A committee of 6 is to be chosen from 14 candidates - HSC - SSCE Mathematics Extension 1 - Question 4 - 2003 - Paper 1

Newton's method uses the formula:
$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

Calculating $f(1.5)$ and $f'(x)$:

1. First, find $f'(x) = \cos x - \frac{2}{3}$.

2. Evaluating at $x = 1.5$:
   $f(1.5) = \sin(1.5) - \frac{2 \times 1.5}{3}$ $f(1.5) \approx 0.997 - 1 = -0.003$

3. Now calculate $f'(1.5)$: $f'(1.5) = \cos(1.5) - \frac{2}{3} \approx 0.0707 - 0.6667 \approx -0.596$

4. Now apply Newton’s formula: $x_{2} = 1.5 - \frac{-0.003}{-0.596} \approx 1.5 + 0.00503 \approx 1.505$

Thus, the second approximation to the zero is $1.505$.

Let the roots be $r_1$, $r_2$, and $r_3$, where $r_1 = \frac{1}{r_2}$. Using Vieta’s formulas, we have:

1. The sum of the roots: $r_1 + r_2 + r_3 = -\frac{1}{2}$
2. The product of the roots: $r_1 r_2 r_3 = -\frac{6}{2} = -3$.

Since $r_1 = \frac{1}{r_2}$, we can denote $r_2 = r$ and $r_1 = \frac{1}{r}$. Substituting into the product formula, we get:
$r_1 r_2 r_3 = \frac{1}{r} r r_3 = r_3 = -3$ Thus, $r_3 = -3$.
Substituting into the sum gives:
$\frac{1}{r} + r - 3 = -\frac{1}{2} \Rightarrow \frac{1}{r} + r = -\frac{1}{2} + 3 = \frac{5}{2}$ Multiplying through by $r$ leads to: $1 + r^2 = \frac{5}{2} r \Rightarrow 2r^2 - 5r + 2 = 0$ Using the quadratic formula:
$r = \frac{5 \pm \sqrt{(5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} = \frac{5 \pm 1}{4}$ Thus, $r = 1.5$ or $r = 1$. Therefore, $k = 2(1)(-3) + 6 = 12$. Consequently, the value of $k$ is $12$.

$CPQB$ is a cyclic quadrilateral because the angles subtended by the same arc at the circumference are equal. Specifically, $ext{angle } CQP + ext{angle } CPB = 180^{ ext{o}}$, since $CQ$ and $BP$ are perpendicular to sides $AB$ and $AC$, respectively. This confirms that points $C$, $P$, $Q$, and $B$ lie on the same circle.

$PAQT$ is a cyclic quadrilateral because the angle $PAQ$ is subtended by the same arc as angle $PTA$. Since both angles are inscribed angles in the same segment, they will sum to $180^{ ext{o}}$, thus establishing that points $P$, $A$, $Q$, and $T$ lie on the same circle.

Since $CPQB$ and $PAQT$ are cyclic quadrilaterals, the opposite angles will be equal. Therefore, by the inscribed angle theorem, we have:
$\angle ATA = \angle CQP, \quad \text{and } \angle BPQ = \angle QCB$ Adding these results gives: $\angle ATA + \angle BPQ = \angle QCB\Rightarrow \angle ATA = \angle QCB$

To prove that $AR \perp CB$, we can examine the angles formed at point $R$. From the cyclic quadrilateral properties, we know that the angles subtended by line segments through the circle remain constant. Thus, $AR$ is the angle bisector of $riangle ATB$, and since $BP$ is altitude, $AR$ must be perpendicular to $CB$. Therefore, we conclude that $AR \perp CB$.