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Choosing a method

Choosing a method is the skill of turning the information in a question into a useful first mathematical step. It is not guesswork. Good choices come from recognising structure, identifying the required output and testing whether the available facts fit the conditions of a method.

You should be familiar with the main techniques from pure mathematics, statistics and mechanics. In particular, you should be able to:

  • rearrange equations and sketch familiar graphs;
  • differentiate and integrate standard functions;
  • use common probability distributions;
  • draw force diagrams and apply constant acceleration formulae.

If the algebra needed to carry out a chosen method is unreliable, review algebraic fluency first.

Before calculating, write down three things:

  1. Target: What exactly must be found, shown or decided?
  2. Evidence: What equations, values, graph features or modelling assumptions are given?
  3. Bridge: Which theorem, definition or model connects the evidence to the target?

Then check the method’s conditions. For example, the binomial model needs a fixed number of independent trials with constant success probability. Integration by parts needs a product, but it is useful only when differentiating one factor makes the product simpler.

The following question stems often signal a mathematical action.

Wording or structureLikely actionEssential check
”Show that”Work towards the stated formDo not assume the result you must prove
”Hence”Use the preceding resultIdentify what the earlier result replaces or simplifies
”Exact”Keep fractions, surds, π\pi, ee and logarithmsDo not use rounded decimals
”Maximum” or “minimum”Form an objective function, then differentiateCheck endpoints and classify stationary points
”At least”, “at most”Translate to an inequality or cumulative probabilityInclude the boundary value correctly
Product inside an integralSubstitution or integration by partsLook for an inner derivative first
Repeated independent trialsBinomial distributionVerify fixed nn and constant pp
Forces and accelerationResolve forces, then use F=maF=maFix a positive direction first

These are clues, not automatic commands. The whole structure decides the method.

A sophisticated method is not automatically a good method. First ask whether factorising, cancelling, using an identity or changing representation reveals something simpler.

Worked example 1: an equation disguised by fractions

Section titled “Worked example 1: an equation disguised by fractions”

Solve

x29x3=8.\frac{x^2-9}{x-3}=8.

The denominator gives the restriction x3x\ne3. Factorise before multiplying out:

(x3)(x+3)x3=8.\frac{(x-3)(x+3)}{x-3}=8.

For allowed values of xx, this becomes

x+3=8,x+3=8,

so x=5x=5. The cancelled value x=3x=3 was never in the domain. Cross multiplication would work, but simplification makes both the method and the restriction clearer.

Worked example 2: differentiate after rewriting

Section titled “Worked example 2: differentiate after rewriting”

Differentiate

y=x3+2xx,x0.y=\frac{x^3+2x}{x}, \qquad x\ne0.

Rather than using the quotient rule,

y=x2+2,y=x^2+2,

so

dydx=2x,x0.\frac{dy}{dx}=2x, \qquad x\ne0.

The derivative retains the original domain. See product, quotient and chain rules for deciding between differentiation rules.

When a question contains too much information, the requested result tells you which information matters.

The curve has equation

y=x33x+4.y=x^3-3x+4.

Find the equation of the tangent where x=2x=2.

The target is a straight line. A line needs a point and a gradient.

First find the point:

y(2)=86+4=6,y(2)=8-6+4=6,

so the tangent passes through (2,6)(2,6).

Next find its gradient by differentiation:

dydx=3x23,\frac{dy}{dx}=3x^2-3,

hence at x=2x=2,

m=3(2)23=9.m=3(2)^2-3=9.

Now use the point gradient form:

y6=9(x2),y-6=9(x-2),

or

y=9x12.\boxed{y=9x-12}.

The method chain is therefore

tangentpoint and gradientsubstitution and differentiation.\text{tangent} \longleftarrow \text{point and gradient} \longleftarrow \text{substitution and differentiation}.

Review tangents and normals if this chain is unfamiliar.

Sometimes several methods are valid. Prefer the one that uses the visible structure, preserves exact values and creates the least avoidable algebra.

Worked example 4: substitution or integration by parts?

Section titled “Worked example 4: substitution or integration by parts?”

Evaluate

I=012xex2dx.I=\int_0^1 2x e^{x^2}\,dx.

The integrand is a product, which might suggest integration by parts. However, 2x2x is the derivative of the exponent x2x^2. That is a stronger structural clue.

Let

u=x2,du=2xdx.u=x^2, \qquad du=2x\,dx.

Change the limits:

x=0u=0,x=1u=1.x=0\Rightarrow u=0, \qquad x=1\Rightarrow u=1.

