08. CSEC ADD MATHS 2019
[Pages:29]1. (a)
CSEC ADD MATHS 2019
SECTION I Answer BOTH questions. ALL working must be clearly shown.
The function f is such that f ( x) = 2x3 + 7x2 + 3x .
(i) Determine all the linear factors of f ( x).
SOLUTION:
Data: f ( x) = 2x3 + 7x2 + 3x Required to determine: All the linear factors of f ( x).
Solution:
f ( x) = 2x3 + 7x2 + 3x
= x(2x2 + 7x + 3)
= x (2x +1)( x + 3)
So the linear factors of f ( x) are x, 2x +1 and x + 3.
(ii) Compute the roots of the function f ( x).
(A function does NOT have roots. An equation may have roots or solutions. So, we let () = 0.)
SOLUTION: Required to find: The roots of () = 0. Solution:
f ( x) = 2x3 + 7x2 + 3x = x (2x +1)( x + 3)
If f ( x) = 0 then x(2x +1)( x + 3) = 0
and the roots will then be x = 0 or - 1 or -3 2
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(b) Two functions are such that g ( x) = x2 - x and h( x) = 2x - 3. (i) Determine gh( x).
SOLUTION:
Data: g ( x) = x2 - x and h( x) = 2x - 3 Required to determine: gh( x)
Solution:
g (x) = x2 - x \ gh ( x) = ??h ( x)??2 - h ( x)
= (2x - 3)2 - (2x - 3)
= 4x2 - 6x - 6x + 9 - 2x + 3
gh ( x) = 4x2 -14x +12
(ii) Given that hg ( x) = 2x2 - 2x - 3, show that the values of x, for which
hg ( x) = 0, can be expressed as 1? 7 .
2 SOLUTION:
Data: hg ( x) = 2x2 - 2x - 3
Required to show: The solutions of hg ( x) = 0 are 1? 7 .
2 Solution:
When hg ( x) = 0
2x2 - 2x -3 = 0
-(-2) ? (-2)2 - 4(2)(-3)
x=
2(2)
= 2 ? 4 + 24 4
= 2 ? 28 4
= 2?2 7 4
2(1? 7 )
=
2(2)
= 1? 7 2
Q.E.D.
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(c) Solve 3x log 2 + log8x = 2.
SOLUTION:
Data: 3x log 2 + log8x = 2
Required to find: x
Solution: 3log2 + log8. = 2 3log 2 + log (20). = 2 3log 2 + log 20. = 2
3log 2 + 3log 2 = 2
(3 + 3) log2 = 2
(6) log2 = 2
2 6 = log2
2
1
=
=
6log2 3log2
A value of x is only possible if the base of the terms in logs is given.
For instance, if the base is 10, then
1
1
=
=
= 1.11
3log342 0.903
For instance, if the base is 2, then
11
=
=
3log72 3
2. (a) (i)
Express f ( x) = -2x2 - 7x - 6 in the form a ( x + h)2 + k.
SOLUTION:
Data: f ( x) = -2x2 - 7x - 6 Required to express: f ( x) in the form a ( x + h)2 + k.
Solution: -27 - 7 - 6
= -2 :7 + 7 ; - 6 2
7 7 49 = -2 - 6
4 16 7 7 49 = -2 : + ; + - 6 48
77 1 = -2 : + ; +
4 8
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So,
-2x2
-
7
x
-
6
=
-2
? ??
x
+
7 4
2
? ? ?
+
1 8
is of the form
a ( x + h)2 + k, where
a = -2, h = 7 and k = 1.
4
8
Alternative Method:
( + )7 + = ( + )( + ) + = (7 + 2 + 7) + = 7 + 2 + 7 +
( ) So -2x2 - 7x - 6 ? ax2 + 2ahx + ah2 + k
Equating coefficients:
= -2,
2 = -7,
2(-2) h = -7
7 + = -6
-2
? ??
7 4
2
? ? ?
+
k
=
-6
h=7 4
-2
? ??
49 16
? ??
+
k
=
-6
- 6 1 + k = -6 8
k=1 8
So,
-2x2
-
7
x
-
6
?
-2
? ??
x
+
7 4
2
? ? ?
+
1 8
(ii) State the maximum value of f ( x).
SOLUTION:
Required to state: The maximum value of f ( x).
Solution:
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f ( x) = -2x2 - 7x - 6
=
-2
? ??
x
+
7 4
2
? ? ?
+
1 8
? 0"x
\The maximum value of f ( x) = -2(0) + 1
8 The maximum value of () is 3.
C
(iii) State the value of x for which f ( x) is a maximum.
SOLUTION:
Required to state: The value of x for which f ( x) is a maximum
Solution:
The maximum value of
f ( x) occurs when
-2
? ??
x
+
7 4
2
? ? ?
=
0
i.e when x = - 7 4
(iv) Use your answer in (a) (i) to determine all values of x when f ( x) = 0.
SOLUTION:
Required to determine: The values of x when f ( x) = 0.
Solution:
() = -2 D + EG7 + 3
F
C
-2 D + EG7 + 3 = 0
F
C
- 2 D + EG7 = - 3
F
C
D + EG7 = 3
F
3H
+ E = ? 3
F
F
= - E ? 3
FF
= JE?3
F
= JEK3 or JEJ3
F
F
= - H or - C
F
F
= -1 3 or - 2
7
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(v) Sketch the function f ( x) and show your solution set to (a) (iv) when f (x) < 0.
SOLUTION:
Required to sketch: The function f ( x) and write the solution of f (x) < 0.
Solution:
f (0) = -6
For f ( x) < 0:
The solution set for
f ( x) < 0 is
? ? ?
x
:
x
<
-2
?
x
>
-1
1 2
??. ?
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(b) A geometric series can be represented by y + y2 + y3 + ... x x3 x5
( ) Prove that
S? = xy
x2 - y
-1
.
SOLUTION:
Data: y + y2 + y3 + ... is a geometric series. x x3 x5
( ) Required to prove: S? = xy x2 - y -1
Proof:
For the geometric series
y y2 y3 x + x3 + x5 + ...
T2 = T3 = y2 ? y T1 T2 x3 x
y2 x = x3 ? y
y = x2 The series is a geometric progression with first term, a = y and with a common
x ratio of y .
x2
a S? = 1- r , r < 1
y
=x
1-
y x2
y
=
x x2 - y
x2
= y ? x2 x x2 - y
xy = x2 - y
( ) = xy x2 - y -1
Q.E.D. SECTION II
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Answer BOTH questions. ALL working must be clearly shown.
3. (a) A circle with center (1, -1) passes through the point (4, 3).
(i) Calculate the radius of the circle. SOLUTION:
Data: A circle has center (1, -1) and passes through (4, 3).
Required to calculate: The radius of the circle Calculation:
Length of the radius = (4 -1)2 + (3 - (-1))2
= (3)2 + (4)2
= 25 = 5 units (ii) Write the equation of the circle in the form x2 + y2 + 2 fx + 2gy + c = 0. SOLUTION: Required to write: The equation of the circle in the form x2 + y2 + 2 fx + 2gy + c = 0. Solution: Recall for
The equation is ( x - a)2 + ( y - b)2 = r2
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