Relationships and Calculus Unit Test Revision
R&C Assessment Standard 1.1
1. A function f is defined by the formula 𝑓(𝑥) = 3𝑥 3 − 14𝑥 2 + 13𝑥 + 6
where x is a real number
(a) Show that x - 2 is a factor of f ( x ).
(2)
(b) Hence factorise f ( x) fully.
(2)
(c) Solve f ( x)  0.
(2)
2. A function f is defined by the formula 𝑓(𝑥) = 2𝑥 3 − 5𝑥 2 − 22𝑥 − 15
where x is a real number
(a) Show that x + 1 is a factor of f ( x ).
(2)
(b) Hence factorise f ( x) fully.
(2)
(c) Solve f ( x)  0.
(2)
3. A function f is defined by the formula 𝑓(𝑥) = 𝑥 3 − 4𝑥 2 + 𝑥 + 6
where x is a real number
4.
(a) Show that x + 1 is a factor of f ( x ). Hence factorise f ( x) fully.
(4)
(b) Hence solve x3 - 4x2 + 5x – 2 = 4x - 8
(#2.1,1)
Solve the cubic equation f ( x)  0 given the following:
 when f ( x) is divided by x – 5 , the remainder is zero
 when the graph of y  f ( x) is drawn, it passes through the point (-4,0)
 x + 1 is a factor of f ( x) .
5.
Solve the cubic equation f ( x)  0 given the following:
 when f ( x) is divided by x + 3, the remainder is zero
 when the graph of y  f ( x) is drawn, it passes through the point (-1,0)
 x + 2 is a factor of f ( x) .
6.
(#2.2)
(#2.2)
The graph of the function 𝑓(𝑥) = 𝑥 2 − 5𝑥 + (𝑘 + 6) does not touch or cross the
x-axis.
What are the possible values for k ?
(#2.1, 2)
7.
The graph of the function 𝑓(𝑥) = 𝑘𝑥 2 + 2𝑥 − 1 does not touch or cross the
x-axis.
What are the possible values for k ?
(#2.1, 2)
R&C Assessment Standard 1.2
1.
Solve 2 cos 2xo = 1 in the interval 0 ≤ x ≤ 180
(3)
2.
Solve 2 sin 2xo = √3 in the interval 0 ≤ x ≤ 180
(3)
3.
Solve cos 2to – cos to = 0 in the interval 0 ≤ t ≤ 180
(4)
4.
Solve sin 2to – cos to = 0 in the interval 0 ≤ t ≤ 180
(4)
5.
Given 3 cos xo - sin xo = √10 cos (x – 18.4) o
solve 3 cos xo - sin xo = 2.4 for 0 < x < 90
(3)
Given 5 sin xo + cos xo = √26 cos (x – 78.7) o
solve 5 sin xo + cos xo = 1.3 for 0 < x < 360
(3)
6
7
Solve sin xo cos 20o + cos xo sin 20o = 0.75 for 0  x  180.
(#2.1, 2)
R&C Assessment Standard 1.3
(Exercise 6F 20-27)
1.
Find f   x  , given that 𝑓(𝑥) = 5√𝑥 +
2. Find f   x  , given that 𝑓(𝑥) = 8√𝑥 +
3
3. 𝑓(𝑥) = √𝑥 − 2𝑥 5 +
4
√𝑥
6
𝑥3
3
𝑥5
, 𝑥 > 0.
, 𝑥 > 0.
- 3, 𝑥 ≥ 0.
(3)
(3)
(3)
Find and expression for f   x 
(Exercise 6R)
4. A bowler throws a cricket ball vertically upwards. The height (in metres) of the ball t
seconds after it is thrown can be represented by the formula h = 10t – 2t2. The velocity of
dh
the ball at time t is given by v 
.
dt
(2)
a) Find the velocity of the cricket ball after three seconds.
b) Explain your answer in terms of the ball’s movement.
(#2.2)
5. A bowler throws a cricket ball vertically upwards. The height (in metres) of the ball t
seconds after it is thrown can be represented by the formula h = 8t – 3t2. The velocity of the
dh
ball at time t is given by v 
.
dt
(2)
a) Find the velocity of the cricket ball after one second.
(#2.2)
b) Explain your answer in terms of the ball’s movement.
6. The vertical displacement of a golf ball can be modelled by s = 20t -5t2, where s metres is
𝑑𝑠
the height of the ball t seconds after being struck. The velocity of the ball is given by 𝑑𝑡 .
