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