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

Click here to start/reset.

Paper 1

Question 1

Hint 1: a stationary point is when the first derivative is equal to zero

Hint 2: differentiate y(x) to obtain y'(x)

Hint 3: set y'(x) equal to zero

Hint 4: factorise y'(x) to allow you to solve the cubic equation

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

Question 2

Hint 1: equal roots means that the discriminant of the quadratic equation is equal to zero

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

Hint 3: substitute these values into b²-4ac and make it equal to zero

Hint 4: solve the quadratic equation to obtain two values of k

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

Question 3

Hint 1: realise that we need the radius of circle C₁

Hint 2: either complete the squares or use the circle equation formulae to obtain the value of C₁'s radius

Hint 3: knowing the centre of C₂ and its radius, write the equation of C₁ in factorised form

Hint 4: remember to calculate the numerical value of r² and don't leave it as a number²

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

Question 4

4a) Hint 1: realise that we need to form two equations in m and c to allow us to solve for m and c

4a) Hint 2: note that u₁ = 6, u₂ = 9 and u₃ = 11

4a) Hint 3: note that u₂ = m × u₁ + c and u₃ = m × u₂ + c

4a) Hint 4: substitute u₁ and u₂ into one equation to obtain the first equation in m and c

4a) Hint 5: substitute u₂ and u₃ into the other equation to obtain the second equation in m and c

4a) Hint 6: solve the sumultaneous equations in m and c

4b) Hint 7: use these values of m and c in u₄ = m × u₃ + c

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

Question 5

5a) Hint 1: write down vector AB in component form

5a) Hint 2: write down vector AC in component form

5a) Hint 3: determine what vector AB must be multiplied by to obtain vector AC

5a) Hint 4: clearly state that as the vectors are scalar multiples of each other, then they are parallel plus they also have a common point A, then the points A, B and C are collinear

5b) Hint 5: draw a diagram of a straight line with A, B and C on it

5b) Hint 6: use the scalar multiple fraction from part (a) to determine the relative lengths of AB to AC

5b) Hint 7: determine what the relative length BC is, compared to AB

5b) Hint 8: the ratio AB:BC should now be able to be read off the diagram

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

Question 6

Hint 1: realise that y(x) is not ready to differentiate in its current form

Hint 2: re-write the fraction as a bracket with a negative power

Hint 3: differentiate y(x) using the chain rule

Hint 4: re-write your answer which has a term with a negative power, back into a fraction

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

Question 7

Hint 1: draw a sketch of a line that slopes upwards at an angle of 30° from the x-axis. Don't worry about where it cuts the x-axis

Hint 2: add to your sketch a perpendicular line that goes through (0,-4)

Hint 3: realise we need the gradient of the first line

Hint 4: use m = tan(θ) and your knowledge of trigonometric exact values

Hint 5: the perpendicular gradient is the negative reciprocal of the first gradient

Hint 6: write the equation of the perpendicular line, knowing its gradient and its y-intercept.

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

Question 8

8a) Hint 1: area between curves involves integrating the expression of (upper function - lower function) between their intersection points

8a) Hint 2: remember to use brackets to ensure that all of the 'lower function' is subtracted.

8b) Hint 3: you are recommended to fully simplify the integrand's expression to a quadratic expression, BEFORE integrating to find the area.

8b) Hint 4: use brackets when substituting the (-1) limit into the integrated expression

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

Question 9

9a) Hint 1: after obtaining __u__.__v__, factorise the resulting quadratic expression

9a) Hint 2: when factorising, first take out the common factor of 2, to make all the coefficients smaller

9a) Hint 3: vectors are perpendicular if their scalar product is zero

9a) Hint 4: set your factorised expression equal to zero and solve for two values of p

9b) Hint 5: vectors are parallel if one vector is a scalar multiple of the other

9b) Hint 6: set up an equation where __u__ = k __v__, where k is a scalar mutliple

9b) Hint 7: use the values of the y or z components to obtain a value for k

9b) Hint 8: use the value of k with the x components to create an equation in p, which can then be solved for p.

