All solutions here are **suggested**. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions

Numerical Answers (workings/explanations are after the numerical answers. Click to expand.)

Question 1:

Question 2:

Question 3:

Question 4: or ;

Question 5:

Question 6:

Question 7:

Question 8:

Question 9:

Question 10:

Question 11:

Question 12: ; cannot see since ; ; no

Total marks: 4 marks

Since is an asymptote, then

At , we have

At , we have

Using GC,

Note: Students can verify this answer by plotting the graph.

Total marks: 3 marks

Since is a unit vector, then .

Since , then

, where is the angle between and

Thus, and are perpendicular to each other.

Total marks: 4 marks

(a) For to be a root, we must be able to apply conjugate root theorem. Thus, all coefficients of the polynomial should be real numbers. We have that .

(b)

Thus,

Note: Students can verify this answer by using the GC’s polynomial root finder function

Total marks: 6 marks

(a)

or

(b)

Replace by

(rejected, or

Note: Students can verify this answer by plotting the graph.

Total marks: 7 marks

(a)

(b) Using GC, since , and

(c) Note that is a GP with first term and common ratio .

Total marks: 8 marks

(a)

When

Tangent at is given by

Thus,

(b) Since the tangent cuts the origin, then

(c)

Required area

, using GC.

Total marks: 8 marks

(a)

(b) Note: there are some issues with the rendering here, you should do method of difference.

(c)

Observe that .

Thus, the possible values are

Note: Students can verify this answer by calculating the sigma with the GC.

Total marks: 9 marks

(a)

Thus,

(b)

When

Thus,

Note: we can verify the answers using standard series in exam.

(c)

By comparing, and .

Total marks: 10 marks

(a)

, and

(b)

Let , then

Thus, coordinates are .

(c) Observe that when

Total marks: 14 marks

(a)

(b) The axial intercepts are , and the asymptotes are and .

(c)

Thus,

(d)

using GC.

Total marks: 12 marks

(a)(i)

Thus,

(a)(ii) We have that

(b)

, where

Let

Thus, and

(c)

When

Let , using GC, seconds

Total marks: 15 marks

(a) Let be the points nearest to the controller and let be the controller.

for some

Thus, the coordinates of is .

(b)

Thus, the controller will not be able to see.

(c) (i) Clearly, the paths are not parallel. Now suppose the path intersect, then

We have that

Solving, and . Thus .

Thus,

(c) (ii)

Required angle

(d) (i)

(d) (ii)

For the distance to be the shortest, the resultant vector should be perpendicular to both the paths.

Since the resultant vector is not perpendicular to both the paths, the distance found is not the shortest.

Hi, for 11(b) B should be -2 if I remember correctly? Because in the question, exponent for e^Bt had no negative sign in front of the B.

I also need to verify if Q9(c) definite integral range is from 0 to Q or P to Q. Because I may have misread the question during the paper but I did from P to Q.

I’m not sure if you have access to the paper exactly as it is, in that case, my information could be wrong.

Hello, actually I can’t recall whether they had a negative sign in from of the exponent too. For 9c, I think it was 0 to Q, but I may have recalled wrongly too.

Hello, for qn 6 (c), was the answer supposed to be in 1dp/2sf? I recall giving mine to 3sf as 5.40

hello. its to 1 decimal places.