Rational functions and graphs

7(a) x=−1 B1 States vertical asymptote. x(x+1)+9 9 M1 Finds oblique asymptote. y= =x+ x+1 x+1 y= x A1 3 7(b) dy M1 A1 Differentiates and sets derivative equal to 0. =1−9(x+1) −2 =0(x+1)2 =9 dx ( ) 2,5 A1 (−4,−7 ) A1 4 © UCLES 2021 Page 12 of 14 7(c) B1 Axes labelled and correct asymptotes drawn. B1 Upper branch with (0,9) stated or shown on diagram. 9 B1 Lower branch correct and good approach to asymptotes throughout, no extra branches. x 3 © UCLES 2021 Page 13 of 14 7(d) B1 FT FT from sketch in (c) with asymptotes shown. 9 x 13 13 M1 M1 Finds critical points, award M1 for each case. x2 +x+9= (x+1) or x2 +x+9=− (x+1) 2 2 May state that x2 +x+9=−13(x+1) has no real 2 solutions since 7>13. x2 −11x+ 5 =0 or x2 +15x+ 31 =0 2 2 2 2 2 x= 1,5 A1 2 x< 1 and x>5. A1 2 5 © UCLES 2021 Page 14 of 14

7(a) x=−1 B1 States vertical asymptote. x(x+1)+9 9 M1 Finds oblique asymptote. y= =x+ x+1 x+1 y= x A1 3 7(b) dy M1 A1 Differentiates and sets derivative equal to 0. =1−9(x+1) −2 =0(x+1)2 =9 dx ( ) 2,5 A1 (−4,−7 ) A1 4 © UCLES 2021 Page 12 of 14 7(c) B1 Axes labelled and correct asymptotes drawn. B1 Upper branch with (0,9) stated or shown on diagram. 9 B1 Lower branch correct and good approach to asymptotes throughout, no extra branches. x 3 © UCLES 2021 Page 13 of 14 7(d) B1 FT FT from sketch in (c) with asymptotes shown. 9 x 13 13 M1 M1 Finds critical points, award M1 for each case. x2 +x+9= (x+1) or x2 +x+9=− (x+1) 2 2 May state that x2 +x+9=−13(x+1) has no real 2 solutions since 7>13. x2 −11x+ 5 =0 or x2 +15x+ 31 =0 2 2 2 2 2 x= 1,5 A1 2 x< 1 and x>5. A1 2 5 © UCLES 2021 Page 14 of 14

7(a) 1− 5 1+ 5 B1 Vertical asymptotes. Must be exact. x= , x= 2 2 y=−1 B1 Horizontal asymptote. 2 7(b) dy (1+x−x2)(2x−1)−(x2 −x−3)(1−2x) M1 dy = Finds . dx ( 1+x−x2 )2 dx (2x−1)(−2)=0 M1 dy Sets their equal to 0 and forms equation. dx ( ) 1,−13 A1 WWW 2 5 3 © UCLES 2021 Page 14 of 16 7(c) y B1 Both axes labelled and correct asymptotes shown. B1 Correct shape and position, with all asymptotic behaviour clear. x ( ) ( ) 1 +1 13,0 , 1 −1 13,0 , (0,−3) B1 States exact coordinates of intersections with axes. 2 2 2 2 3 © UCLES 2021 Page 15 of 16 7(d) B1 FT FT from sketch in (c). B1 Correct shape as x tends to infinity and intersections with x axis. x2 −x−3 x2 −x−3 M2 Finds critical points, award M1 for each case. =3 or =−3 1+x−x2 1+x−x2 4x2 −4x−6=0 or −2x2 +2x=0 x= 1 +1 7, x= 1 −1 7, x=0, x=1 A1 Must be exact. 2 2 2 2 x< 1 −1 7, 0<x<1, x> 1 +1 7 A1 FT Must be three distinct regions and strict 2 2 2 2 inequalities. 6 © UCLES 2021 Page 16 of 16

7(a) x=1, x=−1 B1 Vertical asymptotes. y =0 B1 Horizontal asymptote. 2 7(b) dy (4−4x2)(4)−(4x+5)(−8x) M1 dy = Finds . dx ( 4−4x2 ) 2 dx 16x2 +40x+16=0 M1 Sets equal to 0 and forms equation. ( ) ( ) −2,1 , −1,1 A1 A1 4 2 4 © UCLES 2021 Page 12 of 14 7(c) B1 Axes and asymptotes. B1 Correct shape and position. x 4x(cid:3397)5 y(cid:3404) 4-4x(cid:2870) ( ) (−5,0), 0,5 B1 States coordinates of intersections with axes. 4 4 3 © UCLES 2021 Page 13 of 14 1-(cid:3404)x 1(cid:3404)x 7(d) B1 FT FT from sketch in part (c). B1 Correct shape at infinity. 4x(cid:3397)5 y(cid:3404)(cid:3628) (cid:3628) 4-4x(cid:2870) x 4x+5 5 4x+5 5 M2 Finds critical points, award M1 for each case. = or =− 4−4x2 4 4−4x2 4 5x2 +4x=0 or 5x2 −4x−10=0 x=−4, x=0 or x= 2−3 6, x= 2+3 6 A1 A0 if –1.07, 1.87 5 5 5 5 5 2−3 6<x<−4, 0<x< 2+3 6 , x≠±1 A1 FT Condone exclusion of x=±1 from the range. 5 5 5 5 5 6 © UCLES 2021 Page 14 of 14 1-(cid:3404)x 1(cid:3404)x

6(a) x=3 B1 States vertical asymptote. 9 M1 A1 Finds oblique asymptote. y=x+3+ leading toy=x+3 x−3 3 6(b) yx−3y=x2 leading tox2 −yx+3y=0 M1 A1 Forms quadratic in x. y2 −4(3y)<0 leading to y2 −12y<0 M1 Uses that discriminant is negative. 0< y<12 A1 AG 4 © UCLES 2021 Page 12 of 16 6(c) B1 Axes and asymptotes. B1 Branches correct. 𝑥(cid:2870) 𝑦 (cid:3404) 𝑥(cid:3398)3 2 © UCLES 2021 Page 13 of 16 3 (cid:3404) 𝑥 6(d)(i) B1 FT FT from sketch in (c). 𝑥(cid:2870) 𝑦 (cid:3404) (cid:4708) (cid:4708) 𝑥(cid:3398)3 B1 Correct shape at infinity. B1 Correct shape of y= x −3. B1 Correct intercepts with axes (may be seen on graph). x 4 6(d)(ii) c ⩽ –3 B1 1 © UCLES 2021 Page 14 of 16 3 (cid:3404) 𝑥

