Skill Align

Skill Align HSC Mathematics Extension 2 - Free Online Pack 0

Full-length Mathematics Extension 2 showcase paper

Paper: HSC Paper Showcase
Reading: 10 minutes
Writing: 3 hours
Assessment: 100 marks

NESA-approved calculators may be used. Before starting, open the supplied companion resource “NSW Government Mathematics Advanced, Extension 1 and 2 Reference Sheet”: https://www.nsw.gov.au/education-and-training/nesa/curriculum/hsc-exam-papers/mathematics-advanced-extension-reference-sheet

Section I

Attempt Questions 1-10. Allow about 15 minutes for this section. Select the best answer for each question. Each question is worth 1 mark.

Question 1

1 mark

For z=sqrt3+i, the modulus and principal argument are

|z|=4, arg z=fracpi3
|z|=2, arg z=fracpi6
|z|=2, arg z=fracpi3
|z|=sqrt2, arg z=fracpi6

Question 2

1 mark

The Argand diagram shows the four roots of z⁴=16. Which root has argument 3pi / 2?

Diagram Preview

2i
2
-2
-2i

Question 3

1 mark

For x>0, the minimum value of x+frac1x is

1
0
2
-2

Question 4

1 mark

The value of int_0^infty e^(-2x),dx is

frac12
2
1
infty

Question 5

1 mark

The solution of y'=3y with y(0)=2 is

y=3e^(2x)
y=2e^(-3x)
y=2e^(3x)
y=6x+2

Question 6

1 mark

A particle has acceleration a(t)=6t-4 and initial velocity v(0)=3. Its velocity at t=2 is

Question 7

1 mark

The vectors (1,2,-1) and (2,-1,0) are

parallel
equal
oppositely directed
perpendicular

Question 8

1 mark

The transformation w=iz represents

a rotation through pi / 2 anticlockwise about the origin
a rotation through pi / 2 clockwise about the origin
a reflection in the real axis
a dilation by factor 2

Question 9

1 mark

The value of sum_(n=1)^(infty)frac1(n(n+1)) is

frac12
1
2
infty

Question 10

1 mark

The value of int_0¹((1) / (1+x²)),dx is

fracpi2
ln2
fracpi4
1

Section II

Attempt Questions 11-16. Allow about 2 hours and 45 minutes for this section. Show sufficient reasoning and working to support each answer.

Question 11

14 marks

On an Argand diagram, A represents 3+4i, B represents -1+2i, and w=x+iy is variable.

Diagram Preview

(a) 1 mark

Find the midpoint of AB.

(b) 2 marks

Find |B-A|.

(c) 2 marks

Find the Cartesian equation of the perpendicular bisector of AB.

(d) 2 marks

Find the intercepts of this perpendicular bisector with the coordinate axes.

(e) 4 marks

Show that the line 2x+y=5 is tangent to the circle |w|=sqrt5, and find the point of contact.

(f) 3 marks

Describe the half-plane containing the origin whose points are at least as close to B as to A. Justify the boundary choice.

Question 12

16 marks

This question develops several divisibility results for integers.

(a) 1 mark

Factorise n³-n.

(b) 2 marks

Hence prove that n³-n is divisible by 6 for every integer n.

(c) 2 marks

Prove that n^2equiv1 (mathrm(mod) 8) for every odd integer n.

(d) 2 marks

Deduce that 16mid(n⁴-1) for every odd integer n.

(e) 2 marks

Find the remainder when 2027⁴ is divided by 16, with justification.

(f) 3 marks

Let p>3 be prime. Prove that p^2equiv1 (mathrm(mod) 24).

(g) 4 marks

Prove that 30mid(n⁵-n) for every integer n.

Question 13

15 marks

For xgeq0, a curve satisfies ((dy) / (dx))=x(1-y) and y(0)=0.

Graph Preview

(a) 1 mark

State the equilibrium solution of the differential equation.

(b) 2 marks

Separate variables and integrate to obtain an implicit general solution for yneq1.

(c) 4 marks

Use y(0)=0 to find y explicitly.

(d) 2 marks

Find the exact positive x-coordinate where y=frac12.

