AQA Core 4 - Monday 10th June 2013 (AM) - Official Thread

365 cm

Feeling a little better after letting it sink in. Vectors part d (maybe c) and question 2 is what I did incorrectly, probably a few other small mistakes.

Got the same D value for vectors, then for E I can't remember the coordinates but remember the constant was +/- 3?

Yeah got 1.3 as constant, and something like 12 for k, maybe 12/pi? Just had to guess really



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Unofficial mark scheme guy - and everyone else - I got positive one for the value of B in the partial fractions question

Unofficial mark scheme guy - and everyone else - I got positive one for the value of B in the partial fractions question

Unofficial mark scheme guy - and everyone else - I got positive one for the value of B in the partial fractions question

How many marks do you guys think I'll lose for only getting up to "x^2 = ____" on the last question? I'm pretty sure the value of X^2 was right, but I didn't have time to try to finish the question.

PS: guys, for the last question I swear the initial height was 4m, therefore surely the answer can't be 3.something... you must have to do something to the 3.something value. It probably has something to do with the 75 second oscillating thing mentioned in the question.

Unofficial mark scheme guy - and everyone else - I got positive one for the value of B in the partial fractions question

Same

i got 11/3 ln2 + 2 but there seems to be some dispute over this

i thought if you integrated y=2 between 0 and -1, the area is just a rectangle of base 1 and height 2..., so positive 2
it doesn't go under the x-axis at any point

I probably sound like a total idiot

UNOFFICIAL MARK SCHEME

1a) Partial fractions: A = 3, B = 1

1b) Integral = $\frac {11}{3} ln(2)$

1c) C = 2

1d) $\frac {11}{3} ln(2) -2$

2a) Show $sin \alpha = \frac{2}{3}$ and hence find $cos \alpha = \frac {\sqrt{5}}{3}$

2b) $sin 2\alpha = \frac{4\sqrt{5}}{9}$

2c) $\frac{2}{15}(5-\sqrt{5})$

3a) Binomial expansion: $1 -2x +8x^2$

3b) $(27-6x)^{\frac{1}{3}} = [27(1-\frac{6}{27}x)]^{\frac{1}{3}} = \frac{1}{3}[1-2(\frac{x}{27}) + 8(\frac{1}{27}x)^{2}] = \frac{1}{3} - \frac{2}{81}x + \frac{8}{2187}x^2$

3c) $\frac{2}{\sqrt[3]{28}} = 2[27+6(\frac{1}{6})]^{\frac{1}{3}}$ which gave $\sqrt[3]{\frac{2}{7}} = 0.659...$ (cannot remember to 6 d.p.)

4a) Show (2x + 3) is a factor of f(x)

4b) $f(x) = (2x+3)(2x^2 -3x -1)$

4c) Show that function with $sin(\theta)$ and $cos(2\theta)$ reduces to f(x) where $x = sin \theta$

4d) Solve for theta: $(2x+3)(2x^2 -3x -1) = 0$

$sin\theta = \frac{-3}{2}$ has no solutions

$2x^2 -3x -1 = 0$

$x = sin\theta = \frac {3 \pm\sqrt{17}}{4}$

$\theta = 196^{\circ}, 344^{\circ}$

5a) Parametric equations: $\frac {dy}{dx} = \frac {4e^{2t}}{-16e^{-2t}}$

5b) Gradient of tangent was -4

5c) Co ordinates of P were (-2, 12)

5d) Gradient of normal was $y-12=\frac{1}{4}(x+2)$

When y = 0, x = -50 and therefore Q (which was the point it cuts the x axis) was (-50,0)

6a) Find vector AB (can't quite remember this)

6b) Vector line equation through AB

6c) Point D was at (5, 1, 2)

6d) For the two values of E I got $(\pm\frac{2}{3}, \pm 1, \mp\frac{2}{3})$

7) $\frac{dh}{dt} = a cos(kt)$ where $a=1.3$ and $k=\frac{\pi}{6}$

8a) Indefinite integral: $\int t cos(\frac{\pi}{4}t) dt$

By parts, this comes out to be:

$\frac {4}{\pi}t sin (\frac{\pi}{4}t) + \frac {16}{\pi^2}cos ({\pi}{4}t) + c$

8b) Differential equation (Courtesy of Nebula, post #600):

$16 \, {\rm x}^{2} = \dfrac{4 \, t \sin\left(\frac{1}{4} \, \pi t\right)}{\pi} + \dfrac{16 \, \cos\left(\frac{1}{4} \, \pi t\right)}{\pi^{2}} - \dfrac{16}{\pi^{2}} + 256$

$x = 3.65 m \quad (=3.64621 \ldots)$ or $365 cm$


Do feel free to correct me! I feel I have missed some questions out too