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Paper 1

Question 1

Hint 1: Know that to find a tangent, we shall need to know the gradient at x = 2

Hint 2: Therefore, we shall have to differentiate the function

Hint 3: Differentiate the function, term by term

Hint 4: Evaluate y'(x) when x = 2

Hint 5: We will also need the y coordinate of the function, when x = 2

Hint 6: Therefore, evaluate y(x) when x = 2

Hint 7: Use the coordinates of (2, 5) with the gradient of m = 4 to calculate the equation of the tangent line

Question 2

Hint 1: Sketch a coordinate diagram, plotting the points A(1, 4) and B(9, 10) on it

Hint 2: Add to your diagram the line going through points A and B

Hint 3: Calculate the coordinates of the point, M, that is the mid-point of AB

Hint 4: Mark that point on your diagram, along with the coordinates of (5, 7) that have been calculated from A's and B's coordinates.

Hint 5: Add to your diagram the line through M, that is perpendicular to line AB

Hint 6: Recognise that we need the equation of this last line, for which we know it goes through M(5, 7) but we need the gradient of the line.

Hint 7: Calculate the gradient of line AB, using the coordinates of points A and B

Hint 8: so m_AB = 3/4

Hint 9: Calculate the perpendicular gradient, m_⊥AB, by taking the negative reciprocal of m_AB

Hint 10: Use the coordinates of (5, 7) with the gradient of m = -4/3 to calculate the equation of the perpendicular bisector

Hint 11: Consider multiplying the equation through by 3, to obtain integer coefficients of all terms, and rearranging into Ax + By + C = 0 format.

Question 3

Hint 1: Recognise that the integrand cannot be integrated yet, as the term 12/x² needs to be re-written

Hint 2: Re-write 12/x² as 12xⁿ, where n is a negative number

Hint 3: Now integrate each term, taking care with the negative and fractional values that will inevitably appear

Hint 4: Don't forget the constant of integration!

Hint 5: Consider re-writing terms with negative powers back as fractions, and terms with fractional powers back as roots.

Question 4

Hint 1: Recognise that both logarithmic terms have the same base, namely 3. This is a good thing!

Hint 2: Know that to use the laws of logarithms, we need to have each term saying 1 × log₃ ....

Hint 3: Re-write 3 × log₃2 as log₃2ⁿ, for some value of n.

Hint 4: Know that log₃A + log₃B = log₃(A × B)

Hint 5: By this stage you should have obtained log₃(1/3)

Hint 6: Know that (1/3) can be written as 3^m for some value of m, that is negative

Hint 7: Use the law of logarithms that log₃x^p = p ×: log₃x

Hint 8: Know that log₃3 can be simplified

Question 5

Hint 1: Recognise that y = f(-x) + 3 is made up of two transformational steps

Hint 2: Know that transforming y = f(x) to y = f(-x) is one step, and that the '+3' is the second step

Hint 3: Know that y = f(x) to y = f(-x) is the equivalent of reflecting in the y-axis

Hint 4: Sketch the original diagram provided, with all of the turning points and intercepts labelled on it

Hint 5: Directly below that diagram, sketch a new diagram so that the y-axis is in line with the first diagram's y-axis, with the function curve reflected in the y-axis, labelling all of the new locations of turning points and intercepts

Hint 6: Know that '+3' will translate the function upwards by 3 units in the positive direction of the y-axis

Hint 7: Directly below the last diagram, draw a third diagram with the previous curve translated upwards by 3 units, labelling all of the new locations of turning points and intercepts

Question 6

6a)i) Hint 1: Copy the diagram provided, and mark in the exact length of the hypotenuse

6a)i) Hint 2: Know that this diagram can now give you exact values for sin(q) and cos(q)

6a)i) Hint 3: Know that sin(2q) can be written in terms of sin(q) and cos(q)

6a)i) Hint 4: Expand sin(2q) and replace both sin(q) and cos(q) with their exact values, and then carefully simplify the fractions

6a)ii) Hint 5: Know that cos(2q) can be written in terms of sin(q) and/or cos(q)

6a)ii) Hint 6: Expand cos(2q) using any of the three possible expansions, and replace the trigonometric terms in 'q' with their exact values, and then carefully simplify the fractions

6b) Hint 7: Know that the provided diagram can give you exact values for sin(r) and cos(r)

