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C3 Edexcel A level 19th June 2018 Unofficial Markscheme

CREDZ TO CHILLICHAMPZ

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

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) Exact value was \frac{\pi}{2} + arctan(2)

c) Maximum value for N(\theta ) = \frac{30}{5} = 6

Occured when \theta= 5.266 (3 decimal places) Exact value was \frac{3\pi + arctan(2)}{2}
(edited 5 years ago)

Scroll to see replies

Well I ****ed that up lol
Reply 2
How were you supposed to do the tan(3x - 45) = tan(x + 28) one?
(edited 5 years ago)
Original post by Jimmybee123
Well I ****ed that up lol


Same dw
Got (0,0) for the stationary points :frown:
I think I did alright but that tan question was hard and what was that last question
Reply 6
Original post by MA28613
How were you supposed to do the tan(3x - 45) = tan(x - 28) one?



Turn both tans into sin/cos, then cross multiply. it ends up being a sin identity =0 from sin(a-b)
this is a mess bc cant remember all my answers or their orders but:
Q1
x_<18

k>13

car question
A= 1500
proof
8.28 years (or something)

f-1(x)= 4/x + 5/2
domain x>8/3
asymptote at x=5/2

exponential question
1/xsqrt(lnx)^2 + 1

stationary points (0,0) (sqrt e-1, 1/e)

some tan identity question
66.66° and 100 something°??

tan(3ø + 45) = tan(ø + 28)
36.5°
126.5°


Q8
max value 95
angle 2.68
max value 6
angle 5.27
Reply 8
**** the tan question
Original post by MA28613
How were you supposed to do the tan(3x - 45) = tan(x - 28) one?


I did arctan each side to get 3x-45 = x-28
Reply 10
Much easier to just get rid of the tan on both sides!
Reply 11
how are people getting 95 as a max value for 9, think i got 55 :/
Guys who remembers the last question. I think i accidently left it out...
Original post by JoeDDM
Much easier to just get rid of the tan on both sides!


You don't get all the solutions
Reply 14
Original post by MA28613
How were you supposed to do the tan(3x - 45) = tan(x - 28) one?



Turn both tans into sin/cos, then cross multiply. it ends up being a sin identity =0 from sin(a-b)
Reply 15
same!!
Original post by jaspreetb
Turn both tans into sin/cos, then cross multiply. it ends up being a sin identity =0 from sin(a-b)


😑😑really
G (x)= sqroot 4lnx-1
Who else got this?
Reply 18
Original post by Mazza2000
Guys who remembers the last question. I think i accidently left it out...

It was about Rsin(theta-alpha) one
Original post by cenet0:o
You don't get all the solutions


i used that method, but u can get all of the solutions if you just remember to add 180

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