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