6b) Hint 8: Know that sin(2q - r) can be expanded and written in terms of sin(2q), cos(2q), sin(r) and cos(r)

6b) Hint 9: Expand sin(2q - r) and replace the trigonometric terms with values from part (a), and sin(r) and cos(r), and then carefully simplify the fractions

Question 7

7a) Hint 1: Know that showing that a value is a factor of the function is equivalent to showing that the value is a root of the function

7a) Hint 2: If we say that f(x) is the provided cubic function, evaluate f(-3)

7a) Hint 3: You should find that f(-3) = 0, so then write down the statement 'as f(-3) = 0, then -3 is a root of f(x), and so (x+3) is a factor of f(x)'

7b) Hint 4: Using either polynomial long division, or synthetic division, factorise f(x) to give (x + 3) and a quadratic expression

7b) Hint 5: Look at the quadratic factor, and endeavour to factorise it further into two linear factors

7b) Hint 6: You should now have f(x) = (x + 3)(5x - 4)(x + 1)

7b) Hint 7: In order to solve the cubic equation, write down the list of the three values of x that will make f(x) = 0

Question 8

Hint 1: There are two approaches to tackling this question...

Hint 2: ...EITHER change '2' into '2log_aa' so that all terms have logs in them ... OR rearrange the equation so that both log terms are on the left hand side

Hint 3: Whichever approach is chosen, use the laws of logarithms to simplify and gather terms together to ultimately obtain a² = 25

Hint 4: Know that there are two solutions to this equation, so write them both down

Hint 5: Refer to the question to see the constraint that a > 0, so you know that one value can be discarded

Hint 6: State your final value for 'a', giving the reason for your choice (which is 'as a > 0')

Question 9

Hint 1: Know that to find where a line meets a circle, the equation of the line can be substituted into the equation of the circle

Hint 2: Substitute '(x+1)' for 'y' in the circle's equation, remembering to put in the brackets

Hint 3: Carefully expand out the brackets, being vigilant for positive and negative signs

Hint 4: After simplification, you should have a quadratic expression in the variable x, that is equal to zero

Hint 5: Factorise the quadratic expression, and then state the two solutions for the variable x

Hint 6: Using the equation of the line, calculate the corresponding two solutions for y from the x values

Hint 7: Present your final answer as two sets of coordinates, which should be: (1, 2) and (-4, -3)

Question 10

Hint 1: To help visualise the situation, consider drawing a sketch of two vectors, u and v , tail-to-tail, with an angle of 45° between them

Hint 2: Know that u .v = |u||v|cos(θ) where θ is the angle between the vectors

Hint 3: Calculate the exact value of |u|

Hint 4: Calculate the exact value of |v| that will be an expression in terms of 'k'

Hint 5: Calculate the value of u .v by multiplying components and adding them up

Hint 6: Substitute all of the above values into the equation: u .v = |u||v|cos(θ)

Hint 7: When terms are gathered and simplified, you should have 4 = √(10 + k²)

Hint 8: Squaring both sides of this equation and then rearranging, should then give k² = 6

Hint 9: Know that there are two solutions to this equation, so write them both down

Hint 10: Refer to the question to see the constraint that k > 0, so you know that one value can be discarded

Hint 11: State your final value for 'k', giving the reason for your choice (which is 'as k > 0')

Question 11

Hint 1: Know that 'two real and distinct roots' means that the discriminant is > 0

Hint 2: In order to use b² - 4ac > 0, we need to identify the values of a, b and c from the provided quadratic equation

Hint 3: Two of these values will be in terms of 'k'

Hint 4: Substitute the values of a, b and c into b²: - 4ac > 0, taking care to put brackets around the expression for 'b'

Hint 5: After simplification, you should have the quadratic inequation: 9k² - 36k > 0

Hint 6: Factorise the left side of this expression

Hint 7: After factorisation, you should have the inequation: 9k(k - 4) > 0

Hint 8: On a separate diagram, sketch a graph of the quadratic function f(k) = 9k(k - 4) that will have horizontal k-axis intercepts at 0 and 4, and the function has a minimum turning point

Hint 9: Interpreting this diagram will mean that for f(k) > 0, we need either k < 0 or k > 4

Question 12

Hint 1: Recognise that they have given us y'(x) and they want us to obtain y(x), and they want us to use the second bullet point's information to fix the value of the constant of integration

Hint 2: Know that 6cos(x) must have come from the derivative of 6sin(x)

Hint 3: Know that 8sin(2x) must have come from the derivative of some term involving -8cos(2x), but with adjustments made for the chain rule's effect

Hint 4: Once the constant of integration is included, you should have y = 6sin(x) - 4cos(2x) + c

Hint 5: In order to fix the value of the constant, c, we need to subsitute the values of x = π/6 and y = 4 into this equation.

