EJU Mathematics Course 2 common mistakes
EJU Mathematics Course 2 · Understand the traps and how to avoid them.
When solving a quadratic inequality such as x²−x−6<0, the candidate factors correctly to (x−3)(x+2)<0 but then writes x<−2 or x>3, giving the union of the outer intervals instead of the interval between the roots −2<x<3.
They memorize 'greater than means outside, less than means inside' as a verbal rule tied to the direction of the inequality sign, but lose track of whether the quadratic opens upward or downward and whether the sign of the leading coefficient was already positive. Under exam time pressure they apply the rule mechanically without sketching the parabola or testing a sample point.
What to do instead
Always sketch the upward parabola with the two roots marked, then read off where the graph lies below or above the x-axis, and confirm with one test value (e.g. x=0) before writing the answer. If the leading coefficient is negative, multiply by −1 and flip the inequality first.
When counting arrangements with restrictions, the candidate counts total permutations and subtracts the 'bad' cases, but double-counts arrangements where two different restrictions overlap, or forgets to divide by repeated-element factorials when letters or digits repeat.
Combinatorial counting has no natural feedback signal, so an answer that 'looks plausible' is accepted without verification. Candidates also conflate permutations with combinations, and treat 'at least one' problems by direct casework instead of the complement, which multiplies opportunities for overlap errors.
What to do instead
For every counting problem, decide explicitly whether order matters and whether repetition is allowed before writing any symbol. For 'at least one' use the complement (total minus none). For overlapping conditions use inclusion–exclusion or a case split on the most restrictive condition, and verify small cases by listing.
When computing a conditional probability P(A|B), the candidate divides P(A∩B) by P(A) instead of P(B), or treats two events as independent and multiplies P(A)P(B) in a problem where the events are actually dependent (drawing without replacement).
The formula is remembered as a vague 'probability of the intersection over a probability', and the denominator is chosen by which event is mentioned first in the sentence rather than which event is given. Without-replacement problems look identical to with-replacement ones unless the candidate consciously tracks whether the sample space shrank.
What to do instead
Rewrite the question as 'given B, find A' and write P(A|B)=P(A∩B)/P(B), checking that the denominator is the conditioning event. For drawing problems, state in words whether the item is replaced; if not, recompute the denominator after each draw. Test independence by checking P(A∩B)=P(A)P(B) rather than assuming it.
When differentiating a composite or product expression such as sin(2x), e^{3x}, or x²ln x, the candidate drops the inner derivative or applies only one term of the product rule, producing derivatives like 2cos(2x)→cos(2x) or x²ln x→2x·(1/x).
The chain rule and product rule are learned as separate templates, and under time pressure the candidate pattern-matches the outer function only. The inner derivative is mentally 'obvious' and gets skipped, and product-rule terms are merged rather than added.
What to do instead
Write the rule explicitly before differentiating: for f(g(x)) write f′(g(x))·g′(x) and fill both slots; for a product write u′v+uv′ on two separate lines before simplifying. Practise five composite functions and five products daily until the intermediate line is automatic.
When computing a definite integral to find an area, the candidate integrates between the x-intercepts of a single curve without splitting the interval where the curve crosses the axis, so a region below the axis cancels a region above it and the reported area is too small (or even zero).
The definite integral computes signed area, but 'area' in the question means geometric area. Candidates treat the integral as a black-box area machine and do not sketch the graph, so sign changes inside the interval go unnoticed.
What to do instead
Sketch the curve over the interval first and mark every zero crossing. Split the integral at each crossing, take the absolute value of each piece, and add. When two curves bound a region, integrate (upper − lower) and find intersections before integrating.
When solving a logarithmic equation such as log₂(x−1)+log₂(x−3)=2, the candidate combines logs, solves the resulting quadratic, and reports both roots without discarding the one that makes an argument non-positive.
The algebraic manipulation is correct, so the answer feels finished. The domain restriction of the logarithm (argument > 0) is treated as a preliminary remark rather than a final filter, and candidates rarely substitute answers back into the original equation.
What to do instead
Before solving, write the domain conditions for every logarithm (each argument > 0, base > 0 and ≠1). After solving, test each root against those conditions and explicitly cross out invalid ones. Make substitution-back a fixed final step for all log and radical equations.
When using the sine rule or cosine rule in a triangle, the candidate pairs a side with the wrong opposite angle, or uses the sine rule to find an angle in an ambiguous SSA configuration and reports only the acute solution.
The sine rule is memorized as 'a over sin A' without the pairing being visually checked against the labelled triangle, and the SSA ambiguity (two possible triangles) is a subtle case often skipped in review. Diagrams are drawn roughly, hiding which angle faces which side.
What to do instead
Label every triangle with sides a, b, c opposite angles A, B, C before substituting, and state the pairing aloud. When given two sides and a non-included angle, always check whether the supplementary angle also satisfies the triangle angle sum, and report both valid triangles if they exist.
When finding the power of a point or a length via intersecting chords and tangents, the candidate applies PA·PB=PC·PD with segments measured from the wrong endpoint, or uses the tangent length squared as PT instead of PT².
The theorem is recalled as a product of 'the two parts', but which two parts depends on whether the point is inside or outside the circle and on which lines are drawn. The tangent case is a limiting case of the chord theorem, and the squared relationship is easy to flatten.
What to do instead
Draw the figure and label each segment from the external (or internal) point outward, then write the theorem in words: 'product of the two segments on one secant equals the product on the other; for a tangent, the tangent length squared equals the secant product.' Substitute only after the word statement is written.
When working with vectors, the candidate computes an inner product to test perpendicularity but confuses the condition a·b=0 with parallelism, or solves for an unknown coefficient using the wrong component pairing, and in 3D forgets that a line and a plane require a normal vector rather than a direction vector.
Perpendicular and parallel conditions are both expressed as simple equations (a·b=0 versus a=kb), and under pressure the candidate recalls 'something equals zero' without attaching it to the correct geometric meaning. Normal versus direction vectors are not distinguished when setting up plane equations.
What to do instead
Write a small reference card: parallel ⇔ a=kb (components proportional), perpendicular ⇔ a·b=0. For planes, identify the normal vector from the coefficients and use it for both the plane equation and the angle with a line. Verify any solved parameter by substituting back into the original geometric condition.
When solving an integer problem (linear Diophantine equation, base conversion, or Euclidean algorithm), the candidate finds one particular solution and stops, omitting the general solution, or converts a number to base b by dividing repeatedly but reads the remainders in the wrong order.
School algebra trains 'solve for x' as producing a single answer, so the integer-solution structure (particular solution plus multiples of the coefficient ratio) is unfamiliar. Base conversion is a mechanical procedure where the order of remainders is easy to reverse and rarely checked.
What to do instead
For ax+by=c, after finding one solution (x₀,y₀), write the general solution x=x₀+(b/g)t, y=y₀−(a/g)t with g=gcd(a,b), and state the condition g|c for existence. For base conversion, read remainders bottom-to-top and verify by expanding the result back into base 10.