The Scottish Qualifications Authority owns the copyright to its exam papers and marking
instructions.

Hints offered by N Hopley, with video solutions by DLB Maths.

Click here to start/reset.

Paper 1

Question 1

Hint 1: know that the second term needs to be re-written without a fraction

Hint 2: differentiate each term by 'bringing down the power, and reducing the power
by 1'

Hint 3: carefully simplify the coefficients and watch out for negative values

Hint 4: consider re-writing the term with a fractional power back into a form
involving roots

Hint 5: consider re-writing the term with a negative power back into the form of a
fraction

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

Question 2

Hint 1: sketch a diagram, labelling and plotting the points P(-2, 6) and Q(10, 0) on
it

Hint 2: on the diagram draw in the segment joining P to Q, as well as the
perpendicular bisector of PQ

Hint 3: calculate the midpoint of PQ by taking the mean of their x & y coordinates

Hint 4: calculate the gradient of PQ

Hint 5: use the gradient of PQ to give the gradient that is perpendicular to PQ

Hint 6: use the coordinates of midpoint of PQ and the gradient just calculated,
to work out the equation of the line

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

Question 3

Hint 1: know that having a single log₅ term will be better than having two of
them

Hint 2: use laws of logarithms to combine the two log₅ terms on the left side

Hint 3: re-write the log₅(x/3) = 2 equation as an exponential equation

Hint 4: make x the subject and evaluate

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

Question 4

4a) Hint 1: draw a sketch of the diagram, marking in all the lengths and angles given

4a) Hint 2: notice the right angled triangle on the left with lengths 5 and 4 will
have the third side of length 3 (as they are a Pythagorean triple)

4a) Hint 3: use Pythagoras' theorem to work out the length of hypotenuse of the
right angled triangle on the right, using the lengths 3 and 6

4a) Hint 4: with your fully labelled diagram, you should be able to now read off
cos(p) and cos(q) from it

4b) Hint 5: expand out cos(p + q), referring to the formula sheet if required

4b) Hint 6: this expansion involves terms in sin(p) and sin(q), whose values can
also be read off your diagram

4b) Hint 7: substitute the values for cos(p), cos(q), sin(p) and sin(q) into
your expanded expression

4b) Hint 8: look to simplify where possible, so that both calculations have a
denominator of 5√5

4b) Hint 9: your simplified final answer should be negative and have a
numerator of 1

4b) Hint 10: and here is a video of the solution:

Question 5

Hint 1: recognise that the phrase 'equal roots' means that discriminants are likely to
be needed

Hint 2: know that 'equal roots' means that the discriminant will need to equal zero

Hint 3: identify the values of a, b and c from the quadratic equation. Some of the
values will involve the variable 'p'

Hint 4: substitute the values for a, b and c into b² - 4ac = 0

Hint 5: expand and simplify the expressions, and you should end up with a
quadratic equation in terms of p, that equals zero

Hint 6: factorise the quadratic expression

Hint 7: state the two values of p that solve the quadratic equation

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

Question 6

Hint 1: know that to integrate, the term 6√x needs to be re-written with a
fractional power

Hint 2: integrate each term....

Hint 3: ... and don't forget the constant of integration!

Hint 4: simplify the coefficients of each term

Hint 5: consider re-writing the term with a fractional power back into a form
involving roots

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

Question 7

7a) Hint 1: use the laws of logarithms to combine the two terms to give a single
log₂ term

7a) Hint 2: simplify the value inside the log₂ function, and write it as a
number to a negative power

7a) Hint 3: take this negative power 'outside the log'

7a) Hint 4: know what log₂(8) is equal to

7a) Hint 5: present your final, simplified answer.

