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HIGHER

4.5H Inequalities: Higher Tier Practice

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 75 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Write down the inequality shown on this number line.
A filled circle at -2 and an arrow to the right from -2 to infinity.

(Total for Question 1 is 1 mark)

Solve the inequality 3x - 5 < 10. Show all steps.

(Total for Question 2 is 2 marks)

Solve the pair of inequalities and give the solution as a single inequality:
2x + 1 ≥ 5 and x - 4 < 3.

(Total for Question 3 is 2 marks)

Solve the inequality 4 - 2x ≤ 10. Give your answer in the form x ≥ or x ≤.

(Total for Question 4 is 3 marks)

Solve the quadratic inequality x² - 5x + 6 < 0. Show how you find the critical values and give the solution in interval notation.

(Total for Question 5 is 4 marks)

Solve the inequality (x+1)/(x-2) ≥ 0. State clearly any values excluded from the solution and give the final solution as a single inequality or union of intervals.

(Total for Question 6 is 5 marks)

Priya has two machines. Machine A produces between 20 and 35 items per hour inclusive. Machine B produces at least 15 items per hour but fewer than 28 items per hour.
Let a be the rate of A and b be the rate of B. Write inequalities for a and b, then find the range of possible values for a + b. Give your answer as an inequality.

(Total for Question 7 is 6 marks)

Show that for all real x, if 2x - 3 < 7 then x < 5. Then determine whether the converse statement is true: "If x < 5 then 2x - 3 < 7". Give a brief justification for your answer.

(Total for Question 8 is 9 marks)

The inequality 2x² - 7x - 15 < 0 is given.
(a) Show that 2x² - 7x - 15 = (2x + 3)(x - 5) and hence solve 2x² - 7x - 15 < 0, giving your answer in inequality form.
(b) Hence, or otherwise, find how many integer values of x satisfy 2x² - 7x - 15 < 0. Explain your reasoning clearly.

(a)Show that 2x² - 7x - 15 = (2x + 3)(x - 5) and hence solve 2x² - 7x - 15 < 0, giving your answer in inequality form.(3)

(b)Hence, or otherwise, find how many integer values of x satisfy 2x² - 7x - 15 < 0. Explain your reasoning clearly.(5)

(Total for Question 9 is 8 marks)

Mark scheme · 4.5H Inequalities: Higher Tier Practice

Question 1

B1 x ≥ -2 cao
Answer: x ≥ -2

Question 2

M1 add 5 to both sides: 3x < 15 (method)
A1 x < 5 cao
Answer: x < 5

Question 3

M1 solve each: 2x ≥ 4 -> x ≥ 2; x < 7 (method for each inequality)
A1 2 ≤ x < 7 cao
Answer: 2 ≤ x < 7

Question 4

M1 subtract 4 from both sides: -2x ≤ 6 (method)
M1 divide both sides by -2 and reverse the inequality sign (method)
A1 x ≥ -3 cao
Answer: x ≥ -3

Question 5

M1 factorise: x² - 5x + 6 = (x - 2)(x - 3) (method)
M1 identify critical values x = 2 and x = 3 (method)
M1 use sign analysis on intervals (-infinity,2), (2,3), (3,infinity) to determine where expression is negative
A1 2 < x < 3 cao
Answer: 2 < x < 3

Question 6

M1 identify critical points x = -1 (numerator zero) and x = 2 (denominator zero) and state x = 2 is excluded
M1 write the sign intervals: (-infinity, -1), {-1}, (-1,2), (2, infinity) or equivalent interval partition
M1 test signs or use multiplicative signs to determine where (x+1)(x-2) is ≥ 0
M1 decide whether x = -1 is included since numerator zero gives value 0, and x = 2 is excluded
A1 final solution: (-infinity, -1] U (2, infinity) cao, with x = 2 excluded
Answer: x ≤ -1 or x > 2 (i.e. (-infinity, -1] U (2, infinity)), with x = 2 excluded

Question 7

M1 write inequalities: 20 ≤ a ≤ 35 and 15 ≤ b < 28 (method)
M1 find minimum a + b by adding minima: 20 + 15 = 35 (method)
M1 find maximum a + b by adding maxima as a bound: 35 + 28 = 63 (method)
M1 note the strictness from b < 28 makes the upper bound strict so a + b < 63 (method)
A1 state final inequality 35 ≤ a + b < 63 cao
B1 reason that endpoints 35 and 63 follow from inclusive and exclusive bounds respectively oe
Answer: 35 ≤ a + b < 63

Question 8

M1 add 3 to both sides of 2x - 3 < 7 to get 2x < 10 (method)
M1 divide both sides by 2 (positive) to give x < 5 (method)
A1 conclude x < 5 for the first part cao
M1 state that the converse must be tested: assume x < 5 (method)
M1 multiply the assumed inequality by 2 to get 2x < 10 (method)
M1 subtract 3 to get 2x - 3 < 7 (method)
A1 conclude the converse is true: if x < 5 then 2x - 3 < 7 cao
B1 state the reason concisely: multiplication by a positive number and subtraction preserve strict inequality oe
B1 explicitly note domain is all real x so no hidden exceptions oe
Answer: First part: x < 5. Converse: true. Reason: assuming x < 5, multiply by 2 to obtain 2x < 10 then subtract 3 to get 2x - 3 < 7; operations preserve the strict inequality because 2 > 0.

Question 9

(a) M1 correct factorisation shown: (2x + 3)(x - 5) (method)
(a) M1 identify roots x = -3/2 and x = 5 from factors (method)
(a) A1 solution: -3/2 < x < 5 cao
(a) Answer: -3/2 < x < 5
(b) M1 use result from part (a): -3/2 < x < 5 (follow through)
(b) M1 identify integers strictly greater than -1.5 and strictly less than 5: -1, 0, 1, 2, 3, 4 (method)
(b) M1 count the listed integers correctly (method)
(b) A1 state final answer 6 integers cao
(b) B1 justify endpoints excluded because inequality is strict so x = -3/2 and x = 5 are not allowed oe
(b) Answer: There are 6 integer solutions: x = -1, 0, 1, 2, 3, 4.

Inequalities: Higher Tier Practice - Worksheets, Questions and Revision