Foundation Tier - Year 11

Year 11 Paper 2: Foundation Shape and Graphs

Covers indices, surds and standard form, functions and rates of change, sequences and graphs, geometry, trigonometry and Pythagoras, and mensuration and vectors.

16 questions - 60 marks - calculator allowed

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.

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Questions

Question 1 [2 marks]

Functions and Rates of Change

The function f is defined by f(x) = x^2 - 3x.

Find f(-2).

Question 2 [3 marks]

Mensuration and Vectors

A rectangular garden has length 14 m and width 9 m.

Find its perimeter.

Question 3 [3 marks]

Sequences and Graphs

Find the midpoint of the line segment joining the points (2, -5) and (8, 3).

Question 4 [3 marks]

Indices, Surds and Standard Form

Work out (2.5 x 10^4) x (6 x 10^-6).

Give your answer in standard form.

Question 5 [4 marks]

Functions and Rates of Change

The function f is defined by f(x) = 2x + 9.

Find the value of x for which f(x) = 3x - 5.

Question 6 [4 marks]

Indices, Surds and Standard Form

Simplify (3b^2 c^5)^3.

Give your answer as a single term.

Question 7 [4 marks]

Pythagoras and Trigonometry

In a right-angled triangle, the side opposite angle y is 9 cm and the side adjacent to angle y is 6.5 cm.

Find y to 1 decimal place.

Question 8 [4 marks]

Angles, Polygons and Circle Theorems

Three angles meet at a point. One angle is 130 degrees, another is 3x degrees and the third is 2x degrees.

Find the value of x, then state the size of the smallest of the three angles at the point.

Question 9 [4 marks]

Sequences and Graphs

Find the equation of the straight line that passes through the points (1, 4) and (4, 13),

giving your answer in the form y = mx + c.

Question 10 [3 marks]

Pythagoras and Trigonometry

In a right-angled triangle, angle A = 37 degrees and the hypotenuse is 14 cm.

Find the length of the side adjacent to angle A, to 1 decimal place.

Question 11 [4 marks]

Functions and Rates of Change

The function h is defined by h(x) = (x - 3)/4 for all values of x.

Find the inverse function h^-1(x).

Question 12 [4 marks]

Sequences and Graphs

The straight line L has equation y = 2x - 7.

Find the equation of the line parallel to L that passes through (3, 4).

Question 13 [4 marks]

Pythagoras and Trigonometry

In triangle PQR, angle P = 48 degrees, PQ = 15 cm and QR = 12 cm.

Given that angle R is acute, find angle R, to 1 decimal place.

Question 14 [4 marks]

Functions and Rates of Change

The functions f and g are defined by f(x) = x^2 - 4 and g(x) = 3x.

Find the values of x for which f(x) = g(x).

Question 15 [5 marks]

Sequences and Graphs

A geometric sequence has first term 40 and common ratio 1/2.

Find the 5th term of the sequence, and find the sum to infinity of the sequence.

Question 16 [5 marks]

Functions and Rates of Change

The function h is defined by h(x) = (2x + 9)/5 for all values of x.

Find h^-1(x), and hence find h^-1(3).

Model solutions

Mark scheme for Question 1 [2 marks]
Question 1[2 marks]
Answer or working	Marks
substituting x = -2 into f(x) = x^2 - 3x	M1
f(-2) = 10	A1
Final answer: 10

Mark scheme for Question 2 [3 marks]
Question 2[3 marks]
Answer or working	Marks
adding two lengths and two widths, or using 2(l + w)	M1
2(14 + 9)	M1
46 m	A1

Mark scheme for Question 3 [3 marks]
Question 3[3 marks]
Answer or working	Marks
adding the x-coordinates and dividing by 2	M1
adding the y-coordinates and dividing by 2	M1
(5, -1)	A1

