Differentiation — Paper 1 · A Level Mathematics 9709

9 y 6 y = 3x – 2 x O 1 2 6 The diagram shows part of the curve y = 3x −2. (i) Find the gradient of the curve at the point where x = 2. [3] (ii) Find the volume obtained when the shaded region is rotated through 360◦about the x-axis, giving your answer in terms of π. [5]

8 marks

Mark scheme: 9 (i) dy/dx = − 6(3x − 2)−2 × 3 B1 M1 B1 (without the ×3). Use of chain rule. If x = 2, m = −1⅛ (−1.125) A1 co. [3] 36 d x B1 2 Attempt at π ∫ y 2 - even if π missing. (ii) Vol = ∫ (3 x − 2 ) − 36 [ ÷ 3] B1 B1 (3 x − 2) No need for π here. Use of limits [2] − [1] → 9π M1 Correct use of correct limits. A1 co. [5] GCE A/AS LEVEL – May/June 2009 9709 01

This question in 9709/11 May/June 2009

8 x metres r metres The inside lane of a school running track consists of two straight sections each of length x metres, and two semicircular sections each of radius r metres, as shown in the diagram. The straight sections are perpendicular to the diameters of the semicircular sections. The perimeter of the inside lane is 400 metres. (i) Show that the area, A m2, of the region enclosed by the inside lane is given by A = 400r −0r2. [4] (ii) Given that x and r can vary, show that, when A has a stationary value, there are no straight sections in the track. Determine whether the stationary value is a maximum or a minimum. [5]

9 marks

Mark scheme: 8 (i) A = 2 xr + πr 2 B1 2 x + 2πr = 400 (⇒ x = 200 − πr ) B1 A = 400 r − πr 2 M1A1 Subst & simplify to AG (www) [4] dA (ii) = 400 − 2πr B1 Differentiate dr = 0 M1 Set to zero and attempt to find r 200 r = oe A1 π x = 0 ⇒ no straight sections AG A1 d 2 A = −2π ( < 0 ) Max B1 Dep on − 2π , or use of other valid 2 dr [5] reason GCE AS/A LEVEL – October/November 2013 9709 11 10 ( )

This question in 9709/11 Oct/Nov 2013

9 y y = 8 −ï 4 −x P 3, 7 x O The diagram shows part of the curve y 8 4 and the tangent to the curve at P 3, 7 . = − −x dy (i) Find expressions for and y dx. [5] dx Ó (ii) Find the equation of the tangent to the curve at P in the form y mx c. [2] = + (iii) Find, showing all necessary working, the area of the shaded region. [4]

11 marks

Mark scheme: 9 y = 8 − 4 − x d y 1 − 1 (i) = − (4 − x ) 2 × −1 B1 B1 Without (−1). For (×−1). d x 2 3 (4 − x ) 2 ∫y dx = 8 x − ÷ −1 3 × B1 B1 for "8x" and +c". B1 for all except 3 2 ÷(−1). B1 for ÷(−1). [5] (n.b. these 5 marks can be gained in(ii) or (ii) Eqn y − 7 = 1 2 ( x − 3 ) (iii)) → y = ½x + 5½ M1A1 M1 unsimplified. A1 as y=mx+c [2] (iii) Area under curve = ∫ from 0 to 3 (58/3) M1 Use of limits – needs use of “0” Area under line = ½(5½ + 7)×3 M1 Correct method  1 2 11x  Or 4 x + from 0 to 3 M1 A1 M1 Subtraction. A1 co  2  [4] 58 75 7 → − = 3 4 12

This question in 9709/12 May/June 2014

dy 12 6 A curve is such that = , where a is a constant. The point P 2, 14 lies on the curve and dx 4x + a the normal to the curve at P is 3y + x = 5. (i) Show that a = 8. [3] (ii) Find the equation of the curve. [4]

7 marks

Mark scheme: dy 12 6 = P (2, 14) Normal 3y + x = 44 dx 4 x + a (i) m of normal = 1 B1 co −3 dy 12 = 3 = → a = 8 M1 A1 Use of m1m₂ = −1. AG. dx 4 x + a [3] 1 1 (ii) ∫ y = 12(4x + a) 2 ÷ ÷ 4 (+c) B1 B1 Correct without “÷4”. for “÷4”. 2 Uses (2, 14) M1 Uses in an integral only. Dep ‘c’. c = −10 A1 co All 4 marks can be given in (i) [4] GCE AS/A LEVEL – May/June 2014 9709 13

