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