Review question

# When is this integral less than or equal to $1 - A/(n+B)$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8396

## Solution

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 $x=0$.

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.

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 $n \ge 1$:

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 $n \ge 1$:

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 $\dfrac{1}{8}$.