Find \( \int_{0}^{2} \sin^2{x} \: dx \) - HSC - SSCE Mathematics Extension 1 - Question 3 - 2006 - Paper 1

To demonstrate that ( f(x) = 3 \log_{e} x - x ) has a root between 1.5 and 2, we will evaluate the function at these points:

• At ( x = 1.5 ): [ f(1.5) = 3 \log_{e}(1.5) - 1.5 \approx 3(0.405) - 1.5 \approx 1.215 - 1.5 < 0 ]

• At ( x = 2 ): [ f(2) = 3 \log_{e}(2) - 2 \approx 3(0.693) - 2 \approx 2.079 - 2 > 0 ]

Since ( f(1.5) < 0 ) and ( f(2) > 0 ), by the Intermediate Value Theorem, there exists at least one root ( P ) in the interval ( (1.5, 2) ).

Newton's method formula is given by:

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

First, we need to compute the derivative:

$f'(x) = \frac{3}{x} - 1$

Now applying Newton's method:

1. At ( x_0 = 1.5 ): [ f(1.5) = 3 \log_{e}(1.5) - 1.5 \approx -0.285 ] [ f'(1.5) = \frac{3}{1.5} - 1 \approx 1 ] [ x_1 = 1.5 - \frac{-0.285}{1} \approx 1.785 ]

2. At ( x_1 = 1.785 ): [ f(1.785) \approx 3 \log_{e}(1.785) - 1.785 \approx -0.024 ] [ f'(1.785) \approx 1.677 ] [ x_2 = 1.785 - \frac{-0.024}{1.677} \approx 1.797 ]

Thus, after rounding, the ( x )-coordinate of P is approximately ( 1.80 ) correct to two decimal places.

To determine the number of different towers that are three blocks high, we can use combinations considering that the blocks can repeat. Since there are 5 colours and the order matters:

• The number of combinations can be calculated as: [ 5^3 = 125 ]

Thus, there are 125 different towers that can be formed with three blocks.

To find the total number of towers ranging from two to five blocks high:

1. For two blocks: ( 5^2 = 25 )
2. For three blocks: ( 5^3 = 125 )
3. For four blocks: ( 5^4 = 625 )
4. For five blocks: ( 5^5 = 3125 )

Adding these together gives: [ 25 + 125 + 625 + 3125 = 3900 ]

Therefore, the total number of towers formed is 3900.

To show that ( QKT ) is cyclic, we need to prove that the opposite angles sum to 180 degrees:

Using the theorem of cyclic quadrilaterals, we know: [ \angle QKT + \angle QMT = 180^{\circ} ]

From the tangent-secant theorem, we conclude that if both angles are subtended by the same arc, they will be supplementary. Thus, ( QKT ) is cyclic.

Since ( QM \perp MN ), then:

• ( \angle KMT ) is the angle in the alternate segment.
• Therefore, it follows that: [ \angle KMT = \angle KQT ]

This is a consequence of the properties of angles in a circle. Thus, the statement holds.

From the previous conclusions:

Since ( \angle KMT = \angle KQT ) and they are corresponding angles formed by a transversal cutting through two parallel lines, we can conclude that:

Thus, we have shown that: ( MK \parallel TP ).