Skill Align

Skill Align VCE Specialist Mathematics Units 3&4 - Free Online Pack 0

Examination 1 showcase | Technology-free

Paper: Examination 1 Showcase
Reading: 15 minutes
Writing: 1 hour
Assessment: 40 marks

Materials supplied: this question and response booklet. Use the standard VCE Specialist Mathematics formula sheet available separately. Calculators, software and notes are not permitted.

Questions

Answer all questions. Unless otherwise specified, an exact answer is required for each question. In questions where more than one mark is available, appropriate working must be shown.

Question 1

4 marks

A monic cubic polynomial p(z) has real coefficients. Two of its roots are 1+2i and -2.

(a) 2 marks

State the third root and hence find the real quadratic factor of p(z).

(b) 2 marks

Find p(z) in expanded form.

Question 2

4 marks

The line ell has vector equation mathbf r=(1,-1,2)+t(2,1,-1). The plane Pi has equation x-2y+2z=1.

(a) 2 marks

Find the point where ell meets Pi.

(b) 2 marks

Find the acute angle alpha between ell and Pi, giving an exact expression.

Question 3

4 marks

For a positive integer n, let S_n=sum_(r=1)^(n)r(r+1). Prove by mathematical induction that S_n=((n(n+1)(n+2)) / (3)).

(a) 1 mark

Verify the result for n=1.

(b) 3 marks

Assume the result is true for n=k, and complete the inductive step.

Question 4

4 marks

The Argand diagram shows A = 2 + i and B = -1 + 3i. Let M be the midpoint of AB.

Diagram Preview

(a) 1 mark

Find the complex number represented by the directed displacement from A to B.

(b) 1 mark

Find the affix of M.

(c) 2 marks

Find a Cartesian equation for the locus of points P(x,y) that are equidistant from A and B.

Question 5

4 marks

A drink cools in a room maintained at 18^circmathrm C. Its temperature T, in degrees Celsius, is modelled by ((dT) / (dt))=-k(T-18), where t is measured in minutes and k>0. Initially T=90, and after 10 minutes T=54.

(a) 2 marks

Find k exactly.

(b) 2 marks

Find the time at which the model predicts T=27.

Question 6

5 marks

The complex number z satisfies z⁴=-16.

(a) 2 marks

Write all four solutions in polar form.

(b) 2 marks

Express the four solutions in Cartesian form.

(c) 1 mark

The four roots form a square on an Argand diagram. Find its side length.

Question 7

5 marks

The points A(1,0,2), B(3,1,0) and C(0,2,1) form a triangle.

(a) 2 marks

Find overrightarrow(AB) × overrightarrow(AC).

(b) 1 mark

Find a Cartesian equation of the plane through A, B and C.

(c) 1 mark

Find the area of triangle ABC.

(d) 1 mark

Find the perpendicular distance from the origin to the plane.

Question 8

5 marks

A function y satisfies ((dy) / (dx))=x(1+y²) and y(0)=0.

(a) 2 marks

Separate the variables and integrate to obtain an implicit solution.

(b) 2 marks

Find y explicitly as a function of x.

(c) 1 mark

State the largest open interval containing 0 on which this solution is finite.

Question 9

5 marks

A particle moves on a straight line with velocity v(t)=t²-6t+5 metres per second for 0leq tleq5. Its initial position is x(0)=0.

(a) 1 mark

Find the times at which the particle is at rest.

(b) 1 mark

State the interval on which the particle moves in the negative direction.

(c) 1 mark

Find the displacement during the first 5 seconds.

(d) 2 marks

Find the total distance travelled during the first 5 seconds.

Worked Solutions And Marking Guide

Question 1

(a) The third root is 1-2i, and the quadratic factor is z²-2z+5.

Non-real roots of a polynomial with real coefficients occur in conjugate pairs. Multiplying (z-(1+2i))(z-(1-2i)) gives z²-2z+5.

(b) p(z)=(z+2)(z²-2z+5)=z³+z+10.

Expand carefully: z(z²-2z+5)+2(z²-2z+5)=z³+z+10.

Mark allocation

Award 1 mark for identifying the conjugate root and 1 mark for the correct quadratic factor.
Award 1 mark for using the factor z + 2 and 1 mark for the fully expanded cubic.

Question 2

(a) (7,2,-1).

Substitution gives 7-2t=1, so t=3. Substituting t=3 into the line gives (7,2,-1).

(b) alpha=sin⁻¹(((2) / (3sqrt6))).

The direction vector is (2,1,-1) and a normal to the plane is (1,-2,2). Thus sinalpha=((|(2,1,-1) × (1,-2,2)|) / (sqrt6 × 3))=((2) / (3sqrt6)).

Mark allocation

Award 1 mark for a correct substitution equation and 1 mark for the intersection point.
Award 1 mark for the correct dot-product relationship and 1 mark for the exact acute angle.

Question 3

(a) S_1=1(2)=2, while ((1 × 2 × 3) / (3))=2.

Both sides equal 2, so the statement is true for n=1.

(b) S_(k+1)=((k(k+1)(k+2)) / (3))+(k+1)(k+2)=(((k+1)(k+2)(k+3)) / (3)). Hence the statement is true for every positive integer n.

Use the induction assumption for S_k, add the new term (k+1)(k+2), factor (k+1)(k+2), and identify the required formula with n=k+1.

