TMJC 2024 JC1 H2 Maths Promo Solutions (with Questions and Markers Comments)

H2 MATHEMATICS 9758 3 hours Additional materials: Printed Answer Booklet List of Formulae and Results (MF27) __________________________________________________________________________________ READ THESE INSTRUCTIONS FIRST Answer all the questions. Write your answers on the Printed Answer Booklet. Follow the instructions on the front cover of the answer booklet. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use an approved graphing calculator. Unsupported answers from a graphing calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphing calculator are not allowed in a question, you must present the mathematical steps using mathematical notations and not calculator commands. You must show all necessary working clearly. The number of marks is given in brackets [ ] at the end of each question or part question. TAMPINES MERIDIAN JUNIOR COLLEGE JC1 PROMOTIONAL EXAMINATION

2 1 It is given that   3 2f , x ax bx cx d    where , , and a b c d are non-zero real constants. The graph of  fy x passes through the points  0,3 and  2,1 . It is also known that one of the stationary points of  fy x is  3,3 . Find the values of a, b, c and d. [4]   3 2f x ax bx cx d     f 0 3 3 d     3 2f 3x ax bx cx       2f ' 3 2x ax bx c   Since one of the stationary points of  fy x is 3,3 ,         2 f ' 3 3 3 2 3 0 27 6 0 ------- 1 a b c a b c            Since one of the stationary point of  fy x is  3,3 , it also means that  fy x passes through  3,3 .           3 2 f 3 3 3 3 3 3 3 27 9 3 0 ------- 2 a b c a b c                        3 2 f 2 1 2 2 2 3 1 8 4 2 2 ------- 3 a b c a b c              Using GC,     3 2 1, 6, 9 3 from above f 6 9 3 a b c d x x x x          Note that at stationary point,  f ' 0.x 

3 [Turn Over 2 (a) State a sequence of three transformations that will transform the curve with equation  fy x onto the curve with equation   2f 3 .y x   [3] (b) Without using a calculator, solve exactly the inequality 2 2 2 1.2 1 x x x    [4] (a)  fy x Replace x by 3x  f 3y x  Replace x by x  f 3y x   Replace y by 2 y     f 32 2f 3 y x y x        The sequence of 3 transformations are as follows: I. Translation of 3 units in the negative direction.x II. Reflection in the axis.y III. Stretch by factor 2 parallel to the axis.y Alternative Method  fy x Replace x by x  fy x  Replace x by 3x      f 3 f 3 y x y x        Replace y by 2 y     f 32 2f 3 y x y x        The sequence of 3 transformations are as follows: I. Reflection in the axis.y II. Translation of 3 units in the positive direction.x III. Stretch by factor 2 parallel to the axis.y Descriptions for Transformation of Curves must be technically precise. Do not use your own terms or phrases. There are many possible answers. Just use replacement of variables to check that you can indeed go from  fy x to  2f 3y x   .

4 (b) Without using a calculator, solve exactly the inequality 2 2 2 1.2 1 x x x    [4] (b) 2 2 2 2 2 2 1 1,2 1 2 2 2 1 02 1 2 2 2 1 02 1 4 1 02 1 x x xx x x x x x x x x x x                 Method 1: Completing the Square            2 2 3 02 1 2 3 2 3 02 1 2 3 2 3 02 1 x x x x x x x x              12 3 or 2 3 2x x      Method 2: Quadratic Formula Consider 2 4 1 0x x            2 2 4 1 4 4 4 1 1 2 1 4 12 4 2 3 2 32 2 x x x                  2 3 2 3 02 1 x x x      12 3 or 2 3 2x x      − − + + − − + + We do not cross multiply when we are not sure how the inequality sign should change. So for this question, we cannot cross multiply by since it could be negative or positive.

5 [Turn Over 3 The curve 1C has equation 122 3 , 2. 2y x x x     (a) Sketch 1,C showing clearly the equations of asymptotes, the coordinates of the axial intercepts and the turning points, if any. [4] (b) By completing the square for both x and y, give a geometrical interpretation of the curve 2C with equation 2 2 4 6 3 0.x x y y     Hence write down the number of real roots for the equation   2 2 2122 2 6 4 . 2x x x          [3] (a) 122 3 , 2 2y x x x     Asymptotes: 2 3 and 2y x x   When 0, 3x y   0, 3 y x ݕ= 2ݔ+ 3 ݔ= 2 (4.45, 16.8) (−0.449, −2.80) ݕ= 2ݔ+ 3 + 12 ݔ− 2 O Turning points are (-0.44949,-2.7980) and (4.44948, 16.7980) to 5 significant figures (s.f.), please round off to 3 s.f. (exact not required) Note that maximum point is on the left of the y-axis Axial intercept(s) should be in coordinates, based on question requirements. Graph should tend towards asymptotes (not away/cut) that should be represented by dotted lines 2. Look at table values 3. Adjust window settings 4. Check if asymptotes are correct 1. A part of graph is out of standard zoom window need to know to look for graph for x > 2

6 (b) By completing the square for both x and y, give a geometrical interpretation of the curve 2C with equation 2 2 4 6 3 0.x x y y     Hence write down the number of real roots for the equation   2 2 2122 2 6 4 . 2x x x          [3] (b)         2 2 2 2 2 2 2 4 6 3 0 2 4 3 9 3 0 2 3 4 x x y y x y x y                 2C is a circle with centre (2, –3) and radius 4 units. There are 2 real roots for   2 2 2122 2 3 3 4 . 2x x x           Read question carefully: “write down number of roots”, not “find the roots” “Hence” question, number of real roots for   2 2 2122 2 3 3 4 2x x x           is equivalent to the number of intersections between 1C and 2C . Provide detailed description for geometrical interpretations y x ݕ= 2ݔ+ 3 ݔ= 2 (4.45, 16.8) (−0.449, −2.80) ݕ= 2ݔ+ 3 + 12 ݔ− 2 O x (2, −3) 4

7 [Turn Over 4 A manufacturer designs a rectangular tank. Th e base of the tank is a rectangle with adjacent sides of x metres by y metres and fixed diagonal length of D metres, where D is a real constant. The area of the base of the tank is denoted by A metres2. (a) Express 2A in terms of x and D. [2] (b) By differentiation, show that as x varies, the largest area of the base is achieved when . 2 Dx [4] The manufacturer designs another type of water tank. Water is poured at a constant rate of 10 3metres per minute into the tank. The volume of the water, V metres3, in the tank at time t minutes, is defined by 31 9V H  , where H metres is the height of the water in the tank at time t minutes. (c) Find the rate of change of the height of the water in the tank in terms of  , when 3.H  [3] (a)   2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 A xy D x y y D x A x y x D x D x x           

8 (b) By differentiation, show that as x varies, the largest area of the base is achieved when . 2 Dx [4] (b) 2 2 2 4A D x x  Differentiate wrt x: 2 3 2 3 d2 2 4d d 2d AA D x xx AA D x xx     When area is the largest, d 0d A x    2 3 2 2 2 0 2 0 D x x x D x     0 or or 2