7b) Hint 6: sketch the graph of y = log₈(x), noting the value where it cuts
the x-axis

7b) Hint 7: identify the part of the graph where the y-values are negative

7b) Hint 8: so the y = log₈(x) function has negative values when 0 < x <
1

7b) Hint 9: hence log₈(a) is negative when 0 < a < 1

7b) Hint 10: and here is a video of the solution:

Question 8

Hint 1: know that to find stationary points, you shall need to
differentiate f(x)

Hint 2: after differentiating f(x), you shall need to find when f'(x)
= 0

Hint 3: for each stationary point, you shall need to use either a
nature table, or f''(x), to determine their nature

Hint 4: so, differentiate f(x) and clearly state that f'(x) = 0

Hint 5: solve the quadratic equation that you get, using
factorising (by taking the 3 out first)

Hint 6: for each of the two values of x, -3 and 1, you will
either draw a nature table, or evaluate f''(x)

Hint 7: you will also need the y-coordinates of each point, by
substituting back into f(x)

Hint 8: clearly state each set of coordinates, using brackets,
along with their associated nature

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

Question 9

Hint 1: know that the graph of an inverse function can be obtained by
reflecting the given graph in the line y = x.

Hint 2: first, sketch the graph of y = f^{-1}(x) using the
reflection line y = x. Plot the new point that corresponds to (3, 1)

Hint 3: then consider how y = f^{-1}(x) - 1 will look
different to it

Hint 4: sketch the graph of y = f^{-1}(x) - 1, plotting
what has happened to the point (1, 3)

Hint 5: be sure to make your final graph look as through it does
not go any lower than the horizontal line y = -1

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

Question 10

10a) Hint 1: know that in order to show that (x + 5) is a factor, then
f(-5) should equal zero

10a) Hint 2: evaluate f(-5) taking care to put (-5) inside brackets
before evaluating each power in the polynomial expression

10a) Hint 3: you know it will have all gone correctly when you
obtain the final answer of zero!

10a) Hint 4: be sure to write the sentence 'as f(-5) = 0, then -5
is a root, which means that (x + 5) is a factor', or similar

10b) Hint 5: using what you have found from part (a), perform
polynomial long division or polynomial synthetic division to factorise out (x + 5)

10b) Hint 6: you should have found that f(x) = (x + 5)(x³ -
2x² + 3x - 6)

10b) Hint 7: you now need to factorise x³ - 2x² + 3x -
6

10b) Hint 8: consider the factors of -6 which are ±1,
±2, ±3, ±6

10b) Hint 9: as in part (a), evaluate the cubic expression
using each of these until you find one that gives a final value of zero

10b) Hint 10: use polynomial long, or synthetic, division to
factorise out (x - 2)

10b) Hint 11: recognise that the final bracket which you
will obtain of (x² + 3) cannot be factorised further

10b) Hint 12: pulling everything together, you should now
have 0 = (x + 5)(x - 2)(x² + 3), or similar

10b) Hint 13: read off the two solutions from this
factorised quartic expression

10b) Hint 14: and here is a video of the
solution:

Question 11

11a) Hint 1: integrate each term, being careful with signs

11a) Hint 2: substitute the values of π and π/2 into your
expression, being careful with signs and zeros

11b) Hint 3: know that the limits represent the furthest left and
furthest right boundaries of the area of interest

11b) Hint 4: recognise that the subtraction in the integrand in
part (a) indicates which is the 'top function' and which is the 'bottom function'

11b) Hint 5: and here is a video of the solution:

Question 12

Hint 1: recognise that this will require the method of 'completing the
square' (but other methods are possible)

Hint 2: factorise the -2 out of the first two terms

Hint 3: this gives -2[x² + 6x] + 7

Hint 4: look at the expression inside the [...] and put in extra
terms so that it can be factorised

Hint 5: this gives -2[x² + 6x + 3² - 3²] + 7

Hint 6: now factorise the first three terms in the [...]

