SPM Additional Mathematics 2018, Paper 2 (Question 1 – 3)


Question 1:
The sum of the first n terms of an arithmetic progression, Sn is given by S n = 3n( n33 ) 2 .  
Find
(a) the sum of the first 10 terms,
(b) the first term and the common difference,
(c) the value of q, given that qth term is the first positive term of the progression.

Solution:
(a)
S n = 3n( n33 ) 2 S 10 = 3( 10 )( 1033 ) 2 S 10 =345

(b)
S n = 3n( n33 ) 2 S 1 = 3( 1 )( 133 ) 2 S 1 =48 T 1 = S 1 =48 First term, a= T 1 =48 T n = S n S n1 T 2 = S 2 S 1 T 2 = 3( 2 )( 233 ) 2 ( 48 ) T 2 =45 Common difference, d = T 2 T 1 =45( 48 ) =3

(c)
First positive term,  T q >0 T q >0 a+( q1 )d>0 48+( q1 )3>0 48+3q3>0 3q>51 q>17 Thus, q=18.



Question 2:
It is given that g : x → 2x – 3 and h : x → 1 – 3x.
(a) Find
(i) h (5)
(ii) the value of k if  g( k+2 )= 1 7 h( 5 ),
(iii) hg(x).
(b) Hence, sketch the graph of y = | hg(x) | for –1 ≤ x ≤ 3.
State the range of y.

Solution:
(a)(i)
h( x )=13x h( 5 )=13( 5 )    =14

(a)(ii)
g( x )=2x3 g( k+2 )= 1 7 h( 5 ) 2( k+2 )3= 1 7 ( 14 ) 2k+43=2 2k=3 k= 3 2

(a)(iii)
g( x )=2x3, h( x )=13x hg( x )=h( 2x3 )  =13( 2x3 )  =16x+9  =106x

(b)
y = |hg(x)|,
y = |10 – 6x|
Range of y : 0 ≤ y ≤ 16






Question 3:
Solution by scale drawing is not accepted.
Diagram 1 shows a triangle OCD.
Diagram 1

(a) Given the area of triangle OCD is 30 unit2, find the value of h.
(b) Point Q (2, 4) lies on the straight line CD.
(i) Find CQ : QD.
(ii) Point P moves such that PD = 2 PQ.
  Find the equation of the locus P.

Solution:
(a)
Given Area of  OCD = 30 1 2  | 0  h 6   0  2   8   0 0 |=30 | ( 0 )( 2 )+( h )( 8 )+( 6 )( 0 )( 0 )( h )( 2 )( 6 )( 8 ) ( 0 )|=60 | 0+8h+00+120|=60 | 8h+ 12|=60 8h+12=60 8h=48 h=6 or  8h+12=60 8h=72 h=9( ignore )



(b)(i)

[ 6( m )+( 6 )( n ) m+n ,  2( m )+( 8 )( n ) m+n ]=( 2, 4 ) 6m6n m+n =2 6m6n=2m+2n 4m=8n m n = 8 4 m n = 2 1 2m+8n m+n =4 2m+8n=4m+4n 2m=4n m n = 4 2 m n = 2 1 Thus, CQ=QD=2:1

(b)(ii)
PD=2PQ ( x6 ) 2 + ( y2 ) 2 =2 ( x2 ) 2 + ( y4 ) 2 ( x6 ) 2 + ( y2 ) 2 =4[ ( x2 ) 2 + ( y4 ) 2 ] x 2 12x+36+ y 2 4y+4=4[ x 2 4x+4+ y 2 8y+16 ] x 2 12x+36+ y 2 4y+4=4 x 2 16x+16+4 y 2 32y+64 The equation of locus P: 3 x 2 +3 y 2 4x28y+40=0


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