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

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Paper 1

Question 1

Hint 1: notice that both the product rule and the chain rule will require to be used

Hint 2: and here is a video of the solution:

Question 2

Hint 1: know how to deal with the (x - 1)² term in the denominator, when writing out the partial fractions

Hint 2: the partial fractions will be A/(x + 3) + B/(x - 1) + C/(x - 1)²

Hint 3: multiply the whole equation through by (x + 3)(x - 1)²

Hint 4: substitute in x = 1 to calculate the value of constant C

Hint 5: substitute in x = -3 to calculate the value of constant A

Hint 6: now pick any value for x, (say, x = 0), and substitute in that value along with A and C to help calculate B

Hint 7: present your final answer with all three constants in their correct fractions

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

Question 3

Hint 1: use your standard method for gaussian elimination, keeping track of the row operations that you perform

Hint 2: if done correctly, your final row should be 0 0 0 | 2

Hint 3: interpret this row as meaning 0z = 2 and then think very carefully what value of z might solve this equation

Hint 4: you ought to conclude that there are no real solutions

Hint 5: you then need to clearly state that the original equations are inconsistent

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

Question 4

Hint 1: select your integration by parts technique so that you integrate x⁴ and differentiate ln(x)

Hint 2: remember to include the 'dx' after the first part of the process, and the constant of integration at the end

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

Question 5

Hint 1: first look to find a solution for y'' - 4y' - 5y = 0

Hint 2: try y = Aemx

Hint 3: calculate y' and y'', and substitute into the homogeneous differential equation

Hint 4: obtain the auxiliary equation, which can be factorised to give two real values

Hint 5: use these to give the complementary function, y = Ae-x + Be5x

Hint 6: now consider the 10x² + 11x - 23, and construct a general form of this polynomial

Hint 7: try y = Cx² + Dx + E

Hint 8: calculate y' and y'', and substitute into the original differential equation

Hint 9: compare the coefficients of terms in x² to calculate the value of C

Hint 10: compare the coefficients of terms in x, and use the value of C, to calculate the value of D

Hint 11: compare the constants, and use the values of C and D to calculate the value of E

Hint 12: you should conclude that the particular integral is y = -2x² + x + 3

Hint 13: this gives the general solution of y = Ae-x + Be5x -2x² + x + 3

Hint 14: in order to fix the value of the constants A and B, we need to use the initial conditions given in the question, namely y(0) = 2 and y'(0) = 14

Hint 15: using the general solution, substitute in x = 0 and y = 2, to obtain a linear equation in A and B

Hint 16: using the derivative of the general solution, substitute in x = 0 and y' = 14, to obtain a second linear equation in A and B

Hint 17: solve the two simultaneous linear equations to obtain values for each of A and B

Hint 18: finally, put everything together to give the particular solution of y = -3e-x + 2e5x -2x² + x + 3

Hint 19: phew!

Hint 20: and here is a video of the solution:

Question 6

6a) Hint 1: sketch a right angled triangle with angle θ, adjacent of length 1 and opposite of length √3

6a) Hint 2: know that the modulus, r, will be the hypotenuse of the right angled triangle

6a) Hint 3: use Pythagoras' theorem to calculate r

6a) Hint 4: read off from your diagram that tan(θ) = √3

6a) Hint 5: solve for θ using your knowledge of exact value triangles

6a) Hint 6: write z in polar form, which is z = r[cos(θ) + i.sin(θ)]

6b) Hint 7: realise that you need to work out z³ and that polar form will be easiest to use, because of De Moivre's Theorem

6b) Hint 8: after applying De Moivre's Theorem, carefully simplify the cos(π) and sin(π) terms

6b) Hint 9: make a clear statement, based on your value for z³, for why z is real.

6b) Hint 10: and here is a video of the solution:

Question 7

7a) Hint 1: use the rules to do with sigma notation to split up Σ(r² + 3r) into two separate summations

7a) Hint 2: factor out the '3' from the second summation

7a) Hint 3: replace Σr² and Σr with their standard formulae

7a) Hint 4: do not expand out every bracket!