Hint 6: To help evaluate sin(π/6) and cos(π/3), consider sketching the standard exact value triangles

Hint 7: After all substitutions have been made and the equation solved, you should obtain c = 3

Hint 8: State the final function for y(x) with the constant of integration replaced with its value

Question 13

13a) Hint 1: Know that stationary point calculations will involve differentiation, which has already been done for us

13a) Hint 2: As we want f'(x) = 0, we can read off the two values of x from the factorised expression

13a) Hint 3: In order to determine which might be a maximum or minimum, EITHER construct two nature tables OR calculate the second derivative, f''(x)

13a) Hint 4: By whichever standard method you use, you should find a minimum turning point at x = 5, and a maximum turning point at x = 2

13b) Hint 5: We need to decipher the meaning of each of the bullet point statements, and what it might mean for the sketch

13b) Hint 6: Know that if f(0) < 0, then the value of the function at x = 0 is below the origin (i.e. the y-axis intercept is negative)

13b) Hint 7: Know that 'f(x) = 0 has exactly one solution' means that the function crosses the x-axis at only one location, and it is not a double-root at that location

13b) Hint 8: We also know from part (a) where possible minimums and maximums are located, at least as far as their x-axis values are known

13b) Hint 9: We know it is a cubic, but not 'which way round the cubic is' (is it a positive cubic, which starts in quadrant 3 and ends in quadrant 1; or is it a negative cubic which starts in quadrant 2 and ends in quadrant 4?)

13b) Hint 10: We do know an expression for f'(x), so if this is integrated, we obtain a cubic and we see that the coefficient of the x³ term is negative

13b) Hint 11: Hence, we have a negative cubic, which will start in quadrant 2 and end in quadrant 4.

13b) Hint 12: Sketch a set of axes, and mark in as much of the information that you can, either by where things might possibly be located, or where things most definitely are not located

13b) Hint 13: Draw in a curve for the function that meets all of the required constraints. Note that you will not be able to exactly pinpoint all of the locations of all features.

Paper 2

Question 1

1a) Hint 1: Know that the altitude through B will have a gradient that is perpendicular to AC

1a) Hint 2: Calculate the gradient, m_AC

1a) Hint 3: Calculate the perpendicular gradient, m_⊥AC, by taking the negative reciprocal of m_AC

1a) Hint 4: Use the coordinates of (9, 20) with the gradient m_⊥AC = 3 to calculate the equation of the altitude

1b) Hint 5: Know that the median through A will also go through the mid-point of BC

1b) Hint 6: Calculate the mid-point of BC, by taking the mean of the x-coordinates of points B and C, and similarly for the y-coordinates

1b) Hint 7: Calculate the gradient between point A(-9, -14) and the midpoint of BC

1b) Hint 8: Use the coordinates of (-9, -14) with the gradient m = 1/2 to calculate the equation of the median

1c) Hint 9: Know that intersection of two lines will come from treating them as simultaneous equations

1c) Hint 10: Use your chosen method of solving simultaneous equations to first find one coordinate, and then to find the second coordinate

1c) Hint 11: Clearly state the coordinates of the point of intersection, which should be (-1, -10)

Question 2

Hint 1: Know to first factorise the 2 out of the first two terms only, to give 2[x² + 8x] + 5

Hint 2: Looking at the terms in the square brackets, complete the square by using half of the coefficient of x

Hint 3: You should now have 2[x² + 8x + 4² - 4²] + 5

Hint 4: The first 3 terms in the square brackets can now be factorised

Hint 5: You should now have 2[(x + 4)(x + 4) - 16] + 5

Hint 6: Now expand out the square brackets, using the 2

Hint 7: You should now have 2(x+4)² - 32 + 5

Hint 8: A final gathering of constant terms will give the desired expression

Question 3

Hint 1: Know that we shall be having to calculate a definite integral

Hint 2: Observe that the lower limit will be 2 and the upper limit will be 4

Hint 3: Know that the integrand will be x² - 2x + 3

Hint 4: Write this calculation with an integral sign, the two limits, the quadratic expression in a set of brackets, and with a 'dx' on the end.

