( ) ( ) ( ) ( ) Given that and its derivative are continuous when, find th values of and. ( ) ( )

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1 1. The piecewise function is defined by where and are constants. Given that and its derivative are continuous when, find th values of and. When When of of Substitute into ;

2 2. Using the substitution, evaluate the integral Give your answer correct to three decimal places. Let [ e ] [ ] 1 e 1 [ ]

3 3. By putting find the general solution to the equation Half angle formulae (the formulae): where EITHER OR When When

4 4. The function is given by Express in partial fractions. Hence evaluate Giving your answer in the form, where are positive integers. Multiply throughout by; EQN Substitute, into EQN Graph of Compare coefficients of in EQN Compare constant terms in EQN Let [ ] 4 3 [ ] Using the following rules of logs:

5 5. The function is defined by Determine whether is an even or an odd function. The function is defined by where is a positive integer. Determine the set of values of for which is: (i) an even function, (ii) an odd function. For an even function, For an odd function, is an odd function (rotational symmetry about the origin). is an odd function when (odd numbers). is an even function when (even numbers). When multiplying functions together;, because the two negatives multiplied make a positive., because positive multiplied by negative gives negative. For to be even, must be odd. For to be odd, must be even.

6 6. The function is defined by Determine the coordinates of the points where the graph of intersects the coordinate axes. Find the coordinates of the stationary points on the graph of. State the equation of each of the asymptotes on the graph of. Sketch the graph of. When crosses, the -, crosses the - at When crosses, the -, crosses the - at At stationary points, When ( ) ( ) stationary point at( ) When ( ) ( ) stationary point at( )

7 There is a vertical asymptote at, because as There is a vertical asymptote at, because as

8 7. A parabola has equation Find (i) the coordinates of the vertex, (ii) the coordinates of the focus, (iii) the equation of the directrix The line cuts the parabola at the points (i) Obtain a quadratic equation whose roots are the -coordinates of (ii) Hence find the gradients of the two tangents from the origin to the parabola. (i) Complete the square: (ii) The notation for a parabola is. In this parabola, the is the and the is the.. On a parabola with equation the focus is the coordinate of the vertex, - On this parabola the vertex is and (see above) The equation of the directrix of parabola, vertex is the In this parabola, the equation of the directrix is Directrix

9 (i) Substitute into the equation of the parabola, (ii) The gradient of the two tangents are. The quadratic equation above is in. For each point on the line, there is only 1 value of (ie the function is one-to-one), so there is only one root of the quadratic equation. This means that the discriminant of the quadratic formula, must be equal to 0, otherwise there will be two values of. In the above quadratic formula, Substituting these values into, we get;

10 8. Using mathematical induction, prove that for positive integral values of. (i) The complex number is a cube root of the complex number. Show that is another cube root of. (ii) Write down the real cube root of. Using the result in (i), or otherwise, find the two complex cube roots of, giving your answers in the form Let be the proposition: for all positive integers of. Prove that is true: Left Hand Side, Right Hand Side, is true. Assume that is true, where is the proposition: Prove that is true when is true: The assumption, is true when is true. is true when is true, since is true, must also be true, if is true, is true, positive integral values of, by mathematical induction. Using compound angle identities: is proved to be true for all

11 (i) If is a cube root of then, ( ) (D M T ) Since, is a cube root of, is also a cube root of (ii)from part (i), let, the cube root of is, Another cube root of is ( ) another cube root of is i of. is the final cube root The conjugate pairs arise when there is one real cube root and two complex roots, due to the determinant being present in the quadratic formula. Can also be worked out by using ( )

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