The Student Room Group

C2 Trapezium rule and logs easy questions

Trapezium rule:



For a) I know the height is 0.5, but using the formula h = (b-a)/n I don't get 0.5, I use b as 4 and a as 1 and n as 7, but obviously n has to be 6, thought i see 7 intervals. Is it because the last interval is 0?

Logs question:



For part bii) I got the question correct, but I done a complete different method from the mark scheme, and I'm wondering how the mark scheme did it.

My method:

[br]2x=5x1[br]xlog2=(x1)log5[br]x(log2)(log5)=x1[br]xx(log2)(log5)=1[br]x(1log2log5)=1[br]11log2log5=x[br][br][br]2^x = 5^{x-1}[br]xlog2=(x-1)log5[br]\dfrac{x(log2)}{(log5)} = x - 1[br]x - \dfrac{x(log2)}{(log5)} = 1[br]x(1 - \dfrac{log2}{log5} ) = 1[br]\dfrac{1}{1 - \frac{log2}{log5}} = x[br][br]

This is correct, but the mark scheme simply got:

[br]x=log5log5log2[br][br]x = \dfrac{log5}{log5 - log2}[br]

If anyone could explain how they got this would be great.
Reply 1
Original post by thorn0123
Trapezium rule:



For a) I know the height is 0.5, but using the formula h = (b-a)/n I don't get 0.5, I use b as 4 and a as 1 and n as 7, but obviously n has to be 6, thought i see 7 intervals. Is it because the last interval is 0?

Logs question:



For part bii) I got the question correct, but I done a complete different method from the mark scheme, and I'm wondering how the mark scheme did it.

My method:

[br]2x=5x1[br]xlog2=(x1)log5[br]x(log2)(log5)=x1[br]xx(log2)(log5)=1[br]x(1log2log5)=1[br]11log2log5=x[br][br][br]2^x = 5^{x-1}[br]xlog2=(x-1)log5[br]\dfrac{x(log2)}{(log5)} = x - 1[br]x - \dfrac{x(log2)}{(log5)} = 1[br]x(1 - \dfrac{log2}{log5} ) = 1[br]\dfrac{1}{1 - \frac{log2}{log5}} = x[br][br]

This is correct, but the mark scheme simply got:

[br]x=log5log5log2[br][br]x = \dfrac{log5}{log5 - log2}[br]

If anyone could explain how they got this would be great.



11log2log5[br][br]=1log5log2log5[br][br]=log5log5log2\dfrac{1}{1 - \frac{log2}{log5}} [br][br]= \dfrac{1}{\frac{log 5 - log2}{log5}}[br][br]= \dfrac{log5}{log5 - log2}

I'll look into the trapezium rule question if it's not answered yet when I have more time. Most likely it will though, answers come fast in this forum. :tongue:
Reply 2
Original post by thorn0123
Trapezium rule:



For a) I know the height is 0.5, but using the formula h = (b-a)/n I don't get 0.5, I use b as 4 and a as 1 and n as 7, but obviously n has to be 6, thought i see 7 intervals. Is it because the last interval is 0?

Logs question:



For part bii) I got the question correct, but I done a complete different method from the mark scheme, and I'm wondering how the mark scheme did it.

My method:

[br]2x=5x1[br]xlog2=(x1)log5[br]x(log2)(log5)=x1[br]xx(log2)(log5)=1[br]x(1log2log5)=1[br]11log2log5=x[br][br][br]2^x = 5^{x-1}[br]xlog2=(x-1)log5[br]\dfrac{x(log2)}{(log5)} = x - 1[br]x - \dfrac{x(log2)}{(log5)} = 1[br]x(1 - \dfrac{log2}{log5} ) = 1[br]\dfrac{1}{1 - \frac{log2}{log5}} = x[br][br]

This is correct, but the mark scheme simply got:

[br]x=log5log5log2[br][br]x = \dfrac{log5}{log5 - log2}[br]

If anyone could explain how they got this would be great.


Multiply everything through by log 5 to get rid of the fraction.
Reply 3
1-((log2/log5)) = 1/x

(log5/log5)=1

(log5-log2)/log5 =1/x

flip it round to find x (reciprocal)

log5/(log5-log2) = x

It's just neater to write this way than yours.

For trapezium rule question.
b=4 a=1 n=6 (because there are 6 strips and 7 ordinates) there are always one more ordinate than strips, try drawing it out with 6 strips and you'll see what I mean.
Reply 4
Ok I understand the log question, but what is an ordinate? I've drawn 6 strips and I still don't see what you mean.
Reply 5
Original post by thorn0123
Ok I understand the log question, but what is an ordinate? I've drawn 6 strips and I still don't see what you mean.


An ordinate is like a point, there is always one more than the number of strips when using the trapezium rule.

Edit If you still don't get it

Spoiler

(edited 11 years ago)
Reply 6
the x values are what I meant, so you have 1,1.5,2 to 4.
7 values (ordinates there)
Reply 7
thanks
Reply 8
are these from edexcel papers?

Quick Reply

Latest