DHS Graphs & Transformations 1 (9758) (Revision Solutions)

Y5 Topical Revision Package 1 1. Graphs and Transformations (Solutions) 1 1.1 Graphs and Transformations 1 (Suggested Solutions) Curve Sketching 1(i) Asymptotes : y = 3, x = 4 (ii) (iii) 32 4 xkx x += − 2 4 3 2 0kx kx x− − − = 2 ( 4 3) ( 2) 0kx k x+ − − + − = ---(*) For 0,k = (*) is a linear equatio and not possible to have two roots. For 0,k  (*) is a quadratic equation and we wand discriminant > 0 2 2 2 ( 4 3) 4( )( 2) 0 16 24 9 8 0 16 32 9 0 kk k k k kk − − − −  + + +  + +  1.66 0.339 (3 s.f.)k or k −  −  set of values = : 1.66 or 0.339, 0k k k k   −  −  2(i) Asymptotes : 7yx=+ and 2x =− (ii) 2 2 2 2 2 2 Discriminant 0 (9 ) 8)( 5 8) 0 5 8 5 9 16 2 ( 2) 9 16 (9 ) (16 2 ) 0 4(16 2 ) 0 10 17 0 ( 8 5) 8 0 (5 xxk x k x x x x k x k k k k k k k k k ++= + + = + + + − + − = − −  −+  − − +  −  − −  −  − + y = 3 y x O x = 4 -2/3 -0.5

Y5 Topical Revision Package 1 1. Graphs and Transformations (Solutions) 2 (iii) 3(i) Asymptotes: and 12 xyx== (ii) ( ) 2 d1 0d2 1 yA x x = − = − ( ) 2 12xA−= Therefore, for C not to have stationary points, A < 0. 4(i) 1c = When 0, 5, 1x y c= = = , 5 1 b= 5b = 227 2 ( 2) 1 x ax b ay x a x c x + + −= = + − +++ When 1, 1xy= − = , 22 1 2( 1) 2 5 5, 5 y x a a a ab = + − = − + − =  = = (ii) 2225 x ax bx xc +++= + One root. 5(i) y (0,4) (–2,0) O (2,0) x –1 y = –1 y x O

Y5 Topical Revision Package 1 1. Graphs and Transformations (Solutions) 3 (ii) ( ) 222 2 2 2 22 2 2 4 1 4 11x x x x hx h h x    −− = +  + =   ++    −      So insert the graph ( ) 22 2 2 1xy h+= − which is an Ellipse with centre O. For only one real root, the ellipse must meet the graph in (ii) at exactly one point. Hence h = 4 and the corresponding root is x = 0. 6(i) (ii) 2x  (iii) From GC : 2 48 5 3 or 62 xx xx −+ =  =−

Y5 Topical Revision Package 1 1. Graphs and Transformations (Solutions) 4 the range of values of x for which 2 48 52 xx x −+ − is 6x  or 23 x . (iv) 1m  7 For no real solutions to the equation 22 1 ( 1)cos( ), 1x a x bx c x+ = − +  range of values for a is 0.899 0.899a−   8(i) Note that 22 6 16 0x yy− + + = ( )( ) ( ) ( ) 22 2 2 2 2 22 9 16 2 30 3 3 5 ( 0) 155 x x x y y y −+ −= − −= − − − = − The equations of the asymptotes are 5 53 3 or 3y x y x y x− =   = + = − + . O ( 0.225, 0.899)−−

Y5 Topical Revision Package 1 1. Graphs and Transformations (Solutions) 5 (ii) From the graph, it follows that 1m − or 1m  . (iii) The other hyperbola with the same asymptotes are ( ) ( ) ( ) 22 22 22 25 0 3 155 3xy yx = −− −− − = Therefore, 0, 3pq== . (iv) (v) The circle ( ) 222 3 rx y+ − = will intersect 2C either 0, 2 or 4 times. So, 0,2,4n = . 9(i) Since 1x = is a vertical asymptote, we have 1b =− . Also, 2yx=− is an oblique asymptote, so 2 1 kyx x= − + − , where k is a constant. Method 1: 2 4 1 x axy x +−= − ( 1) 4 1 x x x ax x − + + −= − ( 1)( 1) 1 4 1 a x ax x + − + + −=+ − 3( 1) 1 axa x −= + + + − Comparing the form we have earlier, we have 1 2 3aa+ = −  = − Method 2: 2 1 kyx x= − + − 2 32 1 xk x x− + += − Comparing coefficient of x in given equation, 3a =− . Intercepts: Let 0 4x y= =

