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