Click HERE for the 2015 'Revised' Higher Maths Examination

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

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

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Paper 1

Question 1

Hint 1: perpendicular vectors are at right angles to each other

Hint 2: if the scalar product of two vectors is equal to zero, then they are perpendicular

Hint 3: set up __u__.__v__ = 0, and create an equation in terms of t

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

Question 2

Hint 1: the equation of a tangent needs a gradient and a point to go through

Hint 2: the gradient is obtained from differentiating the function, and evaluating it at x = -2

Hint 3: the y coordinate of the point is obtained from evaluating the original function when x = -2

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

Question 3

Hint 1: know that if (x + 3) is a factor, then x = -3 is a root

Hint 2: know that if x = -3 is a root, then evaluating the polynomial when x = -3 should give the answer of zero

Hint 3: make sure to clearly state that the value of zero means that (x + 3) is a factor

Hint 4: use either polynomial long division, or synthetic division, to divide the given polynomial by (x + 3)

Hint 5: remember to factorise the resulting quadratic, to give a final factorised form of three linear factors

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

Question 4

Hint 1: look at the highest and lowest values of the function on the graph. The difference between these is double the amplitude

Hint 2: notice that the 'centre line' of the function will be half way between the highest and lowest values of the function

Hint 3: look at the π/2 and notice that it is the period of the function. So, how many cycles would there be in 2π?

Hint 4: know that p is the amplitude, q is the number of cycles in 2π and r the vertical translation

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

Question 5

5a) Hint 1: use a standard method to work out an inverse function

5b) Hint 2: know that g(g^{-1}(x)) is the function of its inverse and by
definition this should return the original value

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

Question 6

Hint 1: aim to combine all the terms into a single log expression

Hint 2: use the following log rule on the fraction one third: n log_{a}p
= log_{a}p^{n}

Hint 3: know that the power of one third is the same as ∛

Hint 4: use the log rule: log(x) + log(y) = log (xy), so long as the logs have the same base (which they do, here)

Hint 5: keep simplifying the expression to give a single integer as a final answer.

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

Question 7

Hint 1: recognise that the function is not ready to differentiate in its current form

Hint 2: multiply out the brackets, simplify each term and re-write any roots or denominators as terms in x to a power

Hint 3: after differentiating, re-write the x terms with fractional powers back in terms of square roots, and any negative powers back as fractions

Hint 4: when the expression is as simple as possible, substitute the value of x = 4 into the expression for f'(x)

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

Question 8

Hint 1: know that the formula for the area of a rectangle is length × breadth

Hint 2: set up an inequation for this formula to be < 15

Hint 3: expand out the brackets and rearrange terms so that the inequation is of the form ... < 0

Hint 4: factorise the quadratic expression

Hint 5: sketch a graph of the quadratic expression, as if it were a function, noting the x-axis intercepts and whether it has a maximum or minimum turning point

Hint 6: decide from your graph which region(s) represent the values of x when the quadratic expression is < 0

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

Question 9

Hint 1: if AB is parallel to the given equation, then rearrange that equation to then read off its gradient

Hint 2: if BC makes an angle of 150°, then we will use m = tan(θ) to obtain the gradient of BC

Hint 3: know that the value for tan(150°) is related to the value for tan(30°). Draw an accurate sketch of the function y = tan(x) to see why.

Hint 4: use exact value triangle knowledge to obtain the exact value of tan(30°) and thus the exact value of the gradient BC

Hint 5: compare the gradients of AB and BC and clearly state whether this shows if the points A, B and C might be collinear, or not.

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

Question 10

10a) Hint 1: sketch a right-angled triangle with angle '2x' and opposite length 3 and adjacent length 4.

10a) Hint 2: use Pythagoras' Theorem to work out the length of the hypotenuse (or just notice a familiar Pythagorean triplet?)

10a) Hint 3: your diagram can now be used to write down the value of cos(2x)

10b) Hint 4: know that cos(2x) can be written in terms of cos²(x)

10b) Hint 5: use your answer from part (a) to then obtain an expression for cos²(x)

10b) Hint 6: know that this will give two values for cos(x)

10b) Hint 7: from the graph of y = cos(x), realise that the restriction given in the question that 0 < x < π/4 means that we can reject one of the values of cos(x)

10b) Hint 8: simplify the remaining expression for cos(x) as far as possible.