Y5 Topical Revision Package 1 1. Graphs and Transformations (Solutions) 6 Let 2 3 00 4 1 xxy x −−= −= 2 3 4 0xx − −= ( 4)( 1) 0xx− + = 4x = or 1x =− (Note to students: you can just use GC for this.) (ii) Method 1 22 2 342 20 21 0 1 xxxx x  −−− − + =  − 22 2 34( 1) 1 20 21 0 1 xxx x  −−− − − + =  − 22 2 34( 1) 21 21 1 xxx x  −−− + =  − 222( 1) 3 4 121 1 x x x x − − −+= − ( ) 2 2( 1) 121 x y − += The graph to add on is an ellipse centred at (1, 0) with horizontal axis length 21 and vertical axis length 1. From graph, the ellipse intersects curve C at 4 distinct points, therefore it has 4 real distinct roots. Method 2 ( ) 22 2 22 2 22 342 20 21 0 1 2 20 21 0 2 2021 2 20 21 xxxx x x x y xxy x x y   −−− − + =  − − − + = − + += − + + =  The graph to add on is 2 2 20 21 xxy − + += (Students can use the GC for this) From the graph, 2 2 20 21 xxy − + += intersects curve C at 4 distinct points, therefore it has 4 real distinct roots. y = x – 2 O (4,0) 0) ) (-1,0) (0, 4) x = 1 (1,0)

Y5 Topical Revision Package 1 1. Graphs and Transformations (Solutions) 7 10 (a)(i) 2 8xxy xk += + . Since vertical asymptote is x = 1,  k = – 1 22 8 8 9 911 x x x xyx x k x x ++ = = = + ++ − −  Oblique Asymptote is y = x + 9 (a)(ii) (b) By guess and check, r = 7 11(i) 22x a a ay x a x a x a Given the oblique asymptote of C is y = x + 2, therefore a = 2. (ii) 22 2 22 2 2 2 2 2 2 2 2 2 2 OR 1 For turning points, let 0 2 0 2 0 ............ 1 If curve has two turning points, discrimina nt 0 2 4 1 0 44 x a a ay x a x a x a x x a x ady x ax a dy a a dx dx x a x a x a dy dx x ax a xa x ax a C aa a 0 10 1 or 0 set of values = { : 1 or 0} a aa aa a a a x y O

Y5 Topical Revision Package 1 1. Graphs and Transformations (Solutions) 8 (iii) (iv) 2 26 222 xyx xx += = + +−− 12(i) C has a vertical asymptote at 0x = . 0 (ans)r= 2 2 2 2 2 2( ) 2 2px q p x pqx q qy p x pqx x x + + += = = + + As x → , 2 0q x → . 2 2y p x pq→+ . Oblique asymptote: 2 2y p x pq=+ Comparing coefficient of x: 2 9 3 or 3 (rejected is non-negative cons tant)p p p=  = − constant: 2 26pq q p =  = = (shown) (ii) 22 2 d99 36 d 36 yyx x x x = + +  = −

Y5 Topical Revision Package 1 1. Graphs and Transformations (Solutions) 9 To find stationary point(s): d 0d y x = 2 22 29 0 32436 18 xx x    − =  =  =  For 18x = , 2 9218 36 18 y    = + + =   . For 18x = , 2 90 18 36 18 y    = − + + =  − . 22 23 d d 18 y xx = For 18x = , ( ) 2 2 2 32 d 18 0d 18 18 y x   = =   . ,218   is minimum point. For 18x =− , ( ) 2 2 2 32 d 18 0d 18 18 y x   = = −   . ,018 − is maximum point. (iii) For 18 = , vertical asymptote at 0x = , Oblique asymptote: 9 18yx=+ , Stationary points at ( 1, 0)− and (1, 36) . No y-intercept. x-intercept at ( 1, 0)− . ( )1,36

Y5 Topical Revision Package 1 1. Graphs and Transformations (Solutions) 10 1sin 1 sin xxa a −= +  = , 18cos 18 cos yya a −= +  = Using trigonometric identity, ( ) ( ) 22 22 22 2 1 18sin cos 1 1 1 18 (ans) xy aa x y a  −−   + =  + =         − + − = For C and D intersect more than once, 13(i) For 2,C when 0,x = 2 2 3 2e 4e 0 4e 2e 1e 2 1 1 1ln ln 23 2 3 tt tt t t − − −= = = = = − Therefore, 12ln 2 ln 2333e e 4.41 (to 3 s.f.)y − = + = (ii) 22e 4e (1)ttx −=− 23e e (2)tty −=+ 22 (2) 3 (1) :2 3 14e (3) tyx −  − = Note that D is a circle with centre at (1, 18) and radius a unit. by guess and check, least integer value of 19a  =