Let $f$ be a twice differentiable function on $(1,6)$. If $f(2)=8$, $f^{\prime }(2)=5$, $f^{\prime }(x)\ge 1$ and $f^{\prime \prime }(x)\ge 4$ for all $x\in (1,6)$, then:

1. Analyze $f^{\prime }(x)$: By Lagrange's Mean Value Theorem (LMVT) on $f^{\prime }(x)$ over $[2,5]$:
$$\frac{f^{\prime }(5)-f^{\prime }(2)}{5-2}=f^{\prime \prime }(c)\ge 4$$
$$f^{\prime }(5)-5\ge 4(3)⟹f^{\prime }(5)\ge 17$$
2. Analyze $f(x)$: Apply LMVT on $f(x)$ over $[2,5]$:
$$\frac{f(5)-f(2)}{5-2}=f^{\prime }(d)$$
Since $f^{\prime \prime }(x)\ge 4>0$, $f^{\prime }(x)$ is an increasing function.Thus, for $d>2$, $f^{\prime }(d)>f^{\prime }(2)=5$.
$$\frac{f(5)-8}{3}>5⟹f(5)-8>15⟹f(5)>23$$
Wait, let's re-evaluate more precisely:$f^{\prime }(x)\ge f^{\prime }(2)+{\int }_2^x{f}^{\prime \prime }(t)dt\ge 5+4(x-2)$.Then $f^{\prime }(5)\ge 5+4(3)=17$.$f(5)=f(2)+{\int }_2^5{f}^{\prime }(x)dx\ge 8+{\int }_2^5[5+4(x-2)]dx$$f(5)\ge 8+[5(3)+4(x^2/2-2x{)}_2^5]=8+15+4(25/2-10-(2-4))=23+4(2.5+2)=23+18=41$.Checking $f(5)+f^{\prime }(5)\ge 41+17=58$.Option C states $f(5)+f^{\prime }(5)\ge 28$, which is definitely true.