C3 Unofficial Mark Scheme June 2018 (Edexcel)

Enjoy the beauty of c3. What a long and difficult question paper. *I will be doing a similar mark scheme for c4 on Friday after the exam*

My grade boundary predictions are 72 for 100 UMS, 64 for A*, 56 for A, 50 for B, 43 for C, 35 for D . Please do not reply this full mark scheme on the thread, just give a separate message.

Worked solutions by Reesharr : https://drive.google.com/file/d/15fOwphLqlqd5TmWJ38tTDaJHzO-oWVk-/view?usp=sharing

Worked solutions by ward309 : https://1drv.ms/b/s!Art0vQqZgjKygYAyYnBDyT0bPqHvmw

Blank question paper : https://drive.google.com/open?id=0B3cAgWQuEcKPOTBaVVhObTZTWEdac3pRbVpWN0ZxYXlVbi1v

Q1: (Total 6 marks)

Given

y=2x(3x-1)^{5}

,

a) find

dy/dx

, giving your answer as a single fully factorised expression. (4)

Answer :

dy/dx = 2(3x-1)^{4}(18x-1)

b) Hence find the set of values for x for which

dy/dx \leq 0

. (2)

Answer :

x = \frac {1}{3}

, or

x \leq \frac{1}{18}

Q2: (Total 7 marks)

The function

f

is defined by

f(x) = \frac{6}{2x+5} + \frac{2}{2x-5} + \frac{60}{4x^2-25}

,

x > 4

a) Show that

f(x) = \frac{A}{Bx+C}

, where A, B and C are constants to be found. (4)

Answer:

\frac {8}{2x-5}

b) Find

f^{-1}(x)

and state its domain. (3)

Answer:

f^{-1}(x) = \frac{8+5x}{2x}

, domain

0 < x < \frac{8}{3}

Worked solution by me : https://drive.google.com/open?id=0B3cAgWQuEcKPTVJQVXlkRzJhSG1wV0xNLXh3NWlERGdNdGJv

Q3: (Total 9 marks)

The value of a car is modelled by the formula

V = 16000e^{-kt} + A

, where V is the value of the car in pounds and t is the age of the car in years.

Given that the value of the car £17,500 when new and £13500 two years later,

a) find the value of A (1)

Answer :

A = 1500

, because 17500 = 16000 + A

b) show that

k = ln(\frac {2}{\sqrt{3}})

(4)

c) Find the value of the car, in years, when the value of the car is £6000. Give your answer to 2 decimal places. (4)

Answer:

t = 8.82

years (2 decimal places). Exact value was

\frac{ln(\frac{32}{9})}{k}

Worked solution by me : https://drive.google.com/open?id=1GdhH6ZbyRf7gO3qPYhr5faTw1KaYVU4s

Q4: (Total 9 marks)

The curve C has the equation

y = e^{-2x} + x^2 - 3

. The point A is where the curve C crosses the y-axis. The line

l

is the normal to the curve at the point A.

a) Find an equation of the line

l

, in the form

y = mx + c

, where m and are constants. (5)

Answer :

dy/dx = -2e^{-2x} + 2x

. substitute

x = 0

,

dy/dx = -2

. Therefore gradient of the normal was

\frac{1}{2}

.

substitute x = 0 into y to get y-coordinate of A, which was -2.

A(0, -2)

so equation of line was

y = \frac{1}{2} x - 2

b) The line

l

then meets C again at the point B. Show that the x coordinate of B is a solution of

x = \sqrt{1 + \frac{1}{2} x - e^{-2x}}

. (2)

c) Use this as an iterative formula with

x_1 = 1

to find

x_2

and

x_3

to 3 decimal places (2)

x_2 = 1.168

and

x_3 = 1.220

(both to 3 decimal places)

Worked solution by me : https://drive.google.com/open?id=14KCcRd1neFTP0CMoPhqsekkBl-kWGaS8

Q5: (Total 8 marks)

Graph of

y = f(x)

