Skill Align HSC Mathematics Extension 2 - Free Online Pack 0
Original Skill Align simulation aligned with the revised 2026 HSC Mathematics Extension 2 question-paper and Section II Answer Booklet format.
- Paper
- HSC Paper Showcase
- Reading
- 10 minutes
- Writing
- 3 hours
- Assessment
- 100 marks
NESA-approved calculators may be used. A six-page Skill Align-authored Mathematics Advanced, Mathematics Extension 1 and Mathematics Extension 2 reference sheet is included at the back of the question paper.
Section I
Attempt Questions 1-10. Allow about 15 minutes for this section. Use the multiple-choice answer sheet for Questions 1-10. Select the best answer for each question. Each question is worth 1 mark.
Question 1
1 mark- |z|=4, arg z=fracpi3
- |z|=2, arg z=fracpi6
- |z|=2, arg z=fracpi3
- |z|=sqrt2, arg z=fracpi6
Question 2
1 mark- 2i
- 2
- -2
- -2i
Question 3
1 mark- 1
- 0
- 2
- -2
Question 4
1 mark- frac12
- 2
- 1
- infty
Question 5
1 mark- y=3e^(2x)
- y=2e^(-3x)
- y=2e^(3x)
- y=6x+2
Question 6
1 mark- 3
- 7
- 11
- 15
Question 7
1 mark- parallel
- equal
- oppositely directed
- perpendicular
Question 8
1 mark- a rotation through π / 2 anticlockwise about the origin
- a rotation through π / 2 clockwise about the origin
- a reflection in the real axis
- a dilation by factor 2
Question 9
1 mark- frac12
- 1
- 2
- infty
Question 10
1 mark- fracpi2
- ln2
- fracpi4
- 1
Section II
Attempt Questions 11-16. Allow about 2 hours and 45 minutes for this section. Answer in the supplied Section II Answer Booklet. Four pages are allocated to each question. Show relevant mathematical reasoning and/or calculations. Additional writing booklets are available; write the question number on each one.
Question 11
14 marksQuestion 12
16 marksQuestion 13
15 marksQuestion 14
15 marksQuestion 15
15 marksQuestion 16
15 marksSample Responses And Practice Marking Guidance
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,π / 2,π,3π / 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 π / 2 anticlockwise about the origin
Multiplication by i adds π / 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) -2
Differentiate velocity with respect to time.
(b) Writing n=2k gives n²+n+1=2(2k²+k)+1, which is odd.
Represent the even integer as 2k and simplify to the form 2q+1.
(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 -2.
- Part b, mark 1: Writes n=2k and substitutes into n²+n+1.
- Part b, mark 2: Expresses the result as 2(2k²+k)+1 and concludes that it is odd.
- 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 -2.
Acceptable alternatives: Any mathematically equivalent valid method or exact form.
Part b (1 mark)
Writes n=2k and substitutes into n²+n+1.
Acceptable alternatives: Any mathematically equivalent valid method or exact form.
Part b (1 mark)
Expresses the result as 2(2k²+k)+1 and concludes that it is odd.
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 (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 (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) 3
Calculate the square root of the sum of the squared components.
(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: Obtains 3.
- 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)
Obtains 3.
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) -5 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-4.634 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 -5 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 -5 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 |
|---|---|---|---|---|
| Applications of Calculus and Mechanics | Q11(a) | 1 | ___ | Review the methods and reasoning used in Q11(a). |
| Calculus | Q4, Q5, Q10 | 3 | ___ | Review the methods and reasoning used in Q4, Q5, Q10. |
| Complex Numbers | Q1, Q2, Q8, Q11(c), Q11(d), Q11(e), Q11(f) | 14 | ___ | Review the methods and reasoning used in Q1, Q2, Q8, Q11(c), Q11(d), Q11(e), Q11(f). |
| Differential Calculus and Mechanics | Q13(b), Q13(c), Q13(d), Q13(e), Q13(f), Q14(a), Q14(b), Q14(c), Q14(d), Q14(e), Q14(f) | 29 | ___ | Review the methods and reasoning used in 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). |
| Further Work with Vectors | Q13(a) | 1 | ___ | Review the methods and reasoning used in Q13(a). |
| Mechanics | Q6 | 1 | ___ | Review the methods and reasoning used in Q6. |
| Nature of Proof | Q11(b) | 2 | ___ | Review the methods and reasoning used in Q11(b). |
| 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). |