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Condensed Practice Examination Showcase

TASC TASC Mathematics Specialised Level 4 Free Online Pack 0 — Condensed Practice Examination Showcase

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TASC Criteria 4-8 Condensed Practice Examination 2026 Edition - Pack 0 v2.0
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Condensed Practice Examination Showcase

10 questions

100 marks

Estimated duration: 100 minutes

Reading: 15 minutes preparation time · Writing: 100 minutes

Read Condensed Practice Examination Showcase online

Skill Align

Skill Align TASC Mathematics Specialised Level 4 (MTS415118) Condensed Practice Examination - Pack 0 - 2026 Edition

Mathematics Specialised Level 4 | MTS415118 | Original Skill Align 100-mark condensed practice examination covering Criteria 4-8. This booklet does not reproduce the complete five-section, 180-mark TASC external examination.

Paper
Condensed Practice Examination Showcase
Reading
15 minutes preparation time
Writing
100 minutes
Assessment
100 marks

The current TASC Mathematics Specialised Information Sheet may be used. TASC-approved calculators and their functions may be used throughout unless a question states otherwise. Internet access, AI tools, messaging and external communication are not permitted.

Section A - Criterion 4: Sequences and series (20 marks)

Answer both compulsory questions. Show complete proof, exact reasoning and all required working.

Question 1

10 marks
For ninmathbb N, let S_n=sum_(r=1)^(n)((1) / (r(r+2))).
(a) 4 marks
Express ((1) / (r(r+2))) as a difference of two simple fractions.
(b) 6 marks
Derive a closed form for S_n, then find lim_(ntoinfty)S_n.

Question 2

10 marks
Use a Maclaurin polynomial to approximate e^(-0.4).
(a) 4 marks
Derive the Maclaurin polynomial for e^(-2x) through the term in x^3.
(b) 6 marks
Use x=0.2 to approximate e^(-0.4). Compare with a calculator value and comment on the error.

Section B - Criterion 5: Matrices and linear algebra (20 marks)

Answer both compulsory questions. Show row operations, transformation order and three-dimensional reasoning where required.

Question 3

10 marks
Solve the system x+y+z=2, 2x-y+3z=-3, and -x+2y+z=2 using row reduction.
(a) 4 marks
Write the augmented matrix and eliminate x from rows 2 and 3.
(b) 6 marks
Continue to reduced row-echelon form and state the solution.

Question 4

10 marks
A shear S=begin(bmatrix)1&20&1end(bmatrix) is followed by a counter-clockwise rotation through ((π) / (2)), with matrix R=begin(bmatrix)0&-11&0end(bmatrix).
Diagram PreviewOPxy
(a) 5 marks
Find the composite matrix and the image of P(1,2).
(b) 5 marks
Find the image of the line y=x and state the area scale factor.

Section C - Criterion 6: Differential calculus, areas and volumes (20 marks)

Answer both compulsory questions. Give exact values unless an approximation is requested and distinguish signed from geometric area.

Question 5

10 marks
Let f(x)=x+frac4x, for xne0.
(a) 5 marks
Find every stationary point and classify each one.
(b) 5 marks
Find the vertical and oblique asymptotes, and determine the exact range of f.

Question 6

10 marks
For 0le yle2, the region between x=y² and x=4 is rotated about the y-axis.
(a) 4 marks
Set up an exact integral for the volume using washers.
(b) 6 marks
Evaluate the integral and state the exact volume.

Section D - Criterion 7: Integration techniques and differential equations (20 marks)

Answer both compulsory questions. Show substitutions, transformed limits, constants of integration and domain restrictions.

Question 7

10 marks
Evaluate I=int_0¹((5x+7) / ((x+1)(x+2))),dx.
(a) 4 marks
Resolve the integrand into partial fractions.
(b) 6 marks
Evaluate I in exact logarithmic form.

Question 8

10 marks
A curve satisfies ((dy) / (dx))=1+((y) / (x)), x>0, and y(1)=2.
(a) 6 marks
Use v=((y) / (x)) to solve the differential equation.
(b) 4 marks
Find y(e) and the gradient there.

Section E - Criterion 8: Complex numbers (20 marks)

Answer both compulsory questions. State argument intervals and boundary conditions precisely and use exact complex-number notation.

Question 9

10 marks
Solve z⁴=-16.
(a) 4 marks
Find all four roots in polar form.
(b) 6 marks
Hence factor z⁴+16 into real quadratic factors.

Question 10

10 marks
Let R=(zinmathbb C:|z-(1+i)|le2, 0leoperatorname(Arg)zle((π) / (3))).
Diagram Preview CPQRRe(z) Im(z)
(a) 4 marks
Describe every boundary of R and state whether it is included.
(b) 6 marks
Sketch R, and determine whether 2+i, 1+3i, and 3+i belong to it.

TASC courses, assessment and certification are administered by the Tasmanian Assessment, Standards and Certification office. Skill Align is an independent publisher and is not affiliated with, authorised by, sponsored by, approved by or endorsed by TASC or the Tasmanian Government. This is original Skill Align condensed practice material, not an official TASC assessment and not a complete simulation of the 180-mark external examination.

Copyright (c) 2026 Skill Align. Free for personal, non-commercial online viewing at https://skillalign.au. You may share the Skill Align page link. Except as permitted by law or with Skill Align's prior written permission, the pack itself must not be resold, copied, redistributed, republished, automatically extracted, or uploaded to a question bank.

Worked Solutions And Marking Guide

Section A Question 1

(a) ((1) / (r(r+2)))=frac12(frac1r-frac1(r+2)).

