Hints offered by N Hopley, with video solutions by 'DLBmaths'.
Click here to start/reset.
Paper 1
Question 1
Hint 1: know that equal roots means that the discriminant is equal to zero
Hint 2: identify the values of a, b and c from the equation
Hint 3: substitute values into the equation b² - 4ac = 0
Hint 4: simplify and solve for k
Hint 5: and here is a video of the solution:
Question 2
Hint 1: recognise that differentiation of f(x) will require to use the chain rule
Hint 2: after obtaining f'(x), substitute in the value of x = 1
Hint 3: and here is a video of the solution:
Question 3
Hint 1: know that the inverse function, f-1(x), satisfies f(f-1(x)) = x
Hint 2: use your composition of functions skills to write f(f-1(x)) in terms of f-1(x)
Hint 3: rearrange the resulting equation to make f-1(x) the subject
Hint 4: and here is a video of the solution:
Question 4
Hint 1: use the points (-4, 2) and (2, -7) to calculate the gradient, m1, of the line passing through them
Hint 2: rearrange the equation 3y = 2x + 9 to make y the subject
Hint 3: read off the gradient, m2, of this line's equation
Hint 4: with the two gradients obtained, check to see if m1 × m2 = -1
Hint 5: clearly communicate your conclusion, giving your reason for it
Hint 6: and here is a video of the solution:
Question 5
5a) Hint 1: recognise that for each right angled triangle, you will need to calculate its missing length, using Pythagoras' Theorem
5a)i) Hint 2: know that sin(p) = 3/hypotenuse
5a)ii) Hint 3: know that cos(q) = adjacent/hypotenuse
5b) Hint 4: use trigonometric addition formulae to expand cos(p + q) in terms of sin(p), cos(p), sin(q) and cos(q)
5b) Hint 5: use your workings from part (a) to obtain values for cos(p) and sin(q)
5b) Hint 6: substitute in the values for all the trigonometric ratios, helped by them all having a denominator of √10
5b) Hint 7: carefully multiply and subtract the fractions to obtain a single fractional answer
5b) Hint 8: and here is a video of the solution:
Question 6
6a) Hint 1: know that f(g(x)) = f(x² - 2x), in this case
6a) Hint 2: after applying the function f, expand the brackets and simplify the expression
6b) Hint 3: know that g(f(x)) = g(2x + 5), in this case
6b) Hint 4: after applying the function g, expand the brackets and simplify the expression, taking care with the signs
6c) Hint 5: when substituting expressions for g(f(x)) and f(g(x)), remember to put the expression for f(g(x)) in brackets to ensure that the whole expression is subtracted, and not just the first term
6c) Hint 6: after simplifying the expression to a quadratic, factorise out the common factor of 2
6c) Hint 7: with the quadratic expression inside the brackets, complete the square to obtain an expression of the form (x + . . .)² - . . .
6c) Hint 8: multiply this expression by the 2, to obtain the required form of the answer
6c) Hint 9: and here is a video of the solution:
Question 7
Hint 1: recognise that integrating cos(. . .) will give sin(. . .)
Hint 2: be aware that the differentiation of sin(3x + π/4) by the chain rule will generate a multiplier of 3
Hint 3: introduce the fraction 1/3 to 'compensate' for the multiplier of 3
Hint 4: it's always a good idea to check your final answer by differentiating it, to see if you obtain the integrand in the question
Hint 5: and here is a video of the solution:
Question 8
Hint 1: know that m= tan(θ)
Hint 2: know that tan(2π/3) = tan(-π/3)
Hint 3: know that tan(-π/3) = -tan(π/3)
Hint 4: use exact value triangle knowledge to write tan(π/3) as a surd
Hint 5: from y = mx + c, you now know the value of m
Hint 6: the line goes through (4, 0), so replace the values of x and y with these, to determine the value of c
Hint 7: and here is a video of the solution:
Question 9
9a) Hint 1: to find the intersection of the functions, make them equal to each other. i.e. x³ - 7x² + 12x + 3 = x³ - x² - 6x + 3
9a) Hint 2: gather and simplify terms to obtain an expression that is then equal to zero
9a) Hint 3: your expression should contain terms in x² and x only
9a) Hint 4: factorise an x out of this expression
9a) Hint 5: solving the resulting equation will give two values for x. One will correspond with the intersection on the y-axis and the other for point A.