(e) 3 marks

Find the maximum gradient of the curve for xgeq0, and where it occurs.

(f) 3 marks

Find the point of inflection and state the horizontal asymptote.

Question 14

15 marks

A particle moves vertically with upward velocity positive. Time t is measured in seconds and velocity v in text(m s)⁻¹. It satisfies ((dv) / (dt))=-2v-10 and v(0)=15.

Graph Preview

(a) 1 mark

Find the limiting velocity.

(b) 2 marks

Show that u=v+5 satisfies ((du) / (dt))=-2u, and solve for u.

(c) 2 marks

Use the initial condition to find v(t).

(d) 3 marks

Find when the particle reaches its greatest height and explain the change of direction.

(e) 4 marks

Find the displacement s(t), where s(0)=0, and hence find the exact maximum height above the initial point.

(f) 3 marks

Find v(2) to three decimal places and describe how the graph approaches its horizontal asymptote.

Question 15

15 marks

The sphere S has centre C=(1,-2,3) and passes through A=(3,-1,5). The line L has equation mathbf r=(3,-1,5)+lambda(1,2,-1).

Diagram Preview

(a) 1 mark

Find overrightarrow(CA).

(b) 2 marks

Find the radius of S and write its equation.

(c) 2 marks

Substitute a general point of L into the sphere equation and show that the intersection parameters satisfy 2lambda(3lambda+2)=0.

(d) 3 marks

Find the second point B where L meets S, and find the chord length AB.

(e) 3 marks

Find the angle angle ACB.

(f) 4 marks

Find the point H on L closest to the origin, and find OH. Verify the required perpendicularity using a scalar product.

Question 16

15 marks

For ngeq0, define I_n=int_0^infty x^ne^(-x),dx. Also define f(x)=frac12x^2e^(-x) for xgeq0, and f(x)=0 otherwise.

Graph Preview

(a) 1 mark

State I_0.

(b) 2 marks

Evaluate I_1 using integration by parts.

(c) 2 marks

Show that I_n=nI_(n-1) for ngeq1.

(d) 3 marks

Hence find I_2, I_3 and I_4.

(e) 3 marks

Show that f is a probability density and find its mean.

(f) 4 marks

Find the variance of the distribution and the x-coordinate of the mode.

Worked Solutions And Marking Guide

Section I Question 1

Answer: |z|=2, arg z=fracpi6

The modulus is √(3+1)=2, and tantheta=1 / sqrt3 in quadrant I.

Section I Question 2

Answer: -2i

The roots have modulus 2 and arguments 0,pi / 2,pi,3pi / 2.

Section I Question 3

Answer: 2

By AM-GM, x+1 / xgeq2sqrt(x(1 / x))=2, with equality at x=1.

Section I Question 4

Answer: frac12

Use the antiderivative -frac12e^(-2x) and take the limit.

Section I Question 5

Answer: y=2e^(3x)

The general solution is y=Ae^(3x), and the initial condition gives A=2.

Section I Question 6

Answer: 7

Integrating acceleration gives v(t)=3t²-4t+3, so v(2)=7.

Section I Question 7

Answer: perpendicular

Their dot product is 2-2+0=0.

Section I Question 8

Answer: a rotation through pi / 2 anticlockwise about the origin

Multiplication by i adds pi / 2 to the argument without changing the modulus.

Section I Question 9

Answer: 1

Write 1 / [n(n+1)]=1 / n-1 / (n+1) and telescope.

Section I Question 10

Answer: fracpi4

An antiderivative is tan⁻¹x.

Section II Question 11

(a) 1+3i.

Average the real parts and the imaginary parts.

(b) 2sqrt5.

Since B-A=-4-2i, its modulus is √20=2sqrt5.

(c) 2x+y=5.

Equate (x-3)²+(y-4)² and (x+1)²+(y-2)², then simplify.

(d) (5 / 2,0) and (0,5).

Set y=0, then set x=0, in 2x+y=5.

(e) The line is tangent at w=2+i.

Substitute y=5-2x into x²+y²=5. The resulting equation is 5(x-2)²=0, so there is one point of intersection, (2,1).

