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

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Hints offered by N Hopley, with video solutions by 'DLBmaths'.

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Paper 1

Question 1

Hint 1: know that equal roots means that the discriminant is equal to zero

Hint 2: identify the values of a, b and c from the equation

Hint 3: substitute values into the equation b² - 4ac = 0

Hint 4: simplify and solve for k

Hint 5: and here is a video of the solution:

Question 2

Hint 1: recognise that differentiation of f(x) will require to use the chain rule

Hint 2: after obtaining f'(x), substitute in the value of x = 1

Hint 3: and here is a video of the solution:

Question 3

Hint 1: know that the inverse function, f^-1(x), satisfies f(f^-1(x)) = x

Hint 2: use your composition of functions skills to write f(f^-1(x)) in terms of f^-1(x)

Hint 3: rearrange the resulting equation to make f^-1(x) the subject

Hint 4: and here is a video of the solution:

Question 4

Hint 1: use the points (-4, 2) and (2, -7) to calculate the gradient, m₁, of the line passing through them

Hint 2: rearrange the equation 3y = 2x + 9 to make y the subject

Hint 3: read off the gradient, m₂, of this line's equation

Hint 4: with the two gradients obtained, check to see if m₁ × m₂ = -1

Hint 5: clearly communicate your conclusion, giving your reason for it

Hint 6: and here is a video of the solution:

Question 5

5a) Hint 1: recognise that for each right angled triangle, you will need to calculate its missing length, using Pythagoras' Theorem

5a)i) Hint 2: know that sin(p) = 3/hypotenuse

5a)ii) Hint 3: know that cos(q) = adjacent/hypotenuse

5b) Hint 4: use trigonometric addition formulae to expand cos(p + q) in terms of sin(p), cos(p), sin(q) and cos(q)

5b) Hint 5: use your workings from part (a) to obtain values for cos(p) and sin(q)

5b) Hint 6: substitute in the values for all the trigonometric ratios, helped by them all having a denominator of √10

5b) Hint 7: carefully multiply and subtract the fractions to obtain a single fractional answer

5b) Hint 8: and here is a video of the solution:

Question 6

6a) Hint 1: know that f(g(x)) = f(x² - 2x), in this case

6a) Hint 2: after applying the function f, expand the brackets and simplify the expression

6b) Hint 3: know that g(f(x)) = g(2x + 5), in this case

6b) Hint 4: after applying the function g, expand the brackets and simplify the expression, taking care with the signs

6c) Hint 5: when substituting expressions for g(f(x)) and f(g(x)), remember to put the expression for f(g(x)) in brackets to ensure that the whole expression is subtracted, and not just the first term

6c) Hint 6: after simplifying the expression to a quadratic, factorise out the common factor of 2

6c) Hint 7: with the quadratic expression inside the brackets, complete the square to obtain an expression of the form (x + . . .)² - . . .

6c) Hint 8: multiply this expression by the 2, to obtain the required form of the answer

6c) Hint 9: and here is a video of the solution:

Question 7

Hint 1: recognise that integrating cos(. . .) will give sin(. . .)

Hint 2: be aware that the differentiation of sin(3x + π/4) by the chain rule will generate a multiplier of 3

Hint 3: introduce the fraction 1/3 to 'compensate' for the multiplier of 3

Hint 4: it's always a good idea to check your final answer by differentiating it, to see if you obtain the integrand in the question

Hint 5: and here is a video of the solution:

Question 8

Hint 1: know that m= tan(θ)

Hint 2: know that tan(2π/3) = tan(-π/3)

Hint 3: know that tan(-π/3) = -tan(π/3)

Hint 4: use exact value triangle knowledge to write tan(π/3) as a surd

Hint 5: from y = mx + c, you now know the value of m

Hint 6: the line goes through (4, 0), so replace the values of x and y with these, to determine the value of c

Hint 7: and here is a video of the solution:

Question 9

9a) Hint 1: to find the intersection of the functions, make them equal to each other. i.e. x³ - 7x² + 12x + 3 = x³ - x² - 6x + 3

9a) Hint 2: gather and simplify terms to obtain an expression that is then equal to zero

9a) Hint 3: your expression should contain terms in x² and x only

9a) Hint 4: factorise an x out of this expression

9a) Hint 5: solving the resulting equation will give two values for x. One will correspond with the intersection on the y-axis and the other for point A.

