2021 Spec 11 Test 3 Marking Key.docx
Mathematics Specialist Year 11
Student name: _______________________Teacher name: ______________
Date: Friday 23rd July 2021
Task type: Response
Time allowed: 45 minutes
Number of questions: 6
Materials required: Calculator with CAS capability (to be provided by the student)
Standard items: Pens (blue/black preferred), pencils (including coloured), sharpener, correction fluid/tape, eraser, ruler, highlighters
Special items: Drawing instruments, templates, notes on two unfolded sheets of
A4 paper, and up to three calculators approved for use in the WACE examinations
Marks available: 40 marks
Task weighting: 10%
Formula sheet provided: Yes
Note: All part questions worth more than 2 marks require working to obtain full marks.
[7 marks]
Use mathematical induction to prove that
3×5+6×6+9×7+…+3nn+4=nn+12n+132
for all positive integers n.
| Solution |
| Let P(n) denote the proposition ‘3×5+6×6+9×7+…+3nn+4=nn+12n+132’ for all positive integers n.With n=1, LHS of P1= 3×5=15RHS of P(1)=1(1+1)(2×1+13)2=15Hence LHS=RHS, and so P(1) is true.Now assume that P(k) is true for some positive integer k. Then3×5+6×6+9×7+…+3kk+4=kk+12k+132.NowLHS of Pk+1=3×5+6×6+9×7+…+3kk+4+3(k+1)(k+1+4)=kk+12k+132+3(k+1)(k+1+4)=kk+12k+132+3(k+1)(k+5)=kk+12k+132+6k+1k+52=kk+12k+13+6k+1k+52=k+1[k2k+13+6k+5]2=k+1(2k2+19k+30)2=k+1(k+2)(2k+15)2=k+1((k+1)+1)(2(k+1)+13)2=RHS of P(k+1)Hence P(k+1) is true.We have shown that P(1) is true, and that if P(k) is true for some positive integer n then P(k+1) is also true. Hence, by the principle of mathematical induction, P(n) is true for all positive integers n. |
| Specific behaviours |
| proves P(1) by evaluating LHS and RHS separately assumes P(k) is true writes LHS of P(k+1) using RHS of P(k) simplifies expression algebraically to one fraction writes numerator with a factor of (k+1) obtains expression for RHS of P(k+1) written in terms of k+1 writes conclusion for whole proof (accept just the second sentence without the first) |
[2 marks]
A question in a Specialist exam paper asked students to prove the following statement:
‘3n is odd if and only if n is odd (where n is an integer)’.
One student wrote the answer below. Explain clearly why they should not receive full marks for this answer.
Proof:
We prove the contrapositive. Assume that n is an even integer. Then n=2k for some integer k. Now
3n=32k
=23k
which is even since 3k is an integer. Hence if n is even then 3n is even, which implies that 3n is odd if and only if n is odd.
| Solution |
| The student has proved only the statement ‘if 3n is odd then n is odd’. However, since the original statement involves the phrase ‘if and only if’, it is also necessary to prove the statement ‘if n is odd then 3n is odd’. |
| Specific behaviours |
| Notes that statement involves ‘if and only if’, or describes as an equivalence statement Explains that the student should also have proved that ‘if n is odd then 3n is odd’, or refers to the ‘backward direction’ |
[9 = 3+3+3 marks]
Write whether each of the following statements is true or false, and prove or disprove it accordingly.
For all positive real numbers x
x3-x≥x2-x
| Solution |
| The statement is false, and is disproved with the following counterexample:Let x=12. Then LHS = 18-12=-38 and RHS = 14-12=-14, meaning that x3-x<x2-x in this case.Hence the statement is false. |
| Specific behaviours |
| states false states counterexample with a particular value of x shows that for that value of x, x3-x<x2-x.[Alternatively give 2nd and 3rd marks if successfully argues false for any value of x with 0<x<1.] |
There exist distinct prime numbers p and q such that p-q=2.
| Solution |
| The statement is true, and is proved with the following example:Let p=5 and q=3. Then p-q=2. |
| Specific behaviours |
| states true states example with values of p and q such that p-q=2 |
There exist distinct prime numbers p and q such that p2-q2=2.
| Solution |
| The statement is false.Let p and q be distinct prime numbers. Thenp2-q2=p+qp-qSince p and q are distinct primes, p+q≥5 and p-q≥1, and so p2-q2≥5.Hence there do not exist distinct prime numbers p and q with p2-q2=2. |
| Specific behaviours |
| states false factorises p2-q2 using difference of squares argues that p2-q2 cannot equal 2. |
[6 marks]
Find the values of x,y and z in each of the following:
A, B, C and D all lie on the circle with centre O:
| Solution |
| 5x+x=180x=30z=2×30=60y=360-60=300 |
| Specific behaviours |
| 1 mark per correct value |
RS is tangent to the circle with centre O.
| Solution |
| x=2×30=60y=30z=180-90-60=30 |
| Specific behaviours |
| 1 mark per correct value |
[5 marks]
ABCD is a quadrilateral such that each of the four sides is tangent to the same circle, at the points P,Q,R and S, as illustrated below. If AB=15, BC=10 and CD=12, find the length AD.
| Solution |
| Since the sides are tangent to the circle, we may write:w=AS=AP, x=BQ=BP, y=CQ=CR and z=DS=DR.Thus w+x=15 (1)x+y=10 (2)y+z=12 (3)Adding equations (1) and (3) givesw+x+y+z=27and subtracting equation (2) givesw+z=17Hence AD=17. |
| Specific behaviours |
| uses theorem for tangent segments from the same point identifies segments of equal lengths sets up equations for side lengths using sums of segment lengths solves set of equations for w+z states correct value[Accept alternative methods.] |
[11 = 3+4+4 marks]
Solve each of the following trigonometric equations for x in the stated domain.
Show all working to support your answers.
2cosx=3 for 0≤x≤2π
| Solution |
| 2cosx=3cosx=32x=π6 or11π6 |
| Specific behaviours |
| isolates cos(x) states at least one correct solution states two correct solutions[No marks for answers only] |
sinx+π4=-12 for -π≤x≤π
| Solution |
| sinx+π4=-12x+π4=5π4+k2π or x+π4=7π4+k2πHencex=π+k2πor x=3π2+k2πWith k=0, x=π or x=3π2With k=-1, x=-π or x=-π2Hence x=-π,-π2 or π |
| Specific behaviours |
| isolates sinx+π4 states at least one correct solution for x+π4 states at least one correct solution or for x states all three correct solutions for x[No marks for answers only] |
13tan(5x)=1 for 0≤x≤π
| Solution |
| tan5x=35x=π3+kπ x=π15+kπ5Letting k=0,1,2,3 and 4 we obtain:x=π15,4π15,7π15,10π15 and 13π15 |
| Specific behaviours |
| isolates tan5x obtains π3 as a solution for 5x obtains π15 as a solution for x states all five correct solutions for x[No marks for answers only] |