(Core 2) Here's what I can remember (question parts/numbers might be mixed up)

1 (b) sum to infinity = 160

3) (c) Integral = 39/2

For the question that was something like show that the x-coordinate of the stationary point can be written in the form 2^p, where p is a rational number, I got p =6/5

For the long logarithm one where you had to show that the equation had only one solution, I ended up with (x + 3)^2 = 0, so x = -3 was the only solution.

For the last trig equation one, I got x = 16, x = 104, or x = 136.

That's all I can remember for now.

(edited 10 years ago)

Reply 6

stuart_aitken

This is all I want to know:

For core 1

Integration question:
Area under curve was 524/15
Shaded area was 961/15

Final question:
20/7 <= k <= 4

Correct or not...?

Posted from TSR Mobile

Reply 7

Deviceing

a. Write log(a)b = c in index form.
b. Show that only one value of x satisfies 2log(2)(x+7) - log(2)(x-5) = 3

a. a^c = b

b. log(2)x = 3
x = 2^3 = 8
log(2)(x+7)^2 - log(2)(x-5) = log(2)8
log(2)((x+7)^2)/(x-5) = log(2)8
((x-7)^2)/(x-5) = 8
(x+7)^2 = 8(x-5)
x^2 + 14x + 49 = 8x - 40
x^2 + 6x + 9 = 0
(x+3)^2 = 0
x = -3

Reply 8

davon806

Original post by mike256

Here's what I can remember (question parts/numbers might be mixed up)

1 (b) sum to infinity = 160

3) (c) Integral = 39/2

For the question that was something like show that the x-coordinate of the stationary point can be written in the form 2^p, where p is a rational number, I got p =6/5

For the long logarithm one where you had to show that the equation had only one solution, I ended up with (x + 3)^2 = 0, so x = -3 was the only solution.

For the last trig equation one, I got x = 16, x = 104, or x = 136.

That's all I can remember for now.

I got all exactly the same as you!
In addition,for the trapezium rule one,I got 7.12.

For the sum of 12 terms(the last part of the first question),I got 159.96

For the obtuse angle of the sector,I got 1.87(to 3sf)

And for the transformation problem,I guess it is stretching the original graph
surd(8x^3 -1) in the x-direction of scale factor 2

For the g(x) and g(4) question,I got g(4) = 2.3

for the term u3(the one that right after asking you to show p = 2/3),I got 56

For the equation of normal one:

-4/9 = (y-11)/(x-4)
9y-99 = -4x + 16
4x + 9y -115 = 0

Anyone else can share their answers to us :smile:

(edited 10 years ago)

Reply 9

timeteo9

The log question was:

log₂(x+7)² - log₂(x+5) - 3 = 0

→log₂(x²+14x+49) = 3 + log₂(x+5)
→log₂(x²+14x+49) = log₂2³ + log₂(x+5)
→log₂(x²+14x+49) = log₂(8x+40)
→x²+14x+49 = 8x+40
→x²+6x+9 = 0
→(x+3)²

So x = -3

(However, for some reason I thought 14-8=8 so messed up the last bit... :mad:

)

Reply 10

Andrreewwww

Original post by stuart_aitken

This is all I want to know:

For core 1

Integration question:
Area under curve was 524/15
Shaded area was 961/15

Final question:
20/7 <= k <= 4

Correct or not...?

Posted from TSR Mobile

Integration under curve was 63/2 for me and shaded area was 27?

Reply 11

stuart_aitken

Original post by Andrreewwww

Integration under curve was 63/2 for me and shaded area was 27?

Your answer is probably correct. My values do look a bit ridiculous. I checked it so thoroughly but couldn't see anything wrong. Must've made a daft tiny error somewhere. Boooo.

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Reply 12

Andrreewwww

Original post by stuart_aitken

Unfortunately for the last question i forgot to turn back to x values, and got two k vales of 22/7 and 20/7, not sure if i'll get full marks for that question

Reply 13

seajamiet

Solutions for tan x=-1 for me was 180+-45 = 135
360+-45 = 315

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Reply 14

username1086199

Original post by Andrreewwww

Integration under curve was 63/2 for me and shaded area was 27?

I got those too :smile:

I think people didnt realise that it was a trapezium.
Area of that was (6+33)*3/2 ?

Reply 15

timeteo9

The transformation of [2 -0.7] was √((x-2)³ +1) -0.7

And the 2^k question, I got k = 6/5

Reply 16

Andrreewwww

Original post by Alifff

I got those too :smile:

I think people didnt realise that it was a trapezium.
Area of that was (6+33)*3/2 ?

Yeah, I couldn't think what the trapezium rule was, so i simply used a square and a triangle, got 117/2 for the trapezium

Reply 17

username1086199

Original post by stuart_aitken

This is all I want to know:

For core 1

Integration question:
Area under curve was 524/15
Shaded area was 961/15

Final question:
20/7 <= k <= 4

Correct or not...?

Posted from TSR Mobile