Qwertyuiopasdfghjklzxcvbnmqwertyu iopasdfghjklzxcvbnmqwertyuiopasdfg hjklzxcvmqwertyuiopasdfghjklzxcvbn SEKOLAH MENENGAH TEKNIK IPOH mqwertyuiopasdfghjklzxcvbnmqwerty uiopasdfghjklzxcvbnmqwertyuiop...
past paper sample for csec add mathsFull description
Add Math pass yearFull description
This is the handout that i use for class discussion.
CXC CSEC ADD MATHS 2013 P2 SOLVED
Topic 9
differentiation
y
5
(3 x +2 )
=
let u = 3x + 2
y
⇒
=
dy du
dy
=
=
dx
(u ) 5 5u
dy
du
4
×
du
and
du dx
4
= 5u × 3 15u
4
=
=
15(3 x + 2) 4
dx
=
3
y y
y
=
(6 x +7 )
4
3
= − (3 x
4)
3x =1 +
5
2
y
x
=
u
let
du dx
dy dx
3
= x = 3 x
( 3 x +2 )
3
2
and
2
v = (3x + 2) dv dx
2
1
= 2(3 x + 2) (3) = 6(3 x + 2)
= x 3 [6(3 x + 2)] + [(3 x + 2) 2 (3 x 2 )] 2 x (3 x + 2) [ 2 x (1) +1 (1)(3 x + 2)] =3 =
2
3 x (3 x + 2) [2 x + 3 x + 2]
2 x (3 x + 2)(5 x + 2) = 3
y
=
4 x(5 x − 1) 4
y = x (1 + 2 x)
2
6
y = (6 x − 1)(3 x − 4)
3
y =
u
let
=
4 x
4 x 5 x + 3 and
dv
du =4 dx
dy dx
dx
(5 x + 3)( 4)
=
=5
(4 x)(5)
(5 x + 3) 2
=
20 x +12 −20 x (5 x +3)
=
v = 5x + 3
12 (5 x + 3)
2
2
(a)
(b )
3 x + 2
( d )
4 x + 7
4
x + 6 2
6 x 4 x
3 x + 8
3
−
(e)
x −1 3 x + 4
2
x (c ) 3 x + 5
2
( f )
4 x
x + 4
mtan gent =
dy
( x1 , y1 )
dx
y − y1
=
mt × mn = −1
mt ( x − x1 ) ( x1 , y1 )
y y1 −
=
mn( x x1 ) −
2 y = 4 x − 3 x at point (1,1).
dy dx
= 8x − 3
When x = 1,
dy dx
= 8(1) − 3 =
5
Gradient of tangent = 5
y = 2 x 2 − 1 at point (2,7). dy dx
=
4 x
When x = 2,
dy dx
= 4( 2) = 8
y − 7 = 8( x − 2) y − 7 = 8x − 16
y = 8 x − 9
y dy dx
2 x x = 6 −
at point (2,8).
= 6 − 2 x
y − y1
When x = 2,
dy mt = = 6 − 2( 2) = 2 dx
mt × mn = −1 2 × m n = −1
mn = −
1 2
=
mn ( x − x1 )
y − 8 = −
1 2
( x − 2)
2( y − 8) = − 1( x − 2)
2 y − 16 = − x + 2
2 y = − x + 18
y
=
2x
3
+
2
8
y = x − 6 x y
=
4 x
3
+
y
=
x
3
−
9
2
y = x + 6 x y = (2 x − 1)( x + 1)
y
y y
=
4
x
=
=
x x
3
2
+
+
2
1
x
dy dx
2
= 4(3 x ) − 5( 2 x) + 7(1) =
d 2 y dx
2
12 x
2
−
10 x + 7
= 12( 2 x ) − 10(1) = 24 x − 10
−3 x 3
= 1
= 3 x
=
3 x
−2
1
−
+2
−
2
2.
A 60m wire is bent into the shape of a rectangle. The length of one of its sides is x cm and its area is A cm². (ii)Show that A = 30x - x². (ii) Hence, find the value of x such that the area of the rectangle is a maximum. Determine the maximum area.
A 60m wire is bent into the shape of a rectangle. The length of one of its sides is x cm and its area is A cm². (ii) Show that A = 30x - x².
60
−2 x 2
=
30 − x
x
Area of rectangle = length X width A = x ( 30 - x ) = 30x -x² (shown)
A 60m wire is bent into the shape of a rectangle. The length of one of its sides is x cm and its area is A cm². (ii) Hence, find the value of x such that the area of the rectangle is a maximum. Determine the maximum area.
A = 30x -x² For maximum area
dA dx
=
0
dA dx
= 30 − 2 x = 0
x = 15 Maximum area occurred when x = 15
A = 30(15) - 15² = 225 cm²
3
Diagram shows a semicircle with centre O. The diameter PQ can be adjusted and R moves on the circumference that PR + QR = 10 cm. Given that QR = x cm and A is the area of triangle PQR. Find the expression dA of in terms of x. Hence, find the dx maximum value of the area of triangle.
3 Diagram shows a semicircle with centre O. The diameter PQ can be adjusted and R moves on the circumference that PR + QR = 10 cm. Given that QR = x dA cm and A is the area of triangle PQR. Find the expression of dx in terms of x. Hence, find the maximum value of the area of triangle.
A =
Area of triangle
10 - x
x
A A
PR + QR = 10 PR + x PR
= 10 = 10 - x
dx For maximum area
dA dx
=
0
2
x (10 −x )
=
=
dA
base× height
=
1 2 1 2
2 2
(10 x − x ) (10 − 2 x) = 5 − x
5− x = 0 x = 5
Maximum value of the area of triangle
A
=
1 2
[10(5) −5 2 ]= 12.5
4
Diagram shows a cuboid ABCDEFGH with a square base EFGH. The volume of the cuboid is 27cm². (ii)If A is the total area of the cuboid, show that A = 2 x
2
+
108
x
(ii) Find the value of x so that the total surface area is minimum.
5
Diagram shows a composite solid which consists of a cone and a cylinder of radius r cm. Given the slant height of the cone is 3r and the volume of cylinder is 32π cm³. (ii)Prove that the total surface area A of the solid is given by A
=
2
4π r
+
64π r
(iv) calculate the minimum value of the surface area.
=0.2
x=8 =?
=
3(8) 2
= 192
= 192 x 0.2 = 38.4 cm³sˉ¹
r=7
= 0.5
∂ A ∂r
= 2π (7)
= 14π
=
dA dr
=
14π × 0.5
= 7π
If y = f(x), the new value of y,
yb = y a + δ y
yb = y a + δ y
Find the turning point for each of the curve below. Determine whether the turning point is a maximum or a minimum point.
Find the turning point for each of the curve below. Determine whether the turning point is a maximum or a minimum point.
