Lebanese University Faculty of Engineering Entrance Examination common mistakes
Lebanese University Faculty of Engineering Entrance Examination · Understand the traps and how to avoid them.
When solving an inequality such as (x-2)/(x+3) ≥ 0, the candidate multiplies both sides by (x+3) and keeps the inequality sign, producing the single interval x ≥ 2 and discarding the negative-interval solution.
The candidate treats an algebraic inequality like an equation and forgets that multiplying or dividing by a quantity whose sign depends on x reverses the inequality. The habit is reinforced by equation-solving practice where cross-multiplication is always safe. The domain restriction x ≠ -3 is also silently dropped.
What to do instead
Convert to a sign table: move everything to one side, factor numerator and denominator, mark the zeros of each factor on a number line, and test one point per interval. Record excluded values separately. Drill five rational inequalities per session until the sign-chart becomes the automatic first move rather than cross-multiplication.
The candidate finds the domain of a composite or radical-logarithmic function by checking only the outermost operation, for example giving the domain of ln(√(x-1) - 2) as x ≥ 1 because the square root appears first, without requiring the logarithm argument to be strictly positive.
Domain questions are answered by scanning the expression left to right and stopping at the first restriction found. Candidates do not systematically list every operation that imposes a condition (even roots, denominators, logarithms, inverse trig), so inner restrictions are forgotten or outer ones are applied to the wrong sub-expression.
What to do instead
Adopt a fixed checklist applied from the inside out: for each operation write its condition as a separate inequality, then intersect all conditions. Always finish by testing a sample value from the proposed domain in the original expression. Practice with nested expressions where the inner and outer conditions conflict.
On a function-study problem the candidate computes f'(x), sets it to zero, and reports the resulting points as extrema without checking the sign change of f', and without computing the second derivative or the limits at the boundaries of the domain.
The candidate has memorised 'derivative zero means extremum' as a procedure and conflates stationary points with extrema. Inflection points with horizontal tangents (such as x³ at 0) and endpoints of a restricted domain are not considered, and the sign of f' is never tabulated.
What to do instead
Write the full variation table every time: domain, limits at open boundaries, sign of f', monotonicity intervals, local extrema, then sign of f'' for concavity and inflection. Verify each claimed extremum by evaluating f at neighbouring test points. Include at least one exercise per session where f' vanishes without a sign change.
When asked for the intersection of a line and a circle, or the tangency condition, the candidate substitutes y = mx + c into the circle equation, forms the quadratic, and then solves for x without imposing the discriminant condition, reporting two intersection points even when the line does not meet the circle.
Substitution is mechanical and comfortable, so the candidate stops as soon as a quadratic appears. The geometric meaning of the discriminant (positive, zero, negative corresponding to secant, tangent, no intersection) is not linked to the algebra, and arithmetic slips in expanding the quadratic go undetected because no geometric sanity check is applied.
What to do instead
Before solving, state the expected geometric case from a sketch or from the distance from the centre to the line compared with the radius. Then use the discriminant explicitly as the condition, and check the number of solutions against the geometric prediction. Memorise the distance-from-point-to-line formula and use it as the primary tangency test.
The candidate solves a trigonometric equation such as sin(2x) = 1/2 by writing 2x = π/6 and then x = π/12, giving one solution instead of the full solution set on the requested interval.
Inverse-trig functions return a principal value, and candidates treat that single output as the complete answer. The periodic structure and the second family of solutions (π - π/6 here) are not generated, and the argument 2x is not rescaled back to the required interval, so valid solutions are lost.
What to do instead
Use a fixed template: solve for the inner argument with both families (θ and π - θ for sine, θ and -θ for cosine, θ + π for tangent), add 2kπ or kπ, then divide by the coefficient and enumerate k values that land inside the stated interval. Finish by counting solutions and checking them numerically.
When computing the volume of a solid of revolution or the area between two curves, the candidate integrates the difference of the functions without checking which curve is upper, and without squaring the outer and inner radii, producing a negative or otherwise impossible result that is reported as the answer.
The candidate focuses on finding the antiderivative and treats the setup as trivial. The geometric configuration (which curve lies above, where the curves cross, whether the axis of revolution is offset from the region) is not analysed, and the washer formula is recalled as π∫(R - r)² instead of π∫(R² - r²).
What to do instead
Always sketch the region, find intersection points, and determine the upper and lower function on each sub-interval before integrating. Write the radius expressions explicitly (distance from each curve to the axis) and square them separately. Treat a negative area or volume as a signal to stop and re-examine the setup.
In a kinematics problem with constant acceleration, the candidate substitutes numbers into v = u + at and s = ut + ½at² using a mixture of units (speed in km/h, time in seconds, distance in metres) and inconsistent sign conventions for upward and downward motion.
Numbers are grabbed from the statement in the order they appear without a units pass, and the choice of positive direction is never declared. Because the equations are memorised as symbol patterns, the candidate does not notice that a term mixes hours and seconds, and a deceleration entered as positive silently corrupts the result.
What to do instead
Begin every mechanics problem by writing 'positive direction = ...' and converting all quantities to SI before substitution. List the five kinematic quantities, mark the three known and the one sought, then select the equation that omits the unused variable. Check the final answer for order of magnitude and sign against physical intuition.
The candidate analyses a system of blocks connected by a string over a pulley by applying F = ma to the whole system and then reports that acceleration as the tension in the string, or applies the same acceleration to both bodies without writing separate equations for each.
The candidate collapses a multi-body problem into a single-body problem because the shared acceleration suggests a single object. Tension is treated as an external force on the system rather than an internal interaction, and no free-body diagram is drawn, so the distinct forces on each mass are never enumerated.
What to do instead
Draw a separate free-body diagram for each body, choose a positive direction along the motion for each, and write Newton's second law for each body separately. Add the equations to eliminate tension and solve for acceleration, then substitute back to obtain tension. Verify that tension is less than the hanging weight when the system accelerates.
When finding the equivalent resistance of a network, the candidate adds series and parallel branches in the order the resistors appear on the page rather than by electrical node, and when a battery with internal resistance is present the candidate uses the emf in Ohm's law instead of the terminal voltage.
Circuit diagrams are read visually left to right, and candidates do not redraw the network to expose which components share the same two nodes. Internal resistance is treated as an abstract detail, so the distinction between emf, terminal voltage, and the voltage across the external resistor is blurred.
What to do instead
Redraw the circuit marking nodes with letters and colour-code nodes at the same potential; combine resistors only when they share exactly two nodes. Write the loop equation ε = I(R + r) explicitly and compute terminal voltage as ε - Ir. Check that the sum of voltage drops around each loop returns to zero.
In a stoichiometry problem the candidate balances the equation, converts grams to moles for the first reactant, and then uses that reactant's mole ratio directly, without determining the limiting reagent when two reactant quantities are given; yields are then reported as if the reaction were complete.
The first-listed reactant is assumed to be limiting, and the mole-ratio step is applied mechanically without comparing the available moles divided by the stoichiometric coefficients. The distinction between theoretical and actual yield, and the meaning of a percentage yield, is not operationalised.
What to do instead
For every problem with two or more reactant amounts, compute moles divided by the coefficient for each and identify the smallest quotient as limiting before any product calculation. Carry units through each step and state the theoretical yield explicitly, then apply the percentage yield as a separate final multiplication. Practise problems where the obvious reactant is not limiting.