Gce Model Questions Vol1 2025 Further Mathematics Question 45

Practice objective / multiple choice question 45 from the 2025 Gce Model Questions Vol1 Further Mathematics examination.

Gce Model Questions Vol1 2025 Further Mathematics Objective / Multiple Choice Medium Difficulty
Question 45 GCE_MODEL_QUESTIONS_VOL1 • 2025 • FURTHER_MATHEMATICS • objective

Solve for \(x\) in the interval \(0^{\circ}\le x<360^{\circ}\) such that \(\sin x + \sqrt{3}\cos x = 1\). Then give the value of \(\tan\frac{x}{2}\) for the smallest positive solution.

Answer Options

Correct Answer A. x = 90^{\circ} \text{ or } 330^{\circ}; \tan\frac{x}{2}=1 \text{ for the smallest positive solution}
B. x = 30^{\circ} \text{ or } 210^{\circ}; \tan\frac{x}{2}=\frac{1}{\sqrt{3}} \text{ for the smallest positive solution}
C. x = 90^{\circ} \text{ or } 270^{\circ}; \tan\frac{x}{2}=0 \text{ for the smallest positive solution}
D. x = 150^{\circ} \text{ or } 330^{\circ}; \tan\frac{x}{2}=\sqrt{3} \text{ for the smallest positive solution}
Correct Answer
A
Correct Option:
x = 90^{\circ} \text{ or } 330^{\circ}; \tan\frac{x}{2}=1 \text{ for the smallest positive solution}

About This Question

This is Gce Model Questions Vol1 2025 Further Mathematics Question 45. It is one of the objective questions from the 2025 Gce Model Questions Vol1 Further Mathematics examination.

Difficulty level: Medium .

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