- Let \(k\) be a real number, \(k \ne \pm 1\). The function \(f(t)\) satisfies the identity \[f(t) - kf(1-t)=t\] for all values of \(t\). By replacing \(t\) with \(1-t\), determine \(f(t)\).

We have that the function \(f(t)\) satisfies the identity

\[f(t) - kf(1-t)=t,\]

and by replacing \(t\) with \(1-t\) we have

\[f(1-t) -kf(t)=1-t.\]

Rearranging the second identity and substituting into the first gives

\[f(t) - k[kf(t) + 1-t]=t,\]

and solving for \(f(t)\) gives

\[f(t)=\frac{k+(1-k)t}{1-k^2}.\]

- Consider the new identity \[\begin{equation*} f(t) - f(1-t) = g(t). \label{eq:starrepeat}\tag{$*$} \end{equation*}\]

- Show that no function \(f(t)\) satisfies \(\eqref{eq:starrepeat}\) when \(g(t)=t\).

Letting \(g(t)=t\) gives

\[f(t) - f(1-t) = t,\]

and replacing \(t\) with \(1-t\) gives

\[f(1-t)-f(t) = 1-t.\]

Summing these two identities gives \(0=1\), which is clearly a contradiction. So there is no function \(f(t)\) that satisfies \(\eqref{eq:starrepeat}\) when \(g(t)=t\).

- What condition must the function \(g(t)\) satisfy for there to be a solution \(f(t)\) to \(\eqref{eq:starrepeat}\)?

If there is a solution \(f(t)\) to \(\eqref{eq:starrepeat}\) then we have

\[f(t) - f(1-t) = g(t)\]

and

\[f(1-t)-f(t)=g(1-t).\]

Summing these identities gives

\[g(t)+g(1-t)=0,\]

so \(g(t)\) must satisfy this condition for there to be a solution to \(\eqref{eq:starrepeat}\).

- Find a solution \(f(t)\) to \(\eqref{eq:starrepeat}\) when \(g(t) = (2t-1)^3\).

### Approach 1

Letting \(g(t)=(2t-1)^3\) means we require \[f(t)-f(1-t)=8t^3-12t^2+6t -1.\]

First let’s try \(f_1(t)=t^3\). This gives

\[\begin{align*} f_1(t) - f_1(1-t) & = t^3 - (1-t)^3 \\ & = 2t^3 - 3t^2 + 3t-1.\\ \end{align*}\] Multiplying by \(4\) gives \[\begin{equation} 4[f_1(t) - f_1(1-t)] = 8t^3 -12t^2 +12t -4 \label{eq4f1} \end{equation}\]which is close to our required expression.

Now let \(f_2(t)=t\), which gives \[\begin{equation} f_2(t) - f_2(1-t) = 2t-1. \label{eqf2} \end{equation}\] Subtracting \(3\) times \(\eqref{eqf2}\) from \(\eqref{eq4f1}\) gives \[\begin{align*} 4[f_1(t) - f_1(1-t)] -3[f_2(t) - f_2(1-t)] &= 8t^3 -12t^2 +12t -4 -6t+3\\ &= 8t^3 - 12t^2 + 6t - 1 \end{align*}\]as required. Hence, \[f(t) = 4f_1(t)-3f_2(t) = 4t^3-3t\] is a solution.

This is just one solution; there are infinitely many solutions.

### Approach 2

Trying \(f(t) = at^3+bt^2+ct+d\) seems like a sensible guess. Can we find constants \(a,b,c\) and \(d\) that will work?

We have \(f(t)-f(1-t) = (at^3+bt^2+ct+d) -(a(1-t)^3+b(1-t)^2+c(1-t)+d)\) which on multiplying out gives

\[f(t)-f(1-t)=2at^3 - 3at^2 + (3a + 2b + 2c)t - a - b - c.\]

Now \((2t-1)^3 = 8t^3 - 12t^2 + 6t - 1\), so comparing coefficients, \(a = 4, b=-1, c=-2, d=0\) is a solution that works.