Common exam mistakes

Most lost marks come from a small number of habits: changing an expression incorrectly, answering a nearby question instead of the one asked, rounding too early, or giving an unsupported conclusion. These are not merely careless slips. Each has a mathematical cause and a reliable prevention method.

This lesson shows how to recognise, correct and check the most common errors across pure mathematics, statistics and mechanics.

What you should know first

You should be able to:

• use brackets, fractions, indices and function notation;
• rearrange equations and solve elementary equations;
• enter expressions into a scientific calculator;
• distinguish exact and approximate values.

Review algebraic fluency, exact arithmetic or calculator fluency if these skills are uncertain.

Read the command before doing the mathematics

Underline mentally what the answer must contain. Common commands require different outputs.

Command	What a complete answer needs
Find, calculate	A value, usually with working
Show that	A logical derivation ending in the stated result
Hence	Use the preceding result
Prove	A general argument, not selected examples
Sketch	Essential shape, intercepts, asymptotes and labelled features
Interpret	Meaning in the stated context, with units where relevant

Example 1: solving the wrong problem

Given $f(x)=x^3-3x^2+2$, find the coordinates of the stationary points.

Differentiating and solving $f'(x)=0$ finds only the $x$ coordinates:

$$f'(x)=3x^2-6x=3x(x-2)=0,$$

so $x=0$ or $x=2$. The question asks for coordinates, so substitute into $f$:

$$f(0)=2, \qquad f(2)=8-12+2=-2.$$

Therefore the stationary points are

$$\boxed{(0,2)\text{ and }(2,-2)}.$$

Writing only $x=0,2$ shows useful work but does not finish the task.

Preserve equivalence in algebra

An equation may be changed only by applying a valid operation. An expression may be rewritten only as an equal expression. The most frequent failures involve brackets, cancellation and square roots.

Expanding brackets incorrectly

The square of a sum is not the sum of the squares:

$$(a+b)^2=a^2+2ab+b^2 \ne a^2+b^2.$$

For example,

$$(2x-3)^2=4x^2-12x+9.$$

A quick check is to put $x=1$. The original expression gives $(-1)^2=1$, and the expansion gives $4-12+9=1$.

Cancelling terms instead of factors

Cancellation is division by a common non-zero factor. It does not work across addition:

$$\frac{x+3}{x} \ne 3.$$

The correct simplification is

$$\frac{x+3}{x}=1+\frac{3}{x}, \qquad x\ne 0.$$

By contrast,

$$\frac{x^2+3x}{x}=\frac{x(x+3)}{x}=x+3, \qquad x\ne 0,$$

because $x$ is a factor of every term in the numerator.

Dividing by a quantity that might be zero

Consider

$$x(x-4)=0.$$

Dividing both sides by $x$ gives $x-4=0$ but silently loses the solution $x=0$. Use the zero-product rule:

$$x=0 \quad \text{or} \quad x-4=0,$$

so

$$\boxed{x=0\text{ or }x=4}.$$

Before dividing by an expression containing the unknown, ask whether it could equal zero.

Forgetting both square roots

From $x^2=25$,

$$x=\pm 5,$$

because both $5^2$ and $(-5)^2$ equal $25$. However, the symbol $\sqrt{25}$ means the principal, non-negative root, so $\sqrt{25}=5$, not $\pm5$.

This distinction matters:

$$x^2=25 \implies x=\pm5, \qquad \sqrt{x^2}=|x|.$$

Respect domains and restrictions

Algebraic procedures can produce candidates that the original problem does not allow. Check denominators, logarithms, square roots and stated intervals.

Example 2: an extraneous root

Solve

$$\sqrt{x+6}=x.$$

Since a square root is non-negative, any solution must satisfy $x\ge 0$. Squaring gives

$$x+6=x^2,$$

so

$$x^2-x-6=0$$

and hence

$$(x-3)(x+2)=0.$$

The candidates are $x=3$ and $x=-2$. Check them in the original equation:

$$\sqrt{3+6}=3,$$

so $x=3$ works, but $x=-2$ cannot work because the right-hand side is negative. Therefore

$$\boxed{x=3}.$$

Squaring is not reversible without a sign condition. It may introduce solutions.

Other restrictions to record include

$$\frac{1}{g(x)}: g(x)\ne0, \qquad \ln(g(x)): g(x)>0, \qquad \sqrt{g(x)}: g(x)\ge0.$$

See functions and logarithms for fuller domain work.

Use function notation precisely

$f(x+1)$ means replace every $x$ in the formula for $f(x)$ by $x+1$. It does not usually mean $f(x)+1$.