7a) Hint 5: take a common factor of n(n + 1)/6 out of each term

7a) Hint 6: simplify as far a possible, aiming for the form stated in the question

7b) Hint 7: note the lower limit of the summation is 11, and not 1

7b) Hint 8: know that a sum from 11 to 20 is equal to a sum from 1 to 20 subtract a sum from 1 to 10

7b) Hint 9: evaluate the expression from part (a) with each of n = 20 and n = 10

7b) Hint 10: careful arithmethical steps and factoring will lead you to a final value of 2950

7b) Hint 11: and here is a video of the solution:

Question 8

8a) Hint 1: know that propositions involving squares and inequalities are best disproven using at least one negative number

8a) Hint 2: if you set both a and b to be negative, say a = -4 and b = -1, then a < b but that a² is not less than b²

8a) Hint 3: be sure to write a clear, final sentence that the values you chose for a and b give a counterexample, and so the original proposition is false

8b) Hint 4: know that you must construct the general form of an odd integer, n = 2m + 1 where m is an integer

8b) Hint 5: for any new variable that you introduce (such as 'm', here), you must clearly define the set of numbers that it is a member of

8b) Hint 6: now evaluate n² - 1 by replacing n with 2m + 1

8b) Hint 7: expand, simplify and then factor out a 4

8b) Hint 8: clearly communicate that the expression multiplying the 4 is an integer, and know why that is true

8b) Hint 9: clearly communicate that you have the general form of a multiple of 4

8b) Hint 10: clearly state a concluding sentence about what you have proven

8b) Hint 11: and here is a video of the solution:

Question 9

9a) Hint 1: write down the matrix for an anti-clockwise rotation by angle θ, using the formula sheet if required

9a) Hint 2: replace θ with π/2 and simplify each element of the matrix to be either a -1, 0 or 1.

9b)i) Hint 3: perform standard matrix multiplication to obtain AB

9b)ii) Hint 4: compare AB to the matrix you wrote down at the start of part (a)

9b)ii) Hint 5: extract from that comparison what value cos(α) must equal, and what sin(α) must equal

9b)ii) Hint 6: using exact value triangles and sketches of graphs, solve each of the cos(α) and sin(α) equations for all solutions between 0 and 2π

9b)ii) Hint 7: identify the value of α that is common to both sets of solutions, as that will be the value of α you need

9c) Hint 8: from part (b), you know that AB is a rotation by 5π/3 radians

9c) Hint 9: hence (AB)n will be a rotation of 5π/3 × n radians

9c) Hint 10: we are told that (AB)n is the identity matrix, so that means it can be considered as an integer multiple of rotations by 2π radians

9c) Hint 11: set up the equation 5π/3 × n = 2π × k, where n and k are both integers

9c) Hint 12: divide both sides by π and make n the subject

9c) Hint 13: consider what the smallest value of k must be so that n works out to be an integer

9c) Hint 14: work out that value of n and state it clearly

9c) Hint 15: and here is a video of the solution:

Paper 2

Question 1

Hint 1: recognise that you will need the chain rule, plus a standard differential for sin-1(x)

Hint 2: be sure to put the (3x) inside a bracket, so that when it's squared, everything is processed correctly.

