Hint

The two outer terms have a constant sum. Only part (b) selects particular angles.

Worked solution

(a)

\(2k=(1+\cos x)+(1-\cos x)=2\), so \(\displaystyle k=1\), for every real x.

Answer: \(k=1\).

(b)

The common difference is \(\displaystyle k-(1+\cos x)=-\cos x\). Requiring \(\displaystyle d= \frac{1}{2}\) gives \(\displaystyle \cos x= - \frac{1}{2}\). The permitted angles are \(\displaystyle \frac{2\pi }{3}\) and \(\displaystyle \frac{4\pi }{3}\).

Answer: \(\displaystyle x= \frac{2\pi }{3}, \frac{4\pi }{3}\).

(c)

Now \(\displaystyle a= 1+ \cos x= \frac{1}{2}\) and \(\displaystyle d= \frac{1}{2}\). The twelfth term is \(\displaystyle \frac{1}{2}+ \frac{11}{2}= 6\). Therefore \(\displaystyle S_{12}= \frac{12(\frac{1}{2}+ 6)}{2}= 39\). Both angles produce the same first term and difference.

Answer: 39.

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: Use 2k equal to the sum of the outer terms.
• 1 mark: Obtain \(\displaystyle k=1\).

Alternative methods: Equality of differences is accepted.

Common error: Assuming k depends on x after the cosines cancel.

(b) · 2 marks

• 1 mark: Reduce the condition to \(\displaystyle \cos x= - \frac{1}{2}\).
• 1 mark: Give both exact permitted angles.

Alternative methods: A unit-circle or reference-angle argument is accepted.

Common error: Using \(\displaystyle \cos x= \frac{1}{2}\) instead of \(\displaystyle - \frac{1}{2}\).

(c) · 3 marks

• 1 mark: Use \(\displaystyle a= \frac{1}{2}\) and \(\displaystyle d= \frac{1}{2}\) for the continued progression.
• 1 mark: Apply a valid 12-term sum calculation.
• 1 mark: Evaluate 39.

Alternative methods: The terms are \(\displaystyle \frac{1}{2}, 1\),…,6, so half the sum of integers 1 to 12 is accepted.

Common error: Summing only the three originally displayed terms.

Total: 7 marks.