, where

f(x) = 2|5-x| + 3

,

x \geq 0

. Given that the equation

f(x) = k

, where k is a constant, has exactly one real root,

a) state the set of possible values of k. (2)

Answer :

k = 3

or

k > 13

b) Solve the equation

f(x) = \frac{1}{2} x + 10

(4)

Answers :

x = \frac{6}{5}

or

x = \frac{34}{3}

The graph with equation

y = f(x)

is transformed onto the graph with equation

y = 4f(x-1)

. The vertex on the graph with equation

y = 4f(x-1)

has coordinates

(p, q)

.

c) State the value of p and the value of q. (2)

Answer: The original vertex of

f(x)

was

(5, 3)

. Shift one right and then stretch vertically factor 4. So

p = 6

,

q = 12

Worked solution by me (for 5b) : https://drive.google.com/open?id=1nmQKm7h-QRzApWRpQy6wYqdyv-QgXLA6

Q6: (Total 11 marks)

a) Using the identity for tan (A+B), solve, for

-90^{\circ} < x < 90^{\circ}

,

\frac{tan2x + tan32}{1-tan2xtan32} = 5

. Give your answers, in degress, to 2 decimal places. (4)

Answers:

x = -66.65^{\circ} , x = 23.35^{\circ}

(2 decimal places)

b) Using the identity for tan(A + B) , show that

tan(3\theta - 45) = \frac{tan3\theta - 1}{1+tan3\theta}

,

\theta \neq (60n + 45^{\circ}), n \in \mathbb{Z}

(2)

c) Hence solve, for

0^{\circ} < x < 180^{\circ}

, the equation

(1 + tan3\theta)(tan\theta + 28) = (tan3\theta - 1)

. (5)

Answers:

\theta = 36.5^{\circ}, \theta = 126.5^{\circ}

Worked solution by me : https://drive.google.com/open?id=0B3cAgWQuEcKPV3V1UDREN3pTcnRNSUNMUktlTHRYT0FsRUZv

Q7: (Total 9 marks)

The curve C has equation

y = \frac{ln(x^2 + 1)}{(x^2 + 1)}

,

x \in \mathbb{R}

a) Find

dy/dx

as a single fraction, simplifying your answer. (3)

Answer :

dy/dx = \frac{2x(1- ln(x^2 + 1)}{(x^2 + 1)^2}

b) Hence find the exact coordinates of the stationary points of C. (6) - eep the bit I got wrong forgot the negative sq root

Answer:

2x = 0

, therefore

x = 0

. When x = 0, y = 0 so

(0, 0)

is a stationary point.

Also,

ln(x^2 + 1) = 1

. This simplifies to

x^2 = e -1

. So

x = \sqrt{e-1} or x = -\sqrt{e-1}

Both these points give a y coordinate of

\frac{1}{e}

(0, 0) , (\sqrt{e-1}, \frac{1}{e}) , (-\sqrt{e-1}, \frac{1}{e})

Worked solution by me : https://drive.google.com/open?id=1x4DQICTtSIxMeOumjymapDf2Q7WFeOCU

Q8: (Total 7 marks)

a) By writing

sec\theta = \frac{1}{cos\theta}

, show that

d/d\theta (sec\theta) = sec\theta tan\theta

(2)

Given that

Unparseable latex formula:

x = e^\sec y

,

x > e , 0 < y < \frac{\pi}{2}

b) show that

dy/dx = \frac{1}{x\sqrt{g(x)}}

,

x > e

, where

g(x)

is a function of

ln(x)

. (5)

Answer :

dy/dx = \frac{1}{x\sqrt{(lnx)^4 - (lnx)^2}}

Worked solution by me : https://drive.google.com/open?id=0B3cAgWQuEcKPOTJER2Jfa3I3c2pYYnd4QXpjb2ZLbFBuWmRN

Q9: (Total 9 marks)

a) Express

sin\theta - 2cos\theta

in the form

Rsin(\theta-\alpha)