9b) Hint 6: the shaded area will involve integrating the difference in the two functions, between the limits of 0 and your answer from part (a)
9b) Hint 7: be careful about which function is to be subtracted from which. If it helps, do 'upper - lower'
Hint 8: and here is a video of the solution:
Question 10
Hint 1: the possible roots of the cubic are the factors of 4. ie ±1, ±2, ±4
Hint 2: if f(x) = 6x³ - 13x² + 4, then evaluate each of f(-1), f(1), f(-2), f(2), etc to find one where f(x) = 0
Hint 3: you should find that f(2) = 0, which means that (x - 2) is a factor
Hint 4: use synthetic devision or polynomial long division to divide f(x) by (x - 2)
Hint 5: be careful to insert a '0x' term into your calculations to account for the 'missing linear term'
Hint 6: present a fully factorised answer of the form (x - 2)( . . . )( . . .)
Hint 7: and here is a video of the solution:
Question 11
11a) Hint 1: notice that the coefficient of f(x) is a positive number, so the maximum of f(x) will correspond with the maximum of g(x)
11b)i) Hint 2: know that the maximum value of f(x) will remain unchanged if its graph is translated horizontally
11b)i) Hint 3: so the maximum value of f(x) = maximum value of f(x - 4)
11b)ii) Hint 4: we were told at the start of the question that the maximum of f(x) is when x = 6
11b)ii) Hint 5: hence f(x - 4) corresponds with f(6)
11b)ii) Hint 6: so x - 4 = 6, and now solve for x
Hint 7: and here is a video of the solution:
Question 12
12a)i) Hint 1: vector AB = vector OB - vector OA
12a)i) Hint 2: when calculating the magnitude of vector AB, be sure to use brackets carefully. i.e. use (-6)² and not -6²
12a)ii) Hint 3: the ratio will be AB:BC, and you can use the information from the question, and your answer from (a)(i) to replace these line segments with numbers
12b) Hint 4: sketch a diagram showing a line with A, B and C on it, and the ratio from part (a)(ii)
12b) Hint 5: realise that point C is reached from starting at point B and moving in the direction of vector AB, but not the full length of vector AB
12b) Hint 6: from part (a)(ii) the ratio was 3:2, so vector AB needs to be divided by 3, and then multiplied by 2 to obtain the vector BC
12b) Hint 7: after calculating vector OC, be sure to write the coordinates of C, as coordinates
12b) Hint 8: and here is a video of the solution:
Question 13
13a) Hint 1: realise that to obtain u5, you first need to calculate u6 from the value of u7
13a) Hint 2: write out the recurrence relation replacing n with 6, and u7 with 20
13a) Hint 3: solve for u6
13a) Hint 4: write out the recurrence relation replacing n with 5, and u6 with your last answer
13a) Hint 5: solve for u5
13b) Hint 6: use a standard method to calculate the limit of the recurrence relation
13b) Hint 7: and here is a video of the solution:
Question 14
Hint 1: expand the scalar product to obtain u.u + u.v
Hint 2: know that u.u = |u|.|u|.cos(0)
Hint 3: know that u.v = |u|.|v|.cos(120)
Hint 4: replace the |u| and |v| with their values from the question
Hint 5: know that cos(120) = cos(60) and use exact value triangles to write this as a fraction
Hint 6: simplify all numerical calculations for the final answer
Hint 7: and here is a video of the solution:
Question 15
Hint 1: sketch out a copy of the diagram and add in all of the points' coordinates, as well as the dimensions of each circle
Hint 2: deduce the coordinates of point P by considering how far away from point A(2, 1) it is
Hint 3: use (x - a)² + (y - b)² = r², and evaluate r²
Hint 4: and here is a video of the solution:
Question 16
Hint 1: use rules of logarithms to combine the first two terms
Hint 2: use rules of logarithms to rewrite the last term so that the coefficient of 2 is a power inside the log term
Hint 3: use rules of logarithms to combine everything so far into a single log2 term
Hint 4: simplify the arithmetic inside the log
Hint 5: notice that 8 = 2³
Hint 6: simplify the log expression to a single number
Hint 7: and here is a video of the solution:
Question 17
17a) Hint 1: know that the graph of an inverse function is the reflection of the original function in the line y = x
