6.6 Fm Answers due 14.10.15

Total: 43 marks. Show all working out. Those not showing mathematical rigour will be penalised.

1. The third term of an AP is 24 whilst the 7th is -12. Find the sum of the first 20 terms. \[ a+2d=24 \\ a+6d=-12 \\ 4d=-36 \therefore d=-9 \text{ and }a=42 \\s_{20}=\frac{20}{2}(84+19(-9))=-870 \]

   [3 marks]

2. Find \[ \sum_{r=5}^{30} \frac{5-2r}{3} = -\frac{5}{3} + -\frac{7}{3} + -\frac{9}{3} +... \\ = \frac{26}{2}\left(-\frac{5}{3}+-\frac{55}{3}\right)= -260\]

   [3 marks]

3. Find n s.t. \[ \sum_{r=7}^{n} \frac{4r+1}{3} = 600 \\ n=30\]

   [4 marks]

4. Find the $u_n$ if $S_n = 2n^2+3n+1$ \[u_n= S_n - S_{n-1} \\ = 4n+1 \\ \text{ (after simplification)}\]

   [3 marks]

5. Complete the square: $y=5-4x-2x^2$ and use your answer to sketch y, indicating clearly any points where the curve crosses the axes \[-2(x+1)^2+7\]

   Upside down parabola shape
   y intercept is 5
   x intercepts $-1\pm\sqrt{\frac{7}{2}}$

   [4 marks]

6. Sketch $y=(3-x)^2(x+5)^3$ indicating any intersections with the axes

   y intercept 1125
   x touch at 3
   point of inflexion at x=-5
   basic shape: bottom left to top right

   [4 marks]

7. \[ 4y^2+x^2=36 \\ y+kx=3 \\ \]Find the value(s) of k such that the line is a tangent to the ellipse.

   The only way a line which intersects with the y axis at 3 can be tangent to an ellipse "centred" on the origin is if the tangent is parallel to the x axis so i.e. is of the form y= constant. \[k=0\]

   [4 marks]

8. Find the values of k s.t. $2x^2-x+8=kx$ has no solutions.

   SKETCH THE QUADRATIC \[ (k+1)^2-64<0 \\ \therefore -9\lt k \lt 7 \]

   [3 marks]

9. Find the centre and radius of the circle that passes through A(7,11), B(-10,-6) and C(15,-1)

   Find the equations of 2 of the perpendicular bisectors of AB, AC and BC. Solve these 2 lines simultaneously to find the centre of the circle. Centre (2,-1) and radius 13

   [5 marks]

10. Simplify as far as possible: \[ \frac{4}{\sqrt 3} - \frac{6}{\sqrt 5 - \sqrt 3} \\ = -3\sqrt 5 - \frac{5}{3}\sqrt 3 \]

    [3 marks]

11. Describe the series of transformations which transform $y= \frac{1}{x}$ to the curve \[y=5-\frac{2}{x-2}\]

    Translation of 2 units right
    Stretch in the y direction (from the x axis) scale factor 2
    Reflection in the x axis
    Translation of 5 units up

    [3 marks]

12. $y=3x-4$ is a tangent to a circle at the point (2,2). Given the centre has coords (k,1) find k and hence the equation of the circle.

    The gradient of the tangent is 3 so the gradient of radius is $-\frac{1}{3}$ which implies that, as the centre is 1 unit down from (2,2), it is 3 units right i.e. at (2+3, 2-1) = (5,1). The radius is the distance between these two points $\sqrt{10}$ \[(x-5)^2+(y-1)^2=10\]

    [4 marks]