, where

R > 0

and

0 < \alpha < \frac{\pi}{2}

. Give the exact value of R and the value of

\alpha

, in radians, to 3 decimal places. (3)

Answer :

\sqrt{5} sin(\theta-1.107)

b)

M(\theta) = 40 + (3sin\theta - 6cos\theta)^2

. Find the maximum value of

M(\theta)

and the smallest value of

\theta

, in the range

0 < \theta < 2\pi

, at which this maximum value of

M(\theta)

occurs. (3)

Maximum value for

M(\theta) = 85

Occurred when

Unparseable latex formula:

\theta $\approx$ 2.678

. Exact value was

\frac{\pi}{2} + arctan(2)

c)

N(\theta) = \frac{30}{5+2(sin2\theta - 2cos2\theta)^2}

. Find the maximum value of

N(\theta)

, and the largest value of

\theta

, in the range

0 < \theta < 2\pi

, for which this maximum value of

N(\theta)

occurs. (3)

Maximum value for

N(\theta ) = \frac{30}{5} = 6

(3)

Occured when

Unparseable latex formula:

\theta $\approx$ 5.266

. Exact value was

\frac{3\pi + arctan(2)}{2}

Worked solution by me : https://drive.google.com/open?id=0B3cAgWQuEcKPdVhGOGdfcjB4bGxfV1NnRGx5RkpQNTZEQ1ZZ

Explanation for Q5a

Attachment not found

Our graph in question was defined for x > 0.

So a horizontal line should be drawn for f(x) = k. The line should only intersect the main red line once. Which it does for values of k > 13 or where k = 3

(edited 5 years ago)

Reply 1

bazinga19

10

Thanks so much but where's the rest? :smile:

Reply 2

16121999

3

thank you!! but wheres the rest also would you be able to put the marks for each question on there too? thanks

Reply 3

Ellis.Collins

3

so helpful, cant wait for 6, 7 and 8 to be on here!

Reply 4

Swimming137

3

Best markscheme so far - can't follow the answers on the other unofficial markscheme page. Can you upload the rest pls? And marks if you have them would be super super helpful :smile:

Reply 5

mlyke

10

Please can someone explain how they got they're theta value for 8c that's the only one I didn't get

Reply 6

Swimming137

3

Anyone know where I can find out how many marks each question is worth?

Reply 7

ricogordon235

11

Brilliant mark scheme brother, clear and concise so far I have dropped 16 marks by this mark scheme, how many more questions are missing would be a massive help if someone could inform me!

Original post by Chillichamps

Q1:

y=2x(3x-1)^{5}

a) find

dy/dx

dy/dx = 2(3x-1)^{4}(18x-1)

b) Hence find the set of values for x for which

dy/dx \leq 0

x = \frac {1}{3}

, or

x \leq \frac{1}{18}

Q2:

a) Simplify

\frac{6}{2x+5} + \frac{2}{2x-5} + \frac{60}{4x^2-25}

Answer:

\frac {8}{2x-5}

b) find

f^-1(x)

and state its domain

Answer:

f^-1(x) = \frac{8+5x}{2x}

, domain

0 < x < \frac{8}{3}

Q3:

V =

16000e^-kt + A

. When t = 0, V = 17500 and when t=2, V = 13500

a)

A = 1500

, because 17500 = 16000 + A

b) show that

k = ln(\frac {2}{\sqrt{3}})

c) when V = 6000,

t = 8.82

years (3 significant figures)

Q4:

The curve C has the equation

y = e^-2x + x^2 - 3

. The point A is where the curve C crosses the y-axis.

a) Find an equation for the normal at the point A in the form

y = mx + c

dy/dx = -2e^-2x + 2x

. substitute

x = 0

,

dy/dx = -2

. Therefore gradient of the normal was

\frac{1}{2}

.

substitute x = 0 into y to get y-coordinate of A, which was -2. A(0, -2) so equation of line was

y = \frac{1}{2} x - 2

b) The normal then intersects C again at the point B. Show that the x coordinate of B is a solution of

x = \sqrt{1 + \frac{1}{2} x - e^-2x}

.

c) Use this as an iterative formula with

x_1 = 1

x_2 = 1.168

and

x_3 = 1.220

(both to 3 decimal places)

Q5:

Graph of

y = 2|5-x| + 3

a) Find the set of values of k such that

f(x) = k

, where k is a constant, has exactly one root.