7(a) x=1, x=−1 B1 Vertical asymptotes. y =0 B1 Horizontal asymptote. 2 7(b) dy (4−4x2)(4)−(4x+5)(−8x) M1 dy = Finds . dx ( 4−4x2 ) 2 dx 16x2 +40x+16=0 M1 Sets equal to 0 and forms equation. ( ) ( ) −2,1 , −1,1 A1 A1 4 2 4 © UCLES 2021 Page 12 of 14 7(c) B1 Axes and asymptotes. B1 Correct shape and position. x 4x(cid:3397)5 y(cid:3404) 4-4x(cid:2870) ( ) (−5,0), 0,5 B1 States coordinates of intersections with axes. 4 4 3 © UCLES 2021 Page 13 of 14 1-(cid:3404)x 1(cid:3404)x 7(d) B1 FT FT from sketch in part (c). B1 Correct shape at infinity. 4x(cid:3397)5 y(cid:3404)(cid:3628) (cid:3628) 4-4x(cid:2870) x 4x+5 5 4x+5 5 M2 Finds critical points, award M1 for each case. = or =− 4−4x2 4 4−4x2 4 5x2 +4x=0 or 5x2 −4x−10=0 x=−4, x=0 or x= 2−3 6, x= 2+3 6 A1 A0 if –1.07, 1.87 5 5 5 5 5 2−3 6<x<−4, 0<x< 2+3 6 , x≠±1 A1 FT Condone exclusion of x=±1 from the range. 5 5 5 5 5 6 © UCLES 2021 Page 14 of 14 1-(cid:3404)x 1(cid:3404)x

5(a) 30 M1 A1 Finds discriminant or roots of x2 x1, or completes square. y2 B1 Horizontal asymptote. 3 5(b) dy (x2 x1)(4x1)(2x2 x1)(2x1) M1 dy  Finds . dx  x2 x1 2 dx 3x2 6x0 M1 Sets equal to 0 and forms equation. 0,1 2,3 A1 A1 One point correct, or both x values. Other point correct. Alternative method for question 5(b) 2x2 x1y(x2 x1)0 M1 Forms quadratic equation Finds discriminant (y1)2 4(y1)(y2)AND states y exists if discriminant ≥ 0 M1 OR does not exist if discriminant < 0 Finds (0, -1) and (-2, 3) A1 Explains why they are stationary values. A1 Double xroots for y1 andy3 or no vertical asymptote etc 4 © UCLES 2022 Page 10 of 15 5(c) B1 Axes and asymptote. B1 Correct shape and position. 1,0, 1,0, (0,1) B1 States coordinates of intersections with 2 axes. 3 5(d) B1 FT FT from sketch in part (c). There must be cusps on the x axis. 0k1 B1 2 © UCLES 2022 Page 11 of 15

5(a) 30 M1 A1 Finds discriminant or roots of x2 x1, or completes square. y2 B1 Horizontal asymptote. 3 5(b) dy (x2 x1)(4x1)(2x2 x1)(2x1) M1 dy  Finds . dx  x2 x1 2 dx 3x2 6x0 M1 Sets equal to 0 and forms equation. 0,1 2,3 A1 A1 One point correct, or both x values. Other point correct. Alternative method for question 5(b) 2x2 x1y(x2 x1)0 M1 Forms quadratic equation Finds discriminant (y1)2 4(y1)(y2)AND states y exists if discriminant ≥ 0 M1 OR does not exist if discriminant < 0 Finds (0, -1) and (-2, 3) A1 Explains why they are stationary values. A1 Double xroots for y1 andy3 or no vertical asymptote etc 4 © UCLES 2022 Page 10 of 15 5(c) B1 Axes and asymptote. B1 Correct shape and position. 1,0, 1,0, (0,1) B1 States coordinates of intersections with 2 axes. 3 5(d) B1 FT FT from sketch in part (c). There must be cusps on the x axis. 0k1 B1 2 © UCLES 2022 Page 11 of 15

1(a) B1 Axes and asymptotes. B1 Branches correct. x 2 © UCLES 2022 Page 6 of 17 1(b) B1 FT FT from sketch in part (a), symmetrical about x0. B1 Correct shape at x0(reflection not turning point). x 1 M1 Finds critical point(s). 2 leading to x 1 x 1 3 1x1, 1 x1 A1 3 3 4 © UCLES 2022 Page 7 of 17

3(a) x1 B1 States vertical asymptote. x1(axa1)a a M1 Finds oblique asymptote. y axa1 x1 x1 yaxa1 A1 3 © UCLES 2022 Page 9 of 17 3(b) ax2 x1 yx y leading to ax2 (y1)x(y1)0 M1 A1 Forms a three term quadratic in x. y12 4a(y1)0 M1 Correct inequality using discriminant. 1 y14a A1 AG Clear method to reach given answer eg (y1)(y14a)0 Alternative method for question 3(b) ax2 x1 yx y leading to ax2 (y1)x(y1)0 M1 A1 Forms three term quadratic in x. y12 4a(y1)0 M1 Correct inequality using discriminant. Must be clear looking for where values of yexist. 1 y14a A1 AG Clear method to reach given answer eg (y1)(y14a)0 Second alternative method for question 3(b) dy (x1)(2ax1)(ax2 x1) M1  0 dx (x1)2 x0or x2 A1 Proves (0, 1) is local maximum and (2,4a1) is local minimum. M1 d2y Either or gradient either side dx2 Reference to position of asymptote or two distinct branches to justify A1 AG 1 y14a 4 © UCLES 2022 Page 10 of 17 3(c) B1 FT Asymptotes identified. FT from part 3a. B1 Two branches in correct position relative to correct asymptotes and of correct shape. Good asymptotic behaviour. B1 Lower branch in all four quadrants from correct asymptotes. 3 © UCLES 2022 Page 11 of 17

