# 2010_Question1_h

• If n=3, then our equation is (x-1)(x-2)(x-3), and, using that the x^3 coefficient is positive (it’s 1), we see that this describes a cubic curve which starts negative, increases, passes through the x-axis at x=1, turns around and passes back down through x=2, and the turns once more and heads up through x=3. So before x=1 it will pass through every negative number, and after x=3 it will pass through every positive number. So all values of k give solutions, and (a) is false.
• By a similar argument to the above, using the fact that the x^{2n} coefficient is always positive, we see that the curve will be above the x-axis for x outside of the interval [1,2n]. Since the function is polynomial, it cannot go to -\infty in the finite interval [1,2n], and so it must be bounded below, which means that some values k are never reached. So this statement is also false. (More simply, if n=2, then (x-1)(x-2)=-1 has no solutions, since the minimum value of (x-1)(x-2) is reached at x=1.5, and we can see that there it is 0.25.)
• Combining the two things we’ve said above, we see that positive values of k always have a solution, of any value of n, so this statement is true.
• This statement is false: looking at (x-1)(x-2), we see that this equals 2 at x=0 and x=3.

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