Question Paper 1 Question Paper 2 Marking Instructions Paper 1 Marking Instructions Paper 2

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Paper 1

Question 1

Hint 1: know that to get the equation of a line, you need a point and a gradient

Hint 2: to find the gradient, know that m= tan(θ)

Hint 3: use exact value triangles to obtain the exact value of tan(30°)

Hint 4: notice that you have a point of intersection with the y-axis

Hint 5: assemble all the information into the form y = mx + c

Hint 6: consider rearranging the equation into the form ax + by + c = 0

Question 2

2a) Hint 1: know that u₂ = 1/5 × u₁ + 12

2a) Hint 2: substitute in the value for u₁ = 20

2b)i) Hint 3: know the fact about the multiplier in a recurrence relation for there to be a limit

2b)i) Hint 4: clearly communicate that the value of 1/5 meets this criteria

2b)ii) Hint 5: know that as n tends to infinity, then both u_n and u_n+1 will each tend to the limit, L

2b)ii) Hint 6: substitute L in for each of u_n and u_n+1 in the given recurrence relation equation

2b)ii) Hint 7: solve the equation for L

Question 3

Hint 1: recognise that you will have to use the chain rule for differentiation

Hint 2: differentiate the 'outside function' leaving the inside function unchanged ..

Hint 3: .. then multiply by the derivative of the inside function

Hint 4: consider tidying up the resulting expression by reordering the parts of the expression

Question 4

Hint 1: draw a diagram of a line with ends labelled with P and Q

Hint 2: mark point R on the line. You will need to decide whether R is closer to P, or closer to Q

Hint 3: R should be closer to P and see it as '2 units from P' and '3 units from Q'

Hint 4: hence R is two-fifths of the way from P to Q

Hint 5: calculate the vector PQ

Hint 6: know that vector OR = vector OP + vector PR

Hint 7: so that vector OR = vector OP + 2/5 × vector PQ

Hint 8: after obtaining vector OR, re-write the components horizontally to give the coordinates of point R

Question 5

Hint 1: know that one definition of an inverse function is that h(h^-1(x)) = x

Hint 2: use composition of functions to re-write the left side of this equation in terms of h^-1(x)

Hint 3: subtract 7 from both sides of the resulting equation

Hint 4: divide both sides by 2

Hint 5: make h^-1(x) the subject, by taking the cube root of both sides of the equation

Question 6

6a)i) Hint 1: copy out the diagram

6a)i) Hint 2: use Pythagoras' theorem to calculate the missing side, and write it on your diagram

6a)i) Hint 3: use right-angled trigonometry knowledge with your diagram to write down exact values of both sin(p) and cos(p)

6a)i) Hint 4: expand sin(2p), checking against the Formulae Sheet if required

6a)i) Hint 5: substitute in values for sin(p) and cos(p) into your expanded expression

6a)i) Hint 6: simplify the fraction multiplications

6a)ii) Hint 7: to expand cos(2p) you have three possible options to choose from ..

6a)ii) Hint 8: .. so decide which option you want to use, given the information you already know

6a)ii) Hint 9: repeat the same substitution process as already done for the expansion of sin(2p)

6b) Hint 10: notice that 4p = 2 × (2p)

6b) Hint 11: hence sin[4p] = sin[2(2p)]

6b) Hint 12: expand out the formula for sin(2x), but using x = (2p) instead

6b) Hint 13: use your answers from part (a) to substitute in values for each of sin(2p) and cos(2p)

6b) Hint 14: simplify the fraction multiplications

Question 7

Hint 1: recognise that because the line is a tangent to the circle, then the line touches the circumference of the circle

Hint 2: hence y = 2x must satisfy the equation of the circle

Hint 3: in the circle equation, replace all the 'y' terms with '2x'

Hint 4: simplify the resulting equation in x

Hint 5: solve the equation in x, preferably by factorising

Hint 6: think why you now have a repeated root, and why this makes sense

Hint 7: read off the x-coordinate of the point of contact

Hint 8: determine the corresponding y-coordinate

Hint 9: write down the coordinates of the point of contact, in the form (x, y)

Question 8

Hint 1: know that 'no real roots' means that the discriminant is strictly less than zero.

Hint 2: identify 'a', 'b' and 'c' from the given quadratic expression. Each will be either a single number, or an expression in terms of 'm'

Hint 3: substitute the three expressions into b² - 4ac < 0

Hint 4: expand and simplify the expression, that's now in terms of 'm'

Hint 5: factorise the expression in terms of 'm'

Hint 6: after obtaining the equation (m - 2)(m - 14) < 0, draw a quick sketch of the quadratic y=(m - 2)(m - 14)

Hint 7: know that we are interested in when this quadratic is below the horizontal graph axis

Hint 8: this happens when m is between 2 and 14

Hint 9: write this as a single inequality statement, in the form .. < m < ..

