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Marking Scheme of F.5 Additional Mathematics Final Examination

1. coeff. of x11 in [pic] = coeff. of x10 in [pic]( 2 = [pic] = −504 2M+1A

2. [pic] ( [pic], [pic] 1M+1A

Note that ( is non-negative for any real value of m, hence E has distinct real roots for any real

value of m except m = 2 and m = 1. 2A

3. The derivative =[pic] 1M

[pic] 1M

[pic] 1M+1A

4. [pic] ( [pic] 1M+1A

Put x = 2, y = 0 into the above, [pic] ( [pic] 1M

hence the slope of normal to the curve at P is [pic]. 1A

5. Let B(x,y). [pic] = −(5i + 7j) + (xi + yj) = (x − 5)i + (y − 7)j 1M

Hence 3i + 4j = (x − 5)i + (y − 7)j ( 3 = x − 5 and 4 = y − 7 ( x = 8 and y = 11 1M

Thus the coordinates of B are (8,11). 1A

The unit vector required = [pic] = [pic](−3i − 4j) = [pic]i −[pic]j 1M+1A

6. [pic] 1M

[pic] 1M

[pic] 1

Hence the minimum value of [pic] is [pic] = −3 (at sin2x = 1) 1M+1A

7.

(a) [pic] 1M + 1

(b) [pic] ( [pic] 1M

By (a), put n = 2010, then [pic] 1A

Put x = 2, y = 1; 1 = 2(1)2010 + C ( C = −1 1M

Hence the required equation is [pic] 1A 8. [pic] 1M

[pic] 1M

[pic] 1M

[pic] 2A

9. Let P(n) be the proposition that “[pic] is divisible by 7”.

When n = 1, [pic], thus P(1) is true. 1

Suppose P(k) is true, i.e. [pic] where Q is a positive integer. 1

Consider P(k + 1)

[pic]

= [pic] (by M.I. assumption) 1M

= [pic] = [pic] 1M

= [pic] = 7R where R is a positive integer

Therefore P(k + 1) is also true. By the principle of mathematical induction, P(n) is true for all

positive integers n. 1

10. Put [pic] into [pic], yield

[pic] ( [pic] ………. (*) 1M

Since M(h , k) is the mid-point of PQ, [pic] 1M

[pic] 1A

and [pic] ( [pic] 1M

Thus, the equation of the required locus is [pic] 1A

11. To find the upper limit of the integral, we solve sin2x = cosx 1M

2sinxcosx = cosx ( (2sinx − 1)cosx = 0 ( sinx = [pic] or cosx = 0

x = [pic] (refer to the graph) 1A

The shaded area = [pic] 1M

= [pic] = [pic] 1A + 1A

12. (a) slope of AB = [pic], slope of AC = [pic] = 2

[pic] 1M +1A

(b) AI is the angle bisector of (BAC ( (IAC = [pic]= ( (say)

[pic] (by (a)) 1M

[pic] ( [pic] 1M

[pic] (negative value was rejected) 1A

13. [pic]

[pic] 1M

[pic]

[pic]

[pic] 1M

[pic]

[pic] 1M

Set [pic] ( [pic] ( [pic] (for [pic])

For [pic] slightly [pic] hence [pic].

For [pic] slightly [pic] hence [pic]. 1M

( The turning point is ([pic], f([pic])) = ([pic],[pic]) and this is a minimum point. 2A

14.

(a) [pic] CG : GM = 2 : 1, [pic] ( [pic] 2M+1

(b) (i) [pic] and [pic] 2A

[pic] [pic] ( [pic] ( [pic] 1M

[pic]

[pic] ([pic] are unit vectors) 1M

[pic] 1M

[pic] 1A

(b)(ii) [pic] 1M

[pic] CY// AB ( [pic] ( [pic] ( [pic] 1M

Thus, [pic] 1A

15

(a) Let the equation of the circle be [pic].

Since the circle passes (0,0), (5,0) and (1,4), we have

[pic] ( F = 0

[pic]

[pic] 1M

Hence the equation of the circle is [pic]. 1A

(b)(i) Equation of AB is [pic] ( [pic] 1M

Put [pic], [pic]. Hence P3 lies on AB.

Slope of PP3 = [pic] 1M

Hence PP3 ( AB, thus coordinates of P3 are ([pic],[pic]). 1

(b)(ii) P1(h , 0) 1A

Equation of OB is [pic]. Let P2(t, 4t).

Since PP2 ( AB ( slope of PP2 = [pic] ( [pic]

[pic] and the coordinates of P2 are ([pic],[pic]) 1A

slope of P1P2 = [pic]

slope of P1P3 = [pic] 1A

slope of P1P2 = slope of P1P3 ( [pic]

( [pic] 1M

( [pic]

It is true since (h,k) lies on the circle [pic], thus P1, P2 and P3 are collinear.

1A

(b)(iii) If H lies on PP3, then O lies on PP3 and the equation of PP3 is the same as the equation of

OP, i.e. y = x (since the slope of AB = −1 and OP ( AB) 1M

Thus the equation of the family is [pic] for any real value n. 1A

16.

(a) s = [pic]

By Heron’s formula, area of (VBC = [pic] = 84 m2 1M+1A

Since VA ( (ABC, AM is the projection of VM on (ABC, and hence AM ( BC, thus

angle between (ABC and (ABC = [pic]

Area of (ABC = [pic] = Area of (VBC ( cos( 1M

= 84 ( 0.6 = 50.4 m2 1A

(b)(i) Let h m be the length of VA.

[pic], hence AM = hcot( = 0.75h 1M

Area of (ABC = [pic] = 5.25h 1M

By (a), [pic] ( h = 9.6 Thus, the length of VA = 9.6 m 1A

Angle between (VAB and (VAC = (BAC (since AB ( VA and AC ( VA)

[pic] [pic] 1M

Let ( = (BAC. By cosine law, [pic]

[pic] 1M

( angle between (VAB and (VAC = ( = 86.1( (correct to 3 significant figures) 1A

(b)(ii) P is the centre of the circumcircle of (ABC and (BAC is an acute angle (by (b)(i)), and

since (BAC is the largest interior angle of (BAC, (BAC is an acute triangle. Thus, P

lies inside of (ABC. . 1M + 1

17.

(a) t1 = [pic] = 5 (s) 1A

QX = [pic] = [pic]

Suppose XZ cuts RQ at N, [pic] ( [pic] 1M

[pic] m2 1

(b) A = area of △XWZ + area of WQNZ (where N is the intersection of RQ and ZY)

= [pic] + area of WQRZ ( area of △RZN

[pic] (m2) 2M+1A

[pic] 1A

Hence [pic]

= [pic] = [pic] 1A

(c) Figure b is impossible because the graph of A = [pic] should be open downwards. 1M

[pic],

[pic] 2M

Hence the slope of tangents at t = 5 are the same for two arcs corresponding to t ( 5 and t ( 5. 1M

Thus, there is no corner at the junction between two arcs, figure a is the best representation. 1

18.

(a) The shaded area = [pic] 1M

= [pic] = [pic] units square 1A

(b) [pic] 1M

= [pic] 1M

= [pic] 1

[pic] 1M

(c)(i) Put h = 3 0.5 M

A = area of the region shown

= [pic] (by (a))

= [pic] 0.5 M

By (b), capacity of the container

= [pic] 1M

= [pic] cubic units. 1A

(c)(ii) [pic] 1A

[pic] 1M

Let p cm be the water depth.

When [pic], [pic] and [pic], hence at that moment

[pic][pic] cm2/s 1A

END OF SOLUTION

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