Cambridge A Level Mathematics 9709 — 2022 Oct/Nov Paper 2 · Variant 2

1 Solve the equation sec 1 = 5 cosec 1 for 0Å < 1 < 360Å. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 1 1 B1 Must be using sec 2 = 1 + tan 2  and Use sec= and cosec= or other appropriate identities cos sin cosec 2= 1 + cot 2 . Obtain tan= k using correct identities M1 OE For any non-zero constant k, if using other identities, must come from a 3-term quadratic equation. Obtain tan= 5 and hence 78.7 A1 AWRT Obtain 258.7 and no other solutions in the range A1 AWRT 4

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2 The solutions of the equation 4x −1 = x + 3 are x = p and x = q, where p < q. Find the exact values of p and q, and hence determine the exact value of p −2 − q −1 . [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Solve 4 x −=1 x + 3 to obtain x = 43 B1 Attempt solution of linear equation where signs of 4x and x are different M1 Obtain final value x = − 52 A1 Substitute numerical values and apply modulus signs correctly to obtain M1 Allow their p and q, p < q. − 125 − 13 or equivalent, retaining exactness and with no subsequent squaring Obtain 1531 A1 or exact equivalent. Alternative method for Question 2 State or imply non-modulus equation (4 x − 1) 2 = ( x + 3) 2 B1 Attempt solution of 3-term quadratic equation M1 Obtain final values − 52 and 43 A1 2 Substitute numerical values and apply modulus signs correctly to obtain M1 Allow their p  q . − 125 − 13 or equivalent, retaining exactness and with no subsequent squaring Obtain 1531 A1 or exact equivalent. 5

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6 y x O 4 6 The diagram shows the curves y = and y = 3e−x −3 for values of x between 0 and 4. The 3x + 2 shaded region is bounded by the two curves and the lines x = 0 and x = 4. Find the exact area of the shaded region, giving your answer in the form ln a + b + ced. 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Mark scheme: 6 6 *M1 for any constant 1k . Integrate to obtain form k1ln(3 x + 2) 3 x + 2 Obtain correct 2ln(3 x + 2) A1 Apply limits correctly DM1 Obtain 2ln14 − 2ln2 and hence ln49 A1 at this stage or later. Integrate 3e −−x 3 to obtain form k 2 e −+x k 3 x M1 for any non-zero constants 2k , 3k . Obtain correct −3e − x − 3 x A1 ( y1 − y2 )dx approach used. Apply limits to obtain −3e −4 − 12 + 3 A1 OE; implied if  Use correct procedure to find exact total area M1 Obtain ln49 + 9 + 3e−4 A1 9

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7 y x O P Q The diagram shows the curve with parametric equations x = 3 cos 21, y = 4 sin 1, for π ≤1 ≤32π. Points P and Q lie on the curve. The gradient of the curve at P is 2. The straight line 3x + y = 0 meets the curve at Q. (a) Find the value of 1 at P, giving your answer correct to 3 significant figures. 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(b) Find the gradient of the curve at Q, giving your answer correct to 3 significant figures. 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Mark scheme: 7(a) dx dy B1 State −6sin2 and 4cos d= d= dy dy dx M1 Use = / and equate to 2 dx d d Use sin2= 2sincos and attempt value of sin M1 Obtain sin= − 16 A1 7(a) Obtain = 3.31 only A1 AWRT; and no second answer. Alternative method for Question 7(a) 2 2 2 2 M1 Using x = 3 2cos − 1 , x = 3 1 − 2sin  or x = 3 cos − sin  to ( ) ( ) ( ) dx obtain a sincos d=  dy  4cos A1 Obtain  =  = 2  dx  −12sincos Attempt value of sin M1 Obtain sin= − 16 A1 Obtain = 3.31 only in the given range A1 5 7(b) State or imply 9cos2+ 4sin= 0 and use identity to obtain quadratic in M1 sin Obtain 18sin 2 − 4sin− 9 = 0 A1 OE Attempt solution to find negative value of sin DM1 Obtain sin= −0.604... A1 2 − 166 Or , = 3.79... 18 Substitute value of sin (or their  between  and 32 ) in expression for M1 first derivative Obtain 0.551 A1 AWRT 7(b) Alternative method for question 7(b) y 2 x M1 Must be a complete method, allow unsimplified. Cartesian equation of curve 1 − = oe 8 3 Intersection of line and curve 27 x 2 + 8 x − 24 = 0 oe M1 x = 0.8062... A1 = 3.791... A1 Substitute value of  (or their  between  and 32 ) in expression for first M1 derivative Obtain 0.551 A1 AWRT 6

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