This question in 9709/13 May/June 2014

8 A curve y = f x has a stationary point at 3, 7 and is such that f ′′ x = 36x−3. (i) State, with a reason, whether this stationary point is a maximum or a minimum. [1] (ii) Find f ′ x and f x . [7]

8 marks

Mark scheme: 8 (i) Minimum since f ″(3) (= 4/3) > 0 www B1 [1] (ii) f ′( x ) = −18 x − 2 (+ c ) B1 0 = − 2 + c M1 Sub f 3 0. (dep c present) c = 2 (→ f ′( x ) = − 18 x − 2 + 2 ) A1 c = 2 sufficient at this stage f ( x ) = 18 x −1 + 2 x (+ k ) B1 B1 Allow cx at this stage 7 = 6 + 6 + k M1 Sub f(3) = 3 (k present & numeric −1 (or no) c) k = −→5 ( f ( x ) = 18 x + 2 x − 5 ) cao A1 [7] 2 2 ( ) ( )2

This question in 9709/13 Oct/Nov 2014

dy −1 9 A curve passes through the point A 4, 6 and is such that = 1 + 2x 2. A point P is moving along dx the curve in such a way that the x-coordinate of P is increasing at a constant rate of 3 units per minute. (i) Find the rate at which the y-coordinate of P is increasing when P is at A. [3] (ii) Find the equation of the curve. [3] (iii) The tangent to the curve at A crosses the x-axis at B and the normal to the curve at A crosses the x-axis at C. Find the area of triangle ABC. [5]

11 marks

Mark scheme: dy 9 (i) At x = ,4 = 2 B1 dx d y dy d x = × = 2 × 3 = 6 M1A1 Use of Chain rule d t d x d t [3] 1 (ii) ( y ) = x + 4 x 2 ( + c ) B1 1 Sub x = ,4 y = 6 → 6 = 4 + ( 4 × 4 2 ) + c M1 Must include c 1 c = − 6 → ( y = x + 4 x 2 − 6 A1 [3] (iii) Eqn of tangent is y − 6 = 2 ( x − 4 ) or M1 Correct eqn thru (4, 6) & with m = ( 6 − 0 ) /( 4 − x ) = 2 A1 their 2 B = (1, 0) (Allow 1) M1 [Expect eqn of normal: ½ Gradient of normal = −1/2 A1 8] C = (16, 0) (Allow 16) A1 1 [5] Area of triangle = × 15 × 6 = 45 Or AB = 45 , AC = 180 → 2 Area = 45.0 3

This question in 9709/13 Oct/Nov 2015

dy 6 A curve has a stationary point at 3, 91 and has an equation for which = ax2 + a2x, where a is a 2 dx non-zero constant. (i) Find the value of a. [2] … … … … … … … … … … (ii) Find the equation of the curve. [4] … … … … … … … … … … … … … … … … … … … … … … … … (iii) Determine, showing all necessary working, the nature of the stationary point. [2] … … … … … … … … … … … …

8 marks

Mark scheme: 6(i) 2 0 9 3 = + a a M1 Sub d 0 and 3 d y x x = = 3 = − a only A1 2 6(ii) ( ) 2 2 3 d 9 3 9 d 2 y x x x y x c x = − + → = − + + M1A1FT Attempt integration. 3 2 2 13 ½ + ax a x scores M1. Ft on their a. 9½ 27 40½ = − + + c DM1 Sub 3, 9½ = = x y . Dependent on c present 4 = − c A1 Expect 2 3 9 4 2 = − + − x y x 4 6(iii) 2 2 d 6 9 d y x x = − + M1 2 2 + ax a scores M1 At 2 2 d 3, 9 0 d y x x = = − < MAX www A1 Requires at least one of ‒9 or < 0. Other methods possible. 2

This question in 9709/11 Oct/Nov 2018

10 A curve has equation y = 1 4x −3 −1. The point A on the curve has coordinates 1, 1 . 2 2 (i) (a) Find and simplify the equation of the normal through A. [5] … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the x-coordinate of the point where this normal meets the curve again. [3] … … … … … … … … … … … … (ii) A point is moving along the curve in such a way that as it passes through A its x-coordinate is decreasing at the rate of 0.3 units per second. Find the rate of change of its y-coordinate at A. [2] … … … … … … … … … … …

