PH 2022 AMath Prelim Paper 2 Anskey

1 2022 PHSS Prelim AMATH Paper 2 Solutions Question Solutions 1 Solve the equation 2 2 53 e 4 e x x  − =− , giving your answer(s) in exact form. ( ) 2 2 222 53 e 4 e 3 e 4e 15 0 x x xx  − =− + − = Let 2e xy= ( )( ) 23 4 15 0 3 5 3 0 5 33 yy yy y or y + − = − + = = =− 2 2 2 5e e 3 (reject as e 0)3 52 ln 3 15ln23 x x x or x x = =−  = = 2 The equation of a polynomial is given by 32f ( ) 2 8 21x x x x= + − + . (i) Show that 3x+ is a factor of f ( )x . 32f ( ) 2 8 21x x x x= + − + Subst 3x=− into f(x) 32f ( 3) 2( 3) ( 3) 8( 3) 21 54 9 24 21 0 − = − + − − − + =− + + + = Since f ( 3) 0−= , by Factor Theorem, 3x+ is a factor of f ( )x . 2 (ii) Hence, show that the equation f ( ) 0x = has only one real root. By inspection, ( )( ) 3 2 22 8 21 3 2 7x x x x x bx+ − + = + + +

2 By comparing Coeff. of x: 3 7 8 5 b b + =− =− ( )( ) 3 2 22 8 21 3 2 5 7x x x x x x + − + = + − + For ( )( ) 2 f ( ) 0 3 2 5 7 0 x x x x = + − + = 22 5 7 0 3x x or x− + = =− ( ) 2 5 4(2)(7) 31 0 D= − − =−  Since discriminant is less than 0, 22 5 7 0xx− + = has no real roots. Therefore, the equation f ( ) 0x = has only one real root which is 3x=− . 3 (i) Differentiate 3 3ln 2x x with respect to x. ( ) ( ) ( ) ( ) 32 36 22 6 4 3 3 3ln 2d 3ln 2 d 3 9 ln 2 3 9ln 2 x x xx x x x x x x x x x x − = − = −= 3 (ii) Hence show that ( ) 2 41 4ln 2 1d ln 218 x x a b x =+ , where a and b are integer values to be determined.

3 ( ) 2 41 2 41 22 431 1 2 33 1 4ln 2 d 4 9ln 2 d9 4 3 3ln 2 d9 4 1 3ln 2 9 4 1 3ln 4 3ln 2 19 8 8 1 4 7 3ln 2 3ln 29 8 4 4 7 9 ln 29 8 4 7 ln 218 1 7 18ln 2 (Shown)18 x x x x x x xx xx x xx =  =−   = − −  = − − − − −   = − + =+  =+ =+    7 18ab== 4 The fourth term in the binomial expansion of 2 2 n x x  − , where n is a positive integer, is a constant a. (a) Show that 9n= and hence find the value of a. (a) 2 2 n x x  − ( ) ( ) ( ) 1 2 3 2 2 r nr r r n r nTx r x n xr − + −  =−  =− Fourth term is when r = 3 3(3) 0 90 nxx n − = −= 9n= (Shown)

4 ( ) 39 23 672 a =− =− (b) Find the coefficient of 6x in the expansion of ( ) 9 4 2 2 13xx x  −+ . ( ) 9 4 2 2 13xx x  −+ ( ) ( ) ( ) ( )( ) 2 239 8 7 22 9 6 3 2 3 9 4 4229 672 ... 1 4 3 3 3 ...2 2 3 18 144 672 .. 1 12 54 108 ... x x x x x x xx x x x x x x          = + − + − − + + + + +                     = − + − + + + + + Term in 6x = ( )( ) 6 3 318 144 108x x x=− + Therefore, coefficient of 6x 18 15552 15534 =− + = 5 (a) The equation of a quadratic curve is 23 4 5y x x=− + − . The line 2y mx=− is a tangent to the curve at the point Q where m > 0. Find the value of the constant m and hence find the coordinates of Q. ( ) 2 2 3 4 5 2 3 4 3 0 x x mx x m x − + − = − + − + = Since line is a tangent to the curve, D = 0 ( ) ( )( ) ( )( ) 2 2 2 4 4 3 3 0 8 16 36 0 8 20 0 10 2 0 m mm mm mm − − = − + − = − − = − + = 10m= or 2m=− Since m > 0, 10m=

5 ( ) ( ) 2 2 2 2 2 3 4 5 10 2 3 10 4 3 0 3 6 3 0 2 1 0 10 1 x x x xx xx xx x x − + − = − + − + = + + = + + = += =− Sub 1x=− into 10 2yx=− ( )10 1 2 12 y= − − =− Therefore, coordinates of Q is ( )1, 12−− . (b) Find the range of values of p such that the graph 2 59y px x p= − + lies entirely below the x-axis. 2 59y px x p= − + Since the graph lies entirely below the x-axis, there is no real roots, p < 0 and ( ) ( )( ) ( )( ) 2 2 0 5 4 9 0 25 36 0 5 6 5 6 0 D pp p pp  − −  − − +  55 66p or p−  Since p < 0, reject 5 6p 5 6p − 5 6 5 6− p

