NSW BOS Mathematics Extension 2 Solutions 2005

NSW BOS Mathematics Extension 2 Solutions 2005

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Q1a Let

Q1bi . Let then . Let then .

Q1bii

Q1c

Q1d

Q1ei .

Q1eii .

Q1eiii

Q2ai .

Q2aii .

Q2aiii .

Q2bi ,,

Q2bii .

Q2biii .

Q2c Let

i.e. all the complex numbers with imaginary part .

consists of all the complex numbers on or outside the circle with radius 1 unit and centred at 1+0i.

Im z

Re z

Q2di Let . Then the angle between and OP is . .

Q2dii Since and ,

Q2diii When

Since the complex number is a point R on the positive imaginary axis. As increases, R moves in the positive direction. The argument of has no effects on the locus of R.

Q3a Given the graph of .

Q3ai

Q3aii

Q3aiii

Q3aiv

Q3b .

Asymptotes are .

x-intercepts:

or .

Q3c

At , , . Equation of the normal is , i.e.

Q3d



, .

Q4ai Volume of a general layer

Volume of solid

Let Volume

Q4aii As volume.

Q4bi

Q4bii Since

Q4biii For

, i.e. . This is impossible because the sum of squares > 0. Hence the quartic function cannot have four real roots.

Q4biv Let . has at least one real root. According to the fundamental theorem of algebra, has four real/complex roots. Since has real coefficients, the roots are pairs of conjugates if they are complex. From part iii and the fact that has at least one real root, it has exactly two complex roots (conjugates) and two real roots.

Q4ci The normal at passes through point B,

, .

Hence , i.e.

or i.e. because is on the ellipse.

Q4cii If (this includes the possibility that i.e. ), then

Since ,

Q5ai Area of

Since or .

Q5aii ,

, hence the result.

Q5bi Mary is the winner, she is the first to win 5 goals. Ferdinand scores only one goal. There are a total of 6 goals. Mary wins the last goal otherwise Ferdinand would not win any goal at all. Hence Ferdinand scores his only goal in any one of the first five trials.

Q5bii Suppose Mary is the winner, the following table lists a possible outcome of each type and the corresponding number of outcomes.

Type / Number
MMMMM / 1
FMMMMM / 5
FFMMMMM /
FFFMMMMM /
FFFFMMMMM /

Ferdinand could win in the same manner, total number of possible outcomes.

Q5ci The following graphs show and .

0 a 0 b

The shaded areas are equal. From the graph of ,

the shaded area +

Q5cii , , .

Q5di Base length AD

height AB. Area of ABCD

Q5dii Volume of a general vertical slice of thickness

Volume of wedge

Q6ai For

it is true.

Assume it is true for i.e. .

Consider ,

, it is also true for . Hence it is true for all .

Q6aii Consider in the interval .

Since and . Hence

Q6aiii From part ii,

Q6aiv As

Q6bi After expanding,

Since

and given

Hence .

Q6bii Let ,

Since

Q6biii Let , where

Then .

Q6biv From part i,.

When , .

Q6bv

or .

Solve the quadratic equation forwhere to obtain

Q7ai Sum of opposite angles = 180o, is cyclic.

Q7aii , because is cyclic and both angles are subtended by the same arc BM.

, because PT is a tangent and the alternate segment theorem follows.

Hence , i.e. the corresponding angles are equal, and this is a sufficient condition for MN // PA.

Q7aiii A

N Q

s a

T r M s P

Dotted lines are added to the given diagram.

and are similar,

Q7aiv Since p is the longest side of , ,

or .

Q7bi The satellite is in free fall,

Q7bii , . Moving from R to x,

Since at , .

Q7biii i.e.

. Moving from R to 0,

the time required

Q8ai Given ,.

Define . Their stationary points occur at the same x value.

since

or The point at is the only stationary point, it must be the required minimum.

Q8aii is minimum at , ,

Since

. Hence

Q8aiii Let a, b and c be the positive real roots.

Expand and compare coefficients, and .

From part ii, , , .

Q8iv For cubic equation , and . . the equation does not have 3 real roots. According to the fundamental theorem of algebra, and the conjugate roots theorem for real coefficients, the equation has exactly one real root and a pair of conjugate roots.

Q8bi ,

Q8bii C

 

Apply the sine rule to and .

Q8biii

is dependent on but independent of P (does not appear in the expression), b and are constant values that define the hyperbola.

Q8biv The result of part iii suggests that the same is also true for another point, Q, on the line and the hyperbola, i.e.

, .

Q8bv As CD moves closer to UV, , and P, Q and T become the same point when the two lines coincide.

Since , i.e. T is the midpoint of CD.

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