Mark Scheme Pure Mathematics Year 1 (AS) Unit Test 6 .

Mark schemePure Mathematics Year 1 (AS) Unit Test 6: DifferentiationQSchemeMarksAOsPearsonProgression Stepand Progressdescriptor1States or implies the formula for differentiation from firstprinciples.B12.15thf ( x) 5 xComplete a proofof a derivativefunction from firstprinciples.3f ( x) limh 0f ( x h) f ( x )hCorrectly applies the formula to the specific formula andexpands and simplifies the formula.M11.1bA11.1bA1*2.2a5 x h 5 x33f ( x) limh 0f ( x) lim h 5 x 3x 2 h 3 xh 2 h3 5 x33h 0h15 x h 15 xh 5h3f ( x) limh 0h22Factorises the ‘h’ out of the numerator and then dividesby h to simplify.f ( x) limh 0 h 15 x 2 15 xh 5h 2 hf ( x) lim 15 x 15 xh 5h 2h 02 States that as h 0, 15x2 15xh 5h2 15x2 o.e.so derivative 15x2 *(4 marks)NotesUse of δx also acceptable.Student must show a complete proof (without wrong working) to achieve all 4 marks.Not all steps need to be present, and additional steps are also acceptable. Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.1

Mark schemeMarksAOsPearsonProgression Stepand ProgressdescriptorAttempts to differentiate.M11.1a5thf ( x) 3x 2 8 x 35A11.1bStates or implies that f(x) is increasing when f ʹ(x) 0M11.2Attempts to find the points where the gradient is zero.M11.1bA12.2aQ2Pure Mathematics Year 1 (AS) Unit Test 6: DifferentiationScheme(3x 7)(x – 5) 0 (or attempts to solve quadratic inequality)x 73and x 5, so f(x) is increasing when73Use derivatives todeterminewhether afunction isincreasing ordecreasing in agiven interval.73{x : x } {x : x 5} (orx or x 5 )(5 marks)NotesAllow other method to find critical value (e.g. formula or calculator). This may be implied by correctanswers.Correct notation (“or” or “ ”) must be seen for final A mark. Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.2

Mark schemeQ3aPure Mathematics Year 1 (AS) Unit Test 6: DifferentiationSchemeMarksAOsPearsonProgression Stepand ProgressdescriptorAttempts to differentiate.M11.1a4thdy 3x 2 2 x 1dxA11.1bCarry outdifferentiation ofsimple functions.(2)3bSubstitutes into equation for C to find y-coordinate.M11.1bx 2, y 23 22 2 2 4Substitutes x 2 into f ʹ(x) to find gradient of tangent.M11.1bM11.1bA11.1bdy 3(4) 2(2) 1 7dxFinds equation of tangent using y y1 m( x x1 )with (2, 4)5thSolve coordinategeometryproblemsinvolvingtangents andnormals usingfirst orderderivatives.y 4 7( x 2)y 7 x 10 o.e.(4) Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.3

Mark scheme3cPure Mathematics Year 1 (AS) Unit Test 6: DifferentiationStates or implies gradient of tangent is 7, so gradient of1normal is 7M1Finds equation of normal using y y1 m( x x1 )with (2, 4)M11.1aSubstitutes y 0 and attempts to solve for x.M11.1bx 30, A(30,0)A11.1by 4 1.21 x 2 75thSolve coordinategeometryproblemsinvolvingtangents andnormals usingfirst orderderivatives.(4)(10 marks)Notes3bUsing y mx c is acceptable. For example 4 7 2 c, so c 103c30 1 Using y mx c is acceptable. For example 4 2 c , so c 7 7 Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.4

Mark schemePure Mathematics Year 1 (AS) Unit Test 6: DifferentiationMarksAOsPearsonProgression Stepand ProgressdescriptorAttempts to differentiate.M11.1a5thf ( x) 3x 2 14 x 24A11.1bStates or implies that the graph of the gradient function willcut the x-axis when f ʹ(x) 0M12.2aA11.1bM12.2aM12.2a73A11.1b7121into f ʹ(x) to obtain y 33A1ft1.1bA1ft2.2aQ4SchemeSketch graphs ofthe gradientfunction ofcurves.f ( x) 0 3x 2 14 x 24 0Factorises f ʹ(x) to obtain (3x 4)( x 6) 04x ,x 63States or implies that the graph of the gradient function willcut the y-axis at f ʹ(0).Substitutes x 0 into f ʹ(x)Gradient function will cut the y-axis at (0, 24).Attempts to find the turning point of f ʹ(x)by differentiating (i.e. finding f ʹʹ(x))f ( x) 0 6 x 14 0 x Substitutes x A parabola with correctorientation withrequired pointscorrectly labelled.(9 marks) Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.5

