Checking solutions and domains

1. State the real domain of $f\left(x\right)=\frac{1}{x+6}$.

   2 marks
2. State the real domain of $g\left(x\right)=\sqrt{3-x}$.

   2 marks
3. State the real domain of $h\left(x\right)=\frac{\sqrt{x}}{x-9}$.

   2 marks
4. Check whether $x=4$ solves $\frac{3}{x-1}=1$.

   2 marks
5. A bus hire cost is $C=\frac{180}{n}$ dollars per passenger, with at most 30 passengers. State a sensible domain.

   3 marks
6. A calculation for $\frac{1}{x-2}=1$ gives $x=3$. Another student says $x=2$ is also possible because multiplying by $x-2$ removes the denominator. Decide both claims.

   3 marks

Equivalent correct methods earn the corresponding marks. Include units where asked.

1. State the real domain of $f\left(x\right)=\frac{1}{x+6}$.

   2 marks

   The denominator is zero only at $x=-6$.

   All real $x$ except $-6$.
2. State the real domain of $g\left(x\right)=\sqrt{3-x}$.

   2 marks

   Require $3-x\ge 0$.

   $x\le 3$
3. State the real domain of $h\left(x\right)=\frac{\sqrt{x}}{x-9}$.

   2 marks

   The root requires nonnegative $x$ and the denominator excludes $9$.

   $x\ge 0$, $x\ne 9$.
4. Check whether $x=4$ solves $\frac{3}{x-1}=1$.

   2 marks

   The input is allowed and $\frac{3}{4-1}=1$.

   Yes.
5. A bus hire cost is $C=\frac{180}{n}$ dollars per passenger, with at most 30 passengers. State a sensible domain.

   3 marks

   The head count is a positive whole number and cannot exceed the capacity.

   $n\in \{1,2,\dots ,30\}$.
6. A calculation for $\frac{1}{x-2}=1$ gives $x=3$. Another student says $x=2$ is also possible because multiplying by $x-2$ removes the denominator. Decide both claims.

   3 marks

   $\frac{1}{3-2}=1$.

   At $x=2$ the original left side divides by zero.

   $x=3$ is valid; $x=2$ is excluded.