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# question watch

1. http://www.mei.org.uk/files/papers/2011_June_c3.pdf
Please can someone see if my method of going about the question 9ii) is right or wrong? Thanks for all help in advance!

So basically I differentiated it to get dy/dx = 0.5 cos (0.5 x) , then i subbed in the value of 0 for x and got the dy/dx to be 0.5.

So then I wrote 0.5 cos(0.5x) > 0.5

then tried to solve the inequality and ended up with 0 and 4pi. Then I wrote that the x value of 0 leads to the gradient of 1/2. 4pi also leads to 0.5 as gradient , but a value of 4 pi is outside the given domain in the question ....do u think they would give me the marks?

If you scroll the document down further , you can find the mark scheme model answer; obviously they have done it in a diff way? am i right in doing this method?
2. (Original post by laurawoods)
http://www.mei.org.uk/files/papers/2011_June_c3.pdf
Please can someone see if my method of going about the question 9ii) is right or wrong? Thanks for all help in advance!

So basically I differentiated it to get dy/dx = 0.5 cos (0.5 x) , then i subbed in the value of 0 for x and got the dy/dx to be 0.5.

So then I wrote 0.5 cos(0.5x) > 0.5

then tried to solve the inequality and ended up with 0 and 4pi. Then I wrote that the x value of 0 leads to the gradient of 1/2. 4pi also leads to 0.5 as gradient , but a value of 4 pi is outside the given domain in the question ....do u think they would give me the marks?

If you scroll the document down further , you can find the mark scheme model answer; obviously they have done it in a diff way? am i right in doing this method?
No I do not think that you would gain the marks

You had to show that was the maximum gradient, not find other values that give the same gradient

Also ... when you solve an inequality you get an inequality as your answer, not values
3. (Original post by laurawoods)
http://www.mei.org.uk/files/papers/2011_June_c3.pdf
Please can someone see if my method of going about the question 9ii) is right or wrong? Thanks for all help in advance!

So basically I differentiated it to get dy/dx = 0.5 cos (0.5 x) , then i subbed in the value of 0 for x and got the dy/dx to be 0.5.

So then I wrote 0.5 cos(0.5x) > 0.5

then tried to solve the inequality and ended up with 0 and 4pi. Then I wrote that the x value of 0 leads to the gradient of 1/2. 4pi also leads to 0.5 as gradient , but a value of 4 pi is outside the given domain in the question ....do u think they would give me the marks?

If you scroll the document down further , you can find the mark scheme model answer; obviously they have done it in a diff way? am i right in doing this method?
I'm not sure how lenient they are but your answer isn't really correct. You said you tried to solve 0.5cos(0.5x) > 0.5 and got 0 and 4pi. That isn't right. In fact the point is that there is no x that satisfies 0.5cos(0.5x)>0.5

0.5cos(0.5x)>0.5 is the same as cos(0.5x)>1 which has no solution as maximum value of cos(0.5x) = 1
4. (Original post by TenOfThem)
No I do not think that you would gain the marks

You had to show that was the maximum gradient, not find other values that give the same gradient

Also ... when you solve an inequality you get an inequality as your answer, not values
Ok , so pls can tell me how do we show that that was the maximum gradient?
5. (Original post by laurawoods)
Ok , so pls can tell me how do we show that that was the maximum gradient?
By knowing that Cos(a) has a maximum value = 1

So 0.5 Cos(a) has a maximum of 0.5
6. (Original post by TenOfThem)
No I do not think that you would gain the marks

You had to show that was the maximum gradient, not find other values that give the same gradient

Also ... when you solve an inequality you get an inequality as your answer, not values
What I was thinking was 0.5 cos(0.5x)> 0 , 0.5 cos(0.5x)> 4pi

But the second one isnt in the domain of this question.

I cant really see why that wont work?
7. (Original post by laurawoods)
What I was thinking was 0.5 cos(0.5x)> 0 , 0.5 cos(0.5x)> 4pi

But the second one isnt in the domain of this question.

I cant really see why that wont work?
Work for what

The line in red is nonsense

cos(a) cannot be bigger than 1
8. (Original post by TenOfThem)
By knowing that Cos(a) has a maximum value = 1

So 0.5 Cos(a) has a maximum of 0.5
Sorry, i don't get that? where did a come from?
9. (Original post by laurawoods)
Sorry, i don't get that? where did a come from?
a doesn't come from anywhere

Cos(it does not matter what I write in here it is just an angle) is never more than 1
10. (Original post by TenOfThem)
a doesn't come from anywhere

Cos(it does not matter what I write in here it is just an angle) is never more than 1
Hello, pls can u explain once more, i don;t still get it! I don't know how to show that that is the max gradient?
11. (Original post by laurawoods)
Hello, pls can u explain once more, i don;t still get it! I don't know how to show that that is the max gradient?
Ok

What is the maximum value of
12. (Original post by TenOfThem)
Ok

What is the maximum value of
1 is the max value!
13. (Original post by laurawoods)
1 is the max value!
And, therefore, the maximum value of
14. (Original post by laurawoods)
So then I wrote 0.5 cos(0.5x) > 0.5
You can do this if you like but you are hoping to find no solutions.

Divide both sides by 0.5:

but this is impossible as the cosine graph only takes value from -1 to 1.

15. (Original post by Mr M)
You can do this if you like but you are hoping to find no solutions.

Divide both sides by 0.5:

but this is impossible as the cosine graph only takes value from -1 to 1.

cool thanks! just another question (not realted to any of this). ! If there is a - sign in front of the x of a function eg (f(-x)) doesn't this mean a refiction in the x axis?
16. (Original post by laurawoods)
cool thanks! just another question (not realted to any of this). ! If there is a - sign in front of the x of a function eg (f(-x)) doesn't this mean a refiction in the x axis?
No, it is a reflection in the y axis.

-f(x) is a reflection in the x axis.
17. (Original post by Mr M)
No, it is a reflection in the y axis.

-f(x) is a reflection in the x axis.
cool thanks!

Just one more question to ask you:

Please can you have a look at question no 6 and then the last part of it where it says : Hence verify that the curve has a stationary point at (1, 1).. Would you get the marks for simply shoving in the values of x and y into the dy/dx that we have got and concluding that it does come to 0? Or do they require something else to be done(I did look at the mark scheme , but I don't really understand the layout)? Also , for Question no 8 part ii) same verification style question . Are we simply shoving in the value of x into the diff equation and concluding that yes , it does come to 0? Any tips would be helpful, thanks in advance!
18. (Original post by TenOfThem)
And, therefore, the maximum value of
Thanks, I think I get it now! Just one question to ask you; please can you have a look at this question: Question 7 part i )

http://www.mei.org.uk/files/papers/2011_Jan_c3.pdf

I don't really understand how the range is –pi + 1 < f(x) < pi + 1

I did get the same but i got less than or equal to and greater than or equal to .....? why is this not right because for the graph of arctan x the range includes greater than or equal to etc...
19. (Original post by TenOfThem)
And, therefore, the maximum value of
ello, pls can you have a look at this question:
http://www.mei.org.uk/files/papers/c4_june_2012.pdf
Question no 8 ; the format seems to be very diff from edexcel vector questions; do u think we would get anything like this in our paper? A' etc is confusing me?
20. (Original post by laurawoods)
ello, pls can you have a look at this question:
http://www.mei.org.uk/files/papers/c4_june_2012.pdf
Question no 8 ; the format seems to be very diff from edexcel vector questions; do u think we would get anything like this in our paper? A' etc is confusing me?
A' is just another name for a point - you could call it E or W or something if you prefer!

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