A Level

A Level Maths Short Paper A

Covers Proof and Algebraic Methods, Coordinate Geometry, Sequences, Series and the Binomial Expansion and 4 more.

10 questions - 40 marks - calculator allowed

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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

Mark scheme for Question 1 [2 marks]
Question 1[2 marks]
Answer or workingMarks
differentiating each termM1
dy/dx = 12x^2 - 14xA1
Mark scheme for Question 2 [3 marks]
Question 2[3 marks]
Answer or workingMarks
using Pythagoras' theorem (a 3-4-5 triangle) to find the third side = 4M1
cos(theta) = 4/5A1
tan(theta) = 3/4A1
Final answer: cos(theta) = 4/5, tan(theta) = 3/4
Mark scheme for Question 3 [3 marks]
Question 3[3 marks]
Answer or workingMarks
the antiderivative x^3 + 2xM1
substituting the limits x = 4 and x = 1M1
69A1
Mark scheme for Question 4 [3 marks]
Question 4[3 marks]
Answer or workingMarks
completing the square on the x terms and the y termsM1
centre (3, -2)A1
radius 5A1
Final answer: centre (3, -2), radius 5
Mark scheme for Question 5 [4 marks]
Question 5[4 marks]
Answer or workingMarks
using the binomial coefficients C(6,0), C(6,1) and C(6,2), equal to 1, 6 and 15M1
including the correct powers of 2x in each termM1
the x term 12xA1
the x^2 term 60x^2, with the constant term 1 also statedA1
Final answer: 1 + 12x + 60x^2 + ...
Mark scheme for Question 6 [4 marks]
Question 6[4 marks]
Answer or workingMarks
applying the quotient rule with u' = 4 and v' = 1M1
dy/dx = 9/(x + 2)^2 after simplificationA1
substituting x = 1M1
gradient = 1A1
Final answer: dy/dx = 9/(x + 2)^2; gradient at x = 1 is 1
Mark scheme for Question 7 [5 marks]
Question 7[5 marks]
Answer or workingMarks
multiplying out to give the four terms 12 - 3*sqrt(5) + 4*sqrt(5) - 5M1
simplifying sqrt(5) x sqrt(5) to 5M1
combining the constant terms to 7A1
combining the surd terms to sqrt(5)A1
the final answer 7 + sqrt(5), so a = 7 and b = 1A1
Final answer: 7 + sqrt(5), a = 7, b = 1
Mark scheme for Question 8 [5 marks]
Question 8[5 marks]
Answer or workingMarks
substituting t = 20M1
M = 29.4 g (3 sf)A1
setting 80 e^(-0.05t) = 40 and taking logarithmsM1
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)
Mark scheme for Question 9 [5 marks]
Question 9[5 marks]
Answer or workingMarks
expanding (2 - x)^5 up to the x^2 term using binomial coefficientsM1
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^2A1
the coefficient = -160A1
Final answer: -160
Mark scheme for Question 10 [6 marks]
Question 10[6 marks]
Answer or workingMarks
forming the two equations 20 = A(2)^n and 312.5 = A(5)^nM1
dividing the equations to eliminate A, giving 312.5/20 = (5/2)^nM1
simplifying to 15.625 = 2.5^nA1
taking logarithms (or recognising 2.5^3 = 15.625) to find nM1
n = 3A1
substituting back to find A = 2.5A1
Final answer: A = 2.5, n = 3