PH 2021 AMath Prelim Paper 2 Anskey

2021 PHSS Prelim AMATH Paper 2 Solutions Q SOLUTIONS MARK REMARKS 1 Express 23 3 12 3 xx xx −+ + in partial fractions. 23 3 12 3 xx xx −+ + = 2 3 1 3 21 3 xx xx −−+ + 22 32 1 3 2 1 3 2 3 ( 3) x x x x x x x x − − − − =++ 2 22 1 3 2 ( 3) 3 x x A Bx C x x x x − − + =+++ 221 3 2 ( 3) ( )x x A x Bx C x− − = + + + Sub x=0, 1 3A= Sub 1 3A= , 2 2 23 9 6 3 3 3x x x Bx Cx− − = + + + Comparing coefficients of 2x , 6 1 3 B− = + 7 3B=− 3C =− 23 3 12 3 xx xx −+ + = 2 1 7 91 3 3( 3) x xx ++− + 2i Explain clearly how 0N and b can be calculated when a straight line graph of ln N against t is drawn. 0 btN eN = 0 btN N e= 0ln ln( ) btN N e= 0ln ln ln btN N e=+ 0ln lnN N bt=+ Gradient = b Intercept = 0ln N Convert logarithmic to exponential form/ Solve logarithmic equation to find 0N

2ii t 2 4 6 8 10 ln N 4.19 5.09 5.99 6.89 7.79 2iii Use your graph to estimate the values of 0N and of b. 0ln 3.29N = ( 0.05) 0 26.8N = b = 0.5 ( 0.1) 2iv A state of emergency will be announced in Tiger City when the number of the virus reaches 5000. There are rumours online claiming that the announcement will happen after 13 hours. Using your graph, explain clearly if this is true. ln(5000) 8.517= From graph, t = 11.5 The state of emergency will be announced between 11th and 12th hour. The rumours are false. 2v The World Health Organisation (WHO) updated the health advisory and suggests that due to a mutation in the virus, N and t are now related by the equation 3ln 150 6 2 oN N bt− = + , where 0N and b are constants. The new straight line graph of ln N against t has a ln N -intercept of 180.16. Find the new value of 0N .

3ln 150 6 2 oN N bt− = + 2ln 50 2 3 o btNN− = + Intercept 050 2 N=+ 0 180.16 50 2N −= 0N = 65.08 OR Sub t = 0, 3ln 150 6 o NN −= 0N = 65.08 3 The diagram shows part of the curve 32yx=+ , meeting the tangent at P, where x = 3. 3i Find the equation of the tangent. 32yx=+ ( ) ( ) 1 2 d1 3 2 2d2 y xx − =+ ( ) 1 2 d 32d y xx − =+ At P, gradient of tangent, 3 d1 d3 x y x = = At P, y = 3 Equation of the tangent is 1 23yx=+ 3ii The area bounded by PQ, the line x = 3 and the line x = a is given as 24 units2. Show that a = 9. Area under line PQ 1 3 2 ( 3) 2423 a a  + + − =  OR ( )( ) 2 3 2 2432 a x x +=  2 12 189 0aa+−= ( 21)( 9) 0aa+ − =

a = 9 ( 21)a− 3iii Find the area of the shaded region bounded by PQ, the curve and the line x = a. Area under the curve PQ = 9 3 3 2 dxx+ = ( ) ( ) 9 3 2 3 32 3 22 x   +  = ( ) ( ) 33 22 1 21 93  − Shaded region = ( ) ( ) 33 22 124 21 93 −−  0.922 units2 4a A curve has equation 2( 1)yk x− + = and a line has equation 1y kx+= , where k is a constant. Find the set of values of k for which the curve meets the line. 2 ( 1) 1k kxx+ + = − 22 ( 1)k x kx x+ + = − 2 ( 2) 2 0kx k x− + − =   2 ( 2) 4( )( 2) 0kk− + − −  2 4 4 8 0k k k+ + +  2 12 4 0kk+ +  At 2 12 4 0kk+ + = 212 12 4(1)(4) 2k −  −= 6 32k =−  6 32k− − or 6 32k− +