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

Question 11

11a) Hint 1: draw a sketch of the circle and plot point T on its circumference

11a) Hint 2: know that a tangent requires a point (point T) and a gradient

11a) Hint 3: calculate the gradient of the line from the centre of the circle to point T

11a) Hint 4: calculate the perpendicular gradient to this gradient

11a) Hint 5: use a standard method to work out the equation of the tangent using point T and the gradient just calculated

11b) Hint 6: know that a tangent to a parabola will intersect it in the same place, twice

11b) Hint 7: realise that this means that the equation formed from equating the quadratic with the equation of the tangent through T will have equal roots

11b) Hint 8: know that equal roots of a quadratic mean that the discriminant will equal zero

11b) Hint 9: the discriminant equation gives a quadratic in p which can be solved to give two values for p

11b) Hint 10: read the question to see the restriction on values of p, and thus reject one of the calculated solutions, clearly stating the reason why

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

Question 12

Hint 1: know that the graph of a cos(bx) is symmetrical about the x-axis

Hint 2: therefore the shaded region above the x-axis is connected to the non-shaded region below the x-axis

Hint 3: know that the non-shaded region would give an integral value that is negative

Hint 4: realise that the non-shaded region is -2 lots of the shaded region

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

Question 13

13a) Hint 1: point P(1, b) has x = 1 and y = b

13a) Hint 2: substitute these values of x and y into the equation y =
2^{x} + 3, and then solve for b

13b)i) Hint 3: know that the graph of the f^{-1}(x) is geometrically
related to the graph of f(x)

13b)i) Hint 4: the graph of the f^{-1}(x) is the reflection of the
graph of f(x) in the line y = x

13b)ii) Hint 5: the image of point (x, y) is the point (y, x)

13c) Hint 6: know that y = 4 - f(x + 1) has three transformations involved: from the 4, the -1 multiplier and the +1 in the brackets

13c) Hint 7: working from the 'x' term outwards, it is first affected by +1, then by -1, then by 4

13c) Hint 8: know that the +1 moves the graph left one unit

13c) Hint 9: know that the -1 multiplier reflects the function in the x-axis

13c) Hint 10: know that the +4 translates the graph upwards by 4

13c) Hint 11: track the point R(3, 11) through each of these transformations

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

Question 14

Hint 1: either complete the square in the x terms, and the y terms, or use the circle equation formula to determine the centre and the radius

Hint 2: from the previous step, you should know the coordinates of the centre of the circle

Hint 3: sketch a set of axes and draw in a circle that would meet the coordinate axes at exactly 3 points

Hint 4: your sketch should be a circle going through the origin, plus one point on the y-axis and one point on the x-axis

Hint 5: realise that the radius of the circle will be the distance from the origin to the centre of the circle

Hint 6: match this value of the radius with your expression for the radius from the first step, and work out the value of k

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

Question 15

Hint 1: realise that you will need to integrate the expression and then use the given information to determine the constant of integration

Hint 2: integrate T'(t) with respect to t, to give T(t), with a constant of integration

Hint 3: use the information that when t = 0, T(0) = 100 in order to fix the value of the constant

Hint 4: use the information that when t = 10, T(10) = 82 in order to fix the value of k

Hint 5: clearly state the expression for T(t) in terms of t.

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

Paper 2

Question 1

1a) Hint 1: the altitude through C has a gradient that is perpendicular to the gradient of AB

1a) Hint 2: work out m_{AB} and then take the negative reciprocal to give
the gradient of the altitude

1a) Hint 3: use a standard method to work out the equation of the line through T with the required gradient

1b) Hint 4: know that the median from B will go through the midpoint of AC

1b) Hint 5: calculate the coordinates of M, the mid-point of AC

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

1b) Hint 7: use a standard method to work out the equation of the line
through B with the gradient m_{BM}

1c) Hint 8: realise that you have simultaneous equations, using your answers from parts (a) and (b)

1c) Hint 9: be sure to state the coordinates in brackets, with a comma between them.

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

Question 2

2a) Hint 1: know that f(g(x)) means to do g(x) first, then f(…) second

2a) Hint 2: replace g(x) in f(g(x)) with what g(x) is equal to

2a) Hint 3: apply function f(…) to what its input is, then expand and simplify the quadratic expression

2b) Hint 4: use a standard method to complete the square for your answer from part (a)

2c) Hint 5: know that h(x) will not be defined if its denominator has the value of zero.

2c) Hint 6: use your answer from part (b) to determine the values of x that would make f(g(x)) equal to zero.

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

Question 3

3a) Hint 1: use the formula for the t_{n} sequence, replacing n with 1

3b) Hint 2: work out the limit of each sequence by a standard method

3b) Hint 3: compare each limit to the number 50 to decide if either the frog or toad can escape

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

Question 4

4a) Hint 1: set f(x) = g(x) and solve for x

4b) Hint 2: realise that integration will be used

4b) Hint 3: realise that the shield is symmetrical, so we can work out the area between f(x) and h(x), and then double it to obtain the total area

4b) Hint 4: treat f(x) as the 'top function' and h(x) as the 'bottom' function

4b) Hint 5: when subtracting h(x) from f(x), be sure to include brackets so that the correct terms are all subtracted

4b) Hint 6: the limits of integration are from zero to the value you found in part(a)

4b) Hint 7: remember to double the integral answer at the end to get the total area!