Let ((1) / (r(r+2)))=((A) / (r))+((B) / (r+2)). Solving A+B=0 and 2A=1 gives A=frac12, B=-frac12.

(b) S_n=frac34-((1) / (2(n+1)))-((1) / (2(n+2))), so lim_(ntoinfty)S_n=frac34.

Writing the terms in difference form cancels every interior reciprocal except 1, frac12, frac1(n+1) and frac1(n+2).

Detailed marking criteria

Part a (4 marks)

Award one mark for each numbered checkpoint below, to a maximum of 4. A correct answer without relevant working is capped at half marks.

Mark-by-mark checkpoints:

  1. Introduces constants A and B in a partial-fraction decomposition.
  2. Forms 1=A(r+2)+Br.
  3. Solves A=frac12 and B=-frac12.
  4. States the correct difference frac12(frac1r-frac1(r+2)).

Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 4 checkpoints and reaches ((1) / (r(r+2)))=frac12(frac1r-frac1(r+2)). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.

Do not credit by itself: Omitting or contradicting: Introduces constants A and B in a partial-fraction decomposition. Giving the final result without establishing: States the correct difference frac12(frac1r-frac1(r+2)).

Part b (6 marks)

Award one mark for each numbered checkpoint below, to a maximum of 6. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.

Mark-by-mark checkpoints:

  1. Substitutes the difference from part (a) into the summation.
  2. Writes enough expanded terms to show the two-step cancellation pattern.
  3. Retains the uncancelled initial terms 1+frac12.
  4. Retains the uncancelled final terms -frac1(n+1)-frac1(n+2).
  5. Obtains S_n=frac34-((1) / (2(n+1)))-((1) / (2(n+2))).
  6. Uses both remainder terms tending to zero to conclude the limit is frac34.

Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 6 checkpoints and reaches S_n=frac34-((1) / (2(n+1)))-((1) / (2(n+2))), so lim_(ntoinfty)S_n=frac34. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.

Do not credit by itself: Omitting or contradicting: Substitutes the difference from part (a) into the summation. Giving the final result without establishing: Uses both remainder terms tending to zero to conclude the limit is frac34.

Section A Question 2

(a) e^(-2x)approx1-2x+2x²-frac43x^3.

Substitute -2x into e^u=1+u+((u²) / (2!))+((u³) / (3!))+ × s.

(b) The polynomial gives 0.669333ldots. Since e^(-0.4)=0.670320ldots, the absolute error is about 0.000987, an underestimate of less than 0.001.

Substitution gives 1-0.4+0.08-0.010666ldots=0.669333ldots.

Detailed marking criteria

Part a (4 marks)

Award one mark for each numbered checkpoint below, to a maximum of 4. A correct answer without relevant working is capped at half marks.

Mark-by-mark checkpoints:

  1. Starts from the Maclaurin expansion of e^u.
  2. Substitutes u=-2x.
  3. Simplifies the quadratic coefficient to 2.
  4. Simplifies the cubic coefficient to -frac43.

Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 4 checkpoints and reaches e^(-2x)approx1-2x+2x²-frac43x^3. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.

Do not credit by itself: Omitting or contradicting: Starts from the Maclaurin expansion of e^u. Giving the final result without establishing: Simplifies the cubic coefficient to -frac43.

Part b (6 marks)

Award one mark for each numbered checkpoint below, to a maximum of 6. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.

Mark-by-mark checkpoints:

  1. Substitutes x=0.2 into every term of the cubic polynomial.
  2. Evaluates the linear contribution as -0.4.
  3. Evaluates the quadratic contribution as 0.08.
  4. Evaluates the cubic contribution as -0.010666ldots.
  5. Obtains 0.669333ldots and compares it with 0.670320ldots.
  6. Calculates the absolute error and correctly identifies the approximation as an underestimate with error below 0.001.

Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 6 checkpoints and reaches The polynomial gives 0.669333ldots. Since e^(-0.4)=0.670320ldots, the absolute error is about 0.000987, an underestimate of less than 0.001. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.

Do not credit by itself: Omitting or contradicting: Substitutes x=0.2 into every term of the cubic polynomial. Giving the final result without establishing: Calculates the absolute error and correctly identifies the approximation as an underestimate with error below 0.001.

Section B Question 3

(a) [begin(array)(ccc|c)1&1&1&20&-3&1&-70&3&2&4end(array)].

Use R_2arrow R_2-2R_1 and R_3arrow R_3+R_1.

(b) The reduced matrix is [begin(array)(ccc|c)1&0&0&10&1&0&20&0&1&-1end(array)], so (x,y,z)=(1,2,-1).

Adding rows 2 and 3 gives 3z=-3, so z=-1. Back-substitution gives y=2 and x=1.

Detailed marking criteria

Part a (4 marks)

Award one mark for each numbered checkpoint below, to a maximum of 4. A correct answer without relevant working is capped at half marks.

Mark-by-mark checkpoints:

  1. Writes the correct 3 × 4 augmented matrix.
  2. States R_2arrow R_2-2R_1.
  3. Obtains row 2 as [0,-3,1mid-7].
  4. Obtains row 3 as [0,3,2mid4] using R_3arrow R_3+R_1.

Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 4 checkpoints and reaches [begin(array)(ccc|c)1&1&1&20&-3&1&-70&3&2&4end(array)]. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.

Do not credit by itself: Omitting or contradicting: Writes the correct 3 × 4 augmented matrix. Giving the final result without establishing: Obtains row 3 as [0,3,2mid4] using R_3arrow R_3+R_1.

Part b (6 marks)

Award one mark for each numbered checkpoint below, to a maximum of 6. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.

