A-Level Maths: June 2009

Question: Use the Remainder Theorem to find the remainder when $3x^3 + 8x^2 - 3x - 5$ is divided by $3x -1$ . 

Answer:$3x - 1$ ... sub in $x = \frac{1}{3}$$f(\frac{1}{3}) = 3(\frac{1}{3})^3 + 8(\frac{1}{3})^2 - 3(\frac{1}{3}) -5$$= -5$

Question:A curve is defined by the parametric equations $x = \frac{1}{t} , y = t + \frac{1}{2t}$Find $\frac{dy}{dx}$ in terms of $t$

Answer:$x = \frac{1}{t}$     $y = t + \frac{1}{2t}$$\frac{dx}{dt} = - \frac{1}{t^2}$     $\frac{dy}{dt} = 1 - \frac{1}{2t^2}$$\frac{dy}{dx} = \frac{dy}{dt} * \frac{dt}{dx}$$= (1 - \frac{1}{2t^2} ) * - t^2$$= - t^2 + \frac{1}{2}$ or $\frac{1}{2} - t^2$

Question 1) a)

Question 2) a)