Example 3: substitution into a function

If

$$f(x)=x^2-3x,$$

then

$$\begin{aligned} f(x+1) &=(x+1)^2-3(x+1)\\ &=x^2+2x+1-3x-3\\ &=x^2-x-2. \end{aligned}$$

Putting the substituted expression in brackets prevents partial substitution.

Also distinguish inverse functions and reciprocals:

$$f^{-1}(x) \ne \frac{1}{f(x)}$$

in general. The notation $f^{-1}$ means the function that reverses $f$.

Do not apply linear rules to non-linear operations

Several tempting but false rules have the same pattern:

$$\sqrt{a+b}\ne\sqrt a+\sqrt b,$$ $$\ln(a+b)\ne\ln a+\ln b,$$ $$\sin(A+B)\ne\sin A+\sin B.$$

The valid rules are structurally different:

$$\sqrt{ab}=\sqrt a\sqrt b \quad (a,b\ge0),$$ $$\ln(ab)=\ln a+\ln b \quad (a,b>0),$$ $$\sin(A+B)=\sin A\cos B+\cos A\sin B.$$

A numerical test often exposes a false identity. For example,

$$\sqrt{9+16}=5, \qquad \sqrt9+\sqrt{16}=7.$$

One counterexample is enough to disprove a claimed identity.

Keep exact values until the final line

Premature rounding changes later calculations and can move a final answer outside the accepted tolerance. Keep calculator values unrounded or store them in memory.

Example 4: early rounding

The radius of a circle is $r=\sqrt7$ cm. Find its area to $3$ significant figures.

Using the exact value,

$$A=\pi r^2=\pi(\sqrt7)^2=7\pi=21.991\ldots,$$

so

$$\boxed{A=22.0\text{ cm}^2 \text{ to }3\text{ s.f.}}$$

If $r$ is first rounded to $2.65$, then

$$\pi(2.65)^2=22.061\ldots,$$

which gives a different result.

Use $=$ only for equality and $\approx$ for approximation:

$$7\pi \approx 22.0.$$

If the question requests an exact answer, leave forms such as $\frac{3\pi}{4}$, $2\sqrt5$ or $\ln 7$ exact.

Check calculator mode and entry

Trigonometric questions may use degrees or radians. A calculator in the wrong mode can produce plausible-looking nonsense.

For example,

$$\sin 30^\circ=\frac12,$$

whereas $\sin(30)$ in radian mode is approximately $-0.988$. Look for the degree symbol, an interval involving $\pi$, or an explicit instruction.

Use brackets to enter complete numerators, denominators and function arguments. To calculate

$$\frac{3+\sqrt5}{2\sin 40^\circ},$$

enter a structure equivalent to

$$(3+\sqrt5)\div(2\sin 40^\circ).$$

After entering a complicated expression, compare the display with the printed mathematics before pressing equals. Review radians for angle measure.

Keep inequality direction under control

Multiplying or dividing an inequality by a negative number reverses its direction.

Example 5: a linear inequality

Solve

$$5-2x<11.$$

Subtract $5$:

$$-2x<6.$$

Divide by $-2$ and reverse the inequality:

$$\boxed{x>-3}.$$

Check a value. Since $x=0$ satisfies $5-2x<11$, the solution must include $0$. This agrees with $x>-3$.

For quadratic and rational inequalities, finding boundary values is not enough. Use a sign diagram or test values on each interval. See inequalities.

In calculus, differentiate the whole structure

Two errors recur: omitting the derivative of an inner function and treating a product as though derivatives multiply.

Example 6: chain rule and product rule

Differentiate

$$y=x^2\sin(3x).$$

The outer structure is a product. Also, $\sin(3x)$ requires the chain rule:

$$\frac{d}{dx}\sin(3x)=3\cos(3x).$$

Therefore

$$\begin{aligned} \frac{dy}{dx} &=(2x)\sin(3x)+x^2\bigl(3\cos(3x)\bigr)\\ &=\boxed{2x\sin(3x)+3x^2\cos(3x)}. \end{aligned}$$

It is false that $(uv)'=u'v'$. The product rule is

$$(uv)'=u'v+uv'.$$

See product, quotient and chain rules for deliberate practice.

For indefinite integration, include the arbitrary constant:

$$\int 6x^2\,dx=2x^3+C.$$

For definite integration, do not add $C$:

$$\int_a^b f(x)\,dx=F(b)-F(a).$$

Remember that an integral can be negative. Geometric area is non-negative, so split the interval where a graph crosses the axis and use positive areas when the question asks for total area. Review definite integrals and areas.

In statistics, answer in the language of evidence

Hypothesis tests do not prove that a hypothesis is true. They measure how surprising the observed result would be if the null hypothesis were true.

