2024 RI H2 Math Promo (Soln)

Page 1 2024 RI H2 Math Year 5 Promotion Examination: Solutions with Comments 1 A curve D has equation ln , 0, cya x bx x x   where ,a b and c are constants. It is given that D has a stationary point at 3 2x and the tangent to D at the point where 1x is 5.yx Find the values of ,a b and .c [4] Solution Comments [4] Method 1 2 d d yb cax xx At 3 ,2x d 0d y x  24 0( 1 )39abc    The gradient of D at x = 1 is equal to the gradient of the line 5.yx At 1,x d 1d y x  1( 2 )abc     y-coordinate of D at 1x is 15 4  . Substituting 1x and 4y into equation of D, we get 4( 3 )ac      From GC, 2, 7, 6.ab c   Method 2 2 d d yb cax xx At 3 ,2x d 0d y x  24 0( 1 )39abc    Gradient of D at 1x is 1 d d x y abcx   . y-coordinate of D at 1x is ac . So, the equation of tangent to D at 1x is () ( ) ( 1 ) ( ) 2ya c a b c x y a b c x b c      Comparing this line with 5yx , we get 1( 2 )abc     25 ( 3 )bc    From GC, 2, 7, 6.ab c   Most students did well and got full marks for this question. Those who did not get full marks, were able to get the equations (1) and (2). Need to observe that since the tangent line and the curve touches at a point, they should share the same y-coordinate at that point. So, use the equation of the given tangent line to find the y-coordinate and obtain equation (3) There are a few students who did not use the GC to solve the simultaneous equations and made careless mistakes in their workings and thus lost some marks. Common Mistakes:  2 d d yb cxx xx   Substituting 1x and 5y into equation of D, instead of 4y

2024 H2 Math Year 5 Promotion Examination: Solutions with Comments _________________________________________________________________________________ Page 2 2 With respect to the origin O, the fixed points A and B have position vectors a and b respectively, where a and b are non-zero and non-parallel. (a) The variable point R has position vector  1, ra b where  is a real parameter. Describe geometrically the set of all possible positions of R. [1] It is given that the angle AOB is 90 . (b) Explain why 0.a.b [1] (c) Among the set of all possible points R, the point *R is the closest to the origin O. Find the position vector of *.R Hence state the ratio **:AR BR in terms of magnitudes of a and b. [4] Solution Comments (a) [1] The set of all possible positions of R is the line that passes through points A and B, or The set of all possible positions of R is the line that passes through point B (or A) and parallel to the vector .AB  Additional Notes: When 0  , we get ,rb which corresponds to the point B. When 1  , we get ,ra which corresponds to the point A. Note that a vector is NOT a line, and if we wish to talk about the position vector b, we should say the line passes through the point with position vector b (or simply point B). A line cannot pass through a vector, and a point cannot be on a vector as well. Mathematical language has to be accurate. Since R is a point, when describing its possible positions, it should be clear that we are describing possible points. When describing a line, you need 2 points or 1 point and a direction vector which the line is parallel to. The line segment AB refers to the part of the line AB which is between A and B inclusive. Since  is real, R is any point on the line AB. (b) [1] cos90 0ab ab   since cos90 0 Additional Note: The definition of the dot product is cosab ab . B ( ) A ( ) BA produced () AB produced () Line segment AB ()

2024 H2 Math Year 5 Promotion Examination: Solutions with Comments _________________________________________________________________________________ Page 3 That the scalar product is 0 because 2 vectors are perpendicular (or the angle between them is 90 ) is a consequence of this definition. (c) [4]  * 1OR  ab  for some   .      * 22 0 10 11 2 0 OR           ab ab a b ab a b  Since a and b are perpendicular, 0ab 2 22  b ab 22 * 22 22OR   ba ab ab ab  ** 22 22 22 22 :1 : : : AR BR     ab ab ab a b The point F on a line closest to a given point P is the foot of the perpendicular from P to the line. This point is not necessarily the midpoint of the line segment AB. When expanding the scalar product, note that 2 cos 0 aa aa a Note that if 1 ra b , as shown in the diagram, **:1 :AR BR  , not **:: 1AR BR  O a A B b

2024 H2 Math Year 5 Promotion Examination: Solutions with Comments _________________________________________________________________________________ Page 4 3 Do not use a calculator in answering this question. Solve the inequality  262 3 21 .21 xx xx   [4] Hence solve  2623 21 .21 xx xx   [3] Solution Comments [4]     2 2 6 2 32 12 1 10, 21 2 21 021 112 210 22 xx x x xx x x xx x              11 1 or 222 xx   Additional Notes: You are strongly advised to use ( ) if you are making algebraic errors in arriving at 221 021 x x   . For students who multiply by  2 21x in the first step, you should always factorize first before any expansion, as shown below:           22 2 62 3 2 1 21 2 1 0 216 232 1 21 0 xx x x x xx x x x            This will avoid unnecessary algebraic manipulation. As the instruction of not using a calculator is given in the question, you should 1) Give all answers in exact form (no 3 s.f) 2) Show all working clearly This also means that graphical approaches need to be clearly justified to obtain full credit. The use of the OR is required when having 2 solution ranges that are distinct. Using ‘,’ or AND is incorrect. Once you have fully factorized the expression, do feel free to use the GC (you do not need to say that you used it) to verify the shape of the cubic graph to avoid unnecessary mistakes. [3]

2024 H2 Math Year 5 Promotion Examination: Solutions with Comments _________________________________________________________________________________ Page 5 By replacing x with ,x 11 1 or 222 11 1 1or or2222 xx xx x         Additional Note: Perhaps the simplest way to think of solving modulus inequalities like 1 2x  would be to tell yourself if the magnitude* of x is smaller than 1 2 , then x itself should not be too far from the origin, that is 11 22 x . Similarly, if 1 2 x  , then x has to be at least 1 2 from the origin, and thus 11 or 22 xx  . Note that 1 2 x holds for all real x. Thus 11 22 x  is equivalent to 1 2x  . *Note how "" is consistently used in various topics. a denotes the magnitude of a vector a, or equivalently the distance of the point A from the origin. z denotes the magnitude of a complex number z, or equivalently the length OP, if P represents the complex number z on the Argand diagram.

2024 H2 Math Year 5 Promotion Examination: Solutions with Comments _________________________________________________________________________________ Page 6 4 A curve C has parametric equations 31x    and 2 31y    , for 0. The point P is a variable point on .C (a) With reference to the origin O , OP forms the diagonal of the rectangle ,O