f(x) = x3 - 3x2 + 4. Use the factor theorem to state whether (x - 2) is a factor of f(x).
(Total for Question 1 is 1 mark)
2
Find the remainder when f(x) = x3 + 2x2 - 5x + 1 is divided by (x - 1).
(Total for Question 2 is 1 mark)
3
f(x) = x3 - 4x + 9. Find f(-2).
(Total for Question 3 is 1 mark)
4
State the values of x for which 7 / ((x-2)(x+5)) is not defined.
(Total for Question 4 is 1 mark)
5
Solve |x - 4| = 7.
(Total for Question 5 is 1 mark)
6
Simplify fully (x2 - 9) / (x - 3), for x ≠ 3.
(Total for Question 6 is 1 mark)
7
f(x) = 2x - 5, g(x) = x + 3. Find fg(x).
(Total for Question 7 is 1 mark)
8
f(x) = 2x - 5. Find f-1(x).
(Total for Question 8 is 1 mark)
9
Fully factorise x2 - 5x + 6.
(Total for Question 9 is 1 mark)
10
State the domain of f(x) = 1/(x-4), where x is real.
(Total for Question 10 is 1 mark)
11
f(x) = x3 + x2 - 10x + 8. Show that (x - 2) is a factor of f(x).
(Total for Question 11 is 2 marks)
12
Find the remainder when g(x) = 2x3 - 5x2 + 3 is divided by (x + 2).
(Total for Question 12 is 2 marks)
13
Express (5x + 7) / ((x-1)(x+2)) in the form A/(x-1) + B/(x+2), stating the values of A and B.
(Total for Question 13 is 2 marks)
14
f(x) = (x - 3)2 + 5, for x ≥ 3. State the range of f.
(Total for Question 14 is 2 marks)
15
f(x) = x3 - 2x2 - 5x + 6.
(a)Show that (x - 1) is a factor of f(x).(2)
(b)Hence fully factorise f(x).(2)
(Total for Question 15 is 4 marks)
16
h(x) = x3 + kx2 - 7x - 2, where k is a constant. Given that (x + 2) is a factor of h(x), find the value of k.
(Total for Question 16 is 3 marks)
17
f(x) = 2x3 + ax2 - 3x + b, where a and b are constants. When f(x) is divided by (x - 2), the remainder is 15. When f(x) is divided by (x + 1), the remainder is -6. Find the values of a and b.
(Total for Question 17 is 4 marks)
18
Functions f and g are defined by f(x) = 3x/(x-2), x is real, x not equal to 2, and g(x) = x - 1, x is real.
(a)Find fg(x), simplifying your answer fully.(2)
(b)Find f-1(x), stating its domain.(2)
(Total for Question 18 is 4 marks)
19
Solve the equation |3x - 2| = x + 4.
(Total for Question 19 is 3 marks)
20
f(x) = 2x2 - 12x + 7, for x ≤ 1.
(a)Express f(x) in the form a(x-p)2 + q, stating the values of a, p and q.(2)
(b)State the range of f, given the domain x ≤ 1.(2)
(Total for Question 20 is 4 marks)
Mark scheme · P11D Pure: Algebra and Functions Depth: Fluency and Exam Drill
Question 1
B1 f(2) = 0, so (x-2) is a factor
Answer: Yes, (x-2) is a factor
Question 2
B1 remainder = -1 cao
Answer: -1
Question 3
B1 f(-2) = 9 cao
Answer: 9
Question 4
B1 x = 2 and x = -5
Answer: x = 2, x = -5
Question 5
B1 x = 11 or x = -3, both values
Answer: x = 11 or x = -3
Question 6
B1 x + 3 cao
Answer: x + 3
Question 7
B1 2x + 1 cao
Answer: fg(x) = 2x + 1
Question 8
B1 (x+5)/2 cao
Answer: f-1(x) = (x+5)/2
Question 9
B1 (x-2)(x-3) cao
Answer: (x-2)(x-3)
Question 10
B1 x is real, x ≠ 4
Answer: x real, x ≠ 4
Question 11
M1 substitutes x = 2 into f(x)
A1 f(2) = 0, so (x-2) is a factor, cso
Answer: f(2) = 0, so (x-2) is a factor
Question 12
M1 substitutes x = -2 into g(x)
A1 remainder = -33 cao
Answer: -33
Question 13
M1 sets up A(x+2) + B(x-1) = 5x + 7 and substitutes x=1 and x=-2 (or compares coefficients)
A1 A = 4, B = 1
Answer: A = 4, B = 1
Question 14
M1 recognises the vertex of f is at (3, 5)
A1 f(x) ≥ 5 cao
Answer: f(x) ≥ 5
Question 15
(a) M1 substitutes x = 1 into f(x)
(a) A1 f(1) = 0, so (x-1) is a factor, cso
(a) Answer: f(1) = 0, so (x-1) is a factor
(b) M1 divides f(x) by (x-1) (or uses inspection) to find the quadratic factor x2 - x - 6
(b) A1 f(x) = (x-1)(x-3)(x+2) cao
(b) Answer: f(x) = (x-1)(x-3)(x+2)
Question 16
M1 substitutes x = -2 into h(x) and sets the result equal to 0
A1 forms the equation 4k + 4 = 0 oe
A1 k = -1 cao
Answer: k = -1
Question 17
M1 forms the equation f(2) = 15, i.e. 4a + b = 5 oe
M1 forms the equation f(-1) = -6, i.e. a + b = -7 oe
M1 solves the two equations simultaneously
A1 a = 4, b = -11 cao
Answer: a = 4, b = -11
Question 18
(a) M1 substitutes g(x) into f, forming 3(x-1)/((x-1)-2)
(a) A1 (3x-3)/(x-3) oe, x ≠ 3
(a) Answer: fg(x) = (3x-3)/(x-3)
(b) M1 rearranges y = 3x/(x-2) to make x the subject
(b) A1 f-1(x) = 2x/(x-3), x is real, x ≠ 3
(b) Answer: f-1(x) = 2x/(x-3), x real, x ≠ 3
Question 19
M1 solves 3x - 2 = x + 4 to get x = 3
M1 solves 3x - 2 = -(x + 4) to get x = -0.5
A1 both values checked as valid in the original equation, x = 3 or x = -0.5 cao