Mark schemePure Mathematics Year 1 (AS) Unit Test 6: DifferentiationNotesA mistake in the earlier part of the question should not count against the students for the last part. If a studentsketches a parabola with the correct orientation correctly labelled for their values, award the final mark.Note that a fully correct sketch without all the working but with all points clearly labelled implies 8 marks inthis question. Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.6

Mark schemePure Mathematics Year 1 (AS) Unit Test 6: DifferentiationQ5aSchemeMarksAOsPearsonProgression Stepand ProgressdescriptorStates or implies that area of base is x2.M13.36thStates or implies that total surface area of the fish tank isx 2 4 xh 1600M13.3M11.1bM11.1bA1*1.1bUse of a letter other than h is acceptable.h 400 x x4 400 x Substitutes for h in V x 2 h x 2 4 xSimplifies to obtain V 400 x x3*4Apply derivativesand the principleof rate of changeto real-lifecontexts.(5)5bDifferentiates f(x)B13.4dV3x 2 400 dx4Attempts to solve400 x dV 0dxM11.1bA11.1bA11.1b6thApply derivativesand the principleof rate of changeto real-lifecontexts.3x 23x 2 0 or 400 4440 3o.e. (NB must be positive)3Substitutes for x in V 400 x Vmax/ min x3432 000 3o.e. or awrt 61609(4) Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.7

Mark scheme5cPure Mathematics Year 1 (AS) Unit Test 6: DifferentiationDifferentiates f ʹ(x)M11.1bd 2V3x o.e.22dxSubstitutes x StatesA140 3into f ʹʹ(x)32.46thApply derivativesand the principleof rate of changeto real-lifecontexts.d 2V 0 , so V in part b is a maximum value.dx 2(2)(11 marks)Notes5aA sketch of a rectangular prism with a base of x by x and a height of h is acceptable for the first method mark.5cOther complete methods for demonstrating that V is a maximum are acceptable. For example a sketchdVof the graph of V against x or calculation of values of V oron either side.dx Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.8

Mark schemePure Mathematics Year 1 (AS) Unit Test 6: DifferentiationQSchemeMarksAOsPearsonProgression Stepand Progressdescriptor6aStates that the perimeter of the track is 2πr 2 x 300The choice of the variable x is not important, but there shouldbe a variable other than r.M13.36thCorrectly solves for x. Award method mark if this is seen in asubsequent step.A11.1bStates that the area of the shape is A πr 2 2rxB13.3Attempts to simplify this by substituting their expressionfor x.M11.1bA1*1.1bx 300 2πr 150 πr2Apply derivativesand the principleof rate of changeto real-lifecontexts.A πr 2 2r 150 πr A πr 2 300r 2πr 2States that the area is A 300r πr 2 *(5)6bAttempts to differentiate A with respect to rM11.1adAFinds 300 2πrdrA13.4Shows or implies that a maximum value will occur when300 2πr 0M11.1aA11.1bM11.1bA11.1bSolves the equation for r, stating r 150πAttempts to substitute for r in A 300r πr 2 , for example 150 150 writing A 300 π π π Solves for A, stating A 22 500π6thApply derivativesand the principleof rate of changeto real-lifecontexts.2(6) Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.9

Mark schemePure Mathematics Year 1 (AS) Unit Test 6: Differentiation(11 marks)Notes6bIgnore any attempts at deriving second derivative and related calculations. Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.10

Q Scheme Marks AOs Pearson Progression Step and Progress descriptor 6a States that the perimeter of the track is 2 rx The choice of the variable x is not important, but there should be a variable other than r. M1 3.3 6th Apply derivatives and the principle of rate of change to real-life .