TMJC H2 Chapter 1 Graphing Techniques Extra Practice Solution 2024

Chapter 1 Graphing Techniques TMJC 2024 Page 1 of 7 H2 Mathematics (9758) Chapter 1 Graphing Techniques Extra Practice Solutions Q1 Solution Q2 Solution Note: There is no easy way to solve for the intercepts in this question hence you are not required to label the intercepts to obtain the 1 mark. Note: Although the graph looks like a straight line, it is actually not a straight line (it is a curve), and your graph needs to showcase this feature Q3 Solution (i) 2cos and sin cos 2 x t y t xt == = 22 2 2 Since sin cos 1, 12 tt xy +=  + = (ii) (4,7) (0,7) (2,3) 7 4 3 y x 2 O y x (e2 + 4, e2 + 2) (e−3 − 6, e−3 − 3) y x (0, 0) (2, 0) (−2, 0) (0, 1) (0, −1)

Chapter 1 Graphing Techniques TMJC 2024 Page 2 of 7 Q4 Solutions ( ) 22 22 2 2 2 2 22 2( 3) 18 ( 1) 1 17 0 2( 3) ( 1) 2 ( 1) ( 3) ( ( 1))( 3) 1 or 121 2 xy xy y x yx − − + + − + = − + + = + − − −− + = + = Q5 Solution (i) C1 : 2 5 10 4 322 xxyx xx −+= = − +−− C2 : 22 4xy−= (ii) Point of intersection satisfy both equations. Solving simultaneous equations give 22 2 5 10 42 xxx x  −+−= − y x (3, −1) 1 ξ2 y x y = x − 3 x = 2 C1 (0, −5) (4, 3) y = x y = −x x2 – y2 = 22 (2, 0) (−2, 0)

Chapter 1 Graphing Techniques TMJC 2024 Page 3 of 7 ( ) ( ) ( ) 22 2 2 5 10 4 2 xx x x −+ −= − ( )( ) ( ) 2222 4 2 5 10x x x x− − = − + (iii) Use GC, draw ( )( ) ( ) 2222 4 2 5 10y x x x x= − − − − + When y = 0, x = 3.66 Q6 Solution (i) Since C has a vertical asymptote at 1x=− , thus 1a= . (ii) 2 4 4 9 511 xxyx xx −+= = − +++ The oblique asymptote is 5yx=− . (iii) 2 44 1 xxy x −+= + ( ) ( ) 2 2 2 1 4 4 44 4 4 0 y x x x yx y x x x y x y + = − + + = − + + − − + − = For real roots, discriminant  0. ( ) ( )( ) ( ) 2 2 2 4 4 1 4 0 8 16 16 4 0 12 0 12 0 12 or 0 yy y y y yy yy yy − − − −  + + − +  + + −  So C cannot lie between −12 and 0. (iv) y x y = x − 5 x = −1 (2,0) (−4,−12) 𝑦 = 𝑥2 − 4𝑥 + 4 𝑥 + 1 (0,4)

Chapter 1 Graphing Techniques TMJC 2024 Page 4 of 7 (v) ( )( ) ( ) 22224 1 4 4x x x x− + = − + 22 2 2 244 441 xx x y xx  −+ = −  = − + Need to sketch 2 2 2 2xy+= , a circle centred (0, 0) and radius 2. Since there are 2 intersections between the graphs, ( )( ) ( ) 22224 1 4 4x x x x− + = − + has 2 real roots. y x y = x − 5 x = −1 (2,0) (−4,−12) 𝑦 = 𝑥2 − 4𝑥 + 4 𝑥 + 1 x2 + y2 = 22

Chapter 1 Graphing Techniques TMJC 2024 Page 5 of 7 Q7 Solution (i) (0, b a ) (ii) 2 2 () xby xa baxa xa += + += − + + x = – a ; y = x – a are equations of the asymptotes. (ii) (iv) 2 2 ( )( )x b x a kx a xb kx axa y kx a + = + − + =−+  = − For no real roots, from graph, 01 k Q8 Solution (i) Vertical asymptote: 44xc=  = When 4, 6,xy== 6 4 2 11 d da =+ =  = ( )( ) 2 2 22 72 44 24 7 44 2 4 8 7 h x bxx xx x x h x bx xx x x x h x bx +−+ + =−− + − + +−=−− + − − + = + − By comparing coefficient of x, 2b=− y x y = x − a x = −a 𝑦 = 𝑥2 + 𝑏 𝑥 + 𝑎 ൬0, 𝑏 𝑎൰ Note: For this question there is no need to label the turning points as the question did not ask for them. Note: y = kx – a corresponds to the asymptote y = x – a of the original curve with k changing the gradient of the line (y-intercept acts like a pivot)

Chapter 1 Graphing Techniques TMJC 2024 Page 6 of 7 (ii) (iii) 48 k  : 4 8kk   9 Suggested Solutions (i) (ii) Since the gradients of the asymptotes are 3 and –3 respectively and the line ( 1)y k x=− passes through the point (1,0), for ( 1)y k x=− to intersect C at 2 points, set of values of k =  : 3 or 3k k k  − y x y = x + 2 x = 4 𝑦 = 𝑥2 − 2𝑥 − 7 𝑥 − 4 ൬0, 7 4൰ (−1.83, 0) (3.83, 0) (3, 4) (5, 8) ( ) 22 22 22 9( 1) 9 1 131 yx xy − − = −−= Equations of asymptotes: 2 2 2 2 2 2 ( 1)3 ( 1)3 3( 1) 33 o 3 3 0 r y x y x yx yx yx − − = =− = − =− =− + (1, 3) (1, 0) (1, –3) O y x

Chapter 1 Graphing Techniques TMJC 2024 Page 7 of 7 (iii) ( ) ( ) 2 2 2 2 1 1 2 x y rr − += is an ellipse centred at (1,0) with the top and bottom vertices at (1, r) and (1, – r) respectively. Hence, for it to intersect C at 4 distinct points, r > 3. Note: Students should think of how to use the previous part i.e. the sketch. The technique is to think of how to manipulate the expression to involve two graphs one of which has been sketched and the other to be inserted and conclude that we are looking at intersection points between the two graphs.