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

Ran out of time, hardest C4 paper ever. Honestly think it was harder than Jan 2013. What a ****ing joke.

I've been getting high 80/low 90s in practice papers. Even with grade boundaries I'm probably going to miss out on an A by a few marks. Absolutely livid - that exam was absolutely horrific in comparison to past papers. I'm pretty sure I dropped marks on nearly every single question.

easiest paper ever, that means high grade boundaries
100UMS - 75
90UMS - 69
80UMS - 63
etc

very very easy, finished after 35 minutes and think i've got 74/75 at least.
what did everyone else think?

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

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

I also got +2.

The integral had 2x in it.

The integral would be (2(0) + ...) - (2(-1) + ...)

= ... + 2

Think I did too.

Happy looking at that mark scheme. I think I missed out the part where you had to find coordinates of Q unless I'm being an idiot and can't remember doing it. I didn't recheck that question though as I thought I'd done it right so I may have done it, but probably not. If I didn't look for the coordinate of Q would that be around 2 marks lost?

UNOFFICIAL MARK SCHEME

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

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

1c) C = 2

1d) $\frac {11}{3} ln(2) -2$ EDIT: I may be wrong on this, since the integral gives $[2x]^{0}_{-1} + \frac {11}{3} ln(2)$ which gives $\frac {11}{3} ln(2)+2$

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

UNOFFICIAL MARK SCHEME


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

Wasnt it +2 because its 11/3ln2 - (-2)?

Indeed. It was minus minus 2.

It's quite telling about exam nerves that I had to draw a little picture to justify minus minus 2 being positive 2 in my mind

1d. I think this may be +2 , not 100% sure

I also got +2.

The integral had 2x in it.

The limits were x=0 and x=-1, so the integral would be (2(0) + ...) - (2(-1) + ...)

= ... + 2

I also got 11/3 ln 2 + 2 and when I checked it on the calculator it was right, but someone who got 7/2 ln 2 +2 also said they checked it on the calculator and got it right?

I got every single answer on the unofficial mark scheme besides the position vector of E. I got to the stage where I had a quadratic equation with 17u^2 I think it was. So hopefully I picked up 1/2 marks
I don't quite remember the signs of position vector of D but I remember mew=-1 and it involved (5,1,2) in that order.

Oh and for the sin(alpha) question I drew the triangle but I forgot to give a value for cos(alpha). But everything else seems legit. Confidence is sky high now

I got every single answer on the unofficial mark scheme besides the position vector of E. I got to the stage where I had a quadratic equation with 17u^2 I think it was. So hopefully I picked up 1/2 marks
I don't quite remember the signs of position vector of D but I remember mew=-1 and it involved (5,1,2) in that order.

Oh and for the sin(alpha) question I drew the triangle but I forgot to give a value for cos(alpha). But everything else seems legit. Confidence is sky high now