2022 AMath Prelims P2 w Ans

1 Solutions 1 Given the function 1 2y x  , where 1 2x . Find d d y x and show that y has no turning points. 2 3 2 d 1 d 1 2d d 12 2 1 1 12 2 1 2 2 1 y x x x x x x x x                When d 0d y x  , 3 2 1 0 2 2 1x x     has no solution.  y has no turning points. [4] Fraction over a fraction is not complete.

2 (i) Express   3 2 2 4 10 2 2 x x x x x      in partial fractions. [5] 4 3 2 2x x  3 24 10 2x x x    3 2( 4 8 )x x   22 2x x      2 23 2 2 4 10 2 2 2 4 2 2x x x x x x x x x             22 2 2 2 2 2 x x x x x A B C x x           2 22 2 22x x Ax Cxx B x      Sub 0x , 2 2 1 B B    Sub 2x , 8 4 2 C C   Sub 1x , 1 1 2 0 A A         3 2 2 2 4 10 2 2 1 42 2 x x x x x x x      

(ii) Hence find   3 2 2 4 10 2 d2 x x x xx x      . [3]   3 2 2 4 10 2 d2 x x x xx x        2 2 14 2 14 2ln 2 , where is an arbitrary constant. dxx x x x c c x           Deduct mark if did not put + c

3 The diagram shows the graph of 4.1 2.3sinh bt  , for 0 12t  . (i) State the value of b. [1] 2 8 4b    (ii) The height, h m, of the tidal wave at a harbour t hours after 4am is given by 4.1 2.3sinh bt  . A yacht at the harbour ca n only leave and return to the harbour when the height at the wave is at least 2 metres. Calculate the range of time in hours and minutes for the yacht to leave and return in the next 10 hours. [4] 4.1 2.3sin 2 4 2.3sin 2.14 t t              Solve for t, 21sin 4 23 1.1507 t        1.1507, 1.9909, 7.4339 4 1.4651, 2.5348, 9.4651 t t    Yacht to sail out between 4 am to 5.27 am and return between 6.33 am to 1.27 pm. 4.1 2.3sinh bt  t h 2 4 6 8 10 12 0 1.8 6.4 2 4 6 8 0 2 1.4651 2.5348 9.4651 1.4651 h = 1 hour 27.9 min 2.5348 h = 2 hour 32.1 min 9.4651 h = 9 hour 27.9 min h Must be in rad mode.

4 A curve is defined by 2 2 xy x   . Find an expression for d d y x and obtain the exact x – coordinates of the point(s) on the curve at which the gradient is 4 1 . [5]           2 2 2 2 2 2 2 2 32 2 2 2 1 2 2 2 2 2 d dx x x xdy dx dx dx x xx x x x x             When gradient is 4 1 ,         32 32 3 2 2 2 2 2 1 42 2 6 8 2 8 2 4 6 x x x x x x            Fraction over a fraction is incomplete.

5 The line 9 31y x a   is a normal to curve 3 3 5y ax x   at the point  1 , 4  . (i) Find a. [3] Method 1 Sub  1 , 4  into either line or curve: 9( 4) ( 1) 31 36 32 4 a a a          (ii) With the value of a, sho w that the normal at  1 , 4  will not cut the curve again. [5] When 4a , 3 3 3 3 14 3 5 ( 35) 9 36 27 45 35 36 26 10 0 18 13 5 0 x x x x x x x x x x                  At  1 , 4  , ( 1)x is a factor of 318 13 5x x  . 3 218 13 5 ( 1)(18 5)x x x x bx      Compare coefficients of x2 , 0 18 18 b b    Discriminant of 218 18 5x x       2 18 4 18 5 36 0      No real roots for 218 18 5x x  . Normal at  1, 4  will not cut the curve again Method 2 3 2 3 5 3 3 y ax x dy axdx      9 31 1 31 9 9 y x a ay x      At  1, 4  , since grad(normal)  grad(curve) = ̶ 1 Gradient of curve 9 23 ( 1) 3 9 1 3 4 a a a        

6 (a) The diagram shows the cu rve 4 32 3 4y x x   and the li ne 2y x  . The curve and the line intersects at points P 1,1 and Q 2,4 . 4 32 3 4y x x   Find the area bounded by the line 2y x  and the curve 4 32 3 4y x x   . [3]     4 3 25 4 2 1 5 4 2 5 4 2 2 2 1 2 3 4 2 3 65 4 2 2(2) 3(2) 2( 1) 3( 1) ( 1)6 2 2 d (2) (2) 6( 1)5 4 2 5 4 351 or 17.55 units20 x xx x x x x x                                        P Q 0 x y 2y x 

(b) (i) Find the exact value of ln4 ln2 2 dxe x . [2]     ln 4 ln 2 ln 4 ln 2 2 d 2 4 2ln 4 2 2ln 2 2 ln 4 (or 2 2ln 2) x x e x e x              (ii) The diagram below shows part of the curv e of ln( 2)y x  and the line (ln 2)y x . Use your result in 6b(i) to find the area of the shaded region. [3] Method 1 (wrt y-axis) Area of triangle 1 2 2ln 22 2ln 2    area of the shaded region 2ln 2 (2 2 ln 2) 4ln 2 2       ln2y x x y 2 ln( 2)y x  Exact! Method 2 ln( 2) 2 2 y y y x e x x e       area of the shaded region     ln4 ln2 2 1 ln 4 2 2 d2 ln 4 2 ln 4 2ln 4 2 or 4ln 2 2 units ye y            6b(i)

7 The diagram shows a hollow hemispherical bowl. The bowl is initially empty. At time 0t , water flows in to the bowl. When the height of the water is at h m, the volume of water, V m3, is given by 21 (3 4 )12V h h   , 0 0.25h  . Water flows into the bowl at a constant rate of 600  m3/s. (i) Find the rate of change of height when the height of the water is at 15 cm. [3]     2 2 3 2 1 (3 4 )12 1 3 412 1 6 1212 V h h h h dV h hdh          When 0.15h ,   21 6(0.15) 12(0.15)12 2 m/s 600 21 600 4 0 6 0 3 dV dV dh dt dh dt dh dt dh dt dh dt          (ii) Find the time taken for the bowl to be filled. [2] When 0.25h , 3 3 1 4 0.252 3 m96 V         time to fill the bowl = 96 600 =6.25 s   h

8 Solve the equation     3 27 log 1 2 l 8 og 1x x    . [5]                     3 3 3 3 3 3 3 2 3 3 2 3 3 8 27 8 3 0 log 1 2 log 1 log log 1 2 log 1 3 log 1 2 log 1 log 1 24 2log 1 log 1 2log 1 2 ( 4 8) x x x x x x x x x x                                    Let  3log 1 x u  2 2 24 0 ( 6)( 4) 0 6 or 4 u u u u u u               3 3 6 4 log 1 6 log 1 4 1 3 ( 1) 3 80728 81 or x x x x x x           