Let \(0< k <2\). Below is sketched a graph of \(y=f_k(x)\) where \(f_k(x) = x(x-k)(x-2)\). Let \(A(k)\) denote the area of the shaded region.

Graph of the described cubic. It crosses the axis 3 times, at 0, k and 2. The area under the graph for x between 0 and k, and the area above the graph for x between k and 2, are both shaded.
  1. Without evaluating them, write down an expression for \(A(k)\) in terms of two integrals.

We need to consider the two shaded regions separately, because one is above the \(x\)-axis (giving a positive area), and one is below the \(x\)-axis (giving a negative area), so we can’t just take the integral \(\displaystyle\int_0^2 \! f_k(x) \, \mathrm{d}x\).

Instead, we have

\[A(k) = \displaystyle\int_0^k \! f_k(x) \, \mathrm{d}x - \displaystyle\int_k^2 \! f_k(x) \, \mathrm{d}x,\]

as the negative sign in front of the second integral cancels out the negative sign of the area between \(x=k\) and \(x=2\).


\[A(k) = \displaystyle\int_0^k \! x(x-k)(x-2) \, \mathrm{d}x - \displaystyle\int_k^2 \! x(x-k)(x-2) \, \mathrm{d}x.\]

  1. Explain why \(A(k)\) is a polynomial in \(k\) of degree \(4\) or less. [You are not required to calculate \(A(k)\) explicitly.]

\(A(k)\) is certainly a polynomial in \(k\). Could it include a \(k^5\) term?

Only if you have an \(x^5\) term or a \(kx^4\) term after integrating, which means having an \(x^4\) term or a \(kx^3\) term before integrating.

Neither is possible, by looking at \(x(x-k)(x-2)\), and the same argument works for powers of \(k\) higher than \(5\).

Hence \(A(k)\) is a polynomial in \(k\) of degree at most \(4\).

  1. Verify that \(f_k (1+t) = -f_{2-k}(1-t)\) for any \(t\).

Let’s start by thinking about what \(f_k(1+t)\) and \(f_{2-k}(1-t)\) are, given that \(f_{k}(x)=x(x-k)(x-2)\).

To find \(f_{k}(1+t)\), we need to substitute \(1+t\) for \(x\) in the formula for \(f_{k}(x).\) This gives us \[f_k(1+t) = (1+t)(1+t-k)(1+t-2).\]

To find \(f_{2-k}(1-t)\), we need to substitute \(1-t\) for \(x\), but we also need to substitute \(2-k\) for \(k.\) Therefore \[f_{2-k}(1-t) = (1-t)(1-t-(2-k))(1-t-2).\]

Now we can manipulate \(-f_{2-k}(1-t)\) to try to obtain \(f_k(1+t).\) \[\begin{align*} -f_{2-k}(1-t) \hspace{2mm} & =-(1-t)(1-t-(2-k))(1-t-2) \\ & = -(1-t)(-1-t+k)(-1-t) \\ & = (t-1)(1+t-k)(1+t) \quad \text{by taking a factor of $-1$ from each bracket} \\ & = (1+t)(1+t-k)(t-1) \quad \text{by rearranging the order of the brackets.} \\ \end{align*}\]

If we think about what we are aiming for, we can see that the first two brackets involve \(1+t\), which is a good sign. It would be helpful to rewrite the last bracket in a way that also involves \(1+t.\) We can write \(t-1\) as \(1+t-2\) which means that

\[\begin{align*} -f_{2-k}(1-t) \hspace{2mm} & =(1+t)(1+t-k)(1+t-2) \\ & = f_{k}(1+t) \quad \text{as required.}\\ \end{align*}\]

We started from the RHS because it looks messier and we wanted to get it to a simpler expression. We could have started with the LHS.

  1. How can the graph of \(y=f_k(x)\) be transformed to the graph of \(y = f_{2-k}(x)\)?

From part (iii), we have that

\[f_k(1+t) = -f_{2-k}(1-t).\]

If we substitute \(t\) with \(1-x\), the above equation becomes equivalent to

\[-f_k(2-x) =f_{2-k}(x).\]

Now, \(f_k(2-x)\) is the reflection of \(f_k(x)\) in the line \(x=1\).

f k reflected in the line x = 1.

A second transformation is now required because of the minus sign in front of \(f_{k}\). This is a reflection in the \(x\)-axis.

Together these transformations are the same as a rotation \(180^{\circ}\) around the point \((1,0)\).

The newly obtained graph reflected in the x axis.

In fact, combining two reflections always gives you a rotation about the point where the two mirror lines intersect.

Deduce that \(A(k)=A(2-k)\).

As we can see in the sketch above, since we’re rotating \(180^\circ\) around a point on the x axis, the area between the graph and the \(x\)-axis remains the same for \(f_k\) and \(f_{2-k}\), i.e. \(A(k) = A(2-k)\).

Or else we could say that since both of the reflections preserve the shaded area, combining them does too.

  1. Explain why there are constants \(a, b, c\) such that \[A(k) = a(k-1)^4 + b(k-1)^2 + c.\] [You are not required to calculate \(a,b, c\) explicitly.]

We know that \(A(k) = A(2-k)\), so \(A(1-k) = A(2-(1-k)) = A(1+k)\), and so \(y = A(k)\) is symmetrical about the line \(k = 1\).

So if we translate \(y = A(k)\) by \(1\) to the left, then it is even, so if we express \(A(k)\) as powers of \((x-1)\), only even powers are possible.

Therefore we have \(A(k) = a(k-1)^4 + b(k-1)^2 + c\) for some constants \(a, b, c\).