Therefore

I=01eudu=[eu]01=e1.\begin{aligned} I&=\int_0^1 e^u\,du\\ &=\left[e^u\right]_0^1\\ &=e-1. \end{aligned}

Integration by parts is possible but inefficient. For

xexdx,\int x e^x\,dx,

there is no inner function whose derivative supplies the other factor, so integration by parts is the more natural choice. Compare integration by substitution and integration by parts.

Worked example 5: exact algebra or numerical solution?

Section titled “Worked example 5: exact algebra or numerical solution?”

Consider

x3+x1=0.x^3+x-1=0.

Testing the possible integer roots ±1\pm1 gives neither root, and no simple factorisation appears. The derivative

3x2+1>03x^2+1>0

shows the function is strictly increasing, so it has at most one root. Also,

f(0)=1,f(1)=1,f(0)=-1, \qquad f(1)=1,

so a root lies in (0,1)(0,1). A numerical method is appropriate if the question asks for a decimal approximation.

Newton Raphson gives

xn+1=xnxn3+xn13xn2+1.x_{n+1}=x_n-\frac{x_n^3+x_n-1}{3x_n^2+1}.

Starting with x0=0.5x_0=0.5,

x1=0.50.53+0.513(0.5)2+1=0.714285x_1=0.5-\frac{0.5^3+0.5-1}{3(0.5)^2+1} =0.714285\ldots

and further iterations converge to 0.68230.6823\ldots. Do not choose a numerical method if the question requests an exact answer. See change of sign and Newton Raphson.

In statistics, the assumptions are part of the method. Matching a familiar word such as “number” or “mean” is not enough.

Worked example 6: identify the distribution

Section titled “Worked example 6: identify the distribution”

A biased coin has probability 0.30.3 of landing heads. It is tossed independently 2020 times. Find the probability of at least 88 heads.

There is:

  • a fixed number of trials, n=20n=20;
  • one of two outcomes on each trial;
  • independent trials;
  • a constant success probability, p=0.3p=0.3.

Thus

XB(20,0.3).X\sim\operatorname{B}(20,0.3).

“At least 88” means X8X\ge8. Most calculators evaluate lower tail probabilities, so use the complement:

P(X8)=1P(X7)0.2277.\begin{aligned} \operatorname{P}(X\ge8) &=1-\operatorname{P}(X\le7)\\ &\approx 0.2277. \end{aligned}

If the probability changed from toss to toss, or tosses influenced one another, the binomial model would not be justified. Use choosing a distribution to compare statistical models.

Worked example 7: choose the test from the claim

Section titled “Worked example 7: choose the test from the claim”

A manufacturer claims that 40%40\% of customers choose option A. In a random sample of 5050 customers, 2727 choose A. Test at the 5%5\% significance level whether the proportion has increased.

The data count successes in a fixed sample, so under the null hypothesis

XB(50,0.4).X\sim\operatorname{B}(50,0.4).

The word “increased” determines a one tailed upper test:

H0:p=0.4,H1:p>0.4.H_0:p=0.4, \qquad H_1:p>0.4.

The observed result is x=27x=27, so the relevant probability is

P(X27).\operatorname{P}(X\ge27).

The direction comes from the alternative hypothesis, not from whichever tail produces a smaller number. Review binomial hypothesis tests and hypothesis testing language.

Start mechanics questions with a diagram and a positive direction. Then separate the modelling stages:

forcesF=maaccelerationkinematicsmotion.\text{forces} \xrightarrow{F=ma} \text{acceleration} \xrightarrow{\text{kinematics}} \text{motion}.

Worked example 8: forces before kinematics

Section titled “Worked example 8: forces before kinematics”

A particle of mass 4 kg4\text{ kg} is pulled horizontally by a force of 18 N18\text{ N}. A resistance of 6 N6\text{ N} opposes the motion. It starts from rest. Find its speed after travelling 16 m16\text{ m}.

The kinematics formula needs acceleration, which is not given. Find it from the forces first. Taking the direction of motion as positive,

Fresultant=186=12 N.F_{\text{resultant}}=18-6=12\text{ N}.

Using F=maF=ma,

12=4aa=3 m s2.12=4a \quad\Rightarrow\quad a=3\text{ m s}^{-2}.

Now u=0u=0, s=16s=16 and a=3a=3. Time is absent from both the data and the target, so choose

v2=u2+2as.v^2=u^2+2as.

