Year 11 Paper 6: Mixed Practice
Brings together algebraic manipulation, functions and their graphs, coordinate geometry, differentiation, applications of differentiation, trigonometric identities and equations, non-right-angled triangle trigonometry, and sequences in one longer paper.
Year 11 here means a typical teaching order, not a syllabus rule. No exam board defines what belongs to Year 11, and schools sequence the course differently. Check it against your own scheme of work before using it to decide what a class has covered.
Questions
Question 1 [2 marks]
Coordinate Geometry
Find the gradient of the straight line joining the points A(2, 3) and B(6, 11).
Question 2 [2 marks]
Sequences
An arithmetic sequence has first term 12 and common difference -3.
Find the 15th term of the sequence.
Question 3 [2 marks]
Differentiation
Find dy/dx when y = 5x^3 + 2x^2 - 9.
Question 4 [2 marks]
Trigonometric Identities and Equations
Show that 1/sin(x) - sin(x) can be simplified to cos^2(x)/sin(x).
Question 5 [3 marks]
Applications of Differentiation
A curve has equation y = x^2 - 6x + 10.
Find the equation of the tangent to the curve at the point (4, 2), giving your answer in the form y = mx + c.
Question 6 [3 marks]
Trigonometric Identities and Equations
Solve the equation 2sin(x) = 1 for 0 <= x <= 360 degrees, giving all solutions.
Question 7 [3 marks]
Functions and their Graphs
The function g is defined by g(x) = x^2 - 4 for all real x.
Find the values of x for which g(x) = 12.
Question 8 [3 marks]
Algebraic Manipulation
Factorise fully 4x^2 - 20x + 24.
Question 9 [4 marks]
Non-Right-Angled Triangle Trigonometry
In triangle MNO, MN = 11 cm, NO = 6 cm, and angle NOM = 100 degrees.
Find the size of angle NMO, giving your answer to 1 decimal place.
Question 10 [4 marks]
Differentiation
A curve has equation y = x^3 - 6x^2 + 9x + 1.
Find the coordinates of the stationary points and use the second derivative to determine the nature of each one.
Question 11 [4 marks]
Coordinate Geometry
A circle has equation (x - 2)^2 + (y + 3)^2 = 25.
State the centre and radius of the circle, then determine whether the point (5, 1) lies inside, outside, or on the circle.
Question 12 [4 marks]
Applications of Differentiation
A stone is thrown so that its height, h metres, above the ground after t seconds is given by h = 24t - 6t^2, for 0 <= t <= 4.
Find the maximum height reached by the stone and the time at which it occurs.
Question 13 [4 marks]
Sequences
The first three terms of an arithmetic sequence are 4k - 3, 6k + 1, and 10k - 3, where k is a constant.
Find the value of k, and find the common difference of the sequence.
Question 14 [4 marks]
Algebraic Manipulation
Write as a single fraction in its simplest form: 3/(x + 1) + 2/(x - 2).
Question 15 [4 marks]
Trigonometric Identities and Equations
Solve the equation 2cos^2(x) - 1 = 0 for 0 <= x <= 360 degrees, giving all solutions.
Question 16 [4 marks]
Algebraic Manipulation
Simplify fully (x^2 - x - 6)/(x^2 - 9).
Question 17 [4 marks]
Sequences
The nth term of a sequence is given by Un = 2n^2 + 3n.
Find the 6th term of the sequence, and determine whether 200 is a term of the sequence, showing your working.
Question 18 [5 marks]
Applications of Differentiation
A curve has equation y = x^2 + 3x - 5.
Find the equation of the tangent to the curve at the point where x = 2, giving your answer in the form y = mx + c.
Question 19 [5 marks]
Non-Right-Angled Triangle Trigonometry
In triangle LMN, LM = 12 cm, MN = 9 cm, and angle LMN = 55 degrees.
Find the length of LN, giving your answer to 3 significant figures, and hence find the perimeter of triangle LMN.
Question 20 [5 marks]
Algebraic Manipulation
Solve the equation 3/(x - 1) + 2/(x + 1) = 2.
Show that your equation reduces to a quadratic before solving it, and give both solutions.
Question 21 [5 marks]
Functions and their Graphs
The function h is defined by h(x) = x^2 + 8x + 19 for all real x.
Express h(x) in the form (x + a)^2 + b, where a and b are integers.
Hence state the minimum value of h(x) and the value of x at which it occurs.
Question 22 [6 marks]
Algebraic Manipulation
Write as a single fraction in its simplest form: (2x + 1)/(x^2 - 16) - 3/(x + 4).
Question 23 [6 marks]
Functions and their Graphs
f(x) = 3/(x + 1) - 2 for x not equal to -1.
Find f^-1(x), the inverse function of f, stating the value excluded from its domain.
Question 24 [6 marks]
Applications of Differentiation
The number of bacteria, N, in a culture after t hours is modelled by N = 200 + 30t^2 - 2t^3, for 0 <= t <= 10.