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

Question 2

Hint 1: notice that the numerator has a polynomial that is one order less than the denominator's polynomial

Hint 2: hence the answer will involve the natural logarithm of the denominator

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

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

Question 3

3a) Hint 1: decide which row or column you wish to use. It may be helpful to pick the second or third rows or columns, as they have zeroes in them

3a) Hint 2: take great care with all of the negative signs, multiplications, additions and subtractions

3b) Hint 3: know that A-1 exists only if det(A) is non-zero

3b) Hint 4: so, consider setting the answer from part (a) to be equal to zero, to find whether any value of x would make det(A) = 0

3b) Hint 5: after rearranging, you should obtain x² = -2

3b) Hint 6: decide what you can conclude about x

3b) Hint 7: decide what this implies for det(A) and whether it could be zero

3b) Hint 8: write a clear statement about your conclusion

3b) Hint 9: and here is a video of the solution:

Question 4

Hint 1: know that you want to work out y'(x) to obtain the gradient

Hint 2: notice that the given equation would be challenging to make y(x) the subject, so you will have to use implicit differentiation

Hint 3: after implicitly differentiating the equation, do not attempt to make y'(x) the subject just yet

Hint 4: substitute the values of x = 0 and y = 0 into the equation

Hint 5: you should find that many terms disappear, meaning that y'(x) can be more easily made the subject

Hint 6: clearly state the value of the gradient of the tangent at (0, 0)

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

Question 5

5a) Hint 1: first write out the general term for (a + b)n in terms of r, where 0 ≤ r ≤ n and r is an integer

5a) Hint 2: now replace a with 3x, replace b with -2/x² and replace n with 8

5a) Hint 3: separate all of the terms out so that the terms in x can be gathered together

5b) Hint 4: for the gathered terms in x, the power of x will be an expression in terms of r

5b) Hint 5: set this expression equal to -1 and solve for r

5b) Hint 6: then substitute the value of r back into the relevant parts of the answer from (a) to obtain the value of the coefficient

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

Question 6

6a) Hint 1: use the standard Eulidean Algorithm, that starts off with 703 = 1 × 399 + 304

6b) Hint 2: use the extended Euclidean Algorithm to 'reverse the process' from part (a)

6b) Hint 3: the first line will be 19 = 304 - 3 × 95

6b) Hint 4: the final line ought to be 19 = 4 × 703 - 7 × 399

6b) Hint 5: by comparing the form in the question, clearly state the values of a and b, watching carefully for the signs

6c) Hint 6: notice that 76 = 4 × 19, so this equation is a multiple of that obtained in part (b)

6c) Hint 7: multiply the equation from (b) through by 4, scaling up the coefficients of 703 and 399

6c) Hint 8: clearly state the values of p and q

6c) Hint 9: and here is a video of the solution:

Question 7

7a) Hint 1: recognise that an integrating factor will be required

7a) Hint 2: identify P(x) and calculate the integrating factor required

7a) Hint 3: proceed in a standard way until a constant of integration appears

7a) Hint 4: use the initial conditions of x = 0 and y = -1 to fix the value of the constant

7a) Hint 5: with the constant known, rearrange to make y the subject

7b) Hint 6: using y(x) from part (a), work out y'(x), y''(x) and y'''(x)

7b) Hint 7: substitute y'''(x) and y''(x) into the left side of the equation in part (b)

7b) Hint 8: after simplification you should obtain 36e2x

7b) Hint 9: comparing this to the right side of the equation will deliver the value of k

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

Question 8

8a) i) Hint 1: note that we have u4 = 9 and u7 = 243

8a) i) Hint 2: the formula for un = a.rn-1 can be used with each of these statements

8a) i) Hint 3: you need to solve two simultaneous equations in a and r, but they are not linear equations so adding or subtracting them won't work

8a) i) Hint 4: divide one equation by the other to eliminate a, allowing r to be calculated

8a) ii) Hint 5: using the value for r, substitute it into one of your original equations to calculate the value of a

8b) Hint 6: write down the standard formula for the sum to infinity Sn

8b) Hint 7: replace a and r with the values from part (a)

8b) Hint 8: now write down S2n by replacing 'n' with '2n' in the formula you had for Sn

8b) Hint 9: now divide the expression for S2n by the expression for Sn

8b) Hint 10: carefully simplify the expression

8b) Hint 11: re-write 32n as (3n

8b) Hint 12: re-write 1 as 1²

8b) Hint 13: notice that the numerator is the difference of two squares, so that it can be factorised

