1. Differentiate $g(x)$

Given $g(x)=f(x)+f(1-x).$ Differentiate: $g'(x)=f'(x)-f'(1-x).$

Again, $g''(x)=f''(x)+f''(1-x).$ Since $f''(x)>0$ for all $x\in(0,1)$, we get $g''(x)>0 \quad \text{for all } x\in(0,1).$ So $g$ is strictly convex, hence $g'(x)$ is strictly increasing on $(0,1)$.

2. Use symmetry of $g$

Observe: $g(1-x)=f(1-x)+f(x)=g(x).$ So $g$ is symmetric about $x=\tfrac12$.

Now check derivative at $x=\tfrac12$: $g'\!\left(\tfrac12\right)=f'\!\left(\tfrac12\right)-f'\!\left(1-\tfrac12\right)=0.$

Because $g'$ is strictly increasing and vanishes at $x=\tfrac12$, we have:

• $g'(x)<0$ for $x<\tfrac12$ $\Rightarrow$ $g$ is decreasing on $\left(0,\tfrac12\right)$,
• $g'(x)>0$ for $x>\tfrac12$ $\Rightarrow$ $g$ is increasing on $\left(\tfrac12,1\right)$.

Hence, $\alpha=\frac12.$

3. Evaluate the given expression

We need $\tan^{-1}(2\alpha)+\tan^{-1}\left(\frac1\alpha\right)+\tan^{-1}\left(\frac{\alpha+1}{\alpha}\right).$ Substitute $\alpha=\tfrac12$: $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3).$

Now, $\tan^{-1}(1)=\frac\pi4.$ Also, $\tan^{-1}(2)+\tan^{-1}(3)=\pi-\tan^{-1}\left(\frac{2+3}{1-2\cdot 3}\right)$ but it is easier to combine stepwise.

First, $\tan^{-1}(1)+\tan^{-1}(2).$ Since $1\cdot 2=2>1,$ we use the adjusted formula: $\tan^{-1}(1)+\tan^{-1}(2)=\pi+\tan^{-1}\left(\frac{1+2}{1-2}\right)=\pi+\tan^{-1}(-3)=\pi-\tan^{-1}(3).$ Therefore, $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi.$

So the value is $\boxed{\pi}.$

4. Compare with options

Option B is $\boxed{\pi}.$ So the correct option is B.

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.