(f) 2x+yleq5, including the boundary.

The equality boundary is the perpendicular bisector. At the origin, |0-B|=sqrt5<5=|0-A|, and 2(0)+0leq5.

Mark allocation

Part a, mark 1: Obtains the midpoint as 1+3i.
Part b, mark 1: Forms B-A=-4-2i.
Part b, mark 2: Obtains |B-A|=2sqrt5.
Part c, mark 1: Equates the two squared distances from w to A and B.
Part c, mark 2: Simplifies to 2x+y=5.
Part d, mark 1: Obtains the real-axis intercept (5 / 2,0).
Part d, mark 2: Obtains the imaginary-axis intercept (0,5).
Part e, mark 1: Uses x²+y²=5 for the circle.
Part e, mark 2: Substitutes y=5-2x into the circle equation.
Part e, mark 3: Reduces to a repeated root, equivalently 5(x-2)²=0.
Part e, mark 4: Concludes tangency at w=2+i.
Part f, mark 1: Identifies the equality boundary 2x+y=5.
Part f, mark 2: Tests the origin by distances or substitution.
Part f, mark 3: States the closed half-plane 2x+yleq5.

Detailed marking criteria

Part a (1 mark)

Obtains the midpoint as 1+3i.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part b (1 mark)

Forms B-A=-4-2i.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part b (1 mark)

Obtains |B-A|=2sqrt5.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part c (1 mark)

Equates the two squared distances from w to A and B.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part c (1 mark)

Simplifies to 2x+y=5.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part d (1 mark)

Obtains the real-axis intercept (5 / 2,0).

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part d (1 mark)

Obtains the imaginary-axis intercept (0,5).

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part e (1 mark)

Uses x²+y²=5 for the circle.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part e (1 mark)

Substitutes y=5-2x into the circle equation.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part e (1 mark)

Reduces to a repeated root, equivalently 5(x-2)²=0.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part e (1 mark)

Concludes tangency at w=2+i.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part f (1 mark)

Identifies the equality boundary 2x+y=5.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part f (1 mark)

Tests the origin by distances or substitution.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part f (1 mark)

States the closed half-plane 2x+yleq5.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Section II Question 12

(a) n³-n=n(n-1)(n+1).

Take out the common factor n, then use a difference of squares.

(b) 6mid(n³-n).

The factors are three consecutive integers, so one is even and one is divisible by 3.

(c) n^2equiv1 (mathrm(mod) 8).

Write n=2k+1. Then n²-1=4k(k+1), and k(k+1) is even.

(d) 16mid(n⁴-1).

Since n²-1 is divisible by 8 and n²+1 is even, their product is divisible by 16.

(e) The remainder is 1.

The integer 2027 is odd, so part (d) applies.

(f) p^2equiv1 (mathrm(mod) 24).

Such a prime is odd, so 8mid(p²-1). Also p is not divisible by 3, so 3mid(p²-1). Since 3 and 8 are coprime, 24mid(p²-1).

(g) 30mid(n⁵-n).

The expression is divisible by 2 and 3 because n⁵-n=(n³-n)(n²+1). Modulo 5, Fermat's little theorem gives n^5equiv n, including the case 5mid n. The pairwise-coprime factors 2,3,5 give divisibility by 30.

Mark allocation

Part a, mark 1: Obtains n(n-1)(n+1).
Part b, mark 1: Identifies a factor of 2 among three consecutive integers.
Part b, mark 2: Identifies a factor of 3 and concludes divisibility by 6.
Part c, mark 1: Writes n=2k+1 and obtains n²-1=4k(k+1).
Part c, mark 2: Uses the evenness of k(k+1) to conclude the congruence.
Part d, mark 1: Factorises n⁴-1=(n²-1)(n²+1).
Part d, mark 2: Combines the factors of 8 and 2 to conclude divisibility by 16.
Part e, mark 1: Notes that 2027 is odd and applies part (d).
Part e, mark 2: States the remainder 1.
Part f, mark 1: Establishes 8mid(p²-1) because p is odd.
Part f, mark 2: Establishes 3mid(p²-1) because p is not divisible by 3.
Part f, mark 3: Uses coprimality to conclude 24mid(p²-1).
Part g, mark 1: Shows divisibility by 2.
Part g, mark 2: Shows divisibility by 3.
Part g, mark 3: Shows divisibility by 5, with the 5mid n case covered.
Part g, mark 4: Combines the pairwise-coprime divisors to conclude divisibility by 30.