Hint 5: Perform the standard process for integration, taking care with the signs and the fractions that will inevitably happen

Question 4

Hint 1: Know that the definition for an inverse function is: g(g^-1(x)) = x

Hint 2: Using composition of function knowledge, evaluate g(x) where x = g^-1(x)

Hint 3: Rearrange the equation to make g^-1(x) the subject

Question 5

5a) Hint 1: Know that to show that 3 points are collinear, we need to obtain two vectors and show that one vector is a scalar multiple of the other vector, as well as them sharing a common point

5a) Hint 2: For this explanation, we shall obtain vector AB and vector AC and show that one is a multiple of the other, and that they (clearly) share the common point, A

5a) Hint 3: Calculate vector AB, and factorise out a common factor from its three components

5a) Hint 4: Calculate vector AC, and factorise out a common factor from its three components

5a) Hint 5: Set up the equation vector AB = k × vector AC

5a) Hint 6: You should see that the factorised vectors have the same core components, and can therefore be discarded, leaving an equation in k using the scalar multiples, that can be solved

5a) Hint 7: Write down the final statement: 'as vectors AB and AC are parallel, and they share a common point, A, then points A, B and C are collinear'

5b) Hint 8: Using the value of 'k' calculated in part (a), which should have been k = 3/5, we know that vector AB is three-fifths of the length of vector AC

5b) Hint 9: Consider drawing a sketch to show a line with endpoints A and C, with point B between them, nearer to point C than point A

5b) Hint 10: Mark in on the diagram the value of '3' between A and B, and the value of '2' between B and C (thereby making 3 + 2 = 5)

5b) Hint 11: Hence, you can conclude that point B divides AC in the ratio 3:2

Question 6

6a) Hint 1: Write down the expansion of k.cos(x + a)

6a) Hint 2: Compare the terms' coefficients with the provided expression, to obtain k.cos(a) = 5 and k.sin(a) = 9

6a) Hint 3: Use your chosen, standard method to solve these equations for both k and a, noting that 'a' is likely to be a decimal number of radians

6b) Hint 4: Recognise that the work from part (a) can be used to create a new equation, also equal to 7.

6b) Hint 5: After rearranging, you should obtain the trigonometric equation: cos(x + 1.064) = 7/√106

6b) Hint 6: With your calculator in radian mode, calculate cos^-1(7/√106)

6b) Hint 7: There is one value between 0 and 2π, and we need to list several more values, before we then subtract 1.064 from each of these values

6b) Hint 8: From your first value (which should have been 0.82317) calculate 2π - 0.82317 and 2π + 0.82317

6b) Hint 9: You should now have the equation: x + 1.064 = 0.82317, 5.4600, 7.10636

6b) Hint 10: Subtracting 1.064 from each value should give candidates for the value of x, but note that 0 ≤ x < 2π, so at least one value can be discarded

Question 7

Hint 1: Recognise that this is a fairly standard integral, but that the chain rule needs to considered when fixing coefficients

Hint 2: And don't forget the constant of integration!

Question 8

Hint 1: Know that we need to write vector BE as the sum of other, known vectors

Hint 2: Look for a journey from B to E that goes along other edges...

Hint 3: So, vector BE = vector BA + vector AD + vector DE

Hint 4: From the diagram, we can see that vector BA is parallel to vector AD, but in the opposite direction

Hint 5: Hence, vector BE = - vector AD + vector AD + vector DE

Hint 6: With the given vectors from the question, substitute these into the equation.

Hint 7: If you have been working with vertical vector components, be sure to re-write them back into i, j and k notation

Question 9

9a) Hint 1: Know that as n tends to ∞, then both u_n and u_n+1 will tend to the limit of 10

9a) Hint 2: In the given equation u_n+1 = m.u_n + 4, replace both u terms with 10, and solve for m

9b) Hint 3: In the given equation u_n+1 = m.u_n + 4, replace the m with the value just calculated in part (a), and replace n with zero

9b) Hint 4: Rearrange the equation to make u₀ the subject

Question 10

10a) Hint 1: Recognise that the total area is the sum of a rectangle and a right-angled triangle

10a) Hint 2: Write out an expression for the total area in terms of variables x and y, and make it equal to 150

10a) Hint 3: Rearranging this equation to make y the subject, should give y = (150 - 6x²)/(5x)

