RI 9758 2024 Promo Solutions

RAFFLES INSTITUTION H2 Mathematics (9758) 2024 Year 5 Page 1 of 23 Q1 [4] Method 1 2 d d y b cax x x= + − At 3 ,2x= d 0d y x = 24 0 (1)39a b c+ − = −−−− 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)a b c+ − = −−−− y-coordinate of D at 1x= is 1 5 4− =− . Substituting 1x= and 4y=− into equation of D, we get 4 (3)ac+ =− −−−− From GC, 2, 7, 6.a b c= =− =− Method 2 2 d d y b cax x x= + − At 3 ,2x= d 0d y x = 24 0 (1)39a b c+ − = −−−− Gradient of D at 1x= is 1 d d x y a b cx = = + − . y-coordinate of D at 1x= is ac+ . So, the equation of tangent to D at 1x= is ( ) ( )( 1) ( ) 2y a c a b c x y a b c x b c− + = + − −  = + − − + Comparing this line with 5yx=− , we get 1 (2)a b c+ − = −−−− 2 5 (3)bc− + =− −−−− From GC, 2, 7, 6.a b c= =− =−

2024 H2 Math Year 5 Promotion Examination: Solutions _____________________________________________________________________________________ Page 2 of 23 Q2 Solution (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 (b) [1] cos90 0a b a b = = since cos90 0= Additional Note: 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 1 1 2 0 OR      − = + −  − = − − + −  = ab a b a b a b a b Since a and b are perpendicular, 0=ab 2 22= + b ab 22 * 2 2 2 2OR =+ ++ ba ab a b a b ** 22 2 2 2 2 22 : 1 : : : AR BR =− = ++ = ab a b a b a b O a A B b

2024 H2 Math Year 5 Promotion Examination: Solutions _____________________________________________________________________________________ Page 3 of 23 Q3 Solution [4] ( )( ) ( ) 2 2 6 2 3 2 1 2 1 10, 2 1 2 21 021 112 2 1 0 22 x x x x xx x x x x x + − − + − − − −   − + −      1 1 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 6 2 3 2 1 2 1 2 1 0 2 1 6 2 3 2 1 2 1 0 x x x x x x x x x x + − − − + −   − + − − + −  This will avoid unnecessary algebraic manipulation. [3] By replacing x with ,x

2024 H2 Math Year 5 Promotion Examination: Solutions _____________________________________________________________________________________ Page 4 of 23 1 1 1 or 222 1 1 1 1or or 2222 xx x x 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−  .

2024 H2 Math Year 5 Promotion Examination: Solutions _____________________________________________________________________________________ Page 5 of 23 Q4 Solution (a) [5] Area of rectangle OQPR, ( ) 3 231 A xy   == + ( ) ( ) ( )( )( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) ( ) 23 2 3 3 2 43 23 2 3 3 2 43 23 33 1 3 2 1 3d d 1 1 3 2 1 3 1 31 1 A                + − + = + + − + = + − = + or stationary A, d 0d A  = , giving 1= (since >0) MTD 1 : First Derivative Test ( ) ( ) 23 33 31d d 1 A    − = + Since ( ) 2 33 3 0 1    + for all 0,  1 1 1−+ ( ) 31 − ve 0 ve+− d d A  ve 0 ve+− Maximum A when 1.= MTD 2 : Second Derivative Test O Q ( ),P x y R

2024 H2 Math Year 5 Promotion Examination: Solutions _____________________________________________________________________________________ Page 6 of 23 ( ) ( ) 25 33 3d d 1 A    − = + ( ) ( ) ( )( )( ) ( ) ( ) ( ) 323 4 2 5 3 22 62 3 74 43 1 2 5 3 1 3d 3d 1 4 12 23 1 A               + − − − +=  +  −+=  + When 1, = 2 2 d9 0d8 A  =−  Maximum A when 1.= (b) [4] ( )( ) ( ) ( ) ( ) 23 2 2 3 4 2 333 1 2 3 1d d d 2 ..d d d 1 2 1 2 1 yy xx            + − + −= = = −−+ 4 3 d d d 2 d..d d d 1 2 d y y x x t x t t   −== − When 1, = ( )( )d 1 1 1d y t = − =− unit per sec. Rate of change of the y-coordinate of the point P at this instant is 1− unit per second or Rate of decrease of the y-coordinate of the point P at this instant is 1 unit per second.

