Elementary Functions practice questions — A-Level Maths

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

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

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

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

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

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

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

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

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

Take natural logs of both sides:

$$0.5t=\ln 3$$

$$t=2\ln 3$$

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

Growing by $10\mathrm{\%}$ 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 $\mathrm{\$}\mathrm{2,420}$.

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

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

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