2024 NJC Paper 1 H2 Math Prelim (ANS)

Question 1 (Connected Rates of Change) (i) ( )2 2 2 2 kkk r r r rr   = + = +  + =  = − 2 2 211 22 2 2 k krA r r r r = = − = −  d 2d2 Ak rr =− Interestingly, many students could derive the formula 21 2Ar = by considering 2π2πAr = but not many could do so similarly for the arc length formula .sr = As a result, there are many students who could not prove the required result. (ii) 2 dd 2d 2 d 22 3 10 60 A k r rtt k k k k =− = −  =− This part of the question is better performed than the previous part as almost the whole cohort is able to relate the rates of change. However, many could not apply the information that “arc length is equal to the radius” in a useful manner.

Question 2 (Systems of Linear Equations) ( ) ( ) ( ) 32 2 2 2 0 8 4 2 0 4 2 8 --- (1) a b c a b c a b c − + − + − + =  − + − + =  − + = ( ) 322 2 2i π i π i π3 3 3 42i π i πi2 π 33 3e 3e 3e 0 27e 9 e 3 e 0 1 3 1 327 9 i 3 i 02 2 2 2 9 3 9 3 3 327 i 02 2 2 2 a b c a b c a b c a b c a b      − − −              −−   −           + + + =                + + + =    + − + + − − + =           − − + + − =    Comparing real parts, 9327 0 --- (2)22a b c− − + = Comparing imaginary parts, 9 3 3 3 0 3 0 --- (3)22 a b a b− =  − = Solving (1), (2) and (3) with the GC, we get a = 5, b = 15, c = 18. Alternative Method 2i π3 1 3 3 3 33e 3 i i 2 2 2 2 − = − − = − − Since all coefficients of the polynomial are real, then 3 3 3 i22−+ is also a root of the equation by the Conjugate Root Theorem. Therefore, ( ) ( ) ( )( ) 32 2 2 32 3 3 3 3 3 32 i i 2 2 2 2 3 9(3)2 24 2 3 9 5 15 18 z az bz c z z z zz z z z z z z + + +   = + + + + −       = + + + = + + + = + + + Therefore a = 5, b = 15, c = 18 Both methods were fairly common. Many careless mistakes were made in calculations. Common mistakes:  Not stating “Since the coefficients of the polynomial are all real” when using the Conjugate Root Theorem.  Incorrect conversion of complex numbers from exponential form to cartesian form. (Please identify the quadrant the complex number is in.)  Not converting the complex numbers to cartesian form (in the first method).  Writing “ 2i π33ez −= is a root …”. (Should be “ 2i π33e − is a root …” or “ 2i π33ez −= is a solution …”).

Question 3 (Inequalities) (i) Method 1 ( ) ( )( ) ( )( ) 2 2 2 2 2 2 2 2 2 2 6 145 6 1045 6 4 5 045 41 045 41 045 41 04 5 1 4 5 1 0 since 4 1 1 0 for all x xx x xx x x x xx x xx x xx x xx xx xx − +− − − +− − − + −  +− −− +− + +− + +−  + −  +    Method 2 ( )( ) ( ) ( ) ( ) ( ) ( )( ) ( )( )( ) ( )( ) 2 222 22 22 2 2 6 145 6 4 5 4 5 4 5 6 4 5 0 4 5 4 1 0 4 5 1 4 1 0 4 5 1 0 since 4 1 1 0 for all x xx x x x x x x x x x x x x x x x x xx xx − +−  − + −  + −  + − − − + −   + − − −   + − − −   + −  − −    55Therefore, 1 since and 144 x x x−    −  Many candidates correctly simplified the inequality to 2 2 41 045 x xx −− +− . The correct factorisation of 245xx +− into ( )( )4 5 1xx+− is often seen as well. Some candidates made mistakes in the algebraic manipulations e.g. writing the numerator as 24 1.x − There are some candidates who tr ied to factorise 24 1.x + Please note that complex roots are not considered for the determination of the critical values. Common mistakes:  Inclusion of 5 4x =− and 1x = in the solution of the inequality. These values make the denominator in the original inequality 0, so they need to be excluded.  Incorrect evaluation of signs in the number line, leading to incorrect range of solutions 5 4x − or 1 x 1 5 4− 1