Detailed marking criteria

Part a (1 mark)

Obtains n(n-1)(n+1).

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part b (1 mark)

Identifies a factor of 2 among three consecutive integers.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part b (1 mark)

Identifies a factor of 3 and concludes divisibility by 6.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part c (1 mark)

Writes n=2k+1 and obtains n²-1=4k(k+1).

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part c (1 mark)

Uses the evenness of k(k+1) to conclude the congruence.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part d (1 mark)

Factorises n⁴-1=(n²-1)(n²+1).

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part d (1 mark)

Combines the factors of 8 and 2 to conclude divisibility by 16.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part e (1 mark)

Notes that 2027 is odd and applies part (d).

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part e (1 mark)

States the remainder 1.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part f (1 mark)

Establishes 8mid(p²-1) because p is odd.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part f (1 mark)

Establishes 3mid(p²-1) because p is not divisible by 3.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part f (1 mark)

Uses coprimality to conclude 24mid(p²-1).

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part g (1 mark)

Shows divisibility by 2.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part g (1 mark)

Shows divisibility by 3.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part g (1 mark)

Shows divisibility by 5, with the 5mid n case covered.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part g (1 mark)

Combines the pairwise-coprime divisors to conclude divisibility by 30.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Section II Question 13

(a) y=1.

An equilibrium has dy / dx=0 for every x, so 1-y=0.

(b) -ln|1-y|=((x²) / (2))+C.

Integrate dy / (1-y)=x,dx.

(c) y=1-e^(-x² / 2).

The initial condition gives C=0 and 1-y=e^(-x² / 2).

(d) x=√2ln2.

Solve e^(-x² / 2)=1 / 2.

(e) The maximum gradient is e^(-1 / 2) at x=1.

Since y'=xe^(-x² / 2), y''=e^(-x² / 2)(1-x²), which changes from positive to negative at x=1.

(f) The point of inflection is (1,1-e^(-1 / 2)), and the asymptote is y=1.

The concavity changes at x=1, and e^(-x² / 2)to0 as xtoinfty.

Mark allocation

Part a, mark 1: States y=1.
Part b, mark 1: Separates as dy / (1-y)=x,dx.
Part b, mark 2: Integrates to -ln|1-y|=x² / 2+C.
Part c, mark 1: Substitutes the initial condition into the implicit solution.
Part c, mark 2: Obtains the correct integration constant.
Part c, mark 3: Rearranges to 1-y=e^(-x² / 2).
Part c, mark 4: States y=1-e^(-x² / 2).
Part d, mark 1: Obtains x²=2ln2.
Part d, mark 2: Selects the positive solution x=√2ln2.
Part e, mark 1: Differentiates to y''=e^(-x² / 2)(1-x²).
Part e, mark 2: Identifies x=1 as the maximum-gradient location.
Part e, mark 3: Evaluates the maximum gradient as e^(-1 / 2).
Part f, mark 1: Uses the sign change of y'' at x=1.
Part f, mark 2: Obtains the inflection point (1,1-e^(-1 / 2)).
Part f, mark 3: States the horizontal asymptote y=1.

Detailed marking criteria

Part a (1 mark)

States y=1.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part b (1 mark)

Separates as dy / (1-y)=x,dx.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part b (1 mark)

Integrates to -ln|1-y|=x² / 2+C.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part c (1 mark)

Substitutes the initial condition into the implicit solution.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part c (1 mark)

Obtains the correct integration constant.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part c (1 mark)

Rearranges to 1-y=e^(-x² / 2).

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part c (1 mark)

States y=1-e^(-x² / 2).

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part d (1 mark)

Obtains x²=2ln2.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part d (1 mark)

Selects the positive solution x=√2ln2.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part e (1 mark)

Differentiates to y''=e^(-x² / 2)(1-x²).

