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This ebook is stuffed with transparent revision notes and examination perform questions for A2 Maths scholars. It covers the C3, C4, S2 and M2 modules for the Edexcel examination, and it’s effortless to learn and revise from - everything’s defined easily and punctiliously. each few pages there are fast warm-up questions, a few exam-style questions and on the finish of every module there are perform examination papers (answers on the back). There’s additionally a sprinkling of jokes to lighten the temper. It’s precisely what you must organize on your checks!

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Extra resources for A2-Level Mathematics Edexcel: Complete Revision & Practice

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Unless it made you very sleepy instead. Try these exam questions and make sure you can handle them. Exam Questions 1 5 9x A B 1 Given that, for x ! - , + 2 / , where A and B are integers, + 2 3 ^1 + 3xh ^1 + 3xh ^1 + 3xh find the values of A and B. 2 18x 2 - 15x - 62 B C / A+ + ^3x + 4h^ x - 2h ^3x + 4h ^ x - 2h Find the values of the integers A, B and C. f (x) = A B C , Given that f(x) can be expressed in the form f (x) = + + ^3 - xh ^2x - 1h ^2x - 1h2 find the values of A and B and C. (4 marks) 80x 2 + 49x - 9 A B The algebraic fraction can be written in the form 4 + , + (5x - 1) (2 - 4x) (5x - 1) (2 - 4x) where A and B are constants.

Dx dx dx 3sec 2 3y 3 c m dy Remember to use Trig Identities where Necessary Example For y = 2cos2 x + sin 2x, show that dy = 2(cos 2x − sin 2x). dx 1) Writing out the equation in a slightly different way helps with the chain rule: y = 2(cos x)2 + sin 2x. dy = 4u and du = −sin x. du dx You could also use the identity dy For the second bit, y = sin u, u = 2x, so = cos u and du = 2. cos 2x / 2 cos2 x − 1 du dx before differentiating. dy 3) Putting it all in the chain rule formula gives = −4sin x cos x + 2cos 2x.

X + 2h^2x - 1h^ x - 3h 9x 2 + x + 16 A B C . Now to work out A, B and C. / + + ^ x + 2h^2x - 1h^ x - 3h ^ x + 2h ^2x - 1h ^ x - 3h Add the partial fractions and cancel the denominators from both sides A ^ x + 2h + B ^2x - 1h + C ^ x - 3h / A ^2x - 1h^ x - 3h + B ^ x + 2h^ x - 3h + C ^2x - 1h^ x + 2h ^ x + 2h^2x - 1h^ x - 3h So the numerators are equal: 9x 2 + x + 16 / A ^2x - 1h^ x - 3h + B ^ x + 2h^ x - 3h + C ^2x - 1h^ x + 2h 2 Substitute x for values which get rid of all but one of A, B and C...

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