4b The height above the ground, in metres, of the rooftop of a building is modelled as 2( ) 6 11P x x x=− + + , where x is the horizontal distance from the main office. The rooftop in the neighbouring school is modelled by ( ) 17R x x=− . Find the values of x, in metres, for which the rooftop of the building is above that of the school. 2 6 11 17x x x− + +  − 2 7 6 0xx− +  ( 1)( 6) 0xx− −  1 < x < 6 4c The curve with equation 2( 1) 2y a x bx b= + + + + , where a and b are constants, is always above the x-axis. Write down two conditions which apply to a and b. 1a− 2 4( 1)(2 ) 0b a b− + +  5a Express 2 23 x x− in the form 23 ba x+ − where a and b are constants. Hence, 2 d23 x xx− . 2 3 3 3 12 3 2 3 x xx −+ =+−− 1 331 d ln(2 3)2 3 2 x x x Cx+ = + − +− 5b Differentiate ( )ln 2 3 2 xx − with respect to x. ( )ln 2 3d ln(2 3) 1 2d 2 2 2 2 3 xx xx xx − −= +   − ln(2 3) 2 2 3 xx x −=+ − 5c Using the results from (a) and (b), find 2ln(2 3) d3 x x−  . From (b),

2 ln(2 3) ln(2 3) d2 2 3 2 x x x x xCx −− + = +− 2 ln(2 3) ln(2 3)dd2 2 2 3 x x x x x x C x −− = − + − 3 2ln(2 3) 4 ln(2 3) 4dd3 3 2 3 2 3 x x x x x x C x −− = − + − From (a), 1 23 d ln(2 3)2 3 2 x x x x Cx = + − +− 4 2 2 2 d ln(2 3)3 2 3 3 xx x x Cx = + − +− From (a) and (b), 5 2ln(2 3) 2 ln(2 3) 2d ln(2 3)3 3 3 x x x x x x C−− = − − − + 6i A guide brings her visitors to walk along a trail consisting of pathways AB, BM, MC and CB. Show that T m, the distance of the trail, can be expressed in the form 720 360sin 360cosT = + + . ABC is an isosceles triangle. Perpendicular (90o) from B to M, the midpoint of AC. cos 360 CM = 360cosCM = sin 360 BM = 360sinBM = Total distance = 720 360sin 360cos++ 6ii Express T in the form sin( )pR ++ , where R > 0 and α is an acute angle. 720 sin( )TR = + + 22360 360R=+ = 259200 = 360 2 360tan 1 360 ==

o45 = o720 360 2 sin( 45 )T = + + 6iii Given that the trail in (i) is 1.2 km, find the value of θ. o720 360 2 sin( 45 ) 1200+ + = o 480sin( 45 ) 360 2  += oo45 70.5287+= o25.5 = 7 The equation of a curve is 232xxye −= . 7a Find expressions for d d y x and 2 2 d d y x . ( ) 232d 34d xxy exx −=  − ( ) ( ) ( ) 22 2 3 2 3 2 2 d 4 3 4 3 4d x x x xy e e x xx −−=  − +  −  − ( ) 22 2 23 2 3 2 2 d 4 3 4d x x x xy e e xx −−=− +  − 7b Find the exact value of the coordinates of the stationary point. ( ) 232 3 4 0xxex −  − = ( )3 4 0x−= 3 4x= Reject 232 0xxe − = If 3 4x= , 9 8ye= 7c Find the nature of the stationary point.

If 3 4x= , 2 2 d d y x = 9 84e− This is always negative. Therefore, it is a maximum turning point. 8 A circle, C1, has a diameter AB where A is the point (‒1, 1) and the tangent at B is the line 3 4 43 0yx− + = . 8i Find the equation of the diameter AB and hence the coordinates of B. Gradient of diameter = 3 4− 3 4y x c=− + Equation of the diameter AB is 31 44yx=− + Point of intersection , B : 313 4 43 044xx− + − + = 25 175 44x− =− Coordinates of B is (7, ‒5) 8ii Find the equation of the circle, C1. Centre of C1.= 1 7 1 5,22 − + −  ( )3, 2− Radius = 22(3 ( 1)) ( 2 1)− − + − − =5 units Equation is ( ) 2 23 ( 2) 25xy− + + = 8iii A second circle, C2, has equation 22 4 12x y x y+ − + = . Find the coordinates of the centre and the radius of C2, Centre = 1 ,22  − Radius = 2 21 (2) ( 12)2 − + − −

= 65 4 units 8iv Explain if the circle C2 lies entirely within the circle C1 . Both centre lie on the same horizontal line. Radius is 5 units, so (3‒5, ‒2). (‒2, ‒2) lies on the circumference of C1 Similarly, 1 65 ,224  −− lies on the circumference of C2 Since 1 65 24− on the circumference of C2 is out of circle C1 , C2 cannot lie entirely within C1. 9a Use the substitution of 3xU = to solve the equation 212(3 ) 3 5