5.Curves Sketching Solutions 2023 (RVHS)

River Valley High School, Mathematics Department 2023 Curve Sketching T1 Solutions Curve Sketching 1(i) Asymptotes: 7yx and 2x (ii) 5 2 2 5 2 2k    (iii) 2(i) Method 1: Using Differentiation 22 11 (2 ) dy dx a x x   Putting 0,dy dx  22 11 (2 )a x x   22(2 ) (2 )a x x x a x     xa 1 1 2 2y a a a a   2 2 3 3 22 (2 ) dy dx a x x   When xa , 2 2 3 3 22 0dy dx a a    2,a a   is a min point. Method 2: Non-differentiation method 1 1 2 2 (2 ) ay a x x x a x   is the reciprocal graph of quadratic eqn 𝑦 = 𝑥(2𝑎 − 𝑥) and multiplied by scale factor 2a in the y-direction. 7yx 2x  3.41, 2.17  0.586,7.83

River Valley High School, Mathematics Department 2023 Curve Sketching T2 Since the graph of 𝑦 = 𝑥(2𝑎 − 𝑥) is a parabola with a max point at (𝑎 , 𝑎2), the curve 𝐶1 will have a min point at 𝑥 = 𝑎, and 𝑦 = 2𝑎 ( 1 𝑎2) = 2 𝑎. Thus 2,a a   is a min point. (ii) (iii)   2 2 22 1xa y ab   . The graph of 𝐶2 is an ellipse with centre (𝑎, 0), with 2 end-pts at (0,0) and (2a,0). For the ellipse to intersect 𝐶1 at more than one pt, 𝑏 > 2 𝑎. (iv) 2a and 6b 2 2 1 1 ( ) 12 xaba x x a         From the graph, 0 < 𝑥 < 0.330 or 3.67 < 𝑥 < 4 (3 s.f.)

River Valley High School, Mathematics Department 2023 Curve Sketching T3 3 (i) Since C has a vertical asymptote at 0x , thus 1 0, 1    . (ii) 22 4 2 4 3 3 3 xyx xx    ; Oblique asymptote is 2 3yx (iii) 224 3 xy x  23 2 4xy x 22 3 4 0x xy   The equation above has no real roots when discriminant < 0.      2 2 3 4 2 4 0 9 32 0 32 32( )( ) 033 y y yy        M1, for getting b2-4ac <0 correctly 4 2 4 2 33 y   2a . (iv) 22 4 2 2 6 3( 1) 3 3 3( 1) xy k x kxx       (a) x – intercepts are 23 9 3 24 16 2 k k k    and 23 9 3 24 16 2 k k k    4/3 +k -1

River Valley High School, Mathematics Department 2023 Curve Sketching T4 (b) For the graph of 224 3( 1) xyk x   to cut the x-axis at 2 distinct points, 293 216 2 kk  > 0, =>least integer value of k is 4 4 5x is an asymptote 5d    2 555 ax bx c Ry ax b axx      7yx is an asymptote 1, 5 7 2 a b a b             22 2 5 2 2 22d 5d 5 x x x x cx x c yy xx x         At 1x ,     d 0 1 5 2 2 1 2 0 1d y ccx             Sketch 2 21 5 xxy x       21 2 5 1 0xx     22 2 5 1 0x x x x      22 2 1 5x x x x      2 21 5 xx xx   Therefore, sketch the graph of yx  on the same diagram. (i) yx  , 1  From graph, only one pt of intersection  one real root

River Valley High School, Mathematics Department 2023 Curve Sketching T5 (ii) yx  , 01  From graph, two pts of intersection  two real roots 5 (i)       2 2 22 22 2 2 axy xa x a ax axdy ax a x dx x a x a      Set dy dx  0    22 2 0 2 0 0 2ax a x ax x a x or x a        For all negative values of a, there will be two distinct values of x thus two stationary pts. (shown) Or 22 4 4 0B AC a   for all negative values of a. (ii) (iii)    242 1x k x x    a = -1  22 2 2 2 1 x k x y k xx         insert a circle, centre (0,0) , radius k to cut curve twice .  0 < k < 20 23 2ax ay ax ax a x a    Max pt is (-2a, -4a2) Min pt is (0,0)

River Valley High School, Mathematics Department 2023 Curve Sketching T6 6. Suggested solution (i) 23 2 3 4 3111 xxyx xx     (ii) 2C is a horizontal hyperbola with center  1,4 . Asymptotes:  14y b x   For no point of intersection between 1C and 2C , the asymptote of hyperbola must be as steep or less steep than the oblique asymptote of 1C . Note that the gradient of oblique asymptote of 1C is 3. 03 b   For 3b , Asymptotes:  3 1 4 3 1, 3 7y x x x        0, 3  0, 4  2, 4 31yx x y O 1x  2.15,10.9 37yx  31yx 1x x  0, 3  0.155, 2.93  2.15,10.9 y O

River Valley High School, Mathematics Department 2023 Curve Sketching T7 7(i) 2 2 2 22 24x a x ay xa   By long division, 22 22 62 a a xy xa   Vertical asymptotes: xa or xa Horizontal asymptote: 2y (ii) From 22 22 62, a a xy xa                     2 2 2 2 2 222 2 2 2 2 222 2 2 2 222 62d 0d 12 2 2 12 a x a a a x xy x xa a x a x x xa a x x a xa                     2 2 2 222 22 12d 00d 12 0 a x x ay x xa x x a          Given C has two turning points,         222 2 2 4 0 12 4 1 0 144 4 0 36 0 6 6 0 66 b ac a a a aa a                Since a is a positive constant, 06 a . (Shown)

River Valley High School, Mathematics Department 2023 Curve Sketching T8 (iii) (iv)         2 222 2 2 2 6 2 5 1 16 2 5 1 h x y h yx h           It is an ellipse centred at  6 2 5, 1 . Maximum turning point of C occurs at  6 2 5, 3.236 . For the ellipse to intersect the curve C more than once, 2.236.h . Since h is a positive integer, 3.h x y =2 y x y =2 y 4x 4x 0  6 2 5, 2.326  0, 4  6 2 5, 1.236 4x 4x

River Valley High School, Mathematics Department 2023 Curve Sketching T9 8(i)     23 28 213 3 4 3 2 xy xx xx           22 d 2 1 d 3 4 3 2 0 y x xx     Therefore, C1 has no stationary points. (ii) (iii)     2 0 2 0 22 21Required area 3 d 3 4 3 2 213 ln 4 ln 233 2 1 2 16 ln 2 ln 4 ln 4 ln 23 3 3 3 1 1 16 ln 2 ln 4 units or 6 ln 2 units3 3 3 xxx x x x                             y x O 3 –2.12 3.78

River Valley High School, Mathematics Department 2023 Curve Sketching T10 (iv) Using G.C., the coordinates of the points of intersection are (3.50,1.73) and (3.86, – 1.85). 9(i) 2 1 ax bx cy x   Since C passes through the point 233, ,2     2 3323 9 3 232 3 1 a b c a b c     C passes through the point  2, 10 too, so   2 2210 4 2 10 21 a b c a b c     Since (2, 10) is a minimum point, d 0d y x  when 2x .           2 2 2 2 2 1 1d d 1 2 1 ax b x ax bx cy x x ax ax b c x            So 0 4 4 0.a a b c b c       Solving the three equations using the GC, 3, 2, 2.a b c   (ii) 23 2 2 1 xxy x   Performing long division, 2 2 31 1 3 2 2 33 2 1 3 x x x x xx x x        Hence, 331 1yx x    5 x 3 2 y x O 3 –2.12 3.78