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part e (1 mark)

Identifies x=1 as the maximum-gradient location.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part e (1 mark)

Evaluates the maximum gradient as e^(-1 / 2).

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part f (1 mark)

Uses the sign change of y'' at x=1.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part f (1 mark)

Obtains the inflection point (1,1-e^(-1 / 2)).

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part f (1 mark)

States the horizontal asymptote y=1.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Section II Question 14

(a) -5text( m s)⁻¹.

At limiting velocity, dv / dt=0, so -2v-10=0.

(b) u=Ce^(-2t).

Because du / dt=dv / dt=-2(v+5)=-2u, separation gives u=Ce^(-2t).

(c) v(t)=20e^(-2t)-5.

Since u(0)=20, the constant is 20.

(d) t=ln2 seconds; v changes from positive to negative.

Set v=0: 20e^(-2t)=5, so e^(-2t)=1 / 4 and t=ln2.

(e) s(t)=10(1-e^(-2t))-5t; maximum height =((15) / (2))-5ln2 metres.

Integrate v(t), apply s(0)=0, and substitute t=ln2.

(f) v(2)approx-4.634text( m s)⁻¹; the graph decreases, is concave up, and approaches v=-5 from above.

Substitute t=2. Also v'=-40e^(-2t)<0 and v''=80e^(-2t)>0.

Mark allocation

Part a, mark 1: Obtains the limiting velocity -5text( m s)⁻¹.
Part b, mark 1: Transforms the equation to du / dt=-2u.
Part b, mark 2: Solves to obtain u=Ce^(-2t).
Part c, mark 1: Uses u(0)=20 to determine the constant.
Part c, mark 2: States v(t)=20e^(-2t)-5.
Part d, mark 1: Sets v(t)=0.
Part d, mark 2: Solves to obtain t=ln2.
Part d, mark 3: Explains that the sign changes from upward to downward velocity.
Part e, mark 1: Integrates 20e^(-2t)-5 correctly.
Part e, mark 2: Applies s(0)=0 to obtain s(t)=10(1-e^(-2t))-5t.
Part e, mark 3: Substitutes t=ln2 and e^(-2ln2)=1 / 4.
Part e, mark 4: Obtains 15 / 2-5ln2 metres.
Part f, mark 1: Evaluates v(2)=20e⁻⁴-5approx-4.634.
Part f, mark 2: Uses v'<0 to state that the graph decreases.
Part f, mark 3: Uses v''>0 and the limit to describe approach to v=-5 from above.

Detailed marking criteria

Part a (1 mark)

Obtains the limiting velocity -5text( m s)⁻¹.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part b (1 mark)

Transforms the equation to du / dt=-2u.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part b (1 mark)

Solves to obtain u=Ce^(-2t).

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part c (1 mark)

Uses u(0)=20 to determine the constant.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part c (1 mark)

States v(t)=20e^(-2t)-5.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part d (1 mark)

Sets v(t)=0.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part d (1 mark)

Solves to obtain t=ln2.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part d (1 mark)

Explains that the sign changes from upward to downward velocity.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part e (1 mark)

Integrates 20e^(-2t)-5 correctly.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part e (1 mark)

Applies s(0)=0 to obtain s(t)=10(1-e^(-2t))-5t.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part e (1 mark)

Substitutes t=ln2 and e^(-2ln2)=1 / 4.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part e (1 mark)

Obtains 15 / 2-5ln2 metres.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part f (1 mark)

Evaluates v(2)=20e⁻⁴-5approx-4.634.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part f (1 mark)

Uses v'<0 to state that the graph decreases.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part f (1 mark)

Uses v''>0 and the limit to describe approach to v=-5 from above.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Section II Question 15

(a) overrightarrow(CA)=(2,1,2).

Subtract the coordinates of C from those of A.

(b) Radius 3; (x-1)²+(y+2)²+(z-3)²=9.

The radius is |overrightarrow(CA)|=3.

(c) 2lambda(3lambda+2)=0.

Relative to C, the point is (2+lambda,1+2lambda,2-lambda).

(d) B=(7 / 3,-7 / 3,17 / 3); AB=2sqrt6 / 3.

