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SPM Additional Mathematics 2017, Paper 1 (Question 8 – 10)


Question 8 (3 marks):
It is given that the sum of the first n terms of an arithmetic progression is S n = n 2 [ 133n ]  
Find the nth term.

Solution:
S n = n 2 [ 133n ] S n1 = n1 2 [ 133( n1 ) ]   = 1 2 ( n1 )( 133n+3 )   = 1 2 ( n1 )( 163n ) T n = S n S n1    = n 2 ( 133n ) 1 2 ( n1 )( 163n )    = 13n 2 3 n 2 2 1 2 ( 16n3 n 2 16+3n )    = 13n 2 3 n 2 2 1 2 ( 19n3 n 2 16 )    = 13n 2 3 n 2 2 19n 2 + 3 n 2 2 +8    = 6n 2 +8    =83n


Question 9 (3 marks):
Diagram 5 shows the graph of the function f : x → |1 – 2x| for the domain –2 ≤ x ≤ 4.

Diagram 5

State
(a) the object of 7,
(b) the image of 3,
(c) the domain of 0 ≤ f(x) ≤ 5.


Solution:
(a)
The object of 7 is 4.

(b)
f (x) = |1 – 2x|
f (3) = |1 – 2(3)|
= |1 – 6|
= |–5|
= 5

The image of 3 is 5.

(c)
|1 – 2x| = 5
1 – 2x = ±5
Given when f(x) = 5, x = –2.

When f(x) = –5
1 – 2x = –5
2x = 6
x = 3

Domain: –2 ≤ x ≤ 3.



Question 10 (4 marks):
Given the function g : x → 2x – 8, find
( a )  g 1 ( x ), ( b ) the value of p such that  g 2 ( 3p 2 )=30.

Solution:
(a)
Let y=g( x ) =2x8 2x8=y  2x=y+8    x= y+8 2 Thus,  g 1 ( x )= x+8 2

(b)
g( x )=2x8 g 2 ( x )=g[ g( x ) ]  =g( 2x8 )  =2( 2x8 )8  =4x168  =4x24 g 2 ( 3p 2 )=30 4( 3p 2 )24=30 6p=54 p=9

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