Cambridge A Level Mathematics 9709 — 2023 Oct/Nov Paper 1 · Variant 3

2 The circle with equation x −3 2 + y −5 2 = 40 intersects the y-axis at points A and B. (a) Find the y-coordinates of A and B, expressing your answers in terms of surds. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the equation of the circle which has AB as its diameter. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) ( 0 − 3) 2 + ( y − 5 ) 2 = 40 M1 OE. Substitute x = 0, may use y 2 − 10 y − 6 = 0 . y = 5  31 A1 OE. Must be surd form. 2 2(b) 2 2 2 B1FT B1 FT for their 5 and B1 FT for their 31. Don’t allow x + ( y − 5 ) = 31 Allow ( x − 0 )   B1FT surd form. 2

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3 (a) Show that the equation 5 cos 1 −sin 1 tan 1 + 1 = 0 may be expressed in the form a cos2 1 + b cos 1 + c = 0, where a, b and c are constants to be found. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence solve the equation 5 cos 1 −sin 1 tan 1 + 1 = 0 for 0 < 1 < 2π. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) 5cos 2− sin 2+ cos  = 0 M1 Multiply by cos and replace tan by sin . cos 2 2 M1 5cos − 1 − cos  + cos  = 0  ( ) 6cos 2+ cos−=1 0 A1 Missing ‘= 0’ can be condoned if ‘= 0’ appears earlier. 3 3(b) (3cos− 1)(2cos+ 1) = 0 M1 1 1 Must have 3 term quadratic, expect cos= , − . 3 2 Factors (OE) must be shown. 2π 4π A1 A1 For A1 FT is for both 2π ‒ 1st solutions. = 1.23 ; {2.09 or } ; {5.05 and 4.19 (allow ) } A1 FT 3 3 4

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5 The first, second and third terms of a geometric progression are 2p + 6, 5p and 8p + 2 respectively. (a) Find the possible values of the constant p. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) One of the values of p found in (a) is a negative fraction. Use this value of p to find the sum to infinity of this progression. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) 5 p 8 p + 2 M1 OE. Setting up a valid relationship in terms of p. = 2 p + 6 5 p 9 p 2 − 52 p − 12 [ = 0] DM1 OE. Simplifying to a 3 term quadratic equation, only condone sign errors.  leading to p = −92 and 6 A1 ( 9 p + 2 )( p − 6 ) = 0  3 5(b)  2   50  *M1 2 a = 2  −  + 6  =  FT their − , allow any negative non-integer.  9   9  9 10 50  1  *M1 2 r = −  =− Ft their − , allow any negative non-integer.   9 9  5  9 50  1  125 DM1 A1 Can only get DM1 if |r| < 1. S =   1 −−  = Accept AWRT 4.63 . 9  5  27 4

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6 A line has equation y = 6x −c and a curve has equation y = cx2 + 2x −3, where c is a constant. The line is a tangent to the curve at point P. Find the possible values of c and the corresponding coordinates of P. [7] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 6 cx 2 + 2 x −=3 6 x − c leading to cx 2 − 4 x + ( c − 3)  = 0 B1 3-term quadratic. 16 − 4 c ( c − 3 ) = 0 *M1 Apply b 2 − 4 ac = 0 (‘= 0’ may be implied in subsequent work). Their coefficients must be substituted correctly 4c 2 − 12c − 16  = 0 leading to (4 c − 4 )( c + 1)  = 0  leading to A1 Dependent on factorisation oe. c = 4 and − 1 When c = 4, 4 x 2 − 4 x + 1 = 0  ( 2 x − 1) 2 = 0  DM1 OE. Substituting their c = 4 into their quadratic   equation. 1 A1 Both required. x = , y = −1 2 When c = −1, x 2 + 4 x + 4  = 0  ( x + 2 ) 2 = 0  DM1 OE. Substituting their c = -1 into their quadratic   equation. x = −2, y = −11 A1 Both required. Alternative method for Question 6 dy B1 = 2cx + 2 dx 2cx + 2 = 6 M1 Equating their curve gradient and 6. 2 A1 SOI c = x 2 + 2 x −=3 6 x − c . Simplify 2 x 2 + 3x − 2  = 0 DM1 Substitute c = 2 into cx x to 3-term quadratic. 6 1 A1 Dependent on factorisation. Both required. ( 2 x − 1)( x + 2 ) = 0  → x = or −2 2 c = 4 and −1 A1 Both required, if DM0 given SC B1 for both. y = −1 and − 11 A1 Both required, if DM0 given SC B1 for both. SC one correct (x, y). A1 only 7

