Let f(x)= x+ x-√2 x- x,x [0, ]-( )/(4).

Question

Let $f(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x},x\in [0,\pi ]-{\frac{\pi }{4}}$. Then $f(\frac{7\pi }{12}){f}^{''}(\frac{7\pi }{12})$ is equal to

Accepted Answer

$$f(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x}$$

$$
\begin{aligned}
& =\frac{\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x-1}{\frac{1}{\sqrt{2}} \sin x-\frac{1}{\sqrt{2}} \cos x} \\
& =\frac{\cos \left(x-\frac{\pi}{4}\right)-1}{\sin \left(x-\frac{\pi}{4}\right)} \\
& =\frac{-2 \sin ^2\left(\frac{x}{2}-\frac{\pi}{8}\right)}{2 \sin \left(\frac{x}{2}-\frac{\pi}{8}\right) \cos \left(\frac{x}{2}-\frac{\pi}{8}\right)}
\end{aligned}
$$

$$
\begin{aligned}
& \Rightarrow f(x)=-\tan \left(\frac{x}{2}-\frac{\pi}{8}\right) \\
& \Rightarrow f^{\prime}(x)=-\frac{1}{2} \sec ^2\left(\frac{x}{2}-\frac{\pi}{8}\right)
\end{aligned}
$$

$$
\begin{aligned}
& \Rightarrow f^{\prime \prime}(x)=-\frac{1}{2} \cdot 2 \sec \left(\frac{x}{2}-\frac{\pi}{8}\right) \cdot \sec \left(\frac{x}{2}-\frac{\pi}{8}\right)\tan \left(\frac{x}{2}-\frac{\pi}{8}\right) \times \frac{1}{2} \\
& =-\frac{1}{2} \sec ^2\left(\frac{x}{2}-\frac{\pi}{8}\right) \cdot \tan \left(\frac{x}{2}-\frac{\pi}{8}\right)
\end{aligned}
$$

$$
\begin{aligned}
& \text { Now, } f\left(\frac{7 \pi}{12}\right) f^{\prime \prime}\left(\frac{7 \pi}{12}\right) \\
& =-\tan \left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \times \frac{-1}{2} \sec ^2\left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \times \tan \left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \\
& =\frac{1}{2} \tan ^2\left(\frac{\pi}{6}\right) \times \sec ^2 \frac{\pi}{6} \\
& =\frac{1}{2} \times \frac{1}{3} \times \frac{4}{3}=\frac{2}{9}
\end{aligned}
$$

Solutions · 1