7(a) x= 2 B1 Vertical asymptote. 5 (5x−2)(x+ 2)+ 4 M1 A1 Oblique asymptote. y= 5 5 leading to y= x+ 2 5x−2 5 3 7(b) dy (5x−2)(10x)−(5x2)(5) M1 dy = Finds . dx (5x−2)2 dx 5x2 −4x=0 M1 Sets equal to 0 and forms quadratic equation. (0,0), (4,8) A1 A1 5 5 4 7(c) B1 Axes and asymptotes. B1 Correct upper branch and asymptotic behaviour. B1 Correct lower branch. 3 © UCLES 2022 Page 14 of 15 7(d) B1 FT FT from sketch in part (c). B1 Correct shape as x tends to infinity. 5x2 5x2 M2 Finds critical points, award M1 for each case. =2 or =−2 5x−2 5x−2 5x2 −10x+4=0 or 5x2 +10x−4=0 x=−1−3 5, x=−1+3 5 or x=1−1 5, x=1+1 5 A1 Must be exact. 5 5 5 5 −1−3 5x−1+3 5, 1−1 5x1+1 5 A1 FT Follow through on use of decimals. 5 5 5 5 6 © UCLES 2022 Page 15 of 15

7(a) x=−1 B1 States vertical asymptote. (x+1)(x−2)+2 M1 Finds oblique asymptote. y= x+1 y=x−2 A1 3 7(b) dy M1 A1 Differentiates and sets derivative equal to =1−2(x+1)−2 =0(x+1)2 =2 0. dx ( ) ( ) A1 A1 −1+ 2,−3+2 2 , −1− 2,−3−2 2 4 © UCLES 2022 Page 13 of 15 7(c) B1 Axes and asymptotes labelled. B1 Upper branch with (0,0) and (1,0) stated or clear on scale. B1 Lower branch correct and all asymptotic approaches correct. 3 © UCLES 2022 Page 14 of 15 7(d) B1 FT FT from sketch in part (c). x2 −x x2 −x M2 Finds critical points, award M1 for each =6 or =−6 case. x+1 x+1 x2 −7x−6=0 or x2 +5x+6=0 x=−3,−2, 7−1 73, 7+1 73 A1 2 2 2 2 −3x−2 and 7−1 73x7+1 73 A1 2 2 2 2 5 © UCLES 2022 Page 15 of 15

7(a) x= 2 B1 Vertical asymptote. 5 (5x−2)(x+ 2)+ 4 M1 A1 Oblique asymptote. y= 5 5 leading to y= x+ 2 5x−2 5 3 7(b) dy (5x−2)(10x)−(5x2)(5) M1 dy = Finds . dx (5x−2)2 dx 5x2 −4x=0 M1 Sets equal to 0 and forms quadratic equation. (0,0), (4,8) A1 A1 5 5 4 7(c) B1 Axes and asymptotes. B1 Correct upper branch and asymptotic behaviour. B1 Correct lower branch. 3 © UCLES 2022 Page 14 of 15 7(d) B1 FT FT from sketch in part (c). B1 Correct shape as x tends to infinity. 5x2 5x2 M2 Finds critical points, award M1 for each case. =2 or =−2 5x−2 5x−2 5x2 −10x+4=0 or 5x2 +10x−4=0 x=−1−3 5, x=−1+3 5 or x=1−1 5, x=1+1 5 A1 Must be exact. 5 5 5 5 −1−3 5x−1+3 5, 1−1 5x1+1 5 A1 FT Follow through on use of decimals. 5 5 5 5 6 © UCLES 2022 Page 15 of 15

6(a) x2 B1 States vertical asymptote. x2 2x15x2x47 y x4 M1 A1 Finds oblique asymptote. 3 © UCLES 2023 Page 11 of 16 6(b) dy x22x2  x2 2x15  M1 Differentiates.  dx x22   A1 dy x2 4x110 or dy 1 7  Forms quadratic equation or simplifies . dx  dx x22    42 4(1)(11)280 (or y'0 ) There are no turning points. A1 Correct conclusion. 3 © UCLES 2023 Page 12 of 16 6(c) B1 Axes and asymptotes. B1 Branches correct. (0,7.5),5,0,(3,0) B1 States coordinates of intersections with axes. 3 © UCLES 2023 Page 13 of 16 6(d) B1FT FT from sketch in (c). B1 Correct shape at infinity and on x axis. 2 6(e) x2 2x15 x2 2x15 M2 Finds critical points, award M1 for each case. 15 or 15 x2 2 x2 2 x2 11x0 or x2 19x300 2 2 x0,11 or x12, 5 A1 2 2 12x0 or 5 x11 A1FT 2 2 4 © UCLES 2023 Page 14 of 16

6(a) x2 B1 States vertical asymptote. x2 2x15x2x47 y x4 M1 A1 Finds oblique asymptote. 3 © UCLES 2023 Page 11 of 16 6(b) dy x22x2  x2 2x15  M1 Differentiates.  dx x22   A1 dy x2 4x110 or dy 1 7  Forms quadratic equation or simplifies . dx  dx x22    42 4(1)(11)280 (or y'0 ) There are no turning points. A1 Correct conclusion. 3 © UCLES 2023 Page 12 of 16 6(c) B1 Axes and asymptotes. B1 Branches correct. (0,7.5),5,0,(3,0) B1 States coordinates of intersections with axes. 3 © UCLES 2023 Page 13 of 16 6(d) B1FT FT from sketch in (c). B1 Correct shape at infinity and on x axis. 2 6(e) x2 2x15 x2 2x15 M2 Finds critical points, award M1 for each case. 15 or 15 x2 2 x2 2 x2 11x0 or x2 19x300 2 2 x0,11 or x12, 5 A1 2 2 12x0 or 5 x11 A1FT 2 2 4 © UCLES 2023 Page 14 of 16