Mark scheme for Question 4 [3 marks]
Question 4[3 marks]
Answer or working	Marks
2.5 x 6 = 15	M1
10^4 x 10^-6 = 10^-2, then adjusting 15 x 10^-2	M1
1.5 x 10^-1	A1

Mark scheme for Question 5 [4 marks]
Question 5[4 marks]
Answer or working	Marks
forming the equation 2x + 9 = 3x - 5	M1
collecting x terms, such as 9 + 5 = 3x - 2x	M1
simplifying to x = 14	M1
x = 14	A1

Mark scheme for Question 6 [4 marks]
Question 6[4 marks]
Answer or working	Marks
cubing the coefficient: 3^3 = 27	M1
multiplying the power of b: 2 x 3 = 6	M1
multiplying the power of c: 5 x 3 = 15	M1
27b^6 c^15	A1

Mark scheme for Question 7 [4 marks]
Question 7[4 marks]
Answer or working	Marks
using tan y = opposite/adjacent	M1
tan y = 9/6.5	M1
y = tan^-1(9/6.5)	M1
54.2 degrees	A1

Mark scheme for Question 8 [4 marks]
Question 8[4 marks]
Answer or working	Marks
using angles around a point sum to 360 degrees	M1
130 + 3x + 2x = 360	M1
solving to x = 46	M1
the smallest angle = 2 x 46 = 92 degrees	A1
Final answer: x = 46; smallest angle = 92 degrees

Mark scheme for Question 9 [4 marks]
Question 9[4 marks]
Answer or working	Marks
finding the gradient (13 - 4)/(4 - 1) = 3	M1
substituting (1, 4) into y = 3x + c	M1
solving to find c = 1	M1
y = 3x + 1	A1

Mark scheme for Question 10 [3 marks]
Question 10[3 marks]
Answer or working	Marks
using cos A = adjacent/hypotenuse	M1
adjacent = 14 x cos 37	M1
11.2 cm	A1

Mark scheme for Question 11 [4 marks]
Question 11[4 marks]
Answer or working	Marks
writing x = (y - 3)/4, swapping x and y	M1
multiplying both sides by 4 to get 4x = y - 3	M1
rearranging to make y the subject, y = 4x + 3	M1
h^-1(x) = 4x + 3	A1

Mark scheme for Question 12 [4 marks]
Question 12[4 marks]
Answer or working	Marks
identifying gradient 2 from a parallel line	M1
using y = 2x + c	M1
substituting (3, 4) to get 4 = 6 + c	M1
y = 2x - 2	A1

Mark scheme for Question 13 [4 marks]
Question 13[4 marks]
Answer or working	Marks
using the sine rule QR/sin P = PQ/sin R	M1
sin R = 15 x sin 48 / 12	M1
sin R = 0.9289...	M1
68.3 degrees	A1

Mark scheme for Question 14 [4 marks]
Question 14[4 marks]
Answer or working	Marks
forming the equation x^2 - 4 = 3x	M1
rearranging to x^2 - 3x - 4 = 0	M1
factorising to (x - 4)(x + 1) = 0	M1
x = 4 or x = -1	A1

Mark scheme for Question 15 [5 marks]
Question 15[5 marks]
Answer or working	Marks
using the nth term formula a r^(n-1)	M1
5th term = 40 x (1/2)^4	M1
2.5	A1
using the sum to infinity formula a/(1 - r)	M1
sum to infinity = 80	A1
Final answer: 5th term = 2.5, sum to infinity = 80

Mark scheme for Question 16 [5 marks]
Question 16[5 marks]
Answer or working	Marks
writing x = (2y + 9)/5, swapping x and y	M1
multiplying both sides by 5 to get 5x = 2y + 9	M1
rearranging to make y the subject, y = (5x - 9)/2	M1
h^-1(x) = (5x - 9)/2	A1
h^-1(3) = 3	A1
Final answer: h^-1(x) = (5x - 9)/2, h^-1(3) = 3