Higher Tier - Year 11

Year 11 Paper 4: Higher Algebra and Functions

Covers indices, surds and standard form, linear algebra, quadratics and simultaneous equations, functions and rates of change, sequences and graphs, and statistics and probability.

13 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 [3 marks]

Quadratics and Simultaneous Equations

Factorise x^2 + 2x - 35.

Question 2 [3 marks]

Linear Equations and Inequalities

Expand and simplify 3(2x - 5) + 4(x + 6).

Question 3 [3 marks]

Indices, Surds and Standard Form

Simplify sqrt(8) x sqrt(18), giving your answer as an integer.

Question 4 [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 5 [4 marks]

Statistics and Probability

A bag contains 4 red counters and 6 blue counters.

A counter is drawn at random, its colour noted, and then replaced. A second counter is then drawn at random.

Find the probability that both counters are the same colour.

Question 6 [5 marks]

Functions and Rates of Change

The function h is defined by h(x) = 3/(x - 2) for x is not equal to 2.

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

Question 7 [4 marks]

Statistics and Probability

120 students were asked whether they play a musical instrument.

Of the 65 boys, 28 play an instrument. Of the 55 girls, 31 play an instrument.

A student is chosen at random from all 120 students. Find the probability that the student is a girl who does not play an instrument.

Question 8 [5 marks]

Linear Equations and Inequalities

At a cinema, 3 adult tickets and 2 child tickets cost 34 pounds in total.

2 adult tickets and 5 child tickets cost 41 pounds in total.

Find the cost of one adult ticket and the cost of one child ticket.

Question 9 [5 marks]

Quadratics and Simultaneous Equations

Complete the square for x^2 - 10x + 34.

Hence write down the minimum value of x^2 - 10x + 34.

Question 10 [6 marks]

Linear Equations and Inequalities

Solve the simultaneous equations

3x + 4y = 26

5x - 3y = 24

Question 11 [6 marks]

Sequences and Graphs

The first five terms of a quadratic sequence are 6, 15, 28, 45, 66.

Find an expression for the nth term, then use it to find the 10th term.

Question 12 [6 marks]

Functions and Rates of Change

A curve has equation y = x^3 - 4x^2 + 5.

Find the equation of the tangent to the curve at the point where x = 3.

Question 13 [6 marks]

Linear Equations and Inequalities

A cylinder has total surface area S, base radius r and height h, connected by the formula S = 2 pi r^2 + 2 pi r h.

Make r the subject of the formula, using the quadratic formula. Give your answer in terms of S, h and pi.

Model solutions

Mark scheme for Question 1 [3 marks]
Question 1[3 marks]
Answer or workingMarks
factors of -35 that sum to 2M1
identifying 7 and -5M1
(x + 7)(x - 5)A1
Mark scheme for Question 2 [3 marks]
Question 2[3 marks]
Answer or workingMarks
expanding to 6x - 15 + 4x + 24M1
collecting like x terms and constantsM1
10x + 9A1
Mark scheme for Question 3 [3 marks]
Question 3[3 marks]
Answer or workingMarks
multiplying to get sqrt(8 x 18) = sqrt(144)M1
evaluating 8 x 18 = 144M1
12A1
Mark scheme for Question 4 [4 marks]
Question 4[4 marks]
Answer or workingMarks
identifying gradient 2 from a parallel lineM1
using y = 2x + cM1
substituting (3, 4) to get 4 = 6 + cM1
y = 2x - 2A1
Mark scheme for Question 5 [4 marks]
Question 5[4 marks]
Answer or workingMarks
P(red) = 4/10 and P(blue) = 6/10, unchanged after replacementM1
P(both red) = 4/10 x 4/10 = 16/100M1
P(both blue) = 6/10 x 6/10 = 36/100M1
52/100, or 13/25A1
Final answer: 13/25
Mark scheme for Question 6 [5 marks]
Question 6[5 marks]
Answer or workingMarks
writing x = 3/(y - 2), swapping x and yM1
multiplying both sides by (y - 2) to get x(y - 2) = 3M1
expanding and rearranging to make y the subjectM1
h^-1(x) = (3 + 2x)/xA1
h^-1(1) = 5A1
Final answer: h^-1(x) = (3 + 2x)/x; h^-1(1) = 5
Mark scheme for Question 7 [4 marks]
Question 7[4 marks]
Answer or workingMarks
finding the number of girls who do not play an instrument: 55 - 31 = 24M1
identifying the total number of students is 120M1
forming the probability 24/120M1
1/5A1
Mark scheme for Question 8 [5 marks]
Question 8[5 marks]
Answer or workingMarks
forming the equations 3a + 2c = 34 and 2a + 5c = 41M1
multiplying the first equation by 5 and the second by 2 to align the c termsM1
subtracting to eliminate c, giving 11a = 88M1
a = 8 (adult ticket = 8 pounds)A1
c = 5 (child ticket = 5 pounds)A1
Final answer: adult ticket = 8 pounds, child ticket = 5 pounds
Mark scheme for Question 9 [5 marks]
Question 9[5 marks]
Answer or workingMarks
starting (x - 5)^2M1
(x - 5)^2 - 25 + 34M1
(x - 5)^2 + 9A1
recognising the square term is at least zeroM1
minimum value 9A1
Final answer: (x - 5)^2 + 9, minimum 9
Mark scheme for Question 10 [6 marks]
Question 10[6 marks]
Answer or workingMarks
multiplying the first equation by 3 to get 9x + 12y = 78M1
multiplying the second equation by 4 to get 20x - 12y = 96M1
adding the two equations to eliminate yM1
29x = 174M1
x = 6A1
y = 2A1
Final answer: x = 6, y = 2
Mark scheme for Question 11 [6 marks]
Question 11[6 marks]
Answer or workingMarks
first differences 9, 13, 17, 21M1
second difference 4, giving 2n^2M1
comparing 2n^2 with the sequence to find the linear part 3n + 1M1
nth term = 2n^2 + 3n + 1A1
substituting n = 10M1
231A1
Final answer: 2n^2 + 3n + 1; 10th term = 231
Mark scheme for Question 12 [6 marks]
Question 12[6 marks]
Answer or workingMarks
differentiating to get dy/dx = 3x^2 - 8xM1
substituting x = 3 into dy/dx to find the gradientM1
gradient = 3A1
substituting x = 3 into y = x^3 - 4x^2 + 5 to find the y-coordinateM1
y = -4, giving the point (3, -4)A1
y = 3x - 13A1
Mark scheme for Question 13 [6 marks]
Question 13[6 marks]
Answer or workingMarks
rearranging to 2 pi r^2 + 2 pi h r - S = 0M1
identifying a = 2 pi, b = 2 pi h and c = -S for the quadratic formulaM1
substituting into r = (-b +/- sqrt(b^2 - 4ac)) / (2a)M1
simplifying b^2 - 4ac to 4 pi^2 h^2 + 8 pi SM1
r = (-2 pi h +/- sqrt(4 pi^2 h^2 + 8 pi S)) / (4 pi)A1
taking the positive square root, since r must be positiveA1
Final answer: r = (-2 pi h + sqrt(4 pi^2 h^2 + 8 pi S)) / (4 pi)