9b) Hint 6: the shaded area will involve integrating the difference in the two functions, between the limits of 0 and your answer from part (a)

9b) Hint 7: be careful about which function is to be subtracted from which. If it helps, do 'upper - lower'

Hint 8: and here is a video of the solution:

Question 10

Hint 1: the possible roots of the cubic are the factors of 4. ie ±1, ±2, ±4

Hint 2: if f(x) = 6x³ - 13x² + 4, then evaluate each of f(-1), f(1), f(-2), f(2), etc to find one where f(x) = 0

Hint 3: you should find that f(2) = 0, which means that (x - 2) is a factor

Hint 4: use synthetic devision or polynomial long division to divide f(x) by (x - 2)

Hint 5: be careful to insert a '0x' term into your calculations to account for the 'missing linear term'

Hint 6: present a fully factorised answer of the form (x - 2)( . . . )( . . .)

Hint 7: and here is a video of the solution:

Question 11

11a) Hint 1: notice that the coefficient of f(x) is a positive number, so the maximum of f(x) will correspond with the maximum of g(x)

11b)i) Hint 2: know that the maximum value of f(x) will remain unchanged if its graph is translated horizontally

11b)i) Hint 3: so the maximum value of f(x) = maximum value of f(x - 4)

11b)ii) Hint 4: we were told at the start of the question that the maximum of f(x) is when x = 6

11b)ii) Hint 5: hence f(x - 4) corresponds with f(6)

11b)ii) Hint 6: so x - 4 = 6, and now solve for x

Hint 7: and here is a video of the solution:

Question 12

12a)i) Hint 1: vector AB = vector OB - vector OA

12a)i) Hint 2: when calculating the magnitude of vector AB, be sure to use brackets carefully. i.e. use (-6)² and not -6²

12a)ii) Hint 3: the ratio will be AB:BC, and you can use the information from the question, and your answer from (a)(i) to replace these line segments with numbers

12b) Hint 4: sketch a diagram showing a line with A, B and C on it, and the ratio from part (a)(ii)

12b) Hint 5: realise that point C is reached from starting at point B and moving in the direction of vector AB, but not the full length of vector AB

12b) Hint 6: from part (a)(ii) the ratio was 3:2, so vector AB needs to be divided by 3, and then multiplied by 2 to obtain the vector BC

12b) Hint 7: after calculating vector OC, be sure to write the coordinates of C, as coordinates

12b) Hint 8: and here is a video of the solution:

Question 13

13a) Hint 1: realise that to obtain u₅, you first need to calculate u₆ from the value of u₇

13a) Hint 2: write out the recurrence relation replacing n with 6, and u₇ with 20

13a) Hint 3: solve for u₆

13a) Hint 4: write out the recurrence relation replacing n with 5, and u₆ with your last answer

13a) Hint 5: solve for u₅

13b) Hint 6: use a standard method to calculate the limit of the recurrence relation

13b) Hint 7: and here is a video of the solution:

Question 14

Hint 1: expand the scalar product to obtain u.u + u.v

Hint 2: know that u.u = |u|.|u|.cos(0)

Hint 3: know that u.v = |u|.|v|.cos(120)

Hint 4: replace the |u| and |v| with their values from the question

Hint 5: know that cos(120) = cos(60) and use exact value triangles to write this as a fraction

Hint 6: simplify all numerical calculations for the final answer

Hint 7: and here is a video of the solution:

Question 15

Hint 1: sketch out a copy of the diagram and add in all of the points' coordinates, as well as the dimensions of each circle

Hint 2: deduce the coordinates of point P by considering how far away from point A(2, 1) it is

Hint 3: use (x - a)² + (y - b)² = r², and evaluate r²

Hint 4: and here is a video of the solution:

Question 16

Hint 1: use rules of logarithms to combine the first two terms

Hint 2: use rules of logarithms to rewrite the last term so that the coefficient of 2 is a power inside the log term

Hint 3: use rules of logarithms to combine everything so far into a single log₂ term

Hint 4: simplify the arithmetic inside the log

Hint 5: notice that 8 = 2³

Hint 6: simplify the log expression to a single number

Hint 7: and here is a video of the solution:

Question 17

17a) Hint 1: know that the graph of an inverse function is the reflection of the original function in the line y = x

17a) Hint 2: on your sketch, mark the points (1, 3) and (2, 7) clearly

17b) Hint 3: know that where f^-1(x) cuts the y-axis is equivalent to where f(x) cuts the x-axis