10 marks

Mark scheme: 10(i)(a) ( ) [ ] 2 d ½ 4 3 4 d y x x −   = − − ×   B1B1 When 1, 2 = = − x m B1FT Ft from their d d y x Normal is ( ) ½ ½ 1 − = − y x M1 Line with gradient ‒1/m and through A ½ = y x soi A1 Can score in part (b) 5 10(i)(b) ( ) ( ) ( )( ) ( ) 2 1 2 4 3 2 2 4 3 1 0 2 4 3 2 = → − = → − − = − x x x x x x M1A1 x/2 seen on RHS of equation can score previous A1 1/ 4 = − x A1 Ignore 1 = x seen in addition 3 10(ii) Use of chain rule: ( ) ( ) d 2 0.3 0.6 d y their t = − × ± = M1A1 Allow +0.3 or ‒0.3 for M1 2

This question in 9709/11 Oct/Nov 2018

4 A curve has equation y = 2x −1 −1 + 2x. dy d2y (i) Find and . [3] dx dx2 … … … … … … … … … … … … … … … … … … … … … … … … (ii) Find the x-coordinates of the stationary points and, showing all necessary working, determine the nature of each stationary point. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(i) 2 d / d 2 2 1 2 − = − − + y x x B2,1,0 ‘( ) 2 2 1 − − x ’ and ‘2’) ( ) 3 2 2 d / d 8 2 1 − = − y x x B1 Unsimplified form ok 3 Question Answer Marks Guidance 4(ii) Set d / d y x to zero and attempt to solve – at least one correct step M1 x = 0, 1 A1 Expect ( ) 2 2 1 1 − = x When x = 0, 2 2 d / d 8 (or 0) = − < y x . Hence MAX B1 When x = 1, 2 2 d / d 8 (or 0) = > y x . Hence MIN B1 Both final marks dependent on correct x and correct 2 2 d / d y x and no errors May use change of sign of dy / dx but not at 1/ 2 = x 4

This question in 9709/12 Feb/March 2019

11 y 9 M y = 4x + 1 + 4x + 1 x O 9 The diagram shows part of the curve y 4x 1 and the minimum point M. = + + 4x 1 + dy (i) Find expressions for and y dx. [6] dx Ó … … … … … … … … … … … … … … … (ii) Find the coordinates of M. [3] … … … … … … … … … … … … The shaded region is bounded by the curve, the y-axis and the line through M parallel to the x-axis. (iii) Find, showing all necessary working, the area of the shaded region. [3] … … … … … … … … … … …

12 marks

Mark scheme: 11(i) ( ) 1 2 d 1 4 1 d 2 −   = +     y x x [× 4] ( ) 3 2 9 4 1 2 −   − +     x [× 4] B1B1B1 ( ) ( ) 3 3 2 2 18 8 16 4 1 4 1 4 1   −   −   +   + +   x or x x x SC If no other marks awarded award B1 for both powers of (4x +1) correct. ∫ydx = ( ) 3 2 4 1 3 2     +       x [÷ 4] + ( ) 1 2 9 4 1 1 2     +       x [÷ 4] (+C) B1B1B1 B1 B1 for each, without ÷ 4. B1 for ÷4 twice. + C not required. ( ) ( ) 3 4 1 9 4 1 ( ) 6 2 x x C   +   + + +       SC If no other marks awarded , B1 for both powers of (4x +1) correct. 6 11(ii) d 0 d = y x → 2 4 1 + x − ( ) 3 2 18 4 1 + x = 0 M1 Sets their d d y x to 0 (and attempts to solve 4x + 1 = 9 or (4x + 1)2 = 81 A1 Must be from correct differential. x = 2, y = 6 or M is (2, 6) only. A1 Both values required. Must be from correct differential. 3 Question Answer Marks Guidance 11(iii) Realises area is ∫y dx and attempt to use their 2 and sight of 0. *M1 Needs to use their integral and to see ‘their 2’ substituted. Uses limits 0 to 2 correctly → [4.5 + 13.5] – [ 1 6 + 4.5] ( = 13⅓ ) DM1 Uses both 0 and ‘their 2’ and subtracts. Condone wrong way round. (Area =) 1⅓ or 1.33 A1 Must be from a correct differential and integral. 3 13⅓ or 1⅓ with little or no working scores M1DM0A0.

