Flashcards
Core 3 Key Points
Cards in this deck (61)
- 1Core 3Algebraic fractions
- 2When your algebraic expression has fractions in the numerator or denominator, it is sensible to multiply both by the same 'number' to create an equivalent fractionWhen there are factors common to both the numerator and the denominator cancel down first
- 3To divide by a fraction multiply by the reciprocal of that fractionThe remainder theorem:
Any polynomial F(x) can be put in the form
F(x) = Q(x) * divisor + remainder
where Q(x) is the quotient and is how many times the divisor divides into the function - 4Algebraic fractions can be simplified by cancelling down. To do this the numerators and denominators must be fully factorised firstTo multiply fractions, you simply multiply the numerators and multiply the denominators. If possible cancel down.
- 5To add or subtract each fraction must have the same denominator. This is done by finding the lowest common multiple of the denominators*blank*
- 6FunctionsA function is a special mapping such that every element of the domain is mapped to exactly one element in the range
- 7A one-to-one function is a special type of function where every element of the domainMany mappings can be made into functions by changing the domain. For example, the mapping 'positive square root' can be changed into the function f(x) = √x by having a domain of x>=0
- 8If we combine two or more functions we can create a composite function. The function below is written fg(x) as g acts x first, then acts on the result. For example, g(x) = 2x + 3, f(x) = x^2 fg(4) = f(2(4) + 3) = f(11) = 11^2 = 121 Similarly: fg(x) = (2x + 3)^2The inverse of a function f(x) is written f^-1 (x) and performs the opposite operation(s) to the function. To calculate the inverse function you change the subject of the formula. For example, the inverse function of g(x) = 4x - 3 is
g^-1 (x) = {x + 3}/4 - 9The range of the function is the domain of the inverse function and vice versaThe graph of f^-1 (x) is a reflection of f(x) in the line y = x
- 10The exponential and log functionsExponential functions are ones of the form y = a^x. They all pass through the point (0,1)
The domain is all the real numbers. The range is f(x) > 0 - 11The exponential function y = e^x (where e = 2.718) is a special function whose gradient is identical to the functionThe inverse to e^x is In(x)
- 12The natural log function is a reflection of y = e^x in the line y = x. It passes through the point (1,0) The domain is the positive numbers. The range is all the real numbersTo solve an equation using In(x) or e^x you must change the subject of the formula and use the fact that they are inverses of each other.
- 13Growth and decay models are based around the exponential equations (look in textbook for the graph)*blank*
- 14Numerical methodsIf you find an interval in which f(x) changes sign, and f(x) is continuous in that interval, then the interval must contain a root of the equation f(x) = 0
- 15To solve an equation of the form f(x) = 0 by an iterative method, rearrange f(x) = 0 into a form x = g(x) and use the iteration formula x{n+1} = g(x{n})Different rearrangement of the equation f(x) = 0 give iteration formulae that may lead to different roots of the equation
- 16If you choose a value x{0} = a for the starting value in an iteration formula, and x{0} = a is close to a root of the equation f(x) = 0, then the sequence x{0}, x{1}, x{2}, x{3}, x(4), ... does not necessarily converge to that root. In fact it might not converge to a root at all*blank*
- 17Transforming graphs of functionsThe modulus of a number a, written as !a!, is its positive numerical value
For a > 0, |a| = a
