Lesson G3
G3 Tangents, normals, stationary points and increasing or decreasing Quiz: AQA Maths, Unit 7
20 questions
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Lesson G3, Tangents, normals, stationary points and increasing or decreasing: 20 multiple choice questions for the AQA Maths (7357), Unit 7: Differentiation, written with Revision Ninja.
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The 20 questions
-
What is the gradient of the normal to a curve at a point where the tangent has gradient m, with m not equal to 0?
- m
- -m
- -1/m
- 1/m
-
What condition defines a stationary point on a curve y = f(x)?
- y = 0
- dy/dx is undefined
- dy/dx = 0
- d^2y/dx^2 = 0 always
-
When is a function increasing on an interval?
- d^2y/dx^2 > 0 throughout the interval
- dy/dx > 0 throughout the interval
- dy/dx < 0 throughout the interval
- dy/dx = 0 throughout the interval
-
How is a local maximum distinguished from a local minimum at a stationary point using the second derivative?
- A minimum occurs when f'' < 0 and a maximum when f'' > 0
- A maximum occurs when f'' > 0 and a minimum when f'' = 0
- The second derivative cannot distinguish maxima from minima
- A maximum occurs when f'' < 0 and a minimum when f'' > 0
-
What is the equation of the tangent to y = f(x) at (a, b) with gradient m?
- y + b = m(x + a)
- y - a = m(x - b)
- y - b = m(x - a)
- y - b = -(x - a)/m
-
If the tangent to a curve has gradient 2 at a point, what is the gradient of the normal there?
- 1/2
- 2
- -2
- -1/2
-
Which statement about a stationary point where f''(a) = 0 is true?
- It is always a minimum point of the curve
- It is always a maximum point of the curve
- The point may be a maximum, a minimum or a point of inflection, so further testing is needed
- It cannot be a stationary point of the curve
-
Find the stationary point of y = x^2 - 6x + 5.
- (3, -4)
- (6, 5)
- (-3, 14)
- (3, 4)
-
What is the gradient of the tangent to y = x^3 at x = -1?
- -3
- 1
- -1
- 3
-
Find the equation of the tangent to y = x^2 at the point (1, 1).
- y = 2x + 1
- y = -2x + 3
- y = x
- y = 2x - 1
-
Find the equation of the normal to y = x^2 at the point (1, 1).
- y = x/2 + 1/2
- y = 2x - 1
- y = -2x + 3
- y = -x/2 + 3/2
-
Find the x-coordinates of the stationary points of y = x^3 - 3x.
- x = 0 only
- x = 1 and x = -1
- x = 1 only
- x = 3 and x = -3
-
For y = x^3 - 3x, what type of stationary point is at x = 1?
- A local maximum, since f'' = 6x = 6 is positive
- Not a stationary point, since the gradient is nonzero
- A local minimum, since f'' = 6x = 6 is positive
- A point of inflection with zero second derivative
-
For which values of x is y = 2x^3 - 9x^2 increasing?
- All real values of x
- 0 < x < 3
- x > 3 only
- x < 0 or x > 3
-
On which interval is y = x^2 decreasing?
- All real values of x
- x < 0
- x > 0
- No interval, since it is never decreasing
-
Find the equation of the normal to y = 1/x at x = 2.
- y = -x/4 + 1
- y = -4x + 17/2
- y = 4x - 15/2
- y = 4x - 8
-
How many stationary points does y = x^4 - 8x^2 + 1 have?
- 3
- 2
- 4
- 1
-
Find k such that y = x^2 + kx + 4 has a stationary point at x = 3.
- k = 6
- k = -6
- k = 3
- k = -3
-
For f'(x) = (x - 1)^2 (x - 3), which statement about x = 1 is correct?
- It is a stationary point where the gradient does not change sign, so it is neither a maximum nor a minimum
- It is a minimum since f' is non-negative around it
- It is a maximum since the gradient is zero there
- It is not stationary because f'' is zero there
-
For which x is y = x^3 - 3x^2 + 3x decreasing?
- For all x in the interval (0, 2)
- For x > 1 only
- For x < 1 only
- Nowhere, since dy/dx = 3(x - 1)^2 is never negative
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