Makes an attempt to begin solving the equation. For example, states that M1 Uses the identity to write, M1 States or implies use of the inverse tangent. For example, or M1 Shows understanding that there will be further solutions in the given range, by adding 180° to 30° at least once. (ignore any out of range values). M1 Subtracts 20 and divides each answer by 3. (ignore any out of range values). M1 States the correct final answers to 1 decimal place. cao A1 Total: 6 marks Uses appropriate law of logarithms to write M1 Inverse log11 (or 11 to the) both sides. M1 Derives a 3 term quadratic equation. M1 Correctly factorisesor uses appropriate technique to solve their quadratic. M1 Solves to find A1 Understands that stating that this solution would require taking the log of a negative number, which is not possible. B1 Total: 6 marks Equates the i components for the equation a + b = mc o.e. 2p + 6 = 4m B1 Equates the j components for the their equation a + b = mc −5 − 3p = −5m B1 Makes an attempt to find p by eliminating m in some way. For example,o.e. or o.e. M1 p = 5 A1 NOTES: Alternatively, M1: attempt to eliminate p first. A1: m = 4 and p = 5 (4 marks)
Using their value for p from above, makes a substitution into the vectors to form a + b 10i – 5j + 6i – 15j M1ft Correctly simplifies. 16i – 20j A1ft NOTES: OR M1ft: substitute their m = 4 into their a + b = mc. A1ft: correct simplification. (2 marks) Total: 6 marks Makes an attempt to subsitute 7 into the equation, for example, seen. M1 1644 or 1640 only (do not accept non-integeric final answer). A1 (2 marks) It is the initial bacteria population. B1 (1 mark) States that or that M1 Solves to find M1 24 (hours) cao (do not accept e.g. 24.0). A1
(3 marks) Total: 6 marks y = mx − 2 seen or implied. M1 Substitutes their y = mx − 2 into o.e. M1 Rearranges to a 3 term quadratic in x (condone one arithmetic error). M1 Uses , M1 Rearranges to or any multiple of this. A1 Attempts solution using valid method. For example, M1 or o.e. (NB decimals A0). A1 Total: 7 marks NOTES: Elimination of x follows the same scheme. leading t This leads to Use of gives which reduces to m cannot equal 0, so this must be discarded as a solution for the final A mark. could be used implicitly within the quadratic equation formula. Makes an attempt to find the vector. For example, writing or M1 Shows a fully simplified answer: A1 (2 marks) Correctly interprets the meaning of , by writing o.e. M1
States that 5376p3 = − 84 M1ft Correctly solves for p: p3 = − so p = − A1ft Correctly find the coefficient of the x term: 2304 (−) = −576 B1ft Correctly find the coefficient of the x2 term: 4608 (−)2 = 288 B1ft (4 marks) Total: 8 marks NOTES: ft marks – pursues a correct method and obtains a correct answer or answers from their 5376 from part a. States or implies that the angle at P is 74° B1 States or implies the use of the cosine rule. For example, M1
o.e. (NB must be positive) A1 Substitutes for x in o.e. or awrt 6160 A1 (4 marks) Differentiates f ʹ(x) o.e. M1 Substitutes into f ʹʹ(x) States , so V in part b is a maximum value. A1 (2 marks) Total: 11 marks NOTES: (a): A sketch of a rectangular prism with a base of x by x and a height of h is acceptable for the first method mark. (c): Other complete methods for demonstrating that V is a maximum are acceptable. For example a sketch of the graph of V against x or calculation of values of V or on either side. Attempts to take out x or –x. or M1
Figure 1 q(0) = −20, so y = q(x) intersects y-axis at (0, −20) and x-intercepts labelled (accept incorrect values from part a). B1ft y = p(x) intersects y-axis at (0, 3). B1 y = p(x) intersects x-axis at (6, 0). B1 Graphs drawn as shown with all axes intercepts labelled. The two graphs should clearly intersect at two points, one at a negative value of x and one at a positive value of x. These points of intersection do not need to be labelled. B1 (4 marks) Statement indicating that this is the point where p(x) = q(x) or seen. M1 Their equation factorised, or attempt to solve their equation by completing the square. 2x2 −19x – 46 = 0 (2x – 23)(x + 2) = 0 M1 A1 A1 (4 marks) x < – 2 or o.e. B1 NB: Must see “or” or (if missing SC1 for just the correct inequalities). B1