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Ocr c2 maths help

I sat the C2 paper in May and achieved a C, and hoping to retake. I've gone over the unofficial markscheme for this year, and the only question I can't wrap my head around is question 9)ii and iii). This is the question below;

9ii) Given that x=1/5 pi and x=2/5 pi are the 2 smallest +ve solutions of sin(ax)=k where k is a +ve constant, find the values of a and k.

9iii) Given instead that sin(ax)= root 3 cos (ax), find the two smallest +ve solutions for x giving your answers in exact form in terms of a.

Any responses would be helpful. Thanks!
Original post by MoniC255
I sat the C2 paper in May and achieved a C, and hoping to retake. I've gone over the unofficial markscheme for this year, and the only question I can't wrap my head around is question 9)ii and iii). This is the question below;

9ii) Given that x=1/5 pi and x=2/5 pi are the 2 smallest +ve solutions of sin(ax)=k where k is a +ve constant, find the values of a and k.

9iii) Given instead that sin(ax)= root 3 cos (ax), find the two smallest +ve solutions for x giving your answers in exact form in terms of a.

Any responses would be helpful. Thanks!


For the first one I'd suggest starting with a sketch of sin(ax) - doesn't matter you can't draw a "proper" scale, but mark the points you're given.

For the second, rearrange to get a single trig. term. Note that the value comes from one of the standard angles.
(edited 7 years ago)
Reply 2
Original post by ghostwalker
For the first one I'd suggest starting with a sketch of sin(ax) - doesn't matter you can't draw a "proper" scale, but mark the points you're given.

For the second, rearrange to get a single trig. term. Note that the value comes from one of the standard angles.



Hi, thanks for the response. I've done that, but haven't got any further.
Original post by MoniC255
Hi, thanks for the response. I've done that, but haven't got any further.


Which part are you refering to and what have you got?
Reply 4
Original post by ghostwalker
Which part are you refering to and what have you got?

To part 9ii. I've drawn a sine graph, and have draw in the given smallest solutions. Do you know what I could do after this please to obtain a and k? Thanks
Original post by MoniC255
To part 9ii. I've drawn a sine graph, and have draw in the given smallest solutions. Do you know what I could do after this please to obtain a and k? Thanks


I was hoping you'd post your diagram, otherwise I'm just going to have to assume it's correct.

I am assuming that "a" is positive, though your question doesn't state that.

The standard sin(t) graph has its first maximum at t= pi/2, and importantly it is symmetrical about a vertical line through t=pi/2.

Now the graph you have been given is stretched along the x-axis.

By symmetry, its maximum is half way between x= pi/5 and x=2pi/5, i.e. at x=3pi/10.

So, the graph has been scaled along the x-axis to get pi/2 to 3pi/10.

Can you see how to work out a from that?
Reply 6
Original post by ghostwalker
I was hoping you'd post your diagram, otherwise I'm just going to have to assume it's correct.

I am assuming that "a" is positive, though your question doesn't state that.

The standard sin(t) graph has its first maximum at t= pi/2, and importantly it is symmetrical about a vertical line through t=pi/2.

Now the graph you have been given is stretched along the x-axis.

By symmetry, its maximum is half way between x= pi/5 and x=2pi/5, i.e. at x=3pi/10.

So, the graph has been scaled along the x-axis to get pi/2 to 3pi/10.

Can you see how to work out a from that?


Hi, for some reason I can't post anything.
Oh, and sorry, yes a is positive, and so is K
I've done all of the above. So would the a in sin(ax)=k equal 3pi/5? Please?

Thanks for a response
Original post by MoniC255
Hi, for some reason I can't post anything.
Oh, and sorry, yes a is positive, and so is K
I've done all of the above. So would the a in sin(ax)=k equal 3pi/5? Please?

Thanks for a response


Not quite.

In the standard graph you have sin(pi/2) = 1

In your graph you have sin(a 3pi/10) =1

So,

a3π10=π2a \dfrac{3\pi}{10}=\dfrac{\pi}{2}

and solve for "a".
Reply 8
Original post by ghostwalker
Not quite.

In the standard graph you have sin(pi/2) = 1

In your graph you have sin(a 3pi/10) =1

So,

a3π10=π2a \dfrac{3\pi}{10}=\dfrac{\pi}{2}

and solve for "a".


Oh, I get it now! Thank you very much for your help!

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