Binomial theorem and partial fractions quiz: O-Level Additional Mathematics (SEAB 4049) quiz

$n$

$n - 1$

$2n$

$n + 1$

$10$

$20$

$7$

$25$

$\binom{n}{r} a^{r} b^{n-r}$

$\binom{n}{r} a^{n-r} b^{r}$

$\binom{n}{r} a^{n} b^{r}$

$\binom{n}{r+1} a^{n-r} b^{r}$

$15$

$20$

$6$

$30$

$x^{r}$

$x^{4 - r}$

$x^{4 - 2r}$

$x^{4}$

$4$

There is none

$1$

$6$

Only positive integer $n$

Yes, all real $n$

Only fractional $n$

Only even $n$

The numerator degree must be at least the denominator degree

The fraction must be proper (numerator degree less than denominator degree)

The denominator must be a single linear factor

The numerator must be a constant

$\dfrac{Ax + B}{x - 3}$

$\dfrac{A}{x - 3}$

$\dfrac{A}{(x - 3)^2}$

$\dfrac{Ax^2 + B}{x - 3}$

$\dfrac{A}{x + 1} + \dfrac{B}{x - 4}$

$\dfrac{A}{x + 1} + \dfrac{B}{(x + 1)^2} + \dfrac{C}{x - 4}$

$\dfrac{Ax + B}{(x + 1)^2} + \dfrac{C}{x - 4}$

$\dfrac{A}{(x + 1)^2} + \dfrac{B}{x - 4}$

$3$

$\dfrac{4}{3}$

$\dfrac{11}{3}$

$-\dfrac{11}{3}$

$\dfrac{A}{x^2 + 4}$

$\dfrac{Ax^2 + B}{x^2 + 4}$

$\dfrac{A}{x + 2} + \dfrac{B}{x - 2}$

$\dfrac{Ax + B}{x^2 + 4}$

$8 + 12x + 6x^2 + x^3$

$8 + 6x + 12x^2 + x^3$

$2 + 6x + 6x^2 + x^3$

$8 + 12x + 6x^2 + 3x^3$

They split a fraction into simpler pieces that are easier to integrate

They make a fraction easier to differentiate using the quotient rule

They convert a fraction into a binomial expansion

They remove the need for limits in a definite integral