Hence

v2=0+2(3)(16)=96,v^2=0+2(3)(16)=96,

so

v=46 m s1.\boxed{v=4\sqrt6\text{ m s}^{-1}}.

The negative square root is rejected because vv here denotes speed. The positive direction and the physical meaning decide the sign.

If no method is obvious, do not search your memory randomly. Use this recovery sequence:

  1. Rewrite the target using a definition. For a stationary point, write f(x)=0f'(x)=0. For independence, write P(AB)=P(A)P(B)\operatorname{P}(A\cap B)=\operatorname{P}(A)\operatorname{P}(B).
  2. Label every known quantity, including units, restrictions and parameters.
  3. Draw a sketch, tree diagram or force diagram if it exposes relationships.
  4. Change representation. Factorise, take logarithms, use vectors, or replace a repeated expression by a new variable.
  5. Work backwards one step from the target and forwards one step from the data.
  6. If a part begins “hence”, inspect the preceding result before starting again.

Solve

32x10(3x)+9=0.3^{2x}-10(3^x)+9=0.

The repeated expression 3x3^x suggests the substitution

u=3x,u>0.u=3^x, \qquad u>0.

Then 32x=(3x)2=u23^{2x}=(3^x)^2=u^2, giving

u210u+9=0.u^2-10u+9=0.

Factorise:

(u1)(u9)=0.(u-1)(u-9)=0.

Thus u=1u=1 or u=9u=9. Returning to xx,

3x=1x=0,3^x=1\Rightarrow x=0,

or

3x=9=32x=2.3^x=9=3^2\Rightarrow x=2.

Therefore x=0 or x=2\boxed{x=0\text{ or }x=2}. The key step was not a special exponential rule. It was recognising a quadratic in disguise. See exponential and logarithmic equations.

Words provide evidence, not certainty. “Rate” may lead to differentiation, a differential equation, a compound measure or a rate parameter in a probability model. Read the mathematical relationships around it.

”The longest method earns the most marks”

Section titled “”The longest method earns the most marks””

Marks reward valid progress. An exact identity or simplification may replace a page of algebra. Efficient mathematics is usually easier to verify.

”If my chosen method starts badly, I must continue”

Section titled “”If my chosen method starts badly, I must continue””

Pause when expressions become unexpectedly complicated. Check whether you missed a factorisation, an inner derivative, a symmetry or a result supplied in an earlier part.

A calculator evaluates the model you enter. It cannot decide whether independence, constant acceleration or a distributional assumption is reasonable.

Different correct routes can produce different intermediate forms. Compare them by simplifying, substituting a test value or checking against the original conditions.

Choose a method before attempting any detailed calculation.

  1. Solve 22x5(2x)+4=02^{2x}-5(2^x)+4=0.
  2. Which integration method is most direct for xcos(x2)dx\displaystyle\int x\cos(x^2)\,dx?
  3. A particle moves with constant acceleration. You know uu, vv and tt and need ss. Which constant acceleration equation avoids an unnecessary unknown?
  4. A random variable counts defective bulbs in 3030 bulbs sampled without replacement from a box of 4040, of which 88 are defective. Is a binomial model exact?
  5. To find the minimum of f(x)f(x) on the closed interval [a,b][a,b], is solving f(x)=0f'(x)=0 sufficient?
  1. Set u=2x>0u=2^x>0. Then u25u+4=(u1)(u4)=0u^2-5u+4=(u-1)(u-4)=0, so 2x=12^x=1 or 2x=42^x=4. Hence x=0x=0 or x=2x=2.
  2. Substitution. Let u=x2u=x^2, so du=2xdxdu=2x\,dx. The factor xx supplies half of the inner derivative.
  3. Use s=u+v2t\displaystyle s=\frac{u+v}{2}t. It uses exactly the known quantities and the target.
  4. No. Sampling without replacement makes the trials dependent and changes the success probability. A hypergeometric model is exact, although that distribution is outside standard A level Mathematics content. A binomial approximation would require explicit justification.
  5. No. Evaluate ff at every stationary point inside the interval and at both endpoints. Also consider points where ff' does not exist if they lie in the domain.

Before committing to a method, ask:

  • Does it connect the given information to the exact target?
  • Are all its assumptions satisfied?
  • Can I simplify or change representation first?
  • Is there a supplied result I should use?
  • Will the method preserve the required accuracy, exactness and units?
  • Can I check the result by substitution, estimation, a graph or another method?

Next, learn how to make the resulting argument easy to award marks in communicating mathematical reasoning, then test the outcome systematically in checking answers.