Cambridge A Level Mathematics 9709 — 2013 May/June Paper 3 · Variant 3

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

Mark scheme: 1 EITHER: State or imply non-modular inequality (4x + 3)2 > x2, or corresponding equation or pair of equations 4x + 3 = ± x M1 Obtain a critical value, e.g. –1 A1 3 Obtain a second critical value, e.g. − A1 5 3 State final answer x < –1, x > − A1 5 OR: Obtain critical value x = –1, by solving a linear equation or inequality, or from a graphical method or by inspection B1 3 Obtain the critical value − similarly B2 5 3 State final answer x < –1, x > − B1 [4] 5 [Do not condone ≤or ≥ .]

More questions on Algebra

3 Solve the equation tan 2x 5 cot x, for x [5] = 0Å < < 180Å.

Mark scheme: 3 Use correct tan 2A formula and cot x = 1/tan x to form an equation in tan x M1 Obtain a correct horizontal equation in any form A1 Solve an equation in tan2x for x M1 Obtain answer, e.g. 40.2° A1 Obtain second answer, e.g. 139.8°, and no other in the given interval A1 [5] [Ignore answers outside the given interval.] [Treat answers in radians as a misread and deduct A1 from the marks for the angles.] [SR: For the answer x = 90° give B1 and A1 for one of the other angles.]

More questions on Trigonometry

8 The variables x and t satisfy the differential equation tdx k , −x3 dt = 2x2 for t 0, where k is a constant. When t 1, x 1 and when t 4, x 2. > = = = = (i) Solve the differential equation, finding the value of k and obtaining an expression for x in terms of t. [9] (ii) State what happens to the value of x as t becomes large. [1]

Mark scheme: 8 (i) Separate variables correctly and integrate at least one side M1 Obtain term ln t, or equivalent B1 Obtain term of the form a ln(k – x3) M1 2 Obtain term − ln(k − 3x ) , or equivalent A1 3 EITHER: Evaluate a constant or use limits t = 1, x =1 in a solution containing a ln t and b ln(k – x3) M1* 2 3 2 Obtain correct answer in any form e.g. ln t = − ln(k − x ) + ln(k − )1 A1 3 3 Use limits t = 4, x =2, and solve for k M1(dep*) Obtain k = 9 A1 OR: Using limits t = 1, x = 1 and t = 4, x = 2 in a solution containing a ln t and b ln (k – x3) obtain an equation in k M1* 2 2 Obtain a correct equation in any form, e.g. ln 4 = − ln( k − 8) + ln( k − )1 A1 3 3 Solve for k M1(dep*) Obtain k = 9 A1 3 1 − Substitute k = 9 and obtain x = (9 − 8t 2 ) 3 A1 [9] 1 (ii) State that x approaches 9 3 , or equivalent B1 [1]

More questions on Differential equations

9 y M x O 12 p The diagram shows the curve y sin22x cos x for 0 and its maximum point M. = ≤x ≤120, (i) Find the x-coordinate of M. [6] (ii) Using the substitution u sin x, find by integration the area of the shaded region bounded by the curve and the x-axis. = [4]

Mark scheme: 9 (i) Use product rule M1 Obtain correct derivative in any form, e.g. 4sin2x cos2x cos x – sin2 2x sin x A1 Equate derivative to zero and use a double angle formula M1* Reduce equation to one in a single trig function M1(dep*) Obtain a correct equation in any form, e.g. 10 cos3 x = 6 cos x, 4 = 6 tan2 x or 4 = 10 sin2 x A1 Solve and obtain x = 0.685 A1 [6] (ii) Using du = ± cos x dx, or equivalent, express integral in terms of u and du M1 Obtain ∫ 4u 2 1( − u 2 ) du , or equivalent A1 Use limits u = 0 and u = 1 in an integral of the form au3 + bu5 M1 8 Obtain answer (or 0.533) A1 [4] 15

More questions on Differentiation

10 The line l has equation r i j k ai 2j k , where a is a constant. The plane p has equation x 2y 6. Find the =value+ or+ values+ , of+a in+each of the following cases. + + 2Ï = (i) The line l is parallel to the plane p. [2] (ii) The line l intersects the line passing through the points with position vectors 3i 2j k and i j + + [4] + −k. (iii) The acute angle between the line l and the plane p is tan−1 2. [5]

Mark scheme: 10 (i) Equate scalar product of direction vector of l and p to zero M1 Solve for a and obtain a = –6 A1 [2] (ii) Express general point of l correctly in parametric form, e.g. 3i + 2j + k + µ (2i + j + 2k) or (1 – µ )(3i + 2j + k) + µ (i + j – k) B1 Equate at least two pairs of corresponding components of l and the second line and solve for λ or for µ M1 2 1 2 1 Obtain either λ = or µ = ; or λ = or µ = ; or reach λ ( a − 4) = 0 3 3 a − 1 a − 1 or 1( + µ )( a − 4) = 0 A1 Obtain a = 4 having ensured (if necessary) that all three component equations are satisfied A1 [4] GCE AS/A LEVEL – May/June 2013 9709 33 (iii) Using the correct process for the moduli, divide scalar product of direction vector if l and normal to p by the product of their moduli and equate to the sine of the given angle, or form an equivalent horizontal equation M1* Use 2 as sine of the angle A1 5 a + 6 2 State equation in any form, e.g. = A1 ( a 2 + 4 + )1 1( + 4 + 4) 5 Solve for a M1 (dep*) 60 Obtain answers for a = 0 and a = , or equivalent A1 [5] 31 [Allow use of the cosine of the angle to score M1M1.]

More questions on Vectors