3 (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 3 - 2004 - Paper 1

To find ( b ), we can utilize the information that when ( P(x) ) is divided by ( (x + 1) ), the remainder is -11:

Substituting ( x = -1 ):

( P(-1) = (-1 + 1)(-1 - 3)Q(-1) + a(-1 + 1) + b = -11 \implies b = -11 ).

Using the Remainder Theorem, we know the remainders found from dividing by the factors:\

• For ( (x + 1) ), remainder is -11,
• For ( (x - 3) ), remainder is 1.

Thus, the remainder when dividing by ( (x + 1)(x - 3) ) can be expressed as: ( R(x) = Ax + B ) and substituting both conditions will help find A and B.

From the relationship in the right triangle formed, we can apply Pythagoras' theorem:

$x^2 + h^2 = 16$

Thus, solving for ( x ):

$x = \sqrt{16 - h^2}$

Using related rates, we differentiate the expression for ( x ):

$\frac{dx}{dt} = \frac{1}{2(16 - h^2)^{\frac{1}{2}}}(-2h\frac{dh}{dt})$

Then substitute ( h = -1 ) and ( \frac{dh}{dt} = 0.3 \text{ m/h} ) to find ( \frac{dx}{dt} ).

The geometry of the figure indicates that triangle ( FAC ) is equilateral as all sides are equal; hence, all angles, including ( \angle FAC ), measure 60 degrees.

In the right triangle formed by the center O, point F and line segment AC:

Using the coordinates: Each edge being 2 meters, calculate the lengths:

$FO = \sqrt{AO^2 + AF^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 2} = \sqrt{6}\text{ m}.$

Using trigonometry: The relationship between angles ensures that we can apply the tangent function:

$\tan(\angle XYF) = \frac{XY}{FY} = \frac{1}{\sqrt{6}}$

Solving gives the angle as:\n Calculate the angle using Inverse Tangent function and round to the nearest degree.