Answer :

k = 3

or

k > 13

b) Solve the equation

f(x) = \frac{1}{2} x + 10

Answers,

x = \frac{6}{5}

or

x = \frac{34}{3}

c) The graph

f(x)

is transformed into the graph

4f(x-1)

. The new coordinates of the vertex are

(p, q)

. State the values of p and q.

Answer: The original vertex of

f(x)

was

(5, 3)

. Shift one right and then stretch vertically factor 4. So

p = 6

,

q = 12

Q6:

y = \frac{ln(x^2 + 1)}{(x^2 + 1)}

a) find

dy/dx

dy/dx = \frac{2x(1- ln(x^2 + 1)}{(x^2 + 1)^2}

b) Hence find the the stationary points

Answer:

2x = 0

, therefore

x = 0

. When x = 0, y = 0 so

(0, 0)

is a stationary point.

Also,

ln(x^2 + 1) = 1

. This simplifies to

x^2 = e -1

. So

x = \sqrt{e-1} or x = -\sqrt{e-1}

Both these points give a y coordinate of

\frac{1}{e}

(0, 0) , (\sqrt{e-1}, \frac{1}{e}) , (-\sqrt{e-1}, \frac{1}{e})

Q7:

a) Solve the equation

\frac{tan2x + tan32}{1-tan2xtan32} = 5

Answers,

x = -66.65 , x = 23.35

(2 decimal places)

b) Show that

tan(3\theta + 45) = \frac{tan3\theta + 1}{1-tan3\theta}

c) Solve the equation

(1- tan3\theta)(tan\theta + 28) = (tan3\theta + 1)

Answers,

\theta = 36.5 , \theta = 126.5

Q8:

a) Show that

d/d\theta (sec\theta) = sec\theta tan\theta

b) Find

dy/dx

in terms of x

dy/dx = \frac{1}{x\sqrt{lnx^4 - lnx^2}}

Q9:

a) find

Rsin(x-\alpha)

Answer :

\sqrt{5} sin(x-1.107)

b) Minimum value for

M(\theta) = 85

Occurred when

\theta = 2.678

(3 decimal places)

c) Maximum value for

N(\theta ) = \frac{30}{5} = 6

Occured when

\theta= 5.266

(3 decimal places)

Reply 8

Atiki

6

7a - Isn't only one solution? U have to remove the negative cuz it's out of range.

Reply 9

Chillichamps

OP

8

Original post by Atiki

7a - Isn't only one solution? U have to remove the negative cuz it's out of range.

Sorry, but the range was from

-90 < \theta < 90

Reply 10

Chillichamps

OP

8

Original post by ricogordon235

Brilliant mark scheme brother, clear and concise so far I have dropped 16 marks by this mark scheme, how many more questions are missing would be a massive help if someone could inform me!

Not sure if anything is missing, don't think so

Reply 11

viki123

3

Original post by Chillichamps

Sorry, but the range was from

-90 < \theta < 90

Im pretty sure it was -66.66

Reply 12

11ayuban

2

Original post by Swimming137

Anyone know where I can find out how many marks each question is worth?

Yeah that’s what I’m looking for too. I need to know what potential grade I got 😭

Reply 13

Chillichamps

OP

8

Marks for each question I am doing right now, will be on the mark scheme in a few minutes. The reason only half was on it is because I pressed save halfway , and it took some time to do the final few questions. Hope it helps everyone

(edited 5 years ago)

Thank you

Original post by ricogordon235

Brilliant mark scheme brother, clear and concise so far I have dropped 16 marks by this mark scheme, how many more questions are missing would be a massive help if someone could inform me!