The second parameter is lambda=-2 / 3, and the chord displacement is (-2 / 3)(1,2,-1).

(e) angle ACB=cos⁻¹(23 / 27).

Use overrightarrow(CA)=(2,1,2) and overrightarrow(CB)=(4 / 3,-1 / 3,8 / 3), both of length 3.

(f) H=(11 / 3,1 / 3,13 / 3); OH=√291 / 3.

A closest point satisfies mathbf(OH) × (1,2,-1)=0, which gives lambda=2 / 3.

Mark allocation

Part a, mark 1: Obtains (2,1,2).
Part b, mark 1: Calculates the radius as 3.
Part b, mark 2: Writes the sphere equation with radius squared 9.
Part c, mark 1: Forms the displacement from C.
Part c, mark 2: Expands and simplifies the sphere equation.
Part d, mark 1: Solves for the non-zero parameter -2 / 3.
Part d, mark 2: Obtains the coordinates of B.
Part d, mark 3: Calculates AB=2sqrt6 / 3.
Part e, mark 1: Forms both radius vectors.
Part e, mark 2: Calculates their scalar product as 23 / 3.
Part e, mark 3: Obtains cosangle ACB=23 / 27.
Part f, mark 1: Writes a general point of L.
Part f, mark 2: Uses the zero scalar product to obtain lambda=2 / 3.
Part f, mark 3: Finds H=(11 / 3,1 / 3,13 / 3).
Part f, mark 4: Calculates OH=√291 / 3.

Detailed marking criteria

Part a (1 mark)

Obtains (2,1,2).

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part b (1 mark)

Calculates the radius as 3.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part b (1 mark)

Writes the sphere equation with radius squared 9.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part c (1 mark)

Forms the displacement from C.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part c (1 mark)

Expands and simplifies the sphere equation.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part d (1 mark)

Solves for the non-zero parameter -2 / 3.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part d (1 mark)

Obtains the coordinates of B.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part d (1 mark)

Calculates AB=2sqrt6 / 3.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part e (1 mark)

Forms both radius vectors.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part e (1 mark)

Calculates their scalar product as 23 / 3.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part e (1 mark)

Obtains cosangle ACB=23 / 27.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part f (1 mark)

Writes a general point of L.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part f (1 mark)

Uses the zero scalar product to obtain lambda=2 / 3.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part f (1 mark)

Finds H=(11 / 3,1 / 3,13 / 3).

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part f (1 mark)

Calculates OH=√291 / 3.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Section II Question 16

(a) I_0=1.

Evaluate int_0^infty e^(-x),dx.

(b) I_1=1.

Use u=x, dv=e^(-x)dx; the boundary term vanishes.

(c) I_n=nI_(n-1).

Integrate by parts using u=x^n and dv=e^(-x)dx, and justify the zero boundary term.

(d) I_2=2, I_3=6, I_4=24.

Apply the recurrence successively from I_1=1.

(e) int f=1, and the mean is 3.

Non-negativity is immediate; int_0^infty f=I_2 / 2=1, and E(X)=I_3 / 2=3.

(f) operatorname(Var)(X)=3, and the mode occurs at x=2.

Since E(X²)=I_4 / 2=12, the variance is 12-3²=3. Also f'(x)=frac12e^(-x)x(2-x), so the positive maximum is at x=2.

Mark allocation

Part a, mark 1: States I_0=1.
Part b, mark 1: Sets up integration by parts with a valid u and dv.
Part b, mark 2: Evaluates the boundary and remaining integral to obtain I_1=1.
Part c, mark 1: Applies integration by parts to obtain the recurrence plus its boundary term.
Part c, mark 2: Justifies the zero boundary term and states I_n=nI_(n-1).
Part d, mark 1: Obtains I_2=2.
Part d, mark 2: Obtains I_3=6.
Part d, mark 3: Obtains I_4=24.
Part e, mark 1: States or uses f(x)geq0.
Part e, mark 2: Uses I_2 / 2=1 to establish total probability one.
Part e, mark 3: Uses I_3 / 2 to obtain mean 3.
Part f, mark 1: Calculates E(X²)=I_4 / 2=12.
Part f, mark 2: Calculates operatorname(Var)(X)=12-9=3.
Part f, mark 3: Differentiates to a factor proportional to x(2-x)e^(-x).
Part f, mark 4: Uses the sign change to identify the mode at x=2.

