# Additional Mathematics Paper 1 by nklye

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```									ADDITIONAL MATHEMATICS
SPM 2012
PAPER 1
( PART ONE )

LAST KOPEK REVISION
PAPER 1

1. Diagram shows part of the relation f(x)
State                          f(x) f(x) = (x – 2)2 − 3
(a) the type of
the relation,
x
(b) the range.                       .
(2, -3)

(a) Many-to-one     √1

(b) f(x) ≥ -3 √ 1
2.   Given the function f(x) = 3 – x, g(x) = px2 – q and
gf(x) = 3x2 – 18x + 5. Find
(a) gf(- 2),
(b) the value of p and q.

(a) f (-2) = 3 – (−2) = 5 √ M1
gf(- 2) = g(5) = 3(5)2 – 18(5) + 5
= −10 √ 2
(b) gf(x) = p(3 – x)2 – q
= px2 – 6px + 9p - q √ M1
Compare to 3x2 – 18x + 5
p = 3, 9p – q = 5  9(3) – q = 5
q = 22 √ 2 Both correct
3. A function g is defined as g(x) = 7x – 4. Find
(a) g -1,
(b) the value of h if g -1(h) = 2

Using the principle:
x+4      √1                    cx – b
(a) g -1 =                        ax + b
=
7                  c            a

h+4
(b) g   -1(h)   =         =2      √ M1
7

h = 10    √2
4.   A quadratic equation with roots p and q is
x2 + mx + m = 0 where p, q and m are constants.
Express p in terms of q.

(x – p) (x – q) = 0                    OR:
Use SOR/POR
x2 – (p + q) x + pq = 0 √ M1
method:
Compare to x2 + mx + m = 0
SOR = p + q
− (p + q ) = m and pq = m              POR = pq

pq = − (p + q)   √ M2    pq + p = −q
−q
p=            √3
1+q
5.   Find the range of values of x for 4x2 ≥ 3 – 4x

a = 4 > 0  minimum graph

4x2 + 4x – 3 ≥ 0 √ M1
Let 4x2 + 4x – 3 = 0
x
(2x – 1) (2x + 3) = 0             -³   2       ½
√ M2
x=½; x=−³2 ?            √ M2

x≤−³2; x≥½       √3
6.   Diagram shows the graph of a quadratic function y =
f(x). The straight line y = −15 is a tangent of the curve
y = f(x).
(a) State the equation of the axis of symmetry of the
curve.
(b) Express f(x) in the form of (x + p) 2 + q, where p
and q are constants.
−2 + 6    √ M1
the mid-point of the two                y = f(x)
(a) x =                  points on the curve that
2              intersect at x-axis
x
−2   0     6
y = −15
x=2     √2

(x + p)² + q
x + p = 0  x = −p (axis of symmetry)
(b) (x – 2)² − 15 √ 2    q = optimum value (either minimum or
q = −15 √ M1         maximum based on the graph)
2
7.    Solve the equation: 16x =
8 2−x

√ M1
2
24(x)   =                √ M1
23(2 − x)
24x = 2 [2 −3(2 − x)]

24x = 21 – 6 + 3x
Use the concept of the rules of
4x = – 5 + 3x           √ M2    equation of indices
x = − 5 √3
8. Solve the equation: 2 logx4 + logx8 = −7

√ M1
logx 42 + logx 8 = −7           Converting log to indices

logx 16(8) = −7     √ M1   OR   logx 16(8) = logx x −7
Use the concept of the rules of
equation of logarithms.
128 = x −7     √M2
27 = x −7          OR    2 7 = ( 1 /x ) 7
(½ )−7 = x −7

x=½      √3
9.   The nth term of an arithmetic progression is given by
Tn = 12 – 3n. Find the common difference of the
progression.

a = T1 = 12 – 3(1)
= 9 √ M1
T2 = 12 – 3(2)
= 6
d = T2 − a
=6–9       √ M2
= −3       √3
10.   Given that 12, 6, 3 … is a geometric progression, find
the sum of the first 7 terms after the 3 th. term of the
progression

T4 = 12 ( ½ )3 = 12/8
a = 12
S7* = (12/8) [1 – ( ½ )7]
r = ½ √ M1                      = 2.977
S7* = S10 – S3
12 [1 – ( ½ )10] – 12 [1 – ( ½ )3]
=                                          √ M2
1–½                1–½
= 2.977    √3                S7* = Sum from T4 to T10 or
first 7 terms after the 3rd. term
k
11.   Given 0.471 + 0.000471 + 0.000000472 + … = -------
Find the value of k.                        333

a = 0.471
r = 0.001    √ M1

0.471            k            Use the concept of
S∞ =                 =         √ M2
the sum of infinity.
1 – 0.001         333

=    157
333
k = 157     √3
12. Diagram shows a circle with center O. The length of
minor arc AB is 4.8 cm and the angle of minor sector
Using π = 3.142, find
(a) the value of θ,
(b) the length, in cm, of the major arc AC.
4.8
(a) θ =        = 0.75 rad.   √1                    A
6.4                             O   θ

B
(b) θAC (major) = 2π – (1 + 0.75)
√ M1

SAC (major) = 6.4 ( 4.534 )
= 29.02 cm. √ 2
dy
13.       Given that y =   x3   (3x +   1)4,   find the value of
dx
when x = −1.