Find \( \int_0^{\frac{\pi}{2}} \sin^2 x \, dx \) - HSC - SSCE Mathematics Extension 1 - Question 3 - 2006 - Paper 1

To find the intersection points of ( y = 3\log x ) and ( y = x ), we need to solve:

$3\log x - x = 0$

We can use values between 1.5 and 2 to evaluate:

• For ( x = 1.5 ):

  ( f(1.5) = 3\log(1.5) - 1.5 \approx 0.21 ) ( (positive) )

• For ( x = 2 ):

  ( f(2) = 3\log(2) - 2 \approx -0.56 ) ( (negative) )

Since ( f(1.5) > 0 ) and ( f(2) < 0 ), by the Intermediate Value Theorem, there exists a point ( P ) in the interval ( (1.5, 2) ).

Newton's method updates points using the formula:

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

We need the derivative:

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

Starting at ( x_0 = 1.5 ):

1. Calculate ( f(1.5) ) and ( f'(1.5) ):

   • ( f(1.5) \approx 0.21 )
   • ( f'(1.5) \approx 0.67 )

2. Update using Newton's method:

   $$$$

Repeat this process for better approximation until desired accuracy is achieved.

To find the number of different towers of height 3, we choose 3 blocks from 5 available blocks. The formula for combinations is:

$C(n, k) = \frac{n!}{k!(n-k)!}$

For ( n = 5 ) blocks and ( k = 3 ):

$C(5, 3) = \frac{5!}{3!(5-3)!} = \frac{5\times 4}{2\times 1} = 10$

Thus, there are 10 different combinations. However, each block can be arranged in any order:

1. Arrange 3 blocks:
   • Total arrangements: ( 3! = 6 )

Total towers of height 3:

$\text{Total} = 10 \times 6 = 60$

To find the total number of towers of height 2, 3, 4 and 5, we need to calculate:

• For 2 blocks: ( C(5, 2) \times 2! )
• For 3 blocks: ( C(5, 3) \times 3! )
• For 4 blocks: ( C(5, 4) \times 4! )
• For 5 blocks: ( C(5, 5) \times 5! )

Total:

For 2 blocks: $C(5, 2) = 10 \Rightarrow Total = 10 \times 2 = 20$ For 3 blocks: $C(5, 3) = 10 \Rightarrow Total = 10 \times 6 = 60$ For 4 blocks: $C(5, 4) = 5 \Rightarrow Total = 5 \times 24 = 120$ For 5 blocks: $C(5, 5) = 1 \Rightarrow Total = 1 \times 120 = 120$

Adding up all possibilities: $\text{Total} = 20 + 60 + 120 + 120 = 320$

A cyclic quadrilateral has opposite angles that sum up to 180 degrees. To show that ( QKT M ) is cyclic, we can use the angle at point ( K ):

Show that:

$\angle QKM + \angle QTM = 180^\circ$

Using the inscribed angle theorem considering the circle. If these angles sum to 180, ( QKT M ) is cyclic.

To show the angles are equal: Using the properties of cyclic quadrilaterals, we know that angles subtended by the same arc in a circle are equal. Thus:

$\angle KMT \equiv \angle KQT$

Since both angles subtend the same arc ( KT ), this holds true.

From previous results, we have proved that the corresponding angles:

$\angle KMT = \angle KQT$

Since alternate angles are equal, it follows that:

$MK \parallel TP$.

Thus, shown by angle properties of parallel lines.