Find the velocity of the golf ball after 3 seconds.
(2)
(Exercise 14B)
7. Differentiate the function f(x) = 8cosx with respect to x.
(1)
8. Differentiate the function f(x) = 7sinx with respect to x.
(1)
9. Differentiate the function f(x) = 4sinx with respect to x.
(1)
(Exercise 6J)
10. A curve has equation y = 2x2 - 4x +3.
Find the equation of the tangent to the curve at the point where x = -1.
(4)
11. A curve has equation y = 3x2 - 5x +2.
Find the equation of the tangent to the curve at the point where x = 2.
(4)
12. A curve has equation y = 2x2 -14x +53.
Find the equation of the tangent to the curve at the point where x = 8.
(4)
R&C Assessment Standard 1.4
(Exercise 9H Q3, 9I)
1
1. Find ∫(5𝑥 4 +
1
2
𝑥4
2. Find ∫(4𝑥 3 −
) 𝑑𝑥, 𝑥 > 0.
2
𝑥3
) 𝑑𝑥, 𝑥 > 0.
(4)
(4)
(Exercise 9Q, 14J)
3. ℎ′ (𝑥) = (𝑥 − 6)-5, find ℎ (𝑥), 𝑥 ≠ 6.
(2)
ℎ′ (𝑥) = (𝑥 − 2)-4, find ℎ (𝑥), 𝑥 ≠ 2.
(2)
4.
(Exercise 14C)
5. Find ∫ 5𝑠𝑖𝑛𝑥 𝑑𝑥.
6.
Find ∫ 3𝑐𝑜𝑠𝑥 𝑑𝑥
(1)
(1)
(Exercise 14J Q1,4)
3
4
4
3
7.
Find ∫−1( 𝑥 − 2) dx
8.
Find ∫−1( 𝑥 + 2) dx
(3)
(3)
Answers
R&C Assessment Standard 1.1
1.
2.
3
1
know to use x = 2 and state conclusion
2
know to divide by (x – 2)
(b)
(x - 2)(3x2 - 8x - 3)
(x - 2)(3x + 1)(x - 3)
3
obtain quadratic factor

complete factorisation
(c) x =3, 2, - ⅓
5
interpret and solve equation
(a) Remainder = 0 so (x+1) is a factor
1
2
know to use x = 1 and state conclusion
know to divide by (x – 1)
(b)
(x +1)(2x2 - 7x - 15)
(x + 1)(2x + 3)(x - 5)
(c) x = -1, - 3/2 , 5
3
4
5
obtain quadratic factor
complete factorisation
interpret and solve equation
(a) Remainder = 0 so (x - 2) is a factor
4
(a) Remainder = 0 so (x +1) is a factor •1
(x3 - 4x2 + x + 6) ÷ (x + 1)
•2
•3
(x + 1)(x -2)(x – 3) stated explicitly
•4
Know to use x  1 and state conclusion
Know to divide by x + 1
Obtain quadratic factor
Factorise completely
(b) x3 - 4x2 + x + 6 = 0 stated explicitly #2.1
x = -1, x=2, x=3
•5
State solution
4.
x = 5, x = -4, x = -1
#2.2 communicate solution from the context
5.
x = -3, x = -1, x = -2
#2.2 communicate solution from the context
6.
b2 – 4ac< 0
#2.1 know to, and use, discriminant correctly
a = 1 b = -5
c=k+6
2
(-5) – 4 x 1 x (k + 6) < 0
4k > 1
k > 0.25
1
b2 – 4ac< 0
#2.1 know to, and use, discriminant correctly
a=kb=2
c = -1
2
(2) – 4 x k x (-1) < 0
4k < -4
k < -1
1
7.
