Given a positive integer \(n\) and a real number \(k\), consider the following equation in \(x\), \[(x-1)(x-2)(x-3) \times \cdots \times (x-n) = k.\] Which of the following statements about this equation is true?

  1. If \(n=3\), then the equation has no real solution \(x\) for some values of \(k\).

If \(n=3\), we have the equation \((x-1)(x-2)(x-3) = k\), which is a cubic function in \(x\).

We know from the shape of cubics that the function will take on all real values, and so (a) is untrue.

Useful fact: every cubic equation with real coefficients has at least one real root.

Plot of the above cubic. The graph intersects the x axis in 1 2 and 3, goes to infinity for x going to infinity and goes to minus infinity for x going to minus infinity.
  1. If \(n\) is even, then the equation has a real solution \(x\) for any given value of \(k\).

If \(n\) is even, then the leading term of the function will be a positive even power of \(x\), such as a quadratic (\(n=2\)) or a quartic (\(n=4\)) function.

We know from the shape of these functions that they have a finite minimum value.

Thus the equation will not have a solution for sufficiently negative values of \(k\).

Plot of the function when n is 6. It has 3 local minima and does not reach all negative values.

Hence (b) is untrue.

  1. The equation never has a repeated solution \(x\) for any given values of \(k\) and \(n\).

If we set \(n=2\), we have \((x-1)(x-2) = k\), i.e. \(x^2 - 3x + (2-k)=0\). Completing the square gives \[\left(x-\frac{3}{2}\right)^2 -\left(\frac{1}{4} + k\right) = 0,\] so if we set \(k = -\dfrac{1}{4}\), then the equation has a repeated root at \(x = \dfrac{3}{2}\).

Another way of thinking about this is that when \(n=2\), the graph \(y=(x-1)(x-2)\) is a parabola, so it has a minimum point.

When \(y=k\) touches the graph at this minimum point, there will be a repeated root, and so this curve is a counter-example to (d).

Plot of the above parabola and the y = minus 1 over 4 line that touches it at its vertex.

Hence (d) is untrue, and by process of elimination, the answer is (c).

We can also show that (c) is true directly.

  1. If \(k \ge 0\) then the equation has (at least) one real solution \(x\).

If we let \[f(x) = (x-1)(x-2)(x-3) \times \cdots \times (x-n)\] then \(f(n) = 0\).

If we then increase \(x\) from \(n\), \(f(x)\) also increases from \(0\), since every term in the product is positive and increasing.

Hence \(f(x)\) will take every positive real value for \(x>n\), and so the equation will have some solution for every \(k\ge0\), so (c) is true.

This is also clear from the shape of the graphs of \(y=f(x)\).