Mark allocation

Award 1 mark for a valid base case.
Award 1 mark for a clear induction assumption, 1 mark for adding and factorising the next term, and 1 mark for the concluding induction statement.

Question 4

(a) -3+2i.

Subtract the affix of A from the affix of B: (-1+3i)-(2+i)=-3+2i.

(b) frac12+2i.

Average the real and imaginary components of the two endpoints.

(c) 4y-6x-5=0.

Set (x-2)²+(y-1)²=(x+1)²+(y-3)² and simplify to 4y-6x-5=0.

Mark allocation

Award 1 mark each for the directed displacement and midpoint.
Award 1 mark for an equal-distance equation and 1 mark for the simplified Cartesian equation.

Question 5

(a) k=((ln2) / (10)).

The solution is T-18=72e^(-kt). At t=10, 36=72e^(-10k), so e^(-10k)=frac12 and k=((ln2) / (10)).

(b) 30 minutes.

Set 9=72e^(-kt), so e^(-kt)=frac18=2⁻³. Therefore kt=3ln2, and using k=((ln2) / (10)) gives t=30.

Mark allocation

Award 1 mark for the correct exponential model after applying the initial condition and 1 mark for the exact value of k.
Award 1 mark for the equation at 27 degrees and 1 mark for the exact time.

Question 6

(a) z=2,mathrm(cis)(fracpi4+((kpi) / (2))), where k=0,1,2,3.

Write -16=16,mathrm(cis)(pi+2kpi), take fourth roots of the modulus, and divide each argument by 4.

(b) sqrt2(1+i), sqrt2(-1+i), sqrt2(-1-i), sqrt2(1-i).

Evaluate cosine and sine at pi / 4, 3pi / 4, 5pi / 4 and 7pi / 4, then multiply by 2.

(c) 2sqrt2.

Adjacent roots lie on a circle of radius 2 and subtend a right angle, so the chord length is √(2²+2²)=2sqrt2.

Mark allocation

Award 1 mark for the modulus and 1 mark for a complete set of four equally spaced arguments.
Award 2 marks for four correct Cartesian roots.
Award 1 mark for the exact side length with valid chord or coordinate reasoning.

Question 7

(a) (3,4,5).

overrightarrow(AB)=(2,1,-2) and overrightarrow(AC)=(-1,2,-1). Their cross product is (3,4,5).

(b) 3x+4y+5z=13.

Use normal vector (3,4,5) through A(1,0,2): 3(x-1)+4y+5(z-2)=0.

(c) ((5sqrt2) / (2)).

The area is half the magnitude of the cross product: frac12sqrt(3²+4²+5²)=((5sqrt2) / (2)).

(d) ((13) / (5sqrt2)).

Apply the point-to-plane distance formula to 3x+4y+5z-13=0.

Mark allocation

Award 1 mark for both direction vectors and 1 mark for the correct cross product.
Award 1 mark each for the plane, triangle area and perpendicular distance.

Question 8

(a) tan⁻¹y=((x²) / (2)).

Integrate ((1) / (1+y²)),dy=x,dx to obtain tan⁻¹y=((x²) / (2))+C, then use y(0)=0 to get C=0.

(b) y=tan(((x²) / (2))).

Apply tangent to both sides of the implicit solution.

(c) -sqrtpi<x<sqrtpi.

The nearest vertical asymptotes occur when x² / 2=pi / 2, giving x=pmsqrtpi.

Mark allocation

Award 1 mark for correct separation and 1 mark for integration plus the initial condition.
Award 2 marks for the explicit solution.
Award 1 mark for both correct interval endpoints.

Question 9

(a) t=1 and t=5.

Factor v(t)=(t-1)(t-5).

(b) 1<t<5.

The upward-opening quadratic is negative between its two roots.

(c) -((25) / (3)) mathrm m.

Integrate velocity: int_0⁵(t²-6t+5),dt=[t³ / 3-3t²+5t]_0⁵=-25 / 3.

(d) 13 mathrm m.

An antiderivative with x(0)=0 is x=t³ / 3-3t²+5t. Since x(1)=7 / 3 and x(5)=-25 / 3, the distance is 7 / 3+(7 / 3+25 / 3)=13.

Mark allocation

Award 1 mark each for the rest times, negative-velocity interval and signed displacement.
Award 1 mark for splitting at the direction change and 1 mark for the total distance.

Diagnostic Checklist

Topic	Questions	Marks	Marks Lost	Action
Complex numbers	Q1, Q4 and Q6	13	___	Review conjugate roots, polynomial factors, Argand geometry, loci and roots in polar form.
Vectors and planes	Q2 and Q7	9	___	Review line-plane intersections, angles, cross products and triangle geometry.
Proof	Q3	4	___	Review the base case, induction assumption and factorisation in the inductive step.
Differential equations	Q5 and Q8	9	___	Review exponential models, separation of variables and interval restrictions.
Mechanics	Q9	5	___	Review velocity sign changes, displacement and total distance.

VCE Specialist Mathematics Units 3&4 Free Online Pack 0 — Examination 1 Showcase

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Skill Align VCE Specialist Mathematics Units 3&4 - Free Online Pack 0

Questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Worked Solutions And Marking Guide

Question 1

Mark allocation

Question 2

Mark allocation

Question 3

Mark allocation

Question 4

Mark allocation

Question 5

Mark allocation

Question 6

Mark allocation

Question 7

Mark allocation

Question 8

Mark allocation

Question 9

Mark allocation

Diagnostic Checklist

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