Question

Let f be an odd function defined on the set of real numbers such that for \(x \ge 0\), \(f(x) = 3\sin x + 4\cos x\). Then f(x) at \(x = -\frac{11\pi}{6}\) is equal to:

• A. \(\frac{3}{2}+2\sqrt{3}\)
• B. \(-\frac{3}{2}+2\sqrt{3}\)
• C. \(\frac{3}{2}-2\sqrt{3}\)
• D. \(-\frac{3}{2}-2\sqrt{3}\)

Hint:

Use \(f(-x) = -f(x)\) and evaluate at \(x = 11\pi/6 \ge 0\).

Answer 1

The Correct Option is C

Since f is odd, \(f\left(-\frac{11\pi}{6}\right) = -f\left(\frac{11\pi}{6}\right)\).
For \(x = \frac{11\pi}{6} \ge 0\): \(\sin\frac{11\pi}{6} = -\frac{1}{2}\) and \(\cos\frac{11\pi}{6} = \frac{\sqrt{3}}{2}\).
\(f\left(\frac{11\pi}{6}\right) = 3\left(-\frac{1}{2}\right) + 4\cdot\frac{\sqrt{3}}{2} = -\frac{3}{2} + 2\sqrt{3}\).
Therefore \(f\left(-\frac{11\pi}{6}\right) = \frac{3}{2} - 2\sqrt{3}\).
Answer: (3) \(\frac{3}{2}-2\sqrt{3}\)