2016 Edexcel C2 Morning exam[discussion thread]

My answers:

1 b) (2 marks) $256$
c) (2 marks) $1.602$

2 a) (1 mark) $7$
b) (3 marks) $20.75$
c) (2 marks) $5.75$

3 a) (2 marks) $\sqrt{34}$
c) (4 marks) $3x+5y-95=0$

4 a) (2 marks) $5$
b) (2 marks) $f(-2)=0$
c) (4 marks) $f(x)=(x+2)(3x+2)(2x-1)$

5 a) (4 marks) $16-288x+1944x^2$
b) (1 mark) $16$
c) (2 marks) $\frac{7}{2}$

d) (2 marks) $936$

6 i) (3 marks) $\frac{8\pi}{15}$or $\frac{-2\pi}{15}$

ii) (6 marks) or

7 b) (3 marks) $24.3$

8 ii) (4 marks) $-2.19$

9 a) (2 marks) $\frac{\pi x^2}{3}$
d) (5 marks) $x=16.63$ $P= 120m$
c) (2 marks) $f''x = 0.437 > 0 \therefore$ is a minimum at $x$

my unofficial mark scheme

Thanks! But how come the marks don't add up?

I didn't include the proofs.

My answers:

1 b) (2 marks) $256$
c) (2 marks) $1.602$

2 a) (1 mark) $7$
b) (3 marks) $20.75$
c) (2 marks) $5.75$

3 a) (2 marks) $\sqrt{34}$
c) (4 marks) $3x+5y-95=0$

4 a) (2 marks) $5$
b) (2 marks) $f(-2)=0$
c) (4 marks) $f(x)=(x+2)(3x+2)(2x-1)$

5 a) (4 marks) $16-288x+1944x^2$
b) (1 mark) $16$
c) (2 marks) $\frac{7}{2}$

d) (2 marks) $936$

6 i) (3 marks) $\frac{8\pi}{15}$or $\frac{-2\pi}{15}$

ii) (6 marks) or

7 b) (3 marks) $24.3$

8 ii) (4 marks) $-2.19$

9 a) (2 marks) $\frac{\pi x^2}{3}$
d) (5 marks) $x=16.63$ $P= 120m$
c) (2 marks) $f''x = 0.437 > 0 \therefore$ is a minimum at $x$

my unofficial mark scheme

That was so easy😂
Expect grade boundries to be high.
You must be stupid to have not got at least 70+

can anyone remember the total marks for Q9?

I got the difference between u9 and u10 negative does that matter

Guys it asked for P not x, so 17 is wrong because you had to substitute in x=16.633 and then you get 120 to the nearest metre


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I got the difference between u9 and u10 negative does that matter

What did everyone get when substituting their x value into the 2nd differential to show it was a minimum point? (very last question part e)

I got a really small decimal??

I actually laughed at everyone saying they got 17...

CLASSSIICCC mistake, they forgot to substitute back into the original formula for P. I literally came so close to keeping my answer of 17 before realizing i didnt sub it back in, yeah the actual answer was 120.something and rounded to 120.

For the Next part you sub this number into the second differential.

P'' = 2000/x^3 = 0.0*something tiny but bigger than zero*

because P'' > 0 (and P' = 0) we know its a minimum point.

0.4 something something

it was like 1.5 x 10^-3, i remember 0.4 way too big but i dont think the size matters, just as long as its bigger than zero