2023 NJC P1 Solution

1 Suggested Solution Marker’s Comments (a) By GC,  2, 1, 1  Concepts and/or Skills:  Point of intersection between 3 planes  Solving 3 linear equations using the G.C. Learning points:  2 1 1        is the position vector of a point, not the coordinates of a point. (b) By GC, an equation of the line of intersection between 1p and 2p 79 88 31 , 4 4 0 1                           r  For the line to lie on the 3p , line and normal vector are perpendicular, i.e. 0 m n 78 3 3 04 41                      7 3 3 4 08 4 2         For the line to lie on the 3p , a point on the line must lie on the plane, i.e. D a n Concepts and/or Skills:  The normal of a plane is perpendicular to the direction vector of a line that lies on the same plane.  For a line that lies on a plane, all the points on the line lies on the plane. Common mistakes :  Greek letter (read as alpha) is different from English letter a. They should not be used interchangeably. Similarly, for the letter  (read as beta) and the letter b. Learning points:  Given that line l lies on a plane p, when the equation of l is substituted into equation of p, there will be infinitely many solution since there are infinitely many points that lie on the line and the plane. Hence students who are stuck at this working 9 78 8 31 34 4 4                             need to know that they will not be able to

    98 2 1 34 40 9 12 38 4 3                               proceed if they do not know that for every real value substituted as  , there will be an equation formed in terms of  and  .

2 Suggested Solution Marker’s Comments (a)                2 2 2 2 2 2 2 0 2 0 2 2 0 2 0 x k x kx x k x kx x k x k x x k x x k x x k x x k x k x                    + + - - or 0 2x k x k   Concepts and/or Skills:  Solving inequalities Learning points:  Completing the square gives more rooms for mistakes. Do factorization if possible.  For such question with unknowns, substitute a value for k, say 1k  (since question stated that k is a positive constant) to conduct the “sign” test easily. After which, remember to express the answer in terms of k. (b) Let 5k  and replacing x with x . 2 2 2 2 2 10 5 2 10 5 since x xx x x x xx        Hence, 5 (rejected 0) or 0 2 5 x x x      2 5 0 or 0 2 5x x     (Also, 2 5 2 5 and 0x x    ) Concepts and/or Skills:  Identify appropriate substitution to solve questions using a known result.  Solving inequality that involves modulus function. Learning points:  Students should sketch a simple graph of y x to solve for 0 2 5x  . Please note that the skill for solving equations and inequalities is different. Please revise thoroughly