7(a) x3 B1 States vertical asymptote. x3(x5)16 16 M1 Finds oblique asymptote. y x5 x3 x3 yx5 A1 3 7(b) dy 16 M1 Differentiates and sets equal to zero. 1 0(x3)2 16 dx x32 x1,7 A1 Finds x-coordinates (1,0),(7,16) A1 States coordinates of turning points. 3 © UCLES 2023 Page 14 of 16 7(c) B1FT Axes and labelled asymptotes. B1 Upper branch correct. B1 Lower branch correct and no additional branches. 3 © UCLES 2023 Page 15 of 16 7(d) B1FT Clear labels, axes and their vertical asymptote. x2 2x1 y B1FT x2 2x1 x3 y correct, FT from their sketch in (c). x3 B1 x2 2x1 Upper branch of y2  (positive square root). x3 x2 2x1 y2  x3 B1FT x2 2x1 Lower branch of y2  (negative square root). x2 2x1 x3 y FT from previous mark. x3 x22x1 y2 x3 4 © UCLES 2023 Page 16 of 16

7(a) x=−1, x=2 B1 Vertical asymptotes. y=1 B1 Horizontal asymptote. 2 © UCLES 2023 Page 13 of 15 7(b) dy (x2 −x−2)(2x)−(x2 +2)(2x−1) M1* dy = Finds . dx ( x2 −x−2 )2 dx x2 +8x−2=0 DM1 Sets equal to 0 and forms equation. (−8.2,0.9), (0.2,−0.9). A1 A1 Condone ( ) ( ) −4−3 2,2 2 , −4+3 2,−2 2 . 3 3 4 y 7(c) B1 Axes and all three asymptotes. B1 Correct shape and position, crossing horizontal asymptote. B1 States (0,−1)coordinates of intersection with axes, may be seen on diagram. x (0,−1) 3 © UCLES 2023 Page 14 of 15 7(d) B1 FT FT from sketch in (c) B1 All correct. x 2 7(e) x2 +2 x2 +2 M2 Finds critical points, award M1 for each case. =1 or =−1 x2 −x−2 x2 −x−2 x+4=0 or 2x2 −x=0 x=−4 or x=0, x= 1 A1 2 −4x−1, 0 x 1, x2 B1 Must have three distinct regions. Condone −1 2 and 2. 4 © UCLES 2023 Page 15 of 15

7(a) x=−1 B1 Vertical asymptote. (x+1)(x−1)+1 M1 Oblique asymptote. y= x+1 y=x−1 A1 3 7(b) dy x2 +2x M1 dy = =0 Sets =0. dx (x+1)2 dx (0,0), (−2,−4) A1 2 © UCLES 2023 Page 13 of 15 7(c) y B1 Axes and asymptotes. B1 Left branch correct. B1 Right branch correct. x 3 7(d) ( −2,−1) B1 B1 B1 for each correct coordinate. SC B1 4 for ( −2,−1) and (0,0). 4 2 © UCLES 2023 Page 14 of 15 7(e) y B1 Left branch correct. B1 Right branch correct. x x2 x2 M2 Finds critical points, award M1 for =1 or =−1 each case. x+1 x+1 x2 −x−1=0 x=1−1 5 or x=1+1 5 A1 2 2 2 2 x−1, 1−1 5x1+1 5, x0 B1 Condone missing x0. 2 2 2 2 6 © UCLES 2023 Page 15 of 15

7(a) x=−1, x=2 B1 Vertical asymptotes. y=1 B1 Horizontal asymptote. 2 © UCLES 2023 Page 13 of 15 7(b) dy (x2 −x−2)(2x)−(x2 +2)(2x−1) M1* dy = Finds . dx ( x2 −x−2 )2 dx x2 +8x−2=0 DM1 Sets equal to 0 and forms equation. (−8.2,0.9), (0.2,−0.9). A1 A1 Condone ( ) ( ) −4−3 2,2 2 , −4+3 2,−2 2 . 3 3 4 y 7(c) B1 Axes and all three asymptotes. B1 Correct shape and position, crossing horizontal asymptote. B1 States (0,−1)coordinates of intersection with axes, may be seen on diagram. x (0,−1) 3 © UCLES 2023 Page 14 of 15 7(d) B1 FT FT from sketch in (c) B1 All correct. x 2 7(e) x2 +2 x2 +2 M2 Finds critical points, award M1 for each case. =1 or =−1 x2 −x−2 x2 −x−2 x+4=0 or 2x2 −x=0 x=−4 or x=0, x= 1 A1 2 −4x−1, 0 x 1, x2 B1 Must have three distinct regions. Condone −1 2 and 2. 4 © UCLES 2023 Page 15 of 15

6(a) x2 B1 x2(xa2)2a5 52a M1 y xa2 x2 x2 yxa2 A1 3 6(b) dy x22xa  x2 ax1  M1 Differentiates.  dx x22   A1 dy x2 4x2a10 or dy 1 2a5  Forms quadratic equation or simplifies . dx  dx x22    Not from wrong working. 1642a1208a0 (or y'0 )  No stationary points M1 A1 Consideration of discriminant or sign of y' with correct conclusion. 4 © Cambridge University Press & Assessment 2024 Page 11 of 16 6(c) B1 Axes and asymptotes labelled. y  xa2 B1 Branches correct. (Asymptotes may cross above, on or below the x-axis.) x x2 0, 1 B1 May be seen on their diagram. 2 3 © Cambridge University Press & Assessment 2024 Page 12 of 16 6(d)(i) B1 FT from their attempt in part 6(c). B1 Everything correct (approach to vertical asymptote and cusps correct). x 2 6(d)(ii) y B1 ya x 1 © Cambridge University Press & Assessment 2024 Page 13 of 16 6(d)(iii) x2 ax1 x2 ax1 M1 Or direct use of x3 a or a x2 x2 x2 12a0 or x2 2ax12a0 a a2 2a1 x 2a1,a a2 2a1 x 2a1 a5 A1 2 Question Answer Marks Guidance

