Common mistakes

… When simplifying equations, please resist the temptation to work horizontally. For example, for the equation:

3x² + 2x – 5x = 9x²

try to avoid doing this:

“3x² + 2x – 5x = 9x² = 3x² – 3x” … which makes no sense.

It’s better to work line by line:

3x² + 2x – 5x = 9x²

3x² – 3x = 9x²

-6x² – 3x = 0 etc.

… Take care in the final step of solving a quadratic equation via factorisation. For example:

(3x – 5)(x + 2) = 0

So now actually write out the two solutions a step at a time (don’t try to do this in your head!)

3x – 5 = 0 or x + 2 = 0

3x = 5

x = 5/3 or x = -2

Logarithms

… Logs tell us what power we need to raise a number to, to equal another number. E.g. What power do you need to raise 2 to, to give 16? Log(base2) of 16 would tell us the answer.

… So logs are actually powers. Usually ‘log’ as written means log[base 10].

… ‘ln’ is called the natural log and means log[base e], where ‘e’ is Euler’s number (2.71…)

… Using the ‘twirly whirly’ rule (or ‘unravelling’) to understand logs. For example:

log(base2)y = 3

Then:

2³ = y

… Adding or subtracting same base logs: logA + logB = log(AB) or logA – logB = log(A/B)

… The power rule for logs, e.g. log(C^p) = p(logC) … where ^ means ‘to the power of’

… To solve a logarithm equation, try to simplify each side down to a single log (of the same base) with no multiplying coefficients. Then you can remove the log from both sides (see below).

… You can always ‘log’ both sides of an equation, or remove a log from both sides.

For example, if

LogAB = LogCD,

then

AB = CD

Logs are generally used to bring powers ‘down to the normal world’. For example, to solve 3^x = 10 we would log both sides, and use the power rule which would then enable us to solve the equation by making x the subject.

… You may need to ‘convert’ a term in an equation into a log, so that you can combine it with the other logs and simplify. For example if we were using logs of base 10:

log(x) = log(4) + 2

log(x) = log(4) + 2log10 <—- because log(base10)10 = 1

log(x) = log(4) + log(10²)

log(x) = log(4 × 10²)

log(x) = log(400)

x = 400

… This can also apply to powers as you may need to convert an ‘odd one out’ power to get common bases:

4^x + 2^x = 8

(2^2)^x + 2^x = 8

(2^x)^2 + 2^x = 8 … Now substitute using y = 2^x to form a quadratic. Remember to go back to 2^x for your solutions.

… Log(0) or log(negative number) are NOT possible, why not?

… What is logany base ?

… Log (1/a) = log(a^-1) = -log(a)

… Watch out for logs which look positive, but are in fact negatives in disguise! For example, log(0.2). This is particularly important when solving inequalities because if you multiply or divide both sides by a negative number then you must swap the inequality sign around.

… Using the ‘changing the base of a log’ formula. For example, let’s say we want to change the base of a log to ‘c’:

Logbase b = logc / logc …(try writing this out so you can see it more clearly!)

Here’s another example: If ln(a) = 3, then express loga as a simple natural logarithm:

loga

= loge / loge …using the change of base formula

= ln(x²) / ln(a)

= 3ln(x²)

… Note that log(A) / log(B) cannot be simplified, even if the logs are the same base.

–

Reduction to linear form

… In science, we usually want to plot a straight line relationship of the form “y = mx + c” by plotting appropriate quantities on each axis. Curves are not really much use in forming a mathematical relationship between two quantities.

This enables us to find the relationships of the form:

y = ax^n (‘power law’)

or

y = ab^x (exponential)

by taking logs of both sides, for example:

For the power law:

y = ax^n

log(y) = log(a) + nlog(x)

so plot:

log(y) vs log(x)

For the exponential:

y = ab^x

log(y) = log(a) + xlog(b)

so plot:

log(y) against x

Note that it is NOT good practice to extrapolate outside of the given data using the relationship. This is because we don’t know what the data does in these regions.