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

Question 10

10a) Hint 1: recognise that f(x) has been reflected in some fashion in the x-axis and then translated upwards by 3 units

10b) Hint 2: recognise that the point (2,-1) means that f(2) = -1

10b) Hint 3: note that the point (2, -1) transformed to give (2, 5)

10b) Hint 4: using y = k f(x) + a, replace y with 5, x with 2 and a with your answer from (a)

10b) Hint 5: note that f(2) is -1, so you can create an equation in k, which can be solved.

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

Question 11

Hint 1: recognise that you have a composite function being integrated

Hint 2: after integrating the cos(…) term and considering what the (3x - π/6) contributes to the process, check by differentiating back using the chain rule

Hint 3: take care to substitute the 0 into the expressions correctly

Hint 4: note that sin(- π/6) is the negative of sin(π/6), from the symmetry of the sine function.

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

Question 12

12a) Hint 1: write f(g(x)) as f(5-x) first

12a) Hint 2: then proceed to work out what f(5-x) is

12b) Hint 3: the question stated that f(x) needed x>0

12b) Hint 4: so f(5-x) needs (5-x)>0

12b) Hint 5: so if f(5-x) is undefined, then it's values of x will NOT satisfy the inequality you have.

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

Question 13

13a)i) Hint 1: use triangle ABC and cos(p) = AC/AB

13a)i) Hint 2: use pythagoras' theorem to work out length AC in triangle ABC

13a)ii) Hint 3: use triangle ADE and cos(q) = AD/AE

13b) Hint 4: use the compound angle formula for sin(p+q) written in terms of sin(p), cos(p), sin(q) and cos(q)

13b) Hint 5: use triangles ABC and ADE to obtain sin(p) and sin(q) in a similar manner to that in part(a)

13b) Hint 6: carefully substitute each trigonometric term for its fraction value

13b) Hint 7: consider presenting your final answer with a rational denominator

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

Question 14

14a) Hint 1: on the second term, use the log law of 2log(n) = log(n²)

14a) Hint 2: use the log law of log(a) + log(b) = log(ab)

14a) Hint 3: simplify the final log with your knowledge of how the base is connected to the value inside the log

14b) Hint 4: on the left hand side, use the log law of log(a) - log(b) = log(a/b)

14b) Hint 5: re-write the resulting log statement in exponential form, to obtain a linear equation in x, which can then be solved

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

Question 15

15a) Hint 1: recognise that one trig term is in terms of '2x' and one is in terms of 'x', and that these are not the same

15a) Hint 2: re-write sin(2x) as 2sin(x)cos(x)

15a) Hint 3: factorise the left-hand side by taking out a common factor of 2cos(x)

15a) Hint 4: from the factorised expression, write down the two trig equations needing to be satisfied

15a) Hint 5: where possible, solve each trig equation for x. If it can't be solved, then state why not.

15b) Hint 6: recognise that from part (a), all the 'x' terms have been replaced with '2x'

15b) Hint 7: realise that all of the solutions from part (a) for 'x' are therefore now the solutions for '2x'

15b) Hint 8: realise that the solutions from part (a) were only for between 0 and 360

15b) Hint 9: realise that we need to consider all solutions between 0 and 720 if we are now considering '2x' rather than 'x'

15b) Hint 10: if we know all the solutions for 2x, then the solutions for x can come from dividing them all by 2

Hint 11: and here is a video of the solution:

Question 16

16a) Hint 1: identify the coordinates of C, the centre of the circle

16a) Hint 2: write down vector CP

16a) Hint 3: calculate the magnitude of vector CP, remembering to use brackets around each component when they are squared

16b) Hint 4: recognise that if the magnitude of CP is greater than 5, then P lies outside the circle

16b) Hint 5: construct the quadratic inequality in variable k that needs solving

16b) Hint 6: rearrange the quadratic inequality so that it is either > 0, or < 0.