Mark-by-mark checkpoints:

  1. Eliminates y from row 3 to obtain an equation in z.
  2. Obtains z=-1.
  3. Uses row 2 to obtain y=2.
  4. Uses row 1 to obtain x=1.
  5. Presents the reduced row-echelon matrix or equivalent complete row reduction.
  6. States the unique ordered triple (1,2,-1).

Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 6 checkpoints and reaches The reduced matrix is [begin(array)(ccc|c)1&0&0&10&1&0&20&0&1&-1end(array)], so (x,y,z)=(1,2,-1). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.

Do not credit by itself: Omitting or contradicting: Eliminates y from row 3 to obtain an equation in z. Giving the final result without establishing: States the unique ordered triple (1,2,-1).

Section B Question 4

(a) M=RS=begin(bmatrix)0&-11&2end(bmatrix) and P'=(-2,5).

Because the shear occurs first, multiply R by S, then apply the product to the column vector for P.

(b) The image line is y=-3x, and the area scale factor is 1.

A point (t,t) maps to (-t,3t), so y=-3x. Also |det M|=1.

OP'Lxy
Completed solution diagram

Detailed marking criteria

Part a (5 marks)

Award one mark for each numbered checkpoint below, to a maximum of 5. A correct answer without relevant working is capped at half marks.

Mark-by-mark checkpoints:

  1. Identifies the composite as RS, not SR.
  2. Forms the first row of RS correctly.
  3. Forms the second row of RS correctly.
  4. Multiplies Mbegin(bmatrix)12end(bmatrix).
  5. Obtains the image P'=(-2,5).

Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 5 checkpoints and reaches M=RS=begin(bmatrix)0&-11&2end(bmatrix) and P'=(-2,5). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.

Do not credit by itself: Omitting or contradicting: Identifies the composite as RS, not SR. Giving the final result without establishing: Obtains the image P'=(-2,5).

Part b (5 marks)

Award one mark for each numbered checkpoint below, to a maximum of 5. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.

Mark-by-mark checkpoints:

  1. Parameterises the original line as (t,t).
  2. Maps (t,t) to (-t,3t).
  3. Eliminates t to obtain y=-3x.
  4. Calculates det M=1.
  5. Interprets |det M|=1 as the area scale factor.

Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 5 checkpoints and reaches The image line is y=-3x, and the area scale factor is 1. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.

Do not credit by itself: Omitting or contradicting: Parameterises the original line as (t,t). Giving the final result without establishing: Interprets |det M|=1 as the area scale factor.

Section C Question 5

(a) The stationary points are (-2,-4), a local maximum, and (2,4), a local minimum.

Use f'(x)=1-frac4(x²) to locate the stationary points and f''(x)=frac8(x³) to classify them.

(b) The asymptotes are x=0 and y=x. The range is (-infty,-4]cup[4,infty).

For a proposed value y, the equation y=x+frac4x is equivalent to x²-yx+4=0, which has a real solution exactly when y²-16ge0.

Detailed marking criteria

Part a (5 marks)

Award one mark for each numbered checkpoint below, to a maximum of 5. A correct answer without relevant working is capped at half marks.

Mark-by-mark checkpoints:

  1. Differentiates to obtain f'(x)=1-frac4(x²).
  2. Solves f'(x)=0 to obtain x=-2 and x=2.
  3. Calculates the stationary-point coordinates (-2,-4) and (2,4).
  4. Obtains f''(x)=frac8(x³).
  5. Uses f''(-2)<0 and f''(2)>0 to classify the local maximum and local minimum.

Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 5 checkpoints and reaches The stationary points are (-2,-4), a local maximum, and (2,4), a local minimum. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.

Do not credit by itself: Omitting or contradicting: Differentiates to obtain f'(x)=1-frac4(x²). Giving the final result without establishing: Uses f''(-2)<0 and f''(2)>0 to classify the local maximum and local minimum.

Part b (5 marks)

Award one mark for each numbered checkpoint below, to a maximum of 5. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.

Mark-by-mark checkpoints:

  1. Identifies x=0 as the vertical asymptote.
  2. Uses f(x)-x=frac4xto0 as |x|toinfty to identify the oblique asymptote y=x.
  3. Rearranges y=x+frac4x to the quadratic x²-yx+4=0.
  4. Requires discriminant y²-16ge0 for real x.
  5. States the range (-infty,-4]cup[4,infty), including the endpoint values attained at the stationary points.

Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 5 checkpoints and reaches The asymptotes are x=0 and y=x. The range is (-infty,-4]cup[4,infty). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.

Do not credit by itself: Omitting or contradicting: Identifies x=0 as the vertical asymptote. Giving the final result without establishing: States the range (-infty,-4]cup[4,infty), including the endpoint values attained at the stationary points.

Section C Question 6

(a) V=piint_0²(16-y⁴),dy.

The outer radius is 4 and the inner radius is y^2.

(b) V=((128π) / (5)) cubic units.

Integrate 16-y⁴ to obtain 16y-((y⁵) / (5)), then evaluate at 0 and 2.

x = y²x = 4y = 0y = 2Rxy
Completed solution diagram

Detailed marking criteria

Part a (4 marks)

Award one mark for each numbered checkpoint below, to a maximum of 4. A correct answer without relevant working is capped at half marks.

Mark-by-mark checkpoints:

  1. Identifies the outer radius as 4.
  2. Identifies the inner radius as y^2.
  3. Uses the washer area π(R²-r²).
  4. Uses the correct limits 0le yle2.

Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 4 checkpoints and reaches V=piint_0²(16-y⁴),dy. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.

