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 C30, giving your answer to an appropriate accuracy
13b) Hint 3: from the question, notice that Ct = 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
Hint 10: and here is a video of the solution: