Mathematics 2011

$f'(x)=4x^2-4x^3=4x^2(1-x)$, so the sign of $f'$ is decided entirely by $1-x$.

$$\begin{aligned}f'(x)&=4x^2(1-x)=0\\&\Rightarrow x=0\ \text{or}\ x=1\end{aligned}$$

Because $4x^2\ge0$ for all $x$, $f'>0$ on $(-\infty,1)$ and $f'<0$ on $(1,\infty)$; at $x=0$ the repeated factor does not change sign, so no turning point occurs there.

Therefore $f$ attains its maximum value at $x=1$.

The $n$th term is $\dfrac{n}{n^2-(n-1)^2}$, and the denominator collapses to a linear expression.

$$\begin{aligned}n^2-(n-1)^2&=2n-1\\&\Rightarrow a_n=\frac{n}{2n-1}=\frac{1}{2-\frac1n}\end{aligned}$$

As $n\to\infty$, $\frac1n\to0$, so the denominator tends to $2$.

Therefore the limit of the sequence is $\dfrac12$.

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

$$\begin{aligned}V&=\pi\int_0^1\left[(x^2)^2-(x^3)^2\right]dx\\&=\pi\int_0^1\left(x^4-x^6\right)dx\\&=\pi\left[\frac{x^5}{5}-\frac{x^7}{7}\right]_0^1\\&=\pi\left(\frac15-\frac17\right)\end{aligned}$$

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

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

$$\begin{aligned}f'(x)&=-x^{-2}\\f''(x)&=2x^{-3}\\f'''(x)&=-6x^{-4}\\&\Rightarrow f^{(n)}(x)=(-1)^n\,n!\,x^{-(n+1)}\end{aligned}$$

Each differentiation contributes one factor of $-1$ and one more factor in the factorial, and drops the exponent by one, so after $n$ steps the power is $-(n+1)$.

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

The tested sample is $2\%$ of the production, so its size is the base against which the defects are counted.

$$\begin{aligned}\text{sample size}&=0.02\times2500=50\\&\Rightarrow P(\text{defective})=\frac{5}{50}=0.1\end{aligned}$$

The required probability is the complement of this.

Therefore the probability that an item has no defect is $0.90$.