Question 9

Hint 1: know that you have to reorganise each log_a term so that they each have a coefficient of 1

Hint 2: know your laws of logarithms that allow you to combine log(m) + log(n)

Hint 3: know your laws of logarithms that allow you to re-write nlog(m) as log(mⁿ)

Hint 4: know your laws of logarithms that allow you to combine log(m) - log(n)

Hint 5: simplify the fraction inside the logarithm as far as possible, to give the value of 'k'

Question 10

10a) Hint 1: know that if (x - 1) is a factor, then x = 1 is a root

10a) Hint 2: know that if x = 1 is a root, then f(1) should be equal 0

10a) Hint 3: substitute x = 1 into the given polynomial to show that the expression simplifies to zero

10a) Hint 4: clearly communicate that as f(1) = 0, this means that x = 1 is a root

10a) Hint 5: clearly communicate that as x = 1 is a root, then (x - 1) is a factor

10b) Hint 6: use either polynomial long division, or synthetic division, to divide the polynomial by (x - 1)

10b) Hint 7: if successful, you ought to have (x - 1)(2x³ + 5x² + x - 2)

10b) Hint 8: look to factorise the cubic by trying some likely roots

10b) Hint 9: consider trying -1, +1, -2, +2

10b) Hint 10: once you have found a value that works, repeat the polynomial division technique of your choosing

10b) Hint 11: with the resulting quadratic expression, factorise it by your usual technique

10b) Hint 12: if successful, your final answer will be of the form (x - 1)(x + .. )(x + ..)(2x - ..)

Question 11

11a) Hint 1: start off by expanding k.cos(x - a)°

11a) Hint 2: use your standard technique to apply your wave function knowledge to obtain the expression requested

11a) Hint 3: if successful, your ought to have 2cos(x - 60)°

11b) Hint 4: know that the function has an amplitude of 2

11b) Hint 5: know that the function has a phase shift of 60°

11b) Hint 6: decide whether the cosine function shifts to the left, or to the right

11b) Hint 7: make sure to plot the points for where the maximum, the minimum and axes intercepts are, and label them with their coordinates

Question 12

Hint 1: know that 'rate of change' means to work out the derivative of f(x)

Hint 2: before differentiating, prepare f(x) to be ready to be differentiated

Hint 3: know that ∛x can be written as x^1/3

Hint 4: differentiate f(x) by 'bringing the power down and reducing the power by one'

Hint 5: re-write the expression back as a fraction, where x has a positive fractional value

Hint 6: re-write the denominator back in terms of a cube root

Hint 7: you now have an expression for f'(x), so we now use f'(a) = 1

Hint 8: setup an equation in 'a' that equals 1

Hint 9: rearrange equation to be a² = ..

Hint 10: square root both sides, to give a = ±√64

Hint 11: look back at the question to see the restriction on values of x

Hint 12: hence pick the value of 'a' that is > 0.

Question 13

13a) Hint 1: know that to get a perpendicular bisector, we need a midpoint and a gradient

13a) Hint 2: calculate the mid-point of points P and Q

13a) Hint 3: calculate the gradient of the line PQ

13a) Hint 4: obtain the perpendicular gradient by taking the negative reciprocal of m_PQ

13a) Hint 5: substitute this gradient and the coordinates of the midpoint into y = mx + c to determine the value of 'c'

13a) Hint 6: you should have the equation y = (1/4) x + (19/4)

13a) Hint 7: multiply the equation through by 4 to get 4y = x + 19, as this may be easier to use later

13b) Hint 8: know that the centre of a circle is where the perpendicular bisectors of the two chords intersect

13b) Hint 9: we have the equation of one of the perpendicular bisectors, so we need the other equation

13b) Hint 10: the second equation is going to be a vertical line through the midpoint of QR

13b) Hint 11: find the coordinates of where the two equations intersect, as that will be the centre of the circle

13b) Hint 12: know that the points P, Q and R will each lie on the circumference of the circle

13b) Hint 13: we therefore need the distance from the centre of the circle to any of these points

13b) Hint 14: draw a right angled triangle, labelling the lengths of the vertical and horizontal sides and then work out the length of the hypotenuse, as that will be the circle's radius

13b) Hint 15: write the equation of the circle in the form (x - ..)² + (y - ..)² = ..