6(a) x2 B1 x2(xa2)2a5 52a M1 y xa2 x2 x2 yxa2 A1 3 6(b) dy x22xa  x2 ax1  M1 Differentiates.  dx x22   A1 dy x2 4x2a10 or dy 1 2a5  Forms quadratic equation or simplifies . dx  dx x22    Not from wrong working. 1642a1208a0 (or y'0 )  No stationary points M1 A1 Consideration of discriminant or sign of y' with correct conclusion. 4 © Cambridge University Press & Assessment 2024 Page 11 of 16 6(c) B1 Axes and asymptotes labelled. y  xa2 B1 Branches correct. (Asymptotes may cross above, on or below the x-axis.) x x2 0, 1 B1 May be seen on their diagram. 2 3 © Cambridge University Press & Assessment 2024 Page 12 of 16 6(d)(i) B1 FT from their attempt in part 6(c). B1 Everything correct (approach to vertical asymptote and cusps correct). x 2 6(d)(ii) y B1 ya x 1 © Cambridge University Press & Assessment 2024 Page 13 of 16 6(d)(iii) x2 ax1 x2 ax1 M1 Or direct use of x3 a or a x2 x2 x2 12a0 or x2 2ax12a0 a a2 2a1x 2a1,a a2 2a1x 2a1 a5 A1 2 Question Answer Marks Guidance

6(a) x2 30 has no real roots. B1 OE y0 B1 Horizontal asymptote. 2 6(b) dy (x2 3)(x1)(2x) M1 dy  Finds . dx  x2 3 2 dx x2 2x30 M1 Sets equal to 0 and simplifies to quadratic equation.  3, 1 , 1, 1 A1 A1 6 2 4 © Cambridge University Press & Assessment 2024 Page 12 of 16 6(c) y B1 Axes and approach to asymptote. B1 Correct smooth shape and position. x (1, 0),  0, 1 B1 States coordinates of intersections with 3 axes, may be shown on their graph. 3 6(d) y B1 Sections for x0 and x1 correct B1 1x0 correct. x   B1 States coordinates of intersections with (1, 0), 0,  1 3 axes, may be shown on their graph.   B1 May be shown on their graph. 1,  1 2 4 © Cambridge University Press & Assessment 2024 Page 13 of 16

6(a) 1 B1 Vertical asymptotes. x= , x=3 2 y=2 B1 Horizontal asymptote. 2 © Cambridge University Press & Assessment 2024 Page 11 of 15 6(b) dy (2x2 −7x+3)(8x+1)−(4x2 +x+1)(4x−7) M1 dy = Finds . Allow top line only for M1. dx ( 2x2 −7x+3 )2 dx −3x2 +2x+1=0 M1 Sets equal to 0 and forms equation. ( −1,1) , (1,−3) A1 A1 3 5 4 xy 6(c) B1 Axes and asymptotes. Clear identification (label or clear intersection with axes at correct place). B1 x3 correctly approaching asymptotes, not too truncated. B1 1 x3 correct. 2 B1 x 1 correct. 2 ( 0,1) B1 States coordinates of intersection with axis. May 3 be seen on their graph. 5 © Cambridge University Press & Assessment 2024 Page 12 of 15 6(d) B1FT FT from sketch in (c). At least two branches. k3 B1 2 Question Answer Marks Guidance

6(a) x2 +1=0 has no real roots. B1 y=1 B1 Horizontal asymptote. 2 6(b) yx2 + y = x2 +3(1− y)x2 − y+3=0 M1 A1 Forms quadratic in x or uses 2 y =1+ . x2 +1 −4(1−y)(3−y) 0 M1 Uses that discriminant is 0 or 2 0 2. x2 +1 1 y3 A1 Explanation of why y 1 AG. 4 © Cambridge University Press & Assessment 2024 Page 11 of 16 6(c) dy ( x2 +1 )(2x)− ( x2 +3 )(2x) M1 Differentiates = =0 dx ( x2 +1 )2 (0,3) A1 Alternative method for question 6(c) When y =3 2x2 =0 M1 Using inequality from (b) (0,3) A1 2 y 6(d) B1 Axes and correct asymptote labelled. B1 Correct shape and position. B1 States (0,3)coordinates of intersection with axes, may been seen on diagram. x 3 © Cambridge University Press & Assessment 2024 Page 12 of 16 6(e) B1FT FT from sketch in (d) 1 3 x2 +1 1 M1 Finds critical points. = x2 +3 2 x2 =1x=1 A1 −1x1 A1 4 Question Answer Marks Guidance

6(a) 1 B1 Vertical asymptotes. x= , x=3 2 y=2 B1 Horizontal asymptote. 2 © Cambridge University Press & Assessment 2024 Page 11 of 15 6(b) dy (2x2 −7x+3)(8x+1)−(4x2 +x+1)(4x−7) M1 dy = Finds . Allow top line only for M1. dx ( 2x2 −7x+3 )2 dx −3x2 +2x+1=0 M1 Sets equal to 0 and forms equation. ( −1,1) , (1,−3) A1 A1 3 5 4 xy 6(c) B1 Axes and asymptotes. Clear identification (label or clear intersection with axes at correct place). B1 x3 correctly approaching asymptotes, not too truncated. B1 1 x3 correct. 2 B1 x 1 correct. 2 ( 0,1) B1 States coordinates of intersection with axis. May 3 be seen on their graph. 5 © Cambridge University Press & Assessment 2024 Page 12 of 15 6(d) B1FT FT from sketch in (c). At least two branches. k3 B1 2 Question Answer Marks Guidance

7(a) x=−1, x=4 B1 Vertical asymptotes. 2 y=1 B1 Horizontal asymptote. 2 7(b) dy (2x2 −7x−4)(4x−5)−(2x2 −5x)(4x−7) M1 dy = Finds . dx ( 2x2 −7x−4 )2 dx 4x2 +16x−20=0 M1 Sets equal to 0 and forms equation. ( −5,25) ( 1,1) A1 A1 27 3 4 © Cambridge University Press & Assessment 2025 Page 16 of 18 7(c) B1 Axes and asymptotes labelled. B1 Correct shape and position. x (0,0), (5,0 ) B1 States coordinates of intersections with axes. 2 May be seen on their diagram. 3 y 7(d) B1 Correct shape and position. x 1 © Cambridge University Press & Assessment 2025 Page 17 of 18 7(e) 2x2 −5x 1 2x2 −5x 1 M2 Finds critical points, award M1 for each case. = or =− 2x2 −7x−4 9 2x2 −7x−4 9 16x2 −38x+4=0 or 20x2 −52x−4=0 x=19 + 3 33, x=19− 3 33 or x=13+ 3 21, x=13− 3 21 A1 A1 16 16 16 16 10 10 10 10 13 − 3 21 x19 − 3 33, 19+ 3 33x13+ 3 21 A1 10 10 16 16 16 16 10 10 5 © Cambridge University Press & Assessment 2025 Page 18 of 18

