Applications of Derivatives
Higher Derivatives of Trigonometric-Form Function
nta_pyq_2023_apr
Grade 12
Question:
Let $f(x)=\dfrac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x}$, $x\in[0,\pi]-\left\{\dfrac{\pi}{4}\right\}$, then $f\!\left(\dfrac{7\pi}{12}\right)f''\!\left(\dfrac{7\pi}{12}\right)$ is equal to
\dfrac{2}{9}
\dfrac{-2}{3}
\dfrac{-1}{3\sqrt{3}}
\dfrac{2}{3\sqrt{3}}
Step-by-Step Solution
Key Concept: Simplify $f(x)$: write numerator and denominator using sum formulas. Show $f(x)=-\tan\!\left(\frac{x}{2}-\frac{\pi}{8}\right)$ (or similar shifted form).
$f(x)=-\tan\!\left(\frac{x}{2}-\frac{\pi}{8}\right)$. $f(\frac{7\pi}{12})=-\frac{1}{\sqrt{3}}$, $f''(\frac{7\pi}{12})=-\frac{2}{3\sqrt{3}}$. Product $=\frac{2}{9}$.
Correct Answer: 1