A Level Maths Short Paper A
Covers Proof and Algebraic Methods, Coordinate Geometry, Sequences, Series and the Binomial Expansion and 4 more.
Questions
Question 1 [2 marks]
Differentiation
Without using a calculator, differentiate y = 4x^3 - 7x^2 + 2 with respect to x.
Question 2 [3 marks]
Trigonometry
Without using a calculator, given that sin(theta) = 3/5 where theta is acute, find the exact values of cos(theta) and tan(theta).
Question 3 [3 marks]
Integration
Evaluate the definite integral of (3x^2 + 2) with respect to x between x = 1 and x = 4.
Question 4 [3 marks]
Coordinate Geometry
A circle has equation x^2 + y^2 - 6x + 4y - 12 = 0.
Find the centre and radius of the circle.
Question 5 [4 marks]
Sequences, Series and the Binomial Expansion
Without using a calculator, find the first three terms, in ascending powers of x, of the binomial expansion of (1 + 2x)^6, simplifying each coefficient.
Question 6 [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 7 [5 marks]
Proof and Algebraic Methods
Without using a calculator, expand and simplify (3 + sqrt(5))(4 - sqrt(5)), giving your answer in the form a + b*sqrt(5), stating the values of a and b.
Question 8 [5 marks]
Exponentials and Logarithms
The mass, in grams, of a radioactive substance is modelled by M = 80 e^(-0.05t), where t is the time in days after measurements began.
Find the mass after 20 days, and find the time taken for the mass to halve, giving your answer to 1 decimal place.
Question 9 [5 marks]
Sequences, Series and the Binomial Expansion
In the expansion of (1 + 3x)(2 - x)^5 in ascending powers of x, find the coefficient of x^2.
Question 10 [6 marks]
Exponentials and Logarithms
A quantity y is modelled by y = A x^n, where A and n are constants.
Given that y = 20 when x = 2, and y = 312.5 when x = 5, find the values of A and n.
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| differentiating each term | M1 |
| dy/dx = 12x^2 - 14x | A1 |
| Question 2[3 marks] | |
|---|---|
| Answer or working | Marks |
| using Pythagoras' theorem (a 3-4-5 triangle) to find the third side = 4 | M1 |
| cos(theta) = 4/5 | A1 |
| tan(theta) = 3/4 | A1 |
| Final answer: cos(theta) = 4/5, tan(theta) = 3/4 | |
| Question 3[3 marks] | |
|---|---|
| Answer or working | Marks |
| the antiderivative x^3 + 2x | M1 |
| substituting the limits x = 4 and x = 1 | M1 |
| 69 | A1 |
| Question 4[3 marks] | |
|---|---|
| Answer or working | Marks |
| completing the square on the x terms and the y terms | M1 |
| centre (3, -2) | A1 |
| radius 5 | A1 |
| Final answer: centre (3, -2), radius 5 | |
| Question 5[4 marks] | |
|---|---|
| Answer or working | Marks |
| using the binomial coefficients C(6,0), C(6,1) and C(6,2), equal to 1, 6 and 15 | M1 |
| including the correct powers of 2x in each term | M1 |
| the x term 12x | A1 |
| the x^2 term 60x^2, with the constant term 1 also stated | A1 |
| Final answer: 1 + 12x + 60x^2 + ... | |
| Question 6[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 | |
| Question 7[5 marks] | |
|---|---|
| Answer or working | Marks |
| multiplying out to give the four terms 12 - 3*sqrt(5) + 4*sqrt(5) - 5 | M1 |
| simplifying sqrt(5) x sqrt(5) to 5 | M1 |
| combining the constant terms to 7 | A1 |
| combining the surd terms to sqrt(5) | A1 |
| the final answer 7 + sqrt(5), so a = 7 and b = 1 | A1 |
| Final answer: 7 + sqrt(5), a = 7, b = 1 | |
| Question 8[5 marks] | |
|---|---|
| Answer or working | Marks |
| substituting t = 20 | M1 |
| M = 29.4 g (3 sf) | A1 |
| setting 80 e^(-0.05t) = 40 and taking logarithms | M1 |
| t = ln(0.5)/(-0.05) | A1 |
| t = 13.9 days (1 dp) | A1 |
| Final answer: 29.4 g after 20 days; half-life = 13.9 days (1 dp) | |
| Question 9[5 marks] | |
|---|---|
| Answer or working | Marks |
| expanding (2 - x)^5 up to the x^2 term using binomial coefficients | M1 |
| 32 - 80x + 80x^2 (unsimplified terms accepted) | A1 |
| multiplying by (1 + 3x) and identifying the two contributions to the x^2 term: 1 x (80x^2) and 3x x (-80x) | M1 |
| combining to (80 - 240)x^2 | A1 |
| the coefficient = -160 | A1 |
| Final answer: -160 | |
| Question 10[6 marks] | |
|---|---|
| Answer or working | Marks |
| forming the two equations 20 = A(2)^n and 312.5 = A(5)^n | M1 |
| dividing the equations to eliminate A, giving 312.5/20 = (5/2)^n | M1 |
| simplifying to 15.625 = 2.5^n | A1 |
| taking logarithms (or recognising 2.5^3 = 15.625) to find n | M1 |
| n = 3 | A1 |
| substituting back to find A = 2.5 | A1 |
| Final answer: A = 2.5, n = 3 | |