2026 GDLSS P1 MS

Marker’s Report (2026 4Exp Amath Prelim) Paper 1 Q Descriptor Remarks 1 2 d 8d 2 9696 2 2 d 2 d d d d d d d82 d d8 96d 2 ( )2 d1 /d 12 A t Cr rr Ar A rdr A A r t r t rr t r t r cm st         = = =  = = = = = = = Some students still struggle to form the required connected rate of change equations. Errors were also made with multiply/division due to incorrect equation setup. 2a 52 5 2 4 2 4 4 2 44 ( ln ) 2 ln (5 ) 2 5 ln 2 10 ln d xxdx xx x xx x x x x x x =+  =+ =+ Generally okay. Except a few who made mistakes in differentiation of ln x. 2b 2 24 4 5 2 11 22 44 11 2 2 45 1 1 2 4 1 2 4 1 2 4 1 2 10 ln ln 10 ln 2 [32ln 4] 210 ln [32ln 4] 5 1ln [ 12.4 [32ln 4]]10 ln 3.196141... ln 3.20 (3 ) x x x dx x x x x dx x dx x x dx x x x dx x x dx x x dx sf +=  =− + = − + = − + =        Poorly done overall. Many students did not recognise the “hence” requirement or use the answer from the previous part to proceed. Some did not realized that it was a definite integral and incorrectly included +c.

3 ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) 180 180 tan tan 180 tan180 tan 1 tan180 tan 0 tan 1 0 tan tan tan tan 1 tan tan 31 1 3 1 4 2 2 A B C C A B C A B AB AB AB AB AB AB AB + + =  = − + = − + − += ++ −+= ++ =− + +=− − +=− − =− − = Trigonometry: Poorly done overall. Many students failed to recognise and use 𝐴 + 𝐵 + 𝐶 = 180° to proceed. A common error was treating tan𝐶 = tan[180 − (𝐴 + 𝐵)] as tan180 − tan(A + B) instead of applying addition formula. Some students also disregarded the instruction to solve without a calculator and used the inverse tangent function. 4 1 1 1 --- (1)2 2 3 0 --- (2) xy xy += + + = From (2) 2 3 0 2 3 --- (3) xy yx + + = =− − Sub (3) into (1) A handful of students attempted the question without first solving for the coordinates of A and B.(full marks given) Some made errors when solving the quadratic equation, leading to incorrect coordinates. The use of gradients to prove 12 1mm =− was generally well done. However, students need to present their explanations more clearly and logically. Some stated that “the gradient is the same” when they meant that the gradients satisfy the required relationship, which is inconsistent with their proof that 12 1mm =− .

( ) ( ) 2 2 2 1 1 1 2 1 1 1 2 3 2 2 3 1 2 3 2 31 2 3 2 2 6 2 3 2 6 0 32, 2 32,1 , , 62 xy xx xx xx x xx x x x xx xx += +=−− − + + =−− −+ =−− − + =− − + − = =− = −−  From 2 4 0yx− − = , 24 22 yx xy =+ =+ Gradient = 1 2 Gradient of AB ( ) ( ) 61 3 22 2 −−= −− =− Since 1 212− =− , a line joining A and B is perpendicular to the line 2 4 0yx− − = . 5i ( ) 62 1 6 r r r mTx r r−+  =  ( )2 6 0 12 2 0 3 12 4 rr rr r r − − = − − = − =− = Generally well done. Many students were able to use the general term or expand the binomial to find the term independent of x.

( ) ( ) ( ) 24 421 5 4 4 6 4 15 1215 81 3 T x m x m m m −= = = = 5ii ( ) 62 1 6 r r r mTx r r−+  =  ( )2 6 3 12 2 3 39 3 rr rr r r − − = − − = − =− = ( ) ( ) ( ) 33 321 4 3 6 33 20(27) T x x x −= = ( )( ) 33 33 3 1 1215 20(27) 1215 540 1755 xx xx x ++ =+ = Generally well done. However, presentation can be improved, especially as this was a “show that” question. Many students simply wrote (540+1215=1755) instead of showing the terms (1(540x3)+1215(x3)) and explaining how these give the coefficient of (x3). 6a (alt. segment theorem) (alt angle, parallel ) FED EGH FED EDC EB DC EGH EGH  =  =  = Generally well done. However, some students struggled to identify the correct angles for the Alternate Segment Theorem. They often matched angles that were not formed by the tangent and chord, or used angles involving points that were not on the circumference. Students need to remember that the chord must have endpoints on the circumference, so a point such as D, which is not on the circumference, cannot be used as the angle.

