Bac Techniques Mathématiques (Algeria) - independent subject practice common mistakes
Bac Techniques Mathématiques (Algeria) - independent subject practice · Understand the traps and how to avoid them.
When differentiating a composite function, the candidate differentiates only the outer function and drops the inner derivative - writing d/dx sin(3x²) = cos(3x²) or d/dx e^(x²) = e^(x²).
Chain rule is learned as a pattern ('derivative of sin is cos') and under timed conditions the candidate pattern-matches the outer function and stops, treating the argument as a single variable. Nested functions with polynomial or exponential inner parts are where the omission is most likely, because the inner derivative is not visually salient.
What to do instead
Before writing any answer, explicitly name the inner function u and write the chain as outer′(u)·u′ on the page, even for one-line answers. Drill a mixed set of 20 composites (trig, exponential, log, power, inverse trig) and mark each inner derivative in a different colour until the step becomes automatic. On the exam, do a final pass over every derivative checking that each composite has a visible inner-derivative factor.
The candidate computes a definite integral or area but substitutes the limits in the wrong order or forgets to change limits after substitution, producing a sign-flipped or wrong-valued answer, e.g. evaluating ∫₀² with F(0) − F(2).
The upper-minus-lower convention competes in memory with the substitution rule's 'evaluate at old limits', and under pressure the candidate writes the antiderivative values in the order they appear on the page. When using u-substitution the candidate often back-substitutes inconsistently or plugs original limits into the u-antiderivative.
What to do instead
Adopt a fixed written template: bracket the antiderivative, write '= [F(x)]_a^b = F(b) − F(a)', and physically copy b above a before evaluating. For substitution, always either convert the limits to u-values in the same line or write 'back-substitute to x, then apply original limits' - never mix. Practise five substitution integrals where the limits are converted, checking that a positive integrand gives a positive result.
When computing a definite integral with substitution, the candidate keeps the antiderivative in the substituted variable but evaluates it at the original x-limits, or changes the limits to u-values and then substitutes x back before evaluating, producing a numerically wrong value with correct-looking work.
The two valid procedures (change limits and stay in u, or keep limits and back-substitute) are each individually simple, but mixing them is easy because the back-substitution step feels like part of 'finishing the problem'. The error is silent: the algebra is all correct, only the evaluation is wrong.
What to do instead
Choose one convention and write it at the top of the solution. If changing limits, compute the new limits immediately after defining u and cross out the old ones on the integral sign. If keeping limits, write 'back-substitute, then evaluate at x = a and x = b' as a literal step. Sanity-check the sign and rough magnitude against an area estimate.
The candidate states a limit exists and equals a finite number when the one-sided limits disagree, when the function is unbounded, or when the expression is an indeterminate form such as 0/0 that was never resolved; alternatively, the candidate differentiates a function to check continuity.
Limit, continuity and differentiability are studied in adjacent chapters and their definitions blur together. Indeterminate forms tempt the candidate to cancel incorrectly or to declare 'undefined' as if it were a limit value, and the distinction between a limit existing and a function being defined at a point is subtle in notation.
What to do instead
For every limit question, first compute the left-hand and right-hand limits separately and compare them explicitly. Classify the form before doing algebra, and only then choose a technique (factor and cancel, rationalise, or compare growth rates). Write the formal criteria as a three-line checklist: limit exists, value equals f(a), and the two-sided limit matches f(a), for continuity; derivative from the limit definition for differentiability.
When finding eigenvectors, the candidate solves det(A − λI) = 0 for λ correctly but then solves (A − λI)x = 0 by row-reducing A itself rather than A − λI, or picks a single eigenvector for a repeated eigenvalue and misses the second independent one.
Row-reduction of A is a well-practised reflex, and substituting the numeric λ into every diagonal entry is error-prone mental arithmetic. For repeated eigenvalues the candidate stops once one free variable yields a vector, not checking the nullity of A − λI.
What to do instead
Write A − λI explicitly with λ substituted before any row operation, and compute its nullity to know in advance how many independent eigenvectors to expect (nullity = multiplicity for diagonalisable cases). For each eigenvalue, state the dimension of the eigenspace before choosing vectors. Verify each candidate v by computing Av and confirming it equals λv - a 30-second check that catches both errors.
The candidate computes a cross product or scalar triple product and reports a magnitude or a signed value when a vector (or the opposite orientation) was required, or takes the dot product where the cross product is needed for a normal vector.