16b) Hint 7: factorise the quadratic expression in k

16b) Hint 8: solve the quadratic inequality, using a sketch of the quadratic expression in k to help

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

Question 17

17a) Hint 1: expand the brackets to obtain at least three trigonometric terms

17a) Hint 2: use the trig identity sin²x + cos²x = 1

17a) Hint 3: use the trig identity that 2sin(x)cos(x) = sin(2x)

17b) Hint 4: use your expression from part (a) to re-write the integrand as an expression without any powers

17b) Hint 5: integrate each term, noting that you have a compound trig function that would use the chain rule if it were differentiated

17b) Hint 6: remember the constant of integration, as this is an indefinite integral

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

Paper 2

Question 1

1a) Hint 1: calculate the coordinates of point D that's the midpoint of AC

1a) Hint 2: calculate the gradient of BD

1a) Hint 3: use the gradient of BD and the coordinates of either point D or point B to obtain the equation of BD

1b) Hint 4: calculate the gradient of BC

1b) Hint 5: the gradient of AE will be the negative reciprocal of the gradient of BC

1b) Hint 6: use the gradient of AE and the coordinates of point A to obtain the equation of AE

1c) Hint 7: solve the two equations from parts (a) and (b) using a simultaneous equations method

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

Question 2

Hint 1: recognise that the square root in the integrand has to be re-written as x to a power, before it can be integrated

Hint 2: integrate each term, taking care to check each term by differentiating it back to see if it gives the original expression

Hint 3: don't forget the constant of integration

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

Question 3

3a) Hint 1: note that vector BE is vector BA plus vector AE

3a) Hint 2: note that vector BA is the negative of vector AB

3b) Hint 3: note that vector EF is vector EB plus vector BF

3b) Hint 4: note that vector EB is the negative of vector BE, allowing you to use your answer from part (a)

3b) Hint 5: note that vector BF is three quarters of vector BC

3b) Hint 6: note that vector BF is the same as vector AD

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

Question 4

4a) Hint 1: note that a fall of 2.7% equates to a decimal multiplier of 0.973

4b)i) Hint 2: know that the multiplier 'a' needs to be between -1 and 1 for a limit to exist

4b)ii) Hint 3: work out the limit by whatever method, remembering to round your answer to the nearest hundred

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

Question 5

Hint 1: identify on the graph of y=g(x) where the gradient is positive, where it is negative and where it is zero

Hint 2: note that if g(x) is a cubic function, then g'(x) will be a quadratic function

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

Question 6

6a) Hint 1: expand k cos(x + α)

6a) Hint 2: compare that expression with the one given in the question

6a) Hint 3: identify what k cos(α) must be equal to, and what k sin(α) must be equal to

6a) Hint 4: use your standard method to obtain the values of k and α

6b) Hint 5: use your answer from (a) to re-write the given equation

6b) Hint 6: rearrange the trig equation into the form cos(x + α) = some number

6b) Hint 7: take the inverse cosine and list all possible values for (x + α) between 0 and 540

6b) Hint 8: note that we have to look beyond 360 as we shall subtract α from each of these values to obtain the values of x

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

Question 7

7a) Hint 1: use a standard method of either completing the square or expanding the p(x+q)²+r and comparing coefficients

7b) Hint 2: note that a strictly decreasing function has its gradient always less than 0, for all values of x

7b) Hint 3: differentiate f(x) and you should obtain the same expression as you had in part (a)

7b) Hint 4: use your answer from part (a) to write f'(x) in the factorised form

7b) Hint 5: recognise that a (….)² term will always give a value greater than or equal to zero

7b) Hint 6: recognise that a -(….)² term will always give a value less than or equal to zero

7b) Hint 7: recognise that a -(….)² term which then has a value subtracted from it will always give a value less than zero

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

Question 8

8a) Hint 1: use a standard method to work out the inverse of a given function

8b) Hint 2: note that f(1) = 9 and f(1000) = 18

8b) Hint 3: note that f(x) is a strictly increasing function for values from 1 to 1000

