Further Maths Progress Check due Wed 10.2.16

Total: 51 marks. Show all working out. Those not showing mathematical rigour will be penalised.

1. Solve \[ \sin (2x-\frac{\pi}{2}) = -\frac{1}{2}\] for $-\pi \leq x \leq \pi$

   [5 marks]

2. By expressing $\cos x + \sin x$ in the form $R\cos(x- \alpha)$ with $0 \leq \alpha \leq \frac{\pi}{2}$, find the maximum value of \[2 - \cos x - \sin x\] State the smallest positive value of x for which this occurs.

   [5 marks]

3. Prove that \[ \sin 4A + \sin 2A \equiv 2\sin 3A \cos A \]

   [4 marks]

4. Solve \[ \cos \theta + 1 = 2 \sec \theta\] for $-\pi \leq x \leq \pi$

   [4 marks]

5. A is acute and B is obtuse. \[ \text{cosec} A = \frac{5}{3} \\ \sec B = -\frac{13}{5} \] Find $\tan (A+B)$ without a calculator

   [4 marks]

6. \[ f(x) = x^3 - ax^2 + x + b\] $(x-2)$ is a factor of $f(x)$ and the remainder is 5 when $f(x)$ is divided by $(2x+1)$. Find $f(3)$.

   [5 marks]

7. \[ \frac{x^4-x-1}{x^2+2} \equiv ax^2 +bx+c + \frac{dx+e}{x^2+2} \]

   [4 marks]

8. Simply as far as possible \[ 1+ \frac{2x}{x^2-2x-8} - \frac{6}{x^2-16} \]

   [4 marks]

9. Express in partial fractions:
   1. \[ \frac{2x}{(x^2-4)(x+1)} \]
   2. \[ \frac{2-x}{(x^2-4)(x+2)} \]
   3. \[ \frac{3x+2}{(x^2+4)(x+1)} \]
   4. \[ \frac{x^3}{(x^2-1)(x+1)} \]

   [16 marks]

Design your maths camp t shirt

If you are in upper VI and are going to Grinton Maths Camp, why not submit a t-shirt design? Deadline 10th Feb. Email djy

Further Maths Homework due 3.2.16

Total: 27 marks. Show all working out. Those not showing mathematical rigour will be penalised.

1. Find the distance between this line and parallel plane: \[ \underline r = \left(\begin{array}{c} -5\\2\\1\\\end{array}\right) + \lambda \left(\begin{array}{c} -3\\1\\2\\\end{array}\right) \\ \underline r \cdot \left(\begin{array}{c} 1\\3\\0\\\end{array}\right) = 4 \]

   [6 marks]

2. Find the equation of the line where these 2 planes meet in the form $ \underline r \times \underline u = \underline v$ \[x+3y-z=2 \\ 2x-y-z = 1 \]

   [4 marks]

3. Find the eigenvalues and assosciated normalised eigenvectors for this linear transformation: \[ A = \left(\begin{array}{ccc} -2&-4&2 \\ -2&1&2\\ 4&2&5 \\ \end{array}\right) \]

   [8 marks]

4. Find the 3x3 matrix for the transformation represented by T. \[ T: \left(\begin{array}{c} x\\y\\z\\\end{array}\right) \rightarrow \left(\begin{array}{c} x+y\\x-2y\\3z\\\end{array}\right) \] Find the image of the line: \[ \underline r = \left(\begin{array}{c} -5\\2\\1\\\end{array}\right) + \lambda \left(\begin{array}{c} -3\\1\\2\\\end{array}\right) \]

   [4 marks]

5. Find the the shortest distance between these lines \[ \underline r = \left(\begin{array}{c} -1\\1\\0\\\end{array}\right) + \lambda \left(\begin{array}{c} 3\\1\\0\\\end{array}\right) \\ \underline r = \left(\begin{array}{c} -2\\3\\-1\\\end{array}\right) + \mu \left(\begin{array}{c} -3\\4\\2\\\end{array}\right) \]

   [5 marks]

Further Maths Homework Due Mon 7th Dec

Total: 30 marks. Show all working out. Those not showing mathematical rigour will be penalised.
A=24, B=21, C=18

1. Find the gradient function for \[y=\frac{x-2}{x^2}\]

   [3 marks]

2. Find the coordinates of any stationary points and determine their nature \[y=\frac{1+x^2}{x}\]

   [5 marks]

3. Find the equations of the tangent and the normal when $x=4$ in the form $ax+by=c$ where a,b and c are integers \[ y=\frac{4\sqrt{x}-x^2}{x^{\frac{3}{2}}}\]

   [7 marks]

4. Find \[ \int{(x-2)^3} dx \]

   [3 marks]

5. Find $y$ if (1,-2) is a boundary condition\[ \frac{dy}{dx}= \frac{x-1}{x^3} \]

   [4 marks]

6. Find \[ \int_0^2{(2x-3)^4} dx \]

   [4 marks]

7. Find the area enclosed by the curve $y=1-x^2$ and the x axis

   [4 marks]

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]