EJU Mathematics Course 1 common mistakes
EJU Mathematics Course 1 · Understand the traps and how to avoid them.
When solving 2x² − 5x + 1 = 0, the candidate writes the quadratic formula with b = 5 instead of b = −5, or forgets the leading 2 under the radical, producing x = (5 ± √17)/2 instead of x = (5 ± √17)/4.
The formula is memorised as a shape rather than as (−b ± √(b² − 4ac))/2a, so the minus sign on b and the 2a denominator get dropped under time pressure; candidates also confuse b² − 4ac with b² + 4ac when c is positive.
What to do instead
Before substituting, write the general form ax² + bx + c = 0 and label a, b, c explicitly on the page, including signs. Then substitute into a written-out template in one step and compute the discriminant separately. Drill five equations per session where b is negative and a ≠ 1, and check each root by substituting back.
The candidate solves |x − 3| < 2 as x − 3 < 2 giving x < 5, then reports the answer set as x < 5 (or as x > 1 only), instead of 1 < x < 5.
Absolute-value inequalities are treated as ordinary linear inequalities; candidates forget that |A| < k means −k < A < k and that the two-sided bound must be intersected, not just the right-hand side kept.
What to do instead
Convert first, solve second: write |A| < k ⇔ −k < A < k and |A| > k ⇔ A < −k or A > k as an explicit line before touching the expression. Then solve both sides and take the intersection, marking the result on a number line. Practise with k negative and with k = 0 to see when the solution set is empty or all reals.
Faced with 'find the number of integers x satisfying 3x − 1 ≤ 2x + 4 and x² < 20', the candidate solves each inequality, gets x ≤ 5 and −4.47 < x < 4.47, and then counts every integer from −4 to 5 inclusive including endpoints that fail, or forgets to intersect the two sets.
Compound conditions are handled sequentially rather than as a single simultaneous system; candidates also mishandle strict versus non-strict inequality when converting a real bound to an integer count.
What to do instead
Draw one number line and shade each condition in a different direction, then read off only the doubly shaded region. Convert real bounds to integers with explicit reasoning: x < 4.47 gives x ≤ 4, x ≤ 5 stays x ≤ 5, so the count is from −4 to 4. Write the final integer list out to verify the count.
When asked for the axis of symmetry or vertex of y = 2x² − 8x + 3, the candidate answers x = 4 (using −b/a) or x = 8, rather than x = 2, and then reports the vertex y-value computed from the wrong x.
The formula −b/2a is recalled loosely as 'minus b over a' or 'b over 2a'; the vertex is also confused with the y-intercept, and candidates forget that the axis of symmetry is an x-value, not a point.
What to do instead
Always complete the square once to re-derive the result: 2(x − 2)² − 5, so axis x = 2, vertex (2, −5). Compare with −b/2a on the same example so the formula is anchored to a derivation. After finding the vertex, state it as a coordinate pair and check that it lies on the curve.
For a triangle with sides 7 and 9 enclosing 60°, the candidate computes the area as 7 · 9 · sin 60° = 63√3/2, omitting the factor 1/2, or uses cos 60° in place of sin 60°.
The area formula is remembered as 'product of two sides times sine of the included angle' without the half; the sine and cosine rules are also blended, so the cosine rule expression leaks into area work.
What to do instead
Re-derive the formula from (1/2) × base × height with height = b sin C, and write that derivation once in the notes. Before computing, state which two sides and which included angle are being used, and confirm the angle is between them. Sanity-check the answer against a right-triangle estimate.
In a triangle with a = 8, b = 5, A = 40°, the candidate applies the sine rule as sin B = b sin A / a but then reports B = 23.7° without checking the obtuse possibility B = 156.3°, or uses the cosine rule when the sine rule is the appropriate tool.
The ambiguous case of the sine rule is under-practised, and candidates do not systematically check whether the given data determine one triangle or two; the decision rule for choosing sine versus cosine rule is not automatic.
What to do instead
Use a fixed decision checklist: sine rule when you have a matching side–angle pair; cosine rule when you have three sides or two sides plus the included angle. After any sine-rule angle, explicitly test 180° − B and check whether it is compatible with the given angles summing to 180°. Note also that the largest angle faces the longest side as a consistency check.
Asked for sin 15°, the candidate writes sin(45° − 30°) = sin 45° − sin 30° = (√2 − 1)/2, which is wrong, instead of using the correct expansion (√6 − √2)/4.
Trigonometric functions are treated as linear, so f(A − B) = f(A) − f(B) is assumed; the addition formulas are memorised imprecisely and the sign pattern in the sine formula is swapped.
What to do instead
Write the addition formulas in full each time: sin(A ± B) = sin A cos B ± cos A sin B, cos(A ± B) = cos A cos B ∓ sin A sin B. Test linearity with a quick counterexample such as sin 90° ≠ sin 45° + sin 45°. Verify final values numerically against a decimal approximation to catch sign and factor errors.
For data 4, 6, 6, 8, 11 the candidate reports the median as 6 (the mode) or computes the standard deviation by dividing the sum of squared deviations by n − 1 when the question expects the population value, giving a mismatched answer.
Median, mode, and mean blur together when the data set is small and sorted quickly; the distinction between population and sample standard deviation is not stated in the candidate's working, and the deviation is sometimes not squared before summing.
What to do instead
Use a fixed table: list values in order, count n, mark the middle position explicitly (for n = 5, the 3rd value), then compute deviations in a second column and square them in a third. State on the page whether you are dividing by n or n − 1 and why. Recompute the mean from the table before subtracting, since most errors propagate from a wrong mean.
Given a data set where each value is transformed as y = 2x + 3, the candidate states the new standard deviation is 2s + 3 or the new mean is 2x̄ without the +3, mixing up how location and scale changes affect each statistic.
The rules for how linear transformations act on mean, variance, and standard deviation are memorised as a list without the underlying reason, so the additive constant is wrongly applied to spread measures.
What to do instead
Anchor the rule in one sentence: adding a constant shifts every point but changes no spread; multiplying by a constant scales every deviation. So mean becomes 2x̄ + 3, standard deviation becomes 2s, variance becomes 4s². Derive it once symbolically from the definition of mean and deviation, and re-derive on the spot in the exam if unsure.
In a probability question about drawing two balls without replacement from a bag of 4 red and 6 white, the candidate computes the second draw as 4/10 again (with replacement logic) or adds probabilities of successive events instead of multiplying along a branch.
The sample space changes after the first draw but the denominator is not updated; candidates also confuse P(A and B) with P(A) + P(B) and do not distinguish independent from dependent events.
What to do instead
Draw an explicit tree diagram with the reduced totals written on the second-level branches (e.g. 3/9, 6/9). State 'without replacement' or 'with replacement' at the top of the working and adjust denominators accordingly. Use the rule that along a branch you multiply and across mutually exclusive branches you add, and check that all final probabilities sum to 1.