2010 mathematics hsc solutions

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2010 mathematics hsc solutions

  1. 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. 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. 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. 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. 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. 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. 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

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