NECO 2023 · Paper 2 · Q10

(a)
Resolve $\dfrac{2x^3 + 5x^2 - 6x + 4}{(x - 1)(x + 2)}$ into partial fractions.
(b)
In a junior secondary school, 60 students play table tennis or basketball. The number who play table tennis is 7 more than three times the number who play basketball. If 3 students play both games and every student plays at least one game, how many students play table tennis?
(c)
An operation $*$ on the set of real numbers is defined by $p * q = \dfrac{3p + 3q - 5}{3}$ . Find the identity element.

Worked solution (try it first)

The top has degree 3 and the bottom, $(x - 1)(x + 2) = x^2 + x - 2$ , has degree 2, so divide first.
$2x$ times the bottom is $2x^3 + 2x^2 - 4x$ .
Take it away: $3x^2 - 2x + 4$ is left.
$3$ times the bottom is $3x^2 + 3x - 6$ .
Take it away: $-5x + 10$ is left.
So the fraction is $2x + 3 + \dfrac{10 - 5x}{(x - 1)(x + 2)}$ .
Split the remainder: $10 - 5x = A(x + 2) + B(x - 1)$ for $\dfrac{A}{x - 1} + \dfrac{B}{x + 2}$ .
Put $x = 1$ : $5 = 3A$ , so $A = \frac53$ .
Put $x = -2$ : $20 = -3B$ , so $B = -\frac{20}{3}$ .
So the answer is $2x + 3 + \dfrac{5}{3(x - 1)} - \dfrac{20}{3(x + 2)}$ .

Let $B$ play basketball and $T$ play table tennis.
The 3 who play both are counted in both, so $T + B - 3 = 60$ , which gives $T + B = 63$ .
Also $T = 3B + 7$ .
Substitute: $3B + 7 + B = 63$ , so $4B = 56$ and $B = 14$ .
Then $T = 3(14) + 7 = 49$ .
So 49 students play table tennis.