Core 2 Useful Workings For Challenging Questions With Explanations

A quick revision resource I made to help you lot :)

It contains questions from past papers that people may find challenging, with working and explanations at the side :)

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  • Created by: George
  • Created on: 14-05-11 18:51
Preview of Core 2 Useful Workings For Challenging Questions With Explanations

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Mathematics Core 2 Exam Style Questions that are considered challenging
Logarithms
Given that 0 < x < 4 and: log 5 (4 ­ x) ­ 2log 5 (x) = 1, solve for x
Solution
log 5 (4 ­ x) ­ 2log 5 (x) = 1
log 5 (4 ­ x) ­ log 5 (x) 2 = 1
log 5 ((4 ­ x) / x 2) = 1
((4 ­ x) / x 2 = 5 1
5x 2 = 4 ­ x
5x 2 + x ­ 4 = 0
(5x ­ 4)(x + 1) = 0
x = 4 / 5 or ­1
Question states that x must be between 0 and 4, so negative 1 is not a solution. Also negative 1
produces a math error

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Polynomials
f(x) = x 4 + 5x3 + ax + b, where a and b are constants
The remainder when f(x) is divided by (x ­ 2) is equal to the remainder when f(x) is divided by
(x + 1)
(a) Find the value of a
Given that (x + 3) is a factor of f(x)
(b) Find the value of b
Solution
Use the factor theorem to produce two equations, of which you can then solve simultaneously.…read more

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Trigonometrical Identities
Show that the equation
(a) 4sin 2 x + 9cos x ­ 6 = 0, can be written as, 4cos 2x ­ 9cos x + 2 = 0
Hence solve, for 0 x < 720
(b) 4sin 2 x + 9cos x ­ 6 = 0
Solution
4sin 2 x + 9cos x ­ 6 = 0
4(1 ­ cos 2 x) + 9cos x ­ 6 = 0
4 ­ 4cos 2 x + 9cos x ­ 6 = 0
4cos 2…read more

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Look where cos is positive (1/4 is positive) cos and all functions
x = 75.5, 284.5, 435.5, 644.5
Geometric Series
The first three terms of a geometric series are (k + 4), k and (2k ­ 15) respectively, where k is a
positive constant.…read more

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¼
64
Differentiation
A solid right circular cylinder has radius r cm and height h cm.…read more

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