Preparation for HKDSE (S4 S6) Full Solutions

Preparation for HKDSE (S4– S6) Full Solutions

Preparation for HKDSE (S4– S6) Full Solutions

HKDSE Mastering Mathematics 1© Pearson Education Asia Limited 2013

Preparation for HKDSE (S4– S6) Full Solutions

STRUCTURAL QUESTIONS

Suggested SolutionsMarksRemarks

1.Join AB.

(∠in semi-circle)1A

In △ACD,

∵AD = CD

∴(base∠s, isos.△) 1M

(opp.∠s, cyclic quad.)1M

Alternative solution:

Join AB.

(∠in semi-circle)1A

In △ABC,

(∠sum of △)

(opp.∠s, cyclic quad.)1M

In △ACD,

∵AD = CD

∴(base∠s, isos.△) 1M

(∠sum of △)

2.(a)

∴The required solutions are . 1A

(b)There are 15 integers which satisfy both inequalities in (a).1A

3.(a)Inter-quartile range1A

Range1A

(b)At the beginning, there are 22 students and therefore we need to

considerthe 11th and 12th data when finding the median.1M

Now, since the highest mark and the lowest mark are deleted from

the distribution, there are 20 students left and therefore we need to

considerthe 10th and 11th data when finding the new median.

Since the two data considered are the same as before, the new 1M

median remains unchanged.

Therefore, the median will not decrease.1Af.t.

4.(a) 1M

∴ 1A

(b)(i)

∵x– 3 is a factor of f (x).

∴By factor theorem, we have

(ii)

5.(a)Let , where m and n are non-zero constants.1A

So, we have and .1M

By solving, we have and .1Afor both correct

∴ The required area

(b)

∴ The required perimeter is 48 cm.1A

6.(a)Note that the slope of a line segment in the graph represents the

average speed of that part of the journey.1M

Since the slope of the line segment for Part III is the greatest,

Darren rides at the greatest speed for Part III of the journey.1A

(b)Let x h be the time spent in travelling from town A to C.

∴Darren leaves town A at 8:44.1A

Let y km be the distance between town A and B.

y = 11

∵Darren leaves town B at 10:24.

∴The average speed for Part III

7.(a)(i)∵L2 is perpendicular to L1.

∴The slope of L21M

∴The equations of L1 and L2areand

respectively.1Afor both correct

∴By solving, we have x = 3 and y = 4.

i.e. the coordinates of A are (3, 4).1A

(ii)The equation of the circle:

1Aorequivalent

(b)(i)They are concentric circles with centre (3,4).1A + 1A

(ii)Let (x, y) be the coordinates of P.

∵P moves with a constant distance 1 unit with C.

∴The locus of P are two concentric circles with centre (3, 4),

and radii 1 unit and 3 units respectively.

i.e. the equation of P:

and

and1A + 1Aorequivalent

8.(a)Let m be the mean mark of the examination.

∵The standard score of Wendy is 2.

∴1M

m = 46

∴The standard score of Winston

(b)∵5 marks are added to the mark of each student.

∴The mean mark is increased by 5, while the standard

deviation remains unchanged.1M

The new standard score of Winston

∴The standard score of Winston remains unchanged.

i.e.his claim is disagreed.1Af.t.

9.(a)The required probability

1M + 1M

(b)The required probability

1M for using (a)

10.(a)In △ABC,

By sine formula, we have

In △ABD,

(b)(i)In △ABD,

By considering the volume of the tetrahedronABCD, we have

(ii)Since the lengths of AD, BD and CD are constants, the volume

of the tetrahedronABCD varies directly as the value of

.1M

Whendecreases fromto, the value of

increases. Therefore, the volume of the

tetrahedronABCDincreases.1A

11.(a)1M

(b)Total income for the whole year

Total cost for the whole year

∵Total income > total cost

∴ Potato Pet Shop will make a profit for the whole year.1Af.t.

∴ The owner will stop saving in November.1A

MULTIPLE CHOICE QUESTIONS

12.CA

13.CD

The mean of the five numbers

The variance of the five numbers

14.CD

∴

15.CA

16.CA

∴ n= the slope of the graph

i.e.

17.CA

Referring to the figure, the graph of y = g (x) can be obtained by reflecting the graph of

y = f (x) about the x-axis, then translating the graphrightwards by 3 units.

∴ g (x) = –f (x– 3), i.e. the answer is A.