ACJC 2026 JC2 H2 Prelim Paper 2 Markers Report

ANGLO-CHINESE JUNIOR COLLEGE 2026 H2 MATHEMATICS 9758/02 2026 ACJC H2 Math Preliminary Examination Paper 2 Solutions Qn Solution Remarks 1(a) There is a tendency for students to miss out the turning point ( )2,0− . They need to understand why ( )2,0− exists. 1(b) Since the gradient at x = 2 and x = 6 for g( )yx= is zero, therefore ( )g'yx= has (2, 0) and (6, 0). y x O x = 0 y = 0 y x O (2, 0) (6, 0)

ANGLO-CHINESE JUNIOR COLLEGE 2026 H2 MATHEMATICS 9758/02 Qn Solution Remarks 2(a) ( ) ( ) ( ) ( ) ( ) 2 22 22 2 2 2 2 2 2 20 For to be real, 2 4 0 4 4 4 4 0 xky xk y x k x kx k x y k x k yk x y k k yk y ky k k ky −= + + = − + + − − + − = − − − −  + + − +  ( ) 2 80 80 80 Range of values of : 8 or 0 y ky y y k y k or y y y k y + + −  −  Alternative method: ( ) ( ) ( ) ( ) ( ) ( )( ) 2 2 2 2 2 2 2 22 2dLet 0d 2 2 2 0 2 3 0 30 or 3 xky xk x k x k x ky x xk x k x kx k x kx k x k x k x k k −= + + − − +== + − − − − = + − = − + = =− when , 0 when 3 , 8 Range of values of : 8 or 0 x k y x k y k y y k y == =− =− −  Note that 0D , not 0D . Solving 2 80y ky+ was quite a challenge. Mistakes such as • 80y k or y−  • 8 and 0y k y−  • 8 , 0y k y−  are commonly seen. There should be working to indicate which coordinate refers to the maximum or minimum point. 2(b) ( ) ( ) 2 2 2 2 Let f ( ) f ( ) 2 ()2 xkyx xk x k ky x k x k k xy xk kxy k x kxk −== +  +−= = +++ = +  = = ++ f Translation of C by k units in the negative x-direction followed by scaling by factor k parallel to the y-axis OR Scaling of C by factor k parallel to the y-axis followed by translation by k units in the negative x-direction Words that are NOT allowed and will get penalised: • Transform • Shift • Left, right, up, down Key words like “factor of k” is commonly missing.

ANGLO-CHINESE JUNIOR COLLEGE 2026 H2 MATHEMATICS 9758/02 3 (a) ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( ) ( )  ( ) 2 1 1 1 2 3 1 3 1 2 1 13 62 1 2 2 1 1 2 1 2 1 12 1 2( 1) 1 ( )2 n n n r r r r r r r n n n n n n n n n n nn n nn n n n Shown = = = + = + + + +=+  + + +=+ += + +  += + = +    Note: ( ) 1 1 2 n r nnr = += can also be derived using AP Sn formula. 3 (b) Replace k with 3r+ , ( )( ) ( ) ( )( ) ( )( ) ( )( ) ( )( ) ( ) 33 4 3 4 3 1 3 1 2 3 2 2 3 2 33 2 3 3 3 3 83 3 8 31 1 31 1 1 1 1 11 1 nr kr r r n n n rrkk nn rr n rrn nnn nn n nn ++ = + = = = + = + − + −−− = += =+ =+ =+ +   = = +          Replace k with 3r− are seen. 3 (c) When n→ , 1 0n → . ( )( ) 2 3 4 3 3 3 8 1lim lim 1 1 n k n n kka nn→ → = + −−  = = + =  . Hence, 13 1 21 3 S == − . Do NOT write 1a→ 3 (d) Given 1 1 1 n n n uu u + += − and 1 2u = , using GC, it is observed that the sequence has a periodic cycle of 112, 3, , 23−− . Many students are able to observe that there’s a periodic cycle and are able to evaluate the summation.