This question in 9709/12 May/June 2019

6 A point P is moving along a curve in such a way that the x-coordinate of P is increasing at a constant 1 rate of 2 units per minute. The equation of the curve is y = 5x −1 2. (a) Find the rate at which the y-coordinate is increasing when x = 1. [4] … … … … … … … … … … … … … … … … … … … … … … … (b) Find the value of x when the y-coordinate is increasing at 5 units per minute. [3] 8 … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 6(a) ( ) [ ] 1/2 d 1 5 1 5 d 2 −   = − ×     y x x B1 B1 Use d d 2 when 1 d d   = × =     y y their x t x M1 5 2 A1 4 Question Answer Marks 6(b) ( ) 1/2 5 5 2 5 1 2 8 − × − = their x oe M1 ( ) 1/2 5 1 8 − = x A1 13 = x A1 3

This question in 9709/13 May/June 2020

10 y x O M 2 The diagram shows part of the curve y and its minimum point M, which lies on the 2 −x = 3 −2x x-axis. dy d2y (a) Find expressions for dx, dx2 and Ó y dx. [6] … … … … … … … … … … … … … … … … … … (b) Find, by calculation, the x-coordinate of M. [2] … … … … … … … … … … … … (c) Find the area of the shaded region bounded by the curve and the coordinate axes. [2] … … … … … … … … … … … …

10 marks

Mark scheme: 10(a) d (d y x ) = [8] ( 3 [ 3 2 ) − × − +  x [ −1] ( ) 8 1 3 2 ³ x   = −     −   B2, 1, 0 B2 for all three elements correct, B1 for two elements correct, B0 for only one or no elements correct. d² d ² y x = ( ) 4 3 8 (3 2 ) 2 − −× × − × − x ( ) 4 48 3 2x     =   −   B1 FT FT providing their bracket is to a negative power ∫ydx = [(3− 2x)−1 ] [2 ÷ (−1 × −2)] [ − ½x²] (+c) 2 1 1 3 2 2 x c x   = − +   −   B1 B1 B1 Simplification not needed, B1 for each correct element 6 Question Answer Marks Guidance 10(b) d d y x = 0 → (3 – 2x)³ = 8 → 3 – 2x = k → x = M1 Setting their 2-term differential to 0 and attempts to solve as far as x = 1 2 A1 Alternative method for question 10(b) ( ) 2 2 2 0 0 2 (2 1) 0 (3 2 ) y x x x x x = → − = → − − = → = − M1 Setting y to 0 and attempts to solve a cubic as far as x = (3 factors needed) 1 2 A1 2 10(c) Area under curve = their 1 ² 1 1 2 0 1 2 3 2 0 3 2 2               − − −   − ×       − ×        M1 Using their integral, their positive x limit from part (b) and 0 correctly. 1 24 A1 2

This question in 9709/12 Oct/Nov 2020

1 9 The equation of a curve is y = 3x + 1 −4 3x + 1 2 for x > −13. dy d2y (a) Find and . [3] dx dx2 … … … … … … … … … … … … … … … … … … … … … … … (b) Find the coordinates of the stationary point of the curve and determine its nature. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 9(a)  1 1 2 2 d 1 3 4 (3 1) 3 3 6(3 1) d 2 y x x x                          1 differentiation of   1 2 4 3 1 x   . Accept unsimplified.     2 3 3 2 2 2 d 1 6 3 1 3 [ 9 3 1 2 d y x x x               ] B1 WWW. Accept unsimplified. Do not award if d d y x is incorrect. 3 9(b)   1 2 d 0 leading to 3 6 3 1 0 d y x x      M1 Setting their d d y x = 0.   1 2 3 1 2 3 1 4 x x    leading to 1 x  A1 CAO – do not ISW for a second answer. y = −4 [coordinates (1, −4)] A1 Condone inclusion of second value from a second answer.   2 3 2 2 d 9 3 1 1 d y x    = 9 or 8 > 0 so minimum A1 Some evidence of substitution needed but 2 2 d d y x . Do not award if 2 2 d d y x is incorrect or wrongly evaluated. Accept correct consideration of gradients either side of 1 x  . 4