Example 7: writing a valid conclusion

A test of a coin uses

$$H_0:p=0.5, \qquad H_1:p>0.5.$$

Suppose the calculated $p$-value is $0.032$ and the significance level is $5\%$.

Since

$$0.032<0.05,$$

reject $H_0$. A complete contextual conclusion is:

There is sufficient evidence at the $5\%$ significance level to suggest that the probability of heads is greater than $0.5$.

Do not write that $H_0$ is definitely false or that $H_1$ has been proved. A small probability under $H_0$ is evidence, not certainty.

Other frequent statistics errors include:

• assuming correlation proves causation;
• using a regression line outside the observed data range without warning;
• confusing $P(A\mid B)$ with $P(B\mid A)$;
• forgetting that $\operatorname{Var}(X)=E(X^2)-[E(X)]^2$, not $E(X^2)-E(X)$.

Review hypothesis testing language and conditional probability.

In mechanics, model first and keep directions consistent

Write down a positive direction before resolving forces or using constant acceleration formulae. A negative answer then has a clear meaning.

Example 8: signs in vertical motion

A particle is projected vertically upwards at $14\text{ m s}^{-1}$. Find its velocity after $2$ seconds, taking $g=9.8\text{ m s}^{-2}$.

Take upwards as positive. Then

$$u=14, \qquad a=-9.8, \qquad t=2.$$

Using $v=u+at$,

$$v=14+(-9.8)(2)=-5.6\text{ m s}^{-1}.$$

Thus the velocity is $-5.6\text{ m s}^{-1}$ relative to the chosen direction, meaning the particle is moving downwards at speed $5.6\text{ m s}^{-1}$.

Do not confuse speed and velocity. Speed is a non-negative scalar; velocity includes direction.

Also check that a constant acceleration formula is justified. It cannot be used unchanged if acceleration varies with time or displacement.

Show enough working

A correct calculator answer may earn few marks if the question assesses a method. Write the equation or theorem you use, substitute clearly, and retain exact or unrounded values until the final answer.

For a claim such as

$$2x^2-5x-3=0,$$

writing

$$(2x+1)(x-3)=0$$

shows the factorisation and makes the solutions $x=-\frac12$ and $x=3$ defensible.

In a proof, do not assume the statement you are trying to prove. In a “show that” question, begin from known information and transform it towards the required result. See deduction and exhaustion and counterexamples.

A reliable final check

Use the last minute of a question deliberately.

1. Answer: Have you answered the exact command, including every solution?
2. Restrictions: Are solutions in the required domain or interval?
3. Reasonableness: Do sign, size and units make sense?
4. Accuracy: Did you keep full precision and round as requested?
5. Communication: Is the final answer identifiable and supported by working?

Substitution is especially powerful. If you solve an equation, put each candidate into the original equation, not only a rearranged or squared version. For graphs, check intercepts, asymptotes and end behaviour. For probability, check that $0\le P(A)\le1$. For lengths, times, masses and standard deviations, question an unexplained negative answer.

Self-check

Try these without looking back.

1. Simplify $\dfrac{x^2-9}{x-3}$ and state any restriction.
2. Solve $\sqrt{2x+3}=x$.
3. If $f(x)=2x^2+1$, find $f(x-2)$.
4. Differentiate $y=(5x+1)^4$.
5. Solve $7-3x\ge16$.
6. A hypothesis test gives $p=0.081$ at the $5\%$ level. Should $H_0$ be rejected?

Answers

1. Factor first:

   $$\frac{(x-3)(x+3)}{x-3}=x+3, \qquad \boxed{x\ne3}.$$

   The simplified formula does not restore the excluded input.

2. A solution must have $x\ge0$. Squaring gives

   $$2x+3=x^2,$$

   so $(x-3)(x+1)=0$. Checking the candidates in the original equation leaves

   $$\boxed{x=3}.$$

3. Substitute $x-2$ everywhere:

   $$f(x-2)=2(x-2)^2+1=\boxed{2x^2-8x+9}.$$

4. By the chain rule,

   $$\frac{dy}{dx}=4(5x+1)^3\times5=\boxed{20(5x+1)^3}.$$

5. Subtract $7$, then divide by $-3$ and reverse the inequality:

   $$-3x\ge9 \implies \boxed{x\le-3}.$$

6. No. Since $0.081>0.05$, do not reject $H_0$. There is insufficient evidence at the $5\%$ significance level to support the alternative hypothesis.

Next steps

Build a personal error log with three columns: the error, its mathematical cause, and the check that would catch it. Then strengthen the underlying skills through algebraic fluency, calculator fluency, functions, inequalities and hypothesis testing language.