IJC H2 MATH P1 Solution

1 2015 IJC H2 Maths Prelim 2 Paper 1 (Suggested Solution) 1 A water tank is in the shape of a vertically inverted cone with height 2 m and based radius 0.5 m. Water is flowing into the tank at a constant rate of 0.01 m 3 s−1. At the same time, water is leaking out of the tank at a constant rate of 41.2 10  m3 s−1 1through a small hole at the vertex. . Find the rate of change of the depth of the water level in the tank at the instant when the depth of the water is 0.5 m. [4] Q1 Solutions Let the radius of the water level be r m and the depth of the water level be h m at time t seconds. Using similar triangles, 𝑟 ℎ = 0.5 2 = 1 4 , ⇒ 𝑟 = ℎ 4 d 0.01 0.00012 0.00988d V t    2311 3 48V r h h 2d 1 d d 16 d Vh htt  When ℎ = 0.5,  21d0.00988 0.516 d h t d 0.00988 16 0.201d 0.25 h t   (3 s.f.)  the rate of change of the depth of the water level in the tank at the instant when the depth of the water is 0.5 m is 0.201 m s−1.

2 2 Referred to the origin O, the points and are such that OA  a and OB  b . The point M is on AB produced such that : 4:1AM BM  and the point N is on such that : 2:3ON NB  . (i) Find OM  in terms of and . [2] (ii) By considering cross product, find the ratio of the area of triangle ANB to the area of triangle OAM. [4] Q2 Solution (i) 3 4 44 33 OA OMOB OB OAOM   ba (ii) Area of 1 2OAM OA OM     14 23 1 4 6 2 3         baa a b a a a a 0 ab 2 5ON  b Area of ANB             1 2 12 25 12 and 25 13 25 3 10 AN AB               b a b a b a a b a a b b 0 ab ab Ratio of  ANB to  OAM = 9 : 20 A B OB a b

3 3 (i) Express 1 2 3 11r r r in the form ( 1)( 1) Ar B r r r   , where A and B are constants to be determined . [2] (ii) Hence find   2 21 11 n r r r r r   . (There is no need to express your answer as a single fraction.) [3] (iii) Deduce that   2 2 1 5 1 1 4 n r r r r r   . [1] (iv) Using your answer to part (ii), find 1 4 23 ( 1)( 2) n r r r r r     . [2] Q3 Solution (i) 2 2 2 1 2 3 11 ( )( 1) 2( 1)( 1) 3 ( 1) ( 1)( 1) 2 2 3 3 ( 1)( 1) 42 ( 1)( 1) r r r r r r r r r r r r r r r r r r r r r r r r                 A = 4, B = ‒2 (ii) 22 2 1 1 1 2 3 ( 1) ( 1) 2 1 1 1 1 2 3 2 1 2 3 1 2 3 2 3 4 1 2 3 3 4 5 ... ... 1 2 3 3 2 1 1 2 3 21 1 2 3 11 1 1 3 2 322 2 1 1 5 1 3 2 2 1 5 1 3 4 2 2( 1) nn rr r r r r r r r n n n n n n n n n n n n nn nn                                              

4 (iii) Since 0,n 1 02n  and 3 02( 1)n  . Hence, 2 21 ( 1)( 1) 5 1 3 5 4 2 2( 1) 4 n r r r r r nn          (iv) From (ii),   2 2 1 5 1 3 1 1 4 2 2( 1) n r r r r r n n       To find 1 4 23 ( 1)( 2) n r r r r r     , replace r by 1r .    1 43 1 4 2 3 2 1 ( 1)( 2) 1 1 5 1 3 3 4 2 2( 1) (1)(2)(3) 2 3 3 1 3 ( 1)( 2) 4 2 2( 1) nn rr n r rr r r r r r r nn r r r r n n                      

