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Math SL P1


iBzG

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minus because the function is sort of reflected around x axis (starts at low and goes to high, opposite of normal cosine function which starts at high)

hmmm..... right!! but then the there should also exist the horizontal translation in the equation, shouldn't it?? because even if you reflect it then too the does not exactly start from the origin like sin....

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I think it probably depends more on grade boundaries than the grader. keep in mind you should get follow through points if you did a problem right using a wrong value at the beginning, and that you do get points for a correct method even if you didn't quite get to the end of questions. Especially for a question where you just made a sign error, you should get points if you did everything else right.

It's easier said than done but try not to worry, instead focus on studying for your other exams so that even if you maths score isn't a 5, you can pull your score up in other subjects :)

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I am almost sure that the value of A can equal 50 OR -50. That's because the cosine function does not "start" at a specific point (t=0). It's continuous, having multiple revolutions. Therefore, although after t=0, the curve was going upwards (and since it is a cosine function, you would consider this to be negative, since cos goes upwards), after t=2, the curve starts going down, therefore you can also say that A= 50. The only variable being affected by a cosine is the horizontal shift. Otherwise, the amptitude, period and vertical shift are always the same for both cosine and sine functions.

By the way, does anyone remember what question 9 was about? Number 8 was on transforming functions and finding derivatives and question 10 was the ferris wheel.

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By the way, does anyone remember what question 9 was about? Number 8 was on transforming functions and finding derivatives and question 10 was the ferris wheel.

wasn't question 9 the probability dice one?

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What's the formula to obtain the value of A? I just guessed haha.

It's (y-coordinate of min point + y-coordinate of max point) / 2, then you need to add a minus if there's a reflection.

I can't seem to remember that question! Can you please remind me?

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@Ika Lob: no, you're actually not right in that, as there was no horizontal shift of the graph (if there had been, it would've been possible); in this case, however, x=0 is always the max point unless it's been reflected.

I think you're right. Are you saying that since they did not give even a variable for the horizontal shift, it is assumed to be zero?

It was A cosBx + C. So if it had been A cosB(x-D) + C and they asked us to find A, B and C, then it could have been A= 50 or -50 (given that we could alter the horizontal shift).

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Our maths teacher had told us the value of amplitude of sin or cos cannot be negative.. So i got the value of A= 50 and the value of c= -50

Yes but A does not = amplitude. Just like B does not equal the period. 2pie/B equals the period and the absolute value of A is the amplitude.

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the 1st question in section-B related to calculus.. i got the area as 1/6

did anybody else get the same answer? :blink:

Got the same one I think. It seemed suspiciously small but then again the limit was very small as well and I couldn't find a mistake in my calculations...

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