2025 TK Sec 4 Prelim A Maths P1 MS

2025 Add Math Prelims Marking Scheme Qn Solution Remarks 1a = B1 chain Rule B1 diffn of Expo 1b M1 integrate their (a) answer to get A1 o.e 4 2 angle TPR = angle TRP ---(1) (∆TPR is an isosceles) angle TPQ = angle TSP ---(2) (alternate segment thm.) angle QPR = angle TPR – angle TPQ angle TRP = angle TSP + angle RPS (exterior angle of ∆) angle RPS = angle TRP – angle TSP = angle TPR – angle TPQ = angle QPR ⸫ line PR bisects the angle SPQ. (proven) B1 (isosceles) B1 (tangent chord or alternate segment theorem) B1 (exterior angle of ∆) B1 correct answer 4 3a cm/s B1 M1 with correct substn A1 ( )!"#!" !"! ! + ( )! ! !" ! ! " " ! + ( ) ( ) ! ! !" #$ % & !"!" "# '!!" ! !! ! !! " !# "" " !# "" != + ++ !=+ ++ " " ! "!!"+ !!" !#= !" "#!" != !"# "$ ! "" != ! "" " "" " #"#$ %" !! " #" # "" #!! =" #= " ! " #$ " ! " ! =!

3b cm2/s Rate of decreasing = cm2/s M1 with correct substn A1 must be correct 5 4a B2 –1 for each error b Height = Height 62m Since height is 62m, the maximum height reach by the ball is 62 metres. B1 Use of B1 Conclusion c Height = When x = 10, y = 12 >2 The pellet will not hit the object. B1 5 5 Let h be the height of prism B1 V ol of prism o.e B1 correct sq of sides !" #!" != !" !# ! "" != ! "" " """ "# $ % &#%'"! #% !! " #" # ! # ! =" =" # ! "! ! " !=" !!! ! ! " ( ) ( ) ( ) ! ! !! ! ! ! !! " # ! !# " # ! #" #"!# " # ! !! !$ ! $ # ! !$ % ! !! !! !! ! ! !+ + =! ! + "# $%$%=! ! + ! +&' ()()*+*+&',- "#=! ! ! +,- =! ! + ( )! !" # !!!!+ ( ) ( ) ( ) ! ! ! "# !"# ! " $! $! ! ! ! !" !!# !!+ # !!!( )! !" # !!!!+ ( )( )! " # $%&! '" (!" += ° ( )( )! "# !!$ !# % # % & '( ) !=++ ° ( )! "! " ! #$ % $ # ! !!++ = or !"# $%°!"#$ !

cm B1 B1 M1 expansion of denominator A1 6 6a B1 B1 B1 b = Coeff of x = –80 + 32k Coeff of x2 = 52 – 20k –80 + 32k – 20k +52 = –4 –28 + 12k = –4 12k = 24 k = 2 B1 correct coeff x B1 correct coeff x2 M1 their sum = –4 A1 7 7a k = 2 B1 correct expansion of compound angles B1 b P + Q = 2A ---(1) P – Q = 2B ---(2) B1 stating of A or B in term of P and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

B1 correct substn of (1) and (2). c M1 use of (b) or compd ∠ B1 Either special angle correct A1 7 8a Coord B = (5, 5) 5x – y = 4, 5x – x = 4 x = 1 y = 1 Coord A = (1, 1) M1 Sim Eqn to solve for unknown A1 A1 b Coord D = (1 – (5 – 2 ), 1 + 1) = (–2, 2) M1 correct attempt to find x or y A1 c Area M1 determinant method A1 √ 7 9a Let h cm be the height of the casing. Base area = x2 Area of 4 sides = 4xh Total material cost = 2x2 + 4xh $2x2 + 4($1)xh = $600 M1 find area of materials/ find cost for rect piece A1 correct equation for h unsimplified/ Correct base area !"# !"# $!"# %&! $$ !" !"!" +!"# "#+= $% $%&' &' !"# $% !"#&% '( )(*!"# +,!** *-*** ) * °+ ° °°!" !"= #$ #$%& %& !" !"=#$ #$#$ #$%& %& = !"!"!= ! "#!"+= !" # $% $ !! ! " += = = !"!"!= !"#!!# !"$!#=! ! "# $!"% $& %&'()*+, =+ +!! ! = ! ! "## ! $ %##&&&' ! !" ! ! ! != !

V ol of casing, V B1 AG b At stationary value, When x = 10 ⸫ Maximum V olume Maximum V olume B1 M1 set to 0 B1 Test and conclusion of nature √ if min A1 correct volume 7 10a a = 2, b = 3, c = 3 B1 B1 B1 b B1 for 2 complete cycles B1 correct maximum and minimum B1 fully correct curve c When y = 1 Or min point M1 find eqn of y A1 B1 B1 8 ! ! ! " " "## ! "## $%&! !" !! ! !! = !" #= $%&' #= ( )!"# $%% $ "!! " !=! ! "! ! "= !"## " # $# ! ! != = ! ! " # "! " !=! ! ! " # " ! " < ()()! ! !"" #" #" $% &$#"""$'( != !"# $!"=! "#$ % ! ! "#$ % % ! % !% !" !" #" = += + =+ !" # # ! ! += =! !"# $!"=!!=! 0 x y 5– 1–

11a area of the triangle OBX = (7.2)(9) sin θ = 32.4(sin θ) cm2 B1 b Area = 0.5(12)(9) = 54 cm2 B1 M1 use of B1 AG c = = 5 M1 A1 M1 A1 d B1 9 12a When y = 0 x = 0 (n.a) or 4 Coord A = (4, 0) When y = 7.5 x = 1.5 or 2.5 Coord B = (2.5, 7.5) Coord C = (1.5, 7.5) M1 substn to find unknowns A1 A1 A1 12b Area =12 units M1 sum of rectangle and 2 symmetrical parts B1 correct integration to linear x M1 M1 Substn of correct limits A1 9 !"( )() ( )!"# $%&#'() *&% !% '() +,%!!=+ ° " !"# $%&#'() #$&% *+' % *+' ,'() - !! !! =+ += ()() ( )!"#$ !$ %&' ()$ !°" ( )!"# $% &'(&)(!!!°" ! "#$ %$&'!!+( )!"#! !"# !!"#!=+ !"#$ % !&'( ! ! = =° !!!" #=+ !"# ! "!= ( )!" #$ %&'() ! %&'() * %&'+ ! ! ! "° = "° = =° ( )! " ! ! ! ! = ! ( )!" ! ! ! = ! ( ) ( ) ( )( ) ( )( ) !"# $ % ! !"# % !!& !'"# $ $ $ ( $ '"# ) & $ & * ' !"# !"# $ % $ "# !" !! ! ! !! =+ ! ! =+ ! ! !! !! !!=+ ! " ( )!"#$%# &### "! !!! !

13a When x = 1, C = 3 When x = –2, A = 1 3 = A + B B = 2 M1 realising the form (C can be linear) M1 remove denominator A1 A1 A1 –1 mark if no final statement b M1 integrate to ln A1 2 ln (x – 1) B1 M1 substn of limits A1 c When x = 1, the denominator of the integral is 0 and thus does not exist. As such it is not possible to find the value. or Since ln 0 and ln(–3) and ln(–6) do not exist, it is not possible to find the value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