Do not credit by itself: Omitting or contradicting: Identifies the outer radius as 4. Giving the final result without establishing: Uses the correct limits 0le yle2.

Part b (6 marks)

Award one mark for each numbered checkpoint below, to a maximum of 6. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.

Mark-by-mark checkpoints:

  1. Integrates 16 to 16y.
  2. Integrates y⁴ to ((y⁵) / (5)).
  3. Applies the upper limit y=2.
  4. Applies the lower limit y=0.
  5. Simplifies 32-((32) / (5)) to ((128) / (5)).
  6. States ((128π) / (5)) with cubic units.

Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 6 checkpoints and reaches V=((128π) / (5)) cubic units. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.

Do not credit by itself: Omitting or contradicting: Integrates 16 to 16y. Giving the final result without establishing: States ((128π) / (5)) with cubic units.

Section D Question 7

(a) ((5x+7) / ((x+1)(x+2)))=frac2(x+1)+frac3(x+2).

Equating coefficients gives A+B=5 and 2A+B=7.

(b) I=3ln3-ln2=ln(((27) / (2))).

An antiderivative is 2ln(x+1)+3ln(x+2).

Detailed marking criteria

Part a (4 marks)

Award one mark for each numbered checkpoint below, to a maximum of 4. A correct answer without relevant working is capped at half marks.

Mark-by-mark checkpoints:

  1. Writes ((A) / (x+1))+((B) / (x+2)).
  2. Forms 5x+7=A(x+2)+B(x+1).
  3. Solves A=2.
  4. Solves B=3 and states the decomposition.

Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 4 checkpoints and reaches ((5x+7) / ((x+1)(x+2)))=frac2(x+1)+frac3(x+2). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.

Do not credit by itself: Omitting or contradicting: Writes ((A) / (x+1))+((B) / (x+2)). Giving the final result without establishing: Solves B=3 and states the decomposition.

Part b (6 marks)

Award one mark for each numbered checkpoint below, to a maximum of 6. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.

Mark-by-mark checkpoints:

  1. Integrates frac2(x+1) as 2ln|x+1|.
  2. Integrates frac3(x+2) as 3ln|x+2|.
  3. Evaluates the antiderivative at x=1.
  4. Evaluates the antiderivative at x=0.
  5. Simplifies to 3ln3-ln2.
  6. Combines logarithms to obtain ln(((27) / (2))).

Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 6 checkpoints and reaches I=3ln3-ln2=ln(((27) / (2))). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.

Do not credit by itself: Omitting or contradicting: Integrates frac2(x+1) as 2ln|x+1|. Giving the final result without establishing: Combines logarithms to obtain ln(((27) / (2))).

Section D Question 8

(a) y=x(ln x+2), for x>0.

With y=vx, y'=v+xv'. Hence xv'=1, so v=ln x+C, and the initial condition gives C=2.

(b) y(e)=3e and .((dy) / (dx))|_(x=e)=4.

Substitute x=e into y=x(ln x+2), then use y'=1+((y) / (x)).

Detailed marking criteria

Part a (6 marks)

Award one mark for each numbered checkpoint below, to a maximum of 6. A correct answer without relevant working is capped at half marks.

Mark-by-mark checkpoints:

  1. Sets y=vx.
  2. Differentiates to obtain y'=v+x((dv) / (dx)).
  3. Substitutes into the differential equation and simplifies to x((dv) / (dx))=1.
  4. Integrates to obtain v=ln x+C.
  5. Returns to y=x(ln x+C).
  6. Uses y(1)=2 to obtain C=2 and states x>0.

Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 6 checkpoints and reaches y=x(ln x+2), for x>0. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.

Do not credit by itself: Omitting or contradicting: Sets y=vx. Giving the final result without establishing: Uses y(1)=2 to obtain C=2 and states x>0.

Part b (4 marks)

Award one mark for each numbered checkpoint below, to a maximum of 4. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.

Mark-by-mark checkpoints:

  1. Uses ln e=1.
  2. Obtains y(e)=e(1+2)=3e.
  3. Substitutes x=e and y=3e into 1+((y) / (x)).
  4. Obtains the gradient 4.

Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 4 checkpoints and reaches y(e)=3e and .((dy) / (dx))|_(x=e)=4. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.

Do not credit by itself: Omitting or contradicting: Uses ln e=1. Giving the final result without establishing: Obtains the gradient 4.

Section E Question 9

(a) z=2operatorname(cis)((π) / (4)),,2operatorname(cis)((3π) / (4)),,2operatorname(cis)((5π) / (4)),,2operatorname(cis)((7π) / (4)).

Write -16=16operatorname(cis)(π+2kpi), take fourth roots and use k=0,1,2,3.

(b) z⁴+16=(z²-2sqrt2,z+4)(z²+2sqrt2,z+4).

Pair each root with its complex conjugate. Each pair has product 4 and real-part sum pm2sqrt2.

z1z2z3z4Re(z) Im(z)
Completed solution diagram

Detailed marking criteria

Part a (4 marks)

Award one mark for each numbered checkpoint below, to a maximum of 4. A correct answer without relevant working is capped at half marks.

Mark-by-mark checkpoints:

  1. Writes -16 with modulus 16 and argument π+2kpi.
  2. Takes the fourth root of the modulus to obtain 2.
  3. Uses angles ((π+2kpi) / (4)).
  4. Lists the four distinct roots for k=0,1,2,3.

Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 4 checkpoints and reaches z=2operatorname(cis)((π) / (4)),,2operatorname(cis)((3π) / (4)),,2operatorname(cis)((5π) / (4)),,2operatorname(cis)((7π) / (4)). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.

