JEE 2023 — Mathematics PYQ

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

\begin{aligned}
 & =-\frac{1}{2}\mathrm{sec}^{2}\left(\frac{x}{2}-\frac{\pi}{8}\right).\tan\left(\frac{x}{2}-\frac{\pi}{8}\right) \\
 & \mathrm{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}\mathrm{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}