Inverse and Composite Functions - Worksheets, Questions and Revision

21 original exam-style questions - 6 pages of questions with a full mark scheme - free printable PDF.

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7.9 Inverse and Composite Functions

EDEXCEL 1MA1 1MA1 · Calculator allowed · about 85 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

f(x) = 3x + 2. Work out f(4).

(Total for Question 1 is 1 mark)

f(x) = 5x - 3 and g(x) = x + 4. Work out fg(2).

(Total for Question 2 is 2 marks)

f(x) = x² - 1. Work out f(-3).

(Total for Question 3 is 2 marks)

f(x) = x² and g(x) = x + 1. Which of the following correctly represents gf(x)?

A) (x + 1)²
B) x² + 1
C) x² + x
D) 2x² + 1

(Total for Question 4 is 1 mark)

f(x) = 2x + 7. Find f^-1(x).

(Total for Question 5 is 2 marks)

f(x) = (x - 5)/3

(a)Find f^-1(x).(2)

(b)Work out f^-1(4).(1)

(Total for Question 6 is 3 marks)

g(x) = 4x - 1 and h(x) = x²

(a)Find gh(x).(2)

(b)Find hg(x).(2)

(c)Show that gh(2) is not equal to hg(2).(2)

(Total for Question 7 is 6 marks)

f(x) = 1/x and g(x) = x - 4. Find fg(x) and state the value of x for which fg(x) is undefined.

(Total for Question 8 is 3 marks)

f(x) = 3x - 2 and g(x) = x/2 + 1. Solve fg(x) = 13.

(Total for Question 9 is 3 marks)

f(x) = 2x + 3

(a)Find f^-1(x).(2)

(b)Solve f^-1(x) = 7.(2)

(Total for Question 10 is 4 marks)

f(x) = x² + 2 and g(x) = 3x - 1

(a)Find gf(x).(2)

(b)Find fg(x).(2)

(c)Work out gf(2) - fg(2).(2)

(Total for Question 11 is 6 marks)

A gym charges a one-off joining fee of 15 pounds plus 20 pounds per month. The total cost, in pounds, of m months of membership is given by the function C(m) = 20m + 15.

(a)Find C^-1(x) and interpret what it represents.(3)

(b)Priya has budgeted 195 pounds for gym membership. Use your answer to part (a) to work out the maximum number of whole months of membership she can afford.(2)

(Total for Question 12 is 5 marks)

f(x) = 2x - 1. Solve ff(x) = 11.

(Total for Question 13 is 3 marks)

f(x) = 2x² - 3 and g(x) = 5x - 6. Solve f(x) = g(x).

(Total for Question 14 is 4 marks)

f(x) = 2x + 5. Show that f^-1(f(x)) = x.

(Total for Question 15 is 3 marks)

f(x) = (2x + 1)/(x - 3), x is not equal to 3. Find f^-1(x).

(Total for Question 16 is 4 marks)

f(x) = ax + b, where a and b are constants. Given that f(2) = 7 and f(5) = 16, find the values of a and b.

(Total for Question 17 is 4 marks)

g(x) = (x + 1)/(x - 2), x is not equal to 2

(a)Find g^-1(x).(4)

(b)State the value of x for which g(x) is undefined, and the value of x for which g^-1(x) is undefined.(2)

(Total for Question 18 is 6 marks)

f(x) = 2x - 1 and g(x) = x + 3. Show that the equation fg(x) = gf(x) has no solutions.

(Total for Question 19 is 3 marks)

f(x) = 1/(x - 1) + 2, x is not equal to 1. Find f^-1(x), stating the value that must be excluded from its domain.

(Total for Question 20 is 4 marks)

f(x) = (x + 2)/(x - 1), x is not equal to 1. Show that f is self-inverse, that is, show that f^-1(x) = f(x).