Do not credit by itself: Omitting or contradicting: Writes -16 with modulus 16 and argument π+2kpi. Giving the final result without establishing: Lists the four distinct roots for k=0,1,2,3.

Part b (6 marks)

Award one mark for each numbered checkpoint below, to a maximum of 6. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.

Mark-by-mark checkpoints:

  1. Pairs the roots at angles ((π) / (4)) and ((7π) / (4)).
  2. Forms the factor z²-2sqrt2,z+4.
  3. Pairs the roots at angles ((3π) / (4)) and ((5π) / (4)).
  4. Forms the factor z²+2sqrt2,z+4.
  5. Multiplies or otherwise verifies cancellation of the z³ and z terms.
  6. States the complete real quadratic factorisation.

Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 6 checkpoints and reaches z⁴+16=(z²-2sqrt2,z+4)(z²+2sqrt2,z+4). Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.

Do not credit by itself: Omitting or contradicting: Pairs the roots at angles ((π) / (4)) and ((7π) / (4)). Giving the final result without establishing: States the complete real quadratic factorisation.

Section E Question 10

(a) The boundaries are the included circle |z-(1+i)|=2 and the included rays operatorname(Arg)z=0 and operatorname(Arg)z=((π) / (3)), restricted to their common intersection; z=0 is excluded because its argument is undefined.

The non-strict inequalities include their boundaries, but the principal argument is not defined at the origin.

(b) The region is the part of the closed disc lying in the closed sector 0lethetale((π) / (3)), excluding the origin. The points 2+i and 3+i belong to R; 1+3i does not.

Check both the distance from 1+i and the principal argument for each point.

Detailed marking criteria

Part a (4 marks)

Award one mark for each numbered checkpoint below, to a maximum of 4. A correct answer without relevant working is capped at half marks.

Mark-by-mark checkpoints:

  1. Identifies the circle centre 1+i.
  2. Identifies the circle radius 2.
  3. Identifies the two boundary rays at angles 0 and ((π) / (3)).
  4. States that the circle and rays are included but the origin is excluded because operatorname(Arg)0 is undefined.

Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 4 checkpoints and reaches The boundaries are the included circle |z-(1+i)|=2 and the included rays operatorname(Arg)z=0 and operatorname(Arg)z=((π) / (3)), restricted to their common intersection; z=0 is excluded because its argument is undefined. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.

Do not credit by itself: Omitting or contradicting: Identifies the circle centre 1+i. Giving the final result without establishing: States that the circle and rays are included but the origin is excluded because operatorname(Arg)0 is undefined.

Part b (6 marks)

Award one mark for each numbered checkpoint below, to a maximum of 6. A correct answer without relevant working is capped at half marks. Apply consequential marking when a correct later method consistently uses an earlier incorrect result, unless the error materially simplifies the task.

Mark-by-mark checkpoints:

  1. Sketches the disc with centre 1+i and radius 2.
  2. Restricts the sketch to the sector between angles 0 and ((π) / (3)).
  3. Checks 2+i against both conditions and includes it.
  4. Checks 1+3i and excludes it because its argument exceeds ((π) / (3)).
  5. Checks 3+i as lying on the circle and within the sector.
  6. Uses solid boundaries and excludes the origin explicitly.

Acceptable alternatives: Credit an equivalent mathematically correct method only when it establishes all 6 checkpoints and reaches The region is the part of the closed disc lying in the closed sector 0lethetale((π) / (3)), excluding the origin. The points 2+i and 3+i belong to R; 1+3i does not. Accept an algebraically equivalent exact form. Where a decimal is requested, accept the stated rounding or a value within + / -0.01 unless another tolerance is specified.

Do not credit by itself: Omitting or contradicting: Sketches the disc with centre 1+i and radius 2. Giving the final result without establishing: Uses solid boundaries and excludes the origin explicitly.