8b) Hint 14: simplify the expression to obtain the desired 3n + 1

8b) Hint 15: and here is a video of the solution:

Question 9

Hint 1: write down the first 4 powers of 9, from 90 to 9³

Hint 2: identify the largest power of 9 that is less than 572

Hint 3: so 57210 = 7 × 81 + 5

Hint 4: this can be considered as 57210 = 7 × 9² + 0 × 91 + 5 × 90

Hint 5: so 57210 = 7059

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

Question 10

Hint 1: know that to differentiate this function, the power of 5x² will cause problems

Hint 2: take the natural logarithm of both sides so that 5x² can be brought down

Hint 3: use implicit differentiation, watching out for using the product rule in the process

Hint 4: rearrange to make y' the subject, and it will be in terms of x and y

Hint 5: replace the y with the original equation, to obtain y'(x) in terms of just x

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

Question 11

11a) Hint 1: split the isosceles triangle in half to obtain a right angled triangle of dimensions 90cm and 150cm

11a) Hint 2: these values can represent the ratio connecting r and h

11a) Hint 3: simplify this ratio to give r = 3h/5

11a) Hint 4: write out the formula for the volume of a cone

11a) Hint 5: substitute in r = 3h/5 and simplify to obtain the required expression for V

11b) Hint 6: note that 10 litres per second = 10,000 cm³ per second

11b) Hint 7: so dV/dt = 10000

11b) Hint 8: identify that you want dh/dt when h = 125

11b) Hint 9: starting with the formula for V from part (a), implicitly different it with respect to t

11b) Hint 10: replace dV/dt with 10000 and h with 125

11b) Hint 11: rearrange to make dh/dt the subject and evaluate

11b) Hint 12: remember to include the units of cm/s

11b) Hint 13: and here is a video of the solution:

Question 12

Hint 1: proceed with a standard method of proof by induction

Hint 2: write in words at the end the clear logic behind how the base case and the induction step show that the statement is true for all positive integers

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

Question 13

Hint 1: recognise that the method of separation of variables will be required

Hint 2: proceed with the method and when the constant of integration appears, then use m = 807 and P = 1079 to fix that constant

Hint 3: continue rearranging to make P the subject

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

Question 14

Hint 1: write out w² in terms of a, b and i

Hint 2: compare this expression to 8 + 6i, and compare the real parts and compare the imaginery parts

Hint 3: each comparison will give an equation in a and b

Hint 4: solve these equations by making one variable the subject and then substitute it into the other equation

Hint 5: after simplification, this will give a quartic in one variable

Hint 6: with careful factorising, obtain as many real solutions for one variable as you can

Hint 7: use these real solutions to obtain the corresponding solutions for the second variable

Hint 8: you should ultimately have w = 3 + i and w = -3 - i as two solutions

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

Question 15

15a) Hint 1: know that you will need to calculate f''(x) in order to obtain the coefficient of x² in the Maclaurin expansion

15a) Hint 2: first evaluate f'(0) to help obtain the coefficient of x

15a) Hint 3: differentiate f'(x) to give f''(x) using the quotient rule

15a) Hint 4: evaluate f''(0)

15a) Hint 5: assemble the values of f(0), f'(0) and f''(0) together and write down the first 3 terms of the Maclaurin expansion

15b) Hint 6: perform a standard integration by substitution

15b) Hint 7: remember to write the final answer in terms of x, and not u

15c) Hint 8: know that the integral of f'(x) is f(x)

15c) Hint 9: so f(x) = 1/2 tan-1[(x + 1)²] + c

15c) Hint 10: from part (a), we have f(x) = 1 + x/2 - x²/4 + O(x³)

15c) Hint 11: and therefore we have f(0) = 1

15c) Hint 12: using f(0) = 1 to fix the value of the constant, c

15c) Hint 13: hence write down f(x) with the value of the constant that's now known

15c) Hint 14: and here is a video of the solution:

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