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10 A 2.8 rad B r O R r C The diagram shows points A, B and C lying on a circle with centre O and radius r. Angle AOB is 2.8 radians. The shaded region is bounded by two arcs. The upper arc is part of the circle with centre O and radius r. The lower arc is part of a circle with centre C and radius R. (a) State the size of angle ACO in radians. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find R in terms of r. 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(c) Find the area of the shaded region in terms of r. 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Mark scheme: 10(a) Angle ACO = 0.7 B1 Don’t allow AWRT 0.7 . 1 10(b)  R =  1.53 r B1 Allow AWRT 1.53r. 1 10(c) 1 2 2 B1 Sector OAB = r  2.8  = 1.4 r    2 1 2 *M1 Sector CAB = ( their R ) 2 their 0.7 2 1.638 r 2 A1 Allow AWRT 1.64 r 2 . 1 2 1 *M1 2  r sin (− 1.4 ) OR 2  r  theirR sin0.7 2 2 2  0.4927r 2 A1 Allow AWRT 0.98 r 2 to 0.99 r 2 . 2 2 2 DM1 1.4r − their 1.638r − their 0.985r ( ) 0.747r 2 to 0.748r 2 A1 7 10(c) General guidance for alternative methods Finding any useful sector area of the circle radius, r B1 May be ‘nested’ in a segment. Finding the area of sector CAB *M1A1 May be ‘nested’ in a segment. Finding the area of one useful triangle *M1 May be ‘nested’ in a segment. Finding the total area of useful triangles A1 May be ‘nested’ in a segment. A correct plan for the shaded area DM1 0.747r 2 to 0.748r 2 A1 7

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11 y P 2 y = x + 2 2x −1 Q R x O 1 2 2 The diagram shows part of the curve with equation y = x + . The lines x = 1 and x = 2 2x −1 2 intersect the curve at P and Q respectively and R is the stationary point on the curve. (a) Verify that the x-coordinate of R is 3 and find the y-coordinate of R. [4] 2 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the exact value of the area of the shaded region. [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 11(a) dy −3 B1B1 −8 = + 1 Expect + 1 . −2 2 ( 2x − 1)  2 3   dx ( 2 x − 1) 3 dy −8 DB1 AG. Substitute x = leading to = + 1 = 0 . 2 dx 8 dy Or correct solution of = 0 . 3 dx Hence x-coordinate of R is 2 3 2 3 B1 Answer only is acceptable. When x = , y = + = 2 2 4 2 4 11(b) 20 B1 Both required. y-coordinate of P = 3, y-coordinate of Q = 9   2 ( 2 x − 1) −1   1 2  B1 B1 Area below curve.   +  x  −1 2    2   1   1   5  1  M1 13 − + 2 −−+1 = −− Apply limits 1→2 to an integral. Expect .        6  3   2   3  2  1  20  47 M1 Area of trapezium, only allow errors in y-coordinate  3 +  = of Q. 2  9  18 47 13 4 A1 Shaded region. − = 18 6 9 6 Alternative method 1: Changes the award of the first M1 −7 M1 Must be some evidence of use of limits. Their equation of line PQ:[ y = x + 34] . Integrating between 1 and 2. 9 9 Alternative method 2: Changes the award of the first M1, a B1 and the second M1   M1 34 −16 34 2 For area under the line if their is seen integrated Combining line and curve:   x + −  dx  9 9 ( 2 x − 1) 2  correctly and limits used. Correct9 first and 3rd terms. −8 2 34 1 B1 B1 = x + x + 9 9 ( 2 x − 1) Use of limits on the whole integral M1

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