ANGLO-CHINESE JUNIOR COLLEGE 2026 H2 MATHEMATICS 9758/02 ( ) ( ) 4 1 1123 23 7 6 7 6 n r r un n n = = − − + =−  =−  4(a) 13 3 , 2 10 3 1 2 2 3 1 0 1 1 OA OB AB               −               − = −          −      == =   2 3 1 1 0 5 1 1 3            = −  =     −      n Scalar-product form: 1 3 1 5 2 5 13 3 0 3            ==           r  Cartesian form: 5 3 13x y z+ + = (shown). As this is a shown question, using G.C to solve is not allowed. Note: OA OB  is NOT the normal vector. 4(b) Line of intersection of 1π and 2π : 11 33 38 2 , where 15 3 01                 + −                   r=  This answer can be easily obtained using G.C app under “Simult Eqns”. Many various answers can be obtained manually. Students should note that substituting x, y or z to be zero is an easier way to solve for the common point of intersection. 4(c) Let d be the length of projection of AB  onto 2π , which is the same as length of projection of AB  onto line of intersection found in (b), Common misconception of formulae applied: • 3 0 1 10 AB   −•  • 3 0 1 AB   − 

ANGLO-CHINESE JUNIOR COLLEGE 2026 H2 MATHEMATICS 9758/02 3 0 1 10 23 1 10 10 11 1 1 5 10 3 35 10 7= or 1.87 (to 3s.f.)2 d AB   −=        = −       −     =  =  Or 1 2 213 17 1 2 or 1.87 (to 3s.f.) 14 14 14 13 d AB  −        = = − − =             4(d) l: ( )34 3 1 04 pq p q   −+        −+          r = i j+ i j+ k = l does not intersect plane 1π . l is parallel to plane 1π 31 5 0 3 5 12 0 3 43 q q q         =  + + =  =−           l does not lie on plane 1π , hence point B does not lie on plane 1π 1 1 5 13 5 13 18 03 p pp         −   −              Common misconception of formulae applied: • 3q− , 18p • 3q=− , 18p=

ANGLO-CHINESE JUNIOR COLLEGE 2026 H2 MATHEMATICS 9758/02 4(e) Let  be acute angle between the line l and plane 1π . 18 3 : 1 0 04 l         −+          r= 31 05 43sin 25 35 15 5 35 3 35 30.470 30.5 (to 1 d.p.)                     = = =  =  =  Common misconception of formula applied: • 31 05 43sin 25 35                  = 5(a)(i) 1sin (1 )zx −=− sin 1zx=− dcos 1d zz x =− 10, sin 1 2xz −= = = 11, sin 0 0xz −= = = 10 1 0 2 2 0 sin (1 ) d cos d cos d x x z z z z z z   − − = − =   0 and 2ab == Method 2 1sin (1 )zx −=− 22 2 dz 1 1 d 1 (1 ) 1 sin 11 coscos x xz zz −−== − − − −−== 10, sin 1 2xz −= = = 11, sin 0 0xz −= = = As this is “show” question, students should not skip steps. Students need to show the replacement of x values to the corresponding z values in the integral.

ANGLO-CHINESE JUNIOR COLLEGE 2026 H2 MATHEMATICS 9758/02 2 1 01 0 2 0 sin (1 ) d cos d cos d x x z z z z z z   − − = − =   0 and 2ab == (a)(ii)     2 2 2 2 0 0 0 0 cos d sin sin d cos 122 z z z z z z z z     =− = + = −   Students need to present the application of integration by parts formula clearly. (b) (i) Since curve 1sin (1 )yx −=− is symmetrical about (1,0), area A = area R 2 1 0 sin (1 ) dxx− − = = 1 1 0 2 sin (1 ) d xx− − = 2 1 22   − = − Students need to understand how to remove the modulus before integrating. Students DO NOT need to integrate 1sin (1 ) x− − again since the result from (a)(ii) can be used to evaluate 1 1 0 2 sin (1 ) d xx− − . A R O 2 1 x y B

ANGLO-CHINESE JUNIOR COLLEGE 2026 H2 MATHEMATICS 9758/02 (ii) 1sin (1 )yx −=− sin 1yx=− 1 sinxy=− 1 when 2, sin (1 2) 2 x y − = = − =− 1 when 1, sin (1 1) 0 x y − = = − = Volume 2 2 2 2 2 022 022 022 02 0 2 2 2 2 2 (2) d 2 2 (1 sin ) d 2 (1 sin 2sin ) d 312 cos 2 2sin d22 312 sin 2 2cos24 322 4 322 4 5 24 xy yy y y y y y y y yy               − − − − − =−  = − − = − + − = − − −  = − − +   = − − −   = − − =−     Students cannot simply apply the formula, 2 0 2 dxy − , to find the volume generated about the y-axis. 6 Let the random variable X denote the delivery time of an order in minutes. ( ) 2~ N ,X  ( )P 13 0.2X = 13P 0.2Z   − = 13 0.84162 0.84162 13  − =−  − = ( )P 25 0.26X = 25P 0.26Z   − = 25 0.643345 0.643345 25  − =  + = Some students are