The slant (oblique) asymptote of f(x) = (x^2 + 1)/(x) as x → ±∞ is:

f(x) = x + (1)/(x). As x → ±∞, (1)/(x) → 0, so the slant asymptote is y = x.

For the geometric progression (b_n)_n ≥ 1 with b_1 = 2 and q = √(5), compute lfloor b_4 rfloor (the integer part of b_4).

b_4 = 10√(5). Since √(5) ≈ 2.236, b_4 ≈ 22.36, so lfloor b_4 rfloor = 22.

Problems & worked solutions

RO M1

The slant (oblique) asymptote of $f (x) = \frac{x^{2} + 1}{x}$ as $x \to \pm \infty$ is:

Show answer & worked solution

$f (x) = x + \frac{1}{x}$ . As $x \to \pm \infty$ , $\frac{1}{x} \to 0$ , so the slant asymptote is $y = x$ .

RO M1

For the geometric progression $(b_{n})_{n \geq 1}$ with $b_{1} = 2$ and $q = \sqrt{5}$ , compute $⌊ b_{4} ⌋$ (the integer part of $b_{4}$ ).

Show answer & worked solution

$b_{4} = 10 \sqrt{5}$ . Since $\sqrt{5} \approx 2.236$ , $b_{4} \approx 22.36$ , so $⌊ b_{4} ⌋ = 22$ .

RO M1

On $P (X)$ with symmetric difference, the inverse of any set $A$ is:

Show answer & worked solution

Every element is its own inverse: $A △ A = \emptyset$ .

Problems & worked solutions

RO M1

The slant (oblique) asymptote of $f (x) = \frac{x^{2} + 1}{x}$ as $x \to \pm \infty$ is:

Show answer & worked solution

$f (x) = x + \frac{1}{x}$ . As $x \to \pm \infty$ , $\frac{1}{x} \to 0$ , so the slant asymptote is $y = x$ .

RO M1

For the geometric progression $(b_{n})_{n \geq 1}$ with $b_{1} = 2$ and $q = \sqrt{5}$ , compute $⌊ b_{4} ⌋$ (the integer part of $b_{4}$ ).

Show answer & worked solution

$b_{4} = 10 \sqrt{5}$ . Since $\sqrt{5} \approx 2.236$ , $b_{4} \approx 22.36$ , so $⌊ b_{4} ⌋ = 22$ .

RO M1

On $P (X)$ with symmetric difference, the inverse of any set $A$ is:

Show answer & worked solution

Every element is its own inverse: $A △ A = \emptyset$ .