Let $$f$$ be a twice differentiable function on $$(1, 6)$$, If $$f(2) = 8$$, $$f'(2) = 5$$, $$f'(x) \geq 1$$ and $$f''(x) \geq 4$$, for all $$x \in (1, 6)$$, then:

We have a twice-differentiable function on the open interval $$(1,6)$$ with the data

$$f(2)=8,\qquad f'(2)=5,\qquad f'(x)\ge 1,\qquad f''(x)\ge 4\quad\text{for every }x\in(1,6).$$

First we estimate the value of the first derivative at $$x=5$$. The increment of the first derivative is given by the Fundamental Theorem of Calculus applied to the second derivative:

$$f'(5)=f'(2)+\int_{2}^{5}f''(x)\,dx.$$

Because $$f''(x)\ge 4$$ on the whole interval, we may replace $$f''(x)$$ by its lower bound to obtain

$$f'(5)\;\ge\;5+\int_{2}^{5}4\,dx \;=\;5+4(5-2) \;=\;5+12 \;=\;17.$$

So, we already know

$$f'(5)\ge 17.$$

Next we estimate $$f(5)$$ itself. Again we start from the Fundamental Theorem of Calculus, this time for the first derivative:

$$f(5)=f(2)+\int_{2}^{5}f'(x)\,dx.$$

To find a lower bound for the integrand $$f'(x)$$, we integrate the inequality for $$f''(x)$$ once more. For any $$t\in[2,5]$$ we have

$$f'(t)=f'(2)+\int_{2}^{t}f''(x)\,dx \;\ge\;5+\int_{2}^{t}4\,dx \;=\;5+4(t-2).$$

Thus the smallest possible graph of $$f'(x)$$ consistent with all conditions is the straight line

$$f'(x)=5+4(x-2),\qquad x\in[2,5].$$

Using this least value in the integral for $$f(5)$$ we get

$$\begin{aligned} f(5)&=8+\int_{2}^{5}f'(x)\,dx\\ &\ge 8+\int_{2}^{5}\bigl[\,5+4(x-2)\bigr]\,dx\\ &=8+\Bigl[\;5(x-2)+2(x-2)^2\Bigr]_{2}^{5}\\ &=8+\Bigl[\;5(3)+2(3)^2\Bigr]\\ &=8+\bigl[15+18\bigr]\\ &=8+33\\ &=41. \end{aligned}$$

So, combining the two lower bounds we have obtained,

$$f(5)\ge 41\quad\text{and}\quad f'(5)\ge 17,$$

which immediately gives

$$f(5)+f'(5)\;\ge\;41+17\;=\;58.$

This inequality certainly satisfies $$f(5)+f'(5)\ge 28$$ and violates all the other proposed inequalities.

Hence, the correct answer is Option B.