Lyndal hits the target on average 2 out of every 3 shots in archery competitions - HSC - SSCE Mathematics Extension 1 - Question 4 - 2002 - Paper 1

To find the probability that Lyndal hits the target fewer than 9 times, we consider the complementary event:

$P(X < 9) = 1 - P(X = 9) - P(X = 10).$

We already calculated $P(X = 9)$, and now we calculate $P(X = 10)$:

$P(X = 10) = \binom{10}{10} \left(\frac{2}{3}\right)^{10} \left(\frac{1}{3}\right)^{0} = \left(\frac{2}{3}\right)^{10}.$

Thus,

$P(X < 9) = 1 - \left(10 \left(\frac{2}{3}\right)^9 \left(\frac{1}{3}\right) + \left(\frac{2}{3}\right)^{10}\right).$

From Vieta's formulas, we know:

$\alpha + \beta + \gamma = -\frac{\text{coefficient of } x^2}{\text{leading coefficient}} = -(-2) = 2.$

Thus, $\alpha + \beta + \gamma = 2.$

Similarly, using Vieta's formulas, we find:

$\alpha\beta\gamma = -\frac{\text{constant term}}{\text{leading coefficient}} = -\frac{24}{1} = -24.$

Thus, $\alpha\beta\gamma = -24.$

Let the roots be $r, -r, s$. Then:

1. From $\alpha + \beta + \gamma = 2$, we have $r + (-r) + s = 2$, so $s = 2$.
2. From $\alpha\beta\gamma = -24$, we substitute $r(-r)(s) = -24$. Hence: $-r^2(2) = -24 \Rightarrow r^2 = 12 \Rightarrow r = 2\sqrt{3}.$
   Therefore, the roots are $2\sqrt{3}, -2\sqrt{3}, 2$. Finally, substituting back into the polynomial to find $k$, we need to use the identity and coefficients in $P(x)$.

To show this, we start with the second derivative given:

$\frac{d^2x}{dt^2} = -16x.$

The general solution for such a differential equation is given by:

$x(t) = A \cos(kt) + B \sin(kt)$

where $k = 4$. To find the constants $A$ and $B$, we use the initial conditions. As the particle oscillates, it satisfies the equation.

Next, applying conditions at $t = 0, x(0) = 1$ leads to:

$|x| = 4\sqrt{2 - x^2}$ as derived from trigonometric identities of such systems.

The greatest displacement occurs at the maximum amplitude given by the maximum value of $|x|$:

From $|x| = 4 \sqrt{2 - x^2}$, we can see that the maximum occurs when $x = 0$, hence:

$|x|_{max} = 4 \sqrt{2 - 0} = 4\sqrt{2}.$

The general form for $x(t)$ under simple harmonic motion conforms to:

$x(t) = A \cos(wt + \phi)$ where $A$ is the amplitude, $w$ is the angular frequency, and $\phi$ is the phase constant. By substituting known values, you can derive a specific form depending on the given conditions.