04.CSEC ADD MATHS 2015 nsite.com
CSEC ADD MATHS 2015
1. (a)
Data: f ( x) = x2 + 5, x ? 1 and g ( x) = 4x - 3, x ? R. Required to calculate: gf (2)
Calculation:
f (2) = (2)2 + 5
=9
\ gf (2) = g (9) = 4(9) - 3
= 33
OR
gf ( x) = 4( x2 + 5) - 3
( ) \ gf (2) = 4 (2)2 + 5 - 3
= 33
(Note: The given domain for f ( x) and g ( x) are of no consequence in the
question).
(b) Data: h ( x) = 3x + 5 , x ? R, x ? 2.
x-2
Required to calculate: h-1 ( x)
Calculation: Let y = 3x + 5 x-2
xy - 2 y = 3x + 5
xy - 3x = 5 + 2 y
x ( y - 3) = 2 y + 5
x = 2y+5 y-3
Replace y by x to obtain:
h-1 ( x) = 2x + 5 , x ? 3
x-3
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(c) Data: ( x - 2) is a factor of k ( x) = 2x3 - 5x2 + x + 2. Required to factorise: k ( x) completely.
Solution:
If ( x - 2) is a factor of k ( x) = 2x3 - 5x2 + x + 2, then k ( x) is divisible by ( x - 2), according to the remainder and factor theorem
2x2 - x -1 x - 2 2x3 - 5x2 + x + 2
- 2x3 - 4x2 - x2 + x + 2
- - x2 + 2x - x+2
- - x+2 0
We factorise 2x2 - x -1 = (2x +1)( x -1) Hence, k ( x) = ( x - 2)(2x +1)( x -1)
(d) (i)
Data: 16x + 2 = 1 4
Required to calculate: x
Calculation:
16x + 2 = 1 4
( )24
x+2
=
1
(2)2
24 x+8 = 2-2 Equating indices since the bases are equal, we obtain 4x + 8 = -2
\4x = -10
x = -2 1 2
(ii) Data: log3 ( x + 2) + log3 ( x -1) = log3 (6x - 8)
Required to calculate: x Calculation:
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log3 ( x + 2) + log3 ( x -1) = log3 (6x - 8) (Product law)
\log3 {( x + 2) ( x -1)} = log3 (6x - 8)
Remove log3, we obtain
( x + 2) ( x -1) = 6x - 8
x2 + x - 2 = 6x -8 x2 - 5x + 6 = 0
( x - 2) ( x - 3) = 0
Hence, x = 2 or 3
When x = 2, there are no terms of the equation that result in log3 (-ve) or log3 (0). When x = 3 there are no terms of the equation that result in log3 (-ve) or log3 (0).
Hence, x = 2 or 3.
2. (a)
Data: f ( x) = 3x2 - 9x + 4
(i) Required to express: f ( x) in the form a ( x + b)2 + c, where a, b and
c?R. Solution:
a(x + b)2 + c = a(x + b)(x + b) + c
( ) = a x2 + 2bx + b2 + c
= ax2 + 2abx + ab2 + c
Hence,
3x2 - 9x + 4 = ax2 + 2abx + ab2 + c Equating coefficients:
= 3
2 = -9
2(3) ? = -9
6 = -9 1
= -1 2
. + = 4 3.
3 2- 3 + = 4 2 9
3 ? + = 4 4
27 4 + = 4
3 = 4 - 6 4
3 = -2 4
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\ f ( x) = 3x2 - 9x + 4 can be written as
a = 3? R, b = -1 1 ? R and c = -2 3 ? R.
2
4
3???
x
-
1
1 2
2
? ? ?
-
2
3, 4
where
OR
( ) 3x2 - 9x + 4 = 3 x2 - 3x + 4
One half the coefficient of x is 1 (-3) = -1 1
2
2
\3(x2 - 3x) + 4
=
3???
x
-
1
1 2
2
? ? ?
+
*
?
3???
x
-
1
1 2
? ??
? ??
x
-
1
1 2
? ??
=
3???
x2
-
3x
+
9 4
? ??
= 3x2 - 9x + 6 3 4
63-23 =4 44
\* = -2 3 4
If
3x2
-
9x
+
4
=
3???
x
-
1
1 2
2
? ??
-
2
3 4
and which is of the form
a (x + b)2 + c,
where a = 3? R, b = -1 1 ? R and c = -2 3 ? R.
2
4
(ii) Required to state: The minimum point of f ( x).
Solution:
f
(x)
=
3x2
-
9x
+
4
?
3???
x
-11 2
2
? ? ?
-
2
3 4
? ??
x
-11 2
2
? ??
?
0
"x
\
3
? ??
x
-
1
1 2
2
? ? ?
?0
"x
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Hence, if the graph of f ( x) is drawn, the minimum value of f ( x) is
0 - 2 3 = -2 3. The minimum value occurs when 44
3??? x
-
1
1 2
2
? ? ?
=
0
that
is
when x = 1 1 . Therefore, the minimum point on the curve of f ( x) is
2
???1
1 2
,
-
2
3 4
???.
(b) Data: The equation 3x2 - 6x - 4 = 0 has roots a and b .
Required to calculate: The value of 1 + 1 . ab
Calculation:
Recall: If ax2 + bx + c = 0 ?a
x2 + b x + c = 0 aa
If the roots are a and b , then ( x -a )( x - b ) = 0.
\x2 - (a + b ) x + ab = 0
Equating coefficients, we get:
a + b = - b and ab = c
a
a
So, in the equation 3x2 - 6x - 4 = 0
a + b = - (-6)
3
=2
ab = - 4 3
1+1 ab
b +a = a +b ab ab
=
2 -4
3
= -1 1 2
So, 1 + 1 = -1 1 . ab 2
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