2010 Mathematics HSC Solutions

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2010 Mathematics HSC Solutions

2010 Mathematics HSC Solutions  Question 1

(a)

(b)

(b)

( x  4)( x  3)  0

x2  4x  0 x( x  4)  0 x  0 or x  4  0 x4

y

x

3  x  4

 2 5

(c) y  ln  3x  dy 3  dx 3x 1  x

and b  1

(c) ( x  1) 2  ( y  2) 2  25 (d) 2 x  3  9

2x  3  9 2x  6 x3

at x  2, m 

or

(2 x  3)  9

d 2 x tan x  tan x (2 x)  x 2 (sec 2 x) dx  x(2 tan x  x sec2 x)

x 1 2x dx   dx (ii)    2  4 x 2  4  x2 1  ln  4  x 2   c 2

a 1 r 1  1  13 

1 2

1 2 3 (d) (i)    5  5 x  1 2 dx  5 x  1 dx   3 5 2 2 3   5 x  1  c 15

2 x  3  9 2 x  12 x  6

(f) s 

(e)

6

  x  k  dx  30 0

6

 x2   2  kx   30  0

3 2

62  6k  30 2 6k  12 k 2

(g) x  8 Question 2

(a)

4

–3

1 52 52   54 52 52

a  2

(e)

x 2  x  12  0

Question 3

d cos x x( sin x)  cos x(1)  dx x x2  x sin x  cos x  x2

 2  12 4  6  , (a) (i) M    2   2   5, 1

1

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2010 Mathematics HSC Solutions

86 6  12 1  3



(ii) mBC 

3

1

1  0  2 ln 2  ln 3 2  1.24 (2 d.p.)

ln x dx 

(iii) The approximation using the trapezoidal rule is less than the actual value of the integral, because the shaded area of the trapeziums, is less than the actual area below the curve.

2 1 (iii) mMN  25 1  3 since mBC  mMN , BC || MN

y 2

Corresponding angles on parallel lines are equal, so ACB  ANM ABC  AMN  ABC ||| AMN (equiangular)

(iv)

1

1

1 y  2    x  2 3 3y  6  x  2

(v) BC 

12  6    6  8

(a) (i) Forms an AP, a  1 , d  0.75 Tn  1  (n  1)  0.75

2

Tn  0.25  0.75n T9  0.25  0.75  9 T9  7 km Susannah runs 7 km in the 9th week

1 (vi) Area  bh 2 1 44  2 10h 2 22 10 h 5

(ii) Tn  0.25  0.75n 10  0.25  0.75n n  13 In the 13th week.

(iii) S 26 

y

(ii)

x f(x)

26 2

 2 1   26  1  0.75

 269.75 km

3 2 1 -1 -2 -3 -4 -5

2

1

2

1 0

4 x

Question 4

 2 10

(b) (i)

3

-1

x  3y  8  0 2

2

3

2 ln(2)

4

 (b) Area    e 2 x  e x  dx 0

5 x

2

 e2 x    e x   2 0

 e4   e0     e 2     e0  2  2  4 2 e  2e  3  2

3 ln(3)

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2010 Mathematics HSC Solutions

4 r 3  20  0

4 3  12 11 1  11

(c) (i) P (2 mint) 

 r3  5  0 r

1 1 1   11 11 11 3  11

(ii) P (2same) 

(iii) P (2 different)  1  

3

5



d2A 60  4  3 2 dr r 5 when r  3 ,



3 11

2

d A  16  0,  c.c.up, dr 2

8 11

 local minimum at r 

3

5



(d) f  x   f   x   1  e x 1  e  x   1  e x  e x  1

1 1 sin x  cos 2 x cos x cos x 1  sin x  cos 2 x

(b) (i) sec2 x  sec x tan x 

x

 2e e f  x   f   x   1  e x   1  e  x  x

 2  e x  e x

1  sin x cos 2 x 1  sin x  1  sin 2 x 1  sin x  1  sin x 1  sin x 

(ii) sec2 x  sec x tan x 

Question 5

(a) (i)

V   r 2h 10   r 2 h 10 h 2 r



A  2 r 2  2 rh  10   2 r 2  2 r  2  r  20  2 r 2  r

1 1  sin x



4 1 dx (iii) I    0 1  sin x



4    sec 2 x  sec x tan x  dx 0   tan x  sec x 04 

(ii) dA 20  4 r  2 dr r dA let  0 to find stationary points dr

       tan  sec    tan 0  sec 0  4 4      1 2 1  2

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2010 Mathematics HSC Solutions

(iii)

1

1 A1    dx a x

(c)

y

1   ln x a 1

8

1  ln 1  ln  a 

–2

ln  a   1 a

x

1 e

(b) (i) l  r 9  5

b 1 A2    dx 1 x

  1.8 (ii) In OPT and OQT OP = OQ (equal radii of 5 cm) OPT = OQT (both right angles) OT is a common line OPT  OQT (RHS)

1   ln x 1

b

1  ln  b   ln 1 ln  b   1 be

(iii) POT  12 POQ

Question 6

(a) (i)

