Let f(x) = x + cos - 2 x - x , x ∈ [0, π] - \ (π)/(4)\ , then...

Give that:

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

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

$⇒ f\left(x\right) = \dfrac{\sqrt{2}\sin \left(x + \dfrac{\pi}{4}\right) - \sqrt{2}}{\sqrt{2}\left(\sin \left(x - \dfrac{\pi}{4}\right)\right)}$

$⇒ f\left(x\right) = \dfrac{\sin \left(x + \dfrac{\pi}{4}\right) - 1}{\sin \left(x - \dfrac{\pi}{4}\right)}$

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

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

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

$⇒ {f}^{'}\left(x\right) = \dfrac{- 1}{2}\sec^{2}\left(\dfrac{x}{2} - \dfrac{\pi}{8}\right)$

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

$⇒ f\left(\dfrac{7\pi}{12}\right) = - \tan \left(\dfrac{7\pi}{24} - \dfrac{\pi}{8}\right)$= - \tan \left(\dfrac{4\pi}{24}\right)$</p><p>$⇒ - \tan \left(\dfrac{\pi}{6}\right) = - \dfrac{1}{\sqrt{3}}$</p><p>Also,</p><p>$⇒ {f}^{"}\left(\dfrac{7\pi}{12}\right) = - \dfrac{1}{2} \times {\left(\dfrac{2}{\sqrt{3}}\right)}^{2} \times \dfrac{1}{\sqrt{3}} = - \dfrac{2}{3\sqrt{3}}$</p><p>$⇒ f\left(\dfrac{7\pi}{12}\right){f}^{"}\left(\dfrac{7\pi}{12}\right) = \dfrac{2}{9}

Hence this is the correct option.