10a) Hint 4: Write out an expression for the perimeter of the shape, in terms of x and y

10a) Hint 5: Substitute the y values for the expression that was obtained from processing the area

10a) Hint 6: After careful expanding of brackets and gathering of terms, you should then obtain the stated expression for P(x)

10b) Hint 7: Recognise that to obtain the value to make P(x) as small as possible, you will need to differentiate the expression to find the minimum turning point

10b) Hint 8: Obtain P'(x) and set it equal to zero to find a positive value for x (as negative values do not fit the context of this problem)

10b) Hint 9: Use a nature table, or the second derivative P''(x) to check that the value of x = 2.5 does indeed give a minimum

10b) Hint 10: Evaluate P(2.5) to obtain the minimum perimeter value

10b) Hint 11: Present a final answer, with the correct units of length

Question 11

Hint 1: Recognise that we have a trigonometric equation in both sine and cosine terms and we'd like to make it more manageable so there there are no double angles

Hint 2: Accept that there is only one possible expansion of sin(2x) and therefore our hand is forced as to what to do first

Hint 3: After expanding sin(2x), look to factorise out the common term of cos(x)

Hint 4: We then have (6sin(x) + 4).cos(x) = 0 which provides two linear trigonometric equations to solve, namely 6sin(x) + 4 = 0 and cos(x) = 0

Hint 5: Solving cos(x) = 0 ought to be straightforward, after sketching the graph of the y = cos(x) function

Hint 6: Solving 6sin(x) + 4 = 0 requires rearranging and the use of a calculator to generate two values of x between 0° and 360°

Hint 7: Pool all of the solutions together from both equations for a final list of 4 possible solution values for x.

Question 12

12a) Hint 1: Know that h(x) = f(g(x)) means that function g(x) happens first, then f(x) happens thereafter

12a) Hint 2: Write out h(x) by first replacing g(x) with 1 - x³

12a) Hint 3: Then apply function f to the expression (1 - x³)

12b) Hint 4: Recognise that the chain rule for differentiation will be required, as the 'inner function' has a derivative of -3x²

Question 13

13a) Hint 1: Know that the initial mass will be when time, t = 0

13a) Hint 2: Substitute t = 0 into the given formula for M(t), remembering that e⁰ is not zero

13b) Hint 3: Recognise that 120 is now the value for M, and our task is to rearrange the equation to find the value of t

13b) Hint 4: Starting with 120 = 150e^-0.0054t, first divide by 150, and then re-write as a logarithmic expression

13b) Hint 5: Division by -0.0054 ought to give t as the subject and then this can be evaluated

13b) Hint 6: State a final decimal answer, with the correct units of time, for t.

Question 14

14a) Hint 1: Know that the centre coordinates of C₁ can be read off from the two terms in brackets.

14a) Hint 2: Know that the radius of the circle will be the positive square root of 9

14b) Hint 3: Either use the formula provided in the question paper, or the process of completing the square for each of the variables x and y, in order obtain the centre and radius of circle C₂

14b) Hint 4: Sketch an accurate drawing, using the centres and radii of circles C₁ and C₂

Hint 5: Recognise that the distance between the centres of C₁ and C₂ is going to be important

Hint 6: Calculate the vector between these centres, vector C₁C₂, and factorise out a common factor

Hint 7: You should be left with 3 times a vector which has the components of 4 and -3, which you should recognise as a vector that has magnitude 5

Hint 8: Hence the distance between the centres of C₁ and C₂ is 3 × 5 = 15

Hint 9: From this, we can conclude that the diameter of C₃ is 15 - radius of C₂ + radius of C₁

Hint 10: Hence, if the diameter of C₃ is 16, then its radius must be 8

Hint 11: Mark on your diagram, all of the distances that are now known along the line that joins the centre of C₁ to the centre of C₂

Hint 12: Realise that the position of the centre of C₃ must be 5/15 of the way along the line from the centre of C₁ to the centre of C₂

Hint 13: Use a vector equation to obtain the position vector of the centre of C₃ i.e. vector OC₃ = vector OC₁ + 5/15 × vector C₁C₂

Hint 14: With careful arithmetic, you ought to obtain the vector with components -1 and 3

Hint 15: Hence, clearly state the centre of C₃ has coordinates (-1, 3) and radius 8

Hint 16: Then, clearly write the equation of C₃, being sure to evaluate what 8² is equal to.