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