Find \( \int \cos 4x \, dx \) - HSC - SSCE Mathematics Extension 1 - Question 3 - 2004 - Paper 1

To find ( b ), substitute ( x = -1 ) in the polynomial ( P(x) ):

( P(-1) = (-1 + 1)(-4)Q(-1) + a(0) + b = -11 ) gives us:

( b = -11 ).

Using the polynomial remainder theorem, we know:\n( R(-1) = -11 ) and ( R(3) = 1 ).\nTo find the polynomial ( R(x) = Ax + B ):\n[ R(-1) = A(-1) + B = -11 \Rightarrow -A + B = -11][ R(3) = A(3) + B = 1 \Rightarrow 3A + B = 1]\nSolving these equations:

1. ( B = -11 + A )\n2. Substitute in the second equation:\n ( 3A + (-11 + A) = 1 \Rightarrow 4A - 11 = 1 \Rightarrow 4A = 12 \Rightarrow A = 3 )\n ( B = -11 + 3 = -8 )\nThus, the remainder is ( 3x - 8 ).

From the right-angled triangle formed by the pontoon and jetty, we have the Pythagorean theorem:\n ( x^2 + h^2 = 16 ) (since the walkway is 4m long, and the total height difference is ( h )).\nThus, rearranging gives:\n[ x = \sqrt{16 - h^2} ]

Given that ( h ) is changing over time, we find ( \frac{dh}{dt} = 0.3 ) m/hour. Using implicit differentiation:\n[ 2x \frac{dx}{dt} + 2h \frac{dh}{dt} = 0 \Rightarrow \frac{dx}{dt} = -\frac{h}{x} \frac{dh}{dt}]\nSubstituting ( h = 1 ) and ( x = \sqrt{15} ):\n[ \frac{dx}{dt} = -\frac{1}{\sqrt{15}} (0.3) ] results in the rate at which the pontoon moves away.

In cube geometry, by properties of angles and symmetry, the angle ( \angle FAC ) can be shown to be 60 degrees as all edges are equal and the face forms an equilateral triangle with respect to point F.

Using Pythagoras' theorem on triangle ( AFO ):\n[ FO^2 = OA^2 + AF^2 \Rightarrow FO^2 = 2^2 + 2^2 = 4 + 4 = 8 \Rightarrow FO = \sqrt{6} ] due to position of O based on inscribed circle.

Using trigonometric relationships in triangle ( XYF ), with sides calculated as ( XY = 1 ) and ( FY = 2 ):\n[ \tan(\angle XYF) = \frac{XY}{FY} = \frac{1}{2} \Rightarrow \angle XYF = \tan^{-1}(\frac{1}{2}) \approx 28^\circ. ]