8b) Hint 4: so the domain of f(x) is 1 ≤ x ≤ 1000

8b) Hint 5: so the range of f(x) is 9 ≤ f(x) ≤ 18

8b) Hint 6: note that the range of f(x) will be the domain of the inverse of f(x)

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

Question 9

9a) Hint 1: the initial power will be when t = 0

9a) Hint 2: evaluate P(0)

9b) Hint 3: note that reducing by 15% equates to a decimal multiplier of 0.85

9b) Hint 4: set up an equation so that e^{-0.0079t} = 0.85

9b) Hint 5: take natural logarithms of both sides and solve for t

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

Question 10

10a) Hint 1: if (x+3) is a factor, then x = -3 will be a root

10a) Hint 2: substitute -3 into the polynomial

10a) Hint 3: be sure to clearly communicate in words the logic that underpins your conclusion that this shows that (x+3) is a factor

10b) Hint 4: use either synthetic division twice, or polynomial long division twice, to fully factorise the quartic into (linear)(linear)(quadratic)

10b) Hint 5: determine whether the quadratic expression can be factorised into two linear expressions, or not

10b) Hint 6: use the discriminant of the quadratic expression to show that it cannot be factorised further

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

Question 11

11a) Hint 1: construct an expression for the total surface area, A, in terms of both x and h

11a) Hint 2: construct an expression for the volume, V, in terms of both x and h

11a) Hint 3: replace V with 2000 and rearrange the volume equation to make h the subject

11a) Hint 4: substitute this expression for h into the expression for A, and simplify

11b) Hint 5: recognise that the minimum will come from differentiating A(x), finding a stationary point and checking that it's a minimum

11b) Hint 6: recognise that A(x) is not ready to differentiate in its current form

11b) Hint 7: rewrite the 4000/x term to involve x with a negative power

11b) Hint 8: differentiate A(x) and set A'(x) = 0 to then solve for x

11b) Hint 9: use either a nature table, or A''(x), to determine the type of stationary point

11b) Hint 10: work out A(x) for the value of x that will give the minimum.

Hint 11: and here is a video of the solution:

Question 12

Hint 1: start with the given equation for y

Hint 2: take the logs of both sides and use log laws to obtain an equation of the form log(y) = …. x + ….

Hint 3: from the graph note the y-intercept and work out the gradient of the straight line

Hint 4: equate the values of intercept and gradient with the two parts of your log(y) = … equation

Hint 5: solve the two log equations to obtain the values of a and b

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

Question 13

Hint 1: note that 'rate of change' means 'derivative'

Hint 2: recognise that you have been given f'(x) and we need to work out f(x) using integration

Hint 3: note that the constant of integration can be determined from using that f(7) = 0

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

Question 14

Hint 1: expand out __u__.(__u__+__v__)

Hint 2: know that __u__.__u__ is |__u__|²

Hint 3: know that __u__.__v__ = |__u__||__v__|cos(θ)

Hint 4: substitute the values for |__u__| and |__u__| into the equation

Hint 5: rearrange the equation to make cos(θ) the subject

Hint 6: take inverse cosine to obtain the acute angle θ

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

Question 15

15a) Hint 1: work out gradient PC

15a) Hint 2: work out gradient PT that's perpendicular to PC

15a) Hint 3: work out equation of PT using gradient PT and point P

15b)i) Hint 4: use your equation from part (a) to determine the y-intercept, at T

15b)ii) Hint 5: consider triangle CPT that has a right angle at P

15b)ii) Hint 6: know the fact of 'a right-angled triangle inside a semi-circle'

15b)ii) Hint 7: the circle that we want has CT as a diameter and P on its circumference

15b)ii) Hint 8: calculate the mid point of CT, that will be the centre of the circle

15b)ii) Hint 9: calculate the length of CT and half it, to give the radius of the circle

15b)ii) Hint 10: construct the equation of the circle, remembering to calculate the numerical value of r² and don't leave it as a number²

Hint 11: and here is a video of the solution:

Did this hint help?