CAT HIGH 2025 AMATH PRELIM P2 MS

1 Jane bought an electric car in January 2025. After purchase, the value of the car, $V , diminishes over time and can be modelled by 38000 ktVe −= , where k is a positive constant and t is time measured in years. The value of the car is expected to be $29000 after 3 years of driving. Jane intends to sell her car when its value drops to half of its original value. Showing clear mathematical calculations, find the year that Jane is likely to sell her electric car. [5] 3 3 29000 38000 29 38 293 ln 38 1 29ln 0.0900963 38 k k e e k k − − = = −= = −≈ 0.090096 0.090096 19000 38000 1 2 10.090096 ln 2 7.69 years t t e e t t − − = = −= ≈ The year is likely to be 2032. ______________________________________________________________________________________ 2 (i) Given that the curve 2 21y ax bx=+− lies entirely below the x-axis, determine the conditions that must be applied to the constants a and b. [2] 0a< for maximum curve (not required) ( ) ( ) ( ) 2 2 22 2 4 1 0 for no roots 4 40 0 or ba ba ba a b − −< +< + < <− (ii) If a and b are both integers, state an example of the values of a and b which satisfy the conditions found in (i). [2] 5a=− 1b= ______________________________________________________________________________________

3 (a) The variables x and y are defined such that 2 49log 3log log 3xy−= . (i) Give a reason why x and y must be positive numbers. [1] x and y must be positive for 2log x and 2log y to be defined / exist / calculable / computed. (ii) Express x in terms of y. [5] 2 2 2 2 2 22 23 22 2 2 3 2 3 23 3 log 1log 3 log 4 2 3log 1log 22 2 log 3log 1 log log 1 log 1 2 2 2 yx yx xy xy x y x y xy xy −=  −= −= −= = = = = (b) Solve the equation ( ) 15 2 5 11xx+− += , leaving non-exact value(s) of x in the form loga b . [4] ( ) 15 5 2 11 5 x x += Let 5xw= ( )( ) 2 1 5 25 11 5 11 2 0 51 20 1 or 25 5 5 or 2 1 or log 2 x w w ww ww w x − += − += − −= = = =− ______________________________________________________________________________________ A1

4 (a) (i) The polynomials ( ) 32P 10x ax x bx= +++ and ( ) 32Q5x x ax x b=+ −+ leave the same remainder when divided by 2x+ . Show that 44ab+= . [2] ( ) ( ) ( ) ( ) ( ) ( ) 32 3 2 2 2 2 10 2 2 5 2 8 2 14 8 4 10 12 3 12 44 ab a b ab a b ab ab − + − +−+= − +− −−+ − − + = −+ + + − −= − += (ii) If ( )P x′ has a factor of 32x− , find the values of a and b. [4] ( ) 2P 32x ax x b′ = ++ 2 2P0 3 2232 033 44 33 ab ab ′ =  + +=    += − 43 4 4 4 from (i) 28 4 2 ab ab b b a += − += =− =− = (b) Solve the equation 322 4 5 70xxx+ − −= , expressing non-exact solutions in the form 1 2ab± where a and b are constants to be determined. [5] ( ) 32P 2 4 57x xxx= + −− ( ) ( ) ( ) ( ) 32 P1 21 41 5170−= − + − −−−= 1x+ is a factor of ( )P x ( ) ( ) ( ) 2P 12 7x x x Bx=+ +− Term in 2x : 2 2242 2 x Bx x B = + = ( ) ( ) ( ) 2P 12 2 7xx xx=+ +−

For 22 2 70xx+ −= , ( )( ) ( ) 22 2 42 7 22 2 60 4 11 1522 x −± − −= −±= = −± 111, 1522x∴= −−± ______________________________________________________________________________________ 5 The depth of water, h metres, in a shallow port on a particular day is modelled by the formula 7 2sin 6ht π= +  , where t is the number of hours after midnight. (a) State the period of h. [1] 2 12 hours 6 h π π= = (b) Use the model to predict the time when the depth of water is at its lowest. [2] h is lowest when sin 16 tπ =− 3 62 9 t t ππ = = Time is 9 am (or 0900 hrs) A supply boat docks at the port at 5.30 am and it takes the workers 3 hours to unload its cargo immediately after it docks. To leave the port safely, the supply boat requires the depth of water to be at least 5.6 metres. (c) Determine the earliest time, to the nearest minute, that the supply boat can leave the port. [4] ( ) ( ) 1 7 2sin 5.66 7sin 6 10 7basic angle sin 0.7753910 0.77539 3.9169 , 2 0.77539 5.50776 7.4808 ,10.519 t t t t π π π ππ − +=   =−  = ≈   = +≈ −≈ =