For a < 0, |a| = -a - 18To sketch the graph of y = |f(x)|: - Sketch the graph of y = f(x) - Reflect in the x-axis any parts where f(x) < 0 (parts below the x-axis). - Delete the parts below the x-axis.To sketch the graph of y = f(|x|)
- Sketch the graph of y = f(x) for x ≥ 0
- Reflect this in the y-axis - 19To solve an equation of the type |f(x)| = g(x) or |f(x)| = |g(x)|: - Use a sketch to locate the roots - Solve algebratically using -f(x) for reflected parts of y = f(x) and -g(x) for reflected parts of y = g(x)Basic types of transformation are:
- f(x + a) = a horizontal translation of -a
- f(x) + a = a vertical translation of +a
- f(ax) = a horizontal translation of a scale factor 1/a
- af(x) = a vertical stretch of a scale factor a
These may be combined to give, for example bf(x + a), which is a horizontal translation of -a followed by a vertical stretch of scale factor b. - 20For combinations of transformations, the graph can be built up 'one step at a time', starting from a basic or given curve*blank*
- 21Trigonometrylook in the textbook for graphs on sec x, cosec x, and cot x
- 22Sec x1 / cos x
{sec x is undefined when cos x = 0, i.e. at x = (2x + 1) 90⁰, nЄz) - 23cosec x1 / sin x
{cosec x is undefined when sin x = 0, i.e. at x = 180n⁰, nЄz) - 24cot x1 / tan x
or
cos x / sin x
{cot x is undefined when tan x = 0, i.e. at x = 180n^o, nЄz) - 25The inverse of sin x is arcsin x The inverse of cos x is arccos x The inverse of tan x is arctan x Arcsin x isn't equal to sin^-1 x Look in the textbook for the graphs of arcsin x, arccos x, arctan x.Two further Pythagorean identities, derived from sin^2 x + cos^2 x
- 261 + tan^2 xsec^2 x
- 271 + cot^2cosec^2 x
- 28Further trigonometric identities and their applications*blank*
- 29sin(A + B)sin(A)cos(B) + cos(A)sin(B)
- 30cos(A + B)cos(A)cos(B) - sin(A)sin(B)
- 31tan(A + B){tan(A) + tan(B)} / {1 - tan(A)tan(B)}
- 32sin(A - B)sin(A)cos(B) - cos(A)sin(B)
- 33cos(A - B)cos(A)cos(B) + sin(A)sin(B)
- 34tan(A - B){tan(A) - tan(B)} / {1 + tan(A)tan(B)}
- 35sin 2A2sin(A)cos(A)
- 36cos 2Acos^2 (A) - sin^2 (A)
2 cos^2 (A) - 1
1 - 2 sin^2 (A) - 37tan 2A{2tan(A)} / {1 - tan^2 (A)}
- 38a sin(x) ± b cos(x)Rsin(x ± α)
with R > 0 and 0 < α < 90
where R cos(α) = a, R sin(α) = b and R = √{a^2 + b^2} - 39a cos(x) ± b sin(x)R cos(x ± α)
with R > 0 and 0 < α < 90
where R cos(α) = a, R sin(α) = b and R = √{a^2 + b^2} - 402sin(A)cos(B)sin(A + B) + sin(A - B)
- 412cos(A)sin(B)sin(A + B) - sin(A - B)
- 422cos(A)cos(B)cos(A + B) + cos(A - B)
- 432sin(A)sin(B)-(cos(A + B) - cos(A - B))
- 44sin(P) + sin(Q)2sin({P + Q} / 2)cos({P - Q} / 2}
- 45sin(P) - sin (Q)2cos({P + Q} / 2)sin({P - Q} / 2)
- 46cos(P) + cos(Q)2cos({P + Q} / 2)cos({P - Q} / 2)
- 47cos(P) - cos(Q)-2sin({P + Q} / 2)sin({P - Q} / 2)
- 48Differentiation*blank*
- 49y = [f(x)]^ndy/dx = n[f(x)]^{n - 1} f'(x)
- 50y = f[g(x)]dy/dx = f'[g(x)]g'(x)
- 51dy/dx = (dy/du) * (du/dx)dy/dx = 1/(dx/dy)
- 52Product rule = ? (y = uv)y = uv
dy/dx = u(dv/dx) + v(du/dx) - 53Quotient rule = ? (y = u/v)y = u/v
dy/dx = {v(du/dx) - u(dv/dx)} / {v^2} - 54y = e^x y = e^{f(x)}dy/dx = e^x
dy/dx = f'(x)e^{f(x)} - 55y = In(x) y = In[f(x)]dy/dx = 1/x
dy/dx = f'(x)/f(x) - 56y = sin(x) y = sin f(x)dy/dx = cos(x)
dy/dx = f'(x)cos f(x) - 57y = cos(x) y = cos f(x)dy/dx = -sin(x)
dy/dx = -f'(x)sin f(x) - 58y = tan(x) y = tan f(x)dy/dx = sec^2 (x)
dy/dx = f'(x)sec^2 f(x) - 59y = cosec(x) y = cosec f(x)dy/dx = -cosec(x) cot(x)
dy/dx = -f'(x) cosec(x) cot(x) - 60y = sec(x) y = sec f(x)dy/dx = sec(x) tan(x)
dy/dx = f'(x) sec f(x) tan f(x) - 61y = cot(x) y = cot f(x)dy/dx = -cosec^2 (x)
dy/dx = -f'(x) cosec^2 f(x)
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- AS level Maths Flashcards on Core 3 Key Points, created by Benjamin Smith on 11/12/2017.
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- Benjamin Smith
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