Function families and graph transformations practice questions — IB Maths AA

Set $N = 1200$ and divide by $400$:

$$1200 = 400\,e^{0.5t}$$

$$e^{0.5t} = 3$$

Take natural logs of both sides:

$$0.5t = \ln 3$$

$$t = 2\ln 3$$

So $t = 2\ln 3 \approx 2.20$ hours.

The number of doubling periods is $$\frac{12}{3} = 4.$$ Each period multiplies the population by $2$, so $$N = 500 \cdot 2^{4} = 500 \cdot 16 = 8000.$$

For $x \in (0, 1)$, multiplying by $x$ shrinks the value: $x^3 = x \cdot x^2 < x^2$ since $0 < x < 1$.

$f(1) = 1$, $f(2) = \dfrac{1}{2}$. As $x$ grows, $\dfrac{1}{x}$ shrinks — $f$ is strictly decreasing on $(0, +\infty)$.

Growing by $10\%$ each year multiplies the balance by $1.10$, so after $2$ years: $$2000 \times (1.10)^{2}$$ $$2000 \times 1.21 = 2420$$ The balance after $2$ years is $\$2{,}420$.

(Adding a flat $10\%$ of the original $\$200$ each year is *linear* growth and gives only $\$2{,}400$; applying the increase just once gives $\$2{,}200$.)

Set $X = e^{x}$, so $X$ $>$ $0$. The equation becomes $$X^2 - 4X + 3 = 0$$ $$(X - 1)(X - 3) = 0$$ $$X = 1 \quad \text{or} \quad X = 3$$ Both roots are positive, so substitute back: $$e^{x} = 1 \implies x = 0$$ $$e^{x} = 3 \implies x = \ln 3$$ The solution set is $\{0\,;\ \ln 3\}$.

$$A = 2000 \times \left(1 + \frac{5}{100}\right)^{3}$$ $$A = 2000 \times 1.05^{3} = \pounds 2{,}315.25$$

$x^2 + 3 = (x + 1)^2 = x^2 + 2x + 1 \Rightarrow 2 = 2x \Rightarrow x = 1$. Check: $\sqrt{4} = 2 = 1 + 1$ ✓.

$f(3) = 3^{-2} = \dfrac{1}{3^2} = \dfrac{1}{9}$.

$x^2 - 4 \ge 0 \Leftrightarrow x \le -2$ or $x \ge 2$. Domain: $(-\infty, -2] \cup [2, +\infty)$.