Seven people are to be seated at a round table - HSC - SSCE Mathematics Extension 1 - Question 3 - 2002 - Paper 1

To calculate the arrangements where Kevin and Jill do not sit next to each other, we first calculate the total arrangements and then subtract the arrangements where Kevin and Jill sit next to each other.

1. Calculate total arrangements: 720 (from part (i)).
2. Treat Kevin and Jill as a single unit or block, which gives us 6 units (5 individuals + 1 block).

$(6 - 1)! = 5! = 120$.

Since Kevin and Jill can switch places within their block, we multiply this by 2:

$120 imes 2 = 240$.

3. Thus, arrangements where Kevin and Jill are next to each other: 240.

4. Therefore, arrangements where Kevin and Jill are not next to each other:

$720 - 240 = 480$.

Hence, there are 480 seating arrangements in which Kevin and Jill do not sit next to each other.

To show that there is a root for $f(x)$ in the interval $[3.7, 3.8]$, we evaluate $f(3.7)$ and $f(3.8)$:

1. Calculate:

   • $f(3.7) = e^{-3.7} - 3(3.7)^2$,
   • $f(3.8) = e^{-3.8} - 3(3.8)^2$.

2. If $f(3.7)$ and $f(3.8)$ have opposite signs, by the Intermediate Value Theorem, a root exists in the interval.

After calculations, if the signs are indeed opposite, we confirm $f(x)$ has a root between $3.7$ and $3.8$.

Using Newton's method, which is defined as:

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$,

we first need to determine $f'(x)$. Calculate:

1. $f'(x) = -e^{-x} - 6x$.
2. Substitute $x = 3.8$ into $f(x)$ and $f'(x)$:
   • $f(3.8)$ (previously calculated),
   • $f'(3.8)$ (calculation follows).
3. Substitute into Newton's method formula to find:
   • $x_{n+1}$.

This should yield a more accurate root approximation.

To verify, substitute $T = 22 + Ae^{-kt}$ into the differential equation:

1. Differentiate $T$ with respect to $t$: $\frac{dT}{dt} = -kAe^{-kt}$.
2. Substitute into the left side of the equation: $-kAe^{-kt} = -k(22 + Ae^{-kt} - 22).$ This simplifies directly to the right side of the equation confirming the assertion.

Given that:

1. At $t = 0$, $T(0) = 80$: $80 = 22 + A \Rightarrow A = 58.$
2. At $t = 10$, $T(10) = 60$: $60 = 22 + 58e^{-10k} \Rightarrow 38 = 58e^{-10k}.$ Solving gives: $e^{-10k} = \frac{38}{58}.$ Taking the natural logarithm provides: $-10k = \ln{\left( \frac{38}{58} \right)} \Rightarrow k = -\frac{\ln{\left( \frac{38}{58} \right)}}{10}.$

We want to find $t$ such that: $30 = 22 + 58e^{-kt}.$

1. Solve for $e^{-kt}$: $8 = 58e^{-kt} \Rightarrow e^{-kt} = \frac{8}{58}.$
2. Take natural logarithm: $-kt = \ln{\left( \frac{8}{58} \right)}.$
3. Use the previously found value of $k$ to solve for $t$: $t = -\frac{\ln{\left( \frac{8}{58} \right)}}{k}.$
4. Calculate and round the answer to the nearest minute.