subtract 2 both sides:

1/(x-6) - 2 >= 0

We'll multiply by (x-6):

(1 - 2x + 12)/(x - 6) > = 0

We'll combine like terms:

(13 - 2x)/(x - 6)>= 0

A ratio is positive if and only if both numerator and denominator, are positive or negative.

Case 1)

13 - 2x >= 0

-2x >= -13

2x =< 13

x = < 13/2 = 6.5

x - 6 >= 0

x >= 6

The interval of admissible values for x, that makes positive the ratio, is [6 ; 6.5].

Case 2)

13 - 2x = < 0

-2x = < -13

2x >= 13

x >= 13/2 = 6.5

x - 6 =< 0

x =< 6

There is no common interval for admissible values for x, in this case.

**The only admissible interval of valid solutions for x is [6 ; 6.5].**