It is perfectly acceptable, however, to INterpolate within the given data to predict data points.

Trigonometry

Trigonometry identities you’ll need for year 1 are

sin²θ + cos²θ = 1

and

tanθ = sinθ / cosθ

These two identities can be derived (using SOH CAH TOA trigonometry and Pythagoras) from a quadrant graph which has a ‘rotating’ hypotenuse ‘arm’ with radius 1. The angle is always measured from the x axis anti-clockwise.

… Plotting the graphs of sin(A), cos(A) and tan(A) vs angle A.

Sin and cos oscillate between 1 and -1, and repeat every 360 degrees.

Tan has asymptotes every 90 degrees and repeats every 180 degrees.

… When solving trig equations, draw graphs of sin or cos to find the alternative solutions in the range specified using the symmetry of the curves. Tan is much easier: just add or subtract 180 degrees to find the alternative solutions. Always CHECK your alternative solutions by putting them back into the trig function afterwards 🙂

… When solving (for e.g.) sin(2x – 10) = 0.5… First solve in normal way but keep the (2x – 10) ‘packaged up’ until you have found all of the possible solutions in the range given.

Note that you will often need to alter the interval of required solutions to match the bracketed terms, for e.g.:

0 < x < 180

0 < 2x < 360,

-10 < (2x – 10) < 350

Then, once you have found the alternative solutions from your graph and the new interval (using symmetry) you can ‘unpack’ the (2x – 10) by adding 10 and dividing by 2.

… If tanθ = 3, then we can use SOH CAH TOA and Pythagoras to find the value of sinθ and cosθ:

tanθ = 3/1 = Opposite / Adjacent

By drawing a right-angled triangle we can find the hypotenuse:

Hypotenuse² = 3² + 1² = 10

Hypotenuse = √10

=> sinθ = 3/√10

=> cosθ = 1/√10

Radians

… An angle in radians is defined as the ratio of an arc length (s) to the radius (r).

… An angle of 1 radian is a sector which has a radius equal to the arc length. You’ll need to remember the equation:

s = r x (angle in radians)

… There are 2π radians in 360 degrees (a full circle)

… The unit symbol for radians is ‘rad’ (or less often ‘c’ in superscript)

… Watch out for any hint of radians in a question (e.g. angles in terms of π, use of s=rθ)… Remember to change the mode of your calculator and work in radians, which are often (but not always) in terms of π.

… Calculating the AREA of a sector by using the fraction of the circle method:

A = (angle in radians/2π) x πr²

… Note that you can find the area of a sector using degrees as well:

A = (sector angle/360°) x πr²

… The length of an arc can be found in a similar way:

s = (angle in radians/2π) x 2πr

… Always check your calculator mode when using trigonometry functions 🙂

… To convert degrees into radians:

(angle in degrees/360°) x 2π

… To convert radians into degrees:

(angle in radians/2π) x 360°

… You may also need to use the formula for the area of a triangle, but working with an angle in radians:

A = ½ ab(sinC)

… A carefully drawn and labelled diagram is essential when working out circle, arc and sector problems :).

… Small angle approximations, when working in radians:

sin(θ) ≈ θ

tan(θ) ≈ θ

cos(θ) ≈ 1 – θ²/2

These are usually given in your formula booklet.

Polynomials

… The Factor theorem says that if, for example, f(2) = 0 then (x – 2) must be a factor of the function f(x). We can then try to partly or fully factorise a cubic expression by ‘guessing’ values of x which make the function = 0.