6 6 Mark wants to fence out a triangular plot of land for his garden as shown in the diagram. He also intends to build fences along BC to form two different plots of land to plant different vegetables such that they form a pair of similar triangles ABC and ADE . Point B lies on the straight line AD such that 15AB= m and 9BD= m. AE is perpendicular to ED and angle ADE = where 0 90   . (i) Show that m,P the perimeter of the plot of land BCED, is given by 9sin 39cos 9P = + + . 9sinCE = 15cosBC = 9cos 15cos 24cos ED   =+ = 24cos 9sin 15cos 9P   = + + + 9sin 39cos 9P = + + (shown) (b) Express P in the form 9 sin( ),R ++ where 0R and 0 90   . 9sin 39cos 9P = + + 229 39 1602 40.025 R=+ = = 1 39tan 9 77.005   −= = ( )9 1602 sin 77.0P = + +  or ( ) ( ) 9 40.025sin 77.005 9 40.0sin 77.0 P P   = + +  = + +  B D A E 9 m 15 m C 

7 (c) Find the value of P and the corresponding value of  if Mark will like the plot of land BCED to be as large as possible. ( )9 40.025sin 77.005P = + +  Maximum value of P is when ( )sin 77.005 1 +  = 9 40.025 49.025 49.0 (3 P m =+ = = s.f) ( )sin 77.005 1 +  = 77.005 90 + =  12.995 13.0 (3 )   = = s.f Corresponding value of 13.0 = 7 The population of wild Red Pandas has been steadily decreasing over the years, facing the risk of extinction. The table shows the estimated population of wild Red Pandas from 2016 to 2020 where year 2016 is taken to be t = 1 and so on. A wildlife expert believed that these figures can be modelled using the formula 0e ktPP −= , where 0P and k are constants. Year 2016 2017 2018 2019 2020 t 1 2 3 4 5 P 9900 7200 4800 3200 2300 (i) Using the grid below, plot ln P against t and draw a straight line graph. Year 2016 2017 2018 2019 2020 t 1 2 3 4 5 P 9900 7200 4800 3200 2300 ln P 9.20 8.88 8.48 8.07 7.74 0 0 e ln ln ktPP P kt P −= =− + Plot ln P against t (Graph behind) (ii) Use your graph to estimate the value of 0P and of k. 0 0 e ln ln ktPP P kt P −= =− + 0ln 9.55P =

8 0 0 14045 14000 (3 s.f ) P P = = Gradient = 7.74 9.20 51 − − = 0.365− Therefore, k = 0.365 (iii ) The Wildlife Expert uses this model to estimate the population of Red Pandas in 2030. Find the value of this estimation, correct to the nearest whole number, and explain if the estimation obtained reliable. Evidence: 0 0.365 e 14045e kt t PP P − − = = When t = 15, 0.365(15)14045e 58.851 59 (nearest whole number) P −= = = Conclusion & Concept: No, because the value is extrapolated where the linear relationship may no longer hold. 8 A particle traveling in a straight line passes through O with a speed of 5 m/s. The acceleration a m/s2, of the particle, t s after passing through O, is given by 0.53e ta −=− . The particle comes to instantaneous rest at the point X. (i) Find the time taken for the particle to reach X. (i) 0.53e ta −=− 0.5 0.5 0.5 3e d 3 e0.5 6e t t t vt c vc − − − =− =+ =+  When t = 0, v = 5, 05 6e 1 c c =+ =− 0.56e 1 tv −=− At instantaneous rest, v = 0,

9 0.5 0.5 6e 1 0 1e 6 10.5 ln 6 3.5835 3.58 s t t t t t − − −= = −= = = (ii) Calculate the distance OX. ( ) 0.5 0.5 6e 1 12e t t s dt s t d − − =− =− − +  When t = 0, s = 0, 00 12e 0 12 d d =− − + = 0.512e 12tst −=− − + At t = 3.5835, 0.5(3.5835)12e 3.5835 12 6.41648 6.42 s −=− − + = = Distance of OX = 6.42 m (iii ) Show that the particle is again at O at some instant during the twelfth second after passing through O. When t = 11, ( )0.5 11 12e 11 12 0.95096 s − =− − + = When t = 12, ( )0.5 12 12e 12 12 0.029745 s − =− − + =− Since the displacement of the particle changes from a positive value at t = 11 to a negative value at t = 12, the particle passes through O at some instant during the twelfth second.

10 9 The equation of a circle is ( ) ( ) 22 243x r y r kr− + + = where r and k are positive constants. It is given that k = 9. (a) Explain why the x-axis is a tangent to the circle. Evidence: ( ) ( ) 22