SPTA Higher Unit 3
Practice NAB B
PA
1.
3
2
The diagram shows the curve with equation y = x - 5x + 2x + 8
The curve crosses the x-axis at P, Q and R. If Q is (2, 0). Determine the shaded area
4
2.
The line with equation y = 2x + 3 is a tangent to the curve with equation
y = x3 + 3x2 + 2x + 3 at A (0, 3),
as shown in the diagram
The line meets the curve again at B (-3, -3). Find the area enclosed by the line and
the curve
5
3.
Find
∫
+
dx , x ≠ 0
4
4.
Find
∫√
dx , x ≠ 0
2
2.1
2.2
5.
The diagram below shows two right-angled triangles with sides and angles as given
What is the value of sin(p + q)?
4
6.
Show that (3 + 4cosx)(3 - 4cosx) = 16sin2x - 7
3
7.
Solve sin2x + cosx = 0,
1
for 0 ≤ x ≤ 360
4
8.
Solve cos2x – 4sinx + 5 = 0,
for 0 ≤ x ≤ 360
9.
A circle has the equation x2 + y2 + 6x – 2y – 17 = 0
4
Find its centre and radius.
2
10. The diagram shows two congruent circles. One circle has centre the origin and
diameter 20 units.
Find the equation of the other circle which passes through the origin and whose
centre lies on the y-axis.
2
11. Determine algebraically if the line y = 2x + 3 is a tangent to the circle
(x – 8)2 + (y + 2)2 = 45.
3
--------------------End of Assessment----------------------
1
Question
1
Points of expected responses
1
knows to integrate and states
limits
Illustrative scheme
1
∫
2
integrates
2
3
substitute limits
3
(
4
process limits
4
10
NB
candidates who substitute without
integrating, only 1 is available
( )
)
units²
candidates who differentiate at 2, only 1 is
available
2
1
knows to integrate and states
limits
1
∫
2
∫ (
2
use ‘upper – lower’
3
integrate
3
4
substitute limits
4

process limits
NB
(
)
(or equivalent)
(0) –(
5
5
)
(
)
(
) )
or 6 or 6.75 units²
candidates who differentiate at 3,
only 1 and 2 are available
candidates who substitute without
integrating, only 1 and 2 are available
candidates who integrate ‘lower–upper’
lose 2 and also 5 unless final area is
expressed as A = 15.75
3
1
prepares to integrate
1
∫(
2
integrates first term correctly
2
=
3
integrates second term correctly
3
... –
4
includes constant of integration
4
)
+c
4
5
1
prepares to integrate
1
∫(
2
integrates correctly
2
= –2
1
process missing side
1
√5
2
expand
substitution
2
sin(p + q) and begin 
6
complete substitution
or
–
√
+c
= sinpcosq + cospsinq
=
3
+c
√
3
=
√
x
x
√
√
+…
+
x
√
√
4
process
4
= +
1
expand L.H.S.
1
9 – 16 cos2x
2
substitutes for cos2x
2
9 –16(1 – sin2x)
3
simplifies and completes
3
9 – 16 + 16sin2x = 16sin2x – 7 = RHS
#2.1
know to use identities
#2.1
valid strategy
√
=
√
NB
do not award 2 if candidate omits brackets,
ie 16 – 25 + 25cos2x
Alternatively use R.H.S.
7
1
starts to solve
1
2sinxcosx + cosx = 0
2
factorises
2
cosx (2sinx + 1) = 0
3
solves for one factor
3
cosx = 0 ⟹ 90° , 270°
4
solves for second factor
4
2sinx + 1= 0 => sinx = – => x = 210°, 330°
NB 3 and 4 can be marked horizontally or vertically
8
1
starts to solve
1
1 – 2sin2x – 4sinx + 5 = 0
2
factorises
2
2(sinx – 1)(sinx + 3) = 0
3
solves for one factor
3
sinx = –3 ⟹ no solution
4
solves for second factor
4
sinx – 1= 0 => x = 90°
9
10
11
1
finds centre
1
(–3, 1)
2
finds radius
2
√
1
interprets congruent
1
r = 10 units, C(0, 10)
2
correct substitution
2
(x – 0)2 + (y – 10)2 = 100
1
set
2
simplifies
2
5x2 + 4x + 44 = 0
3
solves
3
#2.1
interprets solution
No solution (check with discriminant)
b2 – 4ac ⟹ 42 – 4 x 5 x 44
16 – 880 = –864
up
intersection
equation 1
or
√
(x – 8)2 + (2x + 3 + 2)2 = 45
#2.1 zero points of contact or a negative
discriminant implies line is not a tangent to the curve
Total of 37marks plus 2 x #2.1