Lesson B8
B8 Composite and inverse functions Quiz: AQA Maths, Unit 2
20 questions
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Lesson B8, Composite and inverse functions: 20 multiple choice questions for the AQA Maths (7357), Unit 2: Algebra and functions, written with Revision Ninja.
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The 20 questions
-
Given f(x) = 2x + 3 and g(x) = x^2, find fg(x).
- x^2 + 3
- (2x + 3)^2
- 2x^2 + 3
- 2x^2 + 6
-
With f(x) = 2x + 3 and g(x) = x^2, find gf(x).
- 4x^2 + 12x + 9
- 2x^2 + 3
- 4x^2 + 3
- 2x^2 + 12x + 9
-
What is the inverse of f(x) = 2x + 3?
- (x - 3)/2
- (3 - x)/2
- (x + 3)/2
- 2x - 3
-
What is the inverse of f(x) = x^3 - 1?
- 1/(x^3 - 1)
- (x - 1)^(1/3)
- x^3 + 1
- (x + 1)^(1/3)
-
Which condition must a function satisfy to have an inverse?
- It must be periodic
- It must be an even function
- It must be one-to-one
- It must have a y-intercept
-
What is the inverse of f(x) = x^2 for x >= 0?
- -sqrt(x)
- 1/x^2
- sqrt(x), for x >= 0
- x^2, for x >= 0
-
If f(x) = 3x - 1 and g(x) = x + 2, solve fg(x) = 11.
- x = 11/3
- x = 3
- x = 2
- x = 4
-
For f(x) = 1/(x - 1) and g(x) = x + 2, what restriction applies to the domain of fg(x)?
- x is not equal to 1
- x is not equal to -1
- x is not equal to 0
- x is not equal to 2
-
The graph of f^-1 is obtained by reflecting y = f(x) in which line?
- y = x
- y = -x
- The y-axis
- The x-axis
-
For f(x) = x^2 + 1 with x >= 0, what is f^-1(10)?
- sqrt(10)
- 9
- -3
- 3
-
If f(x) = x^2 - 4 and g(x) = x + 1, find fg(x).
- x^2 - 3
- x^2 + 2x - 4
- x^2 + 2x + 3
- x^2 + 2x - 3
-
With f(x) = 2x - 1 and g(x) = x^2, for which x does f(g(x)) = g(f(x))?
- x = 1 only
- There are no solutions
- x = 0 or x = 1
- x = -1 or x = 1
-
What is the inverse of f(x) = (2x + 1)/(x - 3)?
- (x - 3)/(2x + 1)
- (3x - 1)/(x - 2)
- (3x + 1)/(x - 2)
- (3x + 1)/(2 - x)
-
What must the range of f satisfy for g(f(x)) to be defined?
- It must lie within the domain of g
- It must be negative
- It must contain zero
- It must equal the domain of f
-
If f(x) = 3x + 2, find f^-1(8).
- 8/3
- 10/3
- 2
- 6
-
If f(x) = x^2 - 2 for x >= 0 and g(x) = 2x + 1, find gf(x).
- 4x^2 - 3
- 2x^2 - 3
- 2x^2 - 4x - 3
- 2x^2 - 1
-
Why does f(x) = x^2 have no inverse over all real numbers?
- Its range is empty
- Its graph has no y-intercept
- It is a linear function
- It is not one-to-one, since f(2) = f(-2)
-
For f(x) = 4 - x and g(x) = 1/x, find fg(x).
- 1/(4 - x)
- 4 - 1/x
- (4 - x)/x
- 4 - x
-
Let f(x) = x + 1 and g(x) = 2x. Solve f(g(x)) = g(f(x)).
- No solutions
- All real x
- x = 1
- x = 0
-
If f(x) = 1/(x + 1), find ff(x) in its simplest form.
- 1/(x + 2)
- (x + 1)/(x + 2)
- (x + 2)/(x + 1)
- x + 2
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