SEE Optional Mathematics Nepal common mistakes
SEE Optional Mathematics Nepal · Understand the traps and how to avoid them.
When asked for f∘g(x) or f⁻¹(x), the candidate writes f(x)·g(x) for the composite, and writes 1/f(x) for the inverse, then substitutes a number into that wrong expression.
The notation f⁻¹ looks like an exponent of −1 and juxtaposition fg looks like multiplication, so candidates fall back on familiar algebraic operations instead of the definition of composition and inverse.
What to do instead
Drill the definitions as procedures: for f∘g substitute the whole expression g(x) everywhere x appears in f; for f⁻¹ write y = f(x), swap x and y, then solve for y. Always verify with one numeric check, e.g. f(f⁻¹(2)) must return 2.
In a sequence question asking to insert k arithmetic means between two numbers, or to find the sum of the first n terms, the candidate uses n instead of n + 2 for the total number of terms, or applies the AP sum formula to a series that is actually geometric.
Candidates count the inserted terms but forget the two endpoint terms, and they assume any listed sequence is arithmetic without testing whether the first differences are constant or the ratios are constant.
What to do instead
Before choosing a formula, compute t₂ − t₁ and t₃ − t₂; equal differences mean AP, equal ratios mean GP. For insertion problems, draw the full list with the endpoints marked and count the gaps rather than the numbers.
Solving a quadratic by dividing both sides by x, so the root x = 0 disappears; or forming an equation from given roots as x² + (sum)x + product = 0, giving every sign wrong.
Dividing by a variable is a habit carried over from linear equations, and the sign pattern of the root-coefficient relationship is memorised as a string of symbols rather than derived from (x − α)(x − β) = 0.
What to do instead
Never divide by a variable expression; factor it out instead so each factor can be set to zero. Rebuild the root equation from (x − α)(x − β) = 0 every time, and test by substituting α back into the finished equation.
In a linear programming problem the candidate graphs the constraints but evaluates the objective function only at the vertices formed by intersecting the two oblique lines, ignoring the intercepts on the axes, and shades the wrong side for a ≥ constraint.
Textbook diagrams often show a tidy quadrilateral, so candidates assume the optimum is at the crossing point, and the test-point rule for shading is applied mechanically without checking which side satisfies the inequality at the origin.
What to do instead
List every corner of the feasible region by name, including axis intercepts, then build a table of Z values at all of them before deciding. For each inequality substitute (0,0) explicitly and mark the accepted side with an arrow on the boundary line.
Squaring both sides of a trigonometric equation to make it solvable, then accepting every value obtained, including ones that fail the original equation; or giving a single angle when the question needs a general solution.
Squaring removes the difference between sine and cosine terms and feels like progress, while the loss of information about sign is invisible unless the candidate substitutes back; the periodicity of tangent and sine is also under-practised.
What to do instead
After squaring, substitute every candidate angle into the original equation and discard failures. Write the general solution with its period attached (nπ for tan, 2nπ for sin and cos forms) and state the restriction on the domain given in the question.
In heights-and-distances problems the candidate draws the angle of depression at the ground object instead of at the observer, so the triangle is set up with the wrong angle, and then uses sine where the known side is adjacent to the angle.
The phrase 'angle of depression' is read as a property of the object being looked at, and the alternate-angle relationship between the two horizontal lines is not visualised, so the marked angle ends up in the wrong corner.
What to do instead
Always draw a horizontal line through the observer's eye, mark the angle below it, then mark the equal alternate angle at the lower object. Label each side as opposite, adjacent or hypotenuse relative to the marked angle before choosing tan, sin or cos.
Applying a circle theorem to a figure where its hypothesis is not satisfied, for example treating an angle as a right angle because it 'looks like' it is in a semicircle, or treating a quadrilateral as cyclic without any evidence that all four vertices lie on the circle.
Theorem statements are memorised as pictures rather than as condition–conclusion pairs, and in exam figures the circle is often drawn roughly, so visual appearance is used as a substitute for the given data.
What to do instead
For each theorem write the hypothesis in words (e.g. 'angle subtended by a diameter at the circumference') and quote it in the solution before using the conclusion. Mark the equal angles or arcs on the figure from the given information first, and only then choose a theorem.
In transformation questions the candidate reverses the order of a combined transformation, applying the translation before the enlargement when the question says enlargement followed by translation, or uses a positive scale factor where a negative one is required.
Function-style notation suggests reading left to right, but combined transformations are applied right to left as written, and the meaning of a negative scale factor (image on the opposite side of the centre) is not connected to the coordinates.
What to do instead
Write the combination as a chain of arrows with the object at the start, then apply each step in the stated order. For enlargement, check by computing the vector from centre to image: it must equal k times the vector from centre to object, which forces the sign.
With vectors the candidate writes |a + b| = |a| + |b| and computes the magnitude of a sum by adding magnitudes, or decides three points are collinear because two vectors have the same magnitude rather than because one is a scalar multiple of the other.
Magnitude is treated as an ordinary number that distributes over addition, because the triangle inequality and the distinction between a vector and its length are not reinforced with numerical examples.
What to do instead
Compute a + b component-wise first and only then take the square root of the sum of squares. For collinearity, show AB = k·BC with an explicit value of k, and state that the vectors share a common point.
In grouped statistics the candidate computes the mean by averaging the class marks without weighting by frequency, or takes the median as the class mark of the median class instead of interpolating within it; in probability, draws the second item from the original total after a without-replacement selection.
The formula is recalled as a shape rather than as a weighted average, and in probability the phrase 'without replacement' is not translated into a changed denominator for the second draw.
What to do instead
Write the working table with columns f, x and fx summed explicitly before dividing. For the median, state l, cf, f and h and substitute into the interpolation formula. For probability, redraw the tree with the second-branch denominator reduced by one and add the probabilities of all successful paths.