Q1
Answer
Given that $f(x) = x\sqrt{2x+12}$,
  1. Find $f'(x)$ and show that $f''(x) = \dfrac{3(x+8)}{(2x+12)^{\frac{3}{2}}}$.
  2. Find the coordinates of the turning point of the curve $y=f(x)$ and determine its nature.
  1. $\dfrac{3(x+8)}{(2x+12)^{\frac{3}{2}}$
  2. $(-4,-8)$, second derivative is positive, minimum
Q2
Answer
Given that $f(x) = \dfrac{x}{x-1}$
  1. Find $f'(x)$.
  2. Hence evaluate $\displaystyle\int_2^3 \dfrac{3x^4-4x^3}{(x-1)^2}\ \mathrm{d}x$.
  1. $-\dfrac{1}{(x-1)^2}$
  2. $\dfrac{49}{2}$
Q3
Answer
Use proof by contradiction to prove that there are no positive integers $x$ and $y$ such that $x^2-y^2=1$.
Assume there are, LHS: $(x+y)(x-y)$. Only way for this to work is if $x+y = 1$ and $x - y = 1$. Solve simultaneously, $y > 0$ so no solutions.