Find the rate of growth dN/dt, and find the time t at which the rate of growth is at its maximum, together with this maximum rate.
Question 25 [6 marks]
Coordinate Geometry
The points P(1, 6) and Q(7, -2) are given.
Find the equation of the perpendicular bisector of PQ, giving your answer in the form y = mx + c.
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| using gradient = (y2 - y1)/(x2 - x1) | M1 |
| gradient = 2 | A1 |
| Question 2[2 marks] | |
|---|---|
| Answer or working | Marks |
| using the formula a + (n - 1)d with n = 15 | M1 |
| T15 = -30 | A1 |
| Final answer: -30 | |
| Question 3[2 marks] | |
|---|---|
| Answer or working | Marks |
| differentiating each term using the power rule | M1 |
| dy/dx = 15x^2 + 4x | A1 |
| Question 4[2 marks] | |
|---|---|
| Answer or working | Marks |
| combining over the common denominator sin(x) to give (1 - sin^2(x))/sin(x) | M1 |
| using sin^2(x) + cos^2(x) = 1 to write 1 - sin^2(x) = cos^2(x), giving cos^2(x)/sin(x) | A1 |
| Final answer: cos^2(x)/sin(x), shown using sin^2(x) + cos^2(x) = 1 | |
| Question 5[3 marks] | |
|---|---|
| Answer or working | Marks |
| finding the gradient at x = 4 as dy/dx = 2 | M1 |
| using y - 2 = 2(x - 4) | M1 |
| y = 2x - 6 | A1 |
| Question 6[3 marks] | |
|---|---|
| Answer or working | Marks |
| rearranging to sin(x) = 0.5 | M1 |
| finding the principal value x = 30 degrees | M1 |
| x = 30 degrees or x = 150 degrees | A1 |
| Question 7[3 marks] | |
|---|---|
| Answer or working | Marks |
| forming the equation x^2 - 4 = 12 | M1 |
| rearranging to x^2 = 16 | M1 |
| x = 4 or x = -4 | A1 |
| Question 8[3 marks] | |
|---|---|
| Answer or working | Marks |
| taking out the common factor of 4 | M1 |
| factorising x^2 - 5x + 6 into two brackets | M1 |
| 4(x - 2)(x - 3) | A1 |
| Question 9[4 marks] | |
|---|---|
| Answer or working | Marks |
| using the sine rule MN/sin(O) = NO/sin(M) | M1 |
| rearranging to sin(M) = 6sin(100)/11 | M1 |
| calculating sin(M) = 0.537 (3 s.f.) | M1 |
| angle NMO = 32.5 degrees | A1 |
| Final answer: angle NMO = 32.5 degrees (1 d.p.) | |
| Question 10[4 marks] | |
|---|---|
| Answer or working | Marks |
| differentiating and setting dy/dx = 3x^2 - 12x + 9 = 0 to find x = 1 and x = 3 | M1 |
| the points (1, 5) and (3, 1) | A1 |
| finding the second derivative d2y/dx2 = 6x - 12 and evaluating it at each x value | M1 |
| identifying (1, 5) as a maximum and (3, 1) as a minimum | A1 |
| Final answer: (1, 5) is a maximum; (3, 1) is a minimum | |
| Question 11[4 marks] | |
|---|---|
| Answer or working | Marks |
| centre (2, -3) | B1 |
| radius 5 | B1 |
| calculating the distance from (5, 1) to the centre as sqrt(3^2 + 4^2) | M1 |
| concluding the point lies on the circle since this distance equals 5 | A1 |
| Final answer: Centre (2, -3), radius 5; the point (5, 1) lies on the circle | |
| Question 12[4 marks] | |
|---|---|
| Answer or working | Marks |
| differentiating to get dh/dt = 24 - 12t | M1 |
| setting dh/dt = 0 | M1 |
| t = 2 seconds | A1 |
| maximum height h = 24 metres | A1 |
| Final answer: maximum height 24 m, at t = 2 seconds | |
| Question 13[4 marks] | |
|---|---|
| Answer or working | Marks |
| setting up the equation (T2 - T1) = (T3 - T2) | M1 |
| forming 2k + 4 = 4k - 4 | M1 |
| k = 4 | A1 |
| common difference = 12 | A1 |
| Final answer: k = 4; common difference = 12 | |
| Question 14[4 marks] | |
|---|---|
| Answer or working | Marks |
| using the common denominator (x + 1)(x - 2) | M1 |
| writing 3(x - 2) + 2(x + 1) as the numerator | M1 |
| simplifying the numerator to 5x - 4 | A1 |
| (5x - 4)/((x + 1)(x - 2)) | A1 |
| Question 15[4 marks] | |
|---|---|
| Answer or working | Marks |
| rearranging to cos^2(x) = 0.5 | M1 |