Hint 7: this gives -2[(x + 3)² - 3²] + 7

Hint 8: now expand the -2 into the [...] but do not expand the
(x + 3)² term

Hint 9: this gives -2(x + 3)² + 2 × 3² + 7

Hint 10: simplify the final numerical terms

Hint 11: this gives -2(x + 3)² + 25

Hint 12: and here is a video of the solution:

Question 13

13a)i) Hint 1: know that f(g(π/6)) will require evaluating
g(π/6) first

13a)i) Hint 2: this leads to f(π/3) needing to be evaluated

13a)i) Hint 3: sketch an 'exact value triangle' to help you get the
exact value for sin(π/3)

13a)ii) Hint 4: repeat the same sequence as in (a)(i) but there
will be no numerical evaluation of the functions, so your answer will remain in terms of x

13b)i) Hint 5: note that they are asking for the value of sin(p),
and not actually the value of p itself

13b)i) Hint 6: using your answer from (a)(ii) you should have
2sin(p) = 1/3

13b)i) Hint 7: make sin(p) the subject of this equation

13b)ii) Hint 8: know that g(p) will just be '2p'

13b)ii) Hint 9: so f(g(p)) is just f(2p)

13b)ii) Hint 10: so f(g(p)) = 2sin(2p)

13b)ii) Hint 11: we only know the value of sin(p), but this
expression has a sin(2p) term in it

13b)ii) Hint 12: expand sin(2p) out, referring to a formula
on the formula sheet if required

13b)ii) Hint 13: recognise that we need to know cos(p) in
order to proceed

13b)ii) Hint 14: we already know that sin(p) = 1/6, from
(b)(i)

13b)ii) Hint 15: sketch a right angled triangle,
labelled with angle p, opposite length 1 and hypotenuse length 6

13b)ii) Hint 16: use Pythagoras' theorem to work out
the length of the adjacent side

13b)ii) Hint 17: read off the value of cos(p) from
your diagram

13b)ii) Hint 18: substitute values of sin(p) and
cos(p) into your expression for f(g(p)) and then simplify

13b)ii) Hint 19: and here is a video of the
solution:

Paper 2

Question 1

1a) Hint 1: know that we need a point and a gradient to work out the
equation of the altitude through P

1a) Hint 2: we have the point P(5, -1)

1a) Hint 3: the gradient will be the perpendicular gradient to QR

1a) Hint 4: so calculate the gradient of QR

1a) Hint 5: from that, deduce the gradient of the altitude, using
the negative reciprocal.

1a) Hint 6: now calculate the equation of the altitude by your
standard method

1b) Hint 7: know that you shall need to use m = tan(θ)

1b) Hint 8: calculate the gradient of PR

1b) Hint 9: substitute this value of m into your equation

1b) Hint 10: use the inverse tan function to obtain the
required angle, rounding your answer to an appropriate accuracy

1b) Hint 11: and here is a video of the solution:

Question 2

Hint 1: recognise that 'equation of tangent' will likely need to use
differentation

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

Hint 3: evaluate y(1) and y'(1)

Hint 4: you now have the x and y coordinates of the point, and the
gradient through the point

Hint 5: using your preferred standard method, calculate the
equation of the tangent line

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

Question 3

Hint 1: know that when you integrate cos(...) it gives sin(...)

Hint 2: ... so part of your answer will be sin(4x + π/3)

Hint 3: know that because of the chain rule for differentiation, you
will need to compensate for the '4' that will come outside of the function, were you to
differentiate it

Hint 4: ... so include a 1/4 multiplication term in the coefficient

Hint 5: don't forget the constant of integration!

