This document contains solutions to mathematics questions from the 2010 HSC exam in Australia. Question 1 involves solving equations, inequalities and finding derivatives. Question 2 involves finding derivatives of trigonometric functions. Question 3 involves vectors, gradients and parallel lines. Question 4 involves arithmetic progressions, integrals and area under curves. Question 5 involves volumes, surface areas, maxima and minima. Question 6 involves factorizing polynomials, discriminants and finding angles and areas of figures.

1. http://www.maths.net.au/ 2010 Mathematics HSC Solutions
2010 Mathematics HSC Solutions
Question 1 (b) x 2  x  12  0
(a) ( x  4)( x  3)  0
x2  4x  0
y
x( x  4)  0
x  0 or x  4  0
–3 4
x4 x
(b) 1 52 52
  3  x  4
52 52 54
 2 5 (c) y  ln  3x 
dy 3
a  2 and b  1 
dx 3x
(c) ( x  1) 2  ( y  2) 2  25 1

x
(d) 2 x  3  9 1
at x  2, m 
2
2x  3  9 or (2 x  3)  9
2 3
(d) (i)  5 x  1 dx    5  5 x  1 2 dx
1
2x  6 2 x  3  9 
 3 5 2
x3 2 x  12 2
 5 x  1  c
3

x  6 15
d 2 x 1 2x
(e) x tan x  tan x (2 x)  x 2 (sec 2 x) (ii) 
 dx   dx
dx  4 x 2
2  4  x2
 x(2 tan x  x sec2 x) 1
 ln  4  x 2   c
2
a
(f) s 
(e) 6
1 r
1
  x  k  dx  30
0
  x2 
6
1 1
3
 2  kx   30
3  0

2 62
 6k  30
2
(g) x  8 6k  12
Question 2 k 2
Question 3
d cos x x( sin x)  cos x(1)
(a)   2  12 4  6 
dx x x2 (a) (i) M   , 
 x sin x  cos x  2 2 
   5, 1
x2
1

2. http://www.maths.net.au/ 2010 Mathematics HSC Solutions
86 3 1
(ii) mBC   ln x dx   0  2 ln 2  ln 3
6  12 1 2
1  1.24 (2 d.p.)

3
(iii) The approximation using the
2 1 trapezoidal rule is less than the
(iii) mMN 
25 actual value of the integral, because
1 the shaded area of the trapeziums,

3 is less than the actual area below
since mBC  mMN , the curve.
BC || MN
y
Corresponding angles on parallel 2
lines are equal, so
ACB  ANM 1
ABC  AMN
 ABC ||| AMN (equiangular)
1 2 3 4 x
1
(iv) y  2    x  2
3 -1
3y  6  x  2
Question 4
x  3y  8  0
(a) (i) Forms an AP, a  1 , d  0.75
Tn  1  (n  1)  0.75
12  6    6  8
2 2
(v) BC 
Tn  0.25  0.75n
 2 10
T9  0.25  0.75  9
1 T9  7 km
(vi) Area  bh
2 Susannah runs 7 km in the 9th week
1 (ii) Tn  0.25  0.75n
44  2 10h
2 10  0.25  0.75n
22 10 n  13
h
5 In the 13th week.
(b) (i) y
(iii) S 26  26
2  2 1   26  1  0.75
3  269.75 km
2 2

1 (b) Area    e 2 x  e x  dx
0
-1 1 2 3 4 5 x 2
-2  e2 x 
  e x 
-3  2 0
-4
-5  e4   e0 
   e 2     e0 
(ii) x 1 2 3 2  2 
0 ln(2) ln(3) 2
f(x) e  2e  3
4

2
2

3. http://www.maths.net.au/ 2010 Mathematics HSC Solutions
(c) (i) P (2 mint) 
4 3 4 r 3  20  0

