1. 1
   Given $y = x^3 - 4x^2 + 5$.

   (a)

   Given $y = x^3 - 4x^2 + 5$, what is $\frac{dy}{dx}$?

   [1 mark]

   • A$3x^2 - 8x$
   • B$3x^2 - 4x$
   • C$x^2 - 8x$
   • D$3x^2 - 8x + 5$

   (b)

   What is the value of $\frac{dy}{dx}$ when $x = 2$?

   [1 mark]

   • A$4$
   • B$12$
   • C$-4$
   • D$-12$

   (c)

   Based on your answer to part (b), what does this tell you about the graph of $y = x^3 - 4x^2 + 5$ at $x = 2$?

   [1 mark]

   • AThe curve is increasing at $x = 2$
   • BThe curve is undefined at $x = 2$
   • CThe curve has a stationary point at $x = 2$
   • DThe curve is decreasing at $x = 2$

   Total for question 1: 3 marks

2. 2
   The displacement, in metres, of a particle at time $t$ seconds is given by $s = t^3 - 3t^2$.

   (a)

   Find an expression for the velocity, $v = \frac{ds}{dt}$, of the particle.

   [2 marks]

   (b)

   Find the velocity of the particle at $t = 1$.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   Given $y = x^3 - 12x + 5$.

   (a)

   Given $y = x^3 - 12x + 5$, find $\frac{dy}{dx}$ and hence find the $x$-coordinates of the stationary points of the curve.

   [3 marks]

   (b)

   Find the $y$-coordinates of the two stationary points found in part (a).

   [3 marks]

   Total for question 3: 6 marks

4. 4
   A particle moves so that its displacement, in metres, from a fixed point O after $t$ seconds is $s = t^3 - 9t^2 + 24t$ for $0 \leq t \leq 6$.

   (a)

   Find an expression for the velocity $v = \frac{ds}{dt}$, and find the values of $t$ at which the particle is momentarily at rest ($v = 0$).

   [4 marks]

   (b)

   Find the second derivative $\frac{d^2s}{dt^2}$, and use it to determine whether $t = 2$ gives a maximum or minimum value of the displacement $s$.

   [4 marks]

   (c)

   Find the displacement of the particle at the local maximum (the value of $s$ at $t = 2$), and determine whether $t = 4$ gives a maximum or minimum, stating the corresponding displacement.

   [5 marks]

   Total for question 4: 13 marks

End of questions