Mathematics 2011

$f$ is a polynomial, so its maximum sits where $f'$ turns from positive to negative.

$$\begin{aligned}f'(x)&=4x^2-4x^3\\&=4x^2(1-x)\end{aligned}$$

Since $4x^2\ge0$, the sign of $f'$ is the sign of $1-x$: positive for $x<1$, negative for $x>1$. So $f$ rises up to $x=1$ and falls after it, and the stationary point at $x=0$ is not a turning point.

Therefore the maximum value is attained at $x=1$.

The $n$th term has denominator $n^2-(n-1)^2$, so simplify that first.

$$\begin{aligned}a_n&=\frac{n}{n^2-(n-1)^2}\\&=\frac{n}{2n-1}\\&=\frac{1}{2-\frac{1}{n}}\end{aligned}$$

As $n\to\infty$, $\frac{1}{n}\to0$ and the denominator tends to $2$.

Therefore the limit of the sequence is $\frac{1}{2}$.

On $[0,1]$ we have $x^2\ge x^3$, so revolving about the $x$-axis gives washers with outer radius $x^2$ and inner radius $x^3$.

$$\begin{aligned}V&=\pi\int_0^1\left[(x^2)^2-(x^3)^2\right]dx\\&=\pi\int_0^1\left(x^4-x^6\right)dx\\&=\pi\left[\frac{x^5}{5}-\frac{x^7}{7}\right]_0^1\\&=\pi\left(\frac{1}{5}-\frac{1}{7}\right)\end{aligned}$$

Therefore the volume of the solid is $\frac{2\pi}{35}$ cubic units.

Write $f(x)=x^{-1}$ and differentiate repeatedly, watching the sign and the factorial.

$$\begin{aligned}f'(x)&=-x^{-2}\\f''(x)&=2x^{-3}\\f'''(x)&=-6x^{-4}\end{aligned}$$

Each step brings out one more factor of $-1$ and multiplies by the next integer, so after $n$ steps the coefficient is $(-1)^nn!$ and the power is $x^{-(n+1)}$.

Therefore $f^{(n)}(x)=\frac{(-1)^nn!}{x^{n+1}}$.

The defect rate in the tested sample estimates the rate for the machine, so first find how many items were tested.

$$\begin{aligned}\text{tested}&=0.02\times2500=50\\P(\text{defect})&=\frac{5}{50}=0.1\end{aligned}$$

Therefore $P(\text{no defect})=1-0.1=0.90$. The $2{,}500$ never enters the probability itself; it only fixes the sample size.