# 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} - (x^{2} +1)e^{-x}

= -e^{-x} (x -1)^{2}

**(ii) dy/dx = 0 at stationary points, thus:**

0 = 2xe^{-x} - (x^{2} +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:**

d^{2}y/dx^{2} = 2e^{-x} - 2xe^{-x} - 2xe^{-x} + (x^{2} +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 cm^{2}, 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 = 4^{x}.

**a)** Find the value of x when y = 12, giving your answer to two decimal places

**b)** Show that y = e^{kx}, 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 = e ^{ln4} therefore, multiplying both side by the power of x, gives:**

4^{x} = (e^{ln4})^{x}

From this you can deduce:

y = e^{kx} where k = ln 4.

**(1 mark)**

**c)** dy/dx = ke^{kx} = (ln 4)e^{xln4} = (ln 4).4^{x} = 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)**