If f(3/2) = 0, then:

x = 3/2

2x = 3

2x – 3 = 0

and so the factor would be (2x – 3)

The last (non-x) term of the polynomial gives a clue as to which values of x to try. For example, we could try factors of -6 for the polynomial, such as f(-1):

x³ + 2x² – 5x – 6

Now use brackets when putting values of x into the function:

f(-1) = (-1)³ + 2(-1)² – 5(-1) – 6 = -1 + 2 + 5 – 6 = 0

so (x+1) is a factor

… Once you have found a factor of a polynomial, you can then either:

1) compare coefficients of the different x powers. So for example, if we know that (x+1) is a factor of:

x³ + 2x² – 5x – 6

Then we can write:

(x+1)(ax² + bx + c) = x³ + 2x² – 5x – 6

Now compare the x³ coefficients on the LHS and the RHS first:

a = 1

Now compare the non-x terms on either side (the x^0 coefficients):

c = -6

And finally compare either the x coefficients (take care as there are two ways of multiplying the brackets to get x terms!):

c + b = -5

Therefore

-6 + b = -5

b = 1

or

2) Divide the function by the linear factor using algebraic division DMSB method. Make sure the polynomial is written in descending powers of x. Add in ‘dummy’ terms (e.g. 0x) if there are any powers of x missing.

… Well done for keeping your minus sign in a bubble – easier to spot. Just watch out for those double negatives when subtracting.

… Meaning of a remainder after dividing a polynomial.

polynomial/divisor = quotient + remainder/divisor

… Cubic functions which only have terms in ‘x’ (i.e. no single numbers) can be factorised relatively easily by taking a factor of x outside of a bracket. For example:

x³ + 2x² – 3x

x(x² + 2x – 3)

Now factorise the quadratic inside the bracket:

x(x + 3)(x – 1)

And the cubic is factorised!

… The remainder theorem: If you divide f(x) by, for example, (x + 1), then the remainder (if it is not a factor) will be f(-1). This is a really useful method for questions where they give you the remainder and ask you to find a constant k (used in the polynomial).

Binomial Expansion

… Binomial expansion using Pascal’s triangle or the calculator’s nCr function to find the coefficients of each term.

… The formula for finding the coefficient of the rth power of x is:

nCr = n!/(r!)(n-r)!

We saw how parts of these factorial terms can cancel out hen we put in the values of n and r.

… Use brackets to deal with the more complicated terms which are raised to powers, and go step-by-step to simplify each line.

… Using a column layout is much easier as you can see the pattern of powers and nCr numbers. Make sure you put all the terms together in a single expression once you have simplified everything.

… Writing the coefficients (for example when n = 8), the first coefficient would be 8C0 or 8C8 because Pascal’s triangle is symmetrical.

… Always cancel fraction coefficients down to their simplest form (keep them as fractions)

… Take care with your powers, for example:

(y²)³ = y^6 (Multiply powers)

But…

(y²)(y³) = y^5 (Add powers for common bases which multiply)

… Finding a specific term in an expansion. For example, find the term in x³ in the expansion of (x + y)^5 :

= 5C3 (x)³(y)²

Note that the powers on each term add up to the n value (5 in this case)

… ‘Reverse’ binomial expansion where you know the coefficient of a certain x term, and are required to find the power, n, of the binomial. For example

If we have (1 + x/2)^n and we are given that the coefficient of x² is 7, find the value of n.

In this case, use the factorial equation for the coefficient, using r = 2 (x² is the third term in the expansion, but r = 2) and multiply this by (1/2)^2 to equal the coefficient, 7.

You’ll need to cancel out the factorial terms in the numerator and denominator!

nCr = n!/(r!)(n-r)!

= n(n-1)(n-2)(n-3)… / 2!(n-2)(n-3)…

= n(n-1)/2

… Check that the pattern of signs in your expansion are either all positive, all negative, or alternating. If not, check your working to find the mistake (usually a minus sign goes missing somewhere).

… Another check you can make is to look at the pattern of coefficients for each term. Do any of the coefficients not fit the pattern?

… Using a Binomial expansion to find an approximation to an actual number. For example, by finding the first three terms of the expansion of (1 – x)^15, find an approximation to 0.9^15.

