Hint
Use the Pythagorean identity before cancelling. The cancellation does not change the original denominator restriction.
Worked solution
(a)
The numerator is sin²x, so E(x)=sin x after cancelling on sin x≠0. The excluded values are x=nπ for integer n.
Answer: \(E(x)=\sin x\) for \(x\ne n\pi\), integer n.
(b)
On the permitted domain solve sin x=1/2. The candidates are π/6 and 5π/6; both have nonzero sine and satisfy the original quotient.
Answer: \(x=\pi/6,5\pi/6\).
(c)
A function includes its domain. At x=0 the quotient is 0/0 and is undefined, while the sine function has value zero. They agree on E’s domain, but sine extends it to the missing points.
Answer: The domains differ; E(0) is undefined while sin 0=0.
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 2 marks
- 1 mark: Simplify to sin x.
- 1 mark: Retain the full exclusion x=nπ, n integer.
Alternative methods: A correct statement of the same sine-zero set is accepted.
Common error: Claiming the quotient is defined at all real x after cancellation.
(b) · 3 marks
- 1 mark: Reduce the original equation to sine equal to 1/2 on its valid domain.
- 1 mark: Find both angles π/6 and 5π/6.
- 1 mark: Check that both satisfy the original domain/expression.
Alternative methods: Direct substitution into the quotient is accepted.
Common error: Adding a denominator-zero angle as a solution.
(c) · 2 marks
- 1 mark: Explain the difference in domains.
- 1 mark: Give a correct specific missing point, such as x=0 or π.
Alternative methods: Any integer multiple of π is a valid example.
Common error: Treating 0/0 as zero.
Total: 7 marks.