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

Next, we find the intersection point of the curve and the line by solving:

$f(x) = 0$

Substituting $x = 1.5$ and $x = 2$ into the function:

For ( x = 1.5 ):

$f(1.5) = 3 \log(1.5) - 1.5 \approx 3(0.1761) - 1.5 \approx 0.5283 - 1.5 = -0.9717$

For ( x = 2 ):

$f(2) = 3 \log(2) - 2 \approx 3(0.3010) - 2 \approx 0.9030 - 2 = -1.0970$

Since ( f(1.5) < 0 ) and ( f(2) > 0 ), by the Intermediate Value Theorem, there exists a value of ( x ) between 1.5 and 2 where ( f(x) = 0 ).

Newton's method uses the formula:

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

First, we need to compute the derivative (f'(x) = \frac{3}{x} - 1 ). Now applying Newton's method:

1. Start with ( x_0 = 1.5 ).
2. Calculate ( f(1.5) ) and ( f'(1.5) ):
   • ( f(1.5) \approx -0.9717 )
   • ( f'(1.5) = \frac{3}{1.5} - 1 = 2 - 1 = 1 )
3. Calculate ( x_1 ):
   • ( x_1 = 1.5 - \frac{-0.9717}{1} \approx 2.4717 )
4. Repeat for ( x_1 ) to get closer to the actual root. Continue this process until convergence to two decimal places.
5. After several iterations, you would find ( x \approx 1.91 ).

To find the number of different tower combinations that are three blocks high, we can use the combination formula:

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

For this case, since Sophie has five colored blocks and can choose any three:

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

However, since the arrangement matters (the same three colors can be stacked in different orders), we must also consider permutations:

$P(3) = 3! = 6$

Thus, the total combinations are:

$10 \times 6 = 60$

To calculate the total number of towers she can form with heights of either two, three, or four blocks, we calculate:

1. For two blocks: ( C(5,2) ) with arrangements = ( 10 \times 2! = 20 )
2. For three blocks: ( C(5,3) ) with arrangements = ( 10 \times 6 = 60 )
3. For four blocks: ( C(5,4) ) with arrangements = ( 5 \times 24 = 120 )

Now, adding these:

Total = ( 20 + 60 + 120 = 200 )

To show that quadrilateral ( QKT ) is cyclic, we need to prove that the opposite angles sum to 180 degrees. We can utilize the inscribed angle theorem which states that an angle formed by two chords in a circle is half of the angle subtended by those chords at the circumference.

If ( \angle QKT + \angle QTK = 180^\circ ), then it is cyclic.

Since ( MN ) is tangent to the circle at ( T ), we know that the angle between the tangent line and the chord is equal to the angle subtended by the chord at the opposite side of the circle. Hence:

$\angle KMT = \angle KQT$

Since we established that ( \angle KMT = \angle KQT ) and ( KQT ) implies that corresponding angles being equal leads us to conclude that lines are parallel. Therefore, we can state that:

$MK \parallel TP$