17b) Hint 4: cutting the x-axis means f(x) = 0

17b) Hint 5: write down 0 = log₅(x - 2) + 1

17b) Hint 6: subtract 1 from both sides of the equation

17b) Hint 7: re-write the logarithmic equation as one involving exponentials

17b) Hint 8: solve for x, which will be a fractional answer

17b) Hint 9: write down the coordinates of this point, that is on the graph of y = f(x)

17b) Hint 10: write down the corresponding coordinates that will be on the graph of y = f^-1(x)

17b) Hint 11: and here is a video of the solution:

Paper 2

Question 1

Hint 1: know that y needs to be differentiated and evaluated at x = 3 to obtain the gradient of the tangent

Hint 2: evaluate y at x = 3 to obtain the y-coordinate of the point

Hint 3: use the coordinates and the gradient to obtain the equation of the tangent

Hint 4: and here is a video of the solution:

Question 2

Hint 1: recognise that the integrand needs to be re-written with a negative power before it can be integrated

Hint 2: know that the new power will be (-3/2) + 1

Hint 3: when differentiating the answer, know that the power of -1/2 will need to be 'compensated' by multiplying by -2

Hint 4: remember to include the constant of integration

Hint 5: and here is a video of the solution:

Question 3

Hint 1: know that differentiating h(t) will require the use of the chain rule

Hint 2: when evaluating h(10), remember that the angle is measured in radians, and not degrees

Hint 3: and here is a video of the solution:

Question 4

4a) Hint 1: calculate the gradient of AC, m_AC, using the coordinates of A and C

4a) Hint 2: use the coordinates of B and m_AC to obtain the equation of L₁

4b) Hint 3: notice that AB is horizontal, so a perpendicular line will be vertical, and therefore have an equation of the form x = . . .

4b) Hint 4: find the midpoint of AB, and use its x-coordinate to obtain the equation of L₂

4c) Hint 5: L₁ meets L₂ when the value of x from L₂ is substituted into the equation of L₁

4c) Hint 6: solve for y, and write the final answer as a set of coordinates

Hint 7: and here is a video of the solution:

Question 5

5a) Hint 1: use a standard method for the wave function to re-write the expression with a single trigonometric function

5b)i) Hint 2: know that the minimum value of f(t) will come from knowing the minimum value of sin(t + a)

5b)i) Hint 3: the minimum value of sin(t + a) is -1

5b)ii) Hint 4: know that the minimum value of sin(x) happens when x = 270°

5b)ii) Hint 5: equate t + a = 270, using the value for 'a' from part (a)

Hint 6: and here is a video of the solution:

Question 6

6a) Hint 1: know, and write down, that a stationary point is when f'(x) = 0

6a) Hint 2: differentiate f(x) and equate it to zero to solve for x

6b) Hint 3: the shaded area will be the integral of f(x) between the limits of the answer from part (a) and 9

6b) Hint 4: take great care when integrating the fractional index terms, as well as substituting in the values of the limits

Hint 5: and here is a video of the solution:

Question 7

Hint 1: on the sketch of y = f(x) that is given, annotate which sections of the graph have a positive gradient, a zero gradient and a negative gradient

Hint 2: the two stationary points with x coordinates of -1 and 3 will become zeros on the graph on y = f'(x)

Hint 3: notice that the gradient is positive to the left and the right of the x = 3 stationary point. This will give a minimum on the graph of y = f'(x)

Hint 4: be sure to include on your sketch of y = f'(x) the values of both -1 and 3

Hint 5: and here is a video of the solution:

Question 8

Hint 1: subtract 1 from both sides of the equation

Hint 2: divide both sides of the equation by 2

Hint 3: apply the inverse sine to both sides of the equation

Hint 4: know that there are several values of sin^-1(-½). You should list at least three of them, starting from -30°

Hint 5: these values represent the values of 3x - 60

Hint 6: add 60 to each value

Hint 7: then divide each answer by 3

Hint 8: check that these values of x are all in the domain of between 0 and 180.