2024 H2 Math Year 5 Promotion Examination: Solutions _____________________________________________________________________________________ Page 7 of 23 Q5 Solution (a) [3]   0 ( 2) n r n r n = ++   ( ) ( )( ) 2 1 ( 2)2 1 42 142 n n n n n n nn n nn += + + + +=+ = + + Alternative   ( )( ) 0 0 0 1 ( 2) ( 2) ( 2) ( 1) ( 2) ( 1) ( 1)2 142 n n n r r r n r n r n n r n n r n n nn n n n n nn = = = = + + = + + = + + + = +  + + + = + +     (b) [3] ( ) ( ) ( ) 3 3 3 33 1 2 3 4 1 2 n r r n n = + = + + + + + + ( ) ( ) ( ) 23 3 13 2 3 3 3 1 22 2 12 1 2 3 94 nn rr n r rr r nn + == + = += = − − = + + −   (c) [3] ( ) ( ) 333 3 3 3 3 31 2 3 4 5 6 2 1 2 nn− + − + − + + − − ( ) ( ) ( ) 2 33 11 2 33 11 22 22 22 16 112 2 1 16 ( 1)44 nn rr nn rr rr rr n n n n == == =− =− = + − +   

2024 H2 Math Year 5 Promotion Examination: Solutions _____________________________________________________________________________________ Page 8 of 23 ( ) ( ) ( ) ( ) ( ) 2222 222 2 2 1 4 1 2 1 2 2 43 n n n n n n n nn = + − + = + − + =− + Q6 Solution (a) [5] ( ) 2 2 22 2 e 1 sin 3 ------ (1) Differentiating w.r.t. , de 3cos3d Differentiating w.r.t. , d d de e 9sin 3 9 e 1 (from (1))d d d dde e 9e 9dd y y y y y y y y x x y xx x y y y xx x x yy xx =+ = + =− =− − + + = Dividing throughout by e,y 22 2 dd 9 9edd yyy xx −+ + = (shown) Differentiating w.r.t. x , 32 32 d d d d2 9ed d d d yy y y y x x x x − + = −  23 23 When 0, d d d0; 3; 9; 27d d d x y y yy x x x = = = =− = 23 23 9 270 3 ... 2! 3! 993 ... 22 y x x x y x x x − = + + + +  = − + +

2024 H2 Math Year 5 Promotion Examination: Solutions _____________________________________________________________________________________ Page 9 of 23 (b) [3] e 1 sin 3y x=+ ( ) 3 2333 3 3 2 3 23 ln 1 sin 3 (3 )ln 1 3 ... 3! (3 ) (3 )33 3! 3!(3 )3 ... 3! 2 3 27 9 273 ... 6 2 3 993 ... 22 yx xx xxxx xx x x xx x x x  = +  = + − +     −−       = − − + + = − − + + = − + + which is same as the expansion for y found in part (a), up to and including the term in 3x .

2024 H2 Math Year 5 Promotion Examination: Solutions _____________________________________________________________________________________ Page 10 of 23 Q7 Solution (a) [5] Since , ab  R , then all the coefficients of the polynomial are real, complex roots will occur in conjugate pairs. Given 1 3i+ is a root, then 1 3i− is also a root. ( ) ( ) ( ) ( ) ( ) ( ) 22 2 1 3i 1 3i 1 3i 1 3i 1 3i 2 10 xx xx x xx − + − −     = − − − +     = − − = − + ( )( ) 3 2 2 18 2 10x ax x b x x x p+ + + = − + + Compare coefficients of 2 0 :18 2 10 4 : = 2 6 : 40 x p p x a p a xb =− +  =− −  =− =− The other roots are 4 and 1 3i− . Alternative: Since 1 3ix=+ is a root of the given equation, 32(1 3 ) (1 3 ) 18 0 (1 3 )( 8 6 ) ( 8 6 ) 18(1 3 ) 0 8 24 6 18 8 6 18 24 0 ( 8 8 ) (6 36) 0 i a i x b i i a i i b i i a ai i b a b a i + + + + + = + − + + − + + + + = − − + −