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

Using the Remainder Theorem:

• When divided by ( (x + 1) ), we have: [ P(-1) = -11 ] Substitute ( x = -1 ): [ P(-1) = 0 + 0 + b = -11 \Rightarrow b = -11. ]

Thus, ( b = -11 ).

Using the information given:

• When divided by ( (x - 3) ): [ P(3) = 1 ] From the division, we substitute to find: [ R(3) = (3 + 1)(0) + 3a + b = 1 ] Since we found ( b = -11 ): [ 12 + 3a - 11 = 1 \Rightarrow 3a + 1 = 1 \Rightarrow 3a = 0 \Rightarrow a = 0. ]
• Now, the polynomial can be expressed and the remainder when divided by ( (x + 1)(x - 3) ) is ( -11 + 1 = -10. )

From the geometry, using Pythagoras' theorem:

• The relationship is: [ x^2 + h^2 = 16
  \Rightarrow x^2 = 16 - h^2 \ \Rightarrow x = \sqrt{16 - h^2}. ]

Using related rates, we have:

• ( rac{dh}{dt} = 0.3 , \text{m/hr} ) and the implicit relationship: [ rac{d}{dt} (x^2 + h^2) = 0
  \Rightarrow 2x \frac{dx}{dt} + 2h \frac{dh}{dt} = 0
  \Rightarrow rac{dx}{dt} = -\frac{h}{x} \frac{dh}{dt}. ]
• Substituting the known values when ( h = -1 ): [ x = \sqrt{16 - (-1)^2} = \sqrt{15}. ]
• Thus, the final rate: [ rac{dx}{dt} = -\frac{-1}{\sqrt{15}} (0.3). ]

Since ( A, F, C ) are vertices of a cube and ( O ) is the centre where the circle touches, using the symmetry and properties of the cube, the angles formed by the line to a point on the circle and the edges will create equal segments such that: [ \angle FAC = \angle ABC = 60^{ ext{o}}. ]

Using geometry:

• The distance from the centre ( O ) to each edge (radii) is half of the diagonal: [ FO = \sqrt{AC^2 - AO^2} = \sqrt{2^2 - 1^2} = \sqrt{3} = \sqrt{6}, ]

Using the geometry of the triangle formed:

• The angles related through the relationship of tangents yields: [ \angle XYF = \tan^{-1}(\frac{1}{\sqrt{6}}) \approx 20^{ ext{o}} \text{ to the nearest degree.} ]