Algebra and Functions

13 questions, 99 marks, every one with its mark scheme.

Or open them one at a time, as you finish each question.

Question 1

6 marks

Write each expression below as a single term kxⁿ, with k and n simplified constants.

1(a)[2 marks]

52x√x

Mark scheme

x√x = x^3/2 or 1x^3/2 = x^−3/2M1
52x^−3/2A1

Do not accept: 52x^3/2 (a power of x left in the denominator)

1(b)[2 marks]

(16x⁶)^3/4

Mark scheme

16^3/4 = 8 or (x⁶)^3/4 = x^9/2M1
8x^9/2A1

Do not accept: 8x⁴√x (a root left in the term)

1(c)[2 marks]

(2x^1/3)³4√x

Mark scheme

(2x^1/3)³ = 8x and √x = x^1/2M1
2x^1/2A1

Do not accept: 2√x

Question 2

4 marks

In this question you must show all stages of your working.

2(a)[1 mark]

Write 8√2 in the form 2^k, where k is a constant.

Mark scheme

2^7/2B1

2(b)[3 marks]

Hence solve the equation

4^{3x − 1} = (8√2)^x

Mark scheme

4^{3x − 1} = 2^{6x − 2} and (8√2)^x = 2^7x/2M1
Equates the powers: 6x − 2 = 72xdM1
x = 45A1

Question 3

8 marks

In this question you must show all stages of your working.

3(a)[2 marks]

Write √75 − √12 as k√3, with k an integer.

Mark scheme

√75 = 5√3 or √12 = 2√3M1
3√3A1

3(b)[3 marks]

Hence write

√75 − √122 + √3

in the form a√3 + b, where a and b are integers.

Mark scheme

Multiplies top and bottom by (2 − √3): 3√3(2 − √3)(2 + √3)(2 − √3)M1
6√3 − 94 − 3A1
6√3 − 9A1

Do not accept: a decimal value such as 1.39

3(c)[3 marks]

Solve the equation

2x + 5√3 = x√3 + 4

giving x as p + q√3, with p and q integers.

Mark scheme

Collects the x terms: x(2 − √3) = 4 − 5√3M1
x = 4 − 5√32 − √3 and multiplies top and bottom by (2 + √3)M1
−7 − 6√3A1

Do not accept: a decimal value such as −17.4

Question 4

6 marks

g(x) = 4x² − 24x + 41

4(a)[3 marks]

Write g(x) as p(x + q)² + r, with p, q and r integers.

Mark scheme

4(x² − 6x) + 41, so p = 4B1
4[(x − 3)² − 9] + 41M1
4(x − 3)² + 5A1

4(b)[2 marks]

Hence write down the minimum point of the graph of y = g(x), giving its coordinates.

Mark scheme

x coordinate 3B1ft
(3, 5)B1ft

4(c)[1 mark]

Write down an equation for the axis of symmetry of the graph of y = g(x).

Mark scheme

x = 3B1ft

Question 5

11 marks

5(a)[2 marks]

Solve the inequality

3(2 − x) > 4x − 15

Mark scheme

6 − 3x > 4x − 15 ⇒ 21 > 7xM1
x < 3A1

5(b)[4 marks]

Find the set of values of x for which

2x² + 7 ≤ x(x + 8)

Mark scheme

x² − 8x + 7 ≤ 0 and attempts to solve x² − 8x + 7 = 0M1
Critical values x = 1, 7A1
Chooses the inside region for their two critical valuesM1
1 ≤ x ≤ 7A1

Do not accept: x ≥ 1, x ≤ 7 as two separate inequalities

5(c)[1 mark]

Hence state the values of x satisfying both 3(2 − x) > 4x − 15 and 2x² + 7 ≤ x(x + 8)

Mark scheme

1 ≤ x < 3B1ft

5(d)[4 marks]

Find the set of values of x for which

5x < 2, x ≠ 0

Mark scheme

Multiplies both sides by x²: 5x < 2x²M1
Critical values x = 0, 52A1
Chooses the outside region for their two critical valuesM1
x < 0 or x > 52A1

Do not accept: x > 52 only (from multiplying by x); 0 > x > 52

Question 6

7 marks

After a leak from a tanker, oil spreads over the surface of the sea. The area of the oil slick, A km², t hours after the leak starts is modelled by

A = at + bt²

where a and b are constants.

