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 Q ( ) ( ) !" !# $ "%! $ ! &" &# $' "$ " &" &# $ "" # $' "" # $ "" # $ #!(") & " #$ #! " #$ %* !+ #$ ,& $ ! & #! # ! ! += + += + +!=" +! !+ != ! != !! "=! + ! " ! #$ %&'()*+)'+! += !"#$%&'()*+,-!# !"#$%&'()*+,-!#! () () () ! "!# $ " "# !!" " " %%%&" ## $" "' ! %%% ! !! !! !"#$%&' !" !" !" !"=+ # + # +$% $% $% $%&' &'&' &' =# + + ( )!"!" "++ ! " # !!"#$%&' ( )!"!" "++ ( )!"! !# $ %%%!!!++ ( ) ( )!"# !"# !"# $%! $%! !"# !"# $%! $%! !"# &!"# $%! !" !" !" ! " !" ! " !" ++ ! =++! = ! !"# += ! !"# != ( ) ( )!"# !"# $!"# %&!!" !" ! "++ !=

B1 correct substn of