RVHS 2026 9758-01 Prelims MS

1 ©RIVER VALLEY HIGH SCHOOL 9758/01/2026 Solutions and Comments for H2 Mathematics P1 2026 1 Solutions [6] Maclaurin Series. (a) ( ) 1tany px −= Differentiate with respect to x, ( ) ( ) 22 22 d d 1 d1 Shown d yp x px yp x p x = + += Alternative tan y px= Differentiate with respect to x, ( ) ( ) ( ) 2 2 22 dsec d d1 tan d d1 Shown d yyp x yyp x yp x p x = += += Generally ok. (b) Differentiate with respect to x again, ( ) ( ) 2 2 2 2 2 dd1 2 0 dd yyp x p x xx+ + = . Differentiate with respect to x again, ( ) ( ) ( ) ( ) 3 2 2 2 2 2 2 2 3 2 2 d d d d1 2 2 2 0 d d d d y y y yp x p x p x px x x x+ + + + = At 0x = , ( ) 1tan 0 0y −== , d d y px = , 2 2 d 0d y x = , 3 3 3 d 2d y px =− . By Maclaurin’s Theorem, 3 23 3 3 020 2! 3! 3 py px x x ppx x −= + + + +  − Generally well done. Common errors involve mistakes in the implicit differentiation steps resulting in not having 2 2 d 0d y x = and thus the first 2 non-zero terms involve the terms in x and 2x instead which is wrong. Presentation error: The first 2 non -zero terms should be 3 3 and 3 ppx x , not p and 3 3 p− . 1 It is given that ( ) 1tany px −= , where p is a constant and ( ) 1ππ tan22 px−−   . (a) Show that ( ) 22 d1 d yp x p x+= . [2] (b) By further differentiation of the result in part (a), find the first two non-zero terms in the Maclaurin series for y. [4]

2 ©RIVER VALLEY HIGH SCHOOL 9758/01/2026 2 Solutions [6] Inequalities. (a) ( 1.2581,1.7158)A =− and (1.2770,1.5857)B = From the graphs, for 2 1e2 1 x x − − , 1.26 or 1 1 or 1.28x x x − −    (3 s.f.) Not well attempted as students failed to see the whole graph and hence missed out the answer, 11 x−   . Students need to understand the difference between the use of “or” and “and” in the final answer. (b) 2 1 1 y x = − 22 013 ( 2)( 3) 2( 1) 0( 1)( 3) x xx x x x xx + −−− + − − − −− 2 34 0( 1)( 3) ( 1)( 4) 0( 1)( 3) xx xx xx xx −− −− +− −− 1 or 1 3 or 4x x x  −    Some students went ahead to do cross multiplication and were penalised. Students need to learn to read more carefully as there are several different combinations of the solution. Students need to understand the difference between the use “or” and “and” in the final answer. 2 (a) Solve the inequality 2 1e2 1 x x − − using a graphical approach. [3] (b) Without the use of a graphing calculator, solve the inequality 22 13 x xx + −− . [3] e2xy =− 2 1 1 y x = − A B 1x = 1x =− –1 1 3 4 + + + – –

3 ©RIVER VALLEY HIGH SCHOOL 9758/01/2026 3 Solutions [6] Implicit Differentiation Differentiate 2 3 3e 0 yyx −− + = implicitly with respect to x, ( ) ( ) dd 2 3 3 e 0dd d1 3e 2 3 d d 2 3 d 1 3e y y y yy xx y x y x − − − − + − =  −=  = − Gradient of normal = 1 3e 23 y−−− . ( ) o1 3e tan30 23 11 3e 2 3 3 e1 0 y y y y − − − −−= − + =  = = When 0y = , 2 3 3 0 3 2 x x − + = = 3Coordinates of     is  ,02P   Exact equation of the normal to C at P: 130 23 1 .23 yx xy − = −  =− Not done well. Most are able to do implicit differentiation with a handful unsuccessful in differentiating e y− with respect to x. Some did not understand that otan30 gives the gradient of the normal and simply ignored the given info while many actually equated it to d d y x which is the gradient of the tangent instead. Some assumed y = 0 due to the information given as “made an angle with the x-axis”. While coincidently point P is on the x-axis, this assumption cannot be given credit. Some still are not able to produce the correct equation of a normal, mainly still using the expression for d d y x for its gradient instead of the value of d d y x at point P. 3 A curve C has equation 2 3 3e 0 yyx −− + = , for 0x  . The normal to C at a point P has positive gradient and makes an angle of o30 with the x-axis. Find the exact equation of this normal. [6]