(ii) 2 2 6 145 xx xx − +− Replace x with 1 y and we get 2 2 2 22 2 1 1 66 61 1 1 45451145 y y y y y yyyy yyy     −−     −        +− +−+−   5 5 1Therefore, 1, i.e. 1.44 y x−   −   So 5 1 1 0 or 0 14 4 or 15 xx xx −      −  Many students are able to identify that a suitable replacement is 1 .y Common mistakes:  Solving 51 14 x−   is not simply just taking the reciprocals of the terms and keeping the sign i.e., the inequality above is not equivalent to 41 51 x−   . A reliable way to solve reciprocals is to look at the graph 1y x= .  Some students wrote ‘ 1x x= ’ to mean ‘replace x with 1 x . If the former holds, then 1x =  , which is not part of the solution set.

Question 4 (Vectors I) Using Ratio Theorem, BE ⎯⎯ → ( )( ) 1 23 1 23 12 33 BC BA ⎯⎯ → ⎯⎯ →=+  = − + − = + − c b a b c a b CD ⎯⎯ → ( )( ) 1 23 1 23 12 33 CB CA ⎯⎯ → ⎯⎯ →=+  = − + − = + − b c a c b a c ( ) 12: 33 21 1 , (shown)33 BEl      = + −  = + − +  r b + c a b a b c ( ) 12: 33 21 1 , (shown)33 CDl      = + −  = + + −  r c + b a c a b c At F, ( ) ( )2 1 2 1 113 3 3 3     + − + = + + −a b c a b c 2 2 1 1 113 3 3 3          − + − − + − =           a b + c 0 Many students could find the vectors BE and .CD However, some used vector addition to find the 2 vectors, which is very tedious. Common mistakes:  Some students do not know the difference between the direction vector BE and equation of the line BE.  Some students regarded a, b, c as vectors i, j, k and wrote 1 3 1 2 2 3 3 3 1    + − =  − c a b  Many students did not understand the requirements of the question. They compared the coefficients of non -parallel vectors a, b, c at this stage: ( ) ( ) 21 133 21 133       + − + = + + − a b c a b c As a result, they failed to obtain the required expression. • • A B C D E 1 1 2 2 F

(ii) If OACB is a parallelogram, =+c a b ( ) 2 2 1 1 113 3 3 3 2 2 1 1 113 3 3 3 2 2 1 1 1 1 1 13 3 3 3 3 1 2 213 3 3                              − + − − + − =                − + − − + − + =              − + + − − − + + − =          + − + − =       a b + c 0 a b + a b 0 a + b 0 a b 0 Since a and b are not parallel to each other and non -zero vectors, 1 1 0 (1)3+ − = 22 0 (2)33−= Solving, 3 4== Thus position vector of F is given by ( )3 1 2 4 3 3 31 42  + + − + = b + a b a b = a b This part was usually either not attempted or very badly done. Most students could not recognise that =+c a b . Common mistakes:  Many did not read the question and stopped at th e step 1 2 2 13 3 3      + − + − =       a b 0  Many used =−c a b instead.

Question 5 (Maxima & Minima Problems) Let hV and hv be the volumes of the original cone and cone that was removed. Since both cones are similar, 3 3h h v ar aVr == . Also, it is given that 224hr=− . Hence ( ) ( ) ( ) ( ) ( ) 2 2 3 3 2 3 3 2 3 2 3 2 2 π π 1π 1 π 3 1 π π 16 3 hh hh V ar r V v a r V a V a r a r h a a r r r = + − = + − = + −  − = + − This part was very well done. However, most students did not leave their answers in simplied form.

(a) Given that 0.25a = , 3 2 20.0625π 0.328125π 16V r r r= + − ( ) ( ) 2 22 2 23 2 2 2 2 2 2d 0.1875π 0.328125π 2 16d 2 16 2 16 0.1875π 0.328125π 16 32 30.1875π 0.328125π 16 rrV r r rr r r r r r r rrr r  −= + − +  −  −−=+  −  −=+  − For stationary V, d 0d V r = 2 2 2 32 30.1875π 0.328125π 0 16 rrr r  −+=  − Since 04 r , ( ) 2 2 2 2 42 2 2 4 2 2 4 42 7 32 3 04 16 7 3 32 4 16 49 9 192 1024 16 16 49 9 192 1024 256 16 457 9664 50176 0 rr r r r r rr rr r r r r rr  −+=  −  − = −  −+ = − − + = − − + = Most students could apply the Product rule but experienced great difficulty in