Exam-style Questions: Differentiation
1. A curve C has the equation
a) (i) Show that:
(ii) Hence find the coordinates of the stationary point on the curve C
(iii) Show that this stationary point is a point of inflection.
b) (i) Show that:
where a and b are constants to be determined.
(ii) Deduce that the curve has another point of inflection.
c) Sketch the curve C, indicating the two points of inflection.
(Marks available: 11)
Answer outline and marking scheme for question: 1
Give yourself marks for mentioning any of the points below:
a) (i) Using the product rule gives:
dy/dx = 2xe-x - (x2 +1)e-x
= -e-x (x -1)2
(ii) dy/dx = 0 at stationary points, thus:
0 = 2xe-x - (x2 +1)e-x
This gives stationary point at (1, 2e-1)
(iii) dy/dx (i.e. the gradient) remains the same sign either side of the stationary point. Therefore the stationary point must be a point of inflection.
Mathematically shown below:
dy/dx is less than zero before the stationary point (i.e. at x less than one)
dy/dx equals zero at the stationary point
dy/dx is less than zero after the stationary point (i.e. at x more than one)
(5 marks)
b) (i) Using the product rule on dy/dx gives:
d2y/dx2 = 2e-x - 2xe-x - 2xe-x + (x2 +1)e-x
= e-x (x -1)(x -3)
Therefore a = 1 and b = 3.
(4 marks)
The gradient is always less than zero, so the curve slopes downwards.
From the above there are two points of inflections at x = 1 and x = 3.
The equation never becomes negative, therefore the curve does not cross the x axis.
Therefore the curve looks like:
(2 marks)
(Marks available: 11)
2. A piece of wire, of length 20cm, is to be cut into two parts. One of the parts, of length x cm, is to be formed into a circle and the other part into a square.
a) Show that the sum, A cm2, of the areas of the circle and the square is given by
b) Show that A has a stationary value when
(Marks available: 8)
Answer outline and marking scheme for question: 2
Give yourself marks for mentioning any of the points below:
a) Area of circle
Side of square
Entering these to find a total area, gives:
(3 marks)
b) Differentiating the area equation above, gives:
Solving this equation when dA/dx = 0, gives:
(5 marks)
(Marks available: 8)
3. The variables x and y are related by
y = 4x.
a) Find the value of x when y = 12, giving your answer to two decimal places
b) Show that y = ekx, where k = In 4.
c) Hence find dy/dx.
d) Given that x is a function of a third variable t and that
dy/dx = x, deduce that dy/dx = 12 ln 12, when y = 12.
(Marks available: 7)
Answer outline and marking scheme for question: 3
Give yourself marks for mentioning any of the points below:
a) 4x = 12, using logs on either side, gives:
x log 4 = log 12
Solving this for x gives, x = 1.79 to 2d.p.
(2 marks)
b) We know that 4 = eln4 therefore, multiplying both side by the power of x, gives:
4x = (eln4)x
From this you can deduce:
y = ekx where k = ln 4.
(1 mark)
c) dy/dx = kekx = (ln 4)exln4 = (ln 4).4x = yln4.
(1 mark)
d) dy/dt = dy/dx . dx/dt
= (y ln 4).x
= (y ln 4).(ln 12/ ln 4)
When y = 12, from (a)
= 12 ln 12
(3 marks)
(Marks available: 7)
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