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HIGHER

3.13H Substitution: Higher Tier Practice

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 65 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Evaluate the expression 3x² - 2x + 5 when x = -1.

(Total for Question 1 is 1 mark)

Given f(t) = 2t³ - t² + 4 and t = 0.5, find f(t). Give your answer as a decimal.

(Total for Question 2 is 1 mark)

Substitute x = 3 and y = -2 into the expression 5x - 2y and simplify.

(Total for Question 3 is 2 marks)

A formula for the area A of an isosceles triangle is A = (1/2)bh where b is the base and h is the height. A triangle has b = 8x and h = x + 3. Write an expression for A in terms of x and simplify fully.

(Total for Question 4 is 3 marks)

The temperature T in degrees C is given by T = 20 - 5k, where k is a constant. If k = (2p - 1)/4 and p = 5, find T.

(Total for Question 5 is 3 marks)

Let g(x) = 4 - 3x. Find x such that g(x) = 13. Show your working.

(Total for Question 6 is 4 marks)

H(p) = p² - 6p + 8. Given p = 3 + 2t, expand H(p) fully in terms of t and simplify.

(Total for Question 7 is 5 marks)

Show that if f(x) = ax + b and g(x) = 2x - 5, then the composite function f(g(x)) equals 2.5 when x = 3, given a = 2 and b = 0.5. You must show substitution and evaluation.

(Total for Question 8 is 6 marks)

A scientist models mass m (in kg) by m = c r² where c is a constant and r is the radius in metres. Given c = 0.75 and r = 2k + 1 where k = 2, find m and give your answer in the form a/b or simplified fraction where appropriate.

(Total for Question 9 is 8 marks)

Given the expression E = (2x + 1)/(x - 1). Show that E = 5 when x = 2. Then find the value of x (if any) for which E = 2. You must show all algebraic steps.

(Total for Question 10 is 7 marks)

Mark scheme · 3.13H Substitution: Higher Tier Practice

Question 1

B1 3(-1)² - 2(-1) + 5 = 3 + 2 + 5 = 10 cao
Answer: 10

Question 2

B1 2(0.5)³ - (0.5)² + 4 = 2(0.125) - 0.25 + 4 = 0.25 - 0.25 + 4 = 4.0 cao
Answer: 4.0

Question 3

M1 substitute values: 5*3 - 2*(-2) or equivalent
A1 19 cao
Answer: 19

Question 4

M1 substitute A = (1/2) * 8x * (x+3) or show (1/2)*8x*(x+3)
M1 simplify (1/2)*8x*(x+3) to 4x(x+3) or show distribution step
A1 4x² + 12x cao
Answer: 4x² + 12x

Question 5

M1 substitute p into k: k = (2*5 - 1)/4 = (10 - 1)/4 = 9/4
M1 substitute k into T: T = 20 - 5*(9/4) or 20 - 45/4
A1 35/4 or 8.75 cao
Answer: 35/4 or 8.75

Question 6

M1 set 4 - 3x = 13 and rearrange to linear form
M1 subtract 4: -3x = 9
M1 divide both sides by -3 to get x = -3
A1 x = -3 cao
Answer: x = -3

Question 7

M1 substitute p = 3+2t into p² - 6p + 8
M1 expand (3+2t)² = 9 + 12t + 4t²
M1 expand -6(3+2t) = -18 -12t
M1 collect terms correctly
A1 4t² - 1 cao
Answer: 4t² - 1

Question 8

M1 state g(x)=2x-5 and substitute into f: f(g(x)) = a(2x-5)+b
M1 substitute a=2 and b=0.5 to get 2(2x-5)+0.5
M1 simplify to 4x - 10 + 0.5 = 4x - 9.5
M1 substitute x=3 to evaluate: 4*3 - 9.5
M1 calculate 12 - 9.5 = 2.5
A1 state final evaluation equals 2.5 cao, confirming the given target value
Answer: 2.5

Question 9

M1 substitute k = 2 into r = 2k + 1 to get r = 5
M1 state m = c r² and substitute r: m = 0.75 * 5²
M1 compute r² = 25
M1 compute 0.75 * 25 = 18.75
M1 convert 0.75 to fraction 3/4 and show 3/4 * 25
M1 compute fraction 3/4 * 25 = 75/4
M1 simplify fraction if needed and include units
A1 75/4 kg cao
Answer: 75/4 kg or 18.75 kg

Question 10

M1 substitute x = 2: (2*2 + 1)/(2 - 1) = (4 + 1)/1
M1 calculate numerator and denominator to show 5/1 = 5
M1 for E = 2, multiply both sides by (x - 1): 2x + 1 = 2(x - 1)
M1 expand and rearrange to show contradiction: 2x + 1 = 2x - 2 gives 1 = -2
B1 state domain restriction x ≠ 1 or note that x = 1 would make denominator zero
M1 deduce from 1 = -2 that no solution exists
A1 conclude there is no value of x satisfying E = 2, cao
Answer: E = 5 when x = 2. There is no solution for E = 2 (contradiction 1 = -2), with x ≠ 1 noted.

Substitution: Higher Tier Practice - Worksheets, Questions and Revision