| taking the square root to give cos(x) = +/- 1/sqrt(2) | M1 |
| finding at least two correct solutions | M1 |
| x = 45, 135, 225, 315 degrees | A1 |
| Question 16[4 marks] | |
|---|---|
| Answer or working | Marks |
| factorising the numerator as (x - 3)(x + 2) | M1 |
| factorising the denominator as (x - 3)(x + 3) | M1 |
| cancelling the common factor (x - 3) | M1 |
| (x + 2)/(x + 3) | A1 |
| Question 17[4 marks] | |
|---|---|
| Answer or working | Marks |
| substituting n = 6 into Un = 2n^2 + 3n | M1 |
| the 6th term = 90 | A1 |
| forming the equation 2n^2 + 3n - 200 = 0 and using the discriminant | M1 |
| concluding 200 is not a term since n is not a positive integer, as the discriminant 1609 is not a perfect square | A1 |
| Final answer: 6th term = 90; 200 is not a term of the sequence | |
| Question 18[5 marks] | |
|---|---|
| Answer or working | Marks |
| finding y = 5 at x = 2 | M1 |
| differentiating to get dy/dx = 2x + 3 | M1 |
| substituting x = 2 to find the gradient = 7 | M1 |
| using y - 5 = 7(x - 2) | M1 |
| y = 7x - 9 | A1 |
| Question 19[5 marks] | |
|---|---|
| Answer or working | Marks |
| using the cosine rule LN^2 = LM^2 + MN^2 - 2(LM)(MN)cos(55) | M1 |
| substituting to get LN^2 = 144 + 81 - 216cos(55) | M1 |
| LN = 10.1 cm (3 s.f.) | A1 |
| adding LM + MN + LN | M1 |
| perimeter = 31.1 cm (3 s.f.) | A1 |
| Final answer: LN = 10.1 cm (3 s.f.); perimeter = 31.1 cm (3 s.f.) | |
| Question 20[5 marks] | |
|---|---|
| Answer or working | Marks |
| multiplying both sides by (x - 1)(x + 1) | M1 |
| expanding to 3(x + 1) + 2(x - 1) = 2(x - 1)(x + 1) | M1 |
| rearranging to 2x^2 - 5x - 3 = 0 | M1 |
| solving the quadratic, e.g. by factorising as (2x + 1)(x - 3) = 0 | M1 |
| x = 3 or x = -0.5 | A1 |
| Question 21[5 marks] | |
|---|---|
| Answer or working | Marks |
| attempting to complete the square on x^2 + 8x | M1 |
| (x + 4)^2 as the squared term | M1 |
| (x + 4)^2 + 3 | A1 |
| identifying the minimum value as 3 | B1 |
| stating this occurs at x = -4 | B1 |
| Final answer: (x + 4)^2 + 3; minimum value 3 at x = -4 | |
| Question 22[6 marks] | |
|---|---|
| Answer or working | Marks |
| factorising x^2 - 16 as (x - 4)(x + 4) | M1 |
| identifying the common denominator (x - 4)(x + 4) | M1 |
| writing 3/(x + 4) as 3(x - 4)/((x - 4)(x + 4)) | M1 |
| combining the numerators to give (2x + 1) - 3(x - 4) | M1 |
| simplifying the numerator to 13 - x | A1 |
| (13 - x)/((x - 4)(x + 4)) | A1 |
| Question 23[6 marks] | |
|---|---|
| Answer or working | Marks |
| adding 2 to both sides to give y + 2 = 3/(x + 1) | M1 |
| taking reciprocals of both sides to give 1/(y + 2) = (x + 1)/3 | M1 |
| multiplying both sides by 3 to give 3/(y + 2) = x + 1 | M1 |
| subtracting 1 to give x = 3/(y + 2) - 1 | M1 |
| swapping x and y to give f^-1(x) = 3/(x + 2) - 1 | A1 |
| stating x = -2 is excluded from the domain of f^-1 | B1 |
| Final answer: f^-1(x) = 3/(x + 2) - 1, x not equal to -2 | |
| Question 24[6 marks] | |
|---|---|
| Answer or working | Marks |
| differentiating N to get dN/dt = 60t - 6t^2 | M1 |
| differentiating again to get d2N/dt2 = 60 - 12t | M1 |
| setting d2N/dt2 = 0 to find the stationary point of the rate | M1 |
| t = 5 (hours) | A1 |
| substituting t = 5 into dN/dt | M1 |
| maximum rate = 150 bacteria per hour | A1 |
| Final answer: dN/dt = 60t - 6t^2; rate of growth is greatest at t = 5 hours, when dN/dt = 150 | |
| Question 25[6 marks] | |
|---|---|
| Answer or working | Marks |
| using the midpoint formula | M1 |
| the midpoint of PQ, (4, 2) | A1 |
| finding the gradient of PQ as -4/3 | M1 |
| using the perpendicular gradient rule to find gradient 3/4 | M1 |
| using y - 2 = (3/4)(x - 4) | M1 |
| y = (3/4)x - 1 | A1 |