Swiss Cottage WA2 Revision Paper.Ans

1 2023 4E Additional Mathematics WA2 Marking Scheme 1 Given that     2 3 3 4f 2 3 4 xx x   , find  f . x [3]  f x   3 23 3 4 2 x  f x     1 2 3 3 3 4 42 2 x      [M2]   1 29 3 4 x  [A1] 2 The total surface area of a spherical ball of ice is decreasing at a rate of /scm2 2 . Find the rate of change of the volume when its radius is 0.5 cm. [Volume of sphere = 34 3 r ; total surface area of sphere = 24 r ] [5] 24 rA  d 8d A rr  d d d d d d A A r t r t  d2 8 (0.5) d r t   [M1 – connected rates of change] 2)5.0(8 1  dt dr 2 1 cm/s [M1 - d d r t ] 3 3 4 rV  24 rdr dV  [M1 – d 8d A rr  and 24 rdr dV  ] dt dr dr dV dt dV   2 1)5.0(4 2  [M1 – connected rates of change] scm /2 1 3 The rate of change of volume at r = 0.5 cm is scm /2 1 3 . [A1 – with units and statements]

2 3 It is given that  3 1 2 1.y x x   (i) Show that d d y x can be written in the form 2 1 mx n x   where m and n are integers. [3]  3 1 2 1y x x   d d y x           1 1 2 2 13 1 2 1 2 2 1 32x x x         [M1 – product law]      1 1 2 23 1 2 1 3 2 1x x x             1 22 1 3 1 3 2 1x x x         [M1 – factorisation of   1 2 21x   ]     1 22 1 3 1 6 3x x x           1 22 1 9 2x x     9 2 2 1 x x   [A1] (ii) Show that y is an increasing for 1 .2x [2] For the numerator, when 1 ,2x 99 2x 79 2 2x  For the denominator, when 1 ,2x 2 1x 2 1 0x  2 1 0x  [M1 – for both numerator and denominator] Since 9 2 0x  and 2 1 0x  , d 0d y x  and y is an increasing function. [A1]

3 4 A curve has the equation 2 ,3 4 xy x   4 .3x (i) Find an expression for d .d y x [2] d d y x         2 3 4 1 2 3 3 4 x x x      [M1 – quotient rule]   2 3 4 6 3 3 4 x x x        2 2 3 4x   [A1] (ii) Find the coordinates of the points on the curve where the normal is parallel to the line 2 16 3.y x  [3] 38 2y x  Gradient of normal 8 Gradient of tangent 1 8 d d y x 1 8   2 2 3 4x   1 8 [M1 – equate gradient of tangent to d d y x ] [M1]   2 3 4x 16 3 4 4x  8 3x or 0x [M1] When 8 ,3x 82 13 8 63 43 y          When 0,x   2 0 1 3 0 4 2y   

4 The coordinates are 8 1,3 6     and 10, . 2     [A1 for both] 5 3 2 y cx dx y cx dx     8 5 7 4 1 c     5 1 4 1 d d    [M1] for dividing by x on both sides. [M1] for finding gradient. [A1] [A1]

5 6(i) k e ln ln ln e ln ln ln e ln ln kt t d A d A d A kt d kt A        ݐ 16 32 48 64 80 ݀ 0.0364 0.0603 0.0998 0.165 0.273 ln݀ −3.313 −2.808 −2.305 −1.802 −1.298 [B1] for table of values for ln݀ [B1] for plot of points and labelling of axes. [B1] for straight line passing through the points 1.8 −2.25 57 −3.8 ln d t

6 6(ii) From the graph, the vertical intercept, ln 3.8 0.0224 A A   From the graph, the gradient,  3.25 1.45 18 75 3 95 0.0316 (3 s.f) b       [B1] ܣ accept 0.0213 to 0.0235 [M1] for finding 2 convenient points to find gradient. [A1] for ܾ between 0.0310 to 0.0320 6(iii) Method 1: When ݐ= 50, 3 503.8 95e e 0.108 (3 s.f) d   [M1] [A1] Method 2: From the graph, ln 2.25 0.105 (3 s.f) d d   [M1] for reading from graph [A1] (accept 0.100-0.111)