7(a) x=−1, x=4 B1 Vertical asymptotes. 2 y=1 B1 Horizontal asymptote. 2 7(b) dy (2x2 −7x−4)(4x−5)−(2x2 −5x)(4x−7) M1 dy = Finds . dx ( 2x2 −7x−4 )2 dx 4x2 +16x−20=0 M1 Sets equal to 0 and forms equation. ( −5,25) ( 1,1) A1 A1 27 3 4 © Cambridge University Press & Assessment 2025 Page 16 of 18 7(c) B1 Axes and asymptotes labelled. B1 Correct shape and position. x (0,0), (5,0 ) B1 States coordinates of intersections with axes. 2 May be seen on their diagram. 3 y 7(d) B1 Correct shape and position. x 1 © Cambridge University Press & Assessment 2025 Page 17 of 18 7(e) 2x2 −5x 1 2x2 −5x 1 M2 Finds critical points, award M1 for each case. = or =− 2x2 −7x−4 9 2x2 −7x−4 9 16x2 −38x+4=0 or 20x2 −52x−4=0 x=19 + 3 33, x=19− 3 33 or x=13+ 3 21, x=13− 3 21 A1 A1 16 16 16 16 10 10 10 10 13 − 3 21 x19 − 3 33, 19+ 3 33x13+ 3 21 A1 10 10 16 16 16 16 10 10 5 © Cambridge University Press & Assessment 2025 Page 18 of 18

6(a) x=−a B1 States vertical asymptote. (x+a)(x−a)+a2 +a a2 +a M1 Finds oblique asymptote. y= =x−a+ x+a x+a y=x−a A1 3 © Cambridge University Press & Assessment 2025 Page 14 of 19 6(b) dy M1 Differentiates =1−(a2 +a)(x+a)−2 =0 or x2+2ax−a=0 dx A1 Sets correct derivative to zero or ACF of quadratic equation A1 CAO. No equivalent forms. x=−a a2 +a 3 6(c) y B1FT Axes shown and asymptotes labelled. FT on asymptotes from (a) B1 Lower branch correct. B1 Upper branch with (0,1) stated or 1 marked on 1 vertical axis and no others. x 3 © Cambridge University Press & Assessment 2025 Page 15 of 19 6(d) xy B1FT FT from sketch in (c) 1 1 6(e) x2 +a M1 Finds critical points. =a x+a x2−ax+a−a2 =0 x2 +a B1 Quadratic equation sufficient for B1. ISW Or =−ax2 +ax+a+a2 =0 x+a x2 +a May state that =−a has no real solutions x+a since a0. a2 −4 ( a−a2) =5a2 −4a0 M1 Uses that discriminant is positive to get a range of values for a. Can not be 0 so M0. a 4. A1 5 4 © Cambridge University Press & Assessment 2025 Page 16 of 19

7(a) y=1 B1 1 © Cambridge University Press & Assessment 2025 Page 15 of 18 7(b) x2 +x−4= yx2 + yx+2y M1 (y−1)x2 +(y−1)x+2y+4[=0] Correctly forms quadratic equation in x. (y−1)2 −4(y−1)(2y+4) 0(y−1)(y−1−8y−16) 0 M1 Uses discriminant to obtain condition for real roots. (y−1)(7y+17) 0−17 y 1. A1 7 Explains why y1 −17 y1. A1 AG. 7 4 7(c) dy 12x+6 M1 A1 Sets derivative equal to 0. = =0 dx ( x2 +x+2 )2 Alternative x2 +x−4 Sets y=−17 and forms equation for x. =−17 7 x2 +x+2 7 ( −1,−17) B1 2 7 3 © Cambridge University Press & Assessment 2025 Page 16 of 18 7(d) y B1 Axes and asymptotes drawn and labelled. B1 Correct shape and position. x (0,−2), ( 1 ( −1− 17 ) ,0 ) , ( 1 ( −1+ 17 ) ,0 ) B1 Coordinates of the intersections with the axes. 2 2 Allow (0, -2), (1.56, 0), (-2.56, 0) 3 © Cambridge University Press & Assessment 2025 Page 17 of 18 7(e) y B1FT Symmetrical about x=0, FT their graph from (d). B1 Correct shape, not smooth at (0,−2). x |x|2 +|x|−4 M1 Finds critical point(s). =−1 3|x|2 +3|x|−6=0 |x|2 +|x|+2 2 |x|=1 A1 −1x1 A1 5 © Cambridge University Press & Assessment 2025 Page 18 of 18

7(a) x=−3−1 5, x=−3+1 5 B1 Vertical asymptotes. 2 2 2 2 y=0 B1 Horizontal asymptote. 2 7(b) dy (x2 +3x+1)(1)−(x+2)(2x+3) M1 dy = Finds . dx ( x2 +3x+1 )2 dx x2 +4x+5=0 M1 Sets equal to 0 and forms quadratic equation. Discriminant is 42 −20=−40 so no real roots. A1 AG 3 © Cambridge University Press & Assessment 2025 Page 15 of 17 7(c) y B1 Axes and asymptotes labelled. B1 Correct shape and position. B1 States coordinates of intersections with axes, (0,2)and (−2,0), may be seen on diagram. x 3 © Cambridge University Press & Assessment 2025 Page 16 of 17 7(d) y B1FT FT from sketch in part 7(c) B1 From a correct graph and with correct shape at (−2,0). x 2 7(e) x+2 x+2 M2 Finds critical points, award M1 for each =2 or =−2 x2 +3x+1 x2 +3x+1 case. −2x2 −5x=0 or 2x2 +7x+4=0 x=−5, x=0 or x=−7−1 17, x=−7+1 17 A1 2 4 4 4 4 −7−1 17x−5, −7+1 17x0 , x−31 5 A1 −7−1 17x−3−1 5, 4 4 2 4 4 2 2 4 4 2 2 −3−1 5x−5, 2 2 2 −7+1 17x−3+1 5, 4 4 2 2 −3+1 5x0. 2 2 4 © Cambridge University Press & Assessment 2025 Page 17 of 17

