Hint
The coefficient a and distance d are positive. The modulus vanishes at \(\displaystyle x=h\).
Worked solution
(a)
The branches are \(\displaystyle y=1(h-x)+k\) for \(\displaystyle x\le h\) and \(\displaystyle y=1(x-h)+k\) for \(\displaystyle x\ge h\).
Answer: \(\left( 10, \ 4\right)\)
(b)
\(\left|{x - 10}\right| = 4\) gives both boundary values.
Answer: \(\left\{6, 14\right\}\)
(c)
\(\left|{x - 10}\right| < 4\) describes points less than distance d from h.
Answer: \(\left(6, 14\right)\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternatives and consequential reasoning; an unexplained final answer does not earn a method mark.
(a) · 2 marks
- 1 mark: Locate the vertex and give the correct rule on each side.
- 1 mark: Give all results requested in this part, with the correct signs, exact values, units, branch and boundary inclusion where applicable.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: An exterior solution needs two intervals.
(b) · 2 marks
- 1 mark: Solve both signed branches of the modulus equation.
- 1 mark: Give all results requested in this part, with the correct signs, exact values, units, branch and boundary inclusion where applicable.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Do not treat |x−h| as x−h everywhere.
(c) · 2 marks
- 1 mark: Translate the strict distance inequality into two strict endpoint bounds.
- 1 mark: Give all results requested in this part, with the correct signs, exact values, units, branch and boundary inclusion where applicable.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: An exterior solution needs two intervals.
Total: 6 marks.