Recall that x x1 x2 x1 x2 x1 x2 x1 x2 x1 x2
Problem 1a(iii)
b b2 4ac 2a
b b 4ac 2
2a
g ( x )
y
x
b b 4 ac 2
2a
b b 2 4ac b b2 4 ac 2a
b b x
2a 2b
3
1 2 1 2 1
g
1
3
x
3
y
3
2 1
y
2
2 x 3
2a b
x
y
( x ) 2 x 3
a
therefore
1b
Problem 1a (i)
1 x 2 2
Show that it is not a one to one. 1 x 2 0 x 2 1 x 1 x 1
2
We can see that when x = 1, we
a
-5
-2
1
+2
-1.5
4
-3
-3.5
a3
g et a result of Zero, this would represent a Many to one and not a one-to-one.
Pro blem 1(c) 1 2
x 2
1 2
x 2 x x 48
x 2 2 x 48 0
x 2 8 x 6 x 48 0
x
2
8 x 6 x 48 0
x x 8 6 x 8 0
x 8 x 6 0 x 8 and x 6 y
y
1 2 1 2
(8) 2 32 ( 6) 2 18
Problem 2 (a)
Problem 1a (ii)
f ( x) 2 x 2 12 x 9 expressed in Standard Form
Given : f ( x) 1 x
2
g ( x)
;
1 2
x 3
Standard Form f ( x ) a( x h)2 k . 2 f ( x) 2 x 12 x 9
2( x 2 6 x) 9 factor a -2 from the first two terms.
Evaluate
2 x 2 6 x (32 ) 9 2(32 ), Complete the square,
2
1 fg ( x) 1 x 3 2 1 1 1 x 3 x 3 2 2 3 3 1 1 x 2 x x 9 2 2 4 1
3
3
4
2
2
note we added half of 6 squared then subtracted that same same amount to balance the equation. 2
2 x 3 9 18 2
2 x 3 9 There Therefor foree the vert vertex ex is at ( h, k ) (3,9). (3,9).
1 x 2 x x 9 1
x 2 3 x 8
4 Problem 2b
Problem 3a (i)
x
3 5 x 2 x 2 0
2 x 5 x 3 0
2 x 6 x 1x 3 0
kx 2 y 12 E xpress both lines in the slope -intercept form. form.
2
y
2
2 x
2
6 x 1x 3 0
y
2 x 1 x 3 0 2 x 1 0 x 3 0 x 3
x
x
1
2 ; note note that that the the gra gradie dient woul ould be .
3 kx
3
6 2 Therefore if they are perpen dicular, their gradients
2 x x 3 1 x 3 0
x 3 y 6
Given
must be negative reciprocals.. 1 2
k
3 2 k 6
P ro blem 2c(i)
Problem 3a (ii)
If the Series is Geometric then it must have a common ratio
Simultaneous Equations Elimination Method 6( x 3 y 6) Mutliply by 6
Find the ratio, using the recursive Formula: a1 an r
-6 -6 x 2 y 12
0.02 0.2 r r
6x +18y =36
0.02
-6x + 2y =12 Add vertically
0.2 r 0.1 Proof: 0.2 0.1 0.02
0 0..02 0.1 0.002
0. 0.02 0.1 0.0002
20y = 48 y = 2.4 Therefore x 3 y 6 x 3( 3(2.4) 6
x 6 7.2
P ro robl blem 2c(ii)
x 1.2
S S S
a1 1 r 0.2
1 0.1 0.2
0.9 2 S 10 9 10 S 2 10 10 9 2 S 9 Problem 5(a) y 3 4 x x , Po Point P (3, 6) 6) lies on the curve. 2
To find the equation of tangent. y - x 4 x 3 2
y ' 2 x 4 y '( '(3) 2(3) 4 y '(3) (3) 2 , Gr Gradien ient of tan tangent Equation of tangent line. y mx c y 2 x c 6 2(3) c
c 12
y 2 x 12 In the form of
ax by c 0
2 x y 12 0
The center of the Circle would be at the inter intersec section tion of both both lines lines , that that is at ( 1.2,2.4). Equation of the the line : 2 2 2 ( x - (-1.2)) ( y - 2.4) r
( x + 1.2) ( y - 2.4) r 2
2
2
Problem 6(b)
3
Evalu Evaluate ate
4 cos x 2 sin x dx 4c 0
3
3
0
0
4 cos x dx 2 sin xdx
4
3
3
0
0
cos x dx 2 sin x dx
4 sin x 2 cos x
4 sin x 2 cos x0
3
4 sin 2 cos 4 sin 0 2 cos 0 3 3
3 1 4 2 4(0) 2(1) 2 2 2 3 1 0 2 2 3 1
Problem 6(c)
Pr oblem oblem 5 (b) f ( x) 2x 9 x 24 x 7 3
2
f '( '( x) 6 x 18x 24
f "( "( x) 12 x 18
2
2
dx find the function of the curve. Remember the derivative
6 x 18x 24 0
6 x
24 x 6 x 24 0
6 x x 4 6 x 4 0
x 4 6 x 6 0
3
6x 6 0
x 4
and
x -1
5 2(2)3 2 C
5 16 2 C 5 14 C 5 14 C
f ( x) 2 x 9 x 24 x 7
f ( x) 2 x 9 x 24x 7
f (4) 2(4) 9(4) 24(4) 7
f ( 1) 2( 1) 9( 1) 24( 1) 7
f (4) 105
f (-1) 20
3
2
2
3
C 19
2
(ii)
Stationary Points location:
4, 105
20 1, 20
and
Find the area of the finite region by the curve , the x-axis,
f "(4) 12(4) 18
f "(-1 "(-1) - 30 suggest a maximum
Problem 6(a)
(4)4 (4)2 (3)4 (3)2 19(4) 19(3) 2 2 2 2
(4)4 (4)2 (3)4 (3)2 76 76 57 2 2 2 2
4
2
2
4
2 x dx
2
x x 41 4
2
44 44 21
Area of of the definite region region under the curve curve is 65 units .
2 3
256 16 81 9 76 76 57 2 2 2 2 128 8 76 40.5 4.5 57 65
x x 2 dx
x
2
4
f "(1) 12(1) 18
suggest a minimum
4
x x 3 2 x x 19 dx 19 x 2 2 3 3 4
f "( x) = 12x 18 0
Evaluate
f ( x) 2 x3 x 19, th the equation of the curve. the line x = 3 and the line x = 4. Integrate the function of curve.
Nature Nature of Points Points::
f "(4) 30
1 dx 2 x3 x C y 2 x x C
and
2
2
We can find the value of the constant by using point (x,y) that were given.
x - 4 0
3
,then we need to inte integr gra ate to 6x 2 1,the
takes us back to the original function.
3
dy
2
6 x
if
gives you the gradient f unction. Integrate the gradient function,
f '( x) 0.
At Stationa Stationary ry Point
(i )
4
2
44 42 2 4 2 2 4 1 4 1 256 16 16 4 1 4 1 4 48 0 48
Problem 7(c) (i) media median n n 1
Q2
2 51 1
2 Quar Quartil tile e:
Q1
Problem 7(a) (i) P (Ma (Math and Bio Biolo logy gy)) (ii) P (Biology only)
9 60
Q3
n 1
(b)
6 36 6
36
Prepared by:
51 1 4
3 n 1
60
Problem 7( b) (a)
2
4
11
1 6 1 6
Mr. Leovany López, B.Sc. Mathematics Sacred Heart College, San Ignacio, Cayo, C ayo, Belize
n 1
4
26th term = 71
13th term = 56
3 51 1 4
39th term = 79