This question in 9709/12 May/June 2022

dy10 At the point 4, −1 on a curve, the gradient of the curve is −3 It is given that = x−1 2. 2 + k, where k dx is a constant. (a) Show that k = −2. [1] … … … … (b) Find the equation of the curve. [4] … … … … … … … … … … … … … … … … … … (c) Find the coordinates of the stationary point. [3] … … … … … … … … … … … … (d) Determine the nature of the stationary point. [2] … … … … … … … … … … … …

10 marks

Mark scheme: 110(a) B1 − 3 1 1 − = + k leading to k = −2 2 evaluated as or better. AG Need to see 4 1 2 2 4 2 1 10(b) 1 M1 A1 1 2 y = 2 x 2 − 2 x  + c  x Allow − 2 x . 12 −=1 4 −+c8 M1 Substitute x = 4, y = −1 (c present) Expect c = 3. 1 A1 Allow if f(x) = or y = anywhere in the solution. y = 2 x 2 − 2 x + 3 or y = 2 x − 2 x + 3 4 10(c) x −1/2 − 2 = 0 M1 d y Set their to zero. d x 1 A1 2 x =  1 1  1 1  =  max of M1A1 if If    ,3  seen. 4  2  4  4 2  (¼, 3½) A1 3 10(d) d 2 y 1 − 32 B1 2 = − x d x 2 < 0 (or −)4 hence Maximum DB1 1 WWW Ignore extra solutions from x = − . 4 2

This question in 9709/12 Feb/March 2023

9 Water is poured into a tank at a constant rate of 500cm3 per second. The depth of water in the tank, t seconds after filling starts, is hcm. When the depth of water in the tank is hcm, the volume, V cm3, of water in the tank is given by the formula V = 4 25 + h 3 −62500 . 3 3 (a) Find the rate at which h is increasing at the instant when h = 10cm. [3] … … … … … … … … … … … … … … … … … … … … … … … (b) At another instant, the rate at which h is increasing is 0.075cm per second. Find the value of V at this instant. [3] … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 9(a)   2 d 4 3 25 d 3    V h h [= 4900 when h = 10] d V h . d d d d d d   V h V h t t   2 d d "4 25 10 " 500 d d       h h their t t 500 4900       M1 Use chain rule correctly to find a numerical expression for d d h t . Accept e.g. 500 2500 2000 400   . 1 d 0.102 cms d       h t A1 AWRT OE e.g. 5 49 ISW. 3 9(b) d d d 500 d d d     V V h t h t   2 "4 25 " 0.075   their h *M1 SOI Use chain rule correctly to form equation in h.   2 5000 25 [15.8 3           h h 248…] DM1 Solve quadratic to find h. Exact value of h is 5000 25 3  or 50 6 25 3  25 40.82    h 69900  V cm3 A1 AWRT ISW Look for 698(88.5) . 3

This question in 9709/11 May/June 2023

d2y 10 A curve has a stationary point at 2, −10 and is such that = 6x. dx2 dy (a) Find dx. [3] … … … … … … … … … … (b) Find the equation of the curve. [3] … … … … … … … … … … … … (c) Find the coordinates of the other stationary point and determine its nature. [3] … … … … … … … … … … … … (d) Find the equation of the tangent to the curve at the point where the curve crosses the y-axis. [2] … … … … … … … … … … … …

11 marks

Mark scheme: 10(a) dy 2 B1 = 3 x  + c  dx 3  2 2 + c = 0 M1 dy Substitute x = 2 and = 0 into an integral (c must be dx present). dy 2 A1 = 3 x −12 dx 3 10(b) 3 B1 FT FT on their non-zero c (dependent on c being found at y = x − 12 x  +k  some stage). −10 = 23 − 12  2 + k M1 Substitute x = 2, y = ‒10 (k present). 3 A1 3 y = x − 12 x + 6 Must be y = (unless y = x − 12 x + k stated earlier). 3 10(c) 3 x 2 − 12 = 0 [leading to x =−2 ] M1 dy Set their two term = 0 . Expect x =−2 . dx Ignore x = 2 given in addition. 3 A1 y = ( −2 ) − 12 −( 2 ) + 6 = 22 leading to (‒2, 22) d 2 y A1 dy When x = −2, 2  0 (or ‒12) hence Maximum Can be from correct conclusion from sign diagram dx dx dy if calculated correctly. dx Do not allow concave downward for final A1. Can be awarded if the only error is incorrect or missing y-coordinate. 3 10(d) dy M1 d y At x = 0, = −12, y = 6 Both required. FT on their and y. dx d x y − 6 = −12 x A1 OE 2