17a) Hint 2: on your sketch, mark the points (1, 3) and (2, 7) clearly
17b) Hint 3: know that where f-1(x) cuts the y-axis is equivalent to where f(x) cuts the x-axis
17b) Hint 4: cutting the x-axis means f(x) = 0
17b) Hint 5: write down 0 = log5(x - 2) + 1
17b) Hint 6: subtract 1 from both sides of the equation
17b) Hint 7: re-write the logarithmic equation as one involving exponentials
17b) Hint 8: solve for x, which will be a fractional answer
17b) Hint 9: write down the coordinates of this point, that is on the graph of y = f(x)
17b) Hint 10: write down the corresponding coordinates that will be on the graph of y = f-1(x)
17b) Hint 11: and here is a video of the solution:
Paper 2
Question 1
Hint 1: know that y needs to be differentiated and evaluated at x = 3 to obtain the gradient of the tangent
Hint 2: evaluate y at x = 3 to obtain the y-coordinate of the point
Hint 3: use the coordinates and the gradient to obtain the equation of the tangent
Hint 4: and here is a video of the solution:
Question 2
Hint 1: recognise that the integrand needs to be re-written with a negative power before it can be integrated
Hint 2: know that the new power will be (-3/2) + 1
Hint 3: when differentiating the answer, know that the power of -1/2 will need to be 'compensated' by multiplying by -2
Hint 4: remember to include the constant of integration
Hint 5: and here is a video of the solution:
Question 3
Hint 1: know that differentiating h(t) will require the use of the chain rule
Hint 2: when evaluating h(10), remember that the angle is measured in radians, and not degrees
Hint 3: and here is a video of the solution:
Question 4
4a) Hint 1: calculate the gradient of AC, mAC, using the coordinates of A and C
4a) Hint 2: use the coordinates of B and mAC to obtain the equation of L1
4b) Hint 3: notice that AB is horizontal, so a perpendicular line will be vertical, and therefore have an equation of the form x = . . .
4b) Hint 4: find the midpoint of AB, and use its x-coordinate to obtain the equation of L2
4c) Hint 5: L1 meets L2 when the value of x from L2 is substituted into the equation of L1
4c) Hint 6: solve for y, and write the final answer as a set of coordinates
Hint 7: and here is a video of the solution:
Question 5
5a) Hint 1: use a standard method for the wave function to re-write the expression with a single trigonometric function
5b)i) Hint 2: know that the minimum value of f(t) will come from knowing the minimum value of sin(t + a)
5b)i) Hint 3: the minimum value of sin(t + a) is -1
5b)ii) Hint 4: know that the minimum value of sin(x) happens when x = 270°
5b)ii) Hint 5: equate t + a = 270, using the value for 'a' from part (a)
Hint 6: and here is a video of the solution:
Question 6
6a) Hint 1: know, and write down, that a stationary point is when f'(x) = 0
6a) Hint 2: differentiate f(x) and equate it to zero to solve for x
6b) Hint 3: the shaded area will be the integral of f(x) between the limits of the answer from part (a) and 9
6b) Hint 4: take great care when integrating the fractional index terms, as well as substituting in the values of the limits
Hint 5: and here is a video of the solution:
Question 7
Hint 1: on the sketch of y = f(x) that is given, annotate which sections of the graph have a positive gradient, a zero gradient and a negative gradient
Hint 2: the two stationary points with x coordinates of -1 and 3 will become zeros on the graph on y = f'(x)
Hint 3: notice that the gradient is positive to the left and the right of the x = 3 stationary point. This will give a minimum on the graph of y = f'(x)
Hint 4: be sure to include on your sketch of y = f'(x) the values of both -1 and 3
Hint 5: and here is a video of the solution:
Question 8
Hint 1: subtract 1 from both sides of the equation
Hint 2: divide both sides of the equation by 2
Hint 3: apply the inverse sine to both sides of the equation
Hint 4: know that there are several values of sin-1(-½). You should list at least three of them, starting from -30°
Hint 5: these values represent the values of 3x - 60
Hint 6: add 60 to each value
Hint 7: then divide each answer by 3
Hint 8: check that these values of x are all in the domain of between 0 and 180.
