A Level

A Level Paper 6: Full Course Review

Covers proof and algebraic methods, coordinate geometry, trigonometry, differentiation, integration, the binomial probability distribution, forces and Newton's laws, and moments.

22 questions - 100 marks - calculator allowed

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Questions

Question 1 [3 marks]

Coordinate Geometry

Find the equation of the perpendicular bisector of the line segment joining the points A(2, 5) and B(-4, 1), giving your answer in the form y = mx + c.

Question 2 [3 marks]

Probability and the Binomial Distribution

Events A and B are independent, with P(A) = 0.3 and P(B) = 0.6.

Find the probability that at least one of A and B occurs.

Question 3 [3 marks]

Integration

Evaluate the definite integral of (3x^2 - 2x + 1) with respect to x between x = -1 and x = 2.

Question 4 [3 marks]

Forces and Newton's Laws

A box of mass 4 kg rests on a horizontal surface.

Find the normal reaction force on the box. Use g = 9.8 m/s^2.

Question 5 [3 marks]

Trigonometry

Without using a calculator, find the exact value of sin(45 deg) tan(60 deg) + cos(45 deg), giving your answer in the form (sqrt(a) + sqrt(b))/2.

Question 6 [4 marks]

Proof and Algebraic Methods

The functions f and g are defined for all real x by f(x) = 2x - 1 and g(x) = x^2 + 3.

Find fg(2) and gf(2).

Question 7 [3 marks]

Moments

A see-saw pivots at its centre. A child of weight 300 N sits 2 m from the pivot on one side.

Find the distance from the pivot at which a child of weight 250 N must sit on the other side for the see-saw to balance.

Question 8 [3 marks]

Proof and Algebraic Methods

The function f is defined by f(x) = (2x - 1)/(x + 3) for x != -3.

Find f^-1(x) and state its domain.

Question 9 [4 marks]

Differentiation

A curve has equation y = (4x - 1)/(x + 2).

Find dy/dx using the quotient rule, and find the gradient of the curve at the point where x = 1.

Question 10 [5 marks]

Forces and Newton's Laws

A block of mass 8 kg is pulled along a rough horizontal surface by a horizontal force of 30 N, producing an acceleration of 0.5 m/s^2.

Find the coefficient of friction between the block and the surface. Use g = 9.8 m/s^2.

Question 11 [5 marks]

Probability and the Binomial Distribution

For two events A and B, P(A) = 0.5, P(B) = 0.3 and P(A union B) = 0.65.

Find P(A intersect B), and determine whether A and B are independent.

Question 12 [5 marks]

Trigonometry

Without using a calculator, prove the identity sin(x)/(1 - cos(x)) + sin(x)/(1 + cos(x)) = 2/sin(x).

Question 13 [5 marks]

Forces and Newton's Laws

A particle is in equilibrium under the action of three coplanar forces: a force of 20 N acting due north, a force of 15 N acting due east, and a third force F.

Find the magnitude of F, and find its direction as a bearing.

Question 14 [5 marks]

Probability and the Binomial Distribution

In a class of 30 students, 18 study French, 15 study Spanish, and 7 study both French and Spanish.

A student is selected at random. Find the probability that the student studies French but not Spanish, and find the probability that the student studies neither language.

Question 15 [5 marks]

Differentiation

A curve has equation y = x^3 - 3x + 2.

Find the equation of the normal to the curve at the point where x = 2, giving your answer in the form y = mx + c.

Question 16 [5 marks]

Forces and Newton's Laws

A block of mass 5 kg is pulled up a rough plane inclined at 20 degrees to the horizontal by a force of 40 N acting parallel to the plane. The coefficient of friction between the block and the plane is 0.3.

Find the acceleration of the block. Use g = 9.8 m/s^2.

Question 17 [6 marks]

Coordinate Geometry

A circle passes through the points A(2, 8), B(7, 3) and C(-3, 3).

Using the general equation x^2 + y^2 + 2gx + 2fy + c = 0, find the values of g, f and c, and hence state the centre and radius of the circle.

Question 18 [6 marks]

Moments

A uniform ladder AB of length 6 m and weight 180 N rests with end A on rough horizontal ground and end B against a smooth vertical wall, making an angle of 60 degrees with the ground. A person of weight 700 N stands on the ladder at a point 4 m from A.

Given that the ladder is on the point of slipping, find the coefficient of friction between the ladder and the ground.

Question 19 [6 marks]

Integration

The region R is bounded by the curve y = x^2 - 2x and the line y = 3x, between their two points of intersection.

Find the area of R.

Question 20 [6 marks]

Trigonometry

Solve 2 cos(2x) + 3 cos(x) = 0 for 0 <= x <= 360 deg, giving all solutions to 1 decimal place.

Question 21 [6 marks]

Moments

A uniform L-shaped lamina is formed from a 6 cm by 6 cm square with a 3 cm by 3 cm square removed from one corner. Taking the bottom-left corner of the original square as the origin, with the removed square occupying the top-right 3 cm by 3 cm corner, find the coordinates of the centre of mass of the lamina.