 0.9 PT tan(0.9)  5 PT  5 tan(0.9)

f ( x)  ( x  2)( x  4) 2

f ( x)  x3  2 x 2  4 x  8

PT  6.3 cm (1 d.p.)   (iv) PTQ    1.8  2 2 2 (angle sum of a quadrilateral is 2 )

f ( x)  3 x 2  4 x  4 Consider the discriminate,   42  4(3)(4)  32

PTQ  1.34 1 Area  (6.3) 2 sin(1.34) 2 1    (5) 2 (1.8  sin(1.8))  2 

Therefore there are no zeros, and hence, no stationary points. (the derivative function is positive definite)

 9 cm 2

(ii) f ( x )  6 x  4 Question 7

The graph is concave down when 6x  4  0

(a) (i) x   4 cos 2t dt

2 x 3 The graph is concave up when 2 x 3

 2sin 2t  c when t = 0, x  1 , 1  2sin 2(0)  c

c 1

 x  2sin 2t  1 4

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2010 Mathematics HSC Solutions

(ii) at x  0 0  2sin 2t  1 1 sin 2t   2 2t   t

1   T is the point  , 2  . 2  mBT  4 Eqn BT: y  4  4( x  2) y  4x  4



6

 13

Since this line is not vertical, if there is one simultaneous solution between this line and the parabola, it is a tangent. So, sub y  4 x  4 into

,

12 12

Therefore, the first time it will be at 13 rest is at t =  3.4 s 12

y  x2 4 x  4  x2 x2  4 x  4  0

(iii) x  2sin 2t  1 dt 

 x  2

  cos 2t  t  c

0

x2  BT is a tangent to the parabola

at t = 0, x = 0 0   cos 2(0)  0  c

c 1 x   cos 2t  t  1 dy  2x (b) (i) dx at x = –1, m = –2

Question 8

(a) P  Ae kt P  102e kt when t = 75, P = 200 000 000 200000000  102e 75 k k  0.22

y  1  2( x  1) 2x  y 1  0 1 5 (ii) M   ,  2 2 mAB  1 so, to find the x-value on the curve, where the tangent is 1, let 2x = 1. 1 1 Therefore the point C is  ,  . 2 4 Since the x-values of M and C are the same, then the line MC will be vertical.

(iii)

2

P  102e0.22t P  102e0.22(100) P  539 311 817 787 P  539 billion (b) P ( HH )  0.36 P ( H )  0.6 P (T )  0.4 P (TT )  0.16 (c) (i) A  4 (amplitude)

x–coordinate of T is 0.5.

2 b 2  b b2

2x  y 1  0

(ii) T 

1 2  y 1  0 2 y  2

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(iii)

2010 Mathematics HSC Solutions

(ii)1

y

A1  P (1  0.005)1  2000  P (1.005)1  2000

4 3 2 1

A2  A1 (1.005)1  2000   P (1.005)1  2000  (1.005)1  2000  P (1.005) 2  2000(1  1.005)

 2

-1 -2 -3 -4



x

A3  A2 (1.005)1  2000   P (1.005) 2  2000(1  1.005)  (1.005)1  2000  P (1.005)3  2000(1  1.005  1.0052 ) 

(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

1(1.005n  1) 1.005  1  P (1.005n )  400 000  (1.005n  1)  P (1.005n )  2000 

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

2

Consider the graph of y  3 x 2  6 x with x-intercepts at 0 and 2. Vertex at x = 1, y = –3.  if k  3 , f   x  is positive

An  ( P  400 000) 1.005n  400 000 0  (232 175.55  400 000) 1.005n  400 000 400 000 1.005n  167824.45 n  log1.005 (2.38)

definite and hence f  x  is an

n

increasing function.

log10 2.38 log10 1.005

n  174.1

Question 9

Thus there will be money in the account for the next 175 months

(a) (i) A1  500(1  0.005) 240 A2  500(1  0.005) 239 . . . A240  500(1  0.005)1

(b) (i)

0 x2

(ii) The maximum occurs at x = 2,



2

0

f   x  dx  4

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) 1.005  1  $232 175.55

(iii) f  6   f  4 

 500



4

2

f   x  dx  4

f  4   f  2   4 f  4   4  4 f  4  0

The gradient is –3, so f  6   6

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(iv)

2010 Mathematics HSC Solutions

4

y  a (1  2 cos  )

2

4

y  a (1  2(1)) y  3a OA (b) (i) sin   r OA  r sin 

6

–6

V 

r



r

r sin 

Question 10

r sin 

(a) (i)In ACD, DAC and DCA = 90  12  ( sum of )

y 2 dx

r

2

 x 2  dx r

 x3    r 2 x   3  r sin    r3   r 3 sin 3     r 3     r 3 sin   3  3  3 r 2  3sin   sin 3     3

CDB = 180   (suppl. angles) DBC = 90  12  (ABC is isosc.) DCB = 90  32  ( sum of ) ACB = DCB + DCA = 

  

(ii) 1 Initial depth = r. So, find  , to give depth 12 r. From the diagram, r OA  r sin   2 1 sin   2   30

In ABC and ACD, ACB = ADC (both  ) DAC = DBC (both 90  12  )  ABC ||| ACD (equiangular) AD DC a   AC CB x (ii)Corresponding sides of similar triangles are in the same proportion. AD AC   AC AB a x  x a y

also note

 r3 

3 1 2   3  2 8 2 Fraction  2 r 3 3 5  16

a (a  y )  x 2 x 2  a 2  ay

(iiiIn ACD, by the cosine rule 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.

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