Chp 3 Tutorial 3A APQ Soln (RVHS)
Uploaded by KSKS · 20 December 2023
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Text from the first pagesH2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Chp 3 Graphing Techniques (Part 1) (Solutions to Tutorial and Assignment) 29 Additional Practice Questions 1. Given the graph of y = x2 + a x + b , where a > 0 and b > 0, i) State the coordinates of the intersection(s) of the graph with the axes. When ݔ= 0,ݕ= ()మା ()ା = , so ቀ0, ቁ is the y-intercept. When ݕ= 0, 0 = ௫మା ௫ା ,ݔଶ = −ܽ< 0 so there is no x-intercept. ii) Find the equations of the asymptote(s). x = –b ; y = x – b iii) Draw a sketch of the curve, labelling the equations of its asymptotes and coordinates of any intersection with the axes. 2. [MI PU2 Promo 9758/2019/02/Q2] The curve C has equation 2 4 9 3 x xy x . (i) Find algebraically the set of values that y can take, leaving your answer in exact form. [4] (ii) Sketch C, stating the coordinates of the axial intercept, turning points and the equations of any asymptotes. [3] x = –b y = x – b
H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Chp 3 Graphing Techniques (Part 1) (Solutions to Tutorial and Assignment) 30 M2 JC1 Promo 9758/2019/02/Q2 (Solutions) 2(i) 2 2 2 2 2 2 2 2 4 9 3 ( 3) 4 9 4 3 9 0 (4 ) 3 9 0 ..... (*) For values that can take, the equation (*) has real roots: 4 0 (4 ) 4(1)(3 9) 0 (4 ) 12 36 0 4 20 0 2 1 6 or 2 1 6 x xy x y x x x x x xy y x y x y y b ac y y y y y y y y : 2 1 6 or 2 1 6 y y y 2(ii) 2 4 9 6 13 3 Asymptotes are 1 and 3. x xy x x x y x x 2 1 6 2 1 6
H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Chp 3 Graphing Techniques (Part 1) (Solutions to Tutorial and Assignment) 31 3. The curve C has equation y = 2x 2 x 1 x 1 . (i) Find the equations of the asymptotes of C. Vertical Asymptote: ݔ= −1 By long division: ݕ= 2ݔ− 3 + ଶ ௫ାଵ Therefore, oblique asymptote: ݕ= 2ݔ− 3
H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Chp 3 Graphing Techniques (Part 1) (Solutions to Tutorial and Assignment) 32 (ii) Draw a sketch of the curve C, making clear the main relevant features. (iii) Find the range of possible values of k such that the line y = k (x 1) 5 and the curve C do not intersect. ݕ= ݇(ݔ+ 1) − 5 will pass through the point of intersection of the two asymptotes: (-1, 5) It has gradient k. For the curve and the line to not intersect, ݇≤ 2
H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Chp 3 Graphing Techniques (Part 1) (Solutions to Tutorial and Assignment) 33 (iv) Find the set of values of x for which f(x) = ln 2x 2 x 1 x 1 is defined. ݂(ݔ) is defined for ଶ௫మି௫ିଵ ௫ାଵ > 0. Thus −1 <ݔ< − ଵ ଶ or ݔ> 1 4. Consider the curve )f(xy , where f(x) = 2 12 x x . (i) State the coordinates of any points of intersection with the axes. when x = 0, y = - 0.5 when y = 0, x = 1 coordinates are (0, - 0.5), (1, 0) and (-1, 0) (ii) Find the equations of the asymptotes. Vertical asymptote: x = -2 2 322 3)2(2 2 12)2( 2 12 xxx xxx xxx x xy Oblique asymptote: y = x – 2 (iii) Prove, using an algebraic method, that f(x) a or f(x) b, where a and b are exact values to be determined. Let 0)21(2 1 22 yyxxx xy For real x, 0480)21(4 22 yyyy 124012)4( 2 yy 324or 324 yy
