Let $$f : \mathbf{R} \to \mathbf{R}$$ be a differentiable function such that $$f\left(\frac{\pi}{4}\right) = \sqrt{2}$$, $$f\left(\frac{\pi}{2}\right) = 0$$ and $$f'\left(\frac{\pi}{2}\right) = 1$$ and let $$g(x) = \int_x^{\pi} (f'(t) \sec t + \tan t \sec t \, f(t)) dt$$ for $$x \in \left[\frac{\pi}{4}, \frac{\pi}{2}\right)$$. Then $$\lim_{x \to \left(\frac{\pi}{2}\right)^-} g(x)$$ is equal to

Given,

$$g(x)=\int_x^{\pi/4}\left(f'(t)\sec t+f(t)\sec t\tan t\right)dt$$

Observe that

$$\frac{d}{dt}\left(f(t)\sec t\right)=f'(t)\sec t+f(t)\sec t\tan t$$

Hence,

$$g(x)=\left[f(t)\sec t\right]_x^{\pi/4}$$

$$=f\left(\frac{\pi}{4}\right)\sec\left(\frac{\pi}{4}\right)-f(x)\sec x$$

Using

$$f\left(\frac{\pi}{4}\right)=\sqrt2,\qquad \sec\left(\frac{\pi}{4}\right)=\sqrt2$$

we get

$$g(x)=2-\frac{f(x)}{\cos x}$$

Now,

$$\lim_{x\to\left(\frac{\pi}{2}\right)^-}\frac{f(x)}{\cos x}$$

is of the form $$\frac00$$ since

$$f\left(\frac{\pi}{2}\right)=0,\qquad \cos\left(\frac{\pi}{2}\right)=0$$

Applying L'Hospital's Rule,

$$\lim_{x\to\left(\frac{\pi}{2}\right)^-}\frac{f(x)}{\cos x} = \lim_{x\to\left(\frac{\pi}{2}\right)^-}\frac{f'(x)}{-\sin x}$$

Using

$$f'\left(\frac{\pi}{2}\right)=1,\qquad \sin\left(\frac{\pi}{2}\right)=1$$

we get

$$\lim_{x\to\left(\frac{\pi}{2}\right)^-}\frac{f(x)}{\cos x}=-1$$

Therefore,

$$\lim_{x\to\left(\frac{\pi}{2}\right)^-}g(x)=2-(-1)=3$$

Hence, the correct answer is

$$\boxed{\text{Option B }:3}$$