For a positive whole number \(n\), the function \(f_{n}(x)\) is defined by \[f_{n}(x) = (x^{2n-1}-1)^2.\]

  1. Sketch the graph of \(y=f_{2}(x)\) labelling where the graph meets the axes.

We have \[f_{2}(x) = (x^3 - 1)^2 = x^6 - 2x^3 +1,\] so \(f_{2}(x)\) crosses the axes at \((1,0)\) and \((0,1)\).

Differentiating \(f_{2}(x)\) gives \[f'_{2}(x) = 6x^5 - 6x^2 = 6x^2(x^3 -1),\] so \(f'_{2}(x) = 0\) at \(x=0\) and \(x=1\). We have \(f'_{2}(x) \le 0\) for \(x<1\) and \(f'_{2}(x) > 0\) for \(x>1\).

Note that \(f_{2}(x)\) is large and positive for large positive and negative values of \(x\), and also that \(f_{2}(x) \ge 0\).

So \(f_{2}(x)\) must have a minimum at \(x=1\) and a point of inflection at \(x=0\).

Graph of the described function. It has turning points where described and touches the x axis at (1,0). It tends to positive infinity as x tends to plus or minus infinity.
  1. On the same axes sketch the graph of \(y=f_{n}(x)\) where \(n\) is a large positive integer.

The function \(f_{n}(x)\) still crosses the axes at \((1,0)\) and \((0,1)\), and is large and positive for large positive and negative values of \(x\).

Also, as \(x^{2n-1}-1\) is very close to \(-1\) for \(-1<x<1\), \(f_{n}(x)\) is close to \(1\) in this range.

The previous graph with the graph of f_n for large n drawn. The graph is close to 1 for x inbetween -1 and 1, and decreases very quickly down to 0 as x reaches 1 and then up toward infinity, and increases very quickly toward infinity as x reaches -1.
  1. Determine \[\int_0^1 \! f_{n}(x) \, \mathrm{d}x.\]

\[\int_0^1 \! f_{n}(x) \, \mathrm{d}x = \int_0^1 \! (x^{2n-1}-1)^2 \, \mathrm{d}x = \int_0^1 \! (x^{4n-2} - 2x^{2n-1} + 1) \, \mathrm{d}x = \frac{1}{4n-1} - \frac{1}{n} + 1.\]

  1. The positive constants \(A\) and \(B\) are such that \[\int_0^1 \! f_{n}(x) \, \mathrm{d}x \le 1 - \frac{A}{n+B} \quad \text{for all $n \ge 1$}.\]

    Show that \[(3n-1)(n+B) \ge A(4n-1)n,\] and explain why \(A \le \dfrac{3}{4}\).

From part (iii) and the information given in the question, we have \[\frac{1}{4n-1} - \frac{1}{n} + 1 \le 1 - \frac{A}{n+B},\] Rearranging this gives the required inequality: \[\begin{align} \frac{1}{4n-1} - \frac{1}{n} &\le - \frac{A}{n+B} \notag\\ (n+B)(n - (4n-1)) &\le - A n (4n-1) \notag\\ -(n+B)(3n-1) &\le - A n (4n-1) \notag\\ (3n-1)(n+B) &\ge A(4n-1)n. \end{align}\]

We can be sure that multiplying through by the denominators doesn’t change the direction of the inequality because the denominators are all positive.

Expanding this and collecting powers of \(n\) gives

\[(3-4A)n^2 + (A+3B-1)n - B \ge 0 \quad \text{for $n\ge1$} \]

The left-hand side is a quadratic in \(n\), which we want to be non-negative for all values of \(n \ge 1\).

Graphically this says that we want the parabola described by this quadratic to lie above the \(n\)-axis when \(n \ge 1\):

vertex-down parabola that lies above the n-axis when n is greater than 1

We see that a necessary condition is that the parabola is vertex-down, which requires the coefficient of \(n^2\) to be non-negative.

Hence we deduce that \(3 -4A \ge 0\), that is, \(A \le \dfrac{3}{4}\).

  1. When \(A=\dfrac{3}{4}\), what is the smallest possible value of \(B\)?

When \(A=\dfrac{3}{4}\), the \(n^2\) term in the inequality disappears, and it becomes

\[(12B-1)n-4B \ge 0 \quad \text{for $n\ge1$}.\]

The graph of the left-hand side against \(n\) is a straight line. The inequality says that we want this line to lie above the \(n\)-axis for all \(n \ge 1\):

line which lies above the n-axis when n is greater than 1

This will happen if and only if the gradient is non-negative, and the value at \(n=1\) is non-negative, that is, \(12B-1 \ge 0\) and \(12B-1-4B \ge 0\).

Rearranging these inequalities for \(B\) gives \(B \ge \dfrac{1}{12}\) and \(B \ge \dfrac{1}{8}\). So the smallest possible value of \(B\) is \(\dfrac{1}{8}\).