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

Question 4

Hint 1: sketch the graph of the given function y = f(x), so that you
can build gradually towards the required graph

Hint 2: then sketch the graph of the function y = f(-x) so that you
can focus on what the negative sign does

Hint 3: then sketch the graph of the function y = 2f(-x) sp that you
can focus on what the multiplier of 2 does

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

Question 5

Hint 1: recognise that the phrase 'rate of change' means the same as
'work out the derivative'

Hint 2: recognise that you will need to use the chain rule to
differentiate the function

Hint 3: use the chain rule to obtain f'(x), watching carefully for
the negative sign

Hint 4: evaluate f'(4), again being careful with signs, brackets
and powers, to give you the final answer for the rate of change

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

Question 6

Hint 1: know that the definition of an inverse function,
f^{-1}(x), is that f( f^{-1}(x) ) = x

Hint 2: using knowledge of composition of functions, substitute
f^{-1}(x) into f(x) to give the left side of the above equation

Hint 3: this gives you 2/f^{-1}(x) + 3 = x

Hint 4: subtract 3 from both sides

Hint 5: multiply both sides of the equation by f^{-1}(x),
and put (x - 3) inside brackets, to keep them together

Hint 6: divide both sides by (x - 3) to make f^{-1}(x)
the subject

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

Question 7

Hint 1: notice that there is a sin(x) and a cos(2x) term

Hint 2: it is better to have all trigonometic terms in (x) and not
(2x)

Hint 3: there are three options to choose from when expanding
cos(2x)

Hint 4: because we have a sin(x) term already, pick the option for
cos(2x) the involves only sin(x) terms

Hint 5: after expanding cos(2x), simplifying and rearranging, you
should have a quadratic expression in sin(x) that equals zero.

Hint 6: factorise this equadratic to get something like
[...sin(x) ....][...sin(x) ....] = 0

Hint 7: each bracket could equal zero, so solve each of the two
linear trigonometric equations

Hint 8: use graphs of y = sin(x) to help identify the related
solutions to the ones that your calculator provides

Hint 9: list all four solutions for the values of x that are
between 0° and 360°, in increasing order.

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

Question 8

Hint 1: notice the x-values for the furthest left and further right
ends of the shaded area

Hint 2: identify which function is the 'top function' of the shaded
area

Hint 3: write down and simplify the expression for: top function -
(bottom function)

Hint 4: did you remember to put the brackets around (x - 5) so that
when you subtract the expression, you will have -x + 5 ?

Hint 5: write down what you have so far as an integral, with
limits, an integrand that's in brackets, and a 'dx' on the end

Hint 6: proceed to integrate term-by-term, and then carefully
substitute in the limits of 1 and -2

Hint 7: take your time simplifying all of the fractions, taking
lots of small steps

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

Question 9

9a) Hint 1: this is a standard 'wave function question' for which you
should employ your practiced technique

9a) Hint 2: you should ultimately obtain 7cos(x) - 3sin(x) =
√(58) sin(x + 113.2)

9b)i) Hint 3: know that the maximum value of the expression from (a)
is when sin(x + 113.2) takes on its maximum value of 1. This gives a maximum value of √58

9b)i) Hint 4: notice that the given expression is double that which
was in part (a)

9b)i) Hint 5: conclude that the given expression has a maximum
value that is double that of the function in part (a)

9b)ii) Hint 6: know that the sine function has a maximum when the
angle is 90°, or 450°, or 810°, etc

9b)ii) Hint 7: so sin(x + 113.2) will have its maximum value of
1 when x + 113.2 equals either 90, or 450, or 810

9b)ii) Hint 8: by subtracting 113.2 from the values of 90, 450
and 810, determine the angle that lies between 0° and 360°

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

Question 10

Hint 1: recognise that 'strictly decreasing' means that we shall be
looking at the values of the derivative

Hint 2: differentiate f(x) to get f'(x)

Hint 3: know that 'strictly decreasing' means that f'(x) < 0

Hint 4: set up the quadratic inequality of 6x² + 18x -
24 < 0

Hint 5: factorise the quadratric expression by
first factoring out the 6

Hint 6: sketch a graph of y = (x - 1)(x + 4) to
help you determine the domain of x values where (x - 1)(x + 4) is less than zero

Hint 7: state your final values of x as an
inequality statement: ... < x < ...