H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Chp 3 Graphing Techniques (Part 1) (Solutions to Tutorial and Assignment) 34 (iv) Sketch )f(xy , showing the main relevant features of the curve. 5. (a) ݔ= 2 sinݐ, ݕ= 3 cosݐ→ ቀ௫ ଶቁ ଶ + ቀ௬ ଷቁ ଶ = 1
H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Chp 3 Graphing Techniques (Part 1) (Solutions to Tutorial and Assignment) 35 (b) ݔ= cosec(ݐ)+ 1,ݕ= 1 − cot(ݐ)→ 1 + (1 −ݕ)ଶ = (ݔ− 1)ଶ (c) ݔ= ସ ୱ୧୬ ௧ ୱ୧୬(ଶ௧) ,ݕ= tanݐ+ 2 → 1 + (ݕ− 2)ଶ = ቀ௫ ଶቁ ଶ (d) ݔ= ݐ+ ଵ ௧ ,ݕ= ݐ− ଵ ௧ →ݔଶ =ݐଶ + ቀ ଵ ௧మቁ + 2,ݕଶ =ݐଶ + ቀ ଵ ௧మቁ − 2 →ݔଶ −ݕଶ = 4
H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Chp 3 Graphing Techniques (Part 1) (Solutions to Tutorial and Assignment) 36 6. Sketch the graph of y ax1 x2 4 , a> 1 2 , stating the exact coordinates of any points of intersection with the axes and any stationary points, and the equations of any asymptotes. See-Think-Wonder! (See) There is a restriction on a (Think) Why? To avoid cases where there are repeated roots in both the numerator and denominator. Also, once a > 0.5 or a < -0.5 the sign of the roots will change (i.e. they will be positive and negative in different regions than before). (Wonder) What if the restriction were changed or removed? The graph can look very different, depending on the values of a used.
H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Chp 3 Graphing Techniques (Part 1) (Solutions to Tutorial and Assignment) 37 7. The curve C has equation x2 a2 y2 2a2 1, where a is a positive constant. (i) Sketch the graph of C, indicating clearly the coordinates of point(s) of intersection with the axes, and the equation(s) of any asymptotes. Leave your answer in terms of a. (ii) By sketching a suitable curve on the same diagram, find the solutions to the equation 2 2 2 2 2 12 x a x a a Replace ݕଶ with ܽଶ −ݔଶ ݕଶ =ܽଶ −ݔଶ →ݔଶ +ݕଶ =ܽଶ Therefore, draw a circle, centre (0,0) radius a. The two curves intersect at ݔ= ±ܽ , therefore the solutions are ݔ= ±ܽ ݕଶ = 2ݔଶ ݕ= √2ݔ ݕ= −√2ݔ ݕ= √2ݔ ݕ= −√2ݔ
H2 Mathematics (9758) JC1 River Valley High School, Mathematics Department, 2022 Chp 3 Graphing Techniques (Part 1) (Solutions to Tutorial and Assignment) 38 (iii) Find the range of values of b for which the graph of x b 2 y2 b2 has two points of intersection with C. (ݔ−ܾ)ଶ +ݕଶ =ܾଶ is a circle, centre (ܾ,0) radius ܾ. It won’t have any intersection if |2ܾ| <ܽ It will have only one intersection if 2ܾ= ±ܽ Therefore, ܾ> ଶ or ܾ< − ଶ (iv) The graph of x2 y b 2 c2 has two points of intersection with C for any constant b. Find c in terms of a and b. ݔଶ + (ݕ−ܾ)ଶ =ܿଶ has two points of intersection with ௫మ మ − ௬మ ଶమ = 1 This is a circle, with radius c, centred on (0, b). Therefore, the two points of intersection should have the same y value. ݕ= √2ݔ ݕ= −√2ݔ ݕ= √2ݔ ݕ= −√2ݔ
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