RI 2022 Year 5 H2 Mathematics Common Test (Solutions)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 ___________________________________________________________________ 2022 Y5 H2 Math Common Test: (modified) Solutions with comments Page 1 of 17 2022 Year 5 H2 Mathematics Common Test (modified): Solutions with Comments 1 It is given that 3 2f ( )x ax bx bx c    and g( ) ,x ax d  where a, b, c and d are constants. Given that the curve with equation f ( )y x has a stationary point at  1,1 and it intersects the curve with equation g( )y x at  3,17 , determine the values of a, b, c and d. [5] Solutions Comments 1 [5] Since f ( )y x passes through  1,1 , 1a b b c    1a c  (1) 2f '( ) 3 2x ax bx b   Since f ( )y x has a stationary point at  1 ,1 , f '( ) 0x  at 1x , 3 2 0a b b   3 0a b  (2) Since g( )y x passes through  3,17 , 3 17a d  (3) For the 2 curves to intersect at  3,17 , 3 2 3 2 ( ) 0 27 9 3( ) 0 24 6 0 (4) ax bx bx c ax d ax bx b a x c d a b b a c d a b c d                       Solving (1), (2), (3) and (4) using GC, 2a , 6b  , 1c  and 11d  . Or Since f ( )y x also passes through  3,17 , 27 6 17a b c   (5) Solving any 4 equations using GC, 2a , 6b  , 1c  and 11d  . Always make full use of the conditions given. The fact that there is a stationary point at (1, 1) gives us 2 conditions  The curve passes through the point  f '( ) 0x  at x = 1. Similarly, that the two curves intersect at (3, 17) gives us 2 conditions as well. Note that at least 4 independent linear equations are needed to solve 4 unknowns completely. It is a lot faster (and less error- prone) to use the GC to solve a system of linear equations.

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ ___________________________________________________________________ 2022 Y5 H2 Math Common Test: (modified) Solutions with comments Page 2 of 17 2 Referred to the origin O, the points A and B have position vectors a and b respectively. Point C lies on BA produced such that : 3 : 2.AB AC  (i) Find OC   in terms of a and b and show that the area of triangle OAC can be written as k a b , where k is a constant to be found. [3] (ii) Given that a and b are unit vectors and the magnitude of a b is 3, find the cosine of the angle between a and b. [3] Solutions Comments 2 (i) [3] : 3 : 2 3 2 AB AC AB AC   By ratio theorem,   2 3 5 5 2 3 1 15 2 5 23 3 OB OCOA OA OB OC OC OA OB                             a b Area of OAC  1 2 OA OC       1 1 5 22 3 1 (since )3        a a b a b a a 0 where 1 3k  . If a point is on BA produced, then the order of the points on the line is B, A, C (left to right or right to left).  1 1 5 22 3 1 5 2 2 3 1 2 3 2 3 3 1 2 2 3 2 1 1 2 3                     a a b a a a b a b a b a b a b Note that area cannot be negative. C A B 2 3

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ ___________________________________________________________________ 2022 Y5 H2 Math Common Test: (modified) Solutions with comments Page 3 of 17 (ii) [3]     2 2 2 si( nce 3 3 3 2 3 2 3 1 1 2 3 1) 0.5                           a b a b a b a b a a b b a b a b a b a b a b a b cos 0.5 1 (1)(1) 2 AOB     a b a b Alternative Method: Let θ be the angle between a and b. Using cosine rule,   2 2 2 2 cos      a b a b a b        2 2 23 1 1 2 1 1 cos π θ      1cos π θ 2    1cos θ cos π θ 2   Most students are able to arrive at cos (1)(1)AOB      a b a b a ba b To be able to make progress, you must realise that the condition 3 a b has to be used. To link it with the scalar product, a common way is to use the fact that for any vector x , 2cos 0  x x x x x .

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ ___________________________________________________________________ 2022 Y5 H2 Math Common Test: (modified) Solutions with comments Page 4 of 17 3 The line 1L has equation 63, 3 yx z   and the line 2L has equation 2 6, 7x y z    . (i) Find the acute angle between 1L and 2L . [2] (ii) Show that 1L and 2L are skew lines. [2] (iii) Let 1P and 1Q be two points on 1L such that distance between them is 10 . Let 2P and 2Q be the two respective points on 2L such that 1 2P P and 1 2Q Q are minimum. Find the exact distance between 2P and 2Q . [2] Solutions Comments 3 (i) [2] 1 2 3 0 : 6 3 , , 0 1 2 1 : 6 1 , 7 0 L L                                            r r   Let  be the acute angle between 1L and 2L . 0 0 1 c 3 1 3os 10 2 2 1 0                    47.9 (1 d.p.)   One should be careful when converting cartesian equation of a line to the vector/parametric form. Do it systematically by first introducing the parameter, 63, 3 yx z    Then express x, y and z in terms of the parameter: 3, 6 3 , x y z      before writing it in the vector form as shown. If you have made a mistake here, it will carry over to subsequent parts of the question, which is very costly. (ii) [2] (1) To show 2 lines are non parallel: Since the angle between 1L and 2L is not 0 or 180, the 2 lines are not parallel. OR 1L is parallel to 0 3 1        and 2L is parallel to 1 1 0         . Since 0 1 3 1 1 0 k                   for any real k, the two lines are not parallel. For 2 lines to be skew, they need to be both non parallel and non intersecting. It is not sufficient to just check that there is no intersection between the 2 lines.

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ ___________________________________________________________________ 2022 Y5 H2 Math Common Test: (modified) Solutions with comments Page 5 of 17 (2) To show 2 lines are non intersecting: Using cartesian equation: If the 2 lines intersect, then 3, 7x z  . The first equation implies that 15y   but the second equation implies that 7y  . Hence there are no intersections. Since both lines are not parallel and do not intersect, they are skew lines. Alternative method for checking for intersection between L1 and L2 using vector equations: Consider 3 2 1 (1) 6 3 6 3 (2) 7 7 (3)                                    Substituting (1) and (3), LHS of (2) = 21 and RHS of (2) = 1 This implies equations (1), (2) and (3) have no solutions for  and , so L1 and L2 do not intersect. Since both lines are not parallel and do not intersect, they are skew lines.

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ ___________________________________________________________________ 2022 Y5 H2 Math Common Test: (modified) Solutions with comments Page 6 of 17 (iii) [2] The required distance is given by 2 2 1 1       cosP Q P Q  , where  is the acute angle between the 2 lines. Essentially, 2 2 P Q is the length of projection of 1 1 P