6b (alt. segment theorem) ---- ANGLE (alt. segment theorem) (corr angle, parallel ) ---- (ANGLE) is similar to BEG GHE HGE DEF CBE DEF EB DC HGE CBE BCE GEH  =  =  =  =   Some students gave a list of angles without clearly identifying which were relevant to the question or how they contributed to the proof. Some also struggled to present the final angle equality clearly based on the results established earlier. 6c Since Triangle BCE and Triangle GEH are similar, BC CE GE EH BC ED ED EC EG EH ED EH EC EG = = =   =  Not very well done overall. Most students attempted to prove that triangle BCE is similar to triangle DEC, and hence to triangle GEH. However, many did not match the corresponding vertices correctly, even though they still arrived at the given ratio. Many also failed to use BC=ED to establish the required relationship and complete the proof. 7 ( ) ( ) 2 3 8 3 6 46 3 8 3 6 46 4 6 4 6 3(50 20 6) 16 6 15 6 6 r + = − + += −+ += − =+ 2 22 8 6h =− ( ) ( ) 2 2 2 22 8 6 15 6 6 37 2 6 l h r=+ = − + + =− Not as well done. Some students used the wrong formula, such as 2rh or 1 3 rh . Some also did not see the pie in the volume given (8 3 6) + when substituting into the given volume expression. Others did not realise that that they need to rationalize. However, most students were able to recognise and apply Pythagoras’ theorem to find.

8a 2 2 2 2 2 2 2 128128 22 1282 2 ( ) 2562 V r h r h h r S r rh S r r r Sr r      = =  = =  + =+ =+ Generally well done. 8b 2 2 2 2 3 3 2562 2564 at stationary point, 0 25640 2564 4 256 64 4 Sr r dS rdr r dS dr r r r r r rr      =+ =− =  − = = = =  = Nature of stationary point 2 2 23 2 23 4 256 5124 when 4, 5124 37.699 04 It is a minimum stationary value. dS rrdr dS dr r r dS dr    −=− =+ = = + =   Generally well done. Students used either the first or second derivative test appropriately. Most errors were made during the differentiation process, particularly when handling terms involving . 9ai 22 22 2 2 22 24 bax bx c a x x c a bba x c aa bba x c aa + + = + +    = + − +       = + − + Poorly done. Students need to strengthen their algebraic understanding of the steps required for completing the square. Common mistakes: - not dividing b/a by 2 - wrong expansion of a( 𝑏 2𝑎) 2

9aii The turning point is 2 ,24bb caa− − + Generally ok. Some forgot to include the c. 9b 2 2 41 4 1 0 x x k x x k − − = − + − − = ( )( ) ( ) 2 2 40 (4) 4 1 1 0 16 4 1 0 16 4 4 0 4 12 3 b ac k k k k k − − − − −  + − −  − −  − −  Generally ok. Common mistakes: - expansion of 4(1+k) - D>0 - change of inequality sign wrongly for −4𝑘 < −12 10a 3sin 2x− Amplitude = 3 Period = 2 2  = Generally ok. Common mistake: amplitude = −3 10b Common mistake - wrong shape, not accounting for the −3 - wrong number of cycles 10c 3 sin 2 0 3 sin 2 3sin 2 xx xx xx    − − = −= −= Draw xy = . From the graph, there are 5 solutions Common mistake was - not manipulating 𝑥 𝜋 correctly - writing 4 solutions instead of 5

11a 4sin 2sin 42 3cos 2cos 32 2sin 4 2cos 3 2sin 2cos 7 D x y y y x x D D     =+ −=  = + −=  = +  = + + + = + + Clearer working must be shown. Many students just added 4 and 3 to the final answer without showing clearly how they did it. Some added 4 and 3 to the wrong value, showing their misconception. 11b 22 1 2 2 8 2tan ( ) 452 2sin 2cos 7 8 sin( 45 ) 7 R D D    − = + = == = + + = + +   Generally ok. 11c Max value = 87+ Occurs when sin( 45 ) 1 45 90 45    += += =    Some students still do not know that the maximum value occurs when sin(𝜃 + 45∘) = 1 12a 22 2 2 22 22 2 2 3 2 3 2 ( 1)( 1) ( 1)( 1)( 1) 32 ( 1) ( 1) 1 ( 1) ( 1) 3 2 ( 1) ( 1)( 1) ( 1) 5when 1, = 4 5when 1