Paper 2

Question 1

1a) Hint 1: know that a median requires a midpoint to be calculated

1a) Hint 2: calculate the midpoint of AC

1a) Hint 3: calculate the gradient of the line joining B to the midpoint of AC

1a) Hint 4: substitute this gradient and coordinates of either point B, or the midpoint of AC, into y = mx + c to determine the value of 'c'

1a) Hint 5: write down clearly the equation of the line that is the median through B

1b) Hint 6: calculate the gradient of line BC

1b) Hint 7: obtain the perpendicular gradient by taking the negative reciprocal of m_BC

1b) Hint 8: calculate the equation of the line through B that has the gradient you've just calculated

1c) Hint 9: write down the two equations from parts (a) and (b)

1c) Hint 10: solve these two equations, simultaneously

1c) Hint 11: after obtaining the value of x, or y, then substitute it back to obtain the other variable's value

1c) Hint 12: clearly write down the coordinates of the intersection

Question 2

Hint 1: know that to get the equation of the tangent, we need to differentiate y

Hint 2: prepare y for differentiating, by writing x with a negative power

Hint 3: differentiate y by 'bringing down the power and reducing the power by one'

Hint 4: re-write f'(x) as a fraction, with x in the denominator with a positive power

Hint 5: evaluate f'(2) to give the gradient at the point when x = 2

Hint 6: evaluate f(2) to give the y-coordinate of the point where x = 2

Hint 7: substitute the coordinates (2, 1) and the gradient -3/2 to into y = mx + c to calculate the value of 'c'

Hint 8: clearly write down the equation of the tangent line

Question 3

3a) Hint 1: calculate vector ED = vector OD - vector OE

3a) Hint 2: calculate vector EF = vector OF - vector OE

3b)i) Hint 3: calculate the scalar product by multiplying corresponding components together, and then adding them up

3b)ii) Hint 4: know that ED.EF= |ED|.|EF|.cos(θ)

3b)ii) Hint 5: calculate the magnitudes of vector ED and vector EF

3b)ii) Hint 6: substitute all these values into the previous equation

3b)ii) Hint 7: solve for cos(θ)

3b)ii) Hint 8: solve for θ by using inverse cosine

Question 4

4a) Hint 1: think about what each part of y = f(x - 4) + 2 will do to the graph of y = f(x)

4a) Hint 2: know that (x - 4) will mean the graph is translated to the right by 4 units

4a) Hint 3: know that +2 will mean the graph is translated up by 2 units

4a) Hint 4: note that the original maximum coordinates are (-1, 3), so move this point to the right and then up

4b) Hint 5: know that the sketch of y = f'(x) will show the gradient of y = f(x) for each value of x

4b) Hint 6: determine which parts of y = f(x) have a positive gradient, and which have a negative gradient, and where the gradient is equal to zero

4b) Hint 7: know that the maximum turning point will become an x-axis intercept on the graph of y = f'(x)

4b) Hint 8: know that the inflexion at x = 2 will become an x-axis intercept on the graph of y = f'(x)

4b) Hint 9: sketch the graph of y = f'(x) going through (-1, 0) and (2, 0), with curves above and below the x-axis that correspond to where the gradient of y = f(x) is positive and negative

Question 5

Hint 1: know that when sin(x) is integrated, it gives -cos(x)

Hint 2: think how integrating sin(5x) is different, by considering the 'reverse' of the chain rule

Hint 3: with the integration complete, and you've checked it by differentiating it back, substitute in each of the limits

Hint 4: know that cos(0) = 1

Hint 5: when evaluating cos(5π/7) make sure that your calculator is in radian mode

Question 6

Hint 1: know that we shall try to re-write y = a.x^b by taking logarithms to base 5 of both sides

Hint 2: know your rules of logarithms log(p.q) = log(p) + log(q) and log(pⁿ) = n.log(p)

Hint 3: taking one small step at a time, take log₅ of both sides of the equation

Hint 4: you should eventually get log₅y = log₅a + b × log₅x

Hint 5: know that log₅a represents the y-intercept of the given graph

Hint 6: setup an equation in log₅a, and then solve for 'a' by re-writing it in exponential form

Hint 7: know that b represents the gradient of the given graph

Hint 8: clearly state the final values of 'a' and 'b', and consider substituting them into the original equation y = a.x^b

Question 7

Hint 1: identify the 'top' function, and the 'bottom' function

Hint 2: construct the distance between them by using y = 'top' - 'bottom'

Hint 3: remember to put brackets around the (x³ -6x² + 11x) expression, so that the negative outside the brackets has the required effect

Hint 4: simplify the resulting cubic expression

Hint 5: note the lower limit and upper limit from the diagram, as being the furthest left and furthest right x-coordinates for the shaded area

Hint 6: write down the integral of your expression between the two limits, and don't forget the 'dx' on the end

Hint 7: integrate the expression term by term and substitute in the limits

Hint 8: carefully do small steps of arithmetic to obtain a final exact value as a fraction

Hint 9: as it is an area, put 'unit²' after your final answer

Question 8

8a) Hint 1: know that f(g(x)) is the same as f(x + 1)

8a) Hint 2: write out f(x), but turn each 'x' into an 'x + 1'

8a) Hint 3: you can leave the answer factorised - there's no need to expand and simplify

8b) Hint 4: know that for a function to be undefined, this typically happens when you happen to divide by zero