Hint 9: you should have three values for which this is true, so discard any values that are larger than 180 or smaller than 0
Hint 10: and here is a video of the solution:
Question 9
9a) Hint 1: know that surface area of cylinder = 2 × circle + rectangle
9a) Hint 2: the rectangle has dimensions: radius × circumference
9a) Hint 3: know that the volume of a cyclinder = base area × height
9a) Hint 4: in the formula V = πr²h, replace the V with 450, and rearrange to make h the subject
9a) Hint 5: substitute your formula for h into the Surface Area expression and simplify to obtain the given expression for A(r)
9b) Hint 6: recognise that we need to calculate A'(r)
9b) Hint 7: rewrite the fractional term of A(r) with a negative index, making it ready to differentiate
9b) Hint 8: after differentiating A(r), re-write the negative index term back as a fraction
9b) Hint 9: write down that a stationary point is when A'(r) = 0
9b) Hint 10: make A'(r) equal to zero and solve for r, which will eventually involve taking a cube root
9b) Hint 11: check that A'(r) is a minimum for the value of r just worked out, by either using a nature table or the second derivative
9b) Hint 12: clearly communicate the value of r that gives the minimum surface area
9b) Hint 13: and here is a video of the solution:
Question 10
10a) Hint 1: start with the left hand side of the identity and try to make it look like the right hand side
10a) Hint 2: know that tan(x) = sin(x)/cos(x)
10a) Hint 3: recognise that 2sin(x)cos(x) is a double angle formula
10b) Hint 4: realise that you will need to integrate the given expression for dy/dx to obtain an expression for y
10b) Hint 5: notice that the given expression is just 3 times the expression given in part (a)
10b) Hint 6: proceed to integrate 3sin(2x)
10b) Hint 7: fix the value of the constant of integration by using the information that when x = 0, y = 3
10b) Hint 8: assemble the final expression for y(x)
10b) Hint 9: and here is a video of the solution:
Question 11
11a) Hint 1: know that vector AB = vector OB - vector OA, and vector AC = vector OC - vector OA
11b) Hint 2: after sketching a diagram of triangle ABC, realise that the scalar product will help calculate the required angle
11b) Hint 3: after working about the magnitudes of vectors AB and AC, proceed to use the scalar product to work out angle BAC
11b) Hint 4: and here is a video of the solution:
Question 12
Hint 1: recognise that uk+1 - uk = 1000
Hint 2: notice that uk+1 = 9uk - 440
Hint 3: substitute this expression for uk+1 into the equation uk+1 - uk = 1000
Hint 4: solve this equation for uk
Hint 5: and here is a video of the solution:
Question 13
13a) Hint 1: notice that vector CF = vector CB + vector BF
13a) Hint 2: realise that vector CB = - (vector BC)
13b) Hint 3: notice that vector DF = vector DC + vector CF
13b) Hint 4: realise that vector DC = ½ vector AB
13c) Hint 5: notice that vector QF = vector QD + vector DF
13c) Hint 6: in this equation, replace vectors QF and DF with the given information, and your answer from part (b)
13c) Hint 7: solve the equation for vector QD
Hint 8: and here is a video of the solution:
Question 14
Hint 1: from the given equation of a circle, we need to determine the coordinates of its centre
Hint 2: use a factorising method, or one of the circle formula, to obtain the centre's coordinates
Hint 3: using the coordinates of A and those of the circle's centre, work out the gradient of the line joining them
Hint 4: the tangent line will have a gradient that is the negative reciprocal of the gradient just calculated
Hint 5: use the point A(3, 5) and the gradient (which should be a positive fraction) to determine the equation of the tangent line
Hint 6: and here is a video of the solution:
Question 15
Hint 1: substitute the equation of the line into the equation of the circle, to obtain an equation that is only in terms of x
Hint 2: expand all brackets and carefully simplify terms to obtain a quadratic that is equal to zero
Hint 3: factorise out a 5 from all terms, then factorise the remaining quadratic
Hint 4: this will give two values of x, one for each of the points Q and P
Hint 5: use the equation of the line to work out the y-values for each of the x-values
Hint 6: clearly communicate the coordinates of each of points Q and P
Hint 7: and here is a video of the solution:
Question 16
Hint 1: calculate the gradient of the line through (0,2) and (6, 4)
Hint 2: this means that log8y = mx + c, and replace m and c with the answer just calculated, and using the graph provided
Hint 3: re-write the logarithmic equation as one involving exponentials
Hint 4: know from laws of indices that 8p+q = 8p × 8q
Hint 5: know from laws of indices that 8pq = (8p)q
Hint 6: know that a fractional power means taking a root of some sort Hint 7: compare the final expression for y with the one given in the question, of y = abx Hint 8: clearly state the values of a and of b Hint 9: and here is a video of the solution:
Did this hint help?