Ten hours after the leak starts the area of the slick is 15 km², and twenty hours after the leak starts it is 50 km².

6(a)[4 marks]

Find the value of a and the value of b.

Mark scheme

Substitutes one data pair: 15 = 10a + 100b or 50 = 20a + 400bM1
Both equations correctA1
Solves the two equations simultaneouslyM1
a = 12, b = 110A1

6(b)[3 marks]

Using algebra, find how many hours after the leak starts the model predicts the slick will cover 30 km².

Mark scheme

Sets 12t + 110t² = 30 and rearranges: t² + 5t − 300 = 0M1
Solves their quadratic: (t + 20)(t − 15) = 0dM1
t = 15 (hours), rejecting t = −20 as the time cannot be negativeA1

Do not accept: t = −20 given as a time

Question 7

5 marks

The constant k is such that the equation

(k − 2)x² + 4x + k + 1 = 0, k ≠ 2

has two distinct real roots.

7(a)[5 marks]

Find all the possible values of k.

Mark scheme

Attempts b² − 4ac with a = k − 2, b = 4, c = k + 1M1
4² − 4(k − 2)(k + 1) > 0 ⇒ k² − k − 6 < 0A1
Solves to find critical values k = −2, 3M1
Chooses the inside region for their critical valuesM1
−2 < k < 3, k ≠ 2A1

Do not accept: k < −2 or k > 3

Question 8

6 marks

In this question you must show all stages of your working.

8(a)[3 marks]

Given that p = 5^x, show that the equation

5^{x + 1} + 5^{2 − x} = 126

can be written as

5p² − 126p + 25 = 0

Mark scheme

Uses an index law: 5^{x + 1} = 5p or 5^{2 − x} = 25pM1
Writes the equation in terms of p and multiplies through by p: 5p² + 25 = 126pM1
5p² − 126p + 25 = 0A1*

8(b)[3 marks]

Hence solve the equation

5^{x + 1} + 5^{2 − x} = 126

Mark scheme

(5p − 1)(p − 25) = 0 ⇒ p = 15, 25M1
Sets 5^x = 15 and 5^x = 25dM1
x = −1, x = 2A1

Question 9

5 marks

The curve C has equation y = 3x² − x + 4

The line l has equation y = kx + 2, where k is a constant.

9(a)[2 marks]

Show that the x coordinates of any points where l meets C satisfy the equation

3x² − (k + 1)x + 2 = 0

Mark scheme

Equates: 3x² − x + 4 = kx + 2M1
3x² − (k + 1)x + 2 = 0A1*

9(b)[3 marks]

The line l touches C. Find the exact possible values of k.

Mark scheme

Attempts b² − 4ac = 0: (k + 1)² − 4 × 3 × 2 = 0M1
Solves: k + 1 = ±√24dM1
k = −1 ± 2√6A1

Do not accept: decimal values of k; an inequality such as (k + 1)² − 24 > 0

Question 10

11 marks

f(x) = x³ − 6x² + 9x

10(a)[3 marks]

Factorise f(x) completely.

Mark scheme

x(x² − 6x + 9)B1
Attempts to factorise the quadraticM1
x(x − 3)²A1

10(b)[3 marks]

Sketch the graph of y = f(x). On your sketch, give the coordinates of every point at which it meets the axes.

Mark scheme

Cubic shape rising to the right (positive x³ coefficient)B1
Passes through (0, 0)B1
Touches the x-axis at (3, 0)B1

10(c)[3 marks]

On separate axes, sketch the curve y = f(x + 2). On your sketch, give the coordinates of every point at which it meets the axes.

