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

To apply Newton's method, we first need to find the derivative of $f(x)$:

$f'(x) = \text{cos} \, x - \frac{2}{3}$

Then, using the initial approximation $x_0 = 1.5$, we compute the next approximation using:

$x_{1} = x_{0} - \frac{f(x_{0})}{f'(x_{0})}$

First, evaluate $f(1.5)$ and $f'(1.5)$:

1. Calculate $f(1.5)$: $f(1.5) = \text{sin}(1.5) - \frac{2 \times 1.5}{3}$

2. Calculate $f'(1.5)$: $f'(1.5) = \text{cos}(1.5) - \frac{2}{3}$

Using these values, we substitute back into the equation to find $x_1$ and round to three decimal places.

Let the roots be $r$ and $\frac{1}{r}$. According to Vieta's formulas, the sum of the roots rac{-b}{a} and the product of the roots rac{c}{a} can be used:

1. The sum of the roots: $r + \frac{1}{r} + r_3 = -\frac{1}{2}$

2. The product of the roots: $r \cdot \frac{1}{r} \cdot r_3 = \frac{6}{2} = 3$

Solve these equations to determine the necessary value of $k$.

A cyclic quadrilateral is one where all corners lie on a single circle. This is true if the opposite angles are supplementary. In $CPQB$, since $CP$ and $BQ$ are diameters, $riangle CPQ$ and $riangle BPQ$ share the angle subtended by $PB$ and $CQ$, making them supplementary.

Similarly, for quadrilateral $PAQT$, the angles subtended by the same arc are equal. This means that if $AT$ subtends an angle at points $P$, $A$, and $Q$ that sum up to $180^ ext{o}$, thus confirming it is cyclic.

To prove this, we can use the fact that angles corresponding to cyclic quadrilaterals are equal. As $AQ$ intersects $CTB$ at $A$, we can conclude that $\angle AQT = \angle LQB$ and thus triangles $AQT$ and $LQCB$ are similar by AA similarity.

By construction, since we have established that $CPQB$ and $PAQT$ are cyclic, angles that sum to $90^ ext{o}$ must also hold true for point $R$, confirming $AR$ is perpendicular to line $CB$.