Diagnostic Checklist

TopicQuestionsMarksMarks LostAction
Section A | Criterion 4 | Sequences and series | Decompose a rational term for telescoping | Expected result: \(\frac{1}{r(r+2)}=\frac12\left(\frac1r-\frac1{r+2}\right)\). | Required evidence: Introduces constants \(A\) and \(B\) in a partial-fraction decomposition.; Forms \(1=A(r+2)+Br\).; Solves \(A=\frac12\) and \(B=-\frac12\).; States the correct difference \(\frac12(\frac1r-\frac1{r+2})\). Q1(a) 4 ___ Q1(a): reproduce this process without the solution: Introduces constants A and B in a partial-fraction decomposition; then Forms 1=A(r+2)+Br; then Solves A=frac12 and B=-frac12; then States the correct difference frac12(frac1r-frac1(r+2)).
Section A | Criterion 4 | Sequences and series | Telescope a finite series and justify its limit | Expected result: \(S_n=\frac34-\frac{1}{2(n+1)}-\frac{1}{2(n+2)}\), so \(\lim_{n\to\infty}S_n=\frac34\). | Required evidence: Substitutes the difference from part (a) into the summation.; Writes enough expanded terms to show the two-step cancellation pattern.; Retains the uncancelled initial terms \(1+\frac12\).; Retains the uncancelled final terms \(-\frac1{n+1}-\frac1{n+2}\).; Obtains \(S_n=\frac34-\frac{1}{2(n+1)}-\frac{1}{2(n+2)}\).; Uses both remainder terms tending to zero to conclude the limit is \(\frac34\). Q1(b) 6 ___ Q1(b): reproduce this process without the solution: Substitutes the difference from part (a) into the summation; then Writes enough expanded terms to show the two-step cancellation pattern; then Retains the uncancelled initial terms 1+frac12; then Retains the uncancelled final terms -frac1(n+1)-frac1(n+2); then Obtains S_n=frac34-((1) / (2(n+1)))-((1) / (2(n+2))); then Uses both remainder terms tending to zero to conclude the limit is frac34.
Section A | Criterion 4 | Sequences and series | Derive a cubic Maclaurin polynomial | Expected result: \(e^{-2x}\approx1-2x+2x^2-\frac43x^3\). | Required evidence: Starts from the Maclaurin expansion of \(e^u\).; Substitutes \(u=-2x\).; Simplifies the quadratic coefficient to \(2\).; Simplifies the cubic coefficient to \(-\frac43\). Q2(a) 4 ___ Q2(a): reproduce this process without the solution: Starts from the Maclaurin expansion of e^u; then Substitutes u=-2x; then Simplifies the quadratic coefficient to 2; then Simplifies the cubic coefficient to -frac43.
Section A | Criterion 4 | Sequences and series | Use and assess a Maclaurin approximation | Expected result: The polynomial gives \(0.669333\ldots\). Since \(e^{-0.4}=0.670320\ldots\), the absolute error is about \(0.000987\), an underestimate of less than \(0.001\). | Required evidence: Substitutes \(x=0.2\) into every term of the cubic polynomial.; Evaluates the linear contribution as \(-0.4\).; Evaluates the quadratic contribution as \(0.08\).; Evaluates the cubic contribution as \(-0.010666\ldots\).; Obtains \(0.669333\ldots\) and compares it with \(0.670320\ldots\).; Calculates the absolute error and correctly identifies the approximation as an underestimate with error below \(0.001\). Q2(b) 6 ___ Q2(b): reproduce this process without the solution: Substitutes x=0.2 into every term of the cubic polynomial; then Evaluates the linear contribution as -0.4; then Evaluates the quadratic contribution as 0.08; then Evaluates the cubic contribution as -0.010666ldots; then Obtains 0.669333ldots and compares it with 0.670320ldots; then Calculates the absolute error and correctly identifies the approximation as an underestimate with error below 0.001.
Section B | Criterion 5 | Matrices and linear algebra | Set up and begin Gauss-Jordan reduction | Expected result: \(\left[\begin{array}{ccc|c}1&1&1&2\\0&-3&1&-7\\0&3&2&4\end{array}\right]\). | Required evidence: Writes the correct \(3\times4\) augmented matrix.; States \(R_2\leftarrow R_2-2R_1\).; Obtains row 2 as \([0,-3,1\mid-7]\).; Obtains row 3 as \([0,3,2\mid4]\) using \(R_3\leftarrow R_3+R_1\). Q3(a) 4 ___ Q3(a): reproduce this process without the solution: Writes the correct 3 × 4 augmented matrix; then States R_2arrow R_2-2R_1; then Obtains row 2 as [0,-3,1mid-7]; then Obtains row 3 as [0,3,2mid4] using R_3arrow R_3+R_1.
Section B | Criterion 5 | Matrices and linear algebra | Complete Gauss-Jordan reduction and interpret a unique solution | Expected result: The reduced matrix is \(\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&2\\0&0&1&-1\end{array}\right]\), so \((x,y,z)=(1,2,-1)\). | Required evidence: Eliminates \(y\) from row 3 to obtain an equation in \(z\).; Obtains \(z=-1\).; Uses row 2 to obtain \(y=2\).; Uses row 1 to obtain \(x=1\).; Presents the reduced row-echelon matrix or equivalent complete row reduction.; States the unique ordered triple \((1,2,-1)\). Q3(b) 6 ___ Q3(b): reproduce this process without the solution: Eliminates y from row 3 to obtain an equation in z; then Obtains z=-1; then Uses row 2 to obtain y=2; then Uses row 1 to obtain x=1; then Presents the reduced row-echelon matrix or equivalent complete row reduction; then States the unique ordered triple (1,2,-1).
