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.  Students need to observe from the question that 0x and exclude zero from the final solution. k 0 2k

3 Suggested Solution Marker’s Comments (a) 1 2 1 2 2 1 (1 tan ) Differentiating with respect to , d 1 2(1 tan )d 1 d(1 ) 2(1 tan )d y x x y xx x yx x x               2 2 2 2 2 2 2 2 2 Differentiating with respect to again, d d 1(1 ) 2 2d d 1 d d(1 ) 2 (1 ) 2 (shown)d d x y yx x x x x y yx x x x x            Learning objectives: - Apply 1d f f'fd n n nx     - Apply 1 2 d f' tan fd 1 fx      - Apply implicit differentiation. Unwanted method: Many students took the inefficient way to prove from LHS to RHS, which is not the elegant way. (b) Differentiate 2 2 2 2 2 d d(1 ) 2 (1 ) 2d d y yx x x x x    with respect to x, 3 2 2 2 2 2 2 2 3 2 2 3 2 2 2 2 2 3 2 d d d d(1 ) 4 (1 ) (2 6 ) 2 (1 ) 0d d d d d d d(1 ) 6 (1 ) (2 6 ) 0d d d y y y yx x x x x xx x x x y y yx x x xx x x               Learning objectives: - Apply implicit differentiation. Though many students did not group the common terms together and they left the answer as 3 2 2 2 2 2 2 2 3 2 2 d d d d(1 ) 4 (1 ) (2 6 ) 2 (1 ) 0d d d d y y y yx x x x x xx x x x        , mark is still awarded. Some students went to expand 2 2(1 ) x which is unnecessarily. (c) When 2 3 2 3 d d d0, 1, 2, 2, 4d d d y y yx y x x x     Hence the Maclaurin’s series for y is 2 3 21 2 3y x x x    . Learning objectives: - Apply Maclaurin’s series from MF26. For those who could not get the answer, mainly is due to careless calculation. (d) Using the Maclaurin’s Series for y, 2 3 21 2 3y x x x    . Learning objectives:

It was observed that the gradient of the tangent to the curve at x = 0 is 2. Hence the gradient of normal is 1 2 . From (c), 1y when 0x . Therefore, the equation of normal to the curve at x = 0 is 1 12y x  . - Deduce gradient of tangent to the curve at x = 0 from the Maclaurin’s Series obtained. Unwanted method: Again, many students were not able to deduce the gradient of the normal from the Maclaurin’s Series obtained in part (c). They took the longer method to find the equation of normal.

4 Suggested Solution Marker’s Comments (a)   22 2 2 2 2 2 24 4 2y a x y a x x y a        2 2 2 0 4 d a a x x represents the area of a quadrant of a circle, 2 24y a x  , with centre at the origin and radius 2a units. Thus the area is   2 21 π 2 π4 a a  . Learning objectives: - Relating integral to area bounded by a function and x-axis. - Identify the function Students are to explain in words what the integral represents. (b)        f 8.5 f 5.5 f 2.5 3 2.5 6 1.5a a a a a a     Learning objectives: - Understand and apply what is meant by a periodic function i.e.    f 3 fx a x  for all real values of x. This part is well done. (c) Learning objectives: - Sketch of a periodic and piecewise function. This part was badly performed as there are students who do not know how to sketch a piecewise and periodic function despite that such common questions are found in tutorial and revision packages. It appeared more than once in A Level TYS. Many students forgot to indicate whether the endpoints are inclusive or exclusive, resulting in mark deduction. (d)        599 2 22 0 2 2 2 2 1 1f d 99 π 3 0.5 1.52 2 297 3100π 2 8 1191100π 8 a x x a a a a a a a a a a                 Learning objectives: - To identify and apply formula for finding area of a right-angle triangle. - To deduce that the area of can be found be adding the area of the quadrant from (a) and area of a right-angle triangle without doing any integration. This part was badly performed. Many students were not able to infer. 2a 3a 0 y x 2a 3a 5a 6a 8a –3a –a (8.5a, 1.5a) (-3a, 2a)

5 Suggested Solution Marker’s Comments (a) 4 9 1 25 2 5 2 xy x x     12 5 2y x   Replace y with  2y , we have 1 5 2y x  . Replace x with 1 2 x , we have 1 5y x  . Replace x with  5x , we have   1 1 5 5y x x   . (1) Translate the graph in the positive y-direction by 2 units. (2) Scale the graph parallel to the x-axis by a factor 2. (3) Translate the graph in the negative x-direction by 5 units. Alternatively, Replace y with  2y , we have 1 5 2y x  . Replace x with  2.5x , we have   1 1 1 25 2 2.5 2y y x x x      . Replace y with 1 2 y , we have 1y x . (4) Translate the graph in the positive y-direction by 2 units. (5) Translate the graph in the negative x-direction by 2.5 units. (6) Scale the graph parallel to the y-axis by a factor 2. Concepts and/or Skills:  Transformations of Graphs (Scaling and Translation) Common mistakes :  Associate the replacements with the wrong transformations.  Fail to use the terms accurately.