Lesson G1
G1 Derivative as gradient, first principles and second derivatives Quiz: AQA Maths, Unit 7
20 questions
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Lesson G1, Derivative as gradient, first principles and second derivatives: 20 multiple choice questions for the AQA Maths (7357), Unit 7: Differentiation, written with Revision Ninja.
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The 20 questions
-
What does the derivative f'(x) represent geometrically?
- The area under the curve y = f(x) from 0 to x
- The value of f(x) at the point where x = 0
- The gradient of the tangent to the curve y = f(x) at the point with x-coordinate x
- The y-intercept of the curve y = f(x) at each point
-
Which limit defines the derivative of f(x) from first principles?
- lim as h tends to 0 of (f(x) - f(h))/x
- lim as h tends to 0 of (f(x + h) - f(x))/h
- lim as h tends to 0 of (f(x + h) + f(x))/h
- lim as h tends to 0 of f(x + h) - f(x)
-
Using first principles, what is the derivative of x^2?
- 2x
- x
- 2
- x^2/2
-
What is the second derivative of y = x^3?
- 6
- 6x
- 3x
- 3x^2
-
What does the second derivative of a function measure?
- The rate of change of the gradient, which indicates whether the curve is convex or concave
- The rate of change of the area under the curve
- The value of the function at its maximum point
- The gradient of the normal at each point of the curve
-
What is a point of inflection?
- A point where the curve crosses the x-axis
- A point where the function is undefined
- A point where the gradient of the curve is always zero
- A point where the curve changes between convex and concave, often where the second derivative changes sign
-
What is the derivative of x^3 with respect to x?
- 3x
- x^3/3
- 3x^2
- x^2
-
What is the gradient of y = x^2 at x = 3?
- 6
- 9
- 5
- 3
-
If f(x) = x^3, what is f'(2)?
- 6
- 8
- 4
- 12
-
Find f''(x) for f(x) = 4x^3 - 2x.
- 24x
- 12x^2 - 2
- 24
- 12x
-
What is the gradient of the curve y = x^2 - 4x at x = 1?
- -4
- -2
- 2
- -3
-
Using first principles on f(x) = x^2, what does (f(x + h) - f(x))/h simplify to before h tends to 0?
- x + h
- 2x - h
- 2x + h
- 2x
-
For f(x) = x^4 - 2x^2, what is f''(x)?
- 4x^3 - 4x
- 12x^2 - 2
- 12x^2 - 4
- 12x - 4
-
A curve has f''(x) > 0 for all x in an interval. What can be said about the curve on that interval?
- It is convex (concave up), so its gradient is increasing
- Its gradient is zero throughout the interval
- It is concave down, so its gradient is decreasing
- It has a maximum point somewhere in the interval
-
If f'(x) is increasing on an interval, what is the sign of f''(x) on that interval?
- Zero
- Undefined
- Negative
- Positive
-
Find the second derivative of y = 1/x^2.
- -2/x^3
- 2/x^3
- 6/x^4
- -6/x^4
-
A curve has second derivative y'' = 6x - 12. What is the x-coordinate of its point of inflection?
- 0
- 2
- -2
- 6
-
Using first principles, find f'(1) for f(x) = 3x^2 + x.
- 6
- 7
- 4
- 8
-
Explain why f(x) = x^3 has a point of inflection at the origin.
- f''(0) = 1, so the curve is convex at the origin
- f''(x) = 6 is constant, so there is no point of inflection
- f''(0) = 3, so the origin is a local minimum
- f''(x) = 6x changes sign at x = 0, so the concavity changes there
-
Estimate the gradient of y = x^2 at x = 1 using the chord from x = 1 to x = 1.01, to 2 decimal places.
- 2.50
- 2.10
- 1.99
- 2.01
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