7(a) x=−1 B1 Vertical asymptote. x(x+1)+1 M1 Oblique asymptote. y= x+1 y=x A1 3 7(b) dy x2 +2x M1 dy = =0 Sets =0. dx (x+1)2 dx (0,1), (−2,−3) A1 A1 1 mark for each correct coordinate. SCB1 if just the correct x values listed 3 © Cambridge University Press & Assessment 2025 Page 17 of 19 7(c) B1FT Axes and asymptotes labelled. B1 Left branch correct with all approaches to the asymptotes correct. B1 Right branch correct with a smooth minimum point. 3 © Cambridge University Press & Assessment 2025 Page 18 of 19 7(d) B1FT Symmetrical about x = 0 and follow through from part (c) right branch B1 Correct shape of a curve and smooth at (0,1) from a correct graph in part (c) 2 7(e) x2 +x+1 M1 Finds critical point. =3 x+1 x2 −2x−2=0 x=1+ 3 A1 For selecting just the positive value of x. −1− 3x1+ 3 A1 3 © Cambridge University Press & Assessment 2025 Page 19 of 19

7(a) x=−3−1 5, x=−3+1 5 B1 Vertical asymptotes. 2 2 2 2 y=0 B1 Horizontal asymptote. 2 7(b) dy (x2 +3x+1)(1)−(x+2)(2x+3) M1 dy = Finds . dx ( x2 +3x+1 )2 dx x2 +4x+5=0 M1 Sets equal to 0 and forms quadratic equation. Discriminant is 42 −20=−40 so no real roots. A1 AG 3 © Cambridge University Press & Assessment 2025 Page 15 of 17 7(c) y B1 Axes and asymptotes labelled. B1 Correct shape and position. B1 States coordinates of intersections with axes, (0,2)and (−2,0), may be seen on diagram. x 3 © Cambridge University Press & Assessment 2025 Page 16 of 17 7(d) y B1FT FT from sketch in part 7(c) B1 From a correct graph and with correct shape at (−2,0). x 2 7(e) x+2 x+2 M2 Finds critical points, award M1 for each =2 or =−2 x2 +3x+1 x2 +3x+1 case. −2x2 −5x=0 or 2x2 +7x+4=0 x=−5, x=0 or x=−7−1 17, x=−7+1 17 A1 2 4 4 4 4 −7−1 17x−5, −7+1 17x0 , x−31 5 A1 −7−1 17x−3−1 5, 4 4 2 4 4 2 2 4 4 2 2 −3−1 5x−5, 2 2 2 −7+1 17x−3+1 5, 4 4 2 2 −3+1 5x0. 2 2 4 © Cambridge University Press & Assessment 2025 Page 17 of 17

7(a) x=1.8 B1 (10x−18)(x+ 7 )−27 27 M1 Must be an attempt to find the constant term. y= 10 5 =x+ 7 − 5 10x−18 10 10x−18 y=x+0.7 A1 3 © Cambridge University Press & Assessment 2025 Page 17 of 20 7(b) dy 100x2 −360x+378 M1 Differentiates. = dx (10x−18)2 A1 Forms quadratic equation if using discriminant   dy 54 50x2 −180x+189=0 or =1+  dy  dx (10x−18)2  method. (Or writes in a form to show it is positive)   dx 1802 −4(50)(189)=−54000 (or y'0 )  No stationary points M1 A1 Consideration of discriminant or sign of y' with correct conclusion. 4 © Cambridge University Press & Assessment 2025 Page 18 of 20 7(c) B1 Axes and asymptotes labelled. B1 Branches correct. x (0,1), (−0.9,0), (2,0) B1 May be seen on their diagram. 3 © Cambridge University Press & Assessment 2025 Page 19 of 20 7(d) y B1 FT from their attempt in (c). B1 Everything correct with sharp changes of direction at intersections with xaxis. x 2 7(e) 10x2 −11x−18 −29 4441  =−4x=  10x−18 20   10x2 −11x−18 M1 Forms an equation to find required critical values for =4 x. 10x−18 10x2 −51x+54=0 A1 p= 3, q=18 A1 x= 3,x=18 2 5 2 5 A1 p,q clearly identified in correct order. 4 © Cambridge University Press & Assessment 2025 Page 20 of 20

6(a) x = − 1 B1 Vertical asymptote. (x+1)(x−2)+4 M1 Oblique asymptote - must show attempt to y= find constant term. 4x+4 y = 14 x − 12 A1 Must be a linear equation. 3 Trace ID: PC-1563EA0F9E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 6(b) dy 4x2 +8x−12 = =0 dx (4x+4)2 © Cambridge University Press & Assessment 2026 Page 18 of 23 4 x 2 + 8 x − 1 2 = 0 M1 Sets d d y x = 0 . (1,1), (−3,−7) A1 A1 for x-coordinates. 4 4 A1 A1 for y-coordinates correctly linked to x- coordinates. Allow A1 for one completely correct point. 3 Trace ID: PC-1563EA0F9E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 6(c) B1 FT Axes and asymptotes labelled. Follow through their two asymptotes from (a) for this mark only. B1 Left branch correct and also all approaches to the asymptotes. B1 Right branch correct including a smooth minimum in the first quadrant. 3 © Cambridge University Press & Assessment 2026 Page 19 of 23 y = 14 x − 12 x = − 1 y x Trace ID: PC-1563EA0F9E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 6(d) B1 Must be fully correct, including vertical asymptote. 1 6(e) © Cambridge University Press & Assessment 2026 Page 20 of 23 x 2 4 − x x + + 4 2 = − 2 or x 2 4 − x x + + 4 2 = 2 x 2 + 7 x + 1 0 = 0 x 2 − 9 x − 6 = 0 M2 Finds critical points, award M1 for each case. Must be using equations. x=−5,−2 x = 12 ( 9 − 1 0 5 ) , x = 12 ( 9 + 1 0 5 ) A1 All four correct, exact answers. − 5  x  − 2 12 ( 9 − 1 0 5 )  x  12 ( 9 + 1 0 5 ) y A1 CAO 4 x Trace ID: PC-1563EA0F9E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

