Given that $,f(n) = \left \{\begin{matrix} n(n+1)\;; & \text {if n is even} \\ (n+3)\;; & \text {if n is odd} \end{matrix}\right.$
And $,8f (m+1) – f(m) = 2 $
$ \textbf{Case 1:}$ If $`m\text{’}$ is an even, then $`m+1\text{’}$ must be odd.
$ 8f (m+1) – f(m) = 2 $
$ \Rightarrow 8(m+1+3) – m(m+1) = 2 $
$ \Rightarrow 8(m+4) – m^{2} – m = 2 $
$ \Rightarrow 8m + 32 – m^{2} – m = 2 $
$ \Rightarrow -m^{2} +7m +30 = 0 $
$ \Rightarrow m^{2} – 7m - 30 = 0 $
$ \Rightarrow m^{2} – 10m + 3m - 30 = 0 $
$ \Rightarrow m(m-10) + 3(m-10) = 0 $
$ \Rightarrow (m-10) (m+3) = 0 $
$ \Rightarrow m-10=0, m+3=0 $
$ \Rightarrow \boxed{m=10}, \boxed{m=-3} $ [rejected $`m\text{’}$ is an even positive integer.]
$ \Rightarrow \boxed{m=10} $
$ \textbf{Case 2:}$ If $`m\text{’}$ is an odd, then $`(m+1)\text{’}$ must be an even.
$8f (m+1) – f(m) = 2 $
$ \Rightarrow 8 (m+1) (m+1+1) – (m+3) = 2 $
$ \Rightarrow 8 (m+1) (m+2) – m – 3 – 2 = 0 $
$ \Rightarrow 8 ( m^{2} + 2m + m + 2 ) – m – 5 = 0 $
$ \Rightarrow 8m^{2} + 16m + 8m + 16 – m – 5 = 0 $
$ \Rightarrow 8m^{2} + 24m + 16 – m – 5 = 0 $
$ \Rightarrow 8m^{2} + 23m + 11 = 0 $
We know that, factors for, $ax^{2} + bx + c = 0 \Rightarrow x = \frac{-b \pm \sqrt{b^{2} – 4ac}}{2a}$
Here, $m = \frac{-23 \pm \sqrt{(23)^{2} – 4(8)(11)}}{16}$
$\Rightarrow m = \frac{-23 \pm \sqrt{529 – 352}}{16}$
$\Rightarrow m = \frac{-23 \pm \sqrt{177}}{16}$
Here, the value of $`m\text{’}$ is not integer.
So, no integer solution is possible for $`m\text{’}$ in the case.
Therefore$, \boxed{m = 10} $
Correct Answer $:10$