4 ©RIVER VALLEY HIGH SCHOOL 9758/01/2026 4 Solutions [8] Abstract Vectors (a) 2 3OX = a  and 3 5OY = b  Method 1: : 3 ,5 AYl AY   =+ =+  −  ra a b a   ( ) 3 , for some 5 31 5 OP     = + − = − +   a b a ab    : 2 ,3 BXl BX   =+ =+  −  rb b a b   ( ) 2 , for some 3 2 13 OP     = + −  =+  − b a b ab    Majority was able to use the correct approach, but there was also many who were unable to start the question. The most commonly seen method was the intersection of lines. Some used the ratio theorem method which was slightly more complicated with , 1–, , 1–. Other mistake: Equating AY  and BX  , instead of the vector equations of the 2 lines. AY  and BX  are directional vectors (which are not localised), but the 2 lines are! Method 2: ( ) 3 5 31 5 OP OA AP AY   =+ =+ = + −  = − + a a b a ab   4 With reference to the origin O, the points A and B are such that OA = a  and OB = b  , where a and b are both non-zero and non-parallel. The point X lies on OA such that : 2 :1OX XA = . The point Y lies on OB such that : 3: 2OY YB = . The lines AY and BX intersect at the point P. (a) Express OP  in terms of a and b. [5] (b) Point Z lies on AB such that 43AZ BZ= . Express OZ  in terms of a and b. [1] (c) Show that O, P and Z are collinear. [2] O B A X Y P Z O B A X Y P Z

5 ©RIVER VALLEY HIGH SCHOOL 9758/01/2026 ( ) 2 3 2 13 OP OB BP BX   =+ =+ = + −  = + − b b a b ab   As a and b are non-parallel and non -zero, comparing the coefficients of a and b, 21 ---(1)3 3 1 ---(2)5   −= =− Solving (1) & (2) using GC, 52 and 93== ( ) 31 5 5 3 51 9 5 9 41 93 OP = − +    −+       + ab = a b = a b  After obtaining a correct equation involving  and , students naturally solved the unknowns by comparing the coefficients of a and b. However, this comparison worked because a and b were non-parallel and non -zero. This condition must be stated! (Less than 5 students stated this!) (b) 43 43 77 43 77 AZ BZ OZ OA OB = =+ += a b  Students who did not get this right made mistake with the ratio. 43 () 43 34 AZ BZ AZ BZ = = not (c) 41 93 7 4 3 9 7 7 7 9 OP OZ =+  + = ab = a b   43 77 9 4 1 7 9 3 9 7 OZ OP =+  + = ab = a b   Students generally understood that collinearity required the parallel vectors condition , but few students did not state the common point condition. ... 82 63 21 2 7 PZ OP = + = = a b   accepted //OP OZ  Since O is the common point, O, P and Z are collinear. A Z B 3 4

6 ©RIVER VALLEY HIGH SCHOOL 9758/01/2026 5 Solutions [7] Integration Techniques (Trigo, by parts) (a) ( ) ( ) ( ) ( ) ( ) ( ) 2 32 32 3 2 cos3 cos 2 cos cos 2 sin sin 2 cos   2cos 1 sin   2sin cos 2cos cos 2sin cos 2cos cos 2 1 cos cos 4cos 3cos cos 4cos 3 . Shown x x x x x x x x x x x x x x x x x x x x xx xx =+ =− = − − = − − = − − − =− =− Almost all could show the result. (b) ( ) ( ) 2 2 2 d 1 cos3 tan3 d 11cos3 tan d33 11cos3 tan d33 11cos3 tan 4sin 3ln sec tan33 14cos3 tan sin ln sec tan .3 tan sin 3 1sec cos3 3 1 cos 4cos 3cos 4cos 3sec 3 x xx x x x x x x x x x x x x c x x x x x xx xx xxx xx c − =− − = − + = − + =−  −  − + − + + = + − + +  −     For integration by parts here, u must be tan x. If u = sin3x, the integral will become too complicated to solve ( 3ln sec cos3 )x x dx . As shown in the solution, when u = tan x, cos3x