Cambridge A Level Mathematics 9709 — 2018 Oct/Nov Paper 3 · Variant 2

1 Solve the inequality 3 2x x 4 . [4] −1 > + ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 State or imply non-modular inequality ( ) ( ) 2 2 2 3 2 1 4 − > + x x , or corresponding quadratic equation, or pair of linear equations/inequalities ( ) ( ) 3 2 1 4 − =± + x x 2 35 44 7 0 − − = x x Make reasonable attempt at solving a 3-term quadratic, or solve two linear equations for x M1 Allow for reasonable attempt at factorising e.g. (5x – 7)(7x + 1) Obtain critical values x = 7 5 and x = 1 7 − A1 Accept 1.4 and -0.143 or better for penultimate A mark State final answer 7 5 > x , 1 7 <− x A1 ‘and’ is A0, 7 1 5 7 < < − x is A0. Must be exact values. Must be strict inequalities in final answer Alternative Obtain critical value x = 7 5 from a graphical method B1 or by inspection, or by solving a linear equation or an inequality Obtain critical value 1 7 = − x similarly B2 State final answer 7 5 > x or 1 7 < − x or equivalent B1 [Do not condone ⩾ for >, or ⩽ for <.] 4

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4 Showing all necessary working, solve the equation ex e−x 4, + ex 1 = + giving your answer correct to 3 decimal places. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 Substitute and obtain 3-term quadratic 2 3 4 1 0 + −= u u , or equivalent 2 3 e 4e 1 0 + −= x x Solve a 3 term quadratic for u M1 Must be an equation with real roots Obtain root ( ) 7 2 / 3 − , or decimal in [0.21, 0.22] A1 Or equivalent. Ignore second root (even if incorrect) Use correct method for finding x from a positive value of ex M1 Must see some indication of method: use of ln = x u Obtain answer x = – 1.536 only A1 CAO. Must be 3 dp 5

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3 cos x 7 A curve has equation y for 2 sin x, = −120 ≤x ≤120. + (i) Find the exact coordinates of the stationary point of the curve. 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Find the value of a, giving your answer correct 2 sin x = Ô0 + to 3 significant figures. 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Mark scheme: 7(i) Use correct quotient or product rule M1 Obtain correct derivative in any form A1 ( ) ( ) 2 d 3sin 2 sin 3cos cos d 2 sin − + − = + y x x x x x x Condone invisible brackets if recovery implied later. Equate numerator to zero M1 Use 2 2 cos sin 1 + = x x and solve for sin x M1 6sin 3 0 − − = x ⇒ sinx =…. Obtain coordinates / 6 π = − x and 3 = y ISW A1 + A1 From correct working. No others in range SR: A candidate who only states the numerator of the derivative, but justifies this, can have full marks. Otherwise they score M0A0M1M1A0A0 6 7(ii) State indefinite integral of the form k ln (2 + sin x) M1* Substitute limits correctly, equate result to 1 and obtain 3 ln (2 + sin a) – 3 ln 2 = 1 A1 or equivalent Use correct method to solve for a M1(dep*) Allow for a correct method to solve an incorrect equation, so long as that equation has a solution. 1 3 1 2 1 sin e + = a ( ) 1 3 1 sin 2 e 1 − ⇒ = −     a Can be implied by 52.3° Obtain answer a = 0.913 or better A1 Ignore additional solutions. Must be in radians. 4

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10 The line l has equation r 5i i k . The plane p has equation = −3j −k + , −2j + r . 3i j k 0. −i −2j + + = The line l intersects the plane p at the point A. (i) Find the position vector of A. 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(ii) Calculate the acute angle between l and p. 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[Question 10(iii) is printed on the next page.] (iii) Find the equation of the line which lies in p and intersects l at right angles. 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Mark scheme: 10(i) Substitute for r and expand the scalar product to obtain an equation in λ 3 5 3 2 1 5 λ λ λ + + −− + −+ = 2 5 11 λ = − or ( ) ( ) ( ) 3 4 1 5 2 1 0 λ λ λ + + −− + −+ = Must attempt to deal with i + 2j Solve a linear equation for λ M1(dep*) Obtain λ = – 3 and position vector rA= 2i +3j – 4k for A A1 Accept coordinates 3 10(ii) State or imply a normal vector of p is 3i + j + k, or equivalent B1 Use correct method to evaluate a scalar product of relevant vectors e.g. (i – 2j + k).(3i + j + k) M1 Using the correct process for calculating the moduli, divide the scalar product by the product of the moduli and evaluate the inverse sine or cosine of the result M1 2 cos 6 11 θ = Second M1 available if working with the wrong vectors Obtain answer 14.3° or 0.249 radians A1 Or better Question Answer Marks Guidance 10(ii) Alternative 1 Use of a point on l and Cartesian equation 3 5 + + = x y z to find distance of point from plane e.g. ( ) 5, 3, 1 − − B 3 5 3 1 5 9 1 1 × −−− + + = d M1 ( ) 6 11 1.809... = = A1 Complete method to find angle e.g. sinθ = d AB M1 1 6 sin 0.249 11 54 θ − = =       A1 Or better Alternative 2 State or imply a normal vector of p is 3i + j + k, or equivalent B1 Use correct method to evaluate a vector product of relevant vectors e.g. (i – 2j + k)x(3i + j + k) M1 3 2 7 − + i j k Using the correct process for calculating the moduli, divide the vector product by the product of the moduli and evaluate the inverse sine or cosine of the result M1 2 2 2 3 2 7 sin 11 6 θ + + = . Second M1 available if working with the wrong vectors Obtain answer 14.3° or 0.249 radians A1 Or better 4 Question Answer Marks Guidance 10(iii) Taking the direction vector of the line to be ai + bj + ck , state a relevant equation in a, b, c, e.g. 3a + b + c = 0 B1 State a second relevant equation, e.g. a – 2b + c = 0, and solve for one ratio, e.g. a : b M1 Obtain a : b : c = 3 : – 2 : – 7, or equivalent A1 State answer r = 2i + 3j – 4k + µ (3i – 2j – 7k) A1ft Or equivalent. The f.t. is on rA Requires ‘r = ….’ Alternative Attempt to calculate the vector product of relevant vectors, e.g. (3i + j + k) × (i – 2j + k) M1 Obtain two correct components of the product A1 Obtain correct product, e.g. 3i – 2j – 7k A1 State answer r = 2i + 3j – 4k + µ ( 3i – 2j – 7k) A1ft Or equivalent. The f.t. is on rA. Requires “ r = ….” 4

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