12 11  r3  5  0
1
 5
11 r 3

1 1 1
(ii) P (2same)    d2A 60
11 11 11  4  3
2
3 dr r

11 5
when r  3 ,

3 2
d A
(iii) P (2 different)  1   16  0,  c.c.up,
11 dr 2
8 5
  local minimum at r  3
11 
(d) f  x   f   x   1  e x 1  e  x 
 1  e x  e x  1 1 1 sin x
x (b) (i) sec2 x  sec x tan x  
 2e e x
cos 2 x cos x cos x
f  x   f   x   1  e x   1  e  x  
1  sin x
cos 2 x
 2  e x  e x
1  sin x
Question 5 (ii) sec2 x  sec x tan x 
cos 2 x
1  sin x
(a) (i) V   r 2h 
1  sin 2 x
10   r 2 h 1  sin x

10 1  sin x 1  sin x 
h 2
r 1

1  sin x
A  2 r 2  2 rh
 10 
 2 r 2  2 r  2  
1
(iii) I  
4
r   dx
0 1  sin x
20
 2 r 2 

4
r    sec 2 x  sec x tan x  dx
0
  tan x  sec x 0

(ii) dA 20 4
 4 r  2
dr r     
dA   tan  sec    tan 0  sec 0 
let  0 to find stationary points  4 4  
dr
 1 2 1
 2
3

4. http://www.maths.net.au/ 2010 Mathematics HSC Solutions
1 1
(iii)
A1   dx
(c) y

a x
1   ln x a
1
8
1  ln 1  ln  a 
–2
ln  a   1 x
1
a
e (b) (i) l  r
9  5
b
1
A2   dx
   1.8
1 x
1   ln x 1
b (ii) In OPT and OQT
OP = OQ (equal radii of 5 cm)
1  ln  b   ln 1 OPT = OQT (both right angles)
ln  b   1 OT is a common line
OPT  OQT (RHS)
be
(iii) POT  1 POQ
Question 6 2
 0.9
(a) (i) f ( x)  ( x  2)( x  4)
2
PT
tan(0.9) 
f ( x)  x3  2 x 2  4 x  8 5
PT  5 tan(0.9)
f ( x)  3 x 2  4 x  4 PT  6.3 cm (1 d.p.)
 
(iv) PTQ    1.8  2
Consider the discriminate, 2 2
  42  4(3)(4) (angle sum of a quadrilateral is 2 )
 32
PTQ  1.34
Therefore there are no zeros, and 1
Area  (6.3) 2 sin(1.34)
hence, no stationary points. (the 2
derivative function is positive 1 
definite)   (5) 2 (1.8  sin(1.8)) 
2 
(ii) f ( x )  6 x  4  9 cm 2
Question 7
The graph is concave down when
6x  4  0
(a) (i) x   4 cos 2t dt

2
x
3  2sin 2t  c
The graph is concave up when
2 when t = 0, x  1 ,

x 1  2sin 2(0)  c
3
c 1
 x  2sin 2t  1

4

5. http://www.maths.net.au/ 2010 Mathematics HSC Solutions
(ii) at x  0
 1 
0  2sin 2t  1  T is the point  , 2  .
2 
1 mBT  4
sin 2t  
2 Eqn BT: y  4  4( x  2)
 y  4x  4
2t  
6
 13 Since this line is not vertical, if there
t ,
12 12 is one simultaneous solution between
this line and the parabola, it is a
Therefore, the first time it will be at tangent. So, sub y  4 x  4 into
13 y  x2
rest is at t =  3.4 s
12 4 x  4  x2
(iii) x  2sin 2t  1 dt x2  4 x  4  0

 x  2
2
0
  cos 2t  t  c
x2
at t = 0, x = 0  BT is a tangent to the parabola
0   cos 2(0)  0  c
c 1
x   cos 2t  t  1
dy Question 8
(b) (i)  2x
dx
at x = –1, m = –2 (a) P  Ae kt
P  102e kt
y  1  2( x  1)
when t = 75, P = 200 000 000
2x  y 1  0 200000000  102e 75 k
k  0.22
(ii) M   , 
1 5
 
2 2 P  102e0.22t
mAB  1
P  102e0.22(100)
so, to find the x-value on the curve,
where the tangent is 1, let 2x = 1. P  539 311 817 787
1 1 P  539 billion
Therefore the point C is  ,  .
2 4 (b) P ( HH )  0.36
Since the x-values of M and C are P ( H )  0.6
the same, then the line MC will be P (T )  0.4
vertical.
P (TT )  0.16
x–coordinate of T is 0.5. (c) (i) A  4 (amplitude)
(iii)
2x  y 1  0 2
(ii) T 
1 b
2  y 1  0 2
2 
y  2 b
b2
5