6(a) x = − 1 B1 Vertical asymptote. (x+1)(x−2)+4 M1 Oblique asymptote - must show attempt to y= find constant term. 4x+4 y = 14 x − 12 A1 Must be a linear equation. 3 Trace ID: PC-F469E54369 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 6(b) dy 4x2 +8x−12 = =0 dx (4x+4)2 © Cambridge University Press & Assessment 2026 Page 18 of 23 4 x 2 + 8 x − 1 2 = 0 M1 Sets d d y x = 0 . (1,1), (−3,−7) A1 A1 for x-coordinates. 4 4 A1 A1 for y-coordinates correctly linked to x- coordinates. Allow A1 for one completely correct point. 3 Trace ID: PC-F469E54369 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 6(c) B1 FT Axes and asymptotes labelled. Follow through their two asymptotes from (a) for this mark only. B1 Left branch correct and also all approaches to the asymptotes. B1 Right branch correct including a smooth minimum in the first quadrant. 3 © Cambridge University Press & Assessment 2026 Page 19 of 23 y = 14 x − 12 x = − 1 y x Trace ID: PC-F469E54369 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 6(d) B1 Must be fully correct, including vertical asymptote. 1 6(e) © Cambridge University Press & Assessment 2026 Page 20 of 23 x 2 4 − x x + + 4 2 = − 2 or x 2 4 − x x + + 4 2 = 2 x 2 + 7 x + 1 0 = 0 x 2 − 9 x − 6 = 0 M2 Finds critical points, award M1 for each case. Must be using equations. x=−5,−2 x = 12 ( 9 − 1 0 5 ) , x = 12 ( 9 + 1 0 5 ) A1 All four correct, exact answers. − 5  x  − 2 12 ( 9 − 1 0 5 )  x  12 ( 9 + 1 0 5 ) y A1 CAO 4 x Trace ID: PC-F469E54369 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

6(a) x2 +x+1=0 has no real roots as discriminant is [1 2 − 4 = ] − 3 B1 The discriminant needs be evaluated, compared and this is less than zero, hence and concluded. shown. Or shows that the expression is always positive (i.e. completing the square) and concludes. y =0 B1 Horizontal asymptote. 2 Trace ID: PC-39A0636DD7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 6(b) dy (x2 +x+1)(1)−(x+1)(2x+1) M1 Finds = dx ( x2 +x+1 )2 © Cambridge University Press & Assessment 2026 Page 16 of 20 d d y x . Allow one arithmetic error for this mark. x 2 + 2 x = 0 M1 Sets equal to 0, forms equation and attempts to solve. (0,1) and (−2,−1) A1 Condone clearly paired x and y. 3 3 6(c) B1 Axes and approach to asymptote. B1 Correct shape and position, and maximum apparently on the axis ( 0 , 1 ) and ( − 1 , 0 ) B1 States coordinates of just the two intersections with axes, may be seen on their diagram. 3 6(d) B1FT FT provided sketch in part (c) has both positive and negative y values B1 Correct shape at (−1,0)and must have a correct graph in part (c) 2 y x x y Trace ID: PC-39A0636DD7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 6(e) © Cambridge University Press & Assessment 2026 Page 17 of 20 x 2 x + + x 1 + 1 = 1 3 or x 2 x + + x 1 + 1  1 3 M1 Writing down the equation or inequality x 2 x + + x 1 + 1 = − 1 3 or x 2 x + + x 1 + 1  − 1 3 M1 Writing down the equation or inequality EITHER: x  1 − 3 (A1 Writing both inequalities ISW , x 1+ 3 x  − 2 A1) ISW OR: Alternative for the final 2 marks x  − 2 , − 2  x  1 − 3 , x  1 + 3 (A2,0) All three regions correct. Available marks 4 Trace ID: PC-39A0636DD7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

6(a) x = 0 , x=1 B1 Both vertical asymptotes. y = 1 B1 Horizontal asymptote and no extra asymptotes. 2 Trace ID: PC-4F45C0CD5B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 6(b) © Cambridge University Press & Assessment 2026 Page 20 of 25 d d y x = − 2 ( 2 x x 2 2 − − 4 2 x x + ) 2 2 *M1 Finds d d y x . A method which clearly demonstrates the use of quotient rule, product rule or rearrangement must be seen. x 2 + 2 x − 1 = 0 DM1 Sets equal to 0 and forms quadratic equation. ( − 2 .4 , 0 .9 ) , ( 0 .4 , − 1 .9 ) A1 A1 Accept ( − 1 − 2 , − 12 + 2 ) , ( − 1 + 2 , − 12 − 2 ) A1 A0 for correct x values but y values missing or incorrect. 4 Trace ID: PC-4F45C0CD5B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 6(c) B1FT Axes and asymptotes labelled. Accept labelled by numerical intercepts with asymptotes. Allow FT from (a) for this B mark only. B1 LHS correct shape and position, crossing horizonal asymptote. The minimum should be indicated by the shape of the curve and the asymptotic approach. Do not expect accurate scale. B1 Both branches for © Cambridge University Press & Assessment 2026 Page 21 of 25 x  0 correct. 3 x = 1 x y y=1 Trace ID: PC-4F45C0CD5B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 6(d) B1FT Must be a single curve from LHS to RHS of the page, with no breaks. Follow through the max / min for their graph for (c) as long as it has three branches. Numerical values need not be considered for this mark. B1 Correct shape and intersections with axis and relationship with asymptotes. © Cambridge University Press & Assessment 2026 Page 22 of 25 − 1  x  0 B1 x  1 B1 4 Question Answer Marks Guidance