Question 22 [6 marks]

Coordinate Geometry

A circle C has centre (1, 2) and radius 10.

The line l has equation 3x - 4y - 5 = 0.

Find the length of the chord cut off on C by l.

Model solutions

Mark scheme for Question 1 [3 marks]
Question 1[3 marks]
Answer or working	Marks
finding the midpoint of AB as (-1, 3)	M1
finding the gradient of AB = 2/3, so the perpendicular gradient = -3/2	M1
y = -3/2 x + 3/2	A1

Mark scheme for Question 2 [3 marks]
Question 2[3 marks]
Answer or working	Marks
finding P(neither occurs) = P(not A) x P(not B) = 0.7 x 0.4	M1
P(neither occurs) = 0.28	A1
P(at least one occurs) = 1 - 0.28 = 0.72	A1
Final answer: 0.72

Mark scheme for Question 3 [3 marks]
Question 3[3 marks]
Answer or working	Marks
the antiderivative x^3 - x^2 + x	M1
substituting the limits x = 2 and x = -1	M1
the value 9	A1
Final answer: 9

Mark scheme for Question 4 [3 marks]
Question 4[3 marks]
Answer or working	Marks
identifying vertical equilibrium: R = weight	M1
weight = mg = 4 x 9.8	M1
R = 39.2 N	A1

Mark scheme for Question 5 [3 marks]
Question 5[3 marks]
Answer or working	Marks
stating sin(45 deg) = sqrt(2)/2, tan(60 deg) = sqrt(3) and cos(45 deg) = sqrt(2)/2	M1
computing sin(45 deg) tan(60 deg) = sqrt(6)/2	M1
the total (sqrt(6) + sqrt(2))/2	A1
Final answer: (sqrt(6) + sqrt(2))/2

Mark scheme for Question 6 [4 marks]
Question 6[4 marks]
Answer or working	Marks
finding g(2) = 7	M1
fg(2) = f(7) = 13	A1
finding f(2) = 3	M1
gf(2) = g(3) = 12	A1
Final answer: fg(2) = 13, gf(2) = 12

Mark scheme for Question 7 [3 marks]
Question 7[3 marks]
Answer or working	Marks
taking moments about the pivot	M1
forming the equation 300 x 2 = 250 x d	M1
d = 2.4 m	A1
Final answer: 2.4 m

Mark scheme for Question 8 [3 marks]
Question 8[3 marks]
Answer or working	Marks
setting y = (2x - 1)/(x + 3) and rearranging to y(x + 3) = 2x - 1	M1
collecting x terms to give x(y - 2) = -1 - 3y	M1
f^-1(x) = (1 + 3x)/(2 - x), with domain x != 2	A1
Final answer: f^-1(x) = (1 + 3x)/(2 - x), x != 2

Mark scheme for Question 9 [4 marks]
Question 9[4 marks]
Answer or working	Marks
applying the quotient rule with u' = 4 and v' = 1	M1
dy/dx = 9/(x + 2)^2 after simplification	A1
substituting x = 1	M1
gradient = 1	A1
Final answer: dy/dx = 9/(x + 2)^2; gradient at x = 1 is 1

Mark scheme for Question 10 [5 marks]
Question 10[5 marks]
Answer or working	Marks
finding the normal reaction R = mg = 78.4 N	M1
forming the equation of motion 30 - F = 8 x 0.5	M1
F = 26 N	A1
using mu = F/R	M1
mu = 0.332 (3 sf)	A1

Mark scheme for Question 11 [5 marks]
Question 11[5 marks]
Answer or working	Marks
using P(A union B) = P(A) + P(B) - P(A intersect B)	M1
rearranging to P(A intersect B) = P(A) + P(B) - P(A union B)	M1
P(A intersect B) = 0.15	A1
comparing with P(A) x P(B) = 0.5 x 0.3 = 0.15	M1
concluding A and B are independent, since P(A intersect B) = P(A) x P(B)	A1
Final answer: P(A intersect B) = 0.15; A and B are independent

Mark scheme for Question 12 [5 marks]
Question 12[5 marks]
Answer or working	Marks
writing the left-hand side as a single fraction over (1 - cos(x))(1 + cos(x))	M1
simplifying the denominator to 1 - cos^2(x)	M1
using 1 - cos^2(x) = sin^2(x)	A1
simplifying the numerator sin(x)(1 + cos(x)) + sin(x)(1 - cos(x)) to 2 sin(x)	M1
combining to 2 sin(x)/sin^2(x) = 2/sin(x), completing the proof	A1
Final answer: Identity proved: the left-hand side simplifies to 2/sin(x)