6. http://www.maths.net.au/ 2010 Mathematics HSC Solutions
(iii) y (ii)1 A1  P (1  0.005)1  2000
 P (1.005)1  2000
4
3 A2  A1 (1.005)1  2000
2   P (1.005)1  2000  (1.005)1  2000
 
1
 P (1.005) 2  2000(1  1.005)
x A3  A2 (1.005)1  2000
-1  
-2 2   P (1.005) 2  2000(1  1.005)  (1.005)1  2000
 
-3
 P (1.005)3  2000(1  1.005  1.0052 )
-4

(d) f  x   x3  3 x 2  kx  8 An  P (1.005) n  2000(1  1.005    1.005n 1 )
f   x   3x 2  6 x  k  P (1.005n )  2000 
1(1.005n  1)
1.005  1
 P (1.005n )  400 000  (1.005n  1)
For an increasing function f   x   0 ,  P (1.005n )  400 000 1.005n  400 000
 ( P  400 000) 1.005n  400 000
i.e. 3 x 2  6 x  k  0
Consider the graph of y  3 x 2  6 x with 2 An  ( P  400 000) 1.005n  400 000
0  (232 175.55  400 000) 1.005n  400 000
x-intercepts at 0 and 2. Vertex at x = 1, 400 000
1.005n 
y = –3.  if k  3 , f   x  is positive 167824.45
n  log1.005 (2.38)
definite and hence f  x  is an log10 2.38
n
log10 1.005
increasing function.
n  174.1
Question 9
(a) (i) A1  500(1  0.005) 240 Thus there will be money in the
account for the next 175 months
A2  500(1  0.005) 239
. (b) (i) 0 x2
.
. (ii) The maximum occurs at x = 2,
2
A240  500(1  0.005)1
 f   x  dx  4
0
f  2  f  0  4
A  A1  A2    A240
f  2  4
 500(1.005  1.0052  
The maximum value is f  x   4
 1.005239  1.005240 )
1.005(1.005240  1) (iii) f  6   f  4 
 500
1.005  1 4
 $232 175.55  f   x  dx  4
2
f  4   f  2   4
f  4   4  4
f  4  0
The gradient is –3, so f  6   6
6

7. http://www.maths.net.au/ 2010 Mathematics HSC Solutions
(iv) 4
y  a (1  2 cos  )
y  a (1  2(1))
2 4 6 y  3a
OA
(b) (i) sin  
r
OA  r sin 
–6 r
V  y 2 dx
r sin 
Question 10
r  x 2  dx
r
 2
r sin 
(a) (i)In ACD, DAC and DCA =  x3 
r
  r 2 x  
90  1  ( sum of )
2  3  r sin 
 r3   r 3 sin 3  
CDB = 180   (suppl. angles)    r 3     r 3 sin   
 3  3 
DBC = 90  1  (ABC is isosc.)
2
r 3
DCB = 90  3  ( sum of )
2 
3
 2  3sin   sin 3  
ACB = DCB + DCA = 
(ii) 1 Initial depth = r. So, find  , to give
In ABC and ACD, depth 1 r. From the diagram,
ACB = ADC (both  )
2
r
DAC = DBC (both 90  1  )
2 OA  r sin  
2
 ABC ||| ACD (equiangular) 1
sin  
AD DC a 2
also note     30
AC CB x
(ii) orresponding sides of similar
C
triangles are in the same proportion.  r3 
3 1
2  
AD AC 3  2 8
  2 Fraction 
AC AB 2 r 3
a x 3

x a y 5

a (a  y )  x 2 16
x 2  a 2  ay
(iii n ACD, by the cosine rule
I
x 2  a 2  a 2  2a 2 cos 
a 2  ay  a 2  a 2  2a 2 cos 
ay  a 2  2a 2 cos 
y  a  2a cos 
y  a (1  2 cos  )
(ivTo get the maximum value of y, cos 
must take its minimum value, of –1.
7