The Last Minute C3 Questions Thread!!

Hi! Thanks for replying,

So question 5 from here was tough http://mei.org.uk/files/papers/C3_2014_June.pdf

If they give it in the format of say find dy/dt and give the value of dy/dx and an equation for x in terms of T I can usually do it as it's just a matter of differentiating and cancelling out but I can never do the worded ones and i think thats because i dont actually know whats going on, i just know the technique to getting the answer which doesn't work well in a worded scenario

Hey fellas, struggling to understand last year's mark scheme for the last question.

The second to last part states: show cos (x ) = -5/13

The next part then replaces x with 2x -70

You proceed to find the solutions for cos (2x-70) = -5/13

That's fine, but using sin (x)^2 + cos (x)^2 = 1 and subbing back in, you'll find sin(x) = -24/26, solving this you find another solution in range, 1.3 to 1dp

However this isn't on the mark scheme and the final a1 mark is correct answers in range only, can someone explain why this solution is Invalid?

What did you solve to find your other solution?

Sin(x)^2 +(-5/13)^2 = 1
Sin(x)^2 = 1 - 25/169
Sin(x)^2 = 144/169
Sin(x) = +-(12/13)

Subbing back into (sec(x) - tan(x) = -5)
(1 / cos(x)) - (sin(x) / cos(x)) = -5
(1 / ( -5/13)) - (12/13 / -5/13) = - 1/5 Which isn't correct, hence sin(x) =/ 12/13
(1/ (-5/13)) - (-12/13 / -5/13) = -5 Which is correct, hence sin(x) = -12/13

Sin(2x -70) = -12/13
2x-70 = -67.38, -112.62, (there are more solutions but these would be out of range, (-90<x<90))
2x = 2.62, -42.62
x = 1.31, -21.3

However, x = 1.31 is not in the mark scheme, hence if you wrote that answer down they would take a mark off you, however I can't see why.

Take a look at the mark scheme, it states 'Allow no other extras in interval'

The question starts "Hence solve the equation" meaning you needed to use an answer from a previous part of the question. You needed to solve cos(2x-70) = -5/13.

Although you have correctly found that -1.31 is a solution to sin(2x-70)=-12/13 it does not work for cos(2x-70) = -5/13.

You are not expected to know why but just remember to use a previous answer if the question uses the work hence.

Hi! Thanks for replying,

So question 5 from here was tough http://mei.org.uk/files/papers/C3_2014_June.pdf

If they give it in the format of say find dy/dt and give the value of dy/dx and an equation for x in terms of T I can usually do it as it's just a matter of differentiating and cancelling out but I can never do the worded ones and i think thats because i dont actually know whats going on, i just know the technique to getting the answer which doesn't work well in a worded scenario

does the OCR spec have same content as Edexcel?

So in question 5, you need to find the rate of increase when r = 8. In other words, you need to find $\frac {dr}{dt}$ then subtitute in r = 8.

Using the chain rule, I use this tip that exam solutions taught me which is $\frac {dr}{dt} = \frac {dr}{d } \times \frac {d }{dt}$
Now the $d$ is a variable that we don't know but we can use the information from the question to find it out and it will be the volume because we can calculate $\frac {dV}{dr}$

We also know that the balloon is inflated at a constant rate of . I know that by looking at the units, $cm^3$ is the unit for volume and is the unit per second. "per second" already implies that there is a rate of change. So now $\frac {dV}{dt} = 10$

We're told that $V = \frac {4}{3} \pi r^3$ $\Rightarrow$ $\frac {dV}{dr} = 4 \pi r^2$

Let's go back to the chain rule and putting the V in, $\frac {dr}{dt} = \frac {dr}{dV} \times \frac {dV}{dt}$

So we get $\frac {dr}{dt} = \frac {1}{4 \pi r^2} \times 10$ hence $\frac {dr}{dt} = \frac {5}{2 \pi r^2}$

Now that we have $\frac {dr}{dt}$, substitute in $r = 8$ to find the rate of change at this point.


Best advice I can give is to use the chain rule tip that exam solutions taught me earlier on in this question as I mentioned. Try to break the question down. I went through this example step by step to show that it's not that bad as it looks!

How does one go about getting close to or 100UMS in C3/C4.

I was underprepared for C3 last year and got 88, and made a silly mistake in C4 and got 86.

Need to make sure it's banged this year, tips please?

Anyone know how to do part (B). I got to θ = 2θ - π/3.

Thanks

I'm ready for C3. I'm just hoping I don't misread a question or miscopy anything tomorrow, I always randomly change my 5s into 2s for some reason?

I'm hoping for you that there will be no 5s in the exam tomorrow

If $\cos x = \cos y$ then an obvious solution is $x = y$ but there are others if you consider CAST/the graphs/general solutions. Can you think what they are?

Quick question:
say you've got something like: 4 - x^2 = |2x -1|
you can +/- the RHS to get:
x^2 + 2x - 5 = 0, or x^2 -2x - 3 = 0
But each of these quadratics only gives one solution that works, and I can see from the graph that there are only 2 points of intersection.
The ms says something about x > 1/2 for one of them and x <1/2 for the other; does anyone know what does this means? (ms attached)