Section B | Criterion 5 | Matrices and linear algebra | Compose transformations in the correct order | Expected result: \(M=RS=\begin{bmatrix}0&-1\\1&2\end{bmatrix}\) and \(P'=(-2,5)\). | Required evidence: Identifies the composite as \(RS\), not \(SR\).; Forms the first row of \(RS\) correctly.; Forms the second row of \(RS\) correctly.; Multiplies \(M\begin{bmatrix}1\\2\end{bmatrix}\).; Obtains the image \(P'=(-2,5)\). Q4(a) 5 ___ Q4(a): reproduce this process without the solution: Identifies the composite as RS, not SR; then Forms the first row of RS correctly; then Forms the second row of RS correctly; then Multiplies Mbegin(bmatrix)12end(bmatrix); then Obtains the image P'=(-2,5).
Section B | Criterion 5 | Matrices and linear algebra | Map a curve and interpret a determinant | Expected result: The image line is \(y=-3x\), and the area scale factor is \(1\). | Required evidence: Parameterises the original line as \((t,t)\).; Maps \((t,t)\) to \((-t,3t)\).; Eliminates \(t\) to obtain \(y=-3x\).; Calculates \(\det M=1\).; Interprets \(|\det M|=1\) as the area scale factor. Q4(b) 5 ___ Q4(b): reproduce this process without the solution: Parameterises the original line as (t,t); then Maps (t,t) to (-t,3t); then Eliminates t to obtain y=-3x; then Calculates det M=1; then Interprets |det M|=1 as the area scale factor.
Section C | Criterion 6 | Differential calculus, areas and volumes | Analyse stationary points of a rational function | Expected result: The stationary points are \((-2,-4)\), a local maximum, and \((2,4)\), a local minimum. | Required evidence: Differentiates to obtain \(f'(x)=1-\frac4{x^2}\).; Solves \(f'(x)=0\) to obtain \(x=-2\) and \(x=2\).; Calculates the stationary-point coordinates \((-2,-4)\) and \((2,4)\).; Obtains \(f''(x)=\frac8{x^3}\).; Uses \(f''(-2)<0\) and \(f''(2)>0\) to classify the local maximum and local minimum. Q5(a) 5 ___ Q5(a): reproduce this process without the solution: Differentiates to obtain f'(x)=1-frac4(x²); then Solves f'(x)=0 to obtain x=-2 and x=2; then Calculates the stationary-point coordinates (-2,-4) and (2,4); then Obtains f''(x)=frac8(x³); then Uses f''(-2)<0 and f''(2)>0 to classify the local maximum and local minimum.
Section C | Criterion 6 | Differential calculus, areas and volumes | Determine asymptotes and exact range | Expected result: The asymptotes are \(x=0\) and \(y=x\). The range is \((-\infty,-4]\cup[4,\infty)\). | Required evidence: Identifies \(x=0\) as the vertical asymptote.; Uses \(f(x)-x=\frac4x\to0\) as \(|x|\to\infty\) to identify the oblique asymptote \(y=x\).; Rearranges \(y=x+\frac4x\) to the quadratic \(x^2-yx+4=0\).; Requires discriminant \(y^2-16\ge0\) for real \(x\).; States the range \((-\infty,-4]\cup[4,\infty)\), including the endpoint values attained at the stationary points. Q5(b) 5 ___ Q5(b): reproduce this process without the solution: Identifies x=0 as the vertical asymptote; then Uses f(x)-x=frac4xto0 as |x|toinfty to identify the oblique asymptote y=x; then Rearranges y=x+frac4x to the quadratic x²-yx+4=0; then Requires discriminant y²-16ge0 for real x; then States the range (-infty,-4]cup[4,infty), including the endpoint values attained at the stationary points.
Section C | Criterion 6 | Differential calculus, areas and volumes | Model a y-axis volume with washers | Expected result: \(V=\pi\int_0^2(16-y^4)\,dy\). | Required evidence: Identifies the outer radius as \(4\).; Identifies the inner radius as \(y^2\).; Uses the washer area \(\pi(R^2-r^2)\).; Uses the correct limits \(0\le y\le2\). Q6(a) 4 ___ Q6(a): reproduce this process without the solution: Identifies the outer radius as 4; then Identifies the inner radius as y²; then Uses the washer area π(R²-r²); then Uses the correct limits 0le yle2.
Section C | Criterion 6 | Differential calculus, areas and volumes | Evaluate an exact washer integral | Expected result: \(V=\frac{128\pi}{5}\) cubic units. | Required evidence: Integrates \(16\) to \(16y\).; Integrates \(y^4\) to \(\frac{y^5}{5}\).; Applies the upper limit \(y=2\).; Applies the lower limit \(y=0\).; Simplifies \(32-\frac{32}{5}\) to \(\frac{128}{5}\).; States \(\frac{128\pi}{5}\) with cubic units. Q6(b) 6 ___ Q6(b): reproduce this process without the solution: Integrates 16 to 16y; then Integrates y⁴ to ((y⁵) / (5)); then Applies the upper limit y=2; then Applies the lower limit y=0; then Simplifies 32-((32) / (5)) to ((128) / (5)); then States ((128π) / (5)) with cubic units.
Section D | Criterion 7 | Integration techniques and differential equations | Resolve distinct linear factors | Expected result: \(\frac{5x+7}{(x+1)(x+2)}=\frac2{x+1}+\frac3{x+2}\). | Required evidence: Writes \(\frac{A}{x+1}+\frac{B}{x+2}\).; Forms \(5x+7=A(x+2)+B(x+1)\).; Solves \(A=2\).; Solves \(B=3\) and states the decomposition. Q7(a) 4 ___ Q7(a): reproduce this process without the solution: Writes ((A) / (x+1))+((B) / (x+2)); then Forms 5x+7=A(x+2)+B(x+1); then Solves A=2; then Solves B=3 and states the decomposition.
Section D | Criterion 7 | Integration techniques and differential equations | Integrate partial fractions with exact limits | Expected result: \(I=3\ln3-\ln2=\ln\left(\frac{27}{2}\right)\). | Required evidence: Integrates \(\frac2{x+1}\) as \(2\ln|x+1|\).; Integrates \(\frac3{x+2}\) as \(3\ln|x+2|\).; Evaluates the antiderivative at \(x=1\).; Evaluates the antiderivative at \(x=0\).; Simplifies to \(3\ln3-\ln2\).; Combines logarithms to obtain \(\ln(\frac{27}{2})\). Q7(b) 6 ___ Q7(b): reproduce this process without the solution: Integrates frac2(x+1) as 2ln|x+1|; then Integrates frac3(x+2) as 3ln|x+2|; then Evaluates the antiderivative at x=1; then Evaluates the antiderivative at x=0; then Simplifies to 3ln3-ln2; then Combines logarithms to obtain ln(((27) / (2))).