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

Question 11

11a) Hint 1: recognise that C₁ has
centre (4, -2) and radius √37

11a) Hint 2: for C₂ use completing the
square, or a standard formula, to determine its centre and radius

11a) Hint 3: sketch a neat diagram showing
the two centres of (4, -2) and (-1, 3)

11a) Hint 4: calculate the distance between
the centres using either Pythagoras' theorem, or the magnitude of the
vector that would join the two centres

11b) Hint 5: on your diagram, add in the
circles with the correct radii so that you can see how the circles
overlap, or not.

11b) Hint 6: notice that the distance
between the centres is less than the sum of the circles' radii

11b) Hint 7: for completeness, you ought
to also check that one circle is not entirely inside the other (which
would mean that they have no intersections)

11b) Hint 8: write a clear inequality
statement about d, r₁ and r₂ that aligns with your
observations

11b) Hint 9: and here is a video of
the solution:

Question 12

Hint 1: recognise that you will have to work
back from dy/dx to find out y, along with the value of its associated
constant of integration

Hint 2: integrate each term to give y = ....
+ c

Hint 3: use the fact that y(-1) = 3 to help
fix the value of the constant, c

Hint 4: write out the full function of y,
with the constant included

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

Question 13

13a) Hint 1: from the question, notice that t
= 30

13a) Hint 2: evaluate C_{30}, giving
your answer to an appropriate accuracy

13b) Hint 3: from the question, notice that
C_{t} = 0.66

13b) Hint 4: substitute this value into the
equation of the model, with the objective of making t the subject

13b) Hint 5: once you have 0.06 =
e^{-0.0053t}, consider how to manipulate it so that t is no
longer in the power

13b) Hint 6: rewrite the equation in
terms of natural logarithms

13b) Hint 7: after solving for t,
consider converting the large number of minutes into hours, minutes and
possibly even seconds!

13b) Hint 8: and here is a video of the
solution:

Question 14

14a)i) Hint 1: sketch a 3D model of the open
box, and write on the lengths 3x, 2x and h to it

14a)i) Hint 2: calculate the total area of
the 5 rectangles added together, in terms of x and h

14a)ii) Hint 3: know that the volume of a
cuboid = length × breadth × height

14a)ii) Hint 4: obtain an expression for V
in terms of x and h, using your diagram to help

14a)ii) Hint 5: rearrange the expression
from (a)(i) to make h the subject

14a)ii) Hint 6: substitute the expression
for h (in terms of x) into the the expression for V, thereby obtaining
an expression for V that is only in terms of x

14a)ii) Hint 7: manipulate this
expression to obtain the stated expression, showing all of your steps
clearly.

14b) Hint 8: recognise that a maximum
volume will be found by differentiating V(x) and then setting V'(x) = 0

14b) Hint 9: solve the resulting
quadratic equation to obtain x² = 400

14b) Hint 10: note that this equation
has two solutions - you should find both and then give a reason why one
of them is discarded

14b) Hint 11: for the remaining
solution, use V''(x) or a nature table to verify that it is the location
of a maximum turning point

14b) Hint 12: and here is a video
of the solution:

Question 15

Hint 1: as the centre of the circle lies on
the y-axis, it will have coordinates (0, y)

Hint 2: the centre will also lie on the line
through (2, 5) that is perpendicular to the given tangent

Hint 3: we need the gradient of the tangent,
in order to obtain the gradient of the line perpendicular to it

Hint 4: rearrange x + 3y = 17 to make y the
subject

Hint 5: read off the gradient of the
tangent

Hint 6: take the negative reciprocal to
obtain the perpendicular gradient

Hint 7: calculate the equation of the
line through (2, 5) with gradient of 3

Hint 8: this line will be found to have
a y intercept of -1

Hint 9: now, write down the
coordinates of the centre of the circle, using brackets around the
coordinates