5 4 A sequence 1 2 3, , ,u u u is given by 1 1u  and 1 41 8 n n uu   for 1n . (i) Find the values of 2u and 3u . [1] (ii) It is given that nul  as n . Showing your working, find the exact value of l . [1] (iii) For this value of l , use the method of mathematical induction to prove that 1 15 2 n nul   . [4] (iv) Hence find 1 N n n u   . [3] Q4 Solution (i) 1 2 41 4(1) 1 3 8 8 8 uu     (or 0.375) 2 3 34141 18 8 8 16 uu      (or 0.0625) (ii) nul  as n  1nul  as n 1 41 8 n n uu   , as n , 41 8 ll  8 4 1 1 4 ll l    (iii) Let nP be the statement 1 115 24 n nu   for n  . LHS of 11 1Pu (given) RHS of 2 1 1 1 5 151 2 4 4 4P      1P is true. Assume that kP is true for some k  , ie. 1 115 24 k ku   . Need to prove 1kP , ie 2 1 115 24 k ku   

6 1 1 2 2 LHS = 41 8 11 28 1 1 1 152 2 4 8 1 1 15 2 8 8 115 24 RHS k k k k k k u u u               1kP  is true Since 1P is true and kP is true 1kP is true,  by Mathematical Induction, nP is true for all n  . (iv) 1 11 1 115 24 1151 42 1 1 41 2 5 1 1 5 5 11 or or equivalent2 2 4 2 2 4 nNN n nn N NN u N NN                         

7 5 A graphic calculator is not to be used in answering this question. The complex number z is given by 2 i 3 . (i) Find 4z in the form i,xy showing your working. [2] (ii) Given that z is a root of the equation 422 49 0,w aw bw    find the values of the real numbers and .ab [3] (iii) Using these values of a and b, find all the roots of this equation in exact form. [4] Q5 Solution (i)                44 2324 4 4 12 344 3 2 3i 2 2 3i 2 3i 2 3i 3i 16 32 3i 72 24 3i 9 47 8 3i z CC C             (ii) Method 1 422 49 0w aw bw              2 2 47 8 3i 2 3i 2 3i 49 0 94 16 3i 4 4 3i 3 2 3i+49=0 45 2 4 16 3i=0 ab a a a b b a b a b                      Comparing the coefficient of real part 2 45 (1)ab   Comparing the coefficient of imaginary part 4 16 (2)ab   11a , 28b Method 2 Since the coefficients of the equation are real, * 2 3iz  is also a root                 22 2 2 3i 2 3i 2 3i 2 3i 2 3i 47 ww ww w ww              

8    4 2 2 22 49 4 7 2 7w aw bw w w w pw        By comparing coeff of 3w , (or inspection mtd) 80 8 p p   2comparing coeff of : 7 32 14 11 comparing coeff of : 56 28 28 w a w b        Method 1 Since the coefficients of the equation are real, * 2 3iz  is also a root.                 22 2 2 3i 2 3i 2 3i 2 3i 2 3i 47 ww ww w ww               By long division or inspection,    4 2 2 22 11 28 49 4 7 2 8 7w w w w w w w             22 8 7 0 8 64 4 2 7 22 ww w       The other roots are 22 3i, 2 2   Method 2      22 8 7 0 8 64 4 2 7 222 2 2 ww w         The other roots are 22 3i, 2 2  

9 6 The function f is defined by 2f : e 3, xxx   . (i) Find 1f ( )x and write down the domain and range of 1f  . [3] (ii) Sketch on the same diagram the graphs of f ( )yx , 1f ( )yx  and 1f f ( )yx  , giving the equations of any asymptotes and the exact coordinates of any points where the curves cross the x- and y-axes. [4] (iii) Find the area of the region bounded between f ( )yx and 1f ( )yx  , giving your answer correct to 2 decimal places. [2] Q6 Solutions (i)   2 2 1 Let e 3 3e ln( 3) 2 ln( 3) 2 f ln( 3) 2 x x y y yx xy xx               1f : ln( 3) 2, , 3x x x x     1 ffRD  (ii) (iii) x – coordinates of points of intersection: x y y = −3 x = −3

10 Using GC, x = ‒2.9932 or 3.9368      3.9368 2 2.9932 Area required ln( 3) 2 e 3 d 34.26 (2 d.p.) xxx          