Two critical timings are 7.29 am (rejected since boat is still unloading) and 10.31am. Hence, earliest time boat can leave the port is 10.31am or 10.32am (or 24hr format). ______________________________________________________________________________________ 6 The variables x and y are known to be related by the formula qy px x= + where p and q are constants. (a) The table below shows some experimental values of x and y. x 1 2 3 4 5 y 8.55 6.20 5.32 5.72 6.00 (i) On the grid below, plot xy against 2x and draw a straight line graph. [2] x2 1 4 9 16 25 xy 8.55 12.40 15.96 22.88 30.00 0 15 20 10 5 30 25 xy 15 20 10 5 30 25 0 35

(ii) Use your graph to estimate (a) the value of p and of q. [3] 2xy px q= + ( ) 26 17.5 20 10.5 0.895 acceptable 0.8 1.1 p p −= − ≈ ≤≤ ( )8 acceptable 7.5 9qq= ≤≤ (b) the positive value of x that satisfies the equation 20qpx xx+= . [2] 20 20 y x xy = = from graph, 2 13.25x = ( )3.64 acceptable 3.60 3.75xx= ≤≤ ( b) Explain how another straight line graph can be obtained from qy px x= + by plotting x y against 1 xy , expressing clearly the gradient and vertical intercept in terms of p and/or q. You need not draw this straight line graph. [3] 2 1 1 11 xy px q px q y xy px q y xy xq y p xy p = + = + = −+ = −+ gradient q p=− 1vertical intercept p= ______________________________________________________________________________________

7 The equation of a curve is ln 2xy x= . (i) Find an expression for 2 2 d d y x and show that 2 22 d d1 20dd yyx x xx+ += . [5] 2 2 1 2 ln2d 2 d 1 ln2 xxy x xx x x  ×−= −= ( )( ) 2 2 24 4 3 1 2 1 ln2 2d 2 d 2 2 ln2 2ln2 3 x xxy x xx xxx x x x x − × −−= −− += −= 2 2 2 3 22 2 d d 1 2ln2 3 1 ln2 122dd 2ln2 3 2 2ln2 1 0 yy x xxx x xx x x x xx x −−  + += + +      −+− += = (ii) Find the exact coordinates of the stationary point of the curve and use the second derivative test to determine if the stationary point is a maximum or minimum. [5] 2 d 1 ln 2 0d ln 2 1 2 2 ln 2 2 y x xx x xe ex ey e e −= = = = = = = Stationary point is 2,2 e e   or 1,22 e e− 

( ) ( ) 2 32 2 33 d 2ln 3 d 2 81 or or 0.398 2 0 ex ye x e e e = −=  −− = − < Hence, 2,2 e e   is a maximum point. ______________________________________________________________________________________ 8 The diagram shows two rods AB and BC of length 2 m and 5 m respectively, joined together at B such that angle 90ABC = ° . Small rings are attached to A and C so that A can move along the vertical pole OY and C can move along the horizontal pole OX. The rod BC makes an angle θ with OX. (i) Show that 5sin 2cosOA θθ= − . [2] sin5 5sin BP BP θ θ = = cos2 2cos BQ BQ θ θ = = 5sin 2cosOA BP BQ θθ=−= − (ii) Express OA in the form ( )sinR θα− where 0R> and 0 90α°< < ° . [3] ( )5sin 2cos sin cos cos sinOA Rθ θ θα θα=−= − cos 5 sin 2 R R α α = = ( )29 sin 21.8OA θ= −° O A B C 2 5 Y X P Q 2tan 5 21.801 α α = ≈° 2 22 2 5 29 29 R R =+= = M1 - substitute their x value into their

(iii) Given that A can be no higher than 3.5 m above O, calculate the greatest possible value of θ. [2] ( ) ( ) 1 29sin 21.801 3.5 3.5sin 21.801 29 3.521.801 sin 40.536 29 greatest 62.3 θ θ θ θ − − °= − °= − °= ≈ ° = ° ______________________________________________________________________________________ 9 A particle moves in a straight line with a fixed point O. The velocity of the particle, v m/s, is given by 25 1021