Question

Let \(f(x)=\frac{sinx+cosx-\sqrt2}{sinx-cosx},x∈[0.\pi]-\left[\frac{\pi}{4}\right]. Then f\left(\frac{7\pi}{12}\right)f^n\left(\frac{7\pi}{12}\right)\)is equal to

• A. \(\frac{-2}{3}\)
• B. \(\frac{-1}{3\sqrt3}\)
• C. \(\frac{2}{3\sqrt3}\)
• D. \(\frac{2}{9}\)

Hint:

Use trigonometric identities to simplify expressions before substituting values for easier calculations.

Answer 1

The Correct Option is D

We are given the function: \[ f(x) = - \tan\left( \frac{x}{2} - \frac{\pi}{8} \right) \] Now, differentiate the function \( f(x) \): \[ f'(x) = - \frac{1}{2} \sec^2\left( \frac{x}{2} - \frac{\pi}{8} \right) \] Now, calculate the second derivative: \[ f''(x) = - \sec^2\left( \frac{x}{2} - \frac{\pi}{8} \right) \cdot \tan\left( \frac{x}{2} - \frac{\pi}{8} \right) \cdot \frac{1}{2} \] Now, calculate the value of \( f\left( \frac{7\pi}{12} \right) \): \[ f\left( \frac{7\pi}{12} \right) = - \tan\left( \frac{\pi}{6} \right) = - \frac{1}{\sqrt{3}}. \] Next, calculate the second derivative at \( x = \frac{7\pi}{12} \): \[ f''\left( \frac{7\pi}{12} \right) = - \frac{1}{2} \sec^2\left( \frac{\pi}{6} \right) \cdot \tan\left( \frac{\pi}{6} \right) = - \frac{1}{2} \times \frac{4}{3} \times \frac{1}{\sqrt{3}} = \frac{-4}{3\sqrt{3}}. \] Finally, we calculate the product: \[ f\left( \frac{7\pi}{12} \right) \cdot f''\left( \frac{7\pi}{12} \right) = \left( -\frac{1}{\sqrt{3}} \right) \cdot \left( \frac{-4}{3\sqrt{3}} \right) = \frac{2}{3}. \] Final Answer: \( \frac{2}{3} \).