Section D | Criterion 7 | Integration techniques and differential equations | Solve a homogeneous differential equation by substitution | Expected result: \(y=x(\ln x+2)\), for \(x>0\). | Required evidence: Sets \(y=vx\).; Differentiates to obtain \(y'=v+x\frac{dv}{dx}\).; Substitutes into the differential equation and simplifies to \(x\frac{dv}{dx}=1\).; Integrates to obtain \(v=\ln x+C\).; Returns to \(y=x(\ln x+C)\).; Uses \(y(1)=2\) to obtain \(C=2\) and states \(x>0\). Q8(a) 6 ___ Q8(a): reproduce this process without the solution: Sets y=vx; then Differentiates to obtain y'=v+x((dv) / (dx)); then Substitutes into the differential equation and simplifies to x((dv) / (dx))=1; then Integrates to obtain v=ln x+C; then Returns to y=x(ln x+C); then Uses y(1)=2 to obtain C=2 and states x>0.
Section D | Criterion 7 | Integration techniques and differential equations | Evaluate a solution and its derivative | Expected result: \(y(e)=3e\) and \(\left.\frac{dy}{dx}\right|_{x=e}=4\). | Required evidence: Uses \(\ln e=1\).; Obtains \(y(e)=e(1+2)=3e\).; Substitutes \(x=e\) and \(y=3e\) into \(1+\frac{y}{x}\).; Obtains the gradient \(4\). Q8(b) 4 ___ Q8(b): reproduce this process without the solution: Uses ln e=1; then Obtains y(e)=e(1+2)=3e; then Substitutes x=e and y=3e into 1+((y) / (x)); then Obtains the gradient 4.
Section E | Criterion 8 | Complex numbers | Apply De Moivre's theorem to roots | Expected result: \(z=2\operatorname{cis}\frac{\pi}{4},\,2\operatorname{cis}\frac{3\pi}{4},\,2\operatorname{cis}\frac{5\pi}{4},\,2\operatorname{cis}\frac{7\pi}{4}\). | Required evidence: Writes \(-16\) with modulus \(16\) and argument \(\pi+2k\pi\).; Takes the fourth root of the modulus to obtain \(2\).; Uses angles \(\frac{\pi+2k\pi}{4}\).; Lists the four distinct roots for \(k=0,1,2,3\). Q9(a) 4 ___ Q9(a): reproduce this process without the solution: Writes -16 with modulus 16 and argument π+2kpi; then Takes the fourth root of the modulus to obtain 2; then Uses angles ((π+2kpi) / (4)); then Lists the four distinct roots for k=0,1,2,3.
Section E | Criterion 8 | Complex numbers | Use conjugate root pairs for real factorisation | Expected result: \(z^4+16=(z^2-2\sqrt2\,z+4)(z^2+2\sqrt2\,z+4)\). | Required evidence: Pairs the roots at angles \(\frac{\pi}{4}\) and \(\frac{7\pi}{4}\).; Forms the factor \(z^2-2\sqrt2\,z+4\).; Pairs the roots at angles \(\frac{3\pi}{4}\) and \(\frac{5\pi}{4}\).; Forms the factor \(z^2+2\sqrt2\,z+4\).; Multiplies or otherwise verifies cancellation of the \(z^3\) and \(z\) terms.; States the complete real quadratic factorisation. Q9(b) 6 ___ Q9(b): reproduce this process without the solution: Pairs the roots at angles ((π) / (4)) and ((7π) / (4)); then Forms the factor z²-2sqrt2,z+4; then Pairs the roots at angles ((3π) / (4)) and ((5π) / (4)); then Forms the factor z²+2sqrt2,z+4; then Multiplies or otherwise verifies cancellation of the z³ and z terms; then States the complete real quadratic factorisation.
Section E | Criterion 8 | Complex numbers | Interpret closed modulus and argument boundaries | Expected result: The boundaries are the included circle \(|z-(1+i)|=2\) and the included rays \(\operatorname{Arg}z=0\) and \(\operatorname{Arg}z=\frac{\pi}{3}\), restricted to their common intersection; \(z=0\) is excluded because its argument is undefined. | Required evidence: Identifies the circle centre \(1+i\).; Identifies the circle radius \(2\).; Identifies the two boundary rays at angles \(0\) and \(\frac{\pi}{3}\).; States that the circle and rays are included but the origin is excluded because \(\operatorname{Arg}0\) is undefined. Q10(a) 4 ___ Q10(a): reproduce this process without the solution: Identifies the circle centre 1+i; then Identifies the circle radius 2; then Identifies the two boundary rays at angles 0 and ((π) / (3)); then States that the circle and rays are included but the origin is excluded because operatorname(Arg)0 is undefined.
Section E | Criterion 8 | Complex numbers | Test points against a combined Argand region | Expected result: The region is the part of the closed disc lying in the closed sector \(0\le\theta\le\frac{\pi}{3}\), excluding the origin. The points \(2+i\) and \(3+i\) belong to \(R\); \(1+3i\) does not. | Required evidence: Sketches the disc with centre \(1+i\) and radius \(2\).; Restricts the sketch to the sector between angles \(0\) and \(\frac{\pi}{3}\).; Checks \(2+i\) against both conditions and includes it.; Checks \(1+3i\) and excludes it because its argument exceeds \(\frac{\pi}{3}\).; Checks \(3+i\) as lying on the circle and within the sector.; Uses solid boundaries and excludes the origin explicitly. Q10(b) 6 ___ Q10(b): reproduce this process without the solution: Sketches the disc with centre 1+i and radius 2; then Restricts the sketch to the sector between angles 0 and ((π) / (3)); then Checks 2+i against both conditions and includes it; then Checks 1+3i and excludes it because its argument exceeds ((π) / (3)); then Checks 3+i as lying on the circle and within the sector; then Uses solid boundaries and excludes the origin explicitly.

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