Mark scheme

Same shape translated 2 units to the leftB1
Crosses the x-axis at (−2, 0) and touches it at (1, 0)B1ft
Meets the y-axis at (0, 2)B1

10(d)[2 marks]

On the same axes as your sketch in part (b), sketch the line with equation y = 4x

Hence state, giving a reason, the number of real solutions of the equation

x³ − 6x² + 9x = 4x

Mark scheme

Straight line through (0, 0) with positive gradient, crossing the curve between 0 and 3 and again beyond 3B1
3, because the two graphs intersect each other three timesB1

Question 11

11 marks

The curve C has equation

y = 6x − 1, x ≠ 1

11(a)[4 marks]

Sketch C. On your sketch, give the equation of each asymptote and the coordinates of any point at which C meets the axes.

Mark scheme

Two branches in the top right and bottom left regions formed by the asymptotesB1
x = 1B1
y = 0B1
(0, −6)B1

Do not accept: the asymptote is the x-axis

11(b)[3 marks]

The curve D has equation

y = 6x − 1 + 2, x ≠ 1

Write down the equations of the asymptotes of D, and find the coordinates of the points where D meets the coordinate axes.

Mark scheme

x = 1 and y = 2B1
Sets y = 0: 6x − 1 = −2 ⇒ x = −2M1
(−2, 0) and (0, −4)A1

11(c)[4 marks]

Using algebra, find the exact coordinates of each point at which D meets the line with equation y = x + 1

Mark scheme

Equates and multiplies by (x − 1): 6 + 2(x − 1) = (x + 1)(x − 1)M1
x² − 2x − 5 = 0A1
Solves their quadratic: (x − 1)² = 6 ⇒ x = 1 ± √6dM1
(1 + √6, 2 + √6) and (1 − √6, 2 − √6)A1

Do not accept: decimal coordinates

Question 12

8 marks

The curve with equation y = f(x) has these features:

Feature	Details
Crosses the x-axis	(−6, 0) only
Maximum point	(−2, 8)
Crosses the y-axis	(0, 5)
Horizontal asymptote	y = 3

As x decreases, y decreases without limit. As x increases, the curve approaches its asymptote from above.

12(a)[3 marks]

Sketch the curve with equation y = f(x − 2). On your sketch, give the coordinates of the maximum point and of every point at which the curve meets the axes, and the equation of the asymptote.

Mark scheme

Maximum (0, 8), which is also where the curve meets the y-axisB1
(−4, 0)B1
y = 3B1

12(b)[3 marks]

Sketch the graph of y = f(12x). On your sketch, give the coordinates of the maximum point and of every point at which the curve meets the axes, and the equation of the asymptote.

Mark scheme

Maximum (−4, 8)B1
(−12, 0) and (0, 5)B1
y = 3B1

12(c)[2 marks]

For a particular constant k, the graph of y = f(x) + k touches the x-axis.

State the value of k and the equation of the asymptote to this curve.

Mark scheme

k = −8B1
y = −5B1

Question 13

11 marks

The curve C has equation y = 14 + 3x − 2x²

The line l has equation y = x + 2

Solutions relying entirely on calculator technology are not acceptable.

13(a)[4 marks]

Find, using algebra, the coordinates of the points where l meets C.

Mark scheme

Substitutes: 14 + 3x − 2x² = x + 2M1
2x² − 2x − 12 = 0 oeA1
(x − 3)(x + 2) = 0 ⇒ x = 3, −2 and substitutes to find yM1
(−2, 0) and (3, 5)A1

13(b)[2 marks]

Using your answer to part (a), write down the values of x for which

14 + 3x − 2x² > x + 2

Mark scheme

Chooses the inside region for their critical valuesM1
−2 < x < 3A1

Do not accept: x > −2, x < 3 as two separate inequalities

13(c)[2 marks]

Find where C meets the x-axis, giving coordinates.

Mark scheme

(2x − 7)(x + 2) = 0M1
(−2, 0) and (72, 0)A1

13(d)[3 marks]

The region R, including its boundary, is the finite region above the x-axis that is bounded by l, C and the x-axis.

Use inequalities to define the region R.

Mark scheme

y ≥ 0B1
y ≤ x + 2B1
y ≤ 14 + 3x − 2x²B1

Do not accept: R ≥ 0 (or any inequality written with R in place of y)

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