2016 May 25th Edexcel Core 2 Questions and answers. [Unofficial mark scheme] 2016

Thankyou, sorry I missed that, $\log_3(3b+1)-\log_3(a-2)=-1$, write b in terms of a

What would you say I would get, I wrote a in terms of b?

I am quite pissed. I got full marks in this paper, but for question 1.b. I wrote -1.602. Would that be penalised? And does that mean I can still get full UMS?

see first post/mark scheme 8 i. I have included it.

No I mean I did it the wrong way around?

1 Geometric series question, prove $a=64$ given $S_4=175$ and workout sum to infinity. Then find the difference between the 9th and 10th term

2 Trapezium rule. $y=8-2^{x-1}$ in the interval $[0,4]$ with 4 trapeziums

3 Circle centred at $(7,8)$. Find the equation of it and of a tangent at point $(10,13)$

4 where $f x =6x^3+13x^2-4$ find the remainder when divided by $(2x+3)$ then factorise it fully given $(x+2)$ is a factor.

5 Expansion of $(2-9x)^4$. The using that expand $(1+kx)(2-9x)^4$ in the form $A -232x + Bx^2$ given the coefficient of $x$

6 $1-2\sin(\theta - \frac{\pi}{5})=0$ solve for $\theta$and$4\cos^2 x + 7\sin x - 2 = 0$

7 This was $\int (3x-x^{\frac{3}{2}}) dx$ and then find the limits (where it crossed the $x$ axis.

8 $\log_3(3b+1)-\log_3(a-2)=-1$, write b in terms of a then find $x$ given $2^{2x+5}-7(2^x)=-1$.

9 Find optimum perimeter of a funny shape which comprised a rectangle, sector and a equilateral triangle, need diagram.
Image by Cake_Chan
Equations given, that needed proving are, $y=\frac{500}{x}-\frac{x}{24}(4\pi+3\sqrt{3})$
and $P=\frac{1000}{x}+\frac{x}{24}(4\pi+36-3\sqrt{4})$


My answers:
1 a) (2 marks) proof
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) $\sqrt{34}$
b) (3 marks) $(x-7)^2+(y-8)^2=34$
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 a) (3 marks) $\frac{3}{2}x^2-\frac{2}{5}x^{\frac{5}{2}}+c$
b) (3 marks) $24.3$

8 i) (3 marks) $b= (3a+5)/9$
ii) (4 marks) $-2.19$

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

This is useful and was spread out over another thread so I thought I would post it here. The missed answers were all either proofs or not suited to a single number response.

8i is wrong.

$log_3{(3b+1)} -log_3{(a-2)} = -1$

$log_3{(\frac{3b+1}{a-2})} =-1$

$3^{-1} =\frac{3b+1}{a-2}$

$\frac{a-2}{3} =3b+1$

$a-2 =9b+3$

$b = \frac{a-5}{9}$

see first post/mark scheme 8 i. I have included it.

8.i is wrong. It's b=(a-5)/9

Indeed, I believe 8(i) is (a-5)/9.

Other than that, all of the others seem right. I suppose it's now down to just showing enough on proof questions. Though on Q9 they were only 3 marks each...

Just to point out fort he first question, you knew what r was... I can't remember correctly but I think it was 3/4

8i is wrong.

$log_3{(3b+1)} -log_3{(a-2)} = -1$

$log_3{(\frac{3b+1}{a-2})} =-1$

$3^{-1} =\frac{3b+1}{a-2}$

$\frac{a-2}{3} =3b+1$

$a-2 =9b+3$

$b = \frac{a-5}{9}$

Bro could you write :

1/3 a -5/9 all divided by 3. Is this correct as well?

8 i) (3 marks) $b= \frac{3a-5}{9}$
ii) (4 marks) $-2.19$

for the first part of the logs

I said 1/3 = 3b+1/a-2

then 1/3a -2/3= 3b+1

1/3a-5/3=3b

1/3a-5/3 all divided by 3 = b

Is this correct and how many marks would i lose if incorrect?

No, it would have to be $\frac{1}{9}a -\frac{5}{9}$

I got a different answer to 8i and cant see where I went wrong



I got the same as this