Detailed marking criteria

Part a (1 mark)

States I_0=1.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part b (1 mark)

Sets up integration by parts with a valid u and dv.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part b (1 mark)

Evaluates the boundary and remaining integral to obtain I_1=1.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part c (1 mark)

Applies integration by parts to obtain the recurrence plus its boundary term.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part c (1 mark)

Justifies the zero boundary term and states I_n=nI_(n-1).

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part d (1 mark)

Obtains I_2=2.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part d (1 mark)

Obtains I_3=6.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part d (1 mark)

Obtains I_4=24.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part e (1 mark)

States or uses f(x)geq0.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part e (1 mark)

Uses I_2 / 2=1 to establish total probability one.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part e (1 mark)

Uses I_3 / 2 to obtain mean 3.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part f (1 mark)

Calculates E(X²)=I_4 / 2=12.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part f (1 mark)

Calculates operatorname(Var)(X)=12-9=3.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part f (1 mark)

Differentiates to a factor proportional to x(2-x)e^(-x).

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Part f (1 mark)

Uses the sign change to identify the mode at x=2.

Acceptable alternatives: Any mathematically equivalent valid method or exact form.

Diagnostic Checklist

Topic	Questions	Marks	Marks Lost	Action
Calculus	Q4, Q5, Q10	3	___	Review the methods and reasoning used in Q4, Q5, Q10.
Complex Numbers	Q1, Q2, Q8, Q11(a), Q11(b), Q11(c), Q11(d), Q11(e), Q11(f)	17	___	Review the methods and reasoning used in Q1, Q2, Q8, Q11(a), Q11(b), Q11(c), Q11(d), Q11(e), Q11(f).
Differential Calculus and Mechanics	Q13(a), Q13(b), Q13(c), Q13(d), Q13(e), Q13(f), Q14(a), Q14(b), Q14(c), Q14(d), Q14(e), Q14(f)	30	___	Review the methods and reasoning used in Q13(a), Q13(b), Q13(c), Q13(d), Q13(e), Q13(f), Q14(a), Q14(b), Q14(c), Q14(d), Q14(e), Q14(f).
Further Integration	Q16(a), Q16(b), Q16(c), Q16(d), Q16(e), Q16(f)	15	___	Review the methods and reasoning used in Q16(a), Q16(b), Q16(c), Q16(d), Q16(e), Q16(f).
Mechanics	Q6	1	___	Review the methods and reasoning used in Q6.
Proof	Q3, Q9, Q12(a), Q12(b), Q12(c), Q12(d), Q12(e), Q12(f), Q12(g)	18	___	Review the methods and reasoning used in Q3, Q9, Q12(a), Q12(b), Q12(c), Q12(d), Q12(e), Q12(f), Q12(g).
Vectors	Q7, Q15(a), Q15(b), Q15(c), Q15(d), Q15(e), Q15(f)	16	___	Review the methods and reasoning used in Q7, Q15(a), Q15(b), Q15(c), Q15(d), Q15(e), Q15(f).

HSC Mathematics Extension 2 Free Online Pack 0 — HSC Paper Showcase

Exam-pack paper structure

Read HSC Paper Showcase online

Section I

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Section II

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Worked Solutions And Marking Guide

Section I Question 1

Section I Question 2

Section I Question 3

Section I Question 4

Section I Question 5

Section I Question 6

Section I Question 7

Section I Question 8

Section I Question 9

Section I Question 10

Section II Question 11

Mark allocation

Detailed marking criteria

Section II Question 12

Mark allocation

Detailed marking criteria

Section II Question 13

Mark allocation

Detailed marking criteria

Section II Question 14

Mark allocation

Detailed marking criteria

Section II Question 15

Mark allocation

Detailed marking criteria

Section II Question 16

Mark allocation

Detailed marking criteria

Diagnostic Checklist

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