Here we would find the first three terms up to and including the term in x² and then substitute x = 0.1 into each term because (1 – 0.1)^15 = 0.9^15

Note that you may need to write down the value of each term by hand in order to add them up (sometimes the question may ask for a LOT of decimal places which the calculator may not be able to handle!)

Numerical integration methods

The Trapezium rule:

… h = strip width = (xn – x0)/n where n is the number of strips. It’s usually a good idea to draw up a table of x and y values from x0 to xn, so you can identify the y0 to yn ‘ordinates’.

… During calculations (e.g. with the trapezium rule), go to an extra sig fig than is required by the question. This way, you can avoid rounding errors while not having to write down all the numbers on your calculator display.

… Check your final answer by using the integration function on your calculator, or by making a quick estimate using a large triangle or trapezium of base xn – x0. You should get a roughly similar answer, if not, check your working!

… When using the Trapezium rule for complicated functions or lots of strips: the TABLE function on your calculator is a great way to save time and minimise risk of typo errors:

MENU > TABLE > enter the formula in terms of ‘x’ > ENTER > start x value > ENTER > end x value > ENTER > strip width (step) > ENTER > table of values

The Mid-Ordinate Rule

…The mid-ordinate rule (uses the mid ordinate y values to make rectangular strips).

… Find the ‘halfway’ x values first and then use the function to calculate the corresponding mid ordinate y values.

… The accuracy of any numerical method can be improved by increasing the number of strips.

Rates of change

… Rates of change questions often start with a geometry situation involving a variables such as ‘x’ or ‘y’. If you can draw a large labelled diagram it will help. The clue is usually in the information they give you. For example, if they say that the volume of the container is 4000 metres cubed, then ask yourself, “how can I find an expression for volume using the geometry in terms of x and y?”

You will usually be able to form two starting equations from the given situation. Substitute one of these into the other to form an equation which eliminates one of the variables. You can then differentiate this to find a min or max stationary value.

Integration

… Integration also enables us to find the original function (y = …) if we are given a function for the gradient (dy/dx = …)

For example, if dy/dx = 2x – 3

Then y = ∫(2x – 3)dx

y = x² – 3x + c

We can find the constant of integration, c, if we know some initial conditions, for example, if the curve passes through the point (0,1):

1 = (0)² -3(0) + c

c = 1

So y = x² – 3x + 1 is the original function of the curve.

… Try to split up a complicated integral into component parts, which are easier to integrate separately. For example:

∫( (x² + 3x)/x ) dx

= ∫ (x + 3 ) dx

= x²/2 + 3x + c

… Integrating a function also enables us to calculate the AREA between the graph and the x-axis. You are actually summing up millions of thin strips of width dx and area ydx from a starting x value to higher x value.

… Indefinite integrals do not have actual x values, and so you will need to add a constant of integration, “+c” to your integrated function.

… Definite integrals have ‘limits’: numerical x values which you can sub into the integrated function (sub in top limit) – (sub in bottom limit). You don’t need a constant of integration because the ‘c’s cancel out in the subtraction.

… Using the ‘Table Function’ in your calculator is a useful way to help you sketch difficult functions: MODE > Table

… When finding multiple areas between a curve and the x-axis, be sure to calculate each area separately. If you get any negative areas (beneath the x-axis) then these should be corrected to be positive before you add the areas together :).

… You may first need to calculate the area of a rectangle or triangle in order to work out the area being asked for in the question.

Vectors

… Parallel vectors have the same ‘vector common factor’, so:

3i + 4j

is parallel to

9i + 12j = 3(3i + 4j)

Another way of working with parallel vectors is to equate the vector component ratios. For example:

8a + 2b is parallel to 5a + kb. Find the value of k.

Here, we can say:

8a/5a = 2b/kb

so 8/5 = 2/k

8k = 10

k = 5/4