JEE Main 2023 — Differentiation Question with Solution

Give that:

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

Convert the numerator and denominator in the form of $\sin \left(A\pm B\right)$.

$\Rightarrow f\left(x\right)=\frac{\sqrt{2}s\mathrm{in}\left(x+{\displaystyle \frac{\pi }{4}}\right)-\sqrt{2}}{\sqrt{2}\left(\sin \left(x-{\displaystyle \frac{\pi }{4}}\right)\right)}$

$\Rightarrow f\left(x\right)=\frac{\sin \left(x+{\displaystyle \frac{\pi }{4}}\right)-1}{\sin \left(x-{\displaystyle \frac{\pi }{4}}\right)}$

$\Rightarrow f\left(x+\frac{\pi }{4}\right)=\frac{\cos x-1}{\sin x}$

$\Rightarrow f\left(x+\frac{\pi }{4}\right)=-\tan \frac{x}{2}$

$\Rightarrow f\left(x\right)=-\tan \left(\frac{x}{2}-\frac{\pi }{8}\right)$

$\Rightarrow f^{\text{'}}\left(x\right)=\frac{-1 }{2}{\sec }^2\left(\frac{x}{2}-\frac{\mathrm{\pi }}{8}\right)$

$\Rightarrow f^{"}\left(x\right)=-\frac{1}{2}\left({\sec }^2\left(\frac{x}{2}-\frac{x}{8}\right)\tan \left(\frac{x}{2}-\frac{\pi }{8}\right)\right)$

$\Rightarrow f\left(\frac{7\pi }{12}\right)=-\tan \left(\frac{7\pi }{24}-\frac{\pi }{8}\right)$$=-\tan \left(\frac{4\pi }{24}\right)$

$\Rightarrow -\tan \left(\frac{\pi }{6}\right)=-\frac{1}{\sqrt{3}}$

Also,

$\Rightarrow f^{"}\left(\frac{7\pi }{12}\right)=-\frac{1}{2}\times {\left(\frac{2}{\sqrt{3}}\right)}^2\times \frac{1}{\sqrt{3}}=-\frac{2}{3\sqrt{3}}$

$\Rightarrow f\left(\frac{7\pi }{12}\right)f^{"}\left(\frac{7\pi }{12}\right)=\frac{2}{9}$

Hence this is the correct option.