NSSCAS Mathematics (8227) common mistakes
NSSCAS Mathematics (8227) · Understand the traps and how to avoid them.
In the binomial expansion of (1 - 2x)^6, writes the x^2 term as C(6,2)(-2x)^2 but evaluates it as 15x^2, squaring only the x and dropping the factor (-2)^2.
The candidate separates the binomial coefficient from the bracket term too early and treats the coefficient inside the bracket as if it only multiplies x, not the whole term.
What to do instead
Write the general term as C(n,r)a^(n-r)b^r with the entire bracket term b = -2x, so the x^2 term is C(6,2)(-2)^2 x^2 = 60x^2. Check by expanding the first three terms numerically for a small x value.
Solves x^2 - 5x + 6 < 0 by finding roots 2 and 3 and writing x < 2 or x > 3, giving the outside intervals for a '< 0' quadratic.
The candidate treats the inequality like a quadratic equation and applies a remembered 'outside the roots' rule without checking the direction of the parabola.
What to do instead
Sketch y = x^2 - 5x + 6, an upward parabola, and shade below the x-axis. The solution is 2 < x < 3; verify with a test point such as x = 2.5.
For f(x) = x^2 - 4x + 7 with no domain restriction, finds f^(-1)(x) = 2 + sqrt(x - 3) and states its domain is all real numbers.
The candidate swaps x and y mechanically and ignores that the original quadratic is not one-to-one on all real numbers, so an inverse function does not exist without restricting the domain.
What to do instead
Restrict the original domain, for example x >= 2, giving range y >= 3. Then the inverse is f^(-1)(x) = 2 + sqrt(x - 3) with domain x >= 3, and state both restrictions explicitly.
For x^2 + y^2 - 8x + 6y - 11 = 0, writes the centre as (-4, 3) after completing the square as (x - 4)^2 + (y + 3)^2 = 36.
The candidate reads the numbers from the completed-square brackets but reverses the signs incorrectly, confusing the coefficients in the original equation with the centre coordinates.
What to do instead
Complete the square: x^2 - 8x = (x - 4)^2 - 16 and y^2 + 6y = (y + 3)^2 - 9. The centre is (4, -3) and the radius is 6; always write the standard form (x - h)^2 + (y - k)^2 = r^2 before reading off (h, k).
Solving sin(3x) = 0.5 for 0 <= x <= 360 degrees, lists only 3x = 30 and 150, then gives x = 10 and 50, missing the solutions from 3x = 390, 510, 750 and 870.
The candidate uses only the principal-value angles and forgets that the interval for 3x is 0 <= 3x <= 1080 degrees, so the sine equation has several more solutions.
What to do instead
First solve for 3x over the transformed interval 0 <= 3x <= 1080 degrees, listing 30, 150, 390, 510, 750, 870, then divide every solution by 3 to get x = 10, 50, 130, 170, 250, 290 degrees.
Uses S_infinity = a/(1 - r) for a geometric series with r = 2, or with r = -1.2, and reports a finite sum.
The candidate memorises the infinite geometric sum formula but does not check the convergence condition |r| < 1 before substituting.
What to do instead
Identify a and r, then check |r| < 1 before using S_infinity = a/(1 - r). If |r| >= 1, state that the series diverges and has no finite infinite sum.
Differentiates y = sin(3x^2) as dy/dx = cos(3x^2), omitting the factor 6x from the inner derivative.
The candidate recognises the outer sine function but treats the bracket 3x^2 as if it were simply x, so the chain rule is only partly applied.
What to do instead
Set u = 3x^2, so dy/dx = cos(u) times du/dx = cos(3x^2) times 6x = 6x cos(3x^2). Write the inner derivative explicitly before simplifying.
Computes the definite integral of x(x - 1) from x = -1 to x = 2 and reports the signed value 1.5 as the area between the curve and the x-axis, even though the curve crosses the axis at x = 0 and x = 1.
The candidate confuses the definite integral, which gives a signed total, with geometric area, which requires negative regions to be counted positively.
What to do instead
Find the roots x = 0 and x = 1, split the integral at those points, evaluate each sub-integral, take the absolute value of each piece, and add: total area = integral from -1 to 0 plus |integral from 0 to 1| plus integral from 1 to 2.
For 'choose 3 students from 10 to form a committee', computes 10P3 = 720 instead of C(10,3) = 120 because the candidate uses the permutation function whenever items are being selected.
The candidate focuses on the number of choices rather than on whether rearranging the chosen students would create a different outcome.
What to do instead
Decide whether order matters: a committee is unchanged by reordering, so use combinations C(10,3) = 120. If the three students were chair, secretary and treasurer, order would matter and permutations would be appropriate.
For the vector 2i - 3j + 6k, computes the magnitude as 2 + 3 + 6 = 11 instead of sqrt(2^2 + (-3)^2 + 6^2) = 7.
The candidate applies a one-dimensional absolute-value habit to components and adds their sizes instead of using the three-dimensional distance formula.
What to do instead
Square each component, add the squares, and take the square root: sqrt(4 + 9 + 36) = sqrt(49) = 7. Check the result is not simply the sum of absolute component values.