MATHS METHODS 3 & 4
TRIAL EXAMINATION 2
SOLUTIONS
2016
SECTION A – Multiple-choice answers
1. C 9. C 17. E
2. A 10. D 18. B
3. D 11. E 19. D
4. A 12. C 20. B
5. A 13. C
6. C 14. B
7. E 15. C
8. D 16. D
SECTI
...
MATHS METHODS 3 & 4
TRIAL EXAMINATION 2
SOLUTIONS
2016
SECTION A – Multiple-choice answers
1. C 9. C 17. E
2. A 10. D 18. B
3. D 11. E 19. D
4. A 12. C 20. B
5. A 13. C
6. C 14. B
7. E 15. C
8. D 16. D
SECTION A – Multiple-choice solutions
Question 1
period=π
n
where n=3 π
4
=π÷
3 π
4
=
43
The answer is C.
Question 2
Sketch the graph of y=(x−1)2 .
The range of y=(x−1)2 is y∈[0,∞) .
For the function f, the range is restricted to [1,∞) .
The graph of f could therefore be
So possible domains of f are x∈(−∞,0] or x∈[2,∞) or x∈(−∞,0]∪[2,∞) .
Only the first domain is offered.
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© THE HEFFERNAN GROUP 2016 Maths Methods 3 & 4 Trial Exam 2 solutions
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THE
GROUP
HEFFERNAN
1 2
1 ( 2 , 1 )
x
y
y ( x 1 ) 2
2 2
1 1 ( 2 , 1 ) 1 ( 2 , 1 )
x
O R O R
x x
y y y2
The answer is A.
©THE HEFFERNAN GROUP 2016 Maths Methods 3 & 4 Trial Exam 2 solutions3
Question 3
The graph of g is an upright quartic which touches the origin so we need a positive x2 term.
This eliminates options A, B and C.
Note that a must be a negative number since the point (a,0) lies to the left of the origin.
Intuitively, we think that we need an (x+a) factor but because a is a negative number, for
example –1, this factor would become (x−1) which is incorrect.
So we need an (x−a) factor which eliminates option E.
The answer is D.
Question 4
We have
f :[0 , a]→R , f (x )=cos(2(x+ π 3)) .
The inverse function f −1 exists if the function f is 1:1.
Sketch the graph of
y=cos(2(x+ π3)), x≥0.
Looking at this graph, we see that it is a 1:1 function for
x∈[0, π6] . (It’s actually a 1:1
function for smaller intervals such as
x∈[0,12 π ] etc but we want the largest interval
possible ie. we want the maximum value of a.)
So if the function f with rule
f (x )=cos(2(x+ π 3 )) is to have an inverse, then the largest
possible value of a is
π6
.
The answer is A.
Question 5
y=loge(ax), a>0
dy
dx
=
a
ax
=
1x
When x=1
a
, y=loge(a ×1 a )=loge(1)=0
©THE HEFFERNAN GROUP 2016 Maths Methods 3 & 4 Trial Exam 2 solutions
x
y
1
- 1
- 0 . 5
2
3
6
3
2
6
5
, 0
3
c o s 2
y x x
Equation of tangent is y−0=a(x−1a )
y=ax−1
y-intercept occurs when x=0 y=−1
.
The answer is A.4
At (1 a ,0), dy dx =11
a
=a