Mark scheme for Question 13 [5 marks]
Question 13[5 marks]
Answer or working	Marks
finding the resultant of the two given forces using Pythagoras: sqrt(20^2 + 15^2)	M1
the resultant magnitude = 25 N	A1
finding the angle using tan(theta) = 15/20	M1
the bearing of the resultant = 036.9 deg (1 dp) from north	A1
F equal and opposite to the resultant: magnitude 25 N, bearing 216.9 deg (1 dp)	A1
Final answer: F = 25 N on a bearing of 216.9 deg (1 dp)

Mark scheme for Question 14 [5 marks]
Question 14[5 marks]
Answer or working	Marks
finding the number studying French only = 18 - 7 = 11	M1
P(French only) = 11/30	A1
finding the number studying at least one language = 18 + 15 - 7 = 26	M1
finding the number studying neither = 30 - 26 = 4	M1
P(neither) = 4/30 = 2/15	A1
Final answer: P(French only) = 11/30; P(neither) = 2/15

Mark scheme for Question 15 [5 marks]
Question 15[5 marks]
Answer or working	Marks
finding the y-coordinate at x = 2: y = 4	M1
differentiating to get dy/dx = 3x^2 - 3	M1
the gradient at x = 2 is 9	A1
the normal gradient = -1/9 and forming y - 4 = -1/9(x - 2)	M1
y = -x/9 + 38/9 (or equivalent)	A1
Final answer: y = -x/9 + 38/9

Mark scheme for Question 16 [5 marks]
Question 16[5 marks]
Answer or working	Marks
resolving perpendicular to the plane: R = mg cos(20)	M1
R = 46.0 N (3 sf) and friction = mu R = 13.8 N (3 sf)	A1
the equation of motion along the plane: 40 - mg sin(20) - friction = 5a	M1
mg sin(20) = 16.8 N (3 sf)	A1
a = 1.89 m/s^2 (3 sf)	A1

Mark scheme for Question 17 [6 marks]
Question 17[6 marks]
Answer or working	Marks
substituting each point into the general equation to form three equations	M1
the three equations 4g + 16f + c = -68, 14g + 6f + c = -58 and -6g + 6f + c = -18 (or equivalent)	A1
eliminating c to obtain two equations in g and f	M1
g = -2 and f = -3	A1
substituting back to find c = -12	M1
the centre (2, 3) and radius 5	A1
Final answer: g = -2, f = -3, c = -12; centre (2, 3), radius 5

Mark scheme for Question 18 [6 marks]
Question 18[6 marks]
Answer or working	Marks
vertical equilibrium: R = 180 + 700 = 880 N	M1
horizontal equilibrium F = S	M1
taking moments about A: S x (6 sin(60)) = 180 x (3 cos(60)) + 700 x (4 cos(60))	M1
S = 321 N (3 sf)	A1
using F = S at limiting equilibrium with mu = F/R	M1
mu = 0.365 (3 sf)	A1

Mark scheme for Question 19 [6 marks]
Question 19[6 marks]
Answer or working	Marks
setting x^2 - 2x = 3x to find the intersections	M1
x = 0 and x = 5	A1
recognising the line lies above the curve on (0, 5) and forming the integral of (3x - (x^2 - 2x)) dx	M1
finding the antiderivative (5/2)x^2 - (1/3)x^3	M1
substituting the limits to obtain 125/6	A1
area = 125/6 square units, stated as an exact answer	A1
Final answer: Area = 125/6 square units

Mark scheme for Question 20 [6 marks]
Question 20[6 marks]
Answer or working	Marks
using cos(2x) = 2 cos^2(x) - 1	M1
substituting and rearranging to 4 cos^2(x) + 3 cos(x) - 2 = 0	M1
solving using the quadratic formula to cos(x) = (-3 + sqrt(41))/8, rejecting the root outside [-1, 1]	A1
cos(x) = 0.425 (3 sf)	A1
using inverse cosine and the symmetry x = 360 - x	M1
x = 64.8 deg or x = 295.2 deg (1 dp)	A1

Mark scheme for Question 21 [6 marks]
Question 21[6 marks]
Answer or working	Marks
treating the lamina as the full 6 by 6 square minus the 3 by 3 removed square	M1
the areas 36 and 9 (remaining area 27)	A1
the centroid of the full square (3, 3) and the centroid of the removed square (4.5, 4.5)	M1
using x_cm = (36 x 3 - 9 x 4.5)/27	M1
x_cm = 2.5	A1
y_cm = 2.5 by symmetry (the same calculation applies for y)	A1
Final answer: Centre of mass = (2.5, 2.5)

Mark scheme for Question 22 [6 marks]
Question 22[6 marks]
Answer or working	Marks
using the perpendicular distance formula \|3(1) - 4(2) - 5\| / sqrt(3^2 + (-4)^2)	M1
the perpendicular distance = 2	A1
using half the chord length = sqrt(r^2 - d^2)	M1
half the chord length = sqrt(100 - 4) = sqrt(96) = 4*sqrt(6)	A1
doubling to find the full chord length	M1
the chord length = 8*sqrt(6) (or 19.6 to 3 sf)	A1
Final answer: Chord length = 8*sqrt(6)