SCGS 2025 AM Prelim Paper 1 solutions

2025 A Math Prelim Paper 1 Solutions 1. (a) (b) 2. Sub (1) into (2) Length = = = units ! " #" "$ %%%&%$ !" " " ! ! "#= $%&' () ! ! "# "$ "" " "" " #"%& '$" "% ()%(&*+,-."# ! "" !! " #" # " # " # ! ! ! = =" =" == !" # ! $!"=! ! ! ! ! ! !! "# $ ! %!! " "!!+ = ! ! ! ! ! !! !! ! ! ! "# ! $ "# ! $ % & !# ' '% & (' & (' & " '$" #$ & !! ! ! !! ! !! !! !! !! !! ! ! + = !+ ! + ! += !+ + = !! = !+ = ! "#$$$ % & "#$$$$' ! " =! =! ( ) ( )!! "# $%++ ! ! !"# !!

3(a) (b) 4(a) (b) !"#$ !%&"#& &!%&#&!%&#' ' & & &&& != !+= !+=! !" !" "!""!! ! ! !!! () !! !! ! !! ! !! ! !! "# #!! $"" $ ! ! %&# ' ( & ! ) ' %( % ( & ! ) ' %( & ! ) ' % ( %( & ! ) ' ! %( & ! # ' %(& ! # ' # *#""" !! !! ! !! ! !! ! !! !! "! ! ! " # !" ! " ! ! " # !" ! ! " " ! # !" ! ! " " # !" ! ! " # !" ! ! " " " != ! + != ! + =! + + =! + + =! + "#=!$% =! ! = & && && & & & !" ##!$%&#'()#&*+" ## #,'#,'-'.(%(' !"# !#" !" !# #!$ #!$ !#$ !$# != ! = =! != ! "# ! "#$%& '(&)&%*+),-.#/ &+0&+0'01#.#02 "),-.#0&+,&%3#&0)4#&0#-2 ").%&0#-&%3#'*#42 ! != ! ! " != ! != ! #! = ! !"# "!$ !"$ !"# #%& "!$ "%& #%& "%&

5(a) Since there is only one intersection point between the curve and the line, the line is a tangent to the curve. Or Discriminant = \ line is a tangent to the curve Coordinates = (b) and 6(a) 0 \ is a factor since remainder = 0 (b) Sub x = 2, Compare x2, Compare constants, !!" # $ %!! !!+ = ! !!" ! #!!!+ = ( )! !"#!!= !!=() ()()! "" ! ! #!! = ( )!" #! ( )! "# $ %!"" !+ + > ( )! "# $ % &!"" !+ +  > () ( )( )! "# $ % & '!!!! + ! < !"!+>!"# $! % !& &!!!+ + < !!>!"! # #$ %!! > !"! ! # $!! > ( )( )!" ! #!!+ > !"#$$$ %%!!< > !!> !" ! ! !# !$ ! ! !! ! ! ! +! + ! ( )!"!!!+ !"!!!! ( )!"!!!! ! !!"!+ ( )( ) ! !! "" # "" !! "!! " !! " # ! A !!!! !! + =+!+!+ ( )( )( )!!"" # "" ! " !!! " ! # ! A !!! = + ++ ! !! "!!!= " = ! !" #!" "!= + " = ! !! " #!" "!=! "=

7(a) Amplitude =3 Period = or 720o (b) (c) ( )( ) ! !! "" # "" " # $ !! "!! " !! ! !!!! !! !"= ! + !+!+ !! !"#$ % & '("&& )*+,'-. & '(" & /0N2+3.(4."(50,#($".S.& !! !" += ! =! y x 1 4 -2 2p 4p y x 1 4 -2 4p -2 2p

8(a) Centre of = (2, y) Grad of normal = = 3 \ Centre of = (2, 3) Radius = 5 units Equation of : (b) Let the centre = (x, y) Equation of : 9(a) (b) !!!"!" #! $ !!=! ()!"#"=! +! !!!!( ) ( )!! !" ! #!"!+ != !! ( )!"#$ # $!! !"++!" =#$%& !"# $!"== !!( ) ( )!! "# $ !%!"!+ ! = ( ) ( ) !! ! !! ! ! "#$%&' "#$ (' )& *************+* "#$ (' ,!" )& ,!" (' "#$ )& %% % -*************+* .... %-*************+ .. %*************+ . )( =+ + !"!"!"+ #$#$ #$ #$%& %& %& + + ( ) ( ) !"# $ !"# % !"# !"# #&' #&' $ !"# !"# #&' #&' % % !"# !"# % #&' #&' $ !"# !"# $#&' #&' (!"# !"# )*#&' #&' *!"+ !"+ ( !" !" !" ! " !" ! " !