Hint 9: you should have three values for which this is true, so discard any values that are larger than 180 or smaller than 0

Hint 10: and here is a video of the solution:

Question 9

9a) Hint 1: know that surface area of cylinder = 2 × circle + rectangle

9a) Hint 2: the rectangle has dimensions: radius × circumference

9a) Hint 3: know that the volume of a cyclinder = base area × height

9a) Hint 4: in the formula V = πr²h, replace the V with 450, and rearrange to make h the subject

9a) Hint 5: substitute your formula for h into the Surface Area expression and simplify to obtain the given expression for A(r)

9b) Hint 6: recognise that we need to calculate A'(r)

9b) Hint 7: rewrite the fractional term of A(r) with a negative index, making it ready to differentiate

9b) Hint 8: after differentiating A(r), re-write the negative index term back as a fraction

9b) Hint 9: write down that a stationary point is when A'(r) = 0

9b) Hint 10: make A'(r) equal to zero and solve for r, which will eventually involve taking a cube root

9b) Hint 11: check that A'(r) is a minimum for the value of r just worked out, by either using a nature table or the second derivative

9b) Hint 12: clearly communicate the value of r that gives the minimum surface area

9b) Hint 13: and here is a video of the solution:

Question 10

10a) Hint 1: start with the left hand side of the identity and try to make it look like the right hand side

10a) Hint 2: know that tan(x) = sin(x)/cos(x)

10a) Hint 3: recognise that 2sin(x)cos(x) is a double angle formula

10b) Hint 4: realise that you will need to integrate the given expression for dy/dx to obtain an expression for y

10b) Hint 5: notice that the given expression is just 3 times the expression given in part (a)

10b) Hint 6: proceed to integrate 3sin(2x)

10b) Hint 7: fix the value of the constant of integration by using the information that when x = 0, y = 3

10b) Hint 8: assemble the final expression for y(x)

10b) Hint 9: and here is a video of the solution:

Question 11

11a) Hint 1: know that vector AB = vector OB - vector OA, and vector AC = vector OC - vector OA

11b) Hint 2: after sketching a diagram of triangle ABC, realise that the scalar product will help calculate the required angle

11b) Hint 3: after working about the magnitudes of vectors AB and AC, proceed to use the scalar product to work out angle BAC

11b) Hint 4: and here is a video of the solution:

Question 12

Hint 1: recognise that u_k+1 - u_k = 1000

Hint 2: notice that u_k+1 = 9u_k - 440

Hint 3: substitute this expression for u_k+1 into the equation u_k+1 - u_k = 1000

Hint 4: solve this equation for u_k

Hint 5: and here is a video of the solution:

Question 13

13a) Hint 1: notice that vector CF = vector CB + vector BF

13a) Hint 2: realise that vector CB = - (vector BC)

13b) Hint 3: notice that vector DF = vector DC + vector CF

13b) Hint 4: realise that vector DC = ½ vector AB

13c) Hint 5: notice that vector QF = vector QD + vector DF

13c) Hint 6: in this equation, replace vectors QF and DF with the given information, and your answer from part (b)

13c) Hint 7: solve the equation for vector QD

Hint 8: and here is a video of the solution:

Question 14

Hint 1: from the given equation of a circle, we need to determine the coordinates of its centre

Hint 2: use a factorising method, or one of the circle formula, to obtain the centre's coordinates

Hint 3: using the coordinates of A and those of the circle's centre, work out the gradient of the line joining them

Hint 4: the tangent line will have a gradient that is the negative reciprocal of the gradient just calculated

Hint 5: use the point A(3, 5) and the gradient (which should be a positive fraction) to determine the equation of the tangent line

Hint 6: and here is a video of the solution:

Question 15

Hint 1: substitute the equation of the line into the equation of the circle, to obtain an equation that is only in terms of x

Hint 2: expand all brackets and carefully simplify terms to obtain a quadratic that is equal to zero

Hint 3: factorise out a 5 from all terms, then factorise the remaining quadratic

Hint 4: this will give two values of x, one for each of the points Q and P

Hint 5: use the equation of the line to work out the y-values for each of the x-values

Hint 6: clearly communicate the coordinates of each of points Q and P

Hint 7: and here is a video of the solution:

Question 16

Hint 1: calculate the gradient of the line through (0,2) and (6, 4)

Hint 2: this means that log₈y = mx + c, and replace m and c with the answer just calculated, and using the graph provided

Hint 3: re-write the logarithmic equation as one involving exponentials

Hint 4: know from laws of indices that 8^p+q = 8^p × 8^q

Hint 5: know from laws of indices that 8^pq = (8^p)^q

Hint 6: know that a fractional power means taking a root of some sort

Hint 7: compare the final expression for y with the one given in the question, of y = ab^x

Hint 8: clearly state the values of a and of b

Hint 9: and here is a video of the solution: