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# C3 Solomon H Watch

1. Can anyone help? Question 4b).

In the mark scheme they got that at (pi/2, -1), -1 = a+b.

And at (3pi/2, -5), -5 = a-b.

But I just don't understand why they got that... I know it's to do with -1 shifting the graph downwards, and b stretching the graph by a scale factor of b...

Thanks
2. (Original post by Barryl)
Can anyone help? Question 4b).

In the mark scheme they got that at (pi/2, -1), -1 = a+b.

And at (3pi/2, -5), -5 = a-b.

But I just don't understand why they got that... I know it's to do with -1 shifting the graph downwards, and b stretching the graph by a scale factor of b...

Thanks
Well, what is f:x actually? It's not to do with transformations or anything (it could be, but I could see that being more complicated than the alternative method).

Think about all of the information that you are given.
3. (Original post by SeanFM)
Well, what is f:x actually? It's not to do with transformations or anything (it could be, but I could see that being more complicated than the alternative method).

Think about all of the information that you are given.
Wow, can't believe I missed it. When x = pi/2, csc x = 1. So at (pi/2, -1), a+b(1) = -1

And at (3pi/2, -5), -5 = a + b(-1) = a-b.

Just had to think a little bit more...

Thanks for the push!
4. (Original post by Barryl)
Wow, can't believe I missed it. When x = pi/2, csc x = 1. So at (pi/2, -1), a+b(1) = -1

And at (3pi/2, -5), -5 = a + b(-1) = a-b.

Just had to think a little bit more...

Thanks for the push!
No worries. In any maths exam, or indeed any exam. think about the information you're given and what use it could possibly be.

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