(Total for Question 21 is 3 marks)

Mark scheme · 7.9 Inverse and Composite Functions

Question 1

B1 f(4) = 14 cao
Answer: f(4) = 14

Question 2

M1 correctly finds g(2) = 6, or substitutes g(2) into f, oe
A1 fg(2) = 27 cao
Answer: fg(2) = 27

Question 3

M1 (-3)² = 9 seen, oe
A1 f(-3) = 8 cao
Answer: f(-3) = 8

Question 4

B1 B cao
Answer: B

Question 5

M1 writes y = 2x + 7 and rearranges to make x the subject, e.g. x = (y - 7)/2
A1 f^-1(x) = (x - 7)/2 oe, cao
Answer: f^-1(x) = (x - 7)/2

Question 6

(a) M1 writes y = (x - 5)/3 and multiplies both sides by 3, e.g. 3y = x - 5
(a) A1 f^-1(x) = 3x + 5 oe, cao
(a) Answer: f^-1(x) = 3x + 5
(b) B1 f^-1(4) = 17 cao, ft from part (a)
(b) Answer: f^-1(4) = 17

Question 7

(a) M1 substitutes h(x) into g correctly, e.g. g(x²) = 4(x²) - 1
(a) A1 gh(x) = 4x² - 1 oe, cao
(a) Answer: gh(x) = 4x² - 1
(b) M1 substitutes g(x) into h correctly, e.g. h(4x - 1) = (4x - 1)²
(b) A1 hg(x) = 16x² - 8x + 1 oe, cao
(b) Answer: hg(x) = 16x² - 8x + 1
(c) B1 gh(2) = 15 correctly found, ft from part (a)
(c) B1 hg(2) = 49 correctly found and 15 not equal to 49 stated as conclusion, ft from part (b), cso
(c) Answer: gh(2) = 15, hg(2) = 49, so gh(2) is not equal to hg(2)

Question 8

M1 substitutes g(x) into f correctly, e.g. f(x - 4) = 1/(x - 4)
A1 fg(x) = 1/(x - 4) oe, cao
B1 x = 4 stated (denominator cannot be zero)
Answer: fg(x) = 1/(x - 4); undefined when x = 4

Question 9

M1 forms fg(x) correctly, e.g. 3(x/2 + 1) - 2, oe
M1 simplifies and sets equal to 13, e.g. 3x/2 + 1 = 13, then rearranges, ft
A1 x = 8 cao
Answer: x = 8

Question 10

(a) M1 writes y = 2x + 3 and rearranges to make x the subject, e.g. x = (y - 3)/2
(a) A1 f^-1(x) = (x - 3)/2 oe, cao
(a) Answer: f^-1(x) = (x - 3)/2
(b) M1 sets (x - 3)/2 = 7 (ft from part (a)) or uses x = f(7), and forms a correct equation
(b) A1 x = 17 cao
(b) Answer: x = 17

Question 11

(a) M1 substitutes f(x) into g correctly, e.g. 3(x² + 2) - 1
(a) A1 gf(x) = 3x² + 5 oe, cao
(a) Answer: gf(x) = 3x² + 5
(b) M1 substitutes g(x) into f correctly, e.g. (3x - 1)² + 2
(b) A1 fg(x) = 9x² - 6x + 3 oe, cao
(b) Answer: fg(x) = 9x² - 6x + 3
(c) M1 substitutes x = 2 into both expressions correctly, ft from parts (a) and (b)
(c) A1 -10 cao
(c) Answer: gf(2) - fg(2) = -10

Question 12

(a) M1 writes y = 20m + 15 and rearranges to make m the subject, e.g. m = (y - 15)/20
(a) A1 C^-1(x) = (x - 15)/20 oe, cao
(a) B1 correct interpretation, e.g. C^-1(x) gives the number of months of membership that a total cost of x pounds buys
(a) Answer: C^-1(x) = (x - 15)/20; it gives the number of months of membership for a total cost of x pounds
(b) M1 substitutes x = 195 into C^-1(x), ft from part (a)
(b) A1 9 (months) cao
(b) Answer: 9 months