Area, volume and normal-vector problems all invoke 'cross product' in the candidate's notes, and the operations get conflated. Sign conventions in the triple product (a·(b×c) versus (a×b)·c) and the direction of a×b are also easy to reverse under time pressure.
What to do instead
Write the geometric quantity required first - 'normal vector =', 'area = ½|a×b|', 'volume = |a·(b×c)|' - then perform the operation that matches it. Compute cross products with the determinant expansion written out in full (i, j, k rows) rather than mentally, and check orientation with the right-hand rule or by confirming a·(a×b) = 0 and b·(a×b) = 0.
Writing g = 10 m/s² or 9.8 m/s² into a calculation when the paper's data sheet specifies a value (use the stated exercise value), or dropping the unit of g entirely and then treating a weight as a mass in later steps.
Candidates memorise one value from textbooks or from a different year's paper and never check the front of the question paper; the unit is dropped because g is treated as a bare number in the formula W = mg.
What to do instead
Before starting a calculation, circle the constants printed on the paper (g, e, h, c, mass of electron) and write them at the top of your answer booklet. In every mechanics line, carry units through: write W = mg = 2.0 kg × 10 m/s² = 20 N, not 2.0 × 10 = 20.
Resolving forces on an inclined plane by using the angle with the horizontal in the component that should use the angle with the vertical, e.g. writing the component of weight along a slope of angle θ as mg cos θ instead of mg sin θ.
The formula is memorised as a pattern without drawing the triangle; candidates do not sketch the weight vector and the slope and identify which angle is actually between the weight and the normal.
What to do instead
Always draw the free-body diagram first and mark the angle θ on it. Then resolve along and perpendicular to the plane by geometry, not memory: the component of weight down the slope is mg sin θ, the normal reaction is mg cos θ. Check with the limits θ = 0 (no component down the slope) and θ = 90° (full weight down the slope).
Using the thin-lens or mirror equation with a real-is-positive sign convention, then reporting a virtual image as real (or a negative focal length as if it were positive), and confusing magnification m = v/u with m = −v/u.
Different textbooks and the candidate's own notes use different sign conventions; the substitution is done mechanically and the sign of the answer is never interpreted physically.
What to do instead
State your convention explicitly at the top of the optics answer (e.g. 'real-is-positive, distances measured from the lens'). After computing, interpret the sign in words: a negative v means a virtual image on the same side as the object. For a converging lens, check that an object inside the focal length gives a virtual, magnified, upright image - if your numbers say otherwise, the sign handling is wrong.
Treating e.m.f. and terminal potential difference as interchangeable, so that a circuit with internal resistance is solved as if the terminal p.d. equalled the e.m.f., and the lost volts are never accounted for.
Introductory problems often use ideal cells, and the phrase 'lost volts' is memorised without linking it to the current through the internal resistance.
What to do instead
Write the energy equation every time: ε = I(R + r) = V_terminal + Ir. Label V_terminal as the p.d. across the external resistor and Ir as the lost volts. Check the limiting cases: open circuit (I = 0, V_terminal = ε) and short circuit (R = 0, V_terminal = 0).
Confusing the direction of induced current or force using Fleming's left- and right-hand rules, and applying them with the field, current and motion in the wrong fingers.
The two rules look similar and are memorised as gestures without a physical anchor; candidates do not connect the left-hand rule to motors (force on a current) and the right-hand rule to generators (induced current).
What to do instead
Anchor each rule to its device: left hand = motor effect (current in a field experiences a force); right hand = generator effect (motion in a field induces a current). In the left-hand rule, first finger = field, second finger = current, thumb = force. Test the answer against Lenz's law: the induced current must oppose the change producing it.
Mixing Celsius and kelvin in the gas law or in specific-heat and latent-heat calculations, e.g. substituting 27 into pV = nRT, or using Δθ and ΔT inconsistently.
Temperature differences are numerically equal in °C and K, so candidates assume absolute temperatures are also interchangeable, and the ideal-gas equation is memorised without the 'absolute' condition.
What to do instead
Convert to kelvin as the first line of any gas calculation: T = θ + 273 (use the value on the data sheet). Note that Δθ = ΔT so differences need no conversion, but pV = nRT and Charles's/Gay-Lussac's laws require absolute temperature. Sanity-check: a gas at 0 °C must not give zero volume or zero pressure.