This question in 9709/11 Oct/Nov 2023

d 2 y 6 5 1 9 A curve is such that = - . It is given that the curve has a stationary point at b , 9l. dx 2 x 4 x 3 2 d 2 y (a) Use the expression for to determine whether the stationary point is a maximum or a dx 2 minimum point. [2] … … … … (b) Find the equation of the curve. [7] … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 9(a) 6 5 M1 1 and evaluate second derivative. − Substitute x = 2  1  4  1  3      2   2  96 − 40  = 56   0  Minimum A1 CWO Evidence and conclusion. d 2 y SC B1 for 2  0 without sight of 96 − 40 or 56. d x 2 9(b)  6 −3  5 −2  B1 B1 OE  x  − x   + c1   −3  −2  B1 for each correct { }. −3 1 into two terms of an integrated expression −2 M1 Substitute x = 2 6  1  5  1  0 = −     + c1 (at least one correct power), now with c1, and equate to 0 to −3  2  −2  2  find c1. c1 = 6 A1 k1 x −2 + k 2 x −1 + k3 x  +c 2  M1 d y Integration of their to produce at least two terms with d x correct powers, k1 , k 2  0. 1 5  1  M1 OE 9 = − + 6 + c2 1   2 , 9 ) into integrated expression (at least two  1   1   2  Substitute ( 2 2      2   2  correct powers) to find c2. −2 5 −1 A1 OE y = x − x + 6 x + 7 Condone their final answer being c2 = 7 if a completely 2 correct simplified expression for the equation containing c 2 has been stated previously. 7

This question in 9709/12 Feb/March 2025

dy 2 7 A curve is such that = 3 x + 10 x - 8 . dx (a) Find the set of values of x for which y decreases as x increases. [3] … … … … … … … … … (b) It is given that the maximum point of the curve has y-coordinate 27. Find the equation of the curve. [4] … … … … … … … … … … … … … … …

7 marks

Mark scheme: 7(a) 2 *B1 OE 2  5  49 3x + 10 x −8 0 or 3  x +  −  0 Condone = or ⩽ 0, but use of > or ⩾ can only score B0B1B0.  3  3 If there is no < , = or ⩽ 0, then a correct final answer implies < 0 here. 2 B1 − 4 and 3 2 DB1 FT OE −4 x  3 2 Use of ⩽ sign(s) gets DB0. Condone x −4, x  , 3 2 2 x −4 and x  but not x −4 or x  . 3 3 FT on their critical values. 3 7(b) Identify x = −4 as the x-value of the maximum point. B1 FT SOI FT on their lower critical value from part (a). 3 3 10 2 B1 Correct integration of the three terms given.  y =  x + x − 8 x  + c  x ) = . 3 2 Condone f (  2  M1  2  Use of y = 27, x = their  −4 or  in their integral. Their integral must be a cubic, their  −4 or  must come  3   3  from (a) or a restart. c = −21  y = x 3 + 5 x 2 − 8 x − 21 A1 Do not ISW if they continue to find a straight line. Condone 3 + 5 x 2 − 8 x + c omission of final statement if y or f ( x ) = x has been seen earlier. 4

This question in 9709/13 May/June 2025

dy 3 2 4 The equation of a curve is such that = kx + , where k is a constant. The curve passes through the dx x 2 point S (2, 20) and the gradient of the curve at S is 65. 2 (a) Find the value of k. [1] … … … … … … (b) The coordinates of a point T on the curve are (1, t). Find the value of t. [5] … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 4(a) [ k =] 4 B1 1 4(b)  k 4  2 x −+2 1  B2,1,0FT FT their k from 4(a) or the letter k.  x +    + c  B2 for both correct components and no other x-terms.  4   −+2 1  B1 for one correct. their k 4 2 M1 Substituting x = 2 and y = 20 into their integrated 20 =  2 − + c 4 2 expression (defined by having at least one correct power and including a '+ 'c ). Allow one slip. c = 5 A1  t =  4 A1 5

This question in 9709/12 Oct/Nov 2025