Question 13

M1 forms ff(x) correctly, e.g. 2(2x - 1) - 1
M1 simplifies and sets equal to 11, e.g. 4x - 3 = 11, then rearranges, ft
A1 x = 3.5 oe (e.g. 7/2) cao
Answer: x = 3.5

Question 14

M1 sets 2x² - 3 = 5x - 6 and rearranges to form a quadratic equal to zero, e.g. 2x² - 5x + 3 = 0
M1 correctly factorises or applies the quadratic formula to the equation from the previous mark, e.g. (2x - 3)(x - 1) = 0
A1 x = 1 cao
A1 x = 1.5 oe (e.g. 3/2) cao
Answer: x = 1 or x = 1.5

Question 15

M1 correctly finds f^-1(x) = (x - 5)/2
M1 substitutes f(x) = 2x + 5 into f^-1, e.g. f^-1(2x + 5) = ((2x + 5) - 5)/2
A1 simplifies fully to x, with no errors seen, cso
Answer: f^-1(f(x)) = x (shown)

Question 16

M1 writes y = (2x + 1)/(x - 3) and multiplies both sides by (x - 3), e.g. y(x - 3) = 2x + 1
M1 expands and collects all terms in x on one side, e.g. xy - 2x = 3y + 1
dM1 factorises out x, e.g. x(y - 2) = 3y + 1, dependent on the previous method mark
A1 f^-1(x) = (3x + 1)/(x - 2) oe, cao
Answer: f^-1(x) = (3x + 1)/(x - 2)

Question 17

M1 forms two correct equations, e.g. 2a + b = 7 and 5a + b = 16
M1 eliminates one variable correctly, e.g. subtracts to get 3a = 9
A1 a = 3 cao
A1 b = 1 cao, ft from their value of a
Answer: a = 3, b = 1

Question 18

(a) M1 writes y = (x + 1)/(x - 2) and multiplies both sides by (x - 2), e.g. y(x - 2) = x + 1
(a) M1 expands and collects all terms in x on one side, e.g. xy - x = 2y + 1
(a) dM1 factorises out x, e.g. x(y - 1) = 2y + 1, dependent on the previous method mark
(a) A1 g^-1(x) = (2x + 1)/(x - 1) oe, cao
(a) Answer: g^-1(x) = (2x + 1)/(x - 1)
(b) B1 x = 2 (g(x) undefined)
(b) B1 x = 1 (g^-1(x) undefined), ft from part (a)
(b) Answer: g(x) is undefined at x = 2; g^-1(x) is undefined at x = 1

Question 19

M1 correctly forms fg(x) = 2x + 5
M1 correctly forms gf(x) = 2x + 2
C1 correct reasoning that 2x + 5 = 2x + 2 leads to 5 = 2, which is false for all x, so there is no solution (or equivalent statement that the two expressions represent parallel lines with different intercepts and so never meet), cso
Answer: No solutions (since 2x + 5 = 2x + 2 would require 5 = 2, which is impossible)

Question 20

M1 writes y = 1/(x - 1) + 2 and rearranges to isolate the fraction, e.g. y - 2 = 1/(x - 1)
M1 takes the reciprocal of both sides correctly, e.g. x - 1 = 1/(y - 2)
A1 f^-1(x) = 1 + 1/(x - 2) oe, cao
B1 x is not equal to 2 stated as the excluded value
Answer: f^-1(x) = 1 + 1/(x - 2), x is not equal to 2

Question 21

M1 writes y = (x + 2)/(x - 1) and multiplies both sides by (x - 1), e.g. y(x - 1) = x + 2, then collects terms in x, e.g. x(y - 1) = y + 2
M1 correctly makes x the subject, e.g. x = (y + 2)/(y - 1)
A1 concludes f^-1(x) = (x + 2)/(x - 1), which is identical to f(x), hence f is self-inverse, cso
Answer: f^-1(x) = (x + 2)/(x - 1) = f(x), so f is self-inverse (shown)

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Inverse and Composite Functions - Worksheets, Questions and Revision

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7.9 Inverse and Composite Functions

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Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21