A-Level Maths: June 2009

Page 1

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)