8b) Hint 5: realise that we are interested in when f(g(x)) is equal to zero

8b) Hint 6: set up an equation with your factorised answer from part (a) being equal to zero

8b) Hint 7: add the constant to both sides and divide by the multiplier to obtain (x + 1)² = 9

8b) Hint 8: taking the square root of both sides introduces two possible values, so include a ±

8b) Hint 9: with x + 1 = ±3, subtract 1 from both sides and simplify to give two integer values of x

Question 9

9a) Hint 1: know that stationary points come from when a function has a gradient of zero

9a) Hint 2: know that to find when the gradient is zero, we find when y'(x) = 0

9a) Hint 3: differentiate y(x) term by term to give y'(x), and then facorise your expression for y'(x)

9a) Hint 4: explicitly write down that you are solving y'(x) = 0, and then solve the factorised quadratic equation

9a) Hint 5: for each value of x, substitute it into y(x) to obtain the corresponding y-coordinates

9a) Hint 6: clearly write down the two pairs of coordinates which are where the stationary points are

9b) Hint 7: you are strongly recommended to sketch a diagram that shows an x-axis running from -1 to 6

9b) Hint 8: plot the two points from part (a) relative to this x-axis

9b) Hint 9: notice that one point is already at the left end of the interval

9b) Hint 10: know that we need to find the coordinates of the point at the other end of the interval, where x = 6

9b) Hint 11: evaluate y(6)

9b) Hint 12: plot the point (6, 19) on your diagram

9b) Hint 13: now look across your diagram to see the lowest y-value and the highest y-value, from the points you plotted

9b) Hint 14: present your final answer in the form .. ≤ y ≤ ..

9b) Hint 15: also, clearly state the least value, and the greatest value, to guarantee that you are seen to be answering the question

Question 10

10a) Hint 1: for the given equation of C₁, either complete the square twice, or use the formula sheet, to obtain the required centre coordinates, and radius

10a) Hint 2: clearly state the centre coordinates, and give the radius in exact value form

10b) Hint 3: sketch a large copy of the diagram from the question, and add in the information from part (a)

10b) Hint 4: realise that we are going to need the radius of C₂

10b) Hint 5: realise that knowing the distance between the centres of C₁ and C₂ is likely to help

10b) Hint 6: consider the distance from the centre of C₁ to the point where the two circles touch internally

10b) Hint 7: this distance will be equal to the distance between the two circles' centres plus the radius of C₂

10b) Hint 8: set up an equation for r₂ and solve it, simplifying the final exact value in terms of √10

10b) Hint 9: write down the general expression for C₂ which is (x - a)² + (y - b)² = r²

10b) Hint 10: replace a, b and r with the values that you have for C₂

10b) Hint 11: remember to calculate the value of r² and not leave it as something squared.

Question 11

11a) Hint 1: think what value of 't' corresponds to 'the end of the year 2020'

11a) Hint 2: substitute this value of t into N(t)

11a) Hint 3: know that e⁰ equals 1

11a) Hint 4: state the final answer along with the units 'million vehicles'

11b) Hint 5: think what value of 't' corresponds to 'the end of the year 2030'

11b) Hint 6: read from the question the value that N has, at this time

11b) Hint 7: substitute 125 for N, and 10 for t, into the given formula, and set about solving for 'k'

11b) Hint 8: after some rearranging, you shall have to use natural logarithms to first find out 10k, and then solve for k.

Question 12

Hint 1: notice that we have a sin(2x) term and a sin(x) term, so we need to do something with the sin(2x) term

Hint 2: there is only one way to expand sin(2x), so re-rewrite the equation using that expansion

Hint 3: factorise sin(x) out of both terms

Hint 4: so you now have two possible solutions: either sin(x) = 0, or 4cos(x) - sin(x) = 0

Hint 5: for the first equation, sketch a graph of sin(x) and notice when the graph cuts the x-axis for 0 ≤ x < 360

Hint 6: for the second equation, rearrange it, then divide by cos(x) to create a tan(x) term

Hint 7: use inverse tangent to obtain one value of x

Hint 8: sketch a graph of tan(x) and determine a second value of x that also gives tan(x) = 4

Hint 9: assemble all four values of x that have been found in ascending order and present them as your final answer

Question 13

Hint 1: notice that there are single roots at x = -1 and x = 5

Hint 2: notice that there is a double root at x = 3

Hint 3: know that double roots come from factors that look like (x + p)²

Hint 4: remember that if a root is positive value, then the factor has a subtraction term

Hint 5: substitute in the values for a, b and c into f(x), using the above information

Hint 6: notice that f(0) = -9

Hint 7: replace f(x) with 9 and substitute 0 in for 'x' in the expression, to give an equation in terms of k

Hint 8: solve for k

Hint 9: write the final answer as f(x) = ..