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

Question 5

5a) Hint 1: complete the square in the x and y terms, or use the circle formula,
to obtain the centre and radius of circle C_{1}

5a) Hint 2: work out the distance from the centre of C_{1} to the centre
of C_{2}

5a) Hint 3: this distance is the sum of the two circles' radii, and you know the
radius of C_{1}, so you can work out the radius of C_{2}

5b) Hint 4: copy the given diagram of two circles and draw C_{3} such
that it has its centre on the line joining the other two circles' centres

5b) Hint 5: know that the radius of C_{3} will be the sum of the
__diameters__ of C_{1} and C_{2}

5b) Hint 6: draw another sketch of the line joining the centres of
C_{1} and C_{2}, and extend it out to show the diameters of C_{1} and C_{2}

5b) Hint 7: know that the centre of C_{3} will be in the middle of
this line that the length of the two diameters

5b) Hint 8: work out the ratio of where the centre point for C_{3}
is relative to line segment joining the centres of C_{1} and C_{2}

5b) Hint 9: use your ratio technique to obtain the coordinates of the
centre of C_{3} and therefore the equation of C_{3}

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

Question 6

6a) Hint 1: know that __p__.(__q__ + __r__) = __p__.__q__ +
__p__.__r__

6a) Hint 2: know that the angle between vectors __p__ and __q__ is 60°

6a) Hint 3: use exact value triangle knowledge to give cos(60°) and know what cos(0°) is as well.

6a) Hint 4: know that |__q__| = |__p__| = 3

6b) Hint 5: look for a route from point E to point A that goes along segments that you know about

6b) Hint 6: vector EC = vector EA + vector AB + vector BC

6b) Hint 7: know that vector EA is the negative of vector AE, and that vector BC is the same as vector ED (because it is a parallelogram)

6c) Hint 8: use your answer from part (b) and a similar technique to part
(a) to expand and simplify a series of scalar products involving __p__, __q__ and __r__

6c) Hint 9: you will need to know the angle between __q__ and __r__,
when the vectors are placed 'tail to tail'

6c) Hint 10: you should find that you want the exact value for cos(30°)

6c) Hint 11: substituting all values into the given equation should
allow you to solve an equation in |__r__|

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

Question 7

7a) Hint 1: realise that integrating the 3cos(2x) term will involve the 'opposite' of the chain rule

7a) Hint 2: remember to include the constant of integration

7b) Hint 3: work with the left hand side and chose an identity for cos(2x) that includes terms like those on the right hand side

7c) Hint 4: notice that the integrand is a linear multiple of the right hand side of part (b)

7c) Hint 5: rewrite the integrand in terms of the left hand side of part (b)

7c) Hint 6: use your workings from part (a) to complete the integration

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

Question 8

8a)i) Hint 1: realise that if the crocodile goes by water the whole way, then x = 20

8a)i) Hint 2: evaluate T(x) when x = 20

8a)ii) Hint 3: realise that if the crocodile swims the shortest distance possible, then it will go straight across the river, and therefore x = 0

8a)ii) Hint 4: evaluate T(x) when x = 0

8b) Hint 5: realise that this is an optimisation equation, so it will involve differentiating T(x) to find a stationary point

8b) Hint 6: realise that T(x) is not ready to differentiate, so you need to re-write the √ as a fractional power

8b) Hint 7: when differentiating T(x), you will need to use the chain rule

8b) Hint 8: after differentiating, re-write any terms with fractional powers back in terms of square roots, and any negative powers back as fractions

8b) Hint 9: clearly state that you are seeking T'(x) = 0 in order to find a stationary point

8b) Hint 10: take careful steps to rearrange and solve for x. You should obtain two values: one positive and one negative

8b) Hint 11: from reading the context of the question, give a reason for rejecting one of the values of x

8b) Hint 12: to check the nature of the stationary point, you will need to use a nature table, as to work out T''(x) requires knowledge that is beyond the Higher course

8b) Hint 13: after confirming that you have a minimum turning point, evaluate T(x) for the found value of x.

Hint 14: and here is a video of the solution:

Question 9

Hint 1: use a standard method for re-writing the given expression in the stated form

Hint 2: set the value of h in the first equation to be h = 100

Hint 3: subtract 65 from both sides of the equation and use your earlier work to re-write the equation in terms of a single sine function

Hint 4: check you have your calculator in radian mode, and take inverse sine to
obtain __two__ values for (1.5t - 0.395)

Hint 5: add 0.395 to each of your values

Hint 6: divide each of your answers by 1.5, to give the values of t required.

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

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