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]

6.6 FM Progress Homework 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.

   [3 marks]

2. Find \[ \sum_{r=5}^{30} \frac{5-2r}{3} \]

   [3 marks]

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

   [4 marks]

4. Find $u_n$ if $S_n = 2n^2+3n+1$

   [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

   [4 marks]

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

   [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.

   [4 marks]

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

   [3 marks]

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

   [5 marks]

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

    [3 marks]

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

    [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.

    [4 marks]

6.6 FM Progress answers 5.10.15

Total: 25 marks. Show all working out. Those not showing mathematical rigour will be penalised.

1. Factorise fully $6x^2-7x-20 = (3x+4)(2x-5)$

   [2 marks]

2. Complete the square: \[ y=5-2x-3x^2 = -3(x+\frac{1}{3})^2 + \frac{16}{3} \]

   

   Has max at (-1/3 , 16/3)
   y intercept is 5
   x intercepts are -5/3 and 1

   [4 marks]

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

   

   
   y intercept is 75
   cross at x=3 with touch x=-5/2
   negative cubic shape

   [4 marks]

4. \[ y^2=x^2+x \\ y=2x+k \\ 4x^2+4kx+k^2 = x^2+x \\ 3x^2+x(4k-1)+k^2=0 \\ \therefore (4k-1)^2-12k^2<0 \\4k^2-8k+1<0 \\ \frac{2-\sqrt 3}{2} < k < \frac{2+\sqrt 3}{2} \]Find the values of k such that the hyperbola and the line do not meet.

   [4 marks]

5. Find the centre and radius of the circle that passes through A(9,3), B(13,-5) and C(-5,-11) \[ (x-3)^2 + (y+5)^2 = 100 \]

   [5 marks]

6. Solve: \[ \frac{3}{27^{5-x}}=\sqrt{81^{5-2x}} \\ \\ 3^{1-3(5-x)}=3^{\frac{4(5-2x)}{2}} \\ \\ 1-15+3x=10-4x \\ x= \frac{24}{7}\]

   [3 marks]

7. Simplify as far as possible: \[ \frac{15}{\sqrt 5} - \frac{6}{\sqrt 7 - \sqrt 5} = -3\sqrt 7 \]

   [3 marks]

4.2 Homework Answers Due 28 Sept 15

Complete neatly in the front of your books. Calculator allowed but working must be shown. Total: 21 marks

1. A teacher has a 3% pay increase to £18,200. What was his previous salary? \[18200 \div 1.03 = 17669.90\]

   [3 marks]

2. £3000 is deposited in a bank account with an interest rate of 3.1% p.a., what will the balance be after 8 years? \[3000 \times 1.031^8 = 3829.93 \]

   [2 marks]

3. Without a calculator, find the value of \[ \frac{5.6 \times 10^{-5}}{8 \times 10^{-8}} = 0.7 \times 10^3 = 7 \times 10^2 \] Give your answer in standard form.

   [2 marks]

4. Solve \[ \text{Start by multiplying both sides by 4} \\ \frac{3x-2}{2}=\frac{4-x}{4} \\ 2(3x-2)=4-x \\ 6x-4=4-x \\ 7x=8 \\ x = \frac{8}{7} \]

   [3 marks]

5. Solve \[ 3(2x+3)-4(5-3x)=1 \\ 6x+9-20+12x=1 \\ 18x-11=1 \\ 18x=12 \\ x=\frac{2}{3} \]

   [3 marks]

6. Find the gradient of the line between A(-1,4) and B(3,-8) and hence find the equation of the line \[ \text{Gradient is} -3 \\ y=-3x+c \\ 4=3 +c \\ c=1 \\ \therefore y=1-3x \]

   [4 marks]

7. Pete's son is a third of his age. 8 years ago, Pete was four times his son's age. How old is Pete today? Show your method clearly. Guesses get nothing!! \[ \begin{array}{|c|c|c|} \hline \ & Pete & Son \\ \hline \ \text{Now} & x & \frac{x}{3} \\ \hline \ \text{8 yrs ago} & x-8 & \frac{x}{3} - 8 \\ \hline \end{array} \\ \text{Let Pete's age be }x \\x-8=4(\frac{x}{3} - 8) \\ x-8 = \frac{4x}{3} - 32 \\ \frac{x}{3} = 24 \\ x=72 \text{ years old}\]

   [4 marks]