solve inequality
solve inequality
R&C Assessment Standard 1.2
1
2
start to solve
solve for 2x
3
solve for x
2. 2x = sin-1(√3/2)
2x = 60o
1
2
start to solve
solve for 2x
x = 30o and 60o
3
solve for x
3. Using cos2t=2cos2t-1 and substituting
2cos2t – cost – 1 = 0
(2cost + 1)(cost – 1)=0
cost = -1/2
cost = 1
o
o
t = 0 , 120
1
2
3
4
start to solve
factorise
solve for one factor
solve for second factor
4. Using sin2t=2 sint cost and substituting
2 sint cost – cost = 0
cost(2 sint – 1)=0
cost = 0
sint = 1/2
t = 30o, 90o, 150o
1
2
3
4
start to solve
factorise
solve for one factor
solve for second factor
5. cos(x – 18.4)o= 2.4/√10
x – 18.4o = 40.6
1
2
Form equation and start to solve
solve for x-18.4o
3
obtain angles
1
2
Form equation and start to solve
solve for x-78.7o
3
obtain angles
1. 2x = cos-1(1/2)
2x = 60o
x = 30o and 150o
x = 59o
6. cos(x – 78.7)o= 1.3/√26
x – 78.7o = 75.2o
x = 153.7o
7.
sin (x + 20)=0.75
48.6oand 131.4o
28.6o and 111.4o
#
2.1 Express in standard form sin(x+a)o and equate
1
Solve equation for one value of x+20
2

Process solutions for x
R&C Assessment Standard 1.3
1
1.
𝑓(𝑥) = 5𝑥 2 + 6x-3
5
𝑓 ′ (𝑥) = x-1/2 + …
1 prepare to differentiate
𝑓 ′ (𝑥) = … -18x-4
3 differentiates second term correctly
2 differentiates first term correctly
2
1
1 prepare to differentiate
2 differentiates first term correctly
3 differentiates second term correctly
2. 𝑓(𝑥) = 8𝑥 2 + 3x-5
𝑓 ′ (𝑥) = 4 x-1/2 + …
𝑓 ′ (𝑥) = … -15x-6
1
−1
3. 𝑓(𝑥) = 𝑥 3 - 2x5 + 4𝑥 2 - 3
1
𝑓 ′ (𝑥) = x-2/3 - 10x4 …
3
𝑓 ′ (𝑥) = … - 2x-3/2
4.
10 - 4t
v = -2 m/s
ball is moving downwards
5.
8 - 6t
v = 2 m/s
ball is moving upwards
6.
20 – 10t
v = -10 m/s
1 prepare to differentiate
2 differentiates two terms correctly
3 differentiates third term correctly
1 differentiates
2 substitutes
#2.2 explaining a solution
1 differentiates
2 substitutes
#2.2 explaining a solution
1 differentiates
2 substitutes
7.
d(8cosx) = -8sinx
dy
1 differentiates correctly
8.
d(7sinx) = 7cosx
dy
1 differentiates correctly
9.
d(4sinx) = 4cosx
dy
1 differentiates correctly
𝑑𝑦
1 differentiates correctly
10.
11.
𝑑𝑥
= 4𝑥 − 4
m = -8
y=9
y -9 = -8(x + 1)
2 finds gradient
3 finds y
4 equation of tangent
𝑑𝑦
1 differentiates correctly
2 finds gradient
3 finds y
4 equation of tangent
= 6𝑥 − 5
m=7
y=4
y -4 = 7(x-2)
𝑑𝑥
𝑑𝑦
= 4𝑥 − 14
m = 18
y=5
y - 69 = 18(x -8)
12.
𝑑𝑥
1 differentiates correctly
2 finds gradient
3 finds y
4 equation of tangent
R&C Assessment Standard 1.4
1
1. ∫( 5𝑥 4 + 2𝑥 −4 ) 𝑑𝑥
5
1 prepares to integrate
2 integrates first term correctly
3 integrates second term correctly
4 includes constant of integration
= 4𝑥 4 …
2
…. - 3 𝑥 −3
+c
1
2. ∫( 4𝑥 3 − 2𝑥 −3 ) 𝑑𝑥
4
1 prepares to integrate
2 integrates first term correctly
3 integrates second term correctly
4 includes constant of integration
= 3𝑥 3 …
… . + 𝑥 −2
+c
3. (𝑥 − 6)−4
1
… x - 4 …+ 𝑐
1 start to integrate
2 complete integration
4. (𝑥 − 2)−3
1
… x - 3 …+ 𝑐
1 start to integrate
2 complete integration
5. -5cosx + c
1 integrates correctly
6. 3sinx + c
1 integrates correctly
7.
(𝑥−2)5
5
1
5
(3 − 2)5 −
1 integrates
1
5
(−1 − 2)5
3 evaluate definite integral
48.8
8.
(𝑥+2)4
4
1
4
(4 + 2)